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Topic 20 Organic Chemistry Answers

Student note:

In the suggested answers that follow some formulas are given as a line-bond formula. These show

only the bonds between the carbon atoms, not the hydrogen atoms. Where these lines meet, and at

the ends of these lines, there is a C atom. The H required for the bonds present is assumed,

e.g. C4H10 is shown as , which equates to 4C plus the tetrahedrally arranged 10 H.

The functional groups present are shown with the atoms, e.g. C4H9OH is shown as: OH

Other formulas are given in their condensed form, e.g. C4H10 is CH3CH2CH2CH3.

An expanded structural formula shows all the bonds and all the atoms present, e.g. C4H10 is written

as:

CC

CCH

H

H

H H

H H

H

H

H

Here are some more examples:

C5H12

butan-2-one

O

propan-2-ol

OH

1-bromopropane Br

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20.1 Exercises

1. Give the structural formula for the following functional groups: (a) Amine

H N R

H

H N R

R

R N R

R

Primary Secondary Tertiary (b) Amide

CNR

O

R

R

(c) Ester

CR

O

OR

(d) Nitrile

C NR

2. The amines have three types – primary, secondary and tertiary. Name the type and write a structural formula for the following amines: (a) Methylamine CH3NH2, primary

H3C N H

H (b) Ethylamine C2H5NH2, primary

CH2 N H

H

H3C

3

(c) Di-ethylamine (C2H5)2NH, secondary

CH2 N CH2

H

H3C CH3

(d) Tri-methylamine

(CH3)3N, tertiary

CH3 N CH3

CH3

3. What does the italic letter N imply when used in naming amines and amides? Used for secondary and tertiary amines. It is used to indicate that the second and/or third

substituent is attached to the nitrogen atom. For example in N,N-dimethylbutylamine the two

methyl groups are attached to the nitrogen.

N

4. Write structural formula for the following compounds: (a) N,N-dimethylethanamine

NCH2

H3CCH3

CH3

(b) dipropylamine

HN

CH2

H2C

H3C CH2

H2C

CH3

(c) 1,3-diaminobutane

H2C

CH2

CH

CH3

NH2H2N

4

(d) phenylamine

NH2

(e) N-methylethylamine

NH

H3CH2C

CH3

(f) 1,1-dichloro-4,4-dimethyl-1-pentanamine

C

H2C

CH2

C

H3C

NH2

CH3

CH3Cl

Cl

(g) 5-aminohexan-1-ol

H2C

CH2

H2C

CH2

CH

CH3

HO NH2

(h) 2-methylpropanamine

H2NCH2

CH

CH3

CH3

(i) N-methylethanamine

NH

H2C

CH3

H3C

5. The smell associated with decaying flesh is due to the formation of the diamines, 1,4-butanediamine (common name putrescine) and 1,5-pentanediamine (common name cadaverine). Draw structures for:

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(a) 1,4-butanediamine

H2C

CH2

H2C

CH2

H2NNH2

(b) 1,5-pentanediamine

H2C

CH2

H2C

CH2

H2C

H2N NH2

6. By applying IUPAC rules, name the following amines and label them as primary, secondary or tertiary: (a) CH3(CH2)2NH2 Primary amine, 1-aminopropane or 1-propanamine or 1-propylamine

(b) CH3NHCH3

Secondary amine, N-methylmethanamine or di-methylamine

(c) CH3NHCH2(CH2)2CH3

Secondary amine, N-methylbutanamine

(d) HOCH2CH2NH2 (an ethanolamine used as an emulsifying agent)

Primary amine, 1-hydroxyethanamine

(e) C6H5NH2 (trivial name analine)

Primary amine, phenylamine

(f) (CH3CH2)2NH

Secondary amine, diethylamine

(g) (CH3CH2)3N Tertiary amine, triethylamine

(h)

N

Tertiary amine, N,N-ethyl-methyl-propanamine

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7. Deduce structural formula for the following amides: (a) Ethanamide

C

H3C

O

NH2

(b) A repeating unit in a polyamide

C

O

NH

n (c) 2-methylpropanamide

C

CH

CH3

O

NH2

H3C

(d) N-methylbutanamide

C

H2C CH2

CH3

O

HN

H3C (e) N,N-diethylmethanamide

CH

O

N

H2C CH3

H2C

H3C

8. By applying IUPAC rules, name the following amides: (a) Ethanamide (b) N,N-dimethylpropanamide

(c) heptanamide

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(d) N,N-dimethyl-2-methylpropanamide

(e) 3-methylbutanamide

9. The legal, but very addictive, health damaging drug, nicotine is an amide. Use the IB data book to name another amide that is a drug, or is used in medicine.

Examples illustrated in the IB data book include lidocaine and acyclovir.

Chilli peppers belong to the Capsicum family of plants. They contain amides and are extensively

used in preparing ‘hot’ tasting foods.

10. Deduce structural formula for the following esters: (a) Methyl butanoate

C

H2C CH2

CH3

O

O

H3C (b) Ethyl propanoate

CH2C

H3C

O

O CH2

CH3

(c) 2-propyl ethanoate

CCH3

O

OH3CCH

CH3

(d) 1-hexyl ethanoate

CH3C O

OCH2

H2C

CH2

H2C

CH2

H3C

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11. By applying IUPAC rules, name the following esters: (a) ethylbutanoate

(b) l-pentylethanoate

(c) l-propylethanoate

(d) 3-methylbutanepropanoate

12. Deduce structural formulas for the following nitriles: (a) Butanenitrile

C CH2

H2C CH3

N (b) Propenenitrile

C CH

CH2

N (c) 2-methylpropanenitrile

C

CH CH3

N

H3C (d) 3-hydroxybutanenitrile

CCH2

CH

CH3

N

OH

13. By applying IUPAC rules, name the following: (a) pentanenitrile

(b) pentanamine

(c) N-ethylmethylamide

(d) ethylbutanoate

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20.2 Exercises 1. Define, with an example, the following terms; (a) Halogenoalkane Compounds where one or more halogen atoms replace one or more hydrogen atoms in alkanes,

e.g. C2H5Cl, chloroethane.

(b) Nucleophile Molecules or ions that can form a covalent bond with another molecule because they have a lone

pair of electrons, e.g. NH3. They are electron rich species which are attracted to partial positive

charges, δ+, i.e. nuclei. Nucleophile means nucleus-loving.

(c) Substitution reactions Reactions where atom(s) in molecules are replaced by other atoms or molecules, e.g. in this

reaction the OH group is replaced by a Br:

C3H7OH + HBr → C3H7Br + H2O

(d) Nucleophilic substitution Is a reaction where a nucleophile, e.g. OH‾, replaces another atom or group of atoms, e.g. a Br in

an alkane.

C3H7Br + OH‾ → C3H7OH + Br‾

(e) Leaving group That which leaves a compound during substitution, e.g. Br is the leaving group in (d) above.

(f) Best leaving group If two leaving groups could be involved, e.g. Cl or I from a compound, the more favoured is labelled

the best, i.e. iodine. The C—Cl bond is shorter than the C—I bond, hence stronger.

(g) Worst leaving group Fluorine is a bad leaving group because of its strong covalent bonding to a C atom (smaller atom,

therefore bond distance shorter, therefore stronger bond).

(h) Substrate The molecule onto which substitution occurs, e.g. C3H7Br is the substrate when reacting with OH‾.

The term substrate is more common in biochemistry.

Substrates are the molecules on which enzymes react to catalyse the reaction.

Enzymes are remarkably effective in catalysing bio-chemical reactions.

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2. Some common nucleophiles are listed. Write Lewis structures for each. (a) hydroxide ion (b) water molecule

HO

HO

H (c) cyanide ion (d) ammonia molecule

C N

NH H

H

(e) bromide ion (g) ethoxide ion

Br

O

As their names show nucleophiles are involved in both nucleophilic substitution and nucleophilic

addition reactions.

