G Further organic chemistry - pedagogics.ca - Home · CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011 G FURTHER ORGANIC CHEMISTRY 1 G Further organic chemistry G1

G FURTHER ORGANIC CHEMISTRY 1CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011

G Further organic chemistry

G1 Electrophilic addition reactions

Reactions of alkenesMost reactions of alkenes involve an electrophilic addition mechanism.

An electrophile is a reagent (a positively charged ion or the positive end of a dipole) which is attracted to regions of high electron density and accepts a pair of electrons to form a covalent bond.

The C=C contains four electrons and is thus a region of very high electron density. Electrophiles will be attracted to the electron density in the double bond.

The reaction of ethene with hydrogen bromideThe overall equation is:

Learning objectives

• Explain the electrophilic addition reactions of alkenes

• Predict and explain the formation of the major product when HX reacts with an unsymmetrical alkene

C

H

H

CH Brheat

Br

H

bromoethane

HHC

H

C +

H

H

H

H–Br is polar, with the Hδ+ and the Brδ−. The H is thus attracted to the high electron density in the C=C.

The mechanism is shown in Figure G1.

The curly arrow represents movement of two electrons.

C

CC

H H

H H H

H

H C

H C+

HH Brδ+ δ–

:Br–

H

H

H C

H C

H

Br

Figure G1 The mechanism of the reaction of ethene with hydrogen bromide.

In the � rst stage, a pair of electrons is donated from the C=C to form a bond between C1 and the H of the H–Br. At the same time the H–Br bond breaks – the pair of electrons from the bond goes to the Br. This is shown in terms of electrons in Figure G2.

The π bond breaks.HL

Figure G2 Movement of electrons in reaction of ethene with hydrogen bromide.

H BrC

H H1 C1

CHH

2Br–

electron from C2

electron from H

this C has lost an electron thereforepositively charged

pair of electrons from double bond

HH

H

H

C+2H

2 CHEMISTRY FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2011G FURTHER ORGANIC CHEMISTRY

C2 has lost an electron (this electron is now in the C–H bond) and thus C2 has a positive charge. The intermediate formed is called a carbocation (a positively charged organic ion with a positive charge on C). The Br atom has gained an electron from the H–Br bond and is therefore a negatively charged bromide ion. In the second stage the Br− ion is attracted to the C+ and donates a pair of electrons to form a bond. The electrons in the � nal compound are shown in Figure G3.

The initial type of bond breaking, when the H–Br bond breaks, is called heterolytic � ssion since the bond breaks so that both electrons go to the same atom (Figure G4).

The reaction of propene with HBrWhen HBr adds to propene, two products are possible:

C

Br

HH

H

H

CH

H Br

pair of electronsfrom H–Br bondgoes to Br

minor product1-bromopropane

major product2-bromopropane

C

CH BrH

H

HH

C

H

C BrH

C

HH

H

HH

C

H

C HH

C

BrH

H

HH

CHH

Figure G3 Electrons in the fi nal compound, bromoethane.

Figure G4 Heterolytic fi ssion.

The main/major product (the minor product is usually only present in very small amounts) may be predicted from Markovnikov’s rule:

When H–X adds to the double bond of an alkene, the H atom becomes attached to the C atom that has the larger number of H atoms already attached.

Markovnikov’s rule can be used to predict the major product formed when H–X adds to any unsymmetrical alkene. Thus, when H–Cl adds to 2-methylpropene, the H atom of the HCl becomes attached to C1 as this has two H atoms already attached, rather than C2, which has no H atoms attached.

H

H

2C

1C

CH

H3CClH

HH

H

H

2C

1C

C

H

Cl

H

H

H3C

H

‘Those who have shall get more.’

The explanation for Markovnikov’s rule depends on the stability of the intermediate carbocation.


C+

H

CH3

primary carbocation

HC+

H

CH3

secondary carbocation

increasing stability of carbocation

CH3

C+

CH3

tertiary carbocation

CH3H3C

primarycarbocation

C+H

H

CH H

CH H

H

secondarycarbocation

CH

H

H

C+H

CH H

H

The two possible carbocations that can be formed when propene reacts with H–Br are:

The secondary carbocation is more stable than the primary carbocation. This is due to the electron-releasing e� ect of alkyl groups. There are two alkyl groups next to the positively charged C in the secondary carbocation but only one in the primary carbocation. Two alkyl groups reduce the positive charge on the C more than one alkyl group and thus stabilise the ion more. Since the secondary carbocation is more stable, it is more likely to be formed.

Alkyl groups have an electron-releasing e� ect. This is also called a positive inductive e� ect.

The electron-releasing e� ect is roughly independent of the size of the alkyl group, i.e. a methyl group donates approximately the same amount of electron density as an ethyl group.

In general, the more alkyl groups there are attached to C+, the more stable the carbocation (Figure G5).

Figure G5 Stability of carbocations.

