ORG CHEM

Chemistry of Carbon Compounds

Unit 29 An introduction to the chemistry of carbon compounds

Unit 30 Isomerism

Unit 31 Typical reactions of selected functional groups

Unit 32 Synthesis of carbon compounds

Unit 33 Important organic substances

Topic 8

KeyC o ncepts

An introduction to the chemistry of carbon compounds

• Homologous series• Systematic naming• Effects of functional groups and chain

length on physical properties

Isomerism• Structural isomerism — chain

isomerism, position isomerism and functional group isomerism

• Stereoisomerism — geometrical isomerism and enantiomerism

Synthesis of carbon compounds• Synthetic routes for carbon

compounds• Preparation of simple carbon

compounds

Important organic substances• Aspirin• Soaps and soapless detergents• Nylons and polyesters• Carbohydrates• Lipids• Proteins

Chemistry ofCarbon Compounds

Typical reactions of selected functional groups

• Reactions of alkanes, alkenes, haloalkanes, alcohols, aldehydes, ketones, carboxylic acids, esters and amides

Topic 84 Chemistry of Carbon Compounds 5Unit 29 An introduction to the chemistry of carbon compounds

29.1 – 29.12 & 29.21

Summary1 The following table summarizes the nomenclature of compounds in various

homologous series.29.1 The value of medicines: longer and healthier lives

29.2 Functional groups: centre of reactivity

29.3 Naming alkanes and alkenes

29.4 Naming carbon compounds with one type of functional group

29.5 Naming haloalkanes

29.6 Naming alcohols

29.7 Naming aldehydes and ketones

29.8 Naming carboxylic acids

29.9 Naming esters

29.10 Naming amides

29.11 Naming amines

29.12 Naming compounds with more than one type of functional group

29.13 Intermolecular forces and physical properties of carbon compounds

29.14 Physical properties of haloalkanes

29.15 Physical properties of alcohols

29.16 Physical properties of aldehydes and ketones

29.17 Physical properties of carboxylic acids

29.18 Physical properties of esters

29.19 Physical properties of amides

29.20 Physical properties of amines

29.21 Common names of carbon compounds

Homologous series

General formula

Functional group it contains

NomenclatureExample

Structural formula IUPAC name

Alkanes CnH2n+2 — add appropriate prefix to –ane CH3CHCH2CH3

CH3

2-methylbutane

Alkenes CnH2n C Creplace ‘ane’ of the corresponding alkane with –ene

CH3CH CH2 propene

Haloalkanes RX–X

(X = F, Cl,Br or I)

add the name of halogeno functional group as prefix to the corresponding alkane

CH

Cl

CH3 CH3

2-chloropropane

Alcohols ROH –OH

replace the last letter ‘e’ of the corresponding alkane with –ol

CH3CH2OH ethanol

Aldehydes RCHOC H

O replace the last letter ‘e’ of the corresponding alkane with –al CH3CH2 C H

Opropanal

Ketones RCOR1C

O replace the last letter ‘e’ of the corresponding alkane with –one CH3 CH3C

Opropanone

Carboxylic acids RCOOH

OHC

Oreplace the last letter ‘e’ of the corresponding alkane with –oic acid

CH3CH2CH2 C OH

Obutanoic acid

Esters RCOOR1

C

O

O

the name consists of two separate words, the first word comes from the alcohol, the second word comes from the acid

CH3CH3

O

OC

methylethanoate

Amides RCONH2

(unsubstituted) C

O

N

replace the ‘oic acid’ ending of the corresponding acid by –amide CH3 NH2C

Oethanamide

Amines RNH2

(primary) N

replace the last letter ‘e’ of the corresponding alkane with –amine

CH3NH2 methanamine

An introduction to the chemistry of carbon compoundsUnit 29


2 Sometimes there is more than one functional group in one compound. This kind of compound should be named according to the following order of precedence of functional groups:

–COOH > –COO– > –CONH2 > –CHO > –CO– > –OH > –NH2 > –C=C–

ExampleGive the IUPAC names of the following compounds.

a)

(1 mark)

CH2COOH

b)

(1 mark)CH3 CH2CH(CH3)2C O

O

c) CH3CH=CHCH2OH (1 mark)

d) CH3CH(NH2)CH2COOH (1 mark)

Answer

a) phenylethanoic acid (1)

b) 2-methylpropyl ethanoate (1)

c) but-2-en-1-ol (1)

d) 3-aminobutanoic acid (1)

29.13 – 29.20

Summary1 The physical properties (e.g. the boiling point and water solubility) of a carbon

compound are affected by

a) the functional group it contains;

b) the length of the carbon chain in molecules.

Exam tipsExam tipsExam tipsExam tips ♦ The longest continuous chain containing the carbon bearing the –OH group may NOT appear in a straight line. Questions often ask about the names of such structures.