3. Supply the missing word. All of the above nucleophiles have a common property; they are Lewis bases.

4. Both OH‾ and H2O are nucleophiles. Which of these is the better nucleophile? Give the reasons for your choice. OH‾ is an ion and has the stronger overall negative charge; it has three lone electron pairs; water,

a polar molecule, only two. Therefore, OH‾ is the better nucleophile.

5. The factors affecting the rate of nucleophilic substitution are: the nucleophile involved, the halogen present and the type of halogenoalkane reacting. (a) Explain how the rate of nucleophilic substitution is dependent on which nucleophile is used.

Rate is dependent on the strength of the negative charge on the nucleophile e.g. OH ‾, is a faster

reactant than a polar molecule, e.g. H2O or NH3. This is because the strength of electrostatic

attraction depends on the magnitude of the charge.

(b) Explain how the rate of nucleophilic substitution in the halogenoalkanes is dependent on which halogen is present. The rate is dependent on the ease of breaking the C—X bond. The strongest bond is C—F

(because the F atom is small). The weakest C—I (because the I is the largest halogen atom). The

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bond distance in C—I is the largest, making it the least strong. Therefore, the reaction is the

fastest.

(c) Explain how the rate of nucleophilic substitution in the halogenoalkanes is dependent on the type of halogenoalkane reacting, i.e. primary, secondary or tertiary. The rate is determined by the mechanism. Primary halogenoalkanes proceed by a SN2 mechanism

which is slower than the SN1 mechanism. Tertiary halogenoalkanes proceed by SN1 mechanism.

The activation energy hurdles are lower for SN1 than they are for SN2 m echanisms.

6. (a) Complete these equations and label the nucleophile, the substrate and the leaving group. substrate leaving group nucleophile

(i) CH3C(CH3)HCH2 ―I + OH‾ CH3C(CH3)HCH2OH + I‾

(ii) CH3CH2C(Cl)HCH3 + NH3 CH3CH2(NH2)HCH3 + HCl

Substrate with Cl the leaving group nucleophile

(b) Rewrite the above completed equations using structural formulas, adding electron lone pair electrons where appropriate, δ+ and δ- symbols to show bond polarity and curly arrows in order to show the mechanism. (i)

(ii)

7. Write balanced equations to show the formation of (a) An alcohol from 2-bromo-2-methylbutane CH3C(Br)(CH3)CH2CH3 + OHˉ → CH3C(OH)(CH3)CH2CH3 + Brˉ

(b) pentan-1-ol from an iodoalkane CH3CH2CH2CH2CH2—I + NaOH → CH3CH2CH2CH2CH2—OH + NaI

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If a halogen atom is directly attached to an aromatic ring, such as in bromobenzene, nucleophilic

substitution will not occur unless extreme conditions are used.

8. Draw a structural formula for each of the following and then classify as a primary, secondary or tertiary halogenoalkane. (a) Ethyl bromide

H2C

CH3Br primary halogenoalkane.

(b) 1-chloro-2-methylpropane

CH2

CH

CH3

CH3

Cl

primary halogenoalkane.

(c) 1-chloro-2-methylbutane

H2C CH2 CH2 CH3

Cl

CH3 primary halogenoalkane.

(e) 2-bromo-2-methylbutane

CH3C

H2CCH3

CH3

Br

tertiary halogenoalkane.

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(f) 3-iodo-2,3-dimethylpentane CH3

CH

C CH2

CH3CH3

H3C

I

tertiary halogenoalkane.

(g) 2-bromopentane

CH3

CHCH2

H2C

H3C

Br

secondary halogenoalkane.

9. Consider the following equation: OH- + (CH3)3C—Br (CH3)3C—OH + Br- (a) Identify the nucleophile OHˉ, the substrate (CH3)3C-Br, and the leaving group Brˉ.

(b) Rewrite the above equation, using structural formulas. Now add electron lone pairs, δ+ and δ- to show bond polarity and curly arrows to the equation to show the mechanism.

(c) Is this reaction an example of heterolysis or homolysis? No free radicals formed, no breaking of a bond into two equal parts, therefore heterolysis.

10. Explain what is meant by SN1 and SN2 mechanisms. SN1 (substitution nucleophilic uni-molecular) is a two step mechanism which involves only one

reacting species in the rate determining step. SN2 (substitution nucleophilic bimolecular) involves

two reactants in it rate determining step.

The symbols SN1 and SN2 represent Substitution Nucleophilic 1st order and 2nd order reactions

respectively. Order describes the number of reactants involved in determining the rate of the

reaction.

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11. Substituting reactions (a) Describe, with equations and curly arrows, the substitution reaction between a primary bromoalkane and ammonia.

(b) Describe, with equations and curly arrows, the substitution reaction between a primary bromoalkane and potassium cyanide.

12. Consider the reactions given in question 11. Explain these reactions in terms of the SN2 mechanism. The nucleophile (NH3 or CNˉ) is attracted to the carbon atom and begins to bond to it as the Br

atom begins to leave.

This situation is the transition state. Because experiment shows rate is determined by the

concentration of both the nucleophile, e.g. [CN–] and the halogenoalkane,[R—Br], it is a

bimolecular mechanism, hence SN2.

13. The nitriles are important intermediates in organic syntheses. They are readily hydrolysed and can be reduced using hydrogen and a catalyst, usually nickel. Using equations, describe the reduction of nitriles using hydrogen. Using propanenitrile and a nickel catalyst the reaction is:

Ni catalyst / pressure

CH3CH2CN + 2H2(g) → CH3CH2CH2NH2

propanamine

14. Consider the following reactions: A. CH3CH2Br + H2O CH3CH2OH + HBr B. CH3CH2Br + OH- CH3CH2OH + Br‾

(a) Rewrite these two equations, using full structural formulas.

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(b) Place curly arrows, δ charges and lone pair electrons in the equations to show the mechanisms in the above reactions. (c) Identify each as SN1 or SN2 mechanisms. Give your reason. Both A and B are SN2 as both are primary halogenoalkanes.

(d) Circle the carbocation in A. No carbocation formed since it is an SN2 mechanism!

(e) What is the hybridization of the central C atom in A?

sp3 because the bonding is tetrahedral.

(f) Name the products formed in the reactions above. In A ethanol and hydrogen bromide; in B ethanol and the bromide ion.

(g) Why does the nucleophile in B attack the substrate on the opposite side to the leaving group? Because the δ– on the Br repels the nucleophile into the opposite position making this the only

place from where the nucleophile can attack.

(g) Why is B described as a one-step or concerted process? Because the leaving group leaves as the nucleophile takes its place.

15. Refer to the energy diagrams shown below.

(a) Mark on each the ∆H of reaction and activation energy, EA.

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(b) Classify each reaction as either SN1 or SN2 and explain your choice. The first shows an activated complex the (transition state) at the EA hurdle and is characteristic for

primary haloalkanes, e.g. OHˉ + C3H7Br, therefore SN2 mechanism.

The second shows the formation of an intermediate (at dip after the top for EA) and this suggests

the SN1 mechanism.

(c) What is the difference between an intermediate in a reaction and a transition state (also called an activated complex)? Transition states cannot be detected, it is a theory; intermediates (e.g. free radicals) can be

detected.

(d) Mark any transition states and/or intermediates on the diagrams above.

Discuss with your teacher whether or not this question is appropriate as a knowledge extension

question.

16. (a) Write an equation, using structural formulas, to show the transition state (or the intermediate) in the reaction between sodium hydroxide solution and 1-bromobutane.

H2C

CH2

H2C

H3C Br

OH

H2C

CH2

H2C

H3C OH + BrH3C CH2 CH2 CH

H

Br

OH

(b) Now put in curly arrows.

(c) Which would act the fastest: 1-iodo, 1-bromo or 1-chlorobutane? Why? 1-iodo because the I—C bond, being the longest (I largest atom) makes this bond the weakest,

hence the reaction is the fastest.

(d) Label the functional group carbons in the substrate, product and transition state/intermediate as sp, sp2, or sp3 hybridised. Substrate and product C has 4 covalent bonds, therefore sp3 hybridisation. Transition state 3

covalent + 2 ½ bonds = 4 bonds, therefore sp3.