Markovnikov’s rule may also be used to explain the products obtained when interhalogen compounds react with propene (or other unsymmetrical alkenes). The less electronegative halogen atom behaves like H and, in the major product, becomes attached to the C that has more H atoms attached.

For example, the reaction of Br–Cl with propene:

A primary carbocation has one C atom attached to the C bearing the positive charge.

A secondary carbocation has two C atoms attached to the C bearing the positive charge.

Examiner’s tipWhen asked to explain which is the major product when HX adds to an alkene, always explain it in terms of the stability of the intermediate carbocation – it is never enough to just say ‘due to Markovnikov’s rule.’

Extension

The electron-releasing e� ect of the alkyl group comes from the donation of electron density from a C–H or a C–C bond into the vacant p orbital on the carbocation. This is called σ-conjugation or hyperconjugation.

1C

major product1-bromo-2-chloropropane

minor product2-bromo-1-chloropropane

C

2CHBrδ–Clδ+

H

HH

C

H

C BrH

C

ClH

H

HH

C

H

C ClH

C

BrH

H

HHHH


Br is less electronegative than Cl and becomes attached to C1, which has two H atoms already attached, rather than C2, which only has one H atom already attached.

Reaction of ethene with bromine

HC

H

H HC + Br2

non-polar

solventH

Br Br

H

C

H

C H

Although the bromine is not originally polar, as it approaches the high electron density in the C=C it becomes polarised (there is an induced dipole). The electrons are repelled in the Br2 molecule so that the Br closest to the C=C has a slight positive charge and the Br further away has a slight negative charge.

The mechanism can be shown in two di� erent ways. The � rst is exactly analogous to the reaction of H–Br with ethene, i.e. via a carbocation intermediate (Figure G6).

Examiner’s tipYou only need to learn one of these mechanisms.

C

CC

H H

H H H

H

H C

H C+

BrBr Br

δ+ δ–

:Br–

H

H

H C

H C

Br

Br

Figure G6 The reaction of ethene with bromine.

The mechanism is, however, probably better represented as shown in Figure G7, where a bromonium ion rather than a carbocation is formed.

C

CC

H H

H H H

Hbromonium ion

H C

H CBr+Br Br

δ+ δ–

:Br–

H

H

H C

H C

Br

Br

Figure G7 The reaction of ethene with bromine showing the formation of a bromonium ion.

Test yourself1 State the names of the major products formed

when the following alkenes react with HCl: a but-1-ene b 3-methylbut-1-ene c 2,4-dimethylpent-2-ene

2 State the name of the major product formed when but-1-ene reacts with ClBr.

In the bromonium ion the bridging Br has donated a lone pair to the C+, which stabilises the ion.


G2 Nucleophilic addition reactionsLike alkenes, aldehydes and ketones also contain a double bond and undergo addition reactions. This time, however, the type of addition is nucleophilic addition.

Reaction with hydrogen cyanideThe reaction of ethanal with hydrogen cyanide is shown in Figure G8. Just as with the reaction of alkenes, the second component of the double bond (π component) is broken and the H–CN is added across it; H adds to the O.

2-hydroxypropanenitrile

H

C CHCN(aq) + alkali

+ HCNH

H

H

C CH

H

C

N

H

OO

H

H

Figure G8 The reaction of ethanal with hydrogen cyanide.

This reaction is important as it introduces an extra C atom into the carbon chain.

The product of the reaction is called a cyanohydrin.

The reaction is generally speeded up by addition of a small amount of alkali, which promotes formation of CN−, the nucleophile (HCN is a very weak acid and, therefore only a very small concentration of CN− is present in solution). In the laboratory it is much more convenient to use KCN and add an acid instead of using the dangerous HCN.

The reaction occurs with both aldehydes and ketones. The reaction of propanone with HCN is shown in Figure G9.

Learning objectives

• Describe and explain the nucleophilic addition reactions of aldehydes and ketones with HCN

• Write equations for the hydrolysis of cyanohydrins

2-hydroxy-2-methylpropanenitrile

H

C CHCN(aq) + alkali

+ HCNH H

H

H

C CH

H

C

N

C

O

H

H

H

HO

H

C

H

Figure G9 The reaction of propanone with HCN.

MechanismThe C=O is polar with Cδ+ and therefore attracts nucleophiles. The reaction mixture will contain CN−, either from the dissociation of HCN or from the KCN. The CN− ion has a lone pair of electrons on the C (Figure G10) and this attacks the Cδ+ of the carbonyl (C=O) group.

The mechanism for the nucleophilic addition of HCN to ethanal is shown in Figure G11.

As the CN− attacks, the second (π) component of the C=O bond breaks and the pair of electrons goes to the O. This makes the O

[ –[C N

Figure G10 The electrons of the CN− ion.


negatively charged (it has one extra electron). The O− then attracts H+ from the dissociation of HCN (or the acid that has been added).

The second step may also be shown as in Figure G12, where the H+ is removed from the undissociated HCN.