3,4C2H5

H3C2C 1CH3

OH

2-methylbutan-2-ol

♦ DO NOT spell ‘amine’ as ‘ammine’. ✔ ✘

♦ DO NOT confuse ‘amine’ and ‘amide’.

The general formula of primary amine is RNH2, while that of amide is (H or R)CONH2.

♦ Students may need to identify the functional groups present in unfamiliar compounds.

e.g.

CH2CH3

CH3

CH3CH2

CH2CH3

CHNH2

C

NHO

OC

OO

oseltamivir

Besides the ether linkage, functional groups present in oseltamivir include:

– C=C bond;

– amide functional group;

– amine functional group; and

– ester functional group.

➤ for (a), the group is a phenyl group. It is attached to the ethanoic

acid.

➤ (b)

CH3 CH2CH(CH3)2C O

O

this part comes from ethanoic acid

this part comes from 2-methylpropan-1-ol

➤ (c) The double bond takes the form -en-.

➤ (d) –COOH is the principle functional group. The –NH2 group is named as a prefix.

RemarksRemarks*


Homologousseries Intermolecular forces Physical properties

Haloalkanes

• permanent dipole-permanent dipole attractions between molecules

Clδ+ δ–CH3

Clδ+ δ–CH3

Clδ+ δ–CH3

key:permanent dipole-permanent dipole attraction

• boiling points higher than those of alkanes of similar relative molecular masses

• polar molecules can interact with water molecules

• slightly soluble in water

Alcohols

• h y d r o g e n b o n d i n g b e t w e e n molecules

CH3CH2 H

O

H

CH2CH3

O

key:hydrogen bond

• boiling points much higher than those of alkanes of similar relative molecular masses

• hydrogen bonding between alcohol molecules and water molecules

CH3CH2 H

CH3CH2 H

O

O

H

H

O

key:hydrogenbond

H

H

O

• alcohols with less carbon atoms are miscible with water in all proportions

• alcohols with a long carbon chain in their molecules are much less soluble in water

2 The following table summarizes the physical properties of members of some homologous series.


Aldehydesand

ketones


C Oδ+ δ–

CH3

CH3

key:permanent dipole-permanent dipole attraction

C Oδ+ δ–

CH3

CH3

C Oδ+ δ–

CH3

CH3

• boiling points higher than those of alkanes of similar relative molecular masses

• hydrogen bonding between aldehyde / ketone molecules and water molecules

CH3

HH

H3C

OO

C

key:hydrogen bond

HH

O

• aldehydes and ketones with less carbon atoms show appreciable water solubility

Carboxylicacids

• h y d r o g e n b o n d i n g b e t w e e n molecules; more extensive than that in alcohols

CH3

key:hydrogen bond

O

HO

O

C

H

O

CCH3

• boiling points higher than those of alcohols of similar relative molecular masses

• hydrogen bonding between acid molecules and water molecules

CH3

key:hydrogen bond

H

C

O

O

O

HH

O

H

H

O H

H

• the first four acids are miscible with water in all proportions



Esters


• boiling points are about the same as those of aldehydes and ketones of similar molecular masses

• hydrogen bonding between ester molecules and water molecules

CH2CH3

CH3 C

O

O

key:hydrogen bond

HH

O

• simple esters are very soluble in water

Amides

• h y d r o g e n b o n d i n g b e t w e e n molecules

H

CH3 C

O

N

H

key:hydrogen bond

H

CH3 C

O

N

H

• boiling points are high

• hydrogen bonding between amide molecules and water molecules

H

CH3 C

O

N

H

key:hydrogen bond

HH

O

HH

OH

HO

• simple amides are very soluble in water


Amines

• hydrogen bonding between molecules of primary amines; hydrogen bonding less strong than that in alcohols

CH2CH3

CH3CH2

CH2CH3

key:hydrogen bond

N

HH

N

HH

N

HH

• boiling points of primary amines higher than those of alkanes but generally lower than those of alcohols of similar relative molecular masses

• h y d r o g e n b o n d i n g b e t w e e n primary amine molecules and water molecules

key:hydrogen bond

CH2CH

3

NH

H

HH

O

H

H

O

• primary amines with less carbon atoms are very soluble in water

Exam tipsExam tipsExam tipsExam tips ♦ Questions may ask students to arrange carbon compounds in order of increasing boiling point.

e.g.

CH3(CH2)2CH3 < CH3(CH2)3CH3 < CH3(CH2)3Cl < CH3(CH2)3OH Attractions weak instantaneous dipole- permanent dipole- hydrogen between induced dipole attractions permanent dipole bonds molecules attractions

♦ The following carbon compounds are miscible with water:

– methanol, ethanol and propan-1-ol;

– ethanal and propanone;

– methanoic acid, ethanoic acid, propanoic acid and butanoic acid.

Topic 812 Chemistry of Carbon Compounds 13Unit 30 Isomerism

ExampleCompounds W, X, Y and Z are all colourless liquids. Suggest how you would distinguish the four compounds from each other.