The combination of reacting atoms, molecules or ions in the transition state is also called the

activated complex.

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17. For the equation: Nu- + R—X R—Nu + X‾ where Nu‾ is a nucleophile, (a) Write a rate equation for an SN1 pathway Rate = k[R – X]

(b) and an SN2 pathway. Rate = k[Nuˉ][R – X]

18. Which of the following is NOT a good nucleophile? A. OH‾ B. H‾ C. NO2

+ D. NH3‾ E. CN‾

Option C is correct, because of its + charge the NO2+ is not a nucleophile.

19. When 1-chloropentane is refluxed with KOH(aq) a nucleophilic substitution reaction occurs. (a) Write a balanced equation for this reaction and name the products. C5H11Cl + KOH(aq) C4H9CH2OH + KCl(aq)

(b) Explain why this reaction is described as nucleophilic substitution. The reaction mechanism involves the OHˉ seeking out the δ+ carbon attached to the

δ– Cl.

(c) Draw a structure for the attacking nucleophile and show the charge and the lone pair electrons.

OH (d) The reverse reaction is possible. Name the nucleophile in this reaction and give its structure.

Bromide ion from hydrobromic acid, HBr (aq). Structure Br

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(e) Sketch and then label the apparatus used for reflux.

A condenser used to condense and so run back the reactants and products into the

reacting flask is called a reflux condenser. Reflux means to flow or run back.

20. Nitriles and synthesis (a) Why are nitriles useful compounds for an organic synthesis? They add a carbon atom to a molecule which is useful to build up a chain or to react to give amines

or acids.

(b) Nitriles can be reduced to amines. Using butanenitrile as your example, give an equation for the catalytic reduction of a nitrile using hydrogen. Name the catalyst.

Nickel is the catalyst. C3H7CN + 2H2(g) → C3H7CH2NH2

(c) Hydrolysis of a nitrile produces a carboxylic acid. Give an equation for the reaction of propanenitrile when refluxed with an aqueous acid. C2H5CN + H+

(aq) + 2H2O → C2H5COOH + NH4+

(aq)

(d) Why is reflux required for reaction (c) above? Volatile reactants can escape from a reaction mixture when heated and heating is required in the

breaking of the covalent bonds in order to form new bonds. Covalent bonds are strong.

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20.3 Exercises 1. What is the difference between a condensation and elimination reaction, when both can split off water molecules? Both involve the splitting off a small molecule, but condensation gives a joining of molecules;

elimination gives a double bond.

2. Write an equation to show the elimination of HBr from 2-bromobutane. CH3CH(Br)CH2CH3 + OH–(al) → CH3CH=CHCH3 + H2O + Br–

Note: alcoholic NaOH is used. The OH–(al) in ethanol is reacting as a Lewis base. This must be

shown in the equation. Aqueous NaOH gives OH–(aq) which acts as a nucleophile.

3. Describe, and explain the mechanism, for the elimination of HBr from 2-bromopropane. The atoms removed are H and Br; their elimination producing propene. The mechanism involves

attack by a strongly basic nucleophile, e.g. :OH–, in alcoholic solution, with EN2 mechanism

favoured for primary and secondary halogenoalkanes and EN1 favoured for tertiary

halogenoalkanes.

The mechanism shown is EN2 because the rate determining step involves two reactants: [:OH–]

and [C3H7Br].

An EN1 mechanism involves only the one reactant, e.g. rate = k[C3H7Br], and the formation of an

intermediate carbocation.

4. Why are elimination reactions so useful in synthesis? It is useful because it gives a reactive double bond into a molecule and double bonds are reactive.

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5. The conditions present will determine main product from the reaction between a halogenoalkane and an alkali, e.g. NaOH. (a) What conditions are needed for an elimination reaction? The alkali is in an ethanol solution, e.g. NaOH(al), and reflux is required.

Nucleophilic substitution, not elimination, occurs with aqueous alkali, NaOH(aq).

(b) Complete this sentence. Elimination is more likely with a secondary or tertiary halogenoalkane.

Elimination is favoured when the halogen is in the middle of the C chain.

(c) Write the equation for NaOH(al) reacting with 2-bromobutane. CH3CH(Br)CH2CH3 + OHˉ(al) → CH2=CHCH2CH3 + H2O + Br ̄

(d) What conditions favour nucleophilic substitution? Using OHˉ(aq), refluxing, and a primary halogenoalkane.

(e) Write the equation NaOH(aq) reacting with 1-bromopropane.

C3H7Br + OHˉ(aq) → C3H7OH + Brˉ(aq)

(f) Outline the mechanism for the reaction in (e) above. The mechanism involves the C – Br bond breaking as the nucleophile forms a new bond with this

C. It is an SN2 mechanism because the rate determining step is

rate = k[OHˉ(aq)][ C3H7Br].

H3C

CH

H3C

BrHO

H3C

CH

H3C

OH + Br

6. Just as nucleophilic substitution reactions can be classified a SN1 or SN2 reaction, so elimination reactions can be classified as EN1 or EN2 (a) Explain, with an example, what is meant by a reaction having an EN1 mechanism. The EN1 mechanism is similar to SN1.

The rate is determined only by the concentration of the reactant. A carbocation, i.e. +C, forms as

an intermediate when the C – Br bond breaks.

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(b) Explain, with an example, what is meant by a reaction having an EN2 mechanism. This is a bimolecular reaction, i.e. rate determined by both the [OH ˉ] and the [halogenoalkane].

(c) In the worked example 2 above is the mechanism suggested EN1 or EN2? Give reasons for your answer. EN2 because the mechanism involves both [OHˉ] and [2-bromopropane] in the rate determining

step. It is a bimolecular reaction, with the rate = k[OHˉ][2-bromopropane].

7. Altering the reaction conditions may cause the same reactant to produce different products. The reaction of a haloalkane with sodium hydroxide is an example. (a) A haloalkane reacting with dilute aqueous sodium hydroxide, NaOH (aq).

(i) Write an equation for the reaction of 2-bromopropane with dilute NaOH. C3H7Br + OHˉ(aq) C3H7OH + Br–

(ii) Using structural formula and curly arrows show the mechanism for the reaction.

H3C

CH

H3C

BrHO

H3C

CH

H3C

OH + Br

(iii) What is the name given to this type of reaction?

Substitution-nucleophilic-bimolecular, i.e. SN2 (iv) Complete this sentence:

In this reaction the OH– is acting as a nucleophile

(b) A haloalkane reacting with hot alcoholic sodium hydroxide under reflux. (i) Write an equation for the reaction of 2-bromopropane with hot NaOH(al).

CH3CH(Br)CH3 + OHˉ(al) H2C = CHCH3 + H2O + Br–

(ii) Using structural formula and curly arrows show the mechanism for the

reaction.

(iii) What is the name given to this type of reaction?

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Elimination, as H and Br have been eliminated from the molecule to leave a double bond.

(iv) Complete this sentence: In this reaction the OH– is acting as a base.

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20.4 Exercises

1. Define, with an example, these terms: (a) Carboxylic acid

Are organic compounds and weak acids with the general formula R – COOH, e.g. ethanoic acid,

CH3COOH.

(b) Ester

Are the organic smells and flavours in nature. Formed from the reaction between and acid and

alcohol and have the general formula RCOOR2, e.g. ethylethanoate CH3COOC2H5.

The ester functional group is a combination of the functional groups from an alcohol and a

carboxylic acid.

(c) Esterification The reaction between an organic acid and an alcohol, with a catalyst (concentrated H2SO4) to give

an ester, e.g.

C2H5OH + CH3COOH → CH3COOC2H5 + H2O

(d) Condensation reaction A reaction where molecules bond together by splitting off a small molecule, e.g. H2O or HBr.

Esterification is an example of a condensation reaction.

(e) Amide

Are nitrogen organic compounds containing the amide functional group, e.g. ethanamide

CH3CONH2.

Polyamides contain many amide groups. Proteins are natural polyamides.