C N

H+ from dissociation of HCN

H

C CH

H

H

C CH

H

C

N

H

O–O

H

H+

H

C CH

H

C

N

H

OH

δ+

δ–

–

Nitrile groups may be hydrolysed to form a carboxylic acidIf a nitrile is boiled with aqueous acid it is converted into a carboxylic acid. The balanced equation for the reaction of ethanenitrile with acid is:

Figure G11 The mechanism for the nucleophilic addition of HCN to ethanol.

C N

H

C CH

HH

C

N

H

O–

H

C CH

H

C

N

H + CN–

OH

Figure G12 H+ is removed from the undissociated HCN.

Hydrolysis – breaking a bond with water.

H2O / H+ reflux

hydrolysisCH3 N + H2O + H3O+C + NH4

+

O HCH3 C

O

heat under reflux

2-hydroxypropanoic acid

H + H2O + H3O+C

O

C

C

N

HH

H

HO O

H + NH4+C

O

C

C

HH

H

HH

The same reaction occurs when the cyanohydrins formed above are heated with acid:

Hydrolysis can also be brought about by re� uxing with aqueous alkali. In this case the product is the salt of the acid and ammonia:

H2O / OH– reflux

hydrolysisH + H2O + NaOHC

OH

H3C

CN

C

OH

H3C

COO–Na+

H + NH3


Test yourself

Examiner’s tipNotes on drawing mechanisms:1 Curly arrows show the movement of an electron pair and

not of anything else.2 Be clear where the electron pair is moving from and where

it is going to. For instance, if OH− is attacking, the curly arrow should start from a lone pair on the O and not from the H.

3 In most mechanisms the movement of electron pairs is such that a bonding pair of electrons becomes a lone pair of electrons or a lone pair of electrons becomes a bonding pair.

C

Hpair of electrons going from being a lone pair on X to a bonding pair between C and X

Curly arrow shows a bonding pair becominga lone pair. The arrow goes from the middleof the bond to the atom.

Oδ–

δ+

X–

C

O

X–

–

CHH

O has extra electron, therefore negatively chargedO

X

Alcohols may be dehydrated to produce alkenesDehyrdation of an alcohol is an elimination reaction, involving the elimination of H2O from the molecule.

The alcohol is heated with either concentrated sulfuric acid (H2SO4) or concentrated phosphoric acid (H3PO4). The H2SO4 or H3PO4 acts as a dehydrating agent.

For example, ethanol can be dehydrated to ethene by heating with excess concentrated sulfuric acid:

G3 Elimination reactionsElimination reactions involve the removal of the components of a small molecule such as HCl or H2O from a molecule without them being replaced by other atoms or groups. This results in the formation of a C=C double bond. This has already been seen in Chapter 10, page 468 of the Coursebook, where the elimination of a hydrogen halide from a halogenoalkane was considered, for example:

conc. KOH in ethanol

heat under refluxC

H

H

C

Cl

CH3

C

H

H

HH + H

2-methylprop-1-ene

C

H

H

H3C

C

C

H

H

H

Cl

Learning objectives

• Describe and explain the elimination of water from alcohols

3 Draw the structural formulae of the products formed when the following carbonyl compounds react with HCN:

a butanal b 4-methylpentan-2-one

4 Draw the structural formulae of the products formed when each of the products formed in Question 3 is boiled with aqueous acid.

5 Write balanced chemical equations using condensed structural formulae to show the reaction of propanal with HCN followed by boiling the product with aqueous acid.


conc. H2SO4

heatH C

O

H

C

H

H

H + H2OC

H

H

H

H

C

H

The H and OH are always removed from adjacent C atoms.

See Higher Level section on page 487 of Chapter 10 in the Coursebook for an explanation of cis and trans.

Elimination by this mechanism is more likely with a tertiary alcohol than with a primary alcohol as a tertiary alcohol forms a more stable carbocation.

H

C CH H

H

H

O

H

CH3

C C

H

H

H

H

C CH H + H2O

H CH3

C C

H

H

Hconc. H3PO4

heat

3-methylbut-1-ene

H

C CH H

H

H

O

H

CH3

C C

H

H

H

H

C CH H + H2O

H CH3

C C

H

H

Hconc. H3PO4

heat

2-methylbut-2-ene

Figure G13 Possible products when 3-methylbutan-2-ol is dehydrated.

With some alcohols more than one organic product is possible, e.g. when 3-methylbutan-2-ol is dehydrated 3-methylbut-1-ene and 2-methylbut-2-ene are possible products (Figure G13).

When butan-2-ol is dehydrated but-1-ene and cis- and trans-but-2-ene are all possible products.

Mechanism for the elimination of waterThe � rst step in the mechanism is protonation of the O of the OH group by the sulfuric acid or phosphoric acid. This creates a group that can more readily leave than the OH group. In the third stage a pair of electrons from the C–H bond is used to form the second component of the C=C.

The H+ is used up in the � rst step but is regenerated in the last step. So it is acting as a catalyst – it is not used up in the reaction.

