Model Answers in Organic Chemistry. For 'A' Level and Ordinary National Certificate Students

T H E C O M M O N W E A L T H A N D I N T E R N A T I O N A L L I B R A R Y

Joint Chairmen of the Honorary Editorial Advisory Board

SIR ROBERT ROBINSON, O.M., F.R.S., London DEAN ATHELSTAN SPILHAUS, Minnesota

Publisher ROBERT MAXWELL, M . C , M.P.

C O M M O N W E A L T H L I B R A R Y O F M O D E L A N S W E R S

Editor C . W . SCHOFIELD

Model Answers in Organic Chemistry

This book is sold subject to the condition that it shall not, by way of trade, be lent, re-sold, hired out or otherwise disposed of without the publisher's consent, in any form of binding or cover other than that in which it is published.

MODEL ANSWERS IN

ORGANIC CHEMISTRY

F O R Ά * LEVEL A N D O R D I N A R Y N A T I O N A L C E R T I F I C A T E S T U D E N T S

BY

A. J. SHOWLER, B.Sc., Ph.D., A.R.I.C.

Lecturer in Organic Chemistry, Medway College of Technology

AND

T. A. BROWN, B.Sc, B.Com. Senior Science Master, St. Bees School

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Copyright © 1965

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First edition 1965

Library of Congress Catalog Card No. 65-18372

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Introduction

IN WRITING this book, the aim has been to present to the student (particularly the student who is working on his own or who is inexperienced in sitting for examinations) a series of answers which show what is required and expected in the General Certificate of Education Advanced Level and Ordinary National Certificate examinations. Opinions differ in many cases as to what is required in answer to a given question, and the authors would not suggest that their answers are perfect. Nor should the student use them as such, or assume that all the organic chemistry he needs is to be found in this book. The answers are intended only to supplement the information in the normal textbook used in preparing for the examination, and to serve as a guide to the reader, in that they show, for example, how long an answer should be, how information should be presented, the type of sketch required, the essential equations and details of a preparation.

Nine examining boards at present set Advanced Level examina-tions in chemistry, and all have substantially the same syllabus. Ordinary National Certificate is of a similar standard, but with generally rather more emphasis on aromatic chemistry, and a wider variation in syllabus, since most technical colleges have their own and set their own examinations. Nevertheless, the Ordinary National Certificate candidate could well find any question repro-duced on the following pages in his examination, though the Ad-vanced Level student would not expect to be called on to answer some of the Ordinary National Certificate questions given here.

For this reason, and since many more students sit for Advanced Level, emphasis has been laid on these questions, which have been reproduced from the past papers of all the Boards prepared to allow this. Ordinary National Certificate questions come from four sources only, three Technical Colleges and the Welsh Joint Education

vii

v i i i INTRODUCTION

Committee. Unless otherwise stated, questions are from the Summer Advanced Level examination in the year stated.

Naturally, the layout of examination papers varies from one examining board to another, but all are set out so that the student answers questions on all three branches of chemistry. It cannot be too strongly stressed that in studying for the examination, no one branch, organic, inorganic or physical, should be sacrificed to learn more of the other two. Such a course may well end in disaster, and even if the student passes, his basic knowledge may then be so poor that further progress at a higher level is found to be impossible. In this respect, Ordinary National Certificate is the more efficient, since such a course is prevented by the setting of three papers, one in each branch of chemistry.

It is assumed that a student will either have a copy of the syllabus of the examination he is to sit for, or will be taught by someone familiar with it, but, nevertheless, a selection of past papers may also be useful, so that the candidate can familiarize himself with the layout of the paper and the type of question set by his board. (Details of appropriate addresses and costs are set out on pages 107-108.) It should be noted that the University of London Advanced Level and the three College Ordinary National Certificate papers (but not the Welsh Joint Education Committee) allow 3 hours for 5 questions, and in all other cases the candidate answers 6 questions in 3 hours, or 5 questions in 2\ hours. Consequently, answers to the former questions are a little longer than the others.

In the course of preparation, many alternative answers to parts of questions presented themselves, and it was felt that these too should be reproduced, but not in the answer itself, which would be unduly lengthened by text which would not normally appear in it. Such alternatives have, therefore, been placed in the notes after the appropriate question, and the student would put one or other of the alternatives (but not both) in his answer.

Other interesting points which arise in connection with an answer are also to be found in the notes, generally because it was felt that in the time allowed they were not sufficiently relevant to the question to be included, even though of importance in the general considera-tion of the topic.

Further questions are reproduced without answers, to enable the student to obtain practice in answering examination questions, to

INTRODUCTION IX

emphasize points already made, to stress the important parts of the syllabus, and to fill gaps which inevitably occur when questions are selected as has been done here.

In answering questions, the student is reminded to take notice of all the advice given to him by his teachers. In particular, when the examination comes, remember that the first marks are the easiest to obtain, so the full number of questions should always be attempted. Do not overrun your time on any question; keep an eye on the clock, and at the end of the appropriate time commence the next question, whether or not the previous one is finished (except perhaps when nearing the end of a calculation). If you have time at the end, come back to the unfinished questions. Always allow time to plan your answer, jotting down a few notes first; this avoids missing out essential items and putting in irrelevant matter. Finally, leave time to read through the finished paper; many a small error can be found and corrected by doing this.

In conclusion, remember that neither this book nor any other will enable you to pass any examination unless you are prepared to do the necessary work to learn the chemistry required.

Acknowledgements

THE authors gratefully acknowledge the help of the Examining Boards and Authorities listed below, in granting permission to reproduce questions set by them in past examinations.

It is expressly to be noted that copyright still, in all cases, rests with the Examining body.

Associated Examining Board for the General Certificate of Education

Southern Universities' Joint Board for School Examinations Welsh Joint Education Committee University of Durham School Examination Board University of London Bournemouth Municipal College of Technology and Commerce Medway College of Technology Whitehaven College of Further Education

These Examining bodies are in no way responsible for the answers contained in this book, which are solely the work of the authors.

xi

Structure and Isomerism

Explain clearly and concisely how the following are carried out in the laboratory: (a) Lassaigne sodium fusion, (b) the Carius method for the quantitative determination of chlorine.

(Welsh Joint Educat ion Commit tee, O . N . C . Syl labus A , 1963, qu . 8)

The most reliable method of detecting nitrogen, sulphur and halogens in organic compounds is by means of the Lassaigne sodium fusion. A small amount of the compound under investiga-tion is gently warmed in a soda-glass tube with a small piece of sodium. When reaction appears to have ceased and charring is complete, the tube and its contents are strongly heated and then plunged into cold water. (During this time the eyes should be carefully shielded, or goggles worn.) In this way any halogen present is converted to sodium halide, any sulphur to sodium sul-phide, and any nitrogen (by combining also with carbon) to cyanide.

The tube is crushed in the water, which is boiled to extract all the soluble material. The solution is then filtered or decanted off, and divided into three parts. Sulphide and cyanide are tested for first, since if present they must subsequently be removed before testing for halide.

(a) A very dilute solution of sodium nitroprusside is added to one portion. A purple colour confirms the presence of sulphur.

(b) To a second portion is added ferrous sulphate solution (extra if sulphide is present). A green precipitate of ferrous hydroxide is produced, since excess sodium will have rendered the solution alkaline, due to the formation of the oxide and peroxide.

F e S 0 4 + 2 NaOH = Fe(OH) 2 + N a 2 S 0 4

The mixture is then boiled, so that any sodium cyanide present will produce ferrous cyanide, and after subsequent acidification

1

2 MODEL ANSWERS IN ORGANIC CHEMISTRY

(with concentrated hydrochloric acid) sodium ferrocyanide.

2 NaCN + F e S 0 4 - Fe(CN) 2 + N a 2 S 0 4

Fe(CN) 2 + 4 NaCN - Na 4Fe(CN) 6

A few drops of ferric chloride are now added, when a blue or green colour (or precipitate) due to the formation of ferric ferro-cyanide results if nitrogen was originally present in the organic compound (see Note 1). Alternatively the presence of blue specks on a filter paper after filtration is also positive.

FeCl 3 + Na 4Fe(CN) e - N a F em[ F e

n( C N ) 6 ] + 3 NaCl

(c) If either or both sulphur and nitrogen are detected, the sulphide and/or cyanide are now removed from the third portion by acidifica-tion with sulphuric acid and boiling in an evaporating basin until no hydrogen sulphide or hydrogen cyanide can be detected in the vapours.

Na 2S + H 2 S 0 4 - N a 2 S 0 4 + H 2SÎ

2 NaCN + H 2 S 0 4 - N a 2 S 0 4 + 2 H C N |

Silver nitrate solution is added to the remaining solution, and any precipitate identified as chloride, bromide or iodide by colour (white, cream or yellow) and ease or otherwise of solubility in ammonium hydroxide.

NaX + A g N 0 3 - N a N O a + A g X |

If chlorine (or other halogen) has been identified qualitatively, and it is then desired to carry out a quantitative estimation, the Carius method is frequently used. This involves heating a weighed amount of the organic material with fuming nitric acid and silver nitrate in a sealed, hard glass tube.

In practice the tube is a long one, and 2-4 ml of acid and about 0-2 g of solid silver nitrate are first placed in the tube. The sample (about 0*15 g) is weighed out in a smaller glass tube which is lowered into the other, and this is then sealed by carefully drawing it out and sealing in a blowpipe flame.

The sealed tube is encased in a length of iron pipe, inverted once to ensure mixing of the sample with the acid and then placed in a furnace and maintained at 200°C for 4 or 5 hours. Stable sub-stances may require higher temperatures and longer times.

The tube is opened by playing a very small flame on the tip, until the glass blows out, after which the end may be safely broken

STRUCTURE AND ISOMERISM 3

off after making a file mark in it. The contents are then washed out quantitatively, filtered through a sintered crucible, and the precipitate washed, dried and weighed. From this the chlorine content of the original sample can be calculated, by

n / , f Wt. of AgCl 35-5 X 100 % of CI = — — χ — — — —

Wt. of compound 143*5

NOTES

1. N o ferric chloride need be added if the ferrous sulphate solution is boiled until yellow (i.e. oxidized) before addition to the original solution.

What is meant by isomerism? Describe four different ways in which isomerism can arise in organic compounds. Where possible, illustrate your answer by examples of each type.

(University of L o n d o n , Jan. 1959, Paper II, qu . 11)

If two or more compounds possess the same molecular formula, but differ in structure or in the arrangement of their functional groups in space, they are said to be isomeric, or to exhibit isomerism. Isomers generally differ from each other in both chemical and physical properties, but optical isomers differ only in optical properties, and crystalline structure in a few cases.

Two main types of isomerism are possible, structural where similar groups are arranged differently on a chain or nucleus, or where the basic structure and functional groups may differ, and spatial or stereoisomerism where the same groups are linked to the same carbon atoms, but with a different arrangement, relative to each other, in space.

Each main type may be subdivided, and examples of some of the ways in which isomerism may arise are given.


(A) STRUCTURAL ISOMERISM

( i ) N U C L E A R O R C H A I N I S O M E R I S M

In this case the same functional groups are present, but arranged differently on the carbon chain or ring of the parent hydrocarbon; chemical properties are therefore in many cases quite similar, and the physical properties differ markedly only in some aromatic compounds.

H OH HO H

CH 3—C—C—H and CH 3—C—C—H I I I I

H H H H propan-l-ol η-propyl alcohol propan-2-oI isopropyl alcohol

N 0 2 N O , N 0 2

E.g.

N 0 2 N 0 2

or/Äo-dinitrobenzene wefa-dinitrobenzene N O ,

/>ara-dinitrobenzene

C H 3—C H—C H 3

CH 3—CH 2—CH 2—CH 3 and | n-butane

2-methylpropane isobutane

( Ü ) M E T A M E R I S M

When different alkyl groups are linked to the same functional group, the isomers are said to be metamers.

E.g. C 2H 5— O — C 2H 5 and CH 3—O—C 3H 7

diethyl ether methyl propyl ether

( Ü i ) F U N C T I O N A L G R O U P I S O M E R I S M

Here the functional group present differs in the different isomers, making them readily distinguishable by chemical (and generally by physical) properties.

E.g. CH 3—CH 2—OH and CH 3—O—CH 3

ethanol dimethyl ether

CH 3—CH 2—CHO and CH 3—CO—CH 3

propionaldehyde acetone


(Β) SPATIAL OR STEREOISOMERISM

(i) G E O M E T R I C A L ISOMERISM

When carbon atoms are linked by double bonds, free rotation is prevented ; thus any pair of groups may be on the same side or on opposite sides of the double bond, giving rise to eis or trans isomers respectively. Here the physical properties are very different, and this has led in some cases to quite different names being given to the two isomers, even though some of the chemical properties are similar.

HOOC COOH H COOH

\ / \ / C = C and C = C

/ \ / \ Η Η HOOC Η

maleic acid (eis) fumaric acid (trans)

( Ü ) O P T I C A L ISOMERISM

If a molecule is asymmetric, and the mirror image is not super-imposable on the original, enantiomorphs or optical isomers are obtained. These differ in that one, the dextrorotatory or ( + ) isomer rotates a beam of plane polarized light to the right, whilst the other, the laevorotatory or (—), rotates it the same amount to the left. Their chemical properties are identical but their rates of reaction with other optical isomers may differ. Four different groups linked to one so-called "asymmetric" carbon atom give rise to molecules which can exist as optical isomers, since each bond is directed to one corner of a regular tetrahedron.

E.g. COOH COOH I I c c

/ j \ / ! \

HO j H H j OH CH3 CH 3

( + ) and (—) isomers of lactic acid

2


Define and write notes on the following terms: (a) Optical isomers, (b) Racemic mixture, (c) Internal compensation, (d) Resolution, (e) Geometrical isomers.

(Medway College of Technology, O.N.C., 1961, qu. 2)

(a) If a molecule is asymmetric, i.e. if its mirror image cannot be superimposed on itself, it is capable of existing in two forms, either one being the mirror image of the other. Such forms are known as optical isomers, or stereoisomers, and are identical in all their chemical properties; they differ physically only in that one rotates the plane of polarized light to the right and the other to the left (to the same extent). The two isomers are known respectively as dextrorotatory or ( + ) and laevorotatory or (—) (see Note 1).

The most common manner in which a molecule becomes asym-metric is to possess an asymmetric carbon atom (but see Note 2), i.e. a carbon atom to which four different atoms or groups are attached. Lactic acid affords an example of a compound of this type and, because the valencies of the carbon atom are directed to the corners of a tetrahedron, any one isomer of lactic acid is non-superimposable on its mirror image.

(b) Normal chemical methods for the preparation of optical isomers result in formation of equimolecular amounts of the ( + ) and (—) isomer and thus the product is optically inactive due to external compensation, i.e. there are an equal number of molecules rotating the plane of the polarized light in each of the two opposed directions and thus the overall rotation is zero. Such products are said to be racemic and dependent on their structure are racemic mixtures (which may be separated by physical methods), racemic compounds or racemic solid solutions (both of which must be separated by chemical methods). All racemic forms are denoted

(c) If one molecule has two similar asymmetric carbon atoms, as has tartaric acid, it is possible for these two centres to confer oppo-site rotations on the molecule. Since the groups on the two atoms

CH. 3 CH 3

HOOC H OH

as ( ± ) .

STRUCTURE AND ISOMERISM

are identical, the magnitude of the rotations is identical and the resultant rotation is zero. Unlike the racemic forms, the cancelling effect here comes from within each molecule, and so is known as internal compensation; the isomer produced is the meso- form.

COOH

HC> I H

M / c

H

COOH

OH

H

C

OH

COOH

HO

C

H

COOH

D ( - ) tartaric acid L ( + ) tartaric acid

COOH

ç

/>.„ COOH

moro-tartaric acid

(d) Such separation of the isomers from the racemic form is known as resolution. Occasionally, where a racemic mixture exists and the crystals themselves are asymmetric, it is possible to separate the crystals by handpicking, as Pasteur did with some of the tartrates, but more frequently all forms, mixture, compound or solid solution, are resolved by forming diastereoisomers. These are optically active compounds with two or more optically active centres, but are not mirror images of each other. These differ in physical properties such as solubility, and so may be separated by fractional crystalliza-tion.

Thus, racemic acids may be resolved by reacting them with optically active bases or, in some cases, alcohols, and crystallizing out the diastereoisomeric salts, or esters, from which the parent acid may be obtained.

( ± ) acid + (—) base -> ( + ) acid (—) base + (—) acid (—) base

(e) Like optical isomers, geometrical isomers differ from each other only in the arrangement of their constituent groups in space but, unlike them, they have a plane of symmetry and so do not exist in ( + ) and (—) forms. Geometrical isomers most frequently arise due to the lack of free rotation about a double bond and thus substituted alkenes of the type Cab=Cab can exist as two isomers, the eis- with the two similar groups on the same side of the double


bond, and the trans- with the two similar groups on opposite sides of the double bond (see Note 3). Such isomers differ markedly in physical properties and so have often been given, in the past, entirely different names, as for example with maleic acid (eis) and fumaric acid (trans).

H—C—COOH HOOC—C—H M maleic acid \> fumaric acid I I (eis) I ! (trans)

H—C—COOH H—C—COOH

NOTES

1. The old symbols d- and /- should no longer be used; neither should D - and L- which refer not to rotation but to the absolute configuration, i.e. arrangement of groups in space.

2. Covalent inorganic compounds can also be optically active, as can molecules which are asymmetric as a whole, without possessing an asymmetric atom of any sort.

3. Thus, a—C—b a—C—b

II and li a—C—b b—C—a

eis trans

whereas in Cab=Cxy the eis- and trans- forms are named arbitrarily

a—C—b a—C—b I I and I I

x—C—y y— C— χ eis and trans

and in Caa=Cab and Cab=Cxx no isomers exist.


C H O

Actual weight 0-264 χ 1? 0-108 χ 1 °*116 ~ ®m™ of element « " + ™

1 2>

- 0-072 g - 0 - 0 1 2 g - 0-023 g

Relative no. of atoms 0-072 Λ _ 0-012 0-032

- 0006 —-— - 0-012 — - - 0-002

\A.W./ 12 1 16

Ratio of atoms (divide by 3 6 1 0-002)

Empirical formula C 3 H 6 0

Molecular formula (C 3H eO),

Molecular weight [(3 χ 12) + (6 χ 1) + (1 χ 16)] β - 58χ

Λ . 0-210 x 1000 . 0-210 x 1000 , Now, density — — g/1. Λ V.D. = —— —— thus J 81-1 Bl 81-1 X 0-09

„ n 2 x 0-210 x 1000 140 ^ since Mol. wt. — 2 χ V.D. — — — ——— — — - = 57-6

81-1 X 0-09 2-43 58x — 57-6 .'. since χ must be an integer χ = 1 and the molecular formula must be C 3H eO.

An organic compound, containing only carbon, hydrogen and oxygen, on complete combustion gave the following data: 0-116 g gave 0·264 g carbon dioxide and 0·108 g water. A further 0·210 g gave 81·1 ml of vapour at S.T.P. Calculate the molecular formula of the compound and write structural formulae for the possible isomers.

Describe FOUR reactions by which these isomers could be dis-tinguished.

(University o f L o n d o n , Jan. 1963, Paper I , qu . 1).

1 0 MODEL ANSWERS IN ORGANIC CHEMISTRY

With one oxygen atom this could be an aldehyde, a ketone, an alcohol or an ether (see Note 1), but the two simplest compounds with this formula would be

H /

CH 3—CH 2—C and CH 3—C—CH 3 \ li ο ο

propionaldehyde acetone

and these could be distinguished by: (1) Warming with Tollen's reagent (ammoniacal silver nitrate).

Only the aldehyde reduces the oxide to give a silver mirror.

CH 3—CH 2—CHO + A g 2 0 - * CH 3—CH 2—COOH + 2 Ag

(2) Warming with Fehling's solution (a cupric salt in alkaline solution). Again, only the aldehyde is a reducing agent, and causes the precipitation of cuprous oxide; a series of colour changes may be observed, green to yellow to red and possibly brown.

CH 3—CH 2—CHO + 2 CuO -> CH 3—CH 2—COOH + C u 2 0

(3) Reduction with lithium aluminium hydride or sodium and water (added to the compound to be reduced). The aldehyde yields the primary alcohol, propan-l-ol, whereas the ketone gives the secondary alcohol, propan-2-ol.