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(f) Monomer The small molecule that is able to form with other molecules to form a polymer, e.g. ethene, is the

monomer for polyethene.

(g) Polymer Large (i.e. very long chained) molecules made by joining monomers, e.g. polyethene (CH2CH2)n

where n is large.

(h) Polyester Polymers made from ester monomers, e.g. PET, polyethyleneterephthalate, is a polyester. PET is

used to make bottles for soft drinks.

(i) Polyamide Polymers made by the linking of amide functional groups, e.g. the very strong kevlar (used in bullet

proof vests and in constructions) is a synthetic polyamide; nylon is another. Proteins are examples

of natural polyamides.

(k) Repeating unit

The basic unit in a polymer that repeats itself thorough out the polymer chain,

e.g. [–R1–NH–CO–R2–CO–NH–] is the repeating unit for polyamides. Nylon is a polyamide.

(l) Natural polymer Found in nature. Proteins are an example, cellulose another.

Some examples of proteins:

keratins – used to make hair, hooves, claws, feathers, beaks and fingernails, enzymes – nature’s so efficient biochemical catalysts,

hormones – the compounds that stimulate organs into action, fibres – e.g. wool and silk fibres, and elastin – the structural protein of skin.

(m) Synthetic polymer Man made polymer, e.g. the very useful nylons.

(n) Plastics A polymer that can be easily moulded into shape, e.g. polythene.

Thermoplastics become ‘plastic’ i.e. can be moulded when hot and harden on cooling.

Thermosetting plastics become hard when heated, because cross linking occurs. Once set they do

not melt.

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Esters are named from the alcohol and acid from which they are made. Thus an ester is given a

two word name: the first word describing the alcohol, the second the acid. However, in writing the

formulas it is usual give the acid component first, then the alcohol.

2. Some revision. Here are some esters: Methyl butanoate (apple flavour) Ethyl butanoate (pineapple flavour) Ethyl methanoate (raspberries) Ethyl ethanoate (nail polish remover)

(a) Give the names of the alcohols and the acids from which they are derived. Methanol and butanoic acid.

Ethanol and butanoic acid.

Ethanol and methanoic acid.

Ethanol and ethanoic acid.

(b) Write structural formulas for these alcohols, acids and esters.

OHMethanol

HOEthanol

O

OHbutanoic acid

OHOmethanoic acid

O

OH

ethanoic acid.

O

O

Methyl butanoate

O

OEthyl butanoate

O OEthyl methanoate

O

OEthyl ethanoate

(c) How are 3-methyl-1-butyl-ethanoate and 2-methyl-1-propyl-butanoate related?

They are isomers.

3. Physical properties and uses (a) Outline the physical properties associated with esters.

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Sweet smelling, slightly soluble, have distinct flavours. Fats are solid esters (triesters made from

triols and long chained saturated fatty acids) while vegetable oils are liquid esters (triesters made

from triols and long chained unsaturated fatty acids).

(b) Outline the uses of esters. In artificial flavours, making solvents (e.g. ethyl ethanoate is a solvent for nail polish), in making a

wide variety of polyesters used in clothing, and in pharmaceuticals.

4. Chemical reactions of esters (a) The main chemical reaction of esters is hydrolysis. This can be acid or alkaline hydrolysis. Give, with equations, an example of each. H+

CH3COOC2H5 + H2O CH3COOH + C2H5OH

OH–

CH3COOC2H5 + NaOH → CH3COONa + C2H5OH + H2O

(b) Why is alkaline hydrolysis the preferred method? The acid hydrolysis is reversible. The second produces a salt, e.g. CH3COONa, and gives a better

yield. The acid CH3COOH can be made by adding HCl(aq).

CH3COONa(aq) + HCl(aq) CH3COOH(aq) + NaCl(aq)

5. Ethanol and acetic acid were heated with a little concentrated sulfuric acid in the flask of the apparatus shown.

(a) Name the organic product expected from the reaction. Ethyl ethanoate

(b) Write the equation for the reaction. C2H5OH + CH3COOH → CH3COOC2H5 + H2O

(c) State the function of the sulfuric acid. Provides the catalyst (H+).

27

(d) Explain why the condenser is used as shown in the diagram. Heat is required for this reaction and the condenser prevents volatile reactants and products

escaping.

(e) Explain why the reactants must be heated for some time to obtain a satisfactory yield of the product.

Covalent bonds are strong are strong bonds. To break them requires considerable energy, hence

longer heating.

6. This question concerns ethyl butanoate. (a) Describe the expected odour of ethyl butanoate. Pineapple odour.

(b) Write an equation for its preparation. C3H7COOH + C2H5OH → C3H7COOC2H5 + H2O

(c) What is the catalyst used?

Sulfuric acid, H2SO4, concentrated.

(d) Compare the water solubility of the reactants with that of ethyl butanoate and hence explain which components of the reaction mixture are mainly in the ‘organic’ layer, and which are mainly in the aqueous layer, when isolating the ester from the reactions in the separating funnel. The acid and alcohol are water soluble; hydrogen bonding possible. Ethyl butanoate, a larger

organic molecule, is more soluble in organic layer.

7. Consider the instructions below (1-9) and then answer the questions. AIM To hydrolyse the ester, ethyl benzoate, under alkaline conditions, and then do

qualitative tests for the products.

PROCEDURE

1. Place about 5 cm3 of ethyl benzoate, 1 r.g. (rice grain equivalent) of pumice (or boiling

chips) and 25 cm3 of 10% sodium hydroxide solution (bench strength) in a flask and reflux for 30-

40 minutes.

2.Set the condenser for distillation and collect the first 5-10 cm3 of distillate. Allow the residue to

cool.

28

3. Add anhydrous magnesium sulfate, MgSO4, to the distillate – enough to cover the

bottom of the container. Stopper and shake. Leave for 5 minutes and then cool.

4. Test some of the liquid with acidified dichromate solution (warm). Note colour

changes.

5. Test some for flammability. A drop on a spatula in the Bunsen flame should ignite.

6. If time permits determine boiling point.

7. Acidify the cooled residue with concentrated sulfuric acid (CARE).

8. If time permits, filter, wash and dry the acidified residue.

9. Determine its melting point, if time permits.

(a) Write a balanced equation for the reaction. C6H5COOC2H5 + H2O → C6H5COOH + C2H5OH

(b) Explain what reflux is and why it is necessary. Organic bonds, being covalent are strong, and these organic compounds are volatile. Therefore,

reflux is used to contain reactants and products while heating and heating is required to break the

strong covalent bonds present in the reactants.

(c) Explain the purpose of the distillation. To separate the alcohol, the product, from unreacted reactants.

(d) Explain the purpose of the anhydrous magnesium sulfate, MgSO4. It is a drying agent. It absorbs water.

(e) Explain the colour changes observed in step 4 with the help of an annotated equation. The alcohol is oxidised to an acid

C2H5OH + (O) → CH3COOH + H2O

The oxidant Cr2O72ˉ(aq) / H+(aq) gives Cr3+, so changing from orange-red Cr2O7

2ˉ(aq) to green

Cr3+(aq).

(f) Write an equation for the combustion. C2H5OH + 3O2 → 2CO2 + 3H2O

(g) Write an equation for the acidification step. Explain why the residue does not distil. 2C6H5COONa + H2SO4 → 2C6H5COOH + Na2SO4

Na2SO4 is a non-volatile ionic compound.

(h) Explain why filtration of the acidified product is possible. Benzoic acid is not very soluble in cold water.

(i) Why is the determination of a melting point much used in organic chemistry? A sharp melting point indicates a pure substance. Impurities lower the melting point of a substance.

29

8. Ethyl ethanoate is an example of an ester. (a) State three important commercial uses of esters. Polymers, flavourings, solvents.

(b) Fats and vegetable oils contain naturally occurring esters. Give the general formula of these esters and explain why they are important commercially.

Where R1, R2 and R3 are C11 to C19 carbon chains in saturated and unsaturated fatty acids.

These esters are called triglycerides. The alcohol in glycerol is a triol; it has three –OH groups per

molecule. They are present in foods, used in cooking, and can be converted into biofuels.