C

C

O+

H H

H H

CC

H

H

H

H

HC

O

H H

H

CC

H

H

H

H+

H

H C

H HO

H HCC

H

H

H

H

H +

H++H+ reformed

H+ from either H2SO4 or H3PO4

C

H

C

H

H H

H

H

+

Test yourself6 Name the organic products formed when the following alcohols

undergo dehydration: a butan-1-ol b butan-2-ol c 3-methypentan-1-ol.


G4 Addition–elimination reactionsThe structure of 2,4-dinitrophenylhydrazine (2,4-DNPH) is shown in Figure G14.

NO2

NO2

N

H

N HH

Figure G14 The structure of 2,4-DNPH.

Hydrazine is N2H4. Phenylhydrazine is a molecule of hydrazine where one H has been replaced by a phenyl (C6H5, benzene ring) group.

Water is eliminated.

Learning objectives

• Write equations for the reaction of 2,4-dinitrophenylhydrazine with aldehydes and ketones

2,4-dinitrophenylhydrazine reacts with aldehydes and ketones in an addition–elimination reaction. The reaction is so-called because the 2,4-DNPH adds to the aldehyde/ketone with the elimination of a molecule of water.

The equation for the reaction between 2,4-DNPH and propanone is shown in Figure G15.

NO2

NO2

N

H

N HH

H

H H

NO2

H

C C OH+

NO2

propanone 2,4-dinitrophenylhydrazone

N

H C

H H

C

H H

H

CNH

+

H

H

C

HO

Figure G15 The reaction between 2,4-DNPH and propanone.

The structure of the product of the reaction can be worked out most easily by putting the C=O group of the aldehyde/ketone next to the NH2 group of the 2,4-DNPH. Water is then removed and the C=O becomes C=N (Figure G16).

NO2

N

H

N H

H NH

NO2

H C

H H

C

H H

H

CN

H C

H H

C

H H

H

CO

Figure G16 Predicting the product of the reaction between 2,4-DNPH and propanone.


The structure of the product formed when butanal (CH3CH2CH2CHO) reacts with 2,4-DNPH is shown in Figure G17.

The reaction with 2,4-DNPH is used as a test for the presence of the carbonyl (C=O) group in aldehydes and ketones. A few drops of the aldehyde or ketone is added to an acidi� ed solution of 2,4-DNPH in methanol (sometimes called Brady’s reagent). The solution is orange and when the aldehyde/ketone is added, a yellow-orange precipitate is formed. The formation of the precipitate indicates the presence of an aldehyde/ketone group.

The product of these reactions is called a hydrazone and is named as an aldehyde/ketone 2,4-dinitrophenylhydrazone, e.g. propanal 2,4-dinitrophenylhydrazone.

If the melting point of these hydrazone crystals is determined and the values compared with those in tables, exactly which aldehyde/ketone was originally present may be determined. For example, the melting point of the 2,4-dinitrophenylhydrazone derivative of propanone is 128 °C, whereas the melting point of the derivative of propanal is 155 °C.

G5 Arenes

Benzene and aromatic compounds

N

NO2

NO2

butanal 2,4-dinitrophenylhydrazone

H C

H

CH

H

C

H

H

H H

C

NH

Figure G17 The structure of the product formed when butanal reacts with 2,4-DNPH.

Test yourself7 Draw the structures of the organic products formed

when the following carbonyl compounds react with 2,4-dinitrophenylhydrazine:

a pentan-3-one b 3-methylbutanal

Determination of the melting point of hydrazone is quite an old-fashioned method of identifying compounds and nowadays infrared/nuclear magnetic resonance spectroscopy would be used.

Compounds that contain a benzene ring are called aromatic compounds or arenes.

Figure G18 Kekulé benzene.

Learning objectives

• Explain the evidence for a delocalised structure for benzene

• Explain the relative rates of substitution of halogens atoms attached directly to a benzene ring or in the side chain

The structure of benzeneBenzene consists of a planar hexagonal ring of C atoms with one hydrogen atom joined to each C.

The structure originally proposed for benzene (C6H6) was due to Kekulé and is usually called Kekulé benzene (Figure G18). In this structure there are alternating single and double bonds between carbon atoms. A more systematic name for this molecule would be cyclohexa-1,3,5-triene or 1,3,5-cyclohexatriene.

The structure was accepted for many years but eventually the weight of evidence against it became too great and a modi� ed structure was proposed. The structure of benzene is nowadays better represented as in Figure G19.

Each C atom in this structure seems to form just three bonds (two to C atoms and one to an H atom) (Figure G20). The remaining electrons form Figure G19 Structure of benzene.


a delocalised system of six electrons, which is represented by the circle in the centre of this structure. These six electrons are not localised between individual C atoms in double bonds but rather spread over the whole ring.

Delocalisation – electrons shared between three or more atoms.

Figure G20 Benzene ring showing all atoms.