CH 3—CH 2—CHO -> CH 3—CH 2—CH 2OH

CH 3—CO—CH 3 -> CH 3—CH(OH)—CH 3

(4) Bubbling dry ammonia into a cooled solution of the aldehyde in ether. A white precipitate of the addition product, propion-aldehyde ammonia is first produced.

CH 3—CH 2—CHO + H N H 2 -> CH 3—CH 2—CH(OH)—NH 2


Acetone, however, gives no comparable product, but instead forms a dimer, to which one molecule of ammonia adds.

CH 3 CH 3 CH 3 CH 3

CH 3—C + C H , — C = 0 — C H 3 — C = C H — C = 0

A CH 3 C H 3 ι ι

CH 3—C=CH—C—OH I

N H 2

diacetoneamine

This may be distinguished from the corresponding aldehyde ammonia by means of a melting-point determination.

NOTES

1. Other possible structures for C 3 H 6 0 are unlikely to be required at 'A' Level but include :

CH 2 CH 3—CH—CH 2 CH 2—CH 2 C H 3 — Ο — C H = C H 2

\ \ / I I methyl vinyl ether

CHOH Ο CH 2—Ο / propylene oxide trimethylene oxide

C H 2

cyclo-propanol

C H 2 = C H — C H 2 O H allyl alcohol

Alkyl Halides and Related Compounds

Describe the essential features of a suitable laboratory method for the preparation of reasonably pure ethyl iodide.

State the reagents and conditions which are necessary for its con-version into any four of the following compounds: (a) n-butane, (b) ethyl benzoate, (c) ethyl η-propyl ether, (d) nitroethane, (e) pro-pionitrile (ethyl cyanide), (f) ethyl benzene.

Write the structural formula of the product in each example chosen.

(Welsh Joint Educat ion Commit tee , 1961, Paper I , qu . 11)

Ethyl iodide is best prepared by the action of phosphorus tri-iodide on ethanol, the former being prepared in situ.

3 C 2H 5OH + PI 3 — 3 C 2H 5I + H 3 P 0 3

ethanol ethyl iodide

Red phosphorus and ethanol are placed in a flask which is then warmed on a waterbath until the ethanol is refluxing. Iodine (either solid or in ethanol solution) is slowly added to the mixture until sufficient to convert all the phosphorus to the tri-iodide has been used. Refluxing is stopped when violet fumes of iodine appear in the condenser, and the mixture is then distilled. Since the iodide is very volatile a double surface condenser is best employed, and the distillate is collected in ice-cold water. The dense layer is then tapped off and washed with (1) sodium carbonate solution, to remove any hydriodic acid, (2) brine, to remove any alcohol, and (3) water. The remaining liquid is dried with calcium chloride (which also serves to remove any ethanol still present), filtered or decanted off, and redistilled.

(a) Conversion to η-butane may be carried out by the Wurtz reaction. The iodide is reacted with sodium wire in dry ether

12

ALKYL HALIDES AND RELATED COMPOUNDS 13

under reflux when η-butane is produced. The gas is carried off

from the top of the condenser and liquefied in a well-cooled receiver. The liquid boils at - 0 - 5 ° C .

(b) Ethyl benzoate can be prepared by refluxing the ethyl iodide with a salt of benzoic acid; if a very pure sample is required the silver salt would be used, otherwise sodium or potassium benzoate would be satisfactory. By fractional distillation, the ester may then be recovered.

C 6 H 5 - C O O A g + C 2H 5I -> C 6 H 5 - C O O - C 2 H 5 + Agi

(c) Williamson's synthesis is a method of preparing ethers from alkyl halides, and ethyl η-propyl ether may be obtained by refluxing a mixture of ethyl iodide and sodium n-propoxide in n-propanol. Again fractional distillation is used to isolate the product.

C 3H 7ONa + C 2H 5I C 3H 7— O — C 2H 5 + Nal

(d) Nitroethane is prepared by refluxing aqueous alcoholic silver nitrite with ethyl iodide.

Some ethyl nitrite is also produced. The nitroethane is almost immiscible with water, and can be tapped off from the reaction mixture (after adding water), washed, dried and distilled.

The structural formula for each of the above products is given below:

2 C 2H 5I + 2 Na -> C 4 H 1 0 + 2 Nal

C 2H 5I + AgONO — C 2 H 5 — N 0 2 + Agi

(a) n-butane

CH3—CH 2—CH 2—CH 3

ethyl benzoate (b)

(c) ethyl η-propyl ether CH 3—CH 2—Ο—ι -CH2—CH2—CH3

(d) nitroethane Ο


NOTES

1. Only four compounds are required in the answer; the other two alternatives are given below.

(e) By refluxing with aqueous alcoholic potassium cyanide, ethyl iodide is converted to ethyl cyanide (propionitrile)- This too may be distilled off, and thus separated from any excess iodide.

(f) If ethyl iodide, benzene and anhydrous aluminium chloride are heated gently on a waterbath, an insoluble complex is produced, and hydrogen iodide evolved. On addition of dilute acid, this breaks down to give an aqueous layer, and above it ethyl benzene dissolved in any excess benzene. The reaction is the Friedel-Crafts reaction and is used for alkylation of aromatic rings.

Unless excess benzene is used over-alkylation may occur and diethyl benzenes will be obtained.

The structural formula for each of the above products is given below :

C 2H 5I + KCN -> C 2H 5C N + KI

C 6H 6 + C 2H 5I A I C I 3

> C 6 H 6 - C 2 H 5 + HI

(e) ethyl cyanide C H 3 — C H 2 — C = N

(f) ethyl benzene

ALKYL HALIDES AND RELATED COMPOUNDS 1 5

Write the structural formulae of the compounds which are pro-duced by the replacement of one or two atoms in a molecule of ethane by bromine. State ONE method by which each of the bromine deriva-tives could be prepared, other than from ethane. What products would be formed in the reaction of each of these compounds with potassium hydroxide (a) in aqueous solution, and (b) in alcoholic solution?

(Southern Universities Joint B o a r d , 1960, Paper I, q u . 9)

If one hydrogen atom in ethane is replaced by bromine, ethyl bromide or bromoethane is produced. Its structure is:

H H I I

H—C—C—Br I I

H H If two hydrogen atoms are replaced, two isomers can be obtained. These are :

H H H H

I I I I Br—C—C—Br and H—C—C—Br

I l I I H H H Br

1,2-dibromoethane 1,1-dibromoethane ethylene dibromide ethylidene dibromide

Ethyl bromide is normally prepared by carefully running bromine in ethanol into a flask containing red phosphorus and ethanol until the molecular ratio of bromine to phosphorus is 3:1. After standing overnight, the mixture is distilled. Ethyl bromide is produced and collected in water to reduce losses due to its high volatility.

3 C 2H 5O H + PBr 3 -> 3 C 2H 5Br + H 3 P 0 3

1,2-Dibromoethane is obtained by bubbling ethylene through bromine water, or bromine covered by a layer of water. A trap is usually inserted in the apparatus to reduce the losses of bromine and the product,

C H 2 = C H 2 + Br 2 — CH 2Br—CH 2Br

most of which, however, remains below the water in the reaction vessel as a dense, colourless liquid, which is tapped off, washed, dried and distilled.


l9l-Dibromoethane is made by reacting acetylene with hydrogen bromide in a suitable vessel. Vinyl bromide is first formed,

HBr HBr

C H = C H T Ô Ô ^ C H 2= C H B r • CH 3—CHBr 2

and further addition gives the asymmetric dibromoethane which remains as a dense, colourless liquid, purified by distillation.

All these compounds react with both aqueous and alcoholic potassium hydroxide. In general, aqueous alkali brings about hydrolysis, the halogen being replaced by —OH, whilst alcoholic alkali causes elimination of hydrogen bromide with the formation of an unsaturated compound.

Thus, if ethyl bromide is refluxed with aqueous sodium hydroxide, hydrolysis occurs and ethanol is produced,

C 2H 5Br + KOH - ^ > C 2 H 5 O H + KBr

but if alcoholic alkali is used some ethylene is obtained

C 2H 5Br + KOH -> C 2H 4 + KBr + H 2 0

together with a large amount of ethanol. Some ether is also produced

by the reactions occurring only in alcoholic solution.

C 2H 5OH + KOH -> C 2H 5OK + H 2 0

C 2H 5OK + C 2H 5Br -> C 2H 5— O — C 2H 5 + KBr

It is often incorrectly stated that this is a method of preparing

ethylene; although an appreciable volume of gas is produced

(224 ml for 1 mole of ethyl bromide) the yield is only 1 per cent. 1,2-Dibromoethane under the same conditions as ethyl bromide

gives the expected products, the dihydric alcohol, ethylene glycol, and (by elimination of two molecules of hydrogen bromide) acetylene respectively,

BrCH 2—CH 2Br + 2 KOH - 2 ^ > HO—CH 2—CH 2—OH + 2 KBr

BrCH 2—CH 2Br + 2 KOH C H = C H + 2 KBr + 2 H 2 0 and \ ̂ dibromoethane also behaves similarly, except that in hy-drolysis, the géw-diol is produced, with two —OH groups on one carbon atom. This is unstable and dehydrates to give acetaldehyde

CH 3—CHBr 2 + 2 KOH

[CH 3—CH(OH) 2] + 2 KBr -> CH 3—CHO + H 2 0

but the elimination reaction proceeds normally again giving acetylene

CH 3—CHBr 2 + 2 KOH ^ > C H = C H + 2 KBr + 2 H 2 0

ALKYL HALIDES AND RELATED COMPOUNDS 17

Describe, with full practical details, the preparation of a pure sample of chloroform, starting with ethyl alcohol. Discuss the possible reactions which occur in the preparation.

What impurities would you expect to find in an old sample of chloroform and why? How would you detect them?

Chloroform is generally prepared in the laboratory by the action of bleaching powder or sodium hypochlorite on acetone, acetalde-hyde, or compounds which can be oxidized by the hypochlorite to yield these or other compounds containing a CH 3CO— group, and thus ethanol is often used for the preparation.

The apparatus sketched above (Fig. 1) is used in the preparation. A distillation flask is not used, since the froth formed is carried over to the distillate more easily than by using an ordinary flask and delivery tube. A splash-head can be inserted here with advantage.

A large excess of bleaching-powder is now ground with the minimum of water, and the suspension decanted off into the flask, the process being repeated only until a little gritty material remains in the mortar used.

About one-tenth of this weight of ethanol is taken, diluted with the same volume of water, and the solution placed in the dropping funnel. It is then run slowly into the flask, which is heated gently.

(Southern Universities Joint B o a r d , 1959, Paper I, q u . 9)

FIG. 1.


The mixture froths and chloroform soon starts to distil over; when this happens more of the ethanol is slowly added until all is used up. A damp cloth is used to cool the flask if frothing becomes too pronounced, and the mixture is heated until the distillate is clear.

The lower layer of chloroform is then tapped off from the upper aqueous layer, washed once with dilute alkali to remove any chlorine and then with water. It is dried over calcium chloride and finally redistilled.

If it is assumed that bleaching powder is a mixture of calcium hydroxide, chloride and hypochlorite and that the latter is capable of liberating chlorine in solution, then it seems probable that the formation of chloroform is the result of a series of reactions, the first being the oxidation of the ethanol to acetaldehyde.

2 CH 3—CH 2—OH + Ca(OCl)2 -> 2 CH 3—CHO + CaCl 2 + 2 H 2 0

Then the available chlorine of the bleaching powder chlorinates the aldehyde forming trichloroacetaldehyde,

CH 3—CHO + 3 Cl 2 -> CC13—CHO + 3 HCl

which is finally hydrolysed by the calcium hydroxide, to give chloroform and calcium formate.

2 CC13—CHO + Ca(OH) 2 -> 2 CHC13 + (H—COO) 2Ca

After standing in air a little chloroform becomes oxidized to give the very poisonous carbonyl chloride and hydrogen chloride.

2 CHC13 + 0 2 -> 2 COCl 2 + 2 HCl

Thus, if after shaking a sample of chloroform with water the aqueous layer gives a precipitate of silver chloride on adding it to silver nitrate solution, decomposition is indicated and the chloroform should not be used for anaesthesia. Some chlorine may also be present after standing, due to the reaction

4 CHC13 + 3 0 2 -> 4 COCl 2 + 2 H 2 0 + 2 Cl 2

and this will also produce hypochlorous acid,

Cl 2 + H 2 0 = HCl + HOC1

ALKYL HALIDES AND RELATED COMPOUNDS 1 9

and so if another part of the aqueous solution is added to potassium iodide solution, iodine will be liberated by the chlorine, if present. Starch may be added, if desired, to make the test more sensitive.

2 KI + Cl 2 = 2 KCl + I 2

If these impurities are present, carbonyl chloride is removed by adding about 1 per cent of ethanol which forms diethyl carbonate

2 C 2H 5O H + COCl 2 -> CO(OC 2H 5) 2 + 2 HCl

and both chlorine and hydrogen chloride can be removed by shaking with dilute alkali, washing and drying. Alternatively, prior addition of ethanol either prevents the formation of the carbonyl chloride or reacts with the traces which are formed.

Ether and Alcohols (a) Starting from ethyl alcohol, explain with full details how you

would prepare a specimen of diethyl ether. (b) How would you convert ethyl alcohol into (i) ethyl bromide,

(ii) normal butane, (iii) normal propyl alcohol, (iv) acetone, (v) ethyl-ene? (Note: Detailed experimental quantities and descriptions are not required in Part (b)).

(University o f D u r h a m , 1962, Paper I , qu . 3)

Diethyl ether is prepared in the laboratory by Williamson's continuous etherification process which involves the dehydration of ethanol using concentrated sulphuric acid.

C 2H 5OH + H 2 S 0 4 -> C 2 H 5 H S 0 4 + H 2 0

C 2 H 5 H S 0 4 + C 2H 5OH -> C 2H 5— O — C 2H 5 + H 2 S 0 4

ethyl hydrogen sulphate diethyl ether

Care is needed to ensure that the temperature is maintained at or below 140°C, otherwise large amounts of ethylene are obtained. The apparatus sketched below (Fig. 2) is used. In the flask is prepared a mixture of equal volumes of ethanol and concentrated sulphuric acid, the latter being slowly added to the former with shaking and cooling. Ethanol (or methylated spirits) is placed in the dropping funnel, which is extended below the surface of the liquid to prevent loss of ethanol on addition.

The flask is now heated until the temperature of the liquid reaches 140°C, by which time ether is distilling over. More alcohol is run in, at the rate at which the ether is distilling over, until all has been added. After this, heating is continued until no more ether is ob-tained. Since ether vapour is very inflammable, the tube from the receiver is carried well clear of and below the apparatus to minimize any danger due to fire.

20

ETHER AND ALCOHOLS 21

FIG. 2.

(b) (i) If a mixture of ethanol and excess concentrated sulphuric acid is carefully prepared, added to potassium bromide and then distilled, ethyl bromide is obtained.

KBr + H 2 S 0 4 = K H S 0 4 + HBr

C 2H 5OH + HBr C 2H 5Br + H 2 0

(ii) If the ethyl bromide thus obtained is then run slowly on to sodium wire in dry ether, a reaction occurs which evolves sufficient heat to cause the ether to boil. After all the bromide has been added, refluxing is continued for some time; during the course of the reaction, η-butane is evolved and may be collected as a gas or condensed out in a freezing mixture. This is the Wurtz reaction.

3

2 C 2H 5Br + 2 Na -> C 4 H 1 0 + 2 NaBr

The ether is washed with dilute alkali or sodium carbonate solu-tion to remove sulphur dioxide (and ethanol) and then thoroughly with water. After tapping off, the ether layer is carefully dried with calcium chloride and redistilled using a water-bath. The fraction boiling between 33 and 37°C is collected.

22 MODEL ANSWERS IN OROANIC CHEMISTRY

(iii) If the ethyl bromide from (i) is refluxed with aqueous-alcoholic potassium cyanide, ethyl cyanide is obtained.

C 2H 5Br + KCN -> C 2 H 5 C N

This is hydrolysed by refluxing with dilute sulphuric acid and the propionic acid thus produced is isolated and dried.

C 2H 5C N + 2 H 2 0 ^ > C 2H 5COOH + N H 3

The acid can then be reduced either with lithium aluminium hydride, or by catalytic hydrogénation using an oxide catalyst, such as copper chromite, to yield n-propanol.

C 2H 5COOH + 4 H L i A 1 H4

> C 3H 7OH + H 2 0

(iv) If ethanol is refluxed with a solution of potassium dichromate and dilute sulphuric acid, it is oxidized to acetic acid, which can be distilled from the reaction mixture.

C 2H 5O H + 2 Ο - * CH 3—COOH + H 2 0

This is converted to the calcium salt with calcium hydroxide

2 CH 3—COOH + Ca(OH) 2 (CH 3—COO) 2Ca + 2 H 2 0

and the solution evaporated to dryness. On heating, the solid salt decomposes to yield acetone.

(CH 3—COO) 2Ca -> CH 3—CO—CH 3 + CaCO a

(v) If an excess (3:1) of concentrated sulphuric acid is carefully and slowly added to ethanol with cooling and shaking and the mixture then heated to about 165°C, ethylene is evolved.

C 2H 5OH + H 2 S 0 4 -> C 2 H 5 H S 0 4 + H 2 0 ethyl hydrogen sulphate

C 2 H 5 H S 0 4 -> C 2 H 4 + H 2 S 0 4

ethylene

Alternatively, the reactions can be given as

C 2H 5OH + H+ — C 2H 5OH 2+ -> C 2 H 4 + H+ + H 2 0

Alcohols and Phenols

Name the isomers of the alcohol which has the molecular formula C 4H 1 0O and give the structural formula of each isomer. How do these isomers behave on (i) oxidation, (ii) with phosphorus penta-chloride?

By what physical property may one of the isomers be distinguished from the others?

(University of London, 1961, Paper I, qu. 1)

Since the compound C 4H 1 0O is an alcohol it must be, more correctly, C 4H 9OH and thus one of the four isomeric butyl alcohols.

These are

(a) CH 3—CH 2—CH 2—CH 2—OH η-butyl alcohol or butan-l-ol

(b) CH 3—CH 2—CH—OH s-butyl alcohol or butan-2-ol

C H 3

(c) CH 3

I CH 3—C—OH t-butyl alcohol or

I 2-methylpropan-2-ol or CH 3 trimethylcarbinol

(d) CH 3

CH 3—C—CH 2—OH isobutyl alcohol or I 2-methylpropan-1 -ol

H

(i) Compounds (a) and (d) are both primary alcohols, and on oxidation with potassium dichromate and dilute sulphuric acid yield first aldehydes and then acids.

23


CH 3—CH 2—CH 2—CH aOH -+ CH 3—CH 2—CH 2—CHO — butyraldehyde; butanal

CH 3—CH 2—CH 2—COOH butyric acid; butanoic acid

CH 3 CH 3 CH 3

\ CH—CH2OH -> CH - C H O > CH--COOH

/ / /

CH 3 CH 3 CH 3

isobutyraldehyde isobutyric acid 2-methylpropanaI 2-methylpropanoic acid

Compound (b) is a secondary alcohol, and is oxidized with little

more difficulty to a ketone, using the same reagents

CH 3—CH 2 CH 3—CH 2

\ CH—OH -> CO

/ / /

CH 3 CH 3

methyl ethyl ketone; butanone,

Compound (c), being a tertiary alcohol, can only be oxidized with stronger reagents, and more difficulty. When oxidation occurs, the molecule is partly broken down to carbon dioxide and water, the remainder yielding a ketone containing fewer carbon atoms than the original alcohol (but see Note 1).

CH 3 CH 3 ι ι CH

3

—C—OH - * CH

3

—CO + C 0

2

+ H

2

0

I acetone CH 3

The aldehydes and ketones produced can be identified by means of

their 2,4-dinitrophenylhydrazones. (ii) All the alcohols react with phosphorus pentachloride to giv

e the corresponding alkyl chlorides. The reaction proceeds mos

t readily with the tertiary alcohol, and least readily with the primary

. PCI5

CH3—CH

2—CH

2—CH

2—OH >· CH

3—CH

2—CH

2—CH

2—C

I n-butyl chloride

CH 3 - C H 2— C H O H — C H 3 - > CH3—CH,—CHC1—CH3

s-butyl chloride

(CH 3) 3C—OH — (CH 3) 3C—CI t-butyl chloride

(CH 3) 2CH—CH 2—OH — ( C H 3) 2C H — C H 2- C 1 isobutyl chloride

ALCOHOLS AND PHENOLS 25

The t-butyl alcohol, being a more symmetrical and therefore "compact" molecule than the other isomers, may be distinguished from them by being a solid at room temperature, whereas the others are liquids.