This question (c) has been included as part of your general chemical knowledge; it is not part of

the examination syllabus.

(c) Describe, giving an outline of the practical details, reaction conditions and equations, how you could produce a sample of ethyl ethanoate using ethanol as the only organic starting material. Separate the C2H5OH into two equal parts.

Oxidised one part using excess Cr2O72ˉ/H+ to ethanoic acid. Distill off the acid. Set up apparatus for

reflux. Add alcohol + acid + drop H2SO4(conc) + boiling chips.

Reflux for 30 minutes. Distil off the ester at its boiling range (77°C)

C2H5OH + (O) → CH3COOH + H2O

C2H5OH + CH3COOH → CH3COOC2H5 + H2O

30

9. Describe, with equations, the following reactions: (a) diethylamine and propanoic acid An acid-base reaction; the product, ionic, is soluble.

(C2H5)2NH + C2H4COOH → (C2H5)NH2+ C2H5COOˉ

(b) 1-propylamine and hexanoic acid

Again an acid-base reaction,

C3H7NH2 + C5H11COOH → C3H7NH3+C5H11COOˉ

An amine plus an acid gives a salt. These salts are generally soluble and medicinal amines must

be soluble to be effective. Usually HCl is used. The anti-histamine drug cetirizine dihydrochloride is

an example. Ecstasy, a hallucinogenic drug, is the hydrochloride of a secondary amine.

10. Synthetic polymers can be classed as addition or condensation polymers. Explain, briefly, the difference.

Addition polymers involve unsaturated monomers linking to form saturated C – C chains; these are

the addition reactions. Condensation polymers form from condensation reactions, where in small

molecules, e.g. H2O, are split off as the polymer chain forms.

11. What is a co-polymer? A polymer formed from 2 or more different monomers, e.g. an alternating co-polymer made from A

and B would be …ABABAB…. A block co-polymer …AAABBBAAA….

Co-polymerization is used to prepare a polymer with specific properties, e.g. very hard.

12. About polyesters (a) What are the polyesters? Are polymers formed by condensation polymerisation between acids with two or more carboxylic

functional groups and alcohols with two or more hydroxyl groups.

(b) Give the structural formula for ethane-1,2-diol. Why is it used as a monomer in condensation polymerisation?

HOOH

It has two –OH functional groups and can act as a monomer with its co-polymer, an acid with two

carboxylic groups.

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(c) Give the structural formula for benzene-1,4-dicarboxylic acid. Why is it used as a monomer in condensation polymerisation?

O

OH

O

HO

It has two carboxylic acid groups per molecule and these can link via condensation reactions.

These numbers are used to show where the functional groups are to be found on

the benzene ring, just as they are in non-aromatic compounds.

O

OH

O

HO

1

2 3

4

56

(d) Now write the equation for the polymerisation reaction between the above two monomers.

O

O

O

O

O

O

O

O

O

REPEATING UNIT

O

OH

O

HO

HOOH

+

(e) Indicate the repeating unit in the polymer formed in (d). (f) What is the commercial name for this polymer? Terylene

32

An early name for benzene-1,4-dicarboxylic acid was terephthalic acid, and for the

ethane -1,2-diol, ethylene glycol. The polymer, when spun into fibres, was called

Terylene. The ter– and –lene giving the name Terylene.

When made into a film it is called Mylar. Mylar, which is very light and strong, was

used on the Albertross – the aeroplane of enormous wing span which was pedalled

across the English Channel in March 1979. Mylar is also used in the sails of yachts.

13. (a) What are the polyamides? Polymers in which the monomers are linked by amide bonds. They are condensation polymers.

(b) Give the structural formula for 1,6-diaminehexane. Why is it used as a monomer in condensation polymerisation?

H2NNH2

It has two amine groups. One at each end of the chain.

(c) Give the structural formula for hexanedioic acid. Why is it used as a monomer in condensation polymerisation?

O

OH

O

HO

It has two carboxylic acid groups; one at each end of the chain.

(d) Now write the equation for the polymerisation reaction between the above two monomers.

33

O

OH

O

HO NH2

H2N

+

O

HN

NH

HN

O

*

O

REPEATING UNIT

+

H2O (e) Indicate the repeating unit in the polymer formed in (d). (f) What is the commercial name for this polymer?

Nylon-6,6

The first nylon prepared was called nylon 6.6 because there were two C6 monomers used, namely

hexanedioic acid and 1,6-hexanediamine.

Its uses are many and varied. Here are some: hosiery, rope, fishing lines. Parachutes and artificial

blood vessels.

Nylon, when mixed with other fibres such as wool, is both warm and very hard wearing. Stretch

fabrics have a nylon base.

34

14. The polymer shown has the trade name Nomex and is used to make fire-proof clothing:

O

NH

O

NH

O

NH

O

REPEATING UNIT (a) Show in the diagram above the repeating unit (b) Supply the missing words. Nomex is a poly amide and the type of polymerisation used to make it is termed condensation polymerisation.

(c) Draw the structural formulas of the monomers used in the manufacture of Nomex.

H2N NH2

O

OH

O

HO

15. Explain why the preparation and use of condensation polymers is of significant economic importance.

Because of their so varied and hence so useful products, vast quantities of these polymers are

made and used. Paints, ropes, clothing, carpets, are but a few of the many 1000’s of products

made.

35

16. Shown is the repeating unit for a condensation polymer.

Write structural formulas for the monomers used to make this polymer.

HO

O

O

OH

H2NNH2

36

20.5 Exercises

Questions 1 (a) and 1 (d) involve more than two steps with chemistry beyond that which is

required at this level. Less demanding and more suitable questions to introduce in this exercise

would have been:

(a) but-2-one from but-2-ene, and (d) butan-1-ol from butane

Do these questions before checking your answer with that given below in (a) and (d)

1. Deduce reaction pathways given the following reactants and products. Include the conditions required and an equation for each step. (a) butan-2-one from but-2-ene

______________________________________________________________________________

______________________________________________________________________________

H2O/ H2SO4

reflux

OH O

Cr2O72-/ H+

reflux (b) propanoic acid from propene

H2O/ H2SO4

refluxOH OHCr2O7

2-/ H+

reflux

O Note that this reaction also gives propanone. A mixture of both propanone and propanoic acid is

formed. OOH

[O]

The mixture can be separated by fractional distillation.

Conclusion: This question has not been well thought through!

37

(c) propanal from 1-bromopropane

Br NaOH OHCr2O7

2- / H+

[O]

H

Oreflux

(aq)

reflux

Note: The oxidation in the final step must NOT be complete. This would produce a carboxylic acid.

Limited oxidation occurs when the Cr2O72- / H+ is added to the alcohol as the aldehyde is distilled

off. There is more alcohol than oxidant. To produce an acid the alcohol is added to excess oxidant.

(d) butan-1-ol from butane

______________________________________________________________________________

______________________________________________________________________________

Br2

UVBr

NaOH(aq)

refluxOH

(e) propanone from propene

HBrBr

NaOHOH

Cr2O72- / H+

[O]

O

reflux (f) butylamine from 1-chloropropane

Cl KCN (al) CN H2 / NiNH2

reflux

(g) N-ethylethanamine from ethanol The steps are:

1. Elimination of H2O from the alcohol using concentrated H2SO4.

2. Add HBr.

3. React with NH3.

OHconc. H2SO4

refluxH2C CH2

HBrH3C CH2

Br

warm

NH3 NC2H5

C2H5

H + 2HBr

38

(h) ethyl ethanoate from ethanol

OHCr2O7

2- / H+

[O]

H

O

OH

H2SO4reflux

O

O

2. (a) The aim in any synthesis is to produce a reaction pathway with the minimum number of steps. Why is this? Firstly, for economic reasons. For example if each step has a 1 hour reflux, then if there are 10

steps then the synthesis will take at least 10 hours. Remember that in business time is money

(staff, overheads, energy used). Secondly, not all steps result in 100% conversion of the reactant

into the desired product and as the next question will show, the more the steps the less the final

yield.