CC

C

CC

C

H

H

H

H

HHC

CC

H

H

The ring of electrons is formed when p orbitals overlap side-on to form a π delocalised system (Figure G21).

Figure G21 Formation of the benzene delocalised system.

HL

Evidence for the structure of benzene

Carbon–carbon bond lengthsAll C–C bond lengths are equal in benzene and intermediate in length between a C–C single bond and a C=C double bond (Table G1).

If the structure of benzene were cyclohexa-1,3,5-triene (Kekulé benzene) then it would be expected that there would be three short C=C bonds (approximately 0.133 nm) and three longer C–C bonds (approximately 0.154 nm).

Bond Compound Bond length / nm

C=C ethene 0.133

C–C ethane 0.154

C–C benzene 0.139

Table G1 Bond lengths in ethene, ethene and benzene.

The delocalised structure of benzene suggests a C–C bond length between that of a C–C bond and a C=C double bond as there are on average three electrons (two from the single bond and one from the delocalised electrons) between each pair of C atoms compared with four electrons in C=C and two electrons in C–C.

Thermochemical evidence When cyclohexene (C6H10) is heated with hydrogen in the presence of a nickel catalyst cyclohexane (C6H12) is formed. This is an addition reaction and hydrogen adds across the C=C double bond of the cyclohexene. The enthalpy change for this reaction is approximately −120 kJ mol−1.

C–C bond lengths may be determined by X-ray crystallography.


Cyclohexene contains one C=C whereas cyclohexa-1,3,5-triene (Kekulé benzene) contains three C=C. It would, therefore be expected that the enthalpy change for the complete hydrogenation of cyclohexa-1,3,5-triene (Kekulé benzene) to cyclohexane would be 3 × −120, i.e. −360 kJ mol−1.

+ H2

cyclohexene cyclohexane (C6H12)

∆H = −120 kJ mol−1

+ 3H2

cyclohexa-1,3,5-triene cyclohexane (C6H

12)

∆H = −360 kJ mol−1

+ 3H2

benzene cyclohexane (C6H12)

∆H = −205 kJ mol−1

The enthalpy change when benzene undergoes complete hydrogenation to cyclohexane is, however, only −205 kJ mol−1.

If an enthalpy level diagram is drawn for the hydrogenation reactions of benzene and cyclohexa-1,3,5-triene (Kekulé benzene) it is seen that benzene is 155 kJ mol−1 more stable than ‘expected’, i.e. 155 kJ mol−1 more stable than would be predicted if it had the structure with alternating single and double carbon–carbon bonds (Figure G22).

The extra stability of benzene compared to the structure with alternating double and single bonds is due to the delocalisation of electrons.

The enthalpy change of hydrogenation of cyclohexa-1,3-diene (1,3-cyclohexadiene) to cyclohexane is approximately −227 kJ mol−1. This is reasonably close to 2 × −120 kJ mol−1 and although there is some

∆H = –205 kJ mol–1

155 kJ mol–1

Enth

alpy

cyclohexane

∆H = –360 kJ mol–1

Figure G22 Comparative enthalpy changes for the hydrogenation reactions of benzene and Kekulé benzene.


delocalisation of electrons in this structure, it is not nearly as extensive as in benzene.

The number of isomers of C6H4Cl2Further evidence for a delocalised structure for benzene comes from examining the number of isomers that are possible for C6H4X2 if we assume a Kekulé or a delocalised structure (Figure G23).

Only three isomers have ever been found for C6H4Cl2 but the structure with alternating double and single bonds would suggest that there should be four. The di� erence comes because two Cl atoms which are next to each other in the ring can either be separated by a C–C single bond or a C=C double bond.

Reactions of benzeneCyclohexa-1,3,5-triene (Kekulé benzene) would be expected to undergo addition reactions (like all other alkenes) and to decolorise bromine water. Benzene does not, however, react like alkenes, i.e. it does not undergo addition reactions under normal conditions and will not decolorise bromine water.

The extra stability associated with the ring of delocalised electrons means that it is regenerated in reactions, therefore benzene undergoes substitution reactions.

For example, benzene reacts with chlorine in the presence of a catalyst such as aluminium chloride to form chlorobenzene, C6H5Cl:

+ 2H2

cyclohexa-1,3-diene cyclohexane (C6H12)

∆H = −227 kJ mol−1

Benzene Kekulé benzene

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Figure G23 Possible isomers for C6H4Cl2.

+ Cl2 + HClAlCl3

Cl

Examiner’s tipThe evidence for the structure of benzene can be classi� ed as either chemical (enthalpy changes of hydrogenation and substitution reactions) or physical (C–C bond lengths, number of isomers).