NOTES

1. Dehydration occurs first, to give an alkene, which is then oxi-dized to the ketone. This in turn may be further oxidized to carbox-ylic acid(s), carbon dioxide and water.

CH 3 CH 3

CH 3—C—OH ^ > CH 8—C - ^ - >

CH 3 CH 2

C H 3 OH

I ο I CH 3—C > CH 3—C + C 0 2 + H 2 0

II IS ο ο


Give (a) two reactions in which ethyl alcohol and phenol behave similarly, (b) two reactions which are given by ethyl alcohol but not by phenol, (c) two reactions which are given by phenol but not by ethyl alcohol.

By means of a labelled sketch and brief notes, give the usual laboratory method of determining the melting-point of an organic compound.

(Associated Examining Board, 1959, Paper II, qu. 10)

Although both alcohols and phenols possess an —OH group the fact that it is linked to an aliphatic chain or an aromatic nucleus respectively tends to confer different properties on the two classes of compounds.

(a) Some reactions are, however, shown by both, e.g. both alcohols and phenols react with acyl halides to produce esters, the former violently, the latter much more slowly and only in the presence of alkali or pyridine.

C 2H 5O H + CH 3—CO—Cl-> CH 3—COO—C 2H 5 + HCl ethanol acetyl chloride ethyl acetate

shake with

C 6H 5OH + C H 3 - C O - C I p y r i d | n e> C H 3 - C O O - C 6 H 5 + HCl phenol phenyl acetate

Similarly both alcohols and phenols react with phosphorus pentahalides, the —OH group being replaced by halogen, but the yield when a phenol is used is poor, triphenyl phosphate being the major product (see Note 1).

C 2H 5OH + PC15 C 2H 5C1 + POCl 3 + HCl ethyl chloride

C e H 5O H + PC15 — C 6H 5C1 + POCl 3 + HCl chlorobenzene

3 C 6 H 5 0 H + PC1S -> ( C eH 50 ) 3P C l 2 — ( C eH 50 ) 3P O triphenyl phosphate

(b) Other reactions only occur with ethanol; thus on heating with excess concentrated sulphuric acid the alcohol is dehydrated and ethylene is obtained,

C 2 H 5 O H + H 2 S 0 4 -> C 2 H 5 H S 0 4 — C 2H 4 + H 2 S 0 4

ethyl hydrogen sulphate ethylene

ALCOHOLS AND PHENOLS 27

Flo. 3.

and on treating ethanol with oxidizing agents such as acid di-chromate, acetaldehyde and then acetic acid are produced.

C H 3 - C H 2 - O H + Ο C H 3 - C H O

CH 3—CHO + Ο :

H 2S 0 4

Reflux

H 2 0

-> CH 3—COOH K 2C r 2 70 + H 2S 0 4 ~ "A

3 ~

Neither of these reactions occur with phenol. (c) Phenol also shows reactions not exhibited by alcohols, for

example, it reacts with aqueous alkali to give salts not hydrolysed by water,

C 6H 5O H + NaOH -> C 6H 5O N a + H 2 0 sodium phenate

and ethers are produced by reacting with alkyl halides in the presence of alkali.

C eH 5OH + CH 3I NaOH

> C 6 H 5 O C H 3 + HI methyl phenyl ether (anisole)

Figure 3 shows the apparatus usually used for the determination of the melting point of an organic compound.


Details are : A—boiling tube containing Β—paraffin oil. C—thin cork (to cover minimum length of thermometer stem)

bored with air vent. D—hand-operated stirrer. Ε—thermometer. F—capillary tube, approx. 1 mm bore, containing compound held

to thermometer by surface tension of paraffin oil, or G—rubber band, well clear of hot oil. H—bunsen burner, raising temperature by 1°C per 2 min in

region of melting-point. The material is finely powdered and tapped into the capillary

tube, which is then put in place. The temperature is raised slowly and steadily until the material is observed to melt. The temperature is noted between the point at which the crystals begin to shrink or move in the tube until they are completely melted.

1. Since the yield is so poor, it might be claimed that this reaction is a difference between the compounds.

A further similarity is with the alkali metals, which react with both alcohols and phenols to liberate hydrogen, and produce a salt.

NOTES

2 C 2H 5OH + 2 Na 2 C 2H 5ONa + H 2Î sodium ethoxide

2 C eH 5O H + 2 Na 2 C 6H 5ONa + H„î sodium phenate

The phenol must be melted first.

Aldehydes and Ketones

Describe a laboratory method for the preparation of a pure sample

of acetaldehyde. How and under what conditions does acetaldehyde react with

(a) hydrogen, (b) sodium hydroxide, (c) hydrogen cyanide, (d) hy-droxylamine?

How would you convert the product formed in (c) to lactic acid?

(University o f D u r h a m , 1962, Paper IT, qu . 6)

Acetaldehyde is generally prepared in the laboratory by the oxidation of ethanol with acid dichromate.

C H 3 - C H 2 - O H + O ~ X C H 3 - C H O + H 2 0

The apparatus sketched below (Fig. 4) is used, a double surface condenser being preferred. A 1:1 (by volume) mixture of ethanol and concentrated sulphuric acid is prepared by carefully adding the latter, with shaking and cooling to the alcohol. This mixture is placed in the dropping funnel and slowly run into the flask which contains a warm 25 per cent aqueous solution of potassium or sodium dichromate.

Oxidation occurs and sufficient heat may be evolved to allow the burner to be removed from beneath the flask. The acetaldehyde distils over. When the volume of distillate is roughly equal to that of the ethanol used, heating is stopped. To the distillate, calcium chloride is added to remove water and alcohol. The upper of the two layers which forms is tapped off and redistilled, preferably using a fractionating column, and the liquid boiling between 20 and 22°C is collected in an ice-cooled receiver. The pure acetalde-hyde thus obtained is very volatile and must be stored in a well-stoppered bottle.

29


(a) Acetaldehyde may be reduced either by catalytic hydrogén-ation, using hydrogen gas and a Raney nickel catalyst, or by nascent hydrogen produced by the action of sodium on a small amount of

FIG. 4.

water added to the aldehyde for this purpose. Alternatively, the nascent hydrogen may be produced by the action of a metal on an acid.

However the reduction is carried out, ethanol is produced.

CH 3—CHO + 2H -> CH 3—CH 2—OH

(b) If acetaldehyde is shaken with dilute sodium hydroxide solution, a pale yellow precipitate of "aldol" is produced by the aldol reaction.

H H

I I CH 3—C + HCH 2—CHO -> CH 3—C—CH 2—CHO

II I Ο OH

If slightly stronger alkali is used the precipitate darkens in colour and resinifies due to dehydration and further polymerization.

ALDEHYDES AND KETONES 31

Heating also promotes this reaction, which is characteristic of aliphatic aldehydes.

CH 3—CH—CH 2—CHO = Ä C H 3— C H = C H — C H O ι — H 20 I crotonaldehyde

OH CH 3—(CH=CH) X—CHO

(c) Hydrogen cyanide reacts readily with acetaldehyde to give an addition compound, the cyanohydrin, which precipitates out when the aqueous solutions of the two reagents are mixed. More com-monly sodium cyanide and dilute sulphuric acid replace the hydrogen cyanide solution.

H H

CH 3—C + HCN -> CH 3 C—CN cyanohydrfn

Ο OH

(d) If acetaldehyde is refluxed with hydroxylamine (or the hydro-chloride and sodium acetate) and the mixture is then cooled, the condensation product, acetaldehyde oxime, crystallizes out as a white solid.

CH 3—CHO + H 2N O H -> C H 3 — C H = N O H + H 2 0

Acetaldehyde cyanohydrin may be converted to lactic acid by hydrolysis. The solid is refluxed with strong aqueous alkali until no further ammonia is evolved and excess calcium hydroxide solution is then added to precipitate the lactic acid as the calcium salt. This is filtered off and decomposed with the calculated amount of sulphuric acid to give the free lactic acid.

CH 3—CH(OH)—CN + H 2 0 + NaOH ->

CH 3—CH(OH)—COONa + N H 3

(CH 3—CH(OH)—COO) 2Ca


(a) If acetaldehyde is oxidized by refluxing with a solution of potassium dichromate in dilute sulphuric acid, acetic acid is pro-d u c e d

- K 2C r 2 0:

C H 3 - C H O + Ο - ^ - ^ CH 3—COOH Η2ί>θ4

Oxidation is in fact so easy that even Tollen's reagent (am-moniacal silver nitrate) is a powerful enough oxidizing agent to react on warming; a silver mirror is deposited, and this serves as a test to distinguish aldehydes from ketones.

CH 3—CHO -f- A g 2 0 -> CH 3—COOH + 2 Ag

Fehling's solution too is reduced in the same way, with the precipitation of red-brown cuprous oxide.

By reducing the aldehyde, using sodium and water, metal and acid or catalytically, a primary alcohol, ethanol, is obtained (see N 0 te

Na/Hg

CH 3—CHO + 2 Η - ί ςο* CH 3—CH 2—OH

(b) Acetone, in contrast, is much more difficult to oxidize than acetaldehyde since the molecule must be broken at the carbonyl group and although the same reagents may be used, the concentra-tion of dichromate may have to be increased and the time of re-fluxing extended. Acetic acid is again produced and perhaps some formic but, since the latter is very easily oxidized, this normally yields carbon dioxide and water and the overall reaction is repre-sented as

K 2Cr«0 7

CH 8—CO—CH, + 4 Ο ) CH 3—COOH + C 0 2 + H 2 0

What compounds are formed by the oxidation and reduction of (a) acetaldehyde, (b) acetone?

State and formulate two other reactions in which aldehydes resemble ketones.

Discuss the reactions of sodium hydroxide with acetaldehyde and benzaldehyde.

How is acetaldehyde manufactured from acetylene?

(University of D u r h a m , 1963, Paper I I , qu. 5)


Reduction, however, can be carried out with the same reagents as for the aldehyde, and with the same ease. Propan-2-ol, a second-ary alcohol, is produced.

CH 3—CO—CH 3 + 2 H ^ > CH 3—CH—CH 3

OH

If zinc amalgam and concentrated hydrochloric acid is used, the ketone is reduced to the hydrocarbon; this is the Clemmensen reduction.

CH 3—CO—CH 3 + 4 H ^ > CH 3—CH 2—CH 3

These reactions then serve to distinguish between the aldehydes and ketones; the former reduce Tollen's reagent and Fehling's solution, while the latter do not. On reduction using sodium amalgam, aldehydes produce a primary alcohol and ketones the more reactive secondary alcohol.

In most other respects aldehydes and ketones are very similar in properties. Both react with sodium bisulphite solution to produce a bisulphite compound,

H H

I I C H 3 — C = 0 + N a H S 0 3 — CH 3—C—OH

I S 0 3 N a

CH 3 CH 3

I I C H 3 — 0 = 0 + NaHSO, -* CH 3—C—OH

S 0 3 N a

acetaldehyde sodium

bisulphite

acetone sodium

bisulphite

and condense with Phenylhydrazine and its derivatives to give phenylhydrazones.

CH 3—CHO + H 2N — N H — C 6H 5 ->

C H 3 — C H - N — N H — C 6 H 5 + H 2 0 acetaldehydephenylhydrazone

(CH 3) 2CO + H 2 N — N H — C e H 5 —

( C H 3 ) 2 C = N — N H — C e H 5 + H 2 0 acetonephenylhydrazone


In most of the addition and condensation reactions involving the —CHO group only, aliphatic and aromatic aldehydes react similarly. However, certain reactions of aliphatic aldehydes involve the a-hydrogen atom, and since the aromatic aldehydes do not possess one, they differ in these reactions.

Thus with dilute aqueous alkali, acetaldehyde undergoes the aldol reaction, a pale yellow precipitate of aldol being obtained by an addition to the — C — C = 0 = O group.

H H

I di i . I CH 3—C + H — C H 2— C H 0 5 i aö H > C H 3 — C — C H 2— C H O

II I Ο OH aldol

Dehydration may occur with stronger alkali, especially on heating, to give crotonaldehyde, and this in turn may condense with more aldehyde molecules. These then eliminate water, to give a long-chain, unsaturated resin.

CH 3—CH(OH)—CH 2—CHO

CH 3—CH=CH—CHO C H

^ ° > CH 3—(CH=CH) œ—CHO crotonaldehyde

Benzaldehyde does not react with dilute alkali, but when shaken for a prolonged period with 50 per cent alkali undergoes the Can-nizzaro reaction, one molecule being oxidized by the other which is consequently reduced.

2 C 6 H 5 - C H O + H 2 0 S > C 6H 5—CH 2—OH + C 6H 5—COOH benzyl alcohol benzoic acid

Acetaldehyde is normally manufactured from acetylene by bubbling the latter through a solution of mercuric sulphate in hot dilute sulphuric acid. (The strength varies from less than 6 per cent to almost concentrated depending on the process). Water is added, probably via a vinyl alcohol intermediate,

C H = C H + H 2 0 -> C H 2= C H — O H ^ CH 3—CHO

and acetaldehyde is produced in solution.


NOTES

1. Other alternatives include lithium aluminium hydride and aluminium isopropoxide, the latter being specific for the reduction of carbonyl groups, and known as the Meerwein-Ponndorf-Verley reduction.

/CH 3 \ \ \

CHO Al + 3 CH 3—CHO ^

VCH3 k CH.

3 C = 0 + (CH 3—CH 2—0) 3A1

C H , ' dil. HCl

τ 3 CH 3—CH 2—OH + A1C13


Outline two methods for the preparation of aromatic aldehydes. How and under what conditions does benzaldehyde react with

(a) sodium hydroxide, (b) potassium cyanide, (c) Phenylhydrazine?

(Bournemouth Mun ic ipa l Col lege of Technology & Commerce , O . N . C . , 1962, qu. 7)

Aromatic aldehydes are often prepared from toluene or toluene derivatives. Two variations in the method are possible and thus initially, if benzaldehyde is to be prepared, chlorine is passed into boiling toluene until the theoretical gain in weight is achieved for the production of benzyl chloride or benzylidene dichloride.

C 6H 5— C H 3 + Cl 2 -> C 6H 5—CH 2C1 + HCl benzyl chloride

C 6H 5— C H 3 -!- 2 CI, - * C f,H 5—CHC1 2 + 2 HCl benzylidene dichloride

If the first compound is produced, as is usual since less chlorine is required, hydrolysis and oxidation are carried out by boiling with lead nitrate solution (in a current of carbon dioxide).

C 6H 5—CHoCl + H 2 0 -> C 6H 5—CH 2OH — C 6 H 5 — C H O benzyl alcohol benzaldehyde

whereas in the second case, hydrolysis only is necessary, and this is usually carried out by using milk of lime.

C eH 6—CHC1 2

2- ^ > [ C 6 H 5 — C H ( O H ) 2 ] -> C eH 6—CHO

In either case the aldehyde may be subsequently tapped oif, washed, dried and distilled; alternatively, the ether extract may be treated with sodium bisulphite solution, and the bisulphite com-pound distilled with dilute sulphuric acid to give the pure benzal-dehyde.

A second method of preparation is the Gattermann-Koch syn-thesis, in which the appropriate aromatic compound is reacted with a mixture of carbon monoxide and hydrogen chloride. A catalyst of aluminium and cuprous chlorides is usually used,

C 6H 6 + CO + HCl -> C 6H 5—CHO + HCl (see Note 1) and benzaldehyde or a substituted benzaldehyde is obtained. The reaction gives the best yields if the ring is activated, for example by a phenol group (see Notes 2 and 3).

(a) If benzaldehyde is shaken with cold strong aqueous sodium hydroxide, a simultaneous oxidation and reduction occurs. This is


the Cannizzaro reaction, and yields a mixture of

2 C 6H 5—CHO + NaOH C 6H 5—COONa + C 6H 5—CH 2OH sodium benzoate benzyl alcohol

benzoic acid and benzyl alcohol. The former is in solution as the sodium salt, the free acid being liberated on acidification. The reac-tion is an example of disproportionation.

(b) Benzaldehyde reacts readily with acidified potassium cyanide solution if the two are mixed and shaken together. A white pre-cipitate of the benzaldehyde cyanohydrin is formed due, in fact, to the reaction of the aldehyde with hydrogen cyanide produced by hydrolysis of the salt (see Note 4).

KCN + H 2 S 0 4 ^ H C N + K H S 0 4

CN

I C 6 H 5 C = 0 + HCN -> C 6H 5—C—OH

I I H H

Alternatively, if benzaldehyde is shaken with alcoholic potassium

cyanide solution, a pale yellow precipitate of benzoin is produced.

KCN

2 C 6H 5—CHO > C 6H 5—CH(OH)—CO—C 6H 5

(c) If a solution of Phenylhydrazine hydrochloride and sodium acetate in water is refluxed with benzaldehyde for a short time and the solution then cooled, the condensation product, benzalde-hyde phenylhydrazone, is obtained as a white solid. C 6H 5—CHO + C 6H 5 — N H — N H 2 -> C 6 H 5 — C H = N — N H — C 6 H 5

NOTES

1. It is suggested that formyl chloride may be an intermediate, but the compound has never been isolated. HCl + CO -> HCOC1

2. The Gattermann synthesis uses hydrogen cyanide in place of hydrogen chloride, but the yield is lower.

3. An alternative method of preparation of benzaldehyde is by dry distillation of a mixture of calcium benzoate and calcium formate.

(C 6H 5—COO)Cai " -> C eH 5—CHO + C a C 0 3

(H—COO)Caj

4. It is safer to add potassium cyanide to the bisulphite compound.


For acetaldehyde and benzaldehyde, describe four reactions which are distinctive of aldehydes and apply to each of these aldehydes, and four reactions in which they behave differently.

(Southern Universities' Joint B o a r d , 1959, Paper I, qu . 12)

The reactions of aldehydes may be divided into two groups, those which involve the aldehyde group only and those which in-volve the rest of the molecule either as well or instead.

In the former group, therefore, are found reactions common to both aliphatic and aromatic aldehydes, and these include the following reactions (but see Note 1):

(a) The aldehyde group is very easily oxidized, and thus both acetaldehyde and benzaldehyde reduce Tollen's reagent (ammoniacal silver nitrate) to a silver mirror on warming. The corresponding acid is produced from each aldehyde.

CH 3—CHO + A g 2 0 -> CH 3—COOH + 2 Ag acetic acid

C e H 5 - C H O + A g 2 0 -> Q H 5 - C O O H + 2 Ag benzoic acid

( b ) On treatment with phosphorus tri- or pentachloride both aldehydes give a gew-dichloride.

CH 3CHO + PCI 5 -> CH 3CHC1 2 + POCl 3

ethylidcne dichloride

C 6H 5CHO + PC15 -> C 6H 5CHC1 2 + POCl 3

benzylidene dichloride

(c) Hydrogen cyanide in aqueous solution reacts with all aldehydes to yield a cyanohydrin. This is best obtained by adding the aldehyde to a solution of sodium cyanide in dilute sulphuric acid, and shaking vigorously. CN

I CH 3—CHO + HCN CH 3—C—OH acetaldehyde cyanohydrin

I H

CN

I C 0H 5—CHO + HCN C eH 5—C—OH benzaldehyde cyanohydrin

I H

ALDEHYDES AND KETONES 3 9

(d) Reduction of both aromatic and aliphatic aldehydes, either catalytically, by the use of nascent hydrogen, or with lithium alu-minium hydride produces a primary alcohol.