CH3CH2OH (CH3)3COH CH3(CH2)2Br CH3(CH2)7OH

W X Y Z (5 marks)

Answer

Distinguishing the liquids by water solubility

Add water to the liquids. Both CH3CH2OH and (CH3)3COH are miscible withwater. (1)

Distinguishing the two liquids which are miscible with water

Warm each of these two liquids with acidified potassium dichromate solution. (1)

Only CH3CH2OH turns the dichromate solution from orange to green. (1)

Distinguishing the two liquids which are not miscible with water

Warm each of the two liquids which are not miscible with water with AgNO3(aq) in ethanol. (1)

Only CH3(CH2)2Br gives a creamy precipitate slowly. (1)

➤ In questions of this type, it is common to distinguish the compounds by their water solubility before other chemical tests.

➤ Hydrocarbons such as cyclohexane and cyclohexene often appear in this type of questions. They are insoluble in water.

RemarksRemarks*

30.1 Isomerism

30.2 Structural isomerism

30.3 Geometrical isomerism

30.4 Physical properties of geometrical isomers

30.5 Chirality

30.6 Enantiomers

30.7 Test for chirality — plane of symmetry

30.8 Properties of enantiomers

IsomerismUnit 30


30.1 – 30.2

SummaryThe following charts show the classification of isomers.

ExampleConsider the isomeric compounds X and Y shown below:

compound X

OH

CHO

compound Y

OH

CHO

a) Name the type of isomerism involved. (1 mark)

b) Which of the above compounds has a higher melting point? Explain. (3 marks)

Answer

a) Position isomerism (1)

b) The melting point of compound Y is higher than that of compound X. (1)

Only compound X can form intramolecular hydrogen bonds. (1)

Compound Y forms more intermolecular hydrogen bonds than compound X does. (1)

CH3CHCH2CH3

CH3

isomersdifferent compounds that have the same molecular formula

structural isomersatoms are linked in different orders

stereoisomersatoms are linked in the same way but with different spatial arrangements

structural isomers

chain isomersisomers with the same functional group but different carbon skeletonse.g.

CH3CH2CH2CH2CH3

and

functional group isomers

isomers with the same molecular formula but different functional groupse.g.

and

position isomersisomers with the same carbon skeleton and functional group, but the position of the functional group is differente.g.

CH3CH2CH2CH2OH

and

CH3CHCH2CH3

OH

CH3C CH3

O

O

CH3CH2C OH

O

Exam tipsExam tipsExam tipsExam tips ♦ When ask about the type of isomerism, give the precise type, instead of just stating structural isomerism.

e.g.

OH andH3C OCH3

The type of isomerism involved is functional group isomerism.

♦ Questions often ask about methods for distinguishing between isomeric compounds.

e.g.

OH andH3C OCH3

can be distinguished by

– a physical method

comparing their boiling points / melting points; OHH3Chas a higher boiling point / melting point.

– a spectroscopic method

comparing their IR spectra; OHH3C has a broad and

strong absorption at 3 230 – 3 670 cm–1.


ExampleConsider the melting points and boiling points of cis-1,2-dichloroethene and trans-1,2-dichloroethene.

Compound Melting point (°C) Boiling point (°C)

cis-1,2-dichloroethene –80 60

trans-1,2-dichloroethene –50 48

Explain why

a) cis-1,2-dichloroethene has a higher boiling point; (3 marks)

b) trans-1,2-dichloroethene has a higher melting point. (2 marks)

Answer

a) The boiling point of a compound depends on its intermolecular attractions. (1)

In a molecule of the cis isomer, the dipole moments of the two polar C–Cl bonds reinforce each other. Thus, the molecule has a net dipole moment. Thus, these molecules are held together by permanent dipole-permanent dipole attractions. (1)

In a molecule of the trans isomer, the dipole moments of the two polar C–Cl bonds cancel each other. Thus, the molecule has no dipole moment. Thus, these molecules are held together by weaker instantaneous dipole-induced dipole attractions. (1)

b) In addition to intermolecular attractions, the melting point of a compound depends also on the degree of compactness of molecules in the solid state. (1)

The cis isomer has a lower degree of symmetry. It fits into a crystalline lattice relatively poor and thus has a lower melting point. (1)

30.3 – 30.4

Summary1 Stereoisomers that have a different arrangement of their atoms in space due to the

restricted rotation about a carbon-carbon double bond are geometrical isomers.

2 Compounds with two different groups attached to each carbon of the double bond have two alternative structures, which are geometrical isomers.

andCC

CH3CH3

HH

cis-isomer

CC

HCH3

CH3H

trans-isomer

3 Geometrical isomers normally have similar chemical properties. However, their physical properties are often quite different.

➤ Questions often ask students to compare the melting points and boiling points of geometrical isomers.

e.g.

m.p. 102 °C m.p. –19 °C

CCCH3OOC

H

H

COOCH3

CCH

CH3OOC

H

COOCH3

Their intermolecular attractions are van der Waals’ forces of comparable strength.