Naming aromatic compounds/arenes The names of some simple aromatic compounds are shown here.

methylbenzene

CH3

nitrobenzene

NO2

phenol

OH

benzoic acid / benzene carboxylic acid

C

O

OH

chlorobenzene

Cl

Relative rates of hydrolysis of halogenated benzene compoundsWe have already seen in Chapter 10, page 468 of the Coursebook that halogenoalkanes will undergo hydrolysis when heated with aqueous sodium hydroxide solution:

NaOH(aq)

heatC

H

H

H C

Cl

H

H + OH– C

H

H

H C

OH

H

H + Cl–

Cl

+ NaOH NO REACTION under normal conditions

NaOH(aq)

heat+ OH– + Cl–C

C Cl

H H

H H

C

C OH

H H

H H

Halogen atom attached directly to the ring.

Here the halogen atom is said to be in the side-chain.

Chlorobenzene does not undergo hydrolysis (nucleophilic substitution) reactions with sodium hydroxide under normal conditions.

To all extents and purposes halogenated benzene compounds with the halogen atom attached directly to the ring are inert to nucleophilic substitution.

When the halogen atom is not attached directly to the benzene ring, but is rather in the side-chain, hydrolysis occurs much more readily, e.g.

Worked exampleWhat is the organic product formed when the compound shown is heated with excess aqueous sodium hydroxide solution?

CClH

C

H

C

Cl

Cl

H

H

H

H


Two of the chlorine atoms are attached directly to the benzene ring and will therefore not be substituted but the third chlorine, which is in the side-chain, will be substituted by an OH.

C

H

H

H

C

OH–

Cl

H

will be substitutedattached directlyto benzene ring

NOT substituted

H

H

Cl

Cl C C

H

H

H

C

OH

H

H + Cl–H

Cl

Cl C

Extension

Halogens atoms attached directly to a benzene ring can be substituted. These reactions occur when there is another group (such as a nitro group) also attached to the ring (at position 2 or 4) and strangely enough the reaction occurs more quickly when it is a � uorine atom attached to the ring rather than chlorine. This reaction occurs by a slightly di� erent mechanism than that discussed in Chapter 10, page 468 of the Coursebook for nucleophilic substitution, and the rate-determining step does not involve breaking the carbon–halogen bond. This suggests that the explanation given here that it is the increased strength of the carbon–halogen bond that is responsible for the lack of substitution may not be strictly accurate.

The reason that halogen atoms attached directly to the benzene ring are not easily substituted is due to overlap of a lone pair of electrons on the halogen atom with the delocalised system of the benzene ring.

HL

OH–

nucleophile repelled byelectrons of delocalised system

nucleophile mustattack this C

overlap of lone pairwith benzene π system

lone pair on CI

This has two e� ects:

• it strengthens the C–Cl bond, so that it is less easily broken (the reaction involves breaking the C–Cl bond)

• nucleophiles (OH− is negatively charged) are repelled by the electrons in the extended delocalised system.


Test yourself8 Give the structure of the organic products formed when the

following molecules are heated with excess aqueous sodium hydroxide:

a b

C

H

H

C

H

H

Cl

Cl

CH2ClClH2C

Cl

G6 Organometallic chemistryOrganometallic compounds are organic molecules which also contain a metal atom. More precisely, they contain a bond between a carbon atom and a metal atom. These are useful in organic synthesis.

The most commonly used organometallic reagents are Grignard reagents. These may be formed when a halogenoalkane is reacted with magnesium turnings using a dry ether (usually ethoxyethane, (C2H5)2O) as the solvent.

H

H H

H

C

H

C O

ethoxyethane

H H

H

C

H

C H

The solvent must be dry as Grignard reagents react vigorously with water.

Learning objectives

• Describe the formation of Grignard reagents

• Write equations for reactions of Grignard reagents

Organolithium compounds are also commonly used.

Grignard is pronounced as grin-yard.

The Grignard reagent cannot be isolated from solution.

This would, however, be rather a waste of a Grignard reagent!

The general equation for the formation of a Grignard reagent is:

dry ethoxyethaneRX + Mg ⎯⎯⎯⎯→ RMgX

For example:

dry ethoxyethaneC2H5Br + Mg ⎯⎯⎯⎯→ C2H5MgBr ethylmagnesium bromide

The structure of the Grignard reagent is shown in Figure G24. There is a covalent bond between the Mg and the C and between the Mg and the Br.

C is signi� cantly more electronegative than Mg and the Grignard reagent acts in reactions as R−, that is, a negatively charged alkyl group. The alkyl group is a nucleophile and will attack positive regions in a molecule.1 Reaction of Grignard reagents with water An alkane is formed, e.g.

C2H5MgI + H2O → C2H6 + Mg(OH)I

C

H

H

H C

H

H

Mg Br

Figure G24 Grignard reagent structure.


2 Reaction of Grignard reagents with carbon dioxideA carboxylic acid is produced. The salt of the carboxylic acid is � rst formed and then the acid is generated by adding a dilute strong acid such as sulfuric acid, e.g.