CH 3—CHO + 2 H -> CH 3—CH 2—OH ethanol

C 6H 5—CHO + 2 H -> C 6H 5—CH 2—OH benzyl alcohol

Where reactions of aromatic and aliphatic aldehydes differ, it is most frequently due to the lack of an α-hydrogen atom in the former. Thus,

(a) Chlorination replaces the α-hydrogen atoms of the alkyl group one by one in acetaldehyde,

CH 3—CHO + Cl 2 - * CH2C1—CHO -> CHC12—CHO — CCl 3CHO chloro- dichloro- trichloro-

acetaldehyde acetaldehyde acetaldehyde

whereas in benzaldehyde (in the absence of a catalyst) the aldehydic hydrogen is replaced.

C 6H 5—CHO + Cl 2 — C 6H 5—CO—CI + HCl benzoyl chloride

Sometimes, however, the aldehydic hydrogen is the more important, so that

(b) When refluxed with aqueous-alcoholic potassium cyanide, benzaldehyde forms benzoin. The reaction is generally known as the benzoin condensation, but since no molecule of water (or anything else) is eliminated the term is not strictly correct.

H I

2 C 6H 5—C—H —> C 6H 5—C C—C 6H 5

II I I I ο OH ο

Aliphatic aldehydes give no reaction under comparable conditions. Also, due to the electron-withdrawing effect of the aromatic

nucleus, benzaldehyde is less easily oxidized than acetaldehyde, and therefore a further difference is that

(c) Only acetaldehyde reduces Fehling's solution on warming to precipitate red-brown cuprous oxide.

CH 3—CHO + Ο F

^ ^ > CH 3—COOH


Amongst the addition reactions, the reaction of the two aldehydes with ammonia differs.

(d) On bubbling the gas into a cooled ethereal solution of acetal-dehyde, the white acetaldehyde ammonia is produced,

H H

I I CH 3—C + N H 3 -> CH 3—C—NH 2

II I Ο OH

whereas benzaldehyde yields a more complex product, hydro-benzamide.

N = C H — C 6 H 5 /

3 C eH 5—CHO + 2 N H 3 -> C 6H 5CH + 3 H 2 0 \

N = C H — Ce

H5

NOTES

1. A further important similarity is (e) Both aldehydes condense with hydroxylamine to produce

oximes.

CH 3—CHO + N H 2O H -> C H 3 — C H = N O H + H 2 0 acetaldehyde oxime

C eH 5—CHO + N H 2O H — C e H 5 — C H = N O H + H 2 0 benzaldehyde oxime

Carboxylic Acids

State and explain the grounds on which its accepted structural formula has been assigned to acetic acid. Outline briefly a method for its commercial preparation and write equations to illustrate THREE ways in which acetic acid (or its salts) differs in chemical properties from formic acid (or its salts).

(Welsh Joint Educat ion Commit tee , 1959, Paper II, qu . 9)

In order to determine the structural formula of an organic compound, the molecular formula must first be determined, and for this a pure sample of the acid is required. Qualitative analysis of acetic acid will show the absence of halogens, nitrogen, sulphur, phosphorus and metals.

Thus it seems likely that the compound contains only carbon and hydrogen, and perhaps oxygen. Quantitative analysis by oxidation and absorption of the water and carbon dioxide produced gives percentage values for hydrogen and carbon content, and the oxygen is assumed to make up the difference between the sum of these two and 100 per cent.

Dividing percentages by atomic weights and then dividing each figure obtained by the smallest, gives an empirical formula, C H 2 0 , and it can be shown that only one sodium salt, C 2 H 3 0 2 N a , exists, and thus the acid is probably monobasic. On this assumption, by preparing and igniting the silver salt the molecular weight of the acid can be determined as 60 (see Note 1), and thus since the molecular formula is (CHgO)^, χ must be 2, and the molecular formula is also C 2 H 4 0 2 .

By consideration of the various reactions of acetic acid, its structure may then be confirmed.

(a) Chlorination yields a product C 2H0 2C1 3, trichloroacetic acid, showing that three hydrogen atoms differ from the fourth (see Note 2).

41


Ο—Η

(b) Sodium hydroxide reacts to give one salt only, sodium acetate, C 2 H 3 0 2 N a , and no other hydrogen atom can be replaced by a metal; thus one hydrogen only is acidic and differs from the other three.

(c) By heating with phosphorus pentachloride, acetic acid yields acetyl chloride, C 2H 30C1. One chlorine atom has therefore replaced an oxygen and a hydrogen atom. This can occur only if an —OH group is present and replaced.

(d) By heating with soda-lime, methane, CH 4, water and sodium carbonate, N a 2 C 0 3 are produced. Thus carbon dioxide is elimi-nated from the molecule, which suggests that both oxygen atoms are linked to the same carbon atom.

Thus we have the following groups present: CH 3—, —OH, Ο—C—O, and —H, which suggests structure (I), and consideration

Η

I H — C — C = 0

I I Η OH

of the other possible structures confirms this, since none of them would be expected to react as above with all four reagents (see Note 3).

Acetic acid is generally manufactured from acetylene. The latter is bubbled through a solution of mercuric sulphate in sul-phuric acid (the strength varies with the process); water adds to give acetaldehyde,

C H = C H + H 2 0 ~y C H 2 = C H — O H ^ CH 3—CHO + H 2 0

and this is distilled off and oxidized in the vapour phase with air. A manganous acetate catalyst is used.

MnAc 2

2 CH 3—CHO + 0 2 > 2 CH 3—COOH

The most outstanding difference between acetic acid and formic acid (and their salts) is shown in the ease of oxidation of the latter, which has, in fact, an aldehyde group in the molecule, as well as an acid group: ,'"""»

/ Ο ) ,' / /

; H—c /

CARBOXYLIC ACIDS 43

Thus formic acid and formates are reducing agents, and reduce ammoniacal silver nitrate (Tollen's reagent) to silver, and mercuric ions to mercurous (see Note 4).

On dehydration, too, the acids differ. Two molecules of acetic acid with sulphuric acid give a little acetic anhydride,

2 CH 3—COOH -> (CH 3—CO) 20 + H 2 0

and phosphorus pentoxide improves the yield, whereas one molecule of formic acid or a formate breaks down with hot concentrated sulphuric acid to give carbon monoxide and water (see Note 5).

H—COOH -> H 2 0 + CO

Finally, when the alkali metal salts are heated carefully, the acetate yields acetone,

2 CH 3—COONa -> CH 3—CO—CH 3 + N a 2 C 0 3

whereas the formate breaks down to give the corresponding oxalate and hydrogen.

2 H—COONa -> H 2 + (COONa) 2

NOTES

1. The vapour density is 30 only if determined well above the boiling-point; otherwise it is much higher, due to association, and is not, therefore, a reliable method for determining the molecular weight.

2. The reaction proceeds in three stages, corresponding to the formation of mono-, di- and tri-chloroacetic acids.

3. N o reactions of a carbonyl group would be expected, due to the electron-donating —OH group and resonance of the ion

Ο o-/ /

—c <-> — c \ ο- ο

4. Fehling's solution is, however, unaffected. 5. Carbon monoxide could be claimed to be an anhydride of formic

acid, but the anhydride corresponding to acetic anhydride would be H—CO—O—CO—H.


Describe in outline the preparation from acetic acid of each of the following: methane, acetone, acetamide, acetic anhydride.

What substances are obtained when acetone is (a) oxidized, (b) reduced?

(University of L o n d o n , A u t u m n 1957, Paper I, qu . 2)

If acetic acid is allowed to react with sodium hydroxide solution, sodium acetate is produced. By evaporation of the water the solid salt can be obtained,

CH 3—COOH + NaOH - CH 3—COONa + H 2 0

and if this is now heated with soda-lime (a mixture of sodium and calcium hydroxides) methane is given off, and may be collected over water in the normal way.

CH 3—COONa + NaOH ~> CH 4Î + N a 2 C 0 3

By reacting the acid with calcium hydroxide instead of sodium hydroxide, a calcium salt is obtained,

2 CH 3—COOH + Ca(OH) 2 - (CH 3—COO) 2Ca + H 2 0

and this, on dry distillation, yields acetone.

(CH 3COO) 2Ca -> CH 3—CO—CH 3 + C a C 0 3

To prepare acetamide, ammonium hydroxide solution is added to an excess of the acid, and this mixture, containing ammonium acetate,

CH 3—COOH + N H 4O H ^ CH 3—COONH 4 + H 2 0

is then heated to drive off excess acid and water. The presence of the excess acid prevents decomposition of the salt to the acid and am-monia, and hydrogen ions catalyse the subsequent dehydration.

The ammonium salt is then heated and refluxed gently (using an air condenser) for several hours. Loss of water occurs, and on distillation (again with an air condenser) the amide is produced, solidifying in the receiver.

CH 3—COONH 4 -» CH 3—CO—NH 2 + H 2 0

Acid chlorides are generally prepared by carefully mixing the appropriate acid with thionyl chloride, phosphorus trichloride or phosphorus pentachloride and distilling. To prepare acetyl chloride,

CARB0XYL1C ACIDS 45

phosphorus trichloride is used, since it is easy to separate from the product (which has a lower boiling point) by distillation.

3 CH 3—COOH + PC13 -> 3 CH 3—CO—CI + H 3 P 0 3

If a 1:1 (molecular) mixture of the acetyl chloride and sodium acetate is now made, then this on distillation yields acetic anhydride. Care must be taken to ensure that the sodium acetate is completely anhydrous before addition.

CH 3—CO—CI + CH 3—COONa -> (CH 3—CO) 20 + NaCl

(a) On oxidation, which proceeds with difficulty, acetone is broken down to give acetic acid, carbon dioxide and water; a little formic acid may also be obtained.

CH 3—CO—CH 3 + 4 Ο --> CH 3—COOH + C 0 2 + H 2 0

Nitric acid or acid dichromate is usually used for the oxidation. (b) On reduction of acetone with sodium (or sodium amalgam) and

water, propan-2-ol (isopropyl alcohol) is obtained (see Note 1).

CH 3—C—CH 3 + 2 Η CH 3—CH—CH 3

II H 2° I Ο OH

Complete reduction may be achieved by using red phosphorus and hydrogen iodide,

CHq—CO—CHo > CH->—CHo—CH3 6 0

Red Ρ ύ Δ J

or by the Clemmensen reduction with zinc amalgam and concen-trated hydrochloric acid, when propane results.

NOTES

1. If magnesium amalgam is used, the reduction is less complete, and pinacol (tetramethylglycol) is also produced.

2 (CH 3) 2CO + 2 H M g / H g

> (CH 3) 2C—OH in C 6H 6 J

(CH 3) 2C—OH


Give a series of reactions which would lead to the synthesis of acetic acid from acetylene.

How may (a) methane, (b) monochloracetic acid and (c) glycine (aminoacetic acid) be prepared from acetic acid?

What action has (i) hydrochloric acid, (ii) sodium hydroxide and (iii) nitrous acid on glycine?

(University of Durham, 1962, Paper II, qu. 3)

At the present time, one of the most important methods of manu-facturing acetic acid is from acetylene. The gas is bubbled through a hot solution of mercuric sulphate in sulphuric acid, often prepared by adding mercuric oxide to the acid. Water is present, and the amount can vary considerably, but it is this which adds to the acetylene, probably to produce vinyl alcohol.

CHEEECH + H 2 0 H g S

°4> C H 2 = C H O H

H 2S 0 4

However, the vinyl alcohol may be considered to be the enol form of acetaldehyde, which is thus produced by migration of a proton (see Note 1).

H H / /

C H 2 = C ^ C H 3 — C

\ \ OH Ο

The acetaldehyde is oxidized with air in the vapour phase, by passing over a heated catalyst, at about 300°C, and the acetic acid then condensed out.

MnAco

2 C H 3 — C H O + 0 2 >2 C H 3 — C O O H

(a) If acetic acid is neutralized with sodium hydroxide solution, and the water evaporated off, sodium acetate is produced. This is heated to obtain the anhydrous salt,

CH 3—COOH + NaOH -> CH 3—COONa + H 2 0

and if this is then mixed with soda-lime and dry distilled, methane is evolved, and may be collected over water.

(CaO)

CH 3—COONa + NaOH > C H J + N a 2 C 0 3

(b) When chlorine is passed into boiling glacial acetic acid,

CARBOXYLIC ACIDS 47

monochloracetic acid is first formed, and no disubstitution occurs until all the acid has been monosubstituted.

Thus by refluxing a weighed amount of the acid, chlorine may be passed in until the theoretical gain in weight is obtained for mono-substitution, and monochloracetic acid is obtained as a white crystalline solid on cooling.

CH 3—COOH + Cl 2 - * CH2C1—COOH + HCl

(c) Glycine is easily prepared from chloracetic acid by adding strong ammonium hydroxide and excess ammonium carbonate and allowing the stoppered reaction vessel to stand for 48 hours.

C1CH2—COOH + ( N H 4 ) 2 C 0 3 - >

NH 2CH 2—COOH + NH 4C1 + C 0 2

The solution remaining is concentrated by distillation, methanol added and the whole cooled. Glycine and ammonium chloride crystallize out; these are filtered off, and the former extracted with hot methanol.

Since glycine is an amino-acid, it will form salts with both acids and bases, and exhibits the properties of both aliphatic primary amines and carboxylic acids.

(i) Thus with hydrochloric acid it forms a hydrochloride,

NH 2CH 2—COOH + HCl -> (NH 3CH 2—COOH)Cl"

(ii) sodium hydroxide gives a sodium salt, sodium glycinate or aminoacetate, when warmed with the acid,

NH 2CH 2—COOH + NaOH -> NH 2CH 2—COO"Na+ + H 2 0

(iii) and nitrous acid reacts, with the liberation of nitrogen, to give the alcohol, hydroxyacetic acid or glycolic acid.

NH 2CH 2—COOH + H N O , -> HOCH 2—COOH + N 2 + H 2 0

NOTES

1. Some authors suggest that the sulphuric acid adds and is then hydrolysed off, and this may be so when the concentration of acid is high, as it is in some of the processes.

2H 20

C H = C H + 2 H — H S 0 4 CH 3—CH(HS0 4) 2 > [CH 3—CH(OH) 2] -> CH 3—CHO + H 2 0


A is a corrosive, colourless, crystalline solid whose aqueous solution reacts with calcium carbonate. When A is boiled with sodium hydroxide solution a compound, B, of molecular formula C 2 H 3 0 3 N a , is formed. When a concentrated solution of A is dissolved in an excess of concentrated ammonia solution and allowed to stand, a salt, C, is formed, having the formula C 2 H 8 0 2 N 2 . If a pure sample of C is dissolved in dilute hydrochloric acid and a solution of sodium nitrite is added, nitrogen is liberated and an ammonium salt, D, is left in solution.

Identify A, B, C, and D and explain the reactions given above. Describe how A reacts with (a) phosphorus trichloride, (b) potassium cyanide.

(Southern Universities Joint B o a r d , 1963, Paper I I , qu . 10)

Since A reacts with calcium carbonate it must be an acid, and therefore possess a carboxyl, —COOH group. It is reasonable to suppose that it is this group that is linked to the sodium in compound B, C 2 H 3 0 3 N a which then becomes (CH 30)COONa. However, since boiling appears to be necessary to convert A to B, hydrolysis as well as salt formation is implied.

A group of molecular formula C H 3 0 — can have as its structure either CH 3—Ο— or HO—CH 2— and since A reacts with con-centrated ammonia to incorporate two nitrogen atoms, the group must have the second structure, because no hydrolysis will yield a C H 3 0 — group, whereas CH 2C1— or CH 2Br— would readily give HO—CH 2—, and either of these halogen-containing groups would react with ammonia to give an amino-compound.

At the same time, the carboxyl group would react to give an ammonium salt, and thus the composition would be expected to be NH 2—CH 2—COONH 4 which agrees with the formula given for compound C. The fact that an amino-group is present is confirmed by the reaction with nitrous acid to evolve nitrogen ; the ammonium ion is unaffected by this reagent.

Thus compound A appears to be a halogen-substituted acetic acid, and from its description, probably chloroacetic acid. The compounds described are therefore:

A—chloroacetic acid, CI—CH 2—COOH Β—sodium glycolate (hydroxyacetate) HO—CH 2—COONa C—ammonium aminoacetate, NH 2—CH 2—COONH 4

D—ammonium glycolate, HO—CH 2—COONH 4

CARBOXYUC ACIDS 4 9

and the reactions are :

2 CI—CH 2—COOH + C a C 0 3 ->

(Cl—CH 2—C00) 2Ca + C 0 2 + H 2 0 calcium chloroacetate

Cl—CH 2—COOH + 2 NaOH ->

HO—CH 2—COONa + NaCl + H 2 0 sodium glycolate

Cl—CH 2—COOH + 3 N H 3 ->

N H 2— C H 2— C O O N H 4 + NH 4C1 ammonium aminoacetate

NH 2—CH 2—COONH 4 + H N 0 2 - *

HO—CH2—COONH4 + N 2 + H 2 0 ammonium glycolate

Compound A would react with phosphorus trichloride to yield chloroacetyl chloride,

3 Cl—CH 2—COOH + PCI3 -> 3 Cl—CH2—COC1 + H 3 P 0 3

and by refluxing with aqueous alcoholic potassium cyanide, cyano-acetic acid is obtained.

Cl—CH 2—COOH + KCN -* (CN)—CH 2—COOH + KCl


If benzene is nitrated, by shaking with nitric acid-sulphuric acid mixture at 60°C, nitrobenzene is produced. This is tapped off, washed, dried and distilled.

N 0 2

H 2S 0 4

+ H N O 3 H II + H 2 0

By reduction with tin and hydrochloric acid, aniline is obtained. This is liberated by adding excess alkali and steam-distilling.

N O , N H ,

+ 6 H | % | II + 2 H 2 0

Outline TWO methods by which benzoic acid may be prepared from benzene, giving the essential reagents and conditions.

Describe briefly the reactions by which benzoic acid may be con-verted into (a) benzene, (b) benzoic anhydride, (c) phenyl cyanide (benzonitrile).

(Southern Universities Joint B o a r d , 1961, Paper I, qu . 12)

Benzoic acid may be prepared by oxidation of toluene or hydrolysis of benzonitrile (phenyl cyanide) and either of these compounds may be prepared from benzene by a suitable series of reactions.

Toluene is prepared by the Friedel-Crafts reaction ; methyl chloride is a gas, and so is dissolved in the anhydrous benzene before being

CH 3

A1C13 r ^ ^ /

+ CH3CI >\ j + HC1 methyl chloride ^ ^ v ^

added slowly to the catalyst, with cooling if necessary. After refluxing, the complex produced is hydrolysed with dilute hydro-chloric acid, and the toluene liberated. This is tapped off, dried and distilled. Oxidation is carried out by refluxing with a mixture of potassium dichromate or permanganate and sulphuric acid, and on cooling the benzoic acid crystallizes out and is filtered off.

CH 3 COOH

+ 3 0 - 4 II + H 2O

CARBOXYLIC ACIDS 51

The purified amine is dissolved in hydrochloric acid, cooled in ice and diazotized by the addition of potassium nitrite solution at 0°C.

N H 3 C I N2C1

phenyl diazonium chloride

The Sandmeyer reaction is then performed, by the addition of cuprous cyanide in potassium cyanide solution, and after refluxing, the benzonitrile is separated and distilled off,

N 2C1 CN

2 |^J | + Cu 2(CN) 2 - * 2 |^J | + 2 N 2 + Cu 2Cl 2

before hydrolysis by refluxing with dilute alkali.

CN COONa

+ NaOH + H 2 0 — f ^ l i + N H 3

Acidification causes the precipitation of the benzoic acid from the solution of the sodium salt (see Note 1 ) .

(a) Benzoic acid may be decarboxylated by heating with soda-lime, when benzene distils off.

COOH COONa

+ NaOH ^ > + H 2 0

COONa

+ N a O H S H II + Na2

COa

(b) Benzoic anhydride is produced by a slow distillation of benzoic acid with acetic anhydride. Acetic acid distils off, and the white solid remaining is benzoic anhydride (see Note 2).

2 C 6H 6COOH + (CH 3CO) 20 -> ( C 6H 5C O ) 20 + 2 CH 3COOH


(c) Benzonitrile is best prepared by heating the ammonium salt of benzoic acid, with excess acid, to yield benzamide.

COOH COONH 4

+ N H 4 O H - > | I I

neat CONH 2

+ H 2 0

By grinding the latter with phosphorus pentoxide and dry-distilling, benzonitrile is obtained.