The trans isomer has a higher melting point because it is more symmetrical.

RemarksRemarks*

➤ Questions often ask students to compare the melting point / volatility of isomeric compounds.

➤ Compound X forms intramolecular hydrogen bonds due to the close proximity of the –OH group and –CHO group.

O

C

H

O

H

hydrogen bond

RemarksRemarks*

Exam tipsExam tipsExam tipsExam tips

♦ andCCH3C

H3C

Br

ClCC

H3C

H3C

Cl

Brare NOT

geometrical isomers. They are identical molecules.

One of the carbon atoms of the double bond has two methyl groups attached to it.


Exam tipsExam tipsExam tipsExam tips ♦ Questions often ask about enantiomers.

♦ Questions may ask students to mark chiral carbons on the structures of unfamiliar compounds.

e.g.

Tamiflu Aspartame

♦ DO NOT confuse the terms ‘chiral’ and ‘achiral’.

30.5 – 30.8

Summary1 A chiral molecule is defined as one that is not superposable on its mirror image. A

chiral molecule and its mirror image are called a pair of enantiomers.

2 Most simple chiral molecules contain one carbon atom bonded to four different atoms or groups of atoms. Such a carbon atom is called a chiral carbon.

3 A pair of enantiomers have the same physical property and chemical property, except

a) their behaviour towards plane-polarized light; and

b) their reactions with chiral reagents.

ExampleCompound X has the following structural formula:

CH(CH3)COOH

The above structural formula can represent two stereoisomers.

a) Draw three-dimensional structures of the two stereoisomers. (2 marks)

b) State a physical property which is different for the two stereoisomers. (1 mark)

CH2CH3

CH3

CH3CH2

CH2CH3

CHNH2

C

NH

O

OC

OO

* **

NH

CH2

CH

CH2

CH

CO2H

H2N OCH3

O

C

O

C* *

➤ Students need to give good drawings of three-dimensional structures. Use the conventions commonly used in the representation of three-dimensional structures.

➤ Questions often ask students to suggest a different physical property between a pair of enantiomers. One of them turns the plane of polarization of a beam of plane-polarized light clockwise, while the other anticlockwise.

RemarksRemarks*

Answer

a)

COOH

CH3

C

HHOOC

CH3

C

H(1) (1)

b) They rotate the plane of polarization of a beam of plane-polarized light to opposite directions. (1)

Topic 820 Chemistry of Carbon Compounds 21Unit 31 Typical reactions of selected functional groups

31.1 – 31.7

Summary1 The figure below summarizes the addition reactions of alkenes.

HBr(g) Br2(aq)

RCH2CH3

alkane

RCHCH2

diol

RCHBrCH2Br

dibromoalkaneOHOH

RCHCH3

major product minor productbromoalkanes

Br

RCHCH2Br

bromoalcohol

RCH2CH2Br RCHBrCH2Br

dibromoalkane

+ +OH

RCH CH2

alkene

Br2

(in organic solvent)

cold alkalinedilute potassiumpermanganate

solution

H2(g)Pt catalyst

2 Markovnikov’s rule for addition reaction of an asymmetric alkene:

When a molecule HA adds to an asymmetric alkene, the major product is the one in which the hydrogen atom attaches itself to the carbon atom already carrying the larger number of hydrogen atoms.

3 Substitution reactions of haloalkanes — alkaline hydrolysis of haloalkanes

R OHR XNaOH(aq)

reflux

31.1 Introduction

31.2 Important reactions of alkanes

31.3 Addition reactions of alkenes

31.4 Addition of hydrogen to alkenes

31.5 Addition of halogens to alkenes

31.6 Addition of hydrogen halides to alkenes

31.7 Substitution reactions of haloalkanes

31.8 Reactions of alcohols

31.9 Reactions of aldehydes and ketones

31.10 Reactions of carboxylic acids

31.11 Hydrolysis of esters

31.12 Hydrolysis of amides

Typical reactions of selected functional groupsUnit 31


ExampleDescribe, by giving reagent(s) and stating observations, how you could distinguish between the following two compounds using a simple test tube reaction.

(4 marks)

CH3CH2CHCH3

compound X

and

Cl

CH3CH2CHCH3

compound Y

I

Exam tipsExam tipsExam tipsExam tips ♦ Questions often ask students to predict the major product of an addition reaction involving an asymmetrical alkene.

e.g.

CH3

CH3

CH3HBr

HC C

C2H5

CH3

C2H5HBr

HC C

HI

CH3

CH3 CH3

Br C

H

C H

CH3

C2H5 C2H5

Br C

H

C H

I

♦ Questions may ask students to deduce the structure of a compound based on the type of isomerism it can exhibit and its reactions.

Given information:

– Molecular formula C6H12

– It has a pair of enantiomers.

– It loses its chiral centre after hydrogenation.

Deductions:

– The compound should be an alkene as it can undergo hydrogenation.