(b) In a four step synthesis each step gives a 50% yield. What is the percentage yield for the final product? Multiply the number of steps, 4, by the % yield in turn. First step gives 50% yield, second step

gives 50% of 50% = 25%, third step gives 50% of 25% = 12.5%, final step gives 50% of 12.5%. In

this case it is a 6.25% yield. Quite a loss!

3. (a) What is the difference between the terms reaction pathways and reaction mechanisms? The reaction pathway indicates the steps in the synthesis, along with all the conditions.

Reaction mechanism gives an insight into how the reaction is actually occurring.

(b) In the following reactions, circle where the curly arrows are correctly used;

The correct mechanisms are the first and third mechanism since both illustrate the movement of

electrons from the nucleophile to the electrophile.

39

20.6 Exercises

1. Explain, with an appropriate example, what is meant by: (a) structural isomers Compounds with the same molecular formula but have different structures, e.g. C4H10 is the

molecular formula for butane and 2-methylpropane.

(b) stereoisomers Compounds with the same molecular formula and same structural formula but have different 3D

shapes, e.g. cis-but-2-ene and trans-but-2-ene.

(c) cis-trans (geometric) isomerism In the cis isomer the functional groups are on the same side of the C = C. In trans they are on

opposite sides.

(d) optical isomerism Results from the opposite effect optical isomers can have on plane polarised light. The amino acid

alanine shows optical isomerism; one isomer will rotate plane polarised light in one direction; the

other isomer will rotate plane polarised light in the opposite direction. This occurs because the

central atom is asymmetric, i.e. has four different groups attached

(e) chiral carbon atom When the C atom in a molecule is attached to four different atoms or group it is described as chiral.

Alanine has a chiral C atom. Chiral carbon atoms are asymmetric.

Lactic acid, 2-hydroxypropanoic acid, is an example of a chiral compound. Its chiral C atom

gives isomers that can exist as mirror images, so making them optically active.

Hint: Draw a mirror as a line before drawing optical isomers. One isomer will be

a reflection of the other.

40

(f) achiral Not attached to 4 different atoms or groups. The C in the –COOH group in amino acids is achiral. It

is attached to only three groups (four bonds, but only three groups).

(g) racemic mixture When there are equal amounts of the optical isomers present in a mixture it is described as

racemic, e.g. lactic acid from sour milk is a racemic mixture.

Racemic mixtures are not optically active because the two optically active isomers

cancel out each other. They have an equal, but opposite effect on polarised light.

(h) enantiomers

Are compounds which react chemically the same, but differ in the way they rotate polarised light –

they are optically active. Lactic acid exists as enantiomers.

Although enantiomers have the same general chemistry and the same

physical properties, their biochemistry can differ. The tragedy of thalidomide is

an example.

Although some reactions produce a racemic mixture, nature seems to prefer

the making of one enantiomer. In biological systems if the molecule does not fit,

then it does not work or, as with thalidomide, does damage.

On a positive side research into enantiomers that do damage may help to

control or remove cancerous cells.

(i) optical activity The ability rotate plane polarised light, shown by optical isomers. Lactic acid shows optical activity.

(j) dextrorotary (no example needed)

An optical isomer that rotates plane polarized light to the right (clockwise).

(k) restricted rotation The inability of the carbon atoms in a carbon-carbon double or triple bond to rotate is described as

restricted rotation. The result is geometric stereoisomerism, e.g. the cis- and trans- isomers of 1,2-

dichloroethene.

41

2. Which of the following can exist as stereoisomers? Explain your choices. (a) CH3CH2CH2CH3

No – because no chiral C atoms present.

(b) CH3CH(OH)COOH Yes, the central atom has four different atom/groups of atoms attached, therefore, chiral.

(c) C4H9Cl No – all C achiral

(d)

H3CHC

CH2CH3

CHO

Yes, central C is chiral.

3. In question 2 above draw and name the structural isomers for (a) and (c). (a)

H3C

CH

H3C

CH3

isobutane

H3CC

H

CH3

2-methylpropane

H3C

(c)

H3CCH

H2C

CH3

Cl

2-chlorobutane

H2CCH2

H2C

CH3

1-chlorobutane

Cl

H3CC

CH3

CH3

2-chloro-2-methylpropane

H2CCH

CH3

Cl

CH3

1-chloro-2-methylpropane

Cl

42

4. 1,2-dichloroethene can exist as a stereoisomer. Explain how this is possible, and draw and name the stereoisomers.

ClCl

Trans 1,2-dichloroethene

Cl

Cis 1,2-dichloroethene

Cl

It contains a C = C. The pi bonds restrict rotation, therefore geometric isomerism possible, giving

cis- and trans- isomers.

5. One form of stereoisomerism is called geometric isomerism. With the help of structural formulas, show which of the following have geometric isomers, and which have not. (a) 1,1-dichloroethene

H2C

Cl

Cl NO GEOMETRIC ISOMERS

(b) 1,2-dichloroethene

ClCl

Trans 1,2-dichloroethene

Cl

Cis 1,2-dichloroethene

Cl

(c) but-2-ene

Trans but-2-ene Cis but-2-ene

(d) 1,2-dichlorocyclopropane

Trans 1,2-dichlorocyclopropane Cis 1,2-dichlorocyclopropaneCl Cl Cl Cl

43

(e) 1,2- dichlorocyclobutane

ClCl Cl Cl

Cis 1,2-dichlorocyclobutane Trans 1,2-dichlorocyclobutane (f) 1,3-dichlorocyclobutane

Cl Cl

Cis 1,3-dichlorocyclobutane Trans 1,3-dichlorocyclobutane

Cl Cl

6. Restricted rotation (a) What is meant by restricted rotation? Rotation is not possible due to the rigidity of the structure. In alkenes and alkynes the double and

triple bonds cannot rotate; their compounds show restricted rotation.

Some inorganic complexes also show restricted rotation, which gives rise to cis- and trans-

isomers. Cisplatin, an anticancer drug, is an example.

(b) Explain, with examples, how restricted rotation forms.

The sigma and pi bond in a double bond are perpendicular to each other and hence for a rotation

to occur, the pi bond would have to break.

Pi bond

Sigma bond

44

7. Hybridisation. (a) Using ethene as an example, explain, with a diagram, sigma (σ) bonding, pi (π) bonding and sp2 hybridisation.

Pi bond

Sigma bond Orbitals that overlap ‘end to end’ give sigma covalent bonds. The C – H bonds are sigma. The

‘side to side’ overlap of orbitals produce pi bonds. Carbon’s p orbitals form these π bonds. The

hybridisation involves a 2s electron being promoted to the 2p orbitals, giving an s with one electron

and 2p orbitals, each with one electron. This is designated as a sp2 hybrid. The remaining p orbital

electron forms the π bond.

(b) Which of the bonding, σ or π, is the stronger and why? The pi bond is weaker than the sigma bond, because there is more extensive overlap, i.e. shorter

distance in the σ bond, so making it the stronger.

This means that the C = C bond (being a σ + π bond) is stronger than a C – C bond

being (σ), but not as strong as two σ bonds.

8. But-2-ene-1,4-dioic acid has stereoisomers. (a) Draw and name these isomers.

O

OH

O

HO

O

HO

OHO

Trans but-2-ene-1,4-dioic acid Cis but-2-ene-1,4-dioic acid (b) How could you distinguish between these isomers? Determine melting point. The trans form has the higher Tm. This is because the trans form is more

open than the cis form. This allows the molecules to come closer together and the resulting

hydrogen bonding gives the trans the higher Tm.

45

(c) When heated one of the above isomers forms a ring compound. Which reacts in this way, the cis or trans form, and why? The cis form because it already has more of a ring structure to it

(d) Write the equation to show the reaction in (c).

O

HO

OHO

Cis but-2-ene-1,4-dioic acid

heat

OO O

+ H2O

furan-2,5-dione

A beam of ordinary light consists of the electromagnetic waves that oscillate in an infinite number

of planes. When a beam of ordinary light is passed through a polariser only the lightwaves

oscillating in a single plane passed through. Hence the name ‘plane polarized light’. A polariser is

used to determine whether or not the substance is optically active. Optical isomers rotate the plane

of polarised light.