C3H7MgBr + CO2 → C3H7COOMgBr propylmagnesium bromide

H+

C3H7COOMgBr + H2O → C3H7COOH + Mg(OH)Br butanoic acid

This can be shown more conveniently in a reaction scheme by putting the reagents for both steps on one arrow:

Note that the length of the carbon chain has increased by one C.

The conditions for the � rst step must be completely free of water – pass dry CO2 through the Grignard dissolved in dry ethoxyethane.

HC3H7MgBr

H

H

C

H H

H

C

H

C CO

O H

(i) CO2

(ii) H+ / H2O

H

H

H

C

H H

H

C

H

C CH

H

H

C

H H

H

C

H

C Mg BrO

O–

C

O

O

δ–

δ–

δ+

H

H

H

C

H H

H

C

H

C– C

O

O

δ–

δ–

δ+

The reactions of Grignard reagents may be more easily understood and remembered if we consider the � rst stage of the mechanism. The electrons from the C–Mg bond are used to form a bond to a δ+ C atom.

We could also show this more simply as a negatively charged alkyl group attacking a δ+ C.

3 Reaction of Grignard reagents with aldehydes and ketonesAlcohols are formed. The product of the initial reaction is the anion of the alcohol and this is protonated by the addition of aqueous acid to form the alcohol. The reaction between ethylmagnesium iodide and propanal forms pentan-3-ol:

Examiner’s tipThe mechanism is not required for the examination.

H

C2H5Mgl

H

H

C

H O

H

C CH

H

H

H

C

H O

H

C

H

C

H

H

Hpentan-3-ol

C

H

H

C H + Mg(OH)l

(i)

(ii) H+ / H2O

Note the increase in the length of the carbon chain.


Ketones form tertiary alcohols when reacted with Grignard reagents:

Aldehydes (with the exception of methanal) will form secondary alcohols when reacted with a Grignard reagent. Methanal forms a primary alcohol:

HHC H

O

H H

H

C

H

C

H

H

C

H

H

C H

O

Mg

Br

H

C

H

H

C C

H

H

HH + Mg(OH)Br

butan-1-ol

(i)

(ii) H+ / H2O

O

C

H

H

C

H

H

C

H

H

H

C

H

H

C HMg

Br

H

C

H

H

C HH + Mg(OH)Br

3-methylhexan-3-ol

(i)

(ii) H+ / H2O

O

C

H

H

C

H

C

H

H

C

H

H

H

HH

C

C H

H

H

H

H

C H

Again, these reactions can be understood more easily from a consideration of the initial stage in the mechanism, which can be regarded as involving a negatively-charged alkyl group attacking the δ+ C of the carbonyl group (Figure G25).Some reactions of Grignard reagents are shown in Figures G26 and G27.

δ+ δ–

H

H

H C

H

H

C–

C

H

H C

H

H

H

C

H

O

Figure G25 The initial stage in the reaction of an ethyl Grignard with propanal. In the reactions of Grignard reagents with CO2, aldehydes and ketones, the alkyl group of the Grignard adds to the C of the C=O and the C=O is converted to an alcohol group.

H2O CO

2

methanal aldehyde

ketone

alkane carboxylic acidhalogenoalkanedry ethoxyethaneMg

secondary alcohol

tertiary alcohol

Grignard reagentprimary alcohol

Figure G26 Some reactions of Grignard reagents.


R2

C

C

H

C

C

H

CH

H

H

C R1

R2

H

H

H C

C

C

H

CH

H

H

C R1

R2

H

H

H

H

C

H

C

CH

HHC

H

C

H

CH

H O

H

H

C R1

R1

R2

H

H tertiary alcohol

secondary alcohol

ketone

aldehydemethanal

primary alcohol

(i)

(i) (i)

conc

. H3P

O4

/ he

at

elim

inat

ion

mec

hani

sm

Mg

dry

etho

xyet

hane

H

C R2

R1

H

H

C

H

O

C

H

(ii) H+/H2O

(ii) H+/H2O (ii) H+/H2O

H2O(ii) H+/H2O

(i) CO2

H

H

C

H

C

H

H

CH

HH

CR2

H

H

C

HO

CH

H

H

C

O

CH

H

O

C H

R2

O

C

R1

CH

H HHC

R1 R2

CH

H H

C

R1 Br

CH H

H R2

C

R1 C

CHH R2

H Mg

Br

HHH

C

R1 C

C R2

H Br

HHH

H C

R1 C

C R2

H

HH

C

R1

C

CH R2

H H

H

OH H

R2C

R1 Mg

C

Br

H

H H

HO

CR1 CH

HH

H

CH

H

C

H

H

conc. H3PO4 / heat

elimination mechanism

conc. H3PO4 / heat

elimination mechanism

Mg

dry ethoxyethane

electrophilicaddition

HBrMarkovnikov’s rule

Figure G27 Reaction pathways for Grignard reagents and dehydration of alcohols.