– The compound should have a chiral carbon.

– Structures of the compound:

H2C=CH

C2H5

CH3

C HHC=CH2

C2H5

H3C

CH

Answer

Put about 2 cm3 of ethanol and 1 cm3 of silver nitrate solution in each of two test tubes. (1)

Place the test tube in a water bath at 60 °C. (1)

Add several drops of compounds X and Y separately to each test tube.

A yellow precipitate forms rapidly in the test tube containing compound Y. (1)

A white precipitate forms slowly in the test tube containing compound X. (1)

➤ Questions often ask about chemical tests for distinguishing haloalkanes.RemarksRemarks*

31.8 – 31.9

Summary

CH3CH CH2

propene

CH3CH2CHO

propanal

C3H7Cl

1-chloropropane

C3H7OH

propan-1-ol

(1° alcohol)

C3H7Br

1-bromopropane

CH3COOC3H7

propyl ethanoate

CH3CH2COOH

propanoic acid

C3H7I

1-iodopropane

• reflux with conc. HCl +

ZnCl2 catalyst; or

• mix with PCl5 ; or

• reflux with SOCl2• reflux with NaBr + conc. H2SO4; or• reflux with red P + Br2

• re

flux

with N

aI +

co

nc. H 3P

O 4; or

• re

flux

with re

d P

+ I 2

CH3COOH +conc. H2SO4

• K2Cr

2O7 / H

3O +, reflux; or

• KMnO

4 / H3O +, reflux

1 LiAlH4 / ethoxyethane

2 H3O +

K 2Cr 2O

7/ H

3O+

disti

l off

the p

ropa

nal

1 Li

AlH4 /

ethox

yeth

ane

2 H 3O

+

• excess conc. H2SO4, 180 °C; or • Al2O3, 300 °C

K2Cr2O7 /H3O

+, heat

CH3CH(OH)CH3

propan-2-ol

CH3COCH3

propanone

(2° alcohol)1 LiAlH4 / ethoxyethane2 H3O

+

K2Cr2O7 / H3O+

reflux

(CH3)3COH

methylpropan-2-olno reaction

(3° alcohol)

K2Cr2O7 / H3O+

reflux


ExampleDescribe, by giving reagent(s) and stating observations, how you could distinguish between the following two compounds using a simple test tube reaction.

(2 marks)

OH

compound X

andCH3

CH2OH

compound Y

Answer

Warm each compound with acidified potassium dichromate solution. (1)

Only compound Y turns the dichromate solution from orange to green. (1)

➤ The oxidation reaction is commonly used to distinguish

– a primary / secondary alcohol from a tertiary alcohol;

– an aldehyde from a ketone.

RemarksRemarks*

Test Compound

Alcohols Aldehydes Ketones

Warm with K2Cr2O7 / H3O

+

1° and 2° alcohols — clear orange solution turns green almost immediately

clear orange solution

turns green

no observable

change

3° alcohols — no observable change

Exam tipsExam tipsExam tipsExam tips ♦ The formation of hydrogen chloride fumes upon reaction with phosphorus pentachloride is a test for the presence of a hydroxyl group in a compound.

♦ The relative ease with which alcohols undergo dehydration shows the following order:

Ease of dehydration: 3° > 2° > 1° alcohol

♦ Questions often ask about the oxidation of alcohols:

– the oxidizing agent required;

– the name(s) of the product(s).

♦ Remember that LiAlH4 will NOT act upon C=C bonds.

♦ The transformation of –COOH to –CH2OH involves 2 steps: reduction by LiAlH4, followed by treatment with H3O

+.

31.10 – 31.12

Summary

CH3CH2OH + conc. H2SO4 CH3CH2COOH

propanoic acid

CH3CH2COO– NH4+

ammonium propanoate

CH3CH2CONH2

propanamide

CH3CH2CH2OH

propan-1-ol

aqueousNH3

heat

1 LiAlH4 / ethoxyethane2 H3O

+

CH3CH2COO– Na+

sodium propanoate

CH3CH2COOH

propanoic acid

CH3CH2COO– Na+

sodium propanoate

CH3CH2COOH

propanoic acid

++

CH3CH2OH

ethanol

CH3CH2OH

ethanol

NaOHsolution

H2O /H3O

+

NaOHsolution

H2O /H3O

+

CH3CH2COOCH2CH3

ethyl propanoate

Exam tipsExam tipsExam tipsExam tips ♦ When propan-1-ol and propanoic acid are heated under reflux to produce an ester, the ester can be separated from the reaction mixture by fractional distillation or using a separating funnel.

♦ Given the structure of an ester, students should be able to analyze the ester and deduce the alcohol and carboxylic acid forming the ester.

e.g.

The active ingredient of a superglue has the following structure:

H

H CN

C C

C

O

OCH3

It is an ester formed from

H

H CN

C

COOH

C and CH3OH.