9. Polarisation (a) What does optically active mean? Plane polarised light is rotated when it is passed through an optically active molecule.

(b) Describe, in brief outline, the action of a polariser. A diagram will help you get an extra mark.

White light, oscillating in many planes, is passed through vertical slits (the polariser) making the

light now oscillating in only one plane. It now passes through the sample. The one plane light is

changed to another plane and the observer adjusts an analyser to measure how much the light has

been rotated (clockwise or anticlockwise).

46

(c) What factors must be held constant to determine the specific rotation? The wavelength of light, the temperature and the concentration of sample used must be kept

constant in order to measure the degree of rotation.

10. Enantiomers may differ may differ in physical properties, e.g. L-aspartame (nutrasweet) is very sweet, while the D-enantiomer is bitter. Or is sweetness a chemical property? (a) What do the L- and D- mean? L is from laevorotary (Latin laevus, left) meaning rotation to the left (anticlockwise). D from

dextrorotatory (dexter, right) meaning clockwise rotation, i.e. moves to the right.

(b) Limonene exists in two enantiomer forms, as does carvone. One form of limonine is found in oranges, and smells like oranges; the other is found in lemons and smells like lemons. With carvone you find one form with a caraway seed smell, the other a spearmint smell.

limonene

O

carvone

Chiral carbon Chiral carbon

Circle the chiral atoms in the structures above.

47

Although you are asked to identify the chiral C atom in an optically active compound, strictly

speaking it is the molecule that is chiral, not the atom. It is the lack of symmetry in the structure of

the molecule that makes it optically active.

11. Compounds with the same molecular formula as pent-3-en-2-ol, show three types of isomerism: structural, geometric and optical. (a) Draw and name a structural isomer of this compound that does not contain a carbon-carbon double bond.

CH3

CCH2

H2C

H3C

O

pentan-2-one (b) Draw and name the geometric isomers of pent-3-en-2-ol.

HO

cis-pent-3-en-2-ol

OH

trans-pent-3-en-2-ol (c) Give the reason why pent-3-en-2-ol shows geometric isomerism. The C = C show restricted rotation, hence the above (cis) and across (trans) forms are possible.

(d) Draw the optical isomers for pent-3-en-2-ol.

OH

CH3

H

Chiral CarbonOH

H3CH

(e) Indicate the chiral atom in (d) above. (f) What is the effect these optical isomers will have on plane polarised light? The plane polarised light will be rotated the same amount but in different directions (L, left and D,

right).

Chiral, pronounced keral, comes from the Greek, kheir, meaning hand. Chiral compounds

48

cannot be superimposed (superimpose means to place over) on each other,

just as your right hand cannot be superimposed on your left.

An Organic Revision Exercise

1. The alkanes (the saturated hydrocarbons) are inert to acids, bases, oxidizing agents and reducing agents. Why is this? The C – C bonds is very strong as is the C – H bond. They are non-polar bonds and C cannot

expand its octet of electrons when combined.

2. Explain: (a) The low bond polarity in the alkanes The electronegativity difference between C and H is small (C = 2.5, H = 2.1) hence limited bond

polarity, limited reaction.

(b) The high bond strength of the alkanes. Include a description of hybrid orbitals in your answer. Because the H atom is very small and the C atom, too, is small, sigma orbital overlap is extensive,

as is the s and p orbital overlap. These sigma bonds are, therefore, strong. The sp3 hybrids are

related to the tetrahedral arrangement of the 4 covalent bonds, presenting a uniform distribution of

electron pairs, so making a non-polar molecule.

3. Reaction mechanisms (a) Although there are many millions of organic compounds four fundamental reaction mechanisms are generally used to account for their chemistry. These are given. Briefly explain what is meant by each. (i) addition

49

In addition reactions functional group atoms are joined to the C atoms at the points of un-

saturation, i.e. at the carbon-carbon double or triple bond.

(ii) elimination Two molecules join by splitting off a small molecule, e.g. H2O or HCl.

(iii) rearrangement

The atoms in a molecule rearrange to produce a new structure. The petroleum industry rearranges

the structure of alkanes to increase the octane number of petrol.

(iv) substitution Are reactions where an atom or a molecule is replaced by another atom or molecule.

(b) Why is it that condensation is not considered to be a fundamental reaction mechanism? Condensation is preceded by addition, and then elimination follows. It is a combination of two

fundamental reactions.

4. Explain what is meant by the terms photochemical and substitution as used in hydrocarbon chemistry. Photochemical refers to reactions initiated by light, e.g. photosynthesis where plants use sunlight

to convert CO2 and H2O into carbohydrates. UV light is used to initiate the production of free

radicals. These, in turn, can substitute, i.e. replace an atom, or group of atoms, in a molecule. The

reaction between Cl2 and CH4 to produce HCl and CH3Cl combines a photochemical reaction and

substitution.

5. Methane-bromine reaction. (a) Write an equation showing the reaction of methane with bromine.

CH4(g) + Br2(g) → CH3Br(g) + HBr(g)

(b) What are the reaction conditions required for (a) above? UV light is required

(c) Give two observations that would show that this reaction has occurred. The brown Br2(l) becomes colourless. The products, CH3Br(g), and HBr(g) are also colourless.

Use an indicator, e.g. moist litmus, to show an acid, HBr, present.

6. Explain, with an example and using fish-hook arrows, what is meant by homolytic fission?

50

The bond breaking (fission) reaction that produces free radicals with unpaired electrons. A Br2

molecule, under UV light, will break its covalent bond to form two free radicals, each radical taking

one electron from the previous shared pair. This electron movement is shown by fish hook arrows.

7. Explain, with an example, using arrow notation, heterolytic fission. This bond breaking produces ion intermediates during the reaction, e.g. HBr → H+ + Brˉ

The electrons are not shared.

8. What is a free radical and what chemical property is associated with such a species? A particle with an unpaired electron. They are formed during homolytic fission. They are very

reactive, e.g. Cl∙.

9. Draw: (a) The Lewis structure for a chlorine atom

Cl

(b) A chlorine free radical

Cl

10. Heterolytic fission tends to occur in polar bonds. Using ethanoic acid, a structural formula and a curly arrow, explain this statement.

O

O H2O

NB: These arrows represent the movement of 2 electrons

H

O

O H3O++

A polar solvent, e.g. water, encourages heterolytic fission; attacking the δ+ H in the –OH of the

acid. The unequal sharing of electrons producing the ion shown.

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11. Using only curly arrow notation show the heterolytic and homolytic fission of HCl.

H Cl

H Cl

H+ + Cl-

H + Cl

HETEROLYTIC FISSION

HOMOLYTIC FISSION

12. Classify the following reactions as heterolysis, homolysis or neither:

(a) CCl2F2 Ultraviolet radiation

CClF2 + Cl homolysis (no ions present)

(b) Br2 Br + Br homolysis (no ions present)

(c) NH3 + CH3COOH CH3COONH4 heterolysis, the acid-base reaction produces ions, CH3COO– and NH4

+.

13. (a) Name the following organic compounds: A. (CH3)3CBr 2-bromo-2-methylpropane

B. CH3CH2CH =CH2

but-1-ene

(b) Give the reagent, conditions and reaction mechanism for the hydrolysis of compound A above. A is a tertiary halogenoalkane. It reacts with aqueous sodium hydroxide to produce a tertiary

alcohol. The mechanism is SN1.

H3C C CH3

Br

CH3

H3C C CH3

CH3

OHH3C C CH3

OH

CH3

+ Br

(c) Give your observations when an alcoholic solution of bromine is added to B above. Addition occurs. The brown alcoholic Br2 becomes colourless as the reaction proceeds.

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15. Outline the reagents, conditions and equations for the two stage conversion of 1-bromopropane to 1-butylamine. Ammonia, 1-bromopropane, mix with gentle heating.