CONH 2 CN

4 - P 2 0 5 - > | Ι + 2 H 3 P O 3

NOTES

1. Another method is to pass chlorine into boiling toluene (pre-pared as in the first method described), and then to hydrolyse the benzotrichloride obtained, but this method is very similar to the first.

Clo Ca(OH)2

C 6H 5— C H 3 — ^ C 6H 5—CCI3 -+ C 6H 5—COOH

2. Alternatively, distillation with thionyl chloride will yield benzoyl chloride, and if this is heated with sodium benzoate, the anhydride is obtained.

C 6H 5—COOH + SOCl2 — C 6H 5—COC1 + S 0 2 + HCl

C eH 5—COC1 + C 6H 5—COONa -> (C 6H 5—CO) 20 + NaCl

Esters

Describe briefly how soap and glycerol are obtained from fats. How are (a) glyceryl trichloride, (b) "nitroglycerine", (c) formic acid obtained by the use of glycerol?


Fats are naturally occurring esters of the trihydric alcohol glycerol with long-chain "fatty" acids, notably palmitic, C 1 5H 3 1—COOH, stearic, C 1 7H 3 5—COOH, and oleic, C 1 7H 3 3—COOH. The sodium and potassium salts of these acids are soaps, and these were originally obtained from the fats by alkaline hydrolysis or "saponification" as the process was called. After treating with hot alkali the soaps were salted out and the upper layer of soap separated off.

It is more usual at the present time to hydrolyse the fats by using superheated steam, sometimes with the addition of very dilute acid.

CH2—O—CO—R CH2—OH R—COOH

CH—O—CO—R' + 3 H 2 0 -> CH—OH + R'—COOH

CH2—Ï—CO—R" CH2—OH R"—COOH

Neutralization of the free acids by the addition of alkali follows ; sodium hydroxide gives a "hard" soap, potassium hydroxide a better quality, more soluble "soft" soap, each containing many fatty acids.

R—COOH + NaOH -> R—COONa + H 2 0

After removal of the soaps, the aqueous layer is concentrated by boiling, and any sodium salts may then be filtered off. Then the remaining water is removed by vacuum distillation before the final purification of the glycerol.

53

5


CH—OH

I CHo—OH

3 S O C l „ \ Ç H2 Cl

CH—Cl + 3 S 0 2 + 3 HCl

I CH 2—Cl

(b) "Nitroglycerine" is an ester of glycerol, not a true nitro com-pound, obtained by running it slowly, or as a spray, into a mixture of concentrated nitric and sulphuric acids in the cold. On pouring the reaction mixture into water, the nitroglycerine separates as an oily layer.

CH 2—OH C H 2 — O — N 0 2

I I CH—OH + 3 HNO3 C H — O — N 0 2 + 3 H 2 0

I I CH 2—OH C H 2 — O — N 0 2

(c) Formic acid is best obtained by heating oxalic acid crystals in glycerol at about 110°C in a distillation apparatus.

Glyceryl monoxalate is first formed, decomposes to the mono-formate and this in turn breaks down to give glycerol and formic acid which distils over; more oxalic acid is then added.

CH 2—OH CH 2—OH I COOH I

CH—OH + I - * CH—OH + H 2 0 — I COOH I

CH 2—OH CH 2—O—CO—COOH glyceryl monoxalate

CH 2—OH CH 2—OH

I (COOH) 2 I CH—OH + C 02

> CH—OH + H—COOHf

I I CH2—O—CO—H CH2—Ο—CO—COOH glyceryl monoformate

(a) Glyceryl trichloride may be obtained from glycerol by reacting the latter with an excess of phosphorus pentachloride and distilling off the product. Alternatively, by using thionyl chloride, gaseous by-products are obtained.

CH 2—CI I

CH—CI + 3 POCI3 + 3 HCl CH 2—OH

C K / CH2—CI

Nitriles

Describe the laboratory preparation of a pure specimen of aceto-nitrile. How does this compound react with (i) water, (ii) sodium hydroxide, (iii) reducing agents? Starting from acetonitrile, how would you prepare (iv) methyl alcohol, (v) ethyl alcohol?

(University o f L o n d o n , Jan. 1963, Paper I I , qu . 2)

If not prepared by substitution, most nitriles are prepared by the dehydration of the corresponding amide. This is the case with acetonitrile, or methyl cyanide, which is prepared from acetamide by treating this compound with phosphorus pentoxide.

CH 3—CO—NH 2 - 5 ? i > CH 3—CN + H 2 0 A convenient amount of phosphorus pentoxide is weighed out,

with great care, as quickly as possible, due to it being very hygro-scopic, and run by means of a paper funnel (to prevent it sticking to the sides) into a flask. It is followed immediately by somewhat less (by weight) acetamide. The mixture is shaken to mix the re-actants as well as possible, and then the apparatus is set up for distillation.

The flask is heated with a small, smoky flame for a few minutes, and then by slightly raising the temperature, the liquid produced, which is mainly acetonitrile, is distilled oif. Some acetic acid is also produced, and this is removed by shaking the distillate with saturated potassium carbonate solution, in which the nitrile is immiscible (due to "salting out"). The lower layer is then run off, and the remaining liquid transferred to a distillation apparatus, from which it is distilled over phosphorus pentoxide. The fraction boiling at 79-82°C is collected.

(i) Acetonitrile is fairly easily hydrolysed, and this is usually carried out by refluxing with dilute acid. Acetamide is produced

55


intermediately, but cannot normally be isolated, and the final product is acetic acid, which may be distilled off together with some water.

CH 3—CN + H 2 0 CH 3—CO—NH 2 reflux

CH 3—CO—NH 2 + H 2 0 -> CH 3—COONH 4

2 CH 3—COONH 4 + H 2 S 0 4 — 2 CH 3—COOH + ( N H ^ S O ,

(ii) With dilute alkali, the same hydrolysis occurs on refluxing with the nitrile, but this time sodium acetate remains, and ammonia is given off until the reaction is completed.

CH 3—CN + H 2 0 ^ > CH 3—CO—NH, °

ώ reflux °

CH 3—CO—NH 2 + H 2 0 -> CH 3—COONH 4

CH 3—COONH 4 + NaOH -> CH 3—COONa + N H 3 + H 2 0

(iii) Since the cyanide group is unsaturated, addition reactions can occur. Thus, by the action of reducing agents, such as sodium amalgam and water (or alcohol) or lithium aluminium hydride in ether, or by catalytic hydrogénation, acetonitrile is reduced to a primary amine, ethylamine, which, being volatile is evolved from the reaction mixture and collected by means of a trap, in hydro-chloric acid, with which it forms a soluble hydrochloride.

CH 3—CN + 4 H - > CH 3—CH 2—NH 2

C H 3— C H 2— N H 2 + HCl -> CH 3—CH 2—NH 3C1~

(iv) In order to prepare methanol from acetonitrile, the latter is first hydrolysed, as above, and acetic acid is thus produced. This is converted to the silver salt, which is dried and then reacted with bromine. The acid breaks down, and methyl bromide is produced.

2 CH 3—COOH + A g 2 0 = 2 CH 3—COOAg + H 2 0

CH 3—COOAg + Br 2 -> CH 3—Br + C 0 2 + AgBr

If this is then reacted with aqueous alkali, hydrolysis occurs, and methanol results (see Note 1).

CH 3—Br + NaOH CH 3—OH + NaBr

NITRILES 57

NOTES

1. An alternative method might be to convert the acid to the am-monium salt, which on heating yields the amide. By heating with bromide and alkali the Hofmann degradation occurs to produce methylamine. Theoretically methanol can then be obtained by treating with nitrous acid, but in practice this is not the case, since no reaction occurs with methylamine (Whitmore and Thorpe, / . Am. Chem. Soc. 63, 1118 (1941)), so that the method is of no value for this amine.

CH 3—CN ^ > CH 3—COOH CH 3—COONH 4 HgSO^

KOH H N 0 2

CH 3—CO—NH 2 — - > C H 3— N H 2 > no reaction Br 2

2. An alternative method is to reduce to the amine and then treat with nitrous acid, but even here the yield of alcohol is poor.

4H HNO 2 CH 3—CN — > C H 3— C H 2— N H 2 >

C H 2 = C H 2 + CH 3—CH 2—OH

(v) The preparation of ethanol also commences with hydrolysis to the acid, and this may then be reduced directly to the alcohol by hydrogénation under pressure with a copper chromite catalyst, or better, by refluxing the anhydrous acid with lithium aluminium hydride in ether, and then adding water to the complex obtained (see Note 2).

CH 3—COOH + 4 H -> CH 3—CH 2—OH

Amines

Give examples of primary, secondary and tertiary amines and show by means of equations the action of nitrous acid on each.

A compound C 4 H U N (A) reacts with nitrous acid to give C 4H 1 0O (B) which when heated with concentrated sulphuric acid gives C 4 H 8 (C). Furthermore, Β on oxidation gives an aldehyde. C reacts with hydro-gen bromide to give C 4H 9Br (D) which is hydrolysed by alkali to C 4 H 1 0O (E). The compound Ε can be synthesized by the action of methyl magnesium iodide on acetone, followed by hydrolysis. Elucidate these reactions and name the compounds A to E.

(Whitehaven College of Further Education, O.N.C., 1961, qu. 7)

(i) Amines are produced when the hydrogen atoms of ammonia are replaced by alkyl or aryl groups. If one atom only is replaced, a primary amine is produced :

C 2 H 5 - N H 2 C 6 H 5 - N H 2

ethylamine aniline (phenylamine)

whereas if two atoms are replaced, the product is a secondary amine:

Q H 5 C 6H 5 C 6H 5

\ \ \ N H N H N H

/ / / C 2H 5 C 6H 5 CH 3

diethylamine diphenylamine N-methylaniline

If all the atoms are replaced, a tertiary amine results:

C 2H 5 C 6H 5 C 6H 5 \ \ \ C 2H 5— Ν C 6H 5— Ν C H 3 - N

/ / / C 2H 5 C 6H 5 CH 3

triethylamine triphenylamine N,JV-dimethylaniline

58

AMINES 59

These amines may be distinguished by their reactions with nitrous acid.

The primary aliphatic amine reacts with evolution of nitrogen to yield an alcohol and some alkene (methylamine does not react),

C 2H 5— N H 2 + H N 0 2 -> C 2H 5—OH + N 2 + H 2 0 ethanol

whereas the primary aromatic amine, if in an ice-cold acid solution as its salt, yields an intermediate diazonium compound.

C 6H 5—NH3CI + H N 0 2 -> C 6H 5—N 2C1 + 2 H 2 0 phenyl diazonium chloride

On warming the aqueous solution, this also yields nitrogen and a hydroxy-compound which in this case is a phenol, not an alcohol.

C 6H 5—N 2C1 + H 2 0 - > C 6H 5—OH + HCl + N 2

phenol

Both aromatic and aliphatic secondary amines react to give an oily, ether-soluble nitrosamine, yellow in colour. N o nitrogen is evolved.

C 6H 5 C 6H 5

\ \

N H + HONO Ν — N = 0 + H2

0

/ /

CH 3 CH 3

nitrosomethylaniline or methylphenylnitrosamine

When warmed with phenol and concentrated sulphuric acid, these nitrosamines give a green solution.

Similarly, most tertiary amines give the same product. They dissolve in cold nitrous acid to give an amine nitrite (but see Note 1), and this decomposes on heating to give a nitrosamine and an alcohol.

( C 2H 5) 3N + H N 0 2 -> ( C 2 H 5 ) 3 - N H + N O -

( C 2 H 5 ) 3 - N H + N O - -> ( C 2 H 5 ) 2 N - N O + C 2 H 5 - O H nitrosodiethylamine or ethanol

diethylnitrosamine


(ii) The action of methyl magnesium iodide on acetone yields a product which, after hydrolysis gives t-butyl alcohol, or trimethyl-carbinol. This then, must be compound E.

CH 3 CH 3

CH 3—C + CH 3—Mgl - * CH 3—C—CH 3 ^ (CH 3) 3C—OH

Ο OMgl Ε

Compound D, C 4H 9Br, which on alkaline hydrolysis produces E, must therefore be t-butyl bromide.

(CH 3) 3C—Br + NaOH - ^ - > (CH 3) 3C—OH + NaBr D Ε

From its formula, compound C, C 4H 8, must almost certainly be an alkene, and if this is so the reaction with hydrogen bromide must be addition. Only one alkene could produce t-butyl bromide, namely isobutylene, and this must be C (see Note 2).

CH 3 CH 3 CH 3

\ \ / C = C H 2 + HBr — C

/ / \ CH 3 CH 3 Br

C D

Now C is obtained from B, C 4H 1 0O , by heating with concentrated sulphuric acid, which suggests that it is an alcohol. Since it gives an aldehyde on oxidation, this is confirmed, and the alcohol must be primary. Since the structure of C is established, Β can only be one primary alcohol, isobutyl alcohol.

CH 3 CH 3

CH—CH 2OH CH—CHO + H 2 0

C H 3 CH 3

Β

C H 3

isobutyraldehyde

s H 2S 0 4

C = C H 2 + H 2 0 /

C H / c

AMINES 61

NOTES

1. Except the dialkylanilines which, being very reactive in the /?ara-positions, give ^-nitroso derivatives.

N = 0

H N 0 2 -> + H 2 0

N(CH 3) 2 N (CH 3) 2

2. Provided air and oxidizing agents are excluded, otherwise the

Kharasch or Peroxide effect may give rise to reverse addition in the

case of hydrogen bromide.

The alcohol Β is obtained from A, C 4 H n N , by treatment with

nitrous acid, and the latter must therefore be a primary amine.

Again there is only one possible compound, and A is isobutylamine.

CH 3

\ CH—CH 2—NH 2 + H N 0 2 ->

/ CH 3 A

CH 3

CH—CH 2—OH + N . + H 2 0

/ CH 3


Describe, with full experimental details, the laboratory preparation of a specimen of methylamine hydrochloride. How, and under what conditions, does methylamine react with (a) acetyl chloride, (b) nitrous acid, (c) methyl iodide?

(University of London, 1958, Paper II, qu. 11)

Methylamine is usually prepared in the laboratory as the hydro-chloride by the Hofmann degradation of acetamide. This involves treating the amide with alkaline hypobromite solution.

CH 3—CO—NH 2 + KOBr + 2 KOH

C H 3— N H 2 + KBr + K 2 C 0 3 + H 2 0

FIG. 5.

The apparatus sketched (Fig. 5) is generally used, and in the flask is placed a 40 per cent solution of potassium or sodium hydroxide. The dropping funnel is filled with a solution containing JV-bromacet-amide (acetobromamide), prepared by adding excess bromine carefully to dry acetamide, and then adding dilute alkali solution until only a pale yellow colour remains in the solution.

CH 3—CO—NH 2 + Br 2 -> CH 3—CO—NHBr + HBr

After warming the solution of alkali to 65°C, the bromacetamide solution is run slowly in, care being taken to prevent the temperature

AMINES 6 3

of the reaction mixture rising above 70°C. When all the solution has been added, the temperature is maintained at 70°C for a further 15 min, and then raised to distil off the methylamine which is absorbed as hydrochloride in the trap.

CH 3—CO—NHBr + NaOH - * CH 3—NCO + NaBr + H 2 0 methyl isocyanate

CH 3—NCO + 2 NaOH — C H 3 — N H 2 + Na 2CO„ methylamine

C H 3— N H 2 + HCl - * CH 3—NH^Cl-methylamine hydrochloride

When the distillate is no longer alkaline, the trap is removed, and the solution evaporated carefully to dryness in a fume cupboard, leaving a white residue. This is heated under reflux with ethanol, and the solution obtained filtered rapidly through a hot funnel. Any residue is ammonium chloride and is rejected, and the filtrate on cooling yields crystals of pure methylamine hydrochloride. These are filtered off, dried, and stored in a tightly stoppered bottle, since they are deliquescent.

(a) Acetyl chloride reacts readily with primary amines to produce the monoacetyl compound (occasionally disubstituted products result). Thus methylamine yields iV-acetylmethylamine or 7V-methyl-acetamide.

C H 3— N H 2 + CH 3—CO—CI -> CH 3—NH—CO—CH 3

(b) Nitrous acid reacts with most primary amines to produce the corresponding alcohol and also a large amount of alkene (often not mentioned),

C H 3— C H 2— N H 2 + H N 0 2 - * CH 3—CH 2—OH + N 2 + H 2 0

C H 3— C H 2— N H 2 + H N 0 2 -> C H 2 = C H 2 + N 2 + 2 H 2 0

but in the case of methylamine little or no methanol is obtained, since no reaction occurs. It is suggested that the methylamine nitrite formed hydrolyses again before it can decompose.

(c) By heating methyl iodide with alcoholic ammonia in a sealed tube, primary, secondary and tertiary amines and quaternary salts


may be obtained. The major product depends on the proportions of the reactants used, but no one compound is ever obtained exclu-sively. The compounds are produced by the following series of reactions :

CH 3—I + N H 3 -> C H 3— N H 2 + HI -> CH 3—NH+I-methylamine methylamine

hydriodide

CH 3 - N H 3 I + N H 3 ^ C H 3— N H 2 + NH 4I

C H 3 - N H 2 -'r CH 3l -> (CH 3) 2NH + HI -> (CH 3) 2NH 2+I-dimethylamine dimethylamine

hydriodide

(CH 3) 2NH 2I + N H 3 ^ (CH 3) 2NH + NH 4I

(CH 3) 2NH + CH3T - * (CH 3) 3N + HI -> (CH 3) 3NH+I trimethylaminc trimethylamine

hydriodide

(CH 3) 3NHI + N H 3 ^ (CH 3) 3N + NH 4I

(CH 3) 3N + CH 3I - (CHa) 4N+l-tetramethylammonium

iodide

AMINES 6 5

Describe the essential experimental steps in the conversion of nitrobenzene into aniline, and of the latter into acetanilide.

Write graphic formulae for the three compounds and equations for all the reactions involved.

(Welsh Joint Educat ion Commit tee , 1963, Paper II, qu . 8)

Aniline is normally prepared by the reduction of nitrobenzene. The overall reaction is represented by

C 6 H 5 - N 0 2 + 6 H -> C 6 H 5 - N H 2 + 2 H 2 0

the reducing agent being granulated tin and hydrochloric acid or, on a larger scale, iron and hydrochloric acid. The latter is used commercially since it is cheaper, even though the yield is poorer.

In the laboratory, nitrobenzene, some water, and an excess of granulated tin are placed in a round-bottomed flask equipped with a reflux condenser. About five times as much by volume of con-centrated hydrochloric acid as nitrobenzene is added little by little down the condenser, with shaking and cooling if necessary. When all the acid has been added, the flask is heated in a water-bath to complete the reduction; this is indicated by the disappearance of the last oily drops of nitrobenzene, since the aniline produced dissolves as the hydrochloride and then as the stannichloride (see Note 1).

Sn + 2 HCl - SnCl2 + 2 H

C 6 H 5 — N 0 2 + 6 H — C 6 H 6 - N H 2 + 2 H 2 0

C 6 H 5 — N 0 2 + 3 SnCl2 + 6 HCl ->

C 6H 5— N H 2 + 3 SnCl4 + 2 H 2 0

C 6H 5— N H 2 + HCl -> C 6H 5—NH 3C1

A solution of alkali is now carefully added, with cooling, until the reaction mixture is alkaline; aniline is liberated and produces an upper oily layer,

C 6H 5—NH3CI + NaOH = C 6 H 5 — N H 2 + NaCl + H 2 0

whilst the tin is precipitated as stannic hydroxide, which dissolves in excess as sodium stannate.

Sn(OH) 4 + 2 NaOH = N a 2 S n 0 3 + 3 H 2 0


Any stannous chloride present will yield stannous hydroxide and sodium stannite.

The mixture is then steam distilled in the apparatus shown (Fig. 6), in order to separate the aniline from the various other reduction products which are not steam volatile, which might otherwise be removed with it by the more obvious method of ether extraction (see Note 2).

When the distillate is clear, distillation is stopped and sodium chloride is added to saturate the aqueous layer and "salt out" the maximum amount of the slightly soluble aniline, which forms the upper layer. This is then tapped off by the use of a separating funnel and dried by shaking with potassium hydroxide pellets, since all the normal drying agents react with amines.