Topic 826 Chemistry of Carbon Compounds 27Unit 32 Synthesis of carbon compounds

ExampleOil of wintergreen is a common ester. Salicylic acid can be obtained from it in two steps.

oil of wintergreen

Step 1a sodium salt of

salicylic acid+

CH3OH

COOCH3

OH

salicylic acid

COOH

OH

Step 2

a) Give the reagent and condition used in Step 1. (2 marks)

b) Suggest a structure for the sodium salt of salicylic acid formed in Step 1. (1 mark)

c) Suggest a reagent that can be used in Step 2. (1 mark)

Answer

a) Sodium hydroxide solution (1)

Heat under reflux (1)

b)

(1)

COO– Na+

OH

c) Dilute sulphuric acid / dilute hydrochloric acid (1)

♦ Questions often give carbon compounds with similar structures (such

as

O

H C O CH3 and OHCH3

O

C ) and compare them.

– Whether they have the same molecular formula / relative molecular mass.

– Whether they are soluble in water.

– Whether they have the same odour.

– Whether they have the same boiling point.

– Whether they have the same chemical properties.

To tackle these questions, first identify the homologous series to which each compound belongs. Then answer according to the general properties of the series concerned.

➤ To draw the structure of sodium salt of salicyclic acid in (b), just use the general equation for the hydrolysis of an ester for deduction.

O– Na+

O

R C O NaOH+ R1 O HR1

O

R C +

➤ Students may also work backwards from the structure of salicyclic acid to deduce the structure of its sodium salt.

RemarksRemarks*

32.1 Introduction

32.2 Planning a synthesis

32.3 Problems in devising a synthesis

32.4 Laboratory preparation of simple carbon compounds

32.5 Preparing 1-bromobutane in the laboratory

Synthesis of carbon compoundsUnit 32


c)OH–(aq)

heat

(1) (0.5)

Cl OH

conc. H2SO4

heat

or conc. H3PO4

heat(1) (0.5)

Br2 (in organic

solvent)

(1)

Br

Br

➤ In part (a), the immediate precursor of the target molecule (an alcohol) may be a carboxylic acid or a haloalkane. As a carboxylic acid can be obtained from the starting molecule (an amide) via hydrolysis, so the synthesis can be done in the two steps shown.

➤ Many conversions involve

– the hydrolysis of haloalkanes to alcohols;

– the oxidation of alcohols to ketones and carboxylic acids;

– the reduction of aldehydes, ketones and carboxylic acids to alcohols.

RemarksRemarks*

32.4 – 32.5

SummaryThe laboratory preparation of a carbon compound involves five main stages:

Stage 1 — planning the preparation;

Stage 2 — carrying out the reaction to produce the desired product;

Stage 3 — separating the crude product from the reaction mixture;

Stage 4 — purifying and drying the product;

Stage 5 — calculating the percentage yield of the product.

Common separation and purification methods for products

Type of product Separation and purification method to employ

Liquid product• simple distillation• fractional distillation• liquid-liquid extraction

Solid product • re-crystallization

32.1 – 32.3

SummarySteps for devising a synthesis:

a) identify an immediate precursor to the target molecule;

b) continue until the starting molecule is reached.

target molecule 1st precursor 2nd precurso starting molecule

Exam tipsExam tipsExam tipsExam tips ♦ The syntheses discussed would NOT involve a change in the length of the carbon chain.

♦ Students should be familiar with the reactions of organic functional groups in order to suggest workable synthetic routes for the transformations.

ExampleOutline a syntheticc route, in NOT MORE THAN THREE STEPS, to accomplish each of the following conversions. For each step, give the reagent(s), the conditions and the structure of the product.

a) CH3CH2CH2CONH2 CH3CH2CH2CH2OH (3 marks)

b)

(4 marks)

CH2CH3 COCH3

c)

(4 marks)

Cl

Br

Br

Answer

a) CH3CH2CH2CONH2 H2O / H3O

+

heat (1) CH3CH2CH2COOH (1)

1 LiAlH4 / ethoxyethane2 H3O

+ (1) CH3CH2CH2CH2OH

b) CH2CH3

(1) (0.5)

CHBrCH3

Br2

UV light or heat

(1) (0.5)

CH(OH)CH3

OH–(aq)

heat

(1)

COCH3

K2Cr2O7 / H3O+

reflux


Example2-chloro-2-methylpropane can be prepared by reacting methylpropan-2-ol with concentrated hydrochloric acid.

(CH3)3COH(l) + HCl(aq) (CH3)3CCl(l) + H2O(l)

This preparation follows the steps outlined below:

Step 1 Shake excess concentrated hydrochloric acid with methylpropan-2-ol in a separating funnel.

Step 2 Separate the organic layer from the aqueous layer.

Step 3 Wash the crude product with 10% sodium hydrogencarbonate solution.

Step 4 Dry the product with a suitable reagent.

Step 5 Purify the dried product to remove the remaining methylpropan-2-ol.