Test yourself9 Draw the structure of the � nal organic products

formed when CH3CH2CH2MgBr reacts with the following compounds:

a CO2

b propanone c CH3CH2CHO

10 a Name two aldehydes that could be reacted with suitable Grignard reagents to produce pentan-2-ol.

b True or false? 2-Methylpropan-2-ol can be made by reacting propanal with a suitable Grignard reagent.

c Give the structural formula of a Grignard reagent that can be reacted with propanone to form 2-methylhexan-2-ol.


G7 Reaction pathwaysWe can use the reactions discussed in the previous section to design reaction pathways.

Worked examplesDesign a reaction pathway showing all reagents and conditions for the conversion of 1-bromobutane to pentanoic acid.

The � rst thing we notice here is that the product has one more C atom than the starting material. This suggests that a Grignard reagent must be used. The � nal product is also a carboxylic acid, therefore it can be formed by the reaction between a Grignard reagent and carbon dioxide.

This synthesis could also have been brought about by reacting the halogenoalkane with CN− (Chapter 10, page 467 of the Coursebook) then hydrolysing the product by boiling with aqueous acid.

Design a reaction pathway showing all reagents and conditions for the conversion of molecule A to molecule B.

C

H

H

C

H Mg

H

CH

H

H H

C

Br

HC

H

H

CH

H

H

H Br

H

C

H

C H Mgdry ethoxyethane

(i) CO2

(ii) H+ / H2O

+

C

H

H

C

H H

H

CH

H

H H

+ Mg(OH)BrCO H

CO

HL

C

H

H

CH

H

H

O H

C

HH C H

C C

H

H

H

H

H

Br

C

A

H

B

Learning objectives

• Deduce reaction pathways using the reactions in the sections above.


Test yourself 11 Give reaction sequences for the following

transformations: a chloroethane to propanoic acid b ethanal to 2-hydroxypropanoic acid c ethanal to butan-2-ol

12 Give reaction sequences for the following transformations:

a 2-chloropropane to propane b 2-chloropropane to 2-methylpropanoic acid c propan-1-ol to 2-chloropropane

G8 Acids and bases

AcidityCarboxylic acids are weak acids that dissociate partially in aqueous solution according to the equation:

Learning objectives

• Explain the relative acidities of substituted carboxylic acids

• Explain the relative acidities of substituted phenols

• Explain the relative basicities of ammonia and amines+ H2O + H3O+R

O

O HC R

O

carboxylate ion

O–

C

The stronger an acid the more it dissociates in aqueous solution (position of equilibrium lies more to the right). Acid strength is sometimes described using the pKa value. The lower the pKa value the stronger the acid. pKa is a log scale and therefore a change in pKa of one unit indicates a ten-fold

The product of the reaction is a tertiary alcohol with more carbon atoms than the original starting material and is therefore formed in the reaction between a Grignard reagent and a ketone.

If we highlight the carbon skeleton of the starting material in the � nal molecule we can see with which ketone we need to react the Grignard reagent.

The Grignard reagent must therefore be reacted with pentan-3-one.

C

H

H

CH H

H

H

O H

C

HH C H

C C

H

H

H

from ketone

from Grignardreagent

C

H

H

CH H

H

H

O H

C

H

C C

H

H

C

H

H

CH H

H

H

O H

C

H

C C

H

H

H

C

H

H

CH

H

H

O H

C

HH C H

C C

H

H

H

H

H

Mg

C

Br

HH

Br

C

A + Mg

H

B + Mg(OH)Brdry ethoxyethane

C

H

H

CH H

H

H

O H

C

H

C C

H

H

(i)

(ii) H+ / H2O


Acid pKa

methanoic acid 3.75

ethanoic acid 4.76

propanoic acid 4.87

butanoic acid 4.82

pentanoic acid 4.86

chloroethanoic acid 2.86

dichloroethanoic acid 1.29

trichloroethanoic acid 0.65

iodoethanoic acid 3.17

2-chlorobutanoic acid 2.84



phenol 10.00

ethanol 16

4-methylphenol 10.26

2,4,6-trinitrophenol 0.42

Table G2 pK a values.

The carboxylate ion actually has both C–O distances equal and between the length of a C–O single bond and a C=O double bond. It is, therefore, better described by the delocalised structure:

RO

OC –

HL

change in acid strength. The pKa value for methanoic acid is approximately one less than that for ethanoic acid and, therefore, methanoic acid is roughly ten times stronger than ethanoic acid (Table G2).

CO

O

HC

ethanoic acid

H

H

H

CO–

OC + H+

ethanoate ion

H

H

H

CO

O

HC

chloroethanoic acid

H

Cl

H

CO–

OC + H+

chloroethanoate ion

H

Cl

H

The di� erence in acidity can be explained by considering the stability of the anion (conjugate base) formed. The Cl is a very electronegative atom and, in the chloroethanoate ion, pulls electron density away from the COO− group (Figure G28). Since the COO group is less negative, it attracts the H+ ion back less strongly. The conjugate base is, therefore, more stable. A more stable conjugate base indicates a greater tendency to dissociate and, therefore a