The aniline is then redistilled under reduced pressure (to minimize atmospheric oxidation) and collected as a colourless liquid which soon darkens on storage. The boiling point is 184°C at 760 mm.

The aniline can be converted to acetanilide by refluxing it with equal volumes of glacial acetic acid and acetic anhydride. Addition of a little zinc dust assists in preventing oxidation which sometimes occurs, giving the product a pink colour. After about 30 min heating is stopped and the mixture poured into a large volume of cold water, with stirring. After thorough cooling, the crude product is filtered off, and recrystallized from aqueous alcohol, to give white plates of acetanilide.

C 6 H 5 — N H 2 + (CH 3—CO) 20 —

C 6H 5—NH—CO—CH 3 + CH 3—COOH

Aniline

- - W a t e r

FIG. 6.

AMINES

The series of reactions is thus

67

NOTES

1. The reactions are SnCl 4 + 2 C 6H 5— N H 3C 1 - > ( C 6 H 5 — N H 3 ) 2

SnCl 6 and (C 6H 5—NH 3) 2SnCl 6 + 6 NaOH -> 2 C 6 H 5 — N H 2 + 6 NaCl + Sn(OH) 4 + 2 H 2 0

2. The weight of the aniline passing over is related to that of the water distilled by

Wt. of aniline V.P. of aniline X Mol. wt. (93)

Wt. of water ~~ V.P. of water χ Mol. wt. (18)


Compare and contrast the properties of aliphatic and aromatic

primary amines.

(Whitehaven College of Further Educat ion , O . N . C . , 1959, qu. 7)

The aliphatic primary amines are typified by such compounds as methylamine, CH 3—NH 2, and ethylamine, C 2H 5— N H 2 , whilst typical aromatic primary amines are aniline, C 6H 5— N H 2 , and C-substituted anilines. Benzylamine, C 6H 5—CH 2—NH 2, due to the possession of an aliphatic side-chain between the aromatic nucleus and the amino- group shows all the reactions of an aliphatic primary amine.

The aromatic amines are much weaker bases than the aliphatic series, since the lone pair of electrons (which make the amine basic) tend to be donated to the ring in the aromatic series (thus the ring is activated) and this is not possible in the aliphatic compounds where, due to the inductive effect, electron shift towards the amino-group tends to occur.

Nevertheless, both series are strong enough bases to form salts with acids, which are solid, water-soluble compounds.

C H 3— N H 2 + HCl -> CH 3—NH+C1-methylamine hydrochloride

C 6 H 5 — N H 2 + HCl - * C eH 5—NH+Cl-aniline

hydrochloride

In a similar way, both react with alkyl halides, to give secondary and tertiary amines, and then quaternary salts.

CH I CH I CH I

C H 3 — N H 2 — V (CH 3) 2NH+I~ (CH^NH+I" — V ( C H ^ N + l " dimethylamine trimethylamine tetramethylammonium

hydriodide hydriodide iodide

The hydrogen atoms on the — N H 2 group are sufficiently active to be replaced by sodium when heated with the metal, or by an acyl group when treated with acid halides or anhydrides.

C 2 H 5 - N H 2 + ( C H 3 - C O ) 2 0 —

C 2H 5—NH—CO—CH 3 + CH 3—COOH iV-acetylethylamine

C 6 H 6 — N H 2 + (CH 3—CO) 20

C 6H 5—NH—CO—CH 3 + CH 3—COOH acetanilide

AMINES 69

Halogens also can replace these atoms, if the amine is in alkaline solution (if aliphatic) or if the aromatic amine is treated with hypo-halite.

C 2H 5— N H 2 + KOBr C 2H 5—NHBr + KOH iV-bromoethylamine

C e H 5 — N H 2 + HOCl -> C 6H 5—NHCl + H 2 0 iV-chloroaniline

When heated with alcoholic potassium hydroxide and chloroform, both series yield the unpleasant smelling carbylamines or isocyanides.

C 3 H 7 — N H 2 + CHCI3 -> C 3H 7— N C + 3 HCl propyl isocyanide

C e H 5 — N H 2 + C H C I 3 ~> C 6H 6— N C + 3 HCl phenyl isocyanide

Lastly, both types of amine condense with aromatic aldehydes to give anils, or Schiff 's bases.

C 2 H 5 — N H 2 + OHC—C 6H 5 -> C 2 H 5 — N = C H — C 6 H 5

benzylidene ethylamine

C 6 H 5 - N H 2 + O H C - C 6 H 5 - C 6 H 5 - N = C H - C 6 H 5

benzylidene aniline

Of the dissimilar reactions, the most important is with nitrous acid. Here the aliphatic compound reacts, even when cold, to give, generally, a mixture of an alcohol (the usually quoted product) and an alkene. Rearrangement of normal to /so-compounds frequently occurs. Methylamine does not react at all.

C 2 H 5 — N H 2 + H N O , -> C 2H 5—OH + N 2 + H 2 0 ethanol

C 2 H 5 — N H 2 + H N 0 2 -> C 2 H 4 + N 2 + 2 H 2 0 ethylene

Aromatic compounds, however, in acid solution below 5°C give a solution of a diazonium compound, without the production of nitrogen.

C6H5—NH3CI + H N 0 2 -> C 6H 5—N 2C1 + 2 H 2 0 phenyl diazonium chloride

6


If this solution is warmed gently, decomposition occurs to give nitrogen and a phenol (hydroxybenzene), so that even here a simi-larity with the aliphatic reaction exists.

Another difference is seen in the ease with which oxidizing agents attack aromatic amines, producing a variety of compounds such as phenylhydroxylamines and nitrosobenzenes,

and, if other groups are present on the nucleus, phenols and quin-ones. Hypochlorites give a purple colour which is characteristic of aniline. Oxidizing agents react with the more stable aliphatic amines much less readily.

So far only reactions involving the — N H 2 group have been mentioned, but since the two types differ in the group to which this is joined, some differences between them are seen in the reactions of the aromatic nucleus or the alkyl group. Thus bromine water gives an instant precipitate of the tribromo-compound with an aromatic amine, whilst little substitution occurs in the aliphatic series.

C 6H 5—N 2C1 + H 2 0 -> C eH 5—OH + N 2 + HCl

C 6 H 5 — N H 2 -> C 6H 5—NHOH -> C 6H 5— N O

NH. 2 N H 2 Br

+ 3 HBr

Br 2,4,6-tribromoaniline

Diazonium Compounds

Describe, with essential practical details, the laboratory method for the preparation of a solution of benzene diazonium chloride from aniline.

How may benzene diazonium chloride be converted into (a) chloro-benzene, (b) benzene, (c) />-hydroxyazobenzene, (d) iodobenzene?

(Welsh Joint Educat ion Commit tee , O . N . C . , 1960, Syl labus B , qu . 2)

To diazotize aniline, the latter is taken and dissolved in the min-imum volume of fairly strong hydrochloric acid. The solution is then diluted, and cooled to 0°C in ice.

C 6 H 5 — N H 2 + HCl -> C 6H 5—NH+C1-aniline hydrochloride

An aqueous solution of potassium nitrite is then prepared, and also cooled to 0°C, before running it very slowly, and with stirring into the solution of aniline hydrochloride in hydrochloric acid. It is run in at such a rate that the temperature never rises above 5°C.

The nitrite reacts with the excess hydrochloric acid to liberate nitrous acid, and at this temperature this is able to react with the amine salt to yield the diazonium salt.

K N 0 2 + HCl = KCl + H N O ,

C eH 5—NH+C1- + H N 0 2 -> C 6H 5—N 2C1 + 2 H 2 0 benzene diazonium chloride

The presence of excess nitrous acid is indicated when starch/ potassium iodide paper turns blue. Addition of nitrite is then stopped.

(a) Chlorobenzene is then prepared from the above solution by the Sandmeyer reaction. A solution of cuprous chloride in con-centrated hydrochloric acid is prepared. The cold diazonium salt solution is then run into the boiling cuprous chloride solution,

71


, N = N — £ y — O N a + HCl

/>-hydroxyazobenzene (sodium salt)

steam being bubbled through the mixture continuously to carry off the chlorobenzene as it is formed. In this way the halogen in the cuprous halide is incorporated in the ring.

C 6 H 5 - N 2 C 1 ^ Ä C 6 H 5 - C 1 + N 2 Î chlorobenzene

Steam distillation separates the chlorobenzene from other com-pounds which may be produced; it is then tapped off from the distillate, washed, dried and redistilled.

(b) If it is desired to obtain the hydrocarbon from a diazonium salt, it is usual to react the latter with a mixture of ethanol and sulphuric acid. The cold solutions are carefully and slowly mixed and then copper powder is added; if sufficient heat is not generated to cause the mixture to reflux, heat is applied until it does. After nitrogen ceases to be evolved, the mixture is cooled and the benzene separated by steam distillation.

The yield is never very good, and considerable amounts of ethers

may be formed (see Note 1).

C 6 H 5 - N 2 C 1 + CH 8—CH 2—OH

C 6H 6 + HCl + C H 3 - C H O + N 2 Î

(c) If a solution of benzene diazonium chloride is allowed to stand, or better, if it is warmed on a water-bath to about 80°C, then when evolution of nitrogen ceases, phenol remains, partly in solution, and partly as a dense layer, in the reaction vessel.

C 6H 5—N 2C1 + H 2 0 -> C 6H 5—OH + N 2 | + HCl

Sufficient sodium hydroxide solution is now added to dissolve the phenol as sodium phenate,

C 6H 5—OH + NaOH -> C 6H 5—ONa + H 2 0

and if then the solution is cooled, and an equivalent amount of the original solution to that already used is added slowly to it, an orange-red precipitate of /?-hydroxyazobenzene is obtained by the coupling of the phenol with the diazonium salt.

DIAZONIUM COMPOUNDS 73

(d) Iodobenzene is not prepared by the Sandmeyer reaction, like other halobenzenes, but by the direct reaction of potassium iodide with benzene diazonium chloride solution.

C 6H 5—N 2C1 + KI -> C 6H 5—I + N 2 + KCl

A solution of the former is slowly added to the latter, with shaking, and allowed to stand. After refluxing, the solution is rendered alkaline and then steam distilled. The iodobenzene is tapped off, washed with sodium carbonate solution and then with water, dried with calcium chloride and redistilled.

NOTES

1. The reaction is

C 6H 5—N 2C1 + C 2H 5—OH — C 6H 5— O — C 2H 5 + N 2 Î + HCl

Benzene and Derivatives

How is benzene obtained industrially? How and under what conditions does benzenereact with (i) chlorine, (ii) hydrogen, (iii) ozone, (iv) sulphuric acid?

(University o f L o n d o n , Jan. 1963, Paper I, qu . 2)

When coal is heated out of contact with air, gas is evolved, and two liquids distil over, a light liquid of low viscosity, known as ammoniacal liquor, and a denser, viscous liquid, coal tar. For the maximum yield of coal tar, distillation is carried out at 600°C under reduced pressure. After running off the former, the latter is used as an industrial source of many aromatic compounds, among them benzene.

The coal tar is first fractionated to give light oil, with a boiling range of 80-170°C, three other higher boiling fractions, and a residue of pitch. The light oil is then allowed to stand (when a further quantity of ammoniacal liquor separates out and is run off) before washing with sodium hydroxide solution, and then with dilute acid to remove respectively acidic compounds (e.g. phenols) and bases (e.g. pyridine). After drying, it is redistilled to give fractions boiling at 80-110°C, 110-140°C, 140-170°C.

The lowest boiling fraction contains mainly benzene and a little toluene, and is known as "benzole"; fractionation or crystallization yields benzene and toluene. About 1 per cent thiophene remains in the benzene and can only be removed by repeated shaking with concentrated sulphuric acid and washing with water. The pure benzene thus obtained is a colourless liquid, boiling at 80-5°C.

(i) With chlorine, benzene may react in one of two ways, de-pending on the conditions. In the presence of a halogen carrier, such as iron powder or aluminium amalgam, and at low temperature, substitution occurs giving chlorobenzene. The reaction should be

74

BENZENE AND DERIVATIVES

carried out in the absence of direct sunlight.

75

C l

+ HC1

Further substitution gives o- and /7-dichlorobenzenes and 1,2,4-trichlorobenzene.

At higher temperatures, in sunlight or ultraviolet light, however, addition occurs, so that if chlorine is passed into boiling benzene, benzenehexachloride or hexachlorocyclohexane is the major product.

Cl

Cl ι Cl

ci Τ ci Cl

(ii) Hydrogen also adds to benzene, to give the saturated cyclo-hexane; the reaction is best carried out by shaking the benzene with hydrogen under pressure, in the presence of a Raney nickel, palladium or platinum catalyst at 200°C.

(iii) Ozone reacts to form a triozonide, when ozonized oxygen is bubbled through benzene.

+ 3 H 2 0 2


The compound produced is unstable and is hydrolysed by cold water (it explodes in hot water) to give glyoxal. The reaction has also been used in establishing the structure of benzene.

(iv) Concentrated sulphuric acid reacts readily with benzene to set up an equilibrium

but appreciable quantities of benzenesulphonic acid can only be obtained by disturbing the equilibrium by removal of water. This may be done by using oleum at 80°C, by heating much more strongly to drive off the water, or by using a catalyst such as boron trifluoride. The reaction mixture is poured into water and neutralized by adding calcium carbonate. Calcium sulphate is filtered off and the calcium salt of benzenesulphonic acid then crystallized out. The calculated quantity of sulphuric acid can be added to liberate the free acid which can be crystallized out after filtering off the calcium sulphate.

2 C 6 H 5 - S 0 3 H + C a C 0 3 — ( C 6 H 5 - S 0 3 ) 2 C a + C 0 2 + H 2 0

(C 6H 5—S0 3) 2Ca + H 2 S 0 4 -> 2 C 6 H 5 — S 0 3 H + C a S 0 4

S 0 3 H

BENZENE AND DERIVATIVES 77

C 6H 4X 2 represents a disubstituted benzene, and three isomers of such compounds exist. These are the 1,2-, 1,3- and 1,4-sub-stituted compounds, known respectively as ortho-, meta- and para-. Thus, if X is, for example, chlorine, the isomers are :

Cl Cl Cl

I c i ι I

Cl 1,2-dichlorobenzene 1,3-dichlorobenzene 1,4-dichIorobenzene o-dichlorobenzene m-dichlorobenzene p-dichlorobenzene

When trisubstitution occurs, as in C 6H 3X 3, three isomers can again be obtained, the 1,2,3,-, 1,2,4- and 1,3,5- positions being substituted. N o common names are given to these three isomers, though the last is sometimes referred to as s- or sym-, since it is symmetrical. Thus, the isomers are :

Cl Cl Cl

Cl ι Cl Cl ι I

1,2,3-trichlorobenzene 1,2,4-trichlorobenzene 1,3,5-trichlorobenzene

To obtain nitrobenzene from benzene, a nitrating mixture of 1:1 nitric and sulphuric acid is used. Benzene is added in small quantities, with shaking, care being taken to keep the temperature below 60°C until a volume equal to about half that of the nitrating

Illustrate fully, by means of structural formulae, the isomerism which is possible in benzene derivatives of the type C 6H 4X 2 and C 6H 3X 3.

What reagents and conditions are used for the nitration of benzene to nitrobenzene? Write the structural formulae and the names of the compounds which can be obtained by the nitration of (a) benzoic acid, (b) toluene, (c) chlorobenzene, (d) phenol.

Suggest a method for the preparation of /i-nitrobenzoic acid giving a reason for your choice of method.

(Welsh Joint Educat ion Commit tee, 1958, Paper II, qu . 8)


H 2 0 f H N 0 3 ^ >

(a) Since the carboxyl group directs meta- (see Note 1), nitration of benzoic acid yields only one product, m-nitrobenzoic acid.

COOH COOH

+ H N 0 3

H 2 S° 4>

NO,

(b) In contrast, alkyl groups direct ortho-para and so toluene yields 0 - and /7-nitrotoluenes under the same conditions (see Note 2).

CH3 CH3 CH3

H N Q 3

H' S° 4>

nitrotoluenc

(c) In chlorobenzene, too, 0 - and p-chloronitrobenzenes are obtained (see Note 3).

CI CI CI

I I N 0 2 I

0 + H N O Ä 0 - - + Q

N O ,

chloronitrobcnzene

(d) Phenolic —OH groups are powerful ortho-para directing groups, due to donation of electrons to the aromatic nucleus, which is so activated that disubstitution occurs with nitric acid-sulphuric acid mixture (trisubstitution also occurs but oxidation lowers the

mixture has been run in. The mixture is then maintained at 60°C for about 40 min, before pouring into cold water. The layer of nitrobenzene can be tapped off, washed, dried and distilled.

N 0 2


OH OH OH I I N O , ι

phenol | N 0 2

o- p-nitrophenol

OH OH I N O . N 0 2 I NO.

I 2,6-dinitrophenol

N 0 2

2,4-dinitrophenol

For the reasons outlined in (a) /7-nitrobenzoic acid cannot be prepared by nitration of benzoic acid. The normal method is to nitrate toluene, as in (b), separate the para-isomev, and oxidize this by boiling with acid dichromate or permanganate solution. The / M i i t r o b e n z o i c acid required can be crystallized out on cooling the solution. N Oa N0 2

CH 3 COOH />-nitrobenzoic acid

NOTES

1. Several rules governing orientation have been put forward, the most useful being that of Hammick and Illingworth, which states : If a group XY is substituted in a benzene nucleus, if Y is more electro-negative than X, the group directs meta- ; if Y is less electronegative than X or is absent, the group directs ortho-para.

2. Further substitution gives 2,4- and 2,6-dinitro- and 2,4,6-trinitro-toluene.

3. Further substitution gives 2,4- and 2,6-dinitro- and 2,4,6-trinitro-chlorobenzene.

4. Further substitution gives 2,4,6-trinitrophenol (picric acid).

yield considerably) and only with cold dilute nitric acid is one nitro-group introduced (see Note 4).


Describe one method by which each of the following preparations can be carried out: (a) chlorobenzene from nitrobenzene, (b) phenol from benzene, (c) benzyl alcohol from toluene.

(Southern Universities Joint B o a r d , 1963, Paper II, qu. 12)

(a) Chlorobenzene is best prepared from nitrobenzene via the amine and diazonium compound.

Nitrobenzene is reduced by reacting it under reflux with granulated tin and concentrated hydrochloric acid. When all the droplets of nitrobenzene have disappeared the reaction can be assumed to be complete,

C 6 H 5 — N 0 2 + 6 H + HCl -> C 6H 5—NH+Cl- + 2 H 2 0 aniline hydrochloride

and the aniline is liberated by addition of excess strong alkali solution (see Note 1).

C eH 5—NH+Cl" + NaOH -> C 6 H 5 — N H 2 + NaCl + H 2 0 Steam distillation separates the base, which is tapped off from

the rest of the distillate and dissolved in dilute hydrochloric acid. The solution is cooled to 0°C in ice, and to it is slowly added, with stirring, a slight excess of potassium nitrite solution at the same temperature. Nitrous acid is liberated, and phenyl diazonium chloride is produced by its reaction with the aniline hydrochloride. Care must be taken to ensure that the temperature remains below

5°

C' C 6H 5—NH+Cl- + H N 0 2 C 6H 5—N 2C1 + 2 H 2 0

Chlorobenzene is then obtained by the Sandmeyer reaction, the solution being run into a solution of cuprous chloride in con-centrated hydrochloric acid and the mixture refluxed (see Note 2).

C 6H 5—N 2C1 C 6 H 5 - C 1 + N 2

The chlorobenzene is then separated by steam distillation. (b) Phenol is generally prepared on a large scale from benzene

by conversion to chlorobenzene followed by hydrolysis. Chlorine is bubbled through benzene under reflux in the presence

of a catalyst such as iron powder (which acts as a "halogen carrier") until the theoretical quantity has been absorbed.

Fe

C 6H 6 + Cl 2 > C 6H 5—Cl + HCl By fractional distillation, the chlorobenzene may be separated

from any benzene and dichlorobenzenes. Hydrolysis is then carried


out by heating with aqueous alkali under pressure at 300°C (see

Note 3).

C 6H 5—CI + NaOH C 6H 5—ONa + HCl b Ö 3 0 0o c t> Ο

The phenol thus formed is present as sodium phenate, and is liberated by acidification. The dense layer is tapped off, dried and distilled.