The table below lists some information of methylpropan-2-ol and 2-chloro-2-methylpropane:

Compound methylpropan-2-ol 2-chloro-2-methylpropane

Density (g cm–3) 0.78 0.84

Boiling point (°C) 82 51

Water solubility miscible very slightly soluble

a) Draw a diagram of the separating funnel after the reaction has taken place in Step 1, labelling clearly the aqueous layer and organic layer. (2 marks)

b) Suggest ONE advantage of using a separating funnel to carry out Step 1. (1 mark)

c) Outline the experimental procedure for washing the crude product in Step 3. Include the necessary safety precautions. (4 marks)

d) Suggest a suitable reagent for drying in Step 4. (1 mark)

e) Name a method for purifying the dried product in Step 5. (1 mark)

Exam tipsExam tipsExam tipsExam tips ♦ DO NOT confuse an experimental set-up for fractional distillation with that for simple distillation.

♦ Questions often ask students to describe the re-crystallization procedure for the purification of a crude solid product.

Answer

a)

(2)

aqueous layer

organic layer

b) Any one of the following:

• Allow better mixing of the reactants by vigorous shaking of the separating funnel. (1)

• Allow easy isolation of the product by draining out the lower aqueous layer from the separating funnel. (1)

c) Add slowly 10% sodium hydrogencarbonate solution. Wait until no effervescence occurs. (1)

Stopper and shake the separating funnel vigorously. (1)

During shaking, open the tap of the separating funnel regularly to release the pressure built inside the funnel. (1)

Remove the aqueous layer. Repeat the washing procedure until no effervescence occurs upon the addition of sodium hydrogencarbonate solution. (1)

d) Anhydrous calcium chloride / sodium sulphate (1)

e) Simple distillation / fractional distillation (1)

➤ Questions often ask about the purposes of different steps in the preparations of carbon compounds.

➤ When giving a reagent for drying, remember to include the word ‘anhydrous’.

RemarksRemarks*

Topic 832 Chemistry of Carbon Compounds 33Unit 33 Important organic substances

33.1 – 33.9 & 33.13

Summary1 a) The structure of the active ingredient of aspirin tablets, acetylsalicylic acid, is

shown below. It contains two functional groups.

CH3O C

O

OHC

Ocarboxyl group

ester group

b) The percentage by mass of acetylsalicylic acid in aspirin tablets can be determined by back titration.

2 a) There are two types of detergents:

i) soap — made from natural fats and oils; and

ii) soapless detergents — made from chemicals derived from petroleum.

b) The structure of a typical anionic detergent is shown below:

hydrophobichydrocarbon ‘tail’

hydrophilic anionic ‘head’

c) A detergent helps water to remove dirt by

i) the ability to act as a wetting agent; and

ii) the emulsifying action.

3 Fats and oils are mixed triglycerides.

4 Soap is made by boiling natural oils or fats with a strong alkali, usually sodium hydroxide or potassium hydroxide solution.

33.1 Introduction

33.2 Aspirin — from herbal remedy to modern drug

33.3 Analyzing aspirin tablets by back titration

33.4 Detergents

33.5 How do detergents help water to clean?

33.6 The wetting and emulsifying properties of detergents in relation to their structures

33.7 The cleaning action of detergents

33.8 Manufacture of soaps and soapless detergents

33.9 The cleaning abilities of soaps and soapless detergents in hard water

33.10 Nylons

33.11 Polyesters

33.12 Carbohydrates

33.13 Lipids

33.14 Proteins and polypeptides

Important organic substancesUnit 33


ExampleThe structure of the main chemical constituent of an animal fat is shown below:

(CH2)nCH3

(CH2)nCH3

(CH2)nCH3

O C

O

O C

O

O C

O

C

C

H C

H

H

H

H

Exam tipsExam tipsExam tipsExam tips ♦ Students should be able to draw the correct structure of the ester linkage in a triglyceride (see the structure shown below). The three carboxylic acids bearing long chains are attached to a glycerol backbone.

♦ When fats and oils are heated with sodium hydroxide solution, they are hydrolyzed first to form glycerol and carboxylic acids. The acids then react with the alkali to form sodium carboxylates, which are soaps. Such reactions are called saponification.

♦ Questions often give the structure of a soap / soapless detergent and ask about its properties.

– Whether it is made from natural fats and oil or from hydrocarbons obtained from petroleum.

– Whether it forms scum in hard water / sea water.

– Whether it is biodegradable.

♦ Hydrogenation of vegetable oils produces margarines.