NH3(g) + CH3CH2CH2Br CH3CH2CH2(NH3)Br (intermediate or transition state)

CH3CH2CH2NH2 + HBr

Now revise your understanding of the SN2 mechanism by using structural formulas and

curly arrows for the above reaction.

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16. Benzene is an aromatic hydrocarbon. (a) Explain the word aromatic when it is used in organic chemistry. It is the traditional term for compounds containing benzene rings. Now these compounds are called

the arenes.

(b) Why is the benzene molecule planar? Each C in the ring is in a trigonal planar arrangement of atoms. The 4th electron from each C is

delocalised into the areas above and below the six C atoms.

(c) Why are all of the C—C bonds in the molecule the same length? Again, these bonds have the same length because the delocalisation of electrons; they have a

covalent bond which is shorter than a C – C bond, but longer than a C = C bond.

(d) Each carbon in benzene is sp2 hybridised. What does this mean? One of the two s electrons from the C atom is promoted to a p orbital. The one s and the two p

electrons form a hybrid sp2 orbital. This gives the trigonal planar structure.

(e) Why are there no alkene groups in benzene? The extra electrons required to form a C = C bond are delocalised in a plane above and below the

ring structure.

(f) What is the evidence for benzene having no alkene groups present? Refer to both physical and chemical properties.

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Benzene is an unsaturated hydrocarbon; it burns with a smoky flame and has low Tb and is soluble

in organic solvents. Thus is like the alkenes. But chemically the differences are significant. It does

not undergo addition reactions nor does it react with oxidising agents, e.g. acidified potassium

permanganate. This is the evidence for no C = C being present.

(g) Which of the following can be used as representing benzene?

A. C6H6 B. C. D. C6H5—

A and B are correct. C is cyclohexane, C6H12, and D is the benzyl radical.

(h) Draw a more acceptable representation of the benzene molecule than any given above.

Question 16 probably extends benzene chemistry outside of the

examination syllabus. Check with your teacher.

17. Substitution reactions (a) Give an example of a substitution reaction using benzene. The reaction with Cl2 or Br2 is substitution.

C6H6 + Br2 → C6H5Br + HBr.

(b) How does your reaction above show the special stability of benzene?

That a catalyst (usually AlCl3 or AlBr3) and energy, e.g. UV light, are required suggest benzene is a

stable compound. The benzene ring is not changed.

(c) Benzene undergoes substitution rather than addition reactions with bromine. Why is this?

The stability of the benzene ring due to the delocalisation of π electrons prevents changes to its

planar structure, therefore substitution is the preferred reaction.

18. Benzene is an unsaturated hydrocarbon. What chemical evidence supports the statement? The C to H ratio is low and it burns with a smoky flame.

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19. Explain the mechanism for the elimination of HBr from bromoalkanes. Reflux the bromoalkane with a concentrated alcoholic solution of KOH. In this case the OH– is

acting as base – it becomes water by adding a proton. The OH– increases the polarity of the

C(δ+)―Br(δ-) bond as it approaches an H (as can be seen in the diagram below, electrons are

pushed towards the bromine). With the higher temperature this H can break off as an H+ and then

combines with the OH– to give water, H+ + OH– H2O. At the same time the Br breaks off as

Br–.

This is illustrated with the following mechanism

As the OH– approaches any H it can, at a high temperature, pull an H+ from the bromopropane to

form water. The pair of electrons from a C―H move to a C―C bond, making it a C=C bond. These

electrons now push the electrons from the C―Br onto Br making it Br–.

20. Monomers and polymers (a) The monomer ethenyl ethanoate (vinyl acetate) is used to make polyvinyl acetate, part of the polymer chain being:

H2C

H2C

H2C

H2C

H2C

H2C

O O O O O OO

CH3 CH3 CH3 CH3 CH3 CH3 CH3

O O O O O O O

Repeating unit (i) Circle the repeating unit in this polymer. (ii) Define monomer. The unit (molecule) from which a polymer may be built, either by addition is condensation

polymerisation.

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(iii) Draw a structural formula for ethenyl ethanoate. O

O

vinyl acetate (b) In industry the following compound is called vinyl chloride: CH2=CHCl. (i) Give the systematic name for CH2=CHCl. chloroethene

(ii) Using structural formulas write an equation to show the formation of the polymer, polyvinyl chloride.

C C

H

H H

Cl

n C C

H

H H

Cl

C

H

H

C

H

Cl

(iii) Explain what is meant by the terms thermoplastic and thermosetting polymers. Thermoplastic means to become plastic on heating. PVC is a thermoplastic.

Thermosetting means that the polymer is resistant to heat. It has set. Bakelite is a thermosetting

polymer.

(c) Polymers are made by addition polymerisation, or by condensation polymerisation. In (a) and (b) above, which of the polymerisations is addition and which condensation? Give your reasons in your answer.

Both are made by addition polymerization since you are adding the same molecule.

21. A compound has the empirical formula, CHBr. It has a relative molecular mass of 185.8. (a) Name an instrument that could used to determine relative molecular mass. The mass spectrometer. (b) Show that the relative molecular mass given above agrees with the molecular formula, C2H2Br2. Calculate the Mr

2 x C = 24.0

2 x H = 2.0

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2 x Br =

159.8

185.8

(c) This compound has two structural isomers. (i) Define structural isomer.

Compounds with the same molecular formula but have different structural formulas.

(ii) Draw structures for and name each isomer. 1,1-dibromoethene

Br

Br

1,2-dibromoethene

BrBr

Trans 1,2-dibromoethene

Br

Cis 1,2-dibromoethene

Br

(d) Halogens can react with alkenes. (i) Name the type of reaction for an alkene combining with a halogen. addition reaction

(ii) What would you observe during the reaction in (i) above if bromine was the halogen used? The loss of colour (brown → colourless) during the reaction.

(e) One of the structural isomers of the compound C2H2Br2 can also show stereoisomerism. (i) Define stereoisomerism. Isomers with the same molecular formulas and structural formulas but with different spatial

(3D) arrangements of their groups of atoms.

(ii) Draw and name the stereoisomers for the isomer you have chosen. cis-1,2-dibromoethene

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Br

Br

trans-1,2-dibromoethene

BrBr

(iii) Explain why stereoisomerism is possible in a haloalkene, but not in a haloalkane. The C = C bond shows restricted rotation; but the C – C bond in haloalkanes allows for

rotation. Stereoisomerism is only possible with restricted rotation.

22. Halogenoalkanes can be hydrolysed to alcohols using hot NaOH(aq). (a) Give the equation for the alkaline hydrolysis of 1-iodopropane. CH3CH2CH2I + OHˉ→ CH3CH2CH2OH + Iˉ

(b) Describe, using curly arrows and relevant dipoles, the mechanism for the reaction in (a) above.

(c) The rates of hydrolysis of 1-chloropropane, 1-bromopropane and 1-iodopropane with hot NaOH(aq) were studied under controlled experimental conditions. (i) Outline three of the conditions that would be require controlling if the rate comparisons obtained can be judged to be scientifically acceptable?

The temperature is controlled, i.e. set and maintained during the experiment. The

concentration of the NaOH(aq) as well as the haloalkane used remain the same in all three

experiments.

(ii) Which of the haloalkanes listed in (c) above would show the fastest, and which would show the slowest rate?

Fastest: 1-iodopropane

Slowest: 1-chloropropane

(iii) Explain the order you have given in (ii) above. The C – l bond length is the longest (I being the largest atom), therefore the weakest bond,

therefore the fastest reaction.

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23. Describe a possible two-step synthesis for the making of petan-2-one starting with 2-iodopentane.

NaOH

Cr2O72- / H+

[O]reflux

OI OH Warm dilute NaOH(aq) gives a secondary alcohol; refluxing with acidified dichromate produces the

ketone.

Here is a summary of the reaction pathways you must know:

Preparing 2-methylpropan-1-amine from propene

H3CCH

CH3

CH2

NH2

HBrBr

NaOHOH

KCN (al) H2 / Ni

reflux

CN

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Your Exam Questions and Answers

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