It should be noted that hydrolysis only occurs under the con-ditions specified, and not by ordinary refluxing (see Note 4).

(c) Benzyl alcohol is prepared from toluene by a similar process to the last. Chlorine is bubbled into boiling toluene in the light (without a catalyst) until the theoretical gain in weight is obtained to yield benzyl chloride.

C 6 H 5 - C H 3 + Cl 2 -> C 6H 5—CH 2C1 + HCl

Fractional distillation will ensure the removal of toluene and any disubstituted compound, and the major fraction is then hydrolysed by refluxing with aqueous alkali; alternatively, moist silver oxide may be used.

NaOH

C 6H 5—CH 2C1 + H 2 0 > C 6H 5—CH 2OH + HCl The benzyl alcohol is extracted with ether, and the solution washed

and dried. The ether, and then the alcohol, are distilled off.

NOTES

1. The aniline in fact dissolves as the stanni-chloride, and this is decomposed by alkali. The stannic hydroxide redissolves as the stannate.

2 C 6H 5—NH+C1- + SnCl4 -> (C 6H 5—NH 3) 2SnCl 6

(C 6H 5—NH 3) 2SnCl 6 + 6 NaOH - *

2 C 6 H 5 — N H 2 + 6 NaCl + Sn(OH)4 + 2 H 2 0 Sn(OH)4 + 2 NaOH N a 2 S n 0 3 + 3 H 2 0

2. The halogen of the cuprous halide, not that of the diazonium salt, is incorporated in the ring.

3. Or with steam over a catalyst at 425°C. 4. In contrast to alkyl halides and benzyl halides, which are readily

hydrolysed.

General Reactions

Explain, giving examples, the following terms as used in organic chemistry: homologous series; condensation reaction; acetylation; carboxyl group; olefine.

(University of London, Summer 1958, Paper I, qu. 1)

If a number of organic compounds of the same type differ suc-cessively from each other in molecular formula by —CH 2— they are said to be members of a homologous series. Such homologues resemble each other closely in chemical properties and their methods of preparation.

The physical properties generally alter relative to molecular weight. Thus, the alkanes form a series in which the lowest members are gases,

CH 4, methane; C 2H 6, ethane; C 3H 8, propane; C 4H 1 0, butane;

whilst higher members are liquids, and further up the series still, solids. This change is also seen, for example, in the alkenes,

C 2H 4 , ethylene; C 3H 6 , propylene; C 4H 8, butylène;

and the primary amines,

C H 3N H 2 , methylamine; C 2 H 5 N H 2 , ethylamine;

C 3 H 7 N H 2 , propylamine.

A condensation reaction is a two-stage process in which two molecules first react by addition. This is followed by elimination of a small molecule such as water (most usually), an alcohol, ammonia or a hydrogen halide.

The aldol reaction is an example of such a process, provided that gentle heat is applied to bring about the elimination of water.

82

GENERAL REACTIONS 8 3

H H H H

CH 3—C + H—C—CHO CH 3—C—C—CHO Il I N a 0 H I I

Ο H OH H "aldol"

H H I I

C H 3— C = C — C H O + H 2 0 crotonaldehyde

Spontaneous elimination occurs in the formation of phenylhydra-zones.

H H

CH 3—C + H 2N — N H — C 6H 5 -> CH 3—C—HN—NH—C 6H 5 ->

II I Ο OH

Η

I C H 3 — C = N — N H — C 6 H 5 + H 2 0

Acetylation is the reaction whereby certain active hydrogen atoms are replaced by the acetyl group, CH 3CO—. Alcohols, phenols and amines (both aliphatic and aromatic) are the most easily acetylated compounds, and the reaction is most generally carried out by refluxing the appropriate compound with a mixture of glacial acetic acid and acetic anhydride (the latter removing water produced during the reaction).

C 6 H 5 — N H 2 + CH 3—COOH -> C 6H 5—NH—CO—CH 3 + H 2 0 aniline acetanilide

Acetyl chloride may also be used, generally in pyridine (to remove the hydrogen chloride formed).

C eH 5—OH + CH 3—CO—Cl -> C 6H 5—O—CO—CH 3 + HCl phenol phenyl acetate

ο y

A carboxyl group has the structure —C and is the func-

\ Ο—H

tional group present in all organic acids, e.g. acetic acid, CH 3—COOH, benzoic acid, C eH 6—COOH, and oxalic acid, (COOH) 2.


The —CO— group of the —COOH shows none of the properties of the carbonyl group, since the attached —OH group lowers the polarity and therefore the reactivity very considerably.

In aqueous solutions, compounds containing carboxyl groups be-have as weak acids, the hydrogen atom being lost as a proton. The electrons which formed the bond between H and Ο remain on the oxygen atom, giving a carboxyl ion, which is a resonance hybrid,

Ο o-/ /

R—C <-> R—C

\ \ o- Ο

and thus more stable that the parent group. In non-polar organic solvents hydrogen bonding occurs, giving

an apparent doubling of the molecular weight, and no acidic prop-erties are exhibited.

Ο . . . Η—Ο / \

R—C C—R

\ / O—H . . . O

Olefine is the old term used for a member of the homologous series now known as alkenes. These are hydrocarbons characterized by the presence of one double bond in the molecule, which gives rise to the typical property of these unsaturated compounds, viz. addition, which may be represented by

\ / I I C = C + AB -> —C—C—

/ \ I I A Β

Examples are :

C H 2 = C H 2 + HBr - * CH 3—CH 2Br ethylene ethyl bromide

C H 3 — C H = C H 2 + Cl 2 CH 3—CHCl—CH 2C1 propylene, propene 1,2-dichloropropane

GENERAL REACTIONS 85

Describe three different examples of reactions which result in an increase in the length of a carbon chain and two different examples of reactions which result in a decrease in the length of a carbon chain.

(Welsh Joint Educat ion Committee, 1961, Paper II, qu . 10)

A large number of organic reactions result in an increase in chain length of one of the reactants. A very simple method of doubling the chain length is by means of the Wurtz reaction. An alkyl halide is run carefully on to sodium wire, in dry ether, in an apparatus set up for reflux. An alkane results, and when the reaction subsides, the ether and alkane are distilled off (unless the latter is gaseous when it is given off as formed and condensed in a cooled trap) (see Note 1).

2 C 4H 9—I + 2 Na ~> C 8 H 1 8 + 2 Nal n-butyl iodide n-octane

Another method is by pyrolysis of calcium salts of acids. The calcium or barium salt is dry distilled, when it breaks down to yield a ketone. If the original acid has η carbon atoms in it, the ketone obtained will have (2n — 1).

CH 3

Heat \

(CH 3—COO) 2Ca — > CO + C a C 0 3

calcium acetate /

CH,

The most important series of reactions is that by which a methylene group is introduced, to give the next homologue to the starting material. This may be an acid or an alcohol (to which it is readily reduced), or an alkyl halide (which is prepared from the alcohol).

The alkyl halide is converted to the cyanide by refluxing with aqueous-alcoholic potassium cyanide, thus increasing the chain length by one carbon atom, and the latter is then either (i) reduced to an amine and converted to an alcohol by treatment with nitrous acid, or (ii) hydrolysed to an acid by refluxing with dilute sulphuric acid.


The scheme is set out below:

LiAIH4

R—COOH — > R—CH a—OH or N a + *

acid EtOH alcohol

KCN

R—CH 2—Br • R—CH,—CN alkyl halide alkyl cyanide

R—CH 2—COOH acid

K 3C r a0 7 H j S 0 4

or Na + >

E t O H X H N O i

R—CH 2—CH 2—NH a—*R—CH 2—CH 2—OH amine alcohol

There are also many methods of decreasing the chain length, one of the simplest being decarboxylation. Here the sodium salt of an acid is heated with soda-lime to give a hydrocarbon.

CH 2—COONa + NaOH — - > C H 4 + N a 2 C 0 8

sodium acetate methane

but the method suffers from the disadvantage that the hydrocarbon is not easily converted into other organic compounds. A more useful method, therefore, is to convert the starting material to an acid, and then to the silver salt. This is treated with bromine, when carbon dioxide is lost and an alkyl bromide formed, which may readily be converted into almost any desired product with one carbon atom less than the original compound.

H.SO A Ae .O Br.

R—CHj—OH , ' *> R—COOH -=-*» R—COOAg — alcohol acid silver salt

R—Br + CO, + AgBr

KOH

R—OH (see Note 2) 3 alcohol


NOTES

1. If two different alkyl halides are used in equimolar proportions, three alkanes are formed.

R—Br + R'—Br + 2 Na - * R—R + R—R' + R — R',

2. A more often quoted method utilizes the Hofmann degradation. In this series of reactions an amide, produced from an acid via the ammonium salt, is heated with alkaline hypobromite solution. The carbonyl group is lost and an amine produced, but the conver-sion of the amine to the alcohol which follows, by treatment with nitrous acid does not generally give a good yield, and methylamine does not react at all.

N H a Heat KOH

R—COOH — > R—COONH 4 > R—CO—NH a R — N H 2

acid ammonium salt amide amine


Outline, with complete equations in all cases, the steps by which you would convert (i) acetylene to acetic acid, (ii) methyl alcohol to ethyl alcohol, (iii) nitrobenzene to phenol, (iv) ethyl alcohol to ethylene dibromide.

(University of L o n d o n , Jan. 1962, Paper I , qu. 1)

(i) If acetylene is bubbled slowly through a solution of mercuric sulphate in warm sulphuric acid (the strength varies with the process), acetaldehyde is obtained, by addition of either water, to give a vinyl alcohol intermediate, or sulphuric acid, to give a disulphate, which is subsequently hydrolysed.

CHo=CH—OH

C H = C H X

* C H a - C H O

CH 3—CH(HS0 4)

If the solution obtained is then oxidized by refluxing with potassium dichromate in 30 per cent sulphuric acid, acetic acid is produced.

H „ S 0 4 4- K„Cr00~

CH 3—CHO -ι- O-1-

1 CH 3—COOH

By collecting the fraction distilling between 100 and 120ÛC an

aqueous solution of the acid is obtained. (ii) Methanol may be converted to methyl iodide by dissolving

in it the appropriate amount of iodine, and slowly adding red phosphorus, or by slowly adding iodine to the other two reactants. The mixture is placed in a stoppered flask and allowed to stand for 12-24 hours, after which the methyl iodide (the only liquid methyl halide) is distilled off.

2 Ρ r 3 12 - 2 PI 3

3 CH 3OH + PT3 — 3 CH3I + P(OH) 3

This may be refluxed with an aqueous alcoholic solution of potassium cyanide to give methyl cyanide or acetonitrile ; this too may be distilled off from the reaction mixture.

CH3—1 -i K.CN —> CH 3—CN j - Kl


After drying, the nitrile is reduced to ethylamine, using sodium and alcohol as the reducing agent, or alternatively lithium aluminium hydride in ether,

CH 3—CN + 4 H -> C H 3— C H 2— N H 2

and this on treatment with nitrous acid yields ethanol (50 per cent yield).

KNO.>

CH 3—CH 2—NH 2 + H N 0 2 "> CH 3—CH 2—OH + H 2 0 + N 2 dil. HCl

(iii) Nitrobenzene and granulated tin are placed in a flask, and an excess of concentrated hydrochloric acid is added portion by portion, allowing the reaction to subside after each addition. After this the mixture is warmed on a water bath for a time to complete the reduction.

C 6 H 5 — N 0 2 + 6 H+ + 3 Sn -> C 6 H 5 — N H 2 + 2 H 2 0 + 3 Sn++

Sodium hydroxide solution is then added until all the precipitate formed has redissolved. This serves to break down the aniline stanni- and stanno-chlorides and liberate the free base.

The aniline is recovered from this mixture by steam distillation, salted out from the distillate and the oily upper layer tapped off, dried and redistilled under reduced pressure. (The last stages are probably unnecessary in the conversion in question.)

Aniline thus obtained is dissolved in the minimum volume of concentrated hydrochloric acid and the solution diluted and cooled in ice. To this is then added, very slowly and with stirring, a solution of potassium nitrite, similarly cooled. Care must be taken during the addition to ensure that the temperature is always below 5°C. A solution of phenyldiazonium chloride is thus obtained.

HCl + K N 0 2 = H N 0 2 + KCl

C6H5—NH3CI h H N 0 2 ^ > C 6H 5—N 2C1 + 2 H 2 0

After addition of water, this is warmed gently and then maintained at 80°C until no further evolution of nitrogen is observed, indicating that the formation of phenol is complete.

C eH 5—NoCl H 2 0 — C 6H 5—OH + N 2 + HCl


(iv) A mixture of ethanol and sulphuric acid is prepared by carefully adding the latter to the former whilst cooling and stirring. Ethyl hydrogen sulphate is thus formed.

C 2H 5—OH + H 2 S 0 4 -> C 2 H 5 — H S 0 4 + H 2 0

The mixture is then heated in the fume cupboard and the temperature maintained at 170°C. As ethylene is evolved, more ethanol and some sulphuric acid is run into the flask to produce a continuous supply of the gas. An excess of sulphuric acid must be maintained to minimize the production of ether.

C 2 H 5 — H S 0 4

1- ^ > C H 2 = C H 2 + H 2 S 0 4

This is bubbled through water and sodium hydroxide solution to remove ethanol and sulphur dioxide respectively, and is finally passed through a wash bottle containing bromine under a layer of water. Considerable heat is evolved as addition takes place and the gas is passed through the bottle until the layer of bromine is replaced by one of ethylene dibromide which is colourless. A second bottle cooled in ice may usefully be employed as a trap.

C H 2 = C H 2 + Br 2 -> CH 2Br—CH 2Br

The dense layer of ethylene dibromide is tapped off, washed with water, dried with calcium chloride, and then redistilled.


For each of the following reagents give one general use for which it is used in organic chemistry: (a) alcoholic potash, (b) sodium amalgam, (c) Phenylhydrazine, (d) potassium dichromate, (e) am-moniacal silver nitrate. In each case give one specific example of the use of the reagent, describing the practical conditions and naming all the organic compounds involved.


(a) Alcoholic potash is used to eliminate hydrogen halide from a molecule containing halogen, since aqueous alkali will normally cause hydrolysis.

Thus, by refluxing η-propyl bromide with alcoholic potash, propylene is obtained.

EtOH

CH 3—CH a—CH 2Br + KOH * C H 3 — C H = C H 2 + H 2 0 + KBr

Similarly, acetylene can be prepared from ethylene (or ethylidene) dibromide.

CH 2Br—CH 2Br + 2 KOH ^ > C H = C H + 2 KBr + 2 H 2 0

(b) Sodium amalgam, due to the presence of the mercury, reacts only slowly with water or alcohols, to yield hydrogen.

2 N a + 2 H 2 0 = 2 NaOH + H 2

2 N a + 2 ROH = 2 RONa + H 2

Due to the slower, steadier evolution of the gas than with sodium itself, the amalgam is a powerful reducing agent, reacting with esters to give first the component alcohol and acid. The latter is then also reduced to an alcohol as in the Bouveault-Blanc reduction:

C 2 H 6 - C O O - C 4 H 9 + 4 H C 2 H 5 - C H 2 - O H + C 4 H 9 - O H butyl propionate n-propanol n-butanol

Here a solution of the ester to be reduced is run slowly on to sodium amalgam in an apparatus set up for refluxing. The solvent is an alcohol or an aqueous alcohol, chosen so that it can easily be separated from the alcohols produced by the reduction. During the


addition, cooling may be necessary, followed by refluxing, after which the reaction mixture is fractionated.

(c) Phenylhydrazine is used to prepare derivatives of aldehydes and ketones, viz. the phenylhydrazones, by which they can be identified. It is most suitable for the aldehydes and ketones of high molecular weight, since the others tend to give low melting solids which are difficult to crystallize ; for these, it is more usual to use the 2,4-dinitro derivative.

The phenylhydrazone is prepared by warming the aldehyde or ketone on a water-bath for 1 to 2 hours with Phenylhydrazine or Phenylhydrazine hydrochloride and sodium acetate in aqueous alcoholic solution, under reflux, when condensation readily occurs.

CH 3 CH 3

\ \ C = 0 + H 2N — N H — C e H 6 -> C = N — N H — C e H 5 + H 2 0

/ / CH 3 C H 3

acetone phenylhydrazone

Dilute acetic acid is then added, and the product filtered off and recrystallized.

(d) Potassium dichromate in dilute sulphuric acid is commonly used as an oxidizing agent in organic chemistry.

Acetaldehyde is generally prepared in the laboratory by the oxidation of ethanol with this reagent. A mixture of ethanol and concentrated sulphuric acid is prepared and run slowly, with shaking, into an aqueous solution of potassium dichromate. Acetaldehyde is produced and distils over (often without any initial heating) a double surface condenser and a receiver cooled in ice being desirable.

3 C 2H 5—OH + K 2 C r 2 0 7 + 4 H 2 S 0 4 -> ethanol

3 CH 3—CHO + K 2 S 0 4 + Cr 2(S04) 3 + 7 H 2 0 acetaldehyde

(e) Ammoniacal silver nitrate, or Tollen's reagent (see Note 1), is used as a mild oxidizing agent and most commonly as a test for aldehydes, which reduce it.

If the reagent is heated slowly in a water-bath with an aldehyde, the silver ions are reduced to silver metal which is deposited as a


mirror on the walls of the vessel containing it, whilst the aldehyde is oxidized to an acid.

CH 3—CHO + A g 2 0 ^ > CH 3—COOH + 2 Ag acetaldehyde acetic acid

(see Note 2)

NOTES

1. Tollen's reagent is prepared by adding dilute ammonium hy-droxide to silver nitrate solution, until the precipitate of silver oxide just redissolves.

2 A g N 0 3 + 2 N H 4O H - A g 2 0 + H 2 0 + 2 N H 4 N 0 3

A g 2 0 + 4 N H 4 O H - 2 Ag(NH 3) 2+OH- + 3 H 2 0

2. The reaction above is usually given, but more correctly is :

CH 3—CHO + 2 Ag(NH 3) 2+OH- ->

CH 3COO- + NH 4+ + 2 Ag + 3 N H 3 + H 2 0

8


What is the reaction, if any, of sodium hydroxide with the following: (a) ethylene dibromide, (b) formaldehyde, (c) acetic anhydride, (d) benzoyl chloride, (e) chloracetic acid, (f) acetamide, (g) methyl-amine, (h) urea, (i) benzoic acid, (j) chlorobenzene?

In each case, state the necessary conditions. (Southern Universities' Joint B o a r d , 1959, Paper I I , q u . 10)

(a) If ethylene dibromide is refluxed with aqueous sodium hydrox-ide, ethylene glycol is obtained by hydrolysis,

BrCH 2—CH 2Br + 2 NaOH HOCH 2—CH 2OH + 2 NaBr

whereas by using alcoholic alkali, dehydrobromination occurs, with the production of acetylene.

BrCH 2—CH 2Br + 2 NaOH ^ > CHEEECH + 2 NaBr + 2 H 2 0

(b) By shaking with 50 per cent aqueous or aqueous alcoholic sodium hydroxide solution, formaldehyde undergoes the Cannizzaro reaction, oxidation and reduction producing sodium formate and methanol respectively.

2 H — C H O + NaOH H—COONa + CH 3—OH

(c) Acetic anhydride is readily hydrolysed by shaking or warming with a dilute sodium hydroxide solution. Sodium acetate is formed.

(CH 3—CO) 20 + 2 NaOH 2 CH 3—COONa + H 2 0

(d) Benzoyl chloride is a comparatively stable acid halide, and is only very slowly hydrolysed with dilute alkali in the cold. The hydrolysis occurs more rapidly, however, on refluxing, with the production of sodium benzoate.

C 6H 5—CO—CI + 2 NaOH C 6H 5—COONa + NaCl + H 2 0

(e) On refluxing with aqueous sodium hydroxide, chloracetic acid undergoes hydrolysis to form hydroxyacetic or glycolic acid, and this may be obtained from the solution on cooling as the sodium salt. CI—CH 2—COOH + 2 NaOH - ^ - >

HO—CH 2—COONa + NaCl + H 2 0

If on the other hand the acid is heated with solid sodium hydroxide, the above reaction will be followed by decarboxylation, and meth-anol will result.

HO—CH 2—COONa + NaOH CH 3—OH + N a 2 C 0 3


(f) Acetamide is hydrolysed by boiling with dilute sodium hydrox-ide solution, the final products being sodium acetate and ammonia. An intermediate compound is ammonium acetate.

CH 3—CO—NH 2 + H 2 0 -> CH 3—COONH 4

CH 3—COONH 4 + NaOH CH 3—COONa + N H 3 + H 2 0 (see Note 1)

(g) Alkali alone gives no reaction with methylamine under normal conditions, but if in ethanolic solution in the presence of chloroform the very unpleasant smelling methyl isocyanide or carbylamine is evolved.

C H 3— N H 2 + CHC13 + 3 NaOH ^ > CH 3—NC + 3 NaCl + 3 H 2 0

(h) Urea, like acetamide, is an amide, and is hydrolysed by sodium hydroxide solution under the same conditions, to yield ammonia and carbon dioxide, which produces sodium carbonate in the solution (see Note 2).

N H 2— C O — N H 2 + 2 NaOH 2 N H 3 + N a 2 C 0 3

(i) If benzoic acid is neutralized with sodium hydroxide solution, sodium benzoate is obtained,

C eH 5—COOH + NaOH -> C 6H 5—COONa + H 2 0

but if the acid is fused with the alkali, decarboxylation of this salt follows, and benzene can be distilled from the mixture.

C 6H 5—COONa + NaOH -> C 6H e + N a 2C O s

(j) Unlike the alkyl halides, the aryl halides are very stable and cannot be hydrolyzed under normal laboratory conditions. Com-mercially, however, chlorobenzene can be converted to phenol by treatment with aqueous sodium hydroxide at 300°C under pressure and subsequent acidification of the sodium phenate solution obtained.

C 6H 5—Cl + 2 NaOH -> C eH 5—ONa + NaCl + H 2 0

NOTES

1. This second reaction will also occur in the cold, and serves to distinguish ammonium acetate from acetamide.

2. An intermediate product is probably ammonium carbamate, NH 2—COONH 4.


Compare and contrast the reactions of the —OH group in ethyl

alcohol, acetic acid and phenol. (Southern Universities' Joint B o a r d , 1961, Paper I , qu . 10)

Ethanol, phenol and acetic acid resemble each other in the pos-session of an —OH group, and thus although each belongs to a different class of compounds, it is not surprising that there are some reactions in which all these compounds behave in a similar manner. For example, all react with alkali metals to form salts, with the evolution of hydrogen.

2 C 2H 5O H + 2 Na -> 2 C 2H 5ONa + H 2 | sodium ethoxide

2 C e H 5 0 H + 2 Na -> 2 C 6H 6O N a + H 2 | sodium phenate

2 C H 3 C O O H + 2 Na -> 2 CH 3COONa + H 2 | sodium acetate

Phosphorus halides react with replacement of the —OH by halogen, though the yield using phenol is very low,

C 2H 5O H + PC15 -> C 2H 5C1 + POCl 3 + HCl ethyl chloride

C eH 5O H + PC1 5 -> C eH 8Cl + POCl 3 + HCl chlorobenzene

CH 3COOH + PC1 5 -> CH 3COCl + POCl 3 + HCl acetyl chloride

and acyl halides react with ethanol, sodium phenate and sodium acetate to produce esters or an anhydride respectively.

C 2H 5O H + C H 3 C O C I CH 3COOC 2H 5 + HCl ethyl acetate

C 6H 5O H + CH 3COCl ->- CH 3COOC eH 5 + NaCl phenyl acetate

CH 3COONa + CH 3COCl -> (CH 3CO) 20 + NaCl acetic anhydride

Alkyl halides also react, to produce ethers or esters, the alkyl group replacing the acyl group in the above series of reactions.

In contrast to these reactions however, only ethanol reacts with hydrogen halides, and then only in the presence of a catalyst such as anhydrous zinc chloride.

C 2H 5OH + HCl ^ > C 2H 5C1 + H 2 0


and this suggests that it is the most easily replaced. Other reactions show it to be least easily ionized, for with indicators, only phenol and acetic acid give an acid reaction, and with aqueous alkali only the two latter form salts; sodium ethoxide exists only in non-aqueous solvents.

C 6H 5O H + NaOH - * C 6H 5O N a + H 2 0

CH3COOH + NaOH -> CH 3COONa + H 2 0

Phenol too is only a very weak acid, since it gives no reaction with carbonates; acetic acid must be somewhat stronger since it reacts with the evolution of carbon dioxide.

2 CH3COOH + N a 2C O a -> 2 CH 3COONa + C 0 2 + H 2 0

It might be expected that an —OH group which is able to ionize would be more stable than one which does not, and this is borne out by oxidation. Only ethanol is easily oxidized,

C H 3 - C H 2 - O H + Ο C H 3 - C H O + H 2 0

phenol and acetic acid being broken down only by much stronger reagents, and then generally to carbon dioxide and water.

Thus it can be seen that the strongest acid is acetic acid and the weakest is ethanol; the latter therefore tends to have the —OH group intact rather than ionized, and so it is generally easier to replace it in an alcohol than in an acid, as has been mentioned.

These differences in reaction depend largely on the nature of the group t o which the —OH is attached. Thus in ethanol, the alkyl group tends to be electron-donating (see Note 1), and this means that there is no possibility of electrons being removed from the

oxygen atom, C 2H 5->— 0 : H and thus the —OH group is stable as

such, and shows little tendency to ionize. In phenol, however, the aromatic nucleus tends to be electron-withdrawing (see Note 2), and so the tendency is for electrons to be withdrawn (at least in part) from the oxygen atom. If this occurs, ionization results,

C 6 H 5 b : H - > ( C 6 H 5 Ö : ) ~ H +

and so phenol behaves as a weak acid, and the properties of the intact —OH group are a little less marked.


NOTES

1. It is said to exhibit a positive inductive, or + / effect. 2. Negative inductive, or —/ effect. 3. The acetate ion is a resonance hybrid structure,

Ο : Ö : -

CH 3—C CH 3—C V · \

O : - O:

often represented as o-*

/ CH 3—C

v .

Acetic acid possesses a very strongly polar carbonyl group, and this serves as a powerful electron-withdrawing group. Thus ioniz-ation occurs even more readily than in phenol, and acetic acid is therefore a stronger acid.

O*- Ö : / /

CH 3—C -> CH 3—C H+ (see Note 3) . \ . .

0 : H O : -

Indeed, except in non-polar solvents the molecule is always ionized, so that the —OH group is very rarely present at all.

Thus although each of the three compounds can be formulated as possessing an —OH group, and this accounts for the similarity of certain properties, the electronic effects account for the many observable differences.

Supplementary Questions

THE questions which follow are intended to supplement those to which answers have already been given. Unless otherwise stated, the questions are taken from the summer examination of the appropriate Board.

STRUCTURE A N D ISOMERISM

60 ml of a mixture of methane, ethylene and hydrogen were mixed with 200 ml (an excess) of oxygen and exploded. On cooling to room temperature 145 ml of gas remained. When this gas was shaken with potassium hydroxide solution its volume decreased by 80 ml. If all volumes were measured at the same temperature and pressure, calculate the percentage composition of the original mixture.

Draw a diagram of the apparatus used to obtain the above results and describe how the experiment would be carried out. How would you modify the experiment to analyse a mixture of the gases nitrogen, carbon monoxide and carbon dioxide ?

(Southern Universities' Joint B o a r d , 1963)

4-444 mg of a monohydric primary alcohol, M, gave on combus-tion 5-45 mg H 2 0 and 11-11 mg C 0 2 . 0-736 mg of M dissolved in 65-25 mg of camphor depressed the melting point of the latter by 5-25°C. What is the molecular formula of M?

How and with what results will M react with (a) phosphorus tribromide, (b) acetyl chloride, (c) sodium?

[Atomic weights: H = 1, C = 12, Ο = 16; cryoscopic constant of camphor is 40° per 1000 mg.]

(University of D u r h a m , 1963)

99


Explain why there exist: (a) two compounds with the formula C 2 H 6 0 ; (b) two acids with the formula C 4 H 4 0 4 ; (c) three lactic acids, C 3 H e 0 3 .

Sketch the structure of all the compounds mentioned above, and name them where possible.

Indicate how you could distinguish experimentally between the compounds (b). What important difference in properties is shown by the compounds (c) ?


ALIPHATIC HYDROCARBONS

Describe suitable methods of preparing: (a) η-propane, (b) propyl-ene, (c) methylacetylene. What chemical tests would you use to distinguish between these three compounds ?

Describe how methylacetylene could be converted into mesitylene and how the latter could be distinguished from propylene.

(Whitehaven Col lege, O . N . C . , 1961)

What is meant by an unsaturated compound? Describe and formulate the reactions of chlorine with (a) ethane, (b) ethylene, (c) acetylene.

Describe and explain the reactions of caustic potash with the product in each case.

(University of L o n d o n , A u t u m n 1955)

By what method is ethylene generally prepared in the laboratory ? Give an account of the reactions of ethylene with (a) hydrogen, (b) bromine, (c) hydrogen bromide. What compounds are obtained when the product of (b) reacts with (i) dilute aqueous potassium carbonate, (ii) alcoholic potash?

(University o f L o n d o n , January 1959)

ALKYL HALIDES

Describe the laboratory preparation of a pure sample of ethyl bromide. Compare and contrast the chemical reactivity of ethyl bromide and bromobenzene.

( M e d w a y Col lege, O . N . C . , 1963)

SUPPLEMENTARY QUESTIONS 101

ETHERS

Outline a suitable method for the preparation of diethyl ether directly from ethyl alcohol. (Practical details are not required.) What are the likely impurities in the crude product and how do they arise?

Describe one general synthetic method which is applicable to the preparation of ethers containing two different alkyl groups.

(Welsh Joint Educat ion Commit tee , 1960)

ALCOHOLS A N D PHENOLS

Outline one method by which ethyl alcohol is obtained com-mercially. How would you attempt to obtain pure (or "absolute") alcohol from an aqueous solution of it, and demonstrate its freedom from water? Upon what evidence is the usual structural formula given to ethyl alcohol ?


Compare and contrast the properties of phenol and benzyl alcohol in their reactions, if any, with sodium carbonate, sodium hydroxide, potassium, bromine, phosphorus pentachloride, sul-phuric acid, acetic acid, and an acidified solution of potassium permanganate.

(Southern Universit ies' Joint B o a r d , 1963)

ALDEHYDES A N D KETONES

Outline the experimental evidence and reasoning on which the structural formula of acetone is based. Compare and contrast its chemical properties with those of the isomeric propionaldehyde.

Give equations for two general methods of preparing aliphatic ke-tones indicating briefly in each case the essential reaction conditions.

(Welsh Joint Educa t ion Commit tee, 1962)


Outline two methods for the preparation of aromatic aldehydes. How and under what conditions does benzaldehyde react with

(a) sodium hydroxide, (b) potassium cyanide, (c) Phenylhydrazine ?

(Bournemouth Col lege, O . N . C . , 1962)

CARBOXYLIC ACIDS

How is formic acid usually prepared on the laboratory scale, and how is the pure acid obtained from its aqueous solution?

Mention two chemical properties of formic acid which are not possessed by acetic acid.

What is the effect of heat on (a) sodium formate, (b) a mixture of calcium formate and calcium acetate?

(University o f L o n d o n , 1956)

How would you convert benzene into benzoic acid? State, with full experimental details, how you would purify the benzoic acid and establish its purity.

How can benzoic acid be converted to (a) benzoyl chloride, (b) ethyl benzoate, (c) benzene ?

(University of D u r h a m , 1963)

ESTERS

Describe the usual laboratory method for the preparation of pure ethyl acetate from ethyl alcohol.

Discuss the reactions of ethyl acetate with water, with aqueous sodium hydroxide and with dilute sulphuric acid.

Mention two other general methods whereby esters may be obtained.

(University o f L o n d o n , A u t u m n 1957)

ACID HALIDES A N D ANHYDRIDES

Describe fully one method of preparation, the physical properties and five important chemical properties of acetyl chloride. Give one simple test to distinguish it from acetic anhydride.



AMIDES

Describe fully one method for the preparation of acetamide, starting from acetic acid. Give an account of the physical properties and five important chemical properties of acetamide.


AMINES

Discuss briefly the isomerism of the primary amines of molecular formula C 4 H n N . Describe three general methods for the prepara-tion of primary aliphatic amines, illustrating your answer in each case by the formation of a different one of the isomers of C 4 H n N .

Identify any isomer(s) which can be converted into a primary alcohol in one step and also any isomer(s) which can be converted into a secondary alcohol in one step. Write the equation for the conversion reaction.

(Welsh Joint Educat ion Committee, 1961)

Give a concise account (experimental details are not required) of one method each for the preparation of aniline and phenol from benzene.

Describe briefly the reactions of aniline with (a) bromine, (b) acetyl chloride, and the reactions of phenol with (c) nitric acid and (d) acetic anhydride. The structural formulae of aromatic products should be given.


Compare and contrast the main physical and chemical properties of ethylamine and aniline.

(Whitehaven Col lege, O . N . C . , 1963)

DIAZONIUM COMPOUNDS

Outline the laboratory preparation of a solution of benzene diazonium chloride. How may this substance be converted into: (a) chlorobenzene, (b) fluorobenzene, (c) benzonitrile, (d) an azo dye, (e) methyl phenyl ether.

( M e d w a y Col lege, O . N . C . , 1963)


BENZENE A N D DERIVATIVES

Outline the separation of benzene from coal tar. Describe the reactions of benzene with (a) methyl chloride, (b) chlorine, (c) sul-phuric acid, (d) nitric acid, mentioning the necessary conditions for each reaction.

(Associated Examin ing B o a r d , 1961)

Outline two methods by which toluene may be obtained from benzene. How does toluene behave on (a) nitration, (b) halogenation, (c) oxidation? Give equations and conditions for each reaction.

(We lsh Joint Educa t ion Commit tee , O . N . C . , 1961, Sy l l . B )

Describe, with practical details, the preparation of a pure sample of nitrobenzene from benzene. How may nitrobenzene be con-verted into (a) meta-dinitrobenzene, and, (b) phenol?

By what tests would you establish the purity of the weta-dinitro-benzene ?


Describe the preparation from benzene of sodium benzene sulphonate.

How may this be converted into phenol ? How and under what conditions does phenol react with the

following? (a) caustic soda (b) zinc dust, (c) bromine water, (d) acetic anhydride.


Explain briefly, with the aid of suitable representative examples, what is meant by the term directive influence in aromatic nuclear substitution reactions. Give one example of a nuclear substitution reaction which proceeds extremely rapidly at room temperature and one which proceeds slowly even with strong heating. Write a structural formula for each aromatic product formed in the examples you give.


GENERAL REACTIONS

Describe the chemical reactions (two in each case) by which you would distinguish between :

(a) a phenol and an organic acid ; (b) an aldehyde and a ketone; (c) an ester and an ether; (d) a paraffin and an olefine; (e) methyl and ethyl alcohols.


The compounds represented by the structural formulae below are in isomeric pairs. Select any four pairs, and, drawing only upon your general knowledge of the chemical properties associated with typical substituent atoms or groups, explain how the one isomer of each pair might be expected to differ from the other in chemical properties. Two illustrative chemical reactions should be given for each pair.

(a) C 6 H 5 - C H 2 - N H 2 , / 7 a r a - C H 3 - C 6 H 4 - N H 2 ; (b) C 6H 5—CH 2—CI, para-CHz—C6H4—CI ; (c) C 6H 5—CH 2—OH, para-CHz—C6H4—OH ; (d) C 6H 5—CO—Cl, para-Cl—C6H4—CHO ; (e) CH 3—CH 2—CHO, CH 3—CO—CH 3; (f) CH 3—CH 2—CH 2—CO—OH, CH 3—CH 2—CO—O—CH 3 ; (g) CH 3—CH 2—CH(OH)—CH 3, CH 3—CH 2—CH 2—O—CH 3.


Give concise explanations of the meaning of any five of the following terms in organic chemistry: (a) hydrolysis, (b) dehydra-tion, (c) esterification, (d) unsaturation, (e) isomerism, (f) diazotiza-tion, (g) decarboxylation.

Outline feasible reaction schemes (experimental details are not required) for two of the following conversions:

(a) EITHER toluene into /?-CH 3—C 6H 4—OH OR toluene into p-Cl—C6H4—CO—OH;

(b) EITHER benzene into p-Br—C6H4—NH2 OR benzene into p-Cl—C6H4—OH.

(Welsh Joint Educat ion Commit tee, 1962)


One illustrative example should be given for each term together with equations and/or structural formulae where they are relevant.


Write graphic formulae for compounds containing an alkyl radical attached to (a) the primary alcoholic group, (b) a secondary alcoholic group, (c) the carboxyl group, (d) the primary amino group. State briefly, giving examples and equations, the character-istic reactions of these compounds.

(University o f L o n d o n , 1957)

Write equations for the reactions that occur between phosphorus pentachloride and (a) acetic acid, (b) ethyl alcohol, (c) acetaldehyde, (d) acetone and (e) diethyl ether. Name the products in each case. What information about the structure of the molecules may be ob-tained from these reactions?


Give an account, with examples, of the use of four of the following reagents in organic chemistry: (a) sulphuric acid, (b) phosphorus pentachloride, (c) sodium (or potassium) hydroxide, (d) sodium hypobromite, (e) an alkaline solution of copper tartrate, (f) sodium nitrite.


State the reagents and conditions required to perform the following syntheses, using the letters over the arrows to show the reactions to which you are referring. (Further notes are not required.)

(i) CaC 2 C 2H 2 -5-» CH 3CHO -^-> CHI3 H.COONa

(ii) CH3COOH CH 2ClCOOH CH 2NH 2COONH 4

CH3CO—NH—CH 2COONH 4

(iii) C 6H 6 - U C 6 H 5 . S 0 3 H C 6H 5ONa - U C 6H 2( N 0 2) 3O H

Write the full equations for the reactions (c) and (d). Reaction (c) is used as a test. Name two other substances which respond to the same test.


Syllabuses and Past Papers

FOR those students who wish to purchase copies of the syllabus and past examination papers, the following details are given, and are correct at the time of publication.

The information applies only to "A" Level, and the O.N.C. of the Welsh Joint Education Committee. Other O.N.C. students will find information available from their own college.

Associated Examining Board for the G.C.E., Hesketh House, Portman Square, W.l .

From the Publications Dept. : Regulations and Syllabus 4s. Od. ; past papers 6d. each paper (not subject). (Order form should first be obtained.)

University of Cambridge Local Examinations Syndicate, Syndicate Buildings, Cambridge.

From the above: Syllabus and Regulations 2s. Od. post free. From E.S.A. Ltd., Pinnacles, Harlow, Essex: Past papers in subject sets, Chemistry, 3s. 6d., plus 5d. post.

Northern Universities Joint Matriculation Board. From J. Sherratt & Son, Park Road, Timperley, Altrincham,

Cheshire: Regulations and Syllabus 2s. Od. post free; past papers 4s. 6d. per set, and 4d. each paper.

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schools.

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From E.S.A. Ltd., Pinnacles, Harlow, Essex: Past papers 3s. 6d. per set, plus 5d. post.

1 0 7


Southern Universities' Joint Board for School Examinations, 22 Berkeley Square, Bristol 8.

From the above: Regulations and Syllabus 2s. 6d. plus 6d. post; past papers 3d. each paper (not subject) plus post.

Welsh Joint Education Committee, 30 Cathedral Road, Cardiff. From the above: Regulations and Syllabus 2s. 6d. post free;

Regulations and Syllabus for O.N.C. ; past O.N.C. papers 3d. each. From E.S.A. Ltd., Pinnacles, Harlow, Essex: Past "A" level

papers 4s. Od. per set, plus 5d. post.

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From the Secretary, above: Regulations and Syllabus 3s. Od. post free; past papers 6d. per subject.

University of London School Examinations Department. From the Publications Dept., 1 Malet Street (First Floor), W.C.I :

Regulations and Syllabus Is. 6d. post free; past papers 6d. each paper, or 3s. 3d. per set, post free.

* The last examinations of this Board were held in Winter 1964.

Documents

Model Answers in Organic Chemistry. For 'A' Level and Ordinary National Certificate Students