R

R

R

OH

OH

OH

O C

O

O C

O

O C

O

C

C

H C

H

H

H

H

3NaOH+

triglyceride in fat or oil

C

C

H C

H

H

H

H

+

glycerol

alkali

3RCOO– Na+

sodiumcarboxylate

(soap)

a) Animal fat is one of the raw materials in the production of detergent X.

i) Name another raw material that is needed in the production of detergent X from an animal fat. (1 mark)

ii) Name the type of reaction that takes place in the production of detergent X.(1 mark)

iii) Write a chemical equation for the reaction involved in the preparation. (1 mark)

b) An oil tanker was wrecked and spilt a lot of crude oil in the sea. State whether detergent X is suitable for treating the oil spill. Explain your answer. (3 marks)

Answer

a) i) Sodium hydroxide (1)

ii) Saponification / alkaline hydrolysis (1)

iii)

(1)

(CH2)nCH3

(CH2)nCH3

(CH2)nCH3

O C

O

O C

O

O C

O

C

C

H C

H

H

H

H

3NaOH+

C

C

H C

H

H

H

H

H

H

HO

O

O

+ 3CH3(CH2)nCOO– Na+

b) Not suitable (1)

Sea water contains a lot of metal ions, such as calcium ions and magnesium ions.(1)

Detergent X will react with these metal ions to form scum. (1)


33.10 – 33.12 & 33.14

Summary1 a) Nylons are polyamides which contain the amide linkage N

H

C

O

.

b) Nylons are formed by condensation polymerization.

c) The repeating unit of nylon-6,6 is

from diamine

H

N

O

C (CH2)4 C

O

H

N (CH2)6

from dicarboxylic acid

2 a) Polyesters contain the ester linkage

O

C O .

b) Polyesters are formed by condensation polymerization.

c) The repeating unit of poly(ethylene terephthalate) is

from dicarboxylic acid from diol

O

C

O

C O CH2 OCH2

3 a) Glucose can exist in open-chain and cyclic forms.

b) The open-chain form of glucose contains an aldehyde group and five hydroxyl groups.

4 a) Fructose can exist in open-chain and cyclic forms.

b) The open-chain form of fructose contains a keto group and five hydroxyl groups.

5 a) When two amino acid molecules undergo condensation reaction, a water molecule is eliminated and a peptide link forms. The product is a dipeptide.

+H N

R

HH

C

O

C OH

H N

R

HH

C

O

C

H

+ H2O

N

R1

HH

C

O

C OH

N

R1

HH

C

O

C OH

amino acid 1 amino acid 2

peptide link (or amide group)

a dipeptide

b) The peptide links in peptides and proteins can be hydrolyzed to release the individual amino acids.

Exam tipsExam tipsExam tipsExam tips ♦ A condensation reaction is a reaction in which two or more molecules react together to form a larger molecule with the elimination of a small molecule such as water.

♦ Nylon and poly(ethylene terephthalate) are thermoplastics as well as condensation polymers. NOT all thermoplastics are addition polymers.

♦ Questions often give the repeating unit of a polymer and ask about information concerning the polymer.

e.g.

– whether it is an addition polymer or a condensation polymer;

– whether it is formed from one monomer or two different monomers.

➤ Questions often ask about the purpose of adding concentrated sodium chloride solution in soap preparation.

Concentrated sodium chloride solution is added to salt out the soap produced.

➤ Questions often ask students to give the products formed from the complete hydrolysis of triglycerides existing in natural fats and oils.

e.g.

OH

OH

OH

CH2

CH

CH2 (CH2)16CH3

(CH2)7CH

(CH2)16CH3

CH(CH2)7CH3

O C

O

O C

O

O C

OCH2

CH

CH2

+ CH3(CH2)16COOH +

+

complete

hydrolysis

CH3(CH2)7

H

H

(CH2)7COOHC C

CH3(CH2)7

H

(CH2)7COOH

HC C

RemarksRemarks*


ExampleTwaron is a heat resistant, high strength fibre used in protective clothing. A short section of the structure of Twaron is shown below.

O

CC

H

N

O N

N

O

CC

H

N

O N

N

O

CC

O

a) Draw the repeating unit of Twaron. (1 mark)

b) Draw the structures of the two monomers that could be used to prepare Twaron. (2 marks)

c) Explain why Twaron has a great strength. (1 mark)

Answer

a)

(1)H

N

N

N

O

CC

O

b) NH2H2N (1)

COOHHOOC (1)

c) There are strong hydrogen bonds between chains of Twaron. (1)

➤ Given the structure of a polymer, students should be able to deduce the structure(s) of monomer(s) used to produce the polymer.

Look at the polymer structure.

– If the polymer backbone contains only carbon atoms, then the polymer is an addition polymer produced from one monomer.

e.g.

C6H5

C

H

H

C

H

C6H5

C

H

H

C

H

C6H5

C

H

H

C

H

is an addition polymer formed

from the following monomer:

C6H5

C

H

H

C

H

– If the polymer backbone contains other atoms, then the polymer is a condensation polymer, probably produced from two monomers (see Twaron shown above).

➤ Some condensation polymers are produced from one monomer.

e.g.

the repeating unit of polylactide is shown below:

CH3

O C

H

C

O

It is a condensation polymer formed from one monomer, lactic acid.

HO OH

lactic acid

CH3

C

H

C

O

RemarksRemarks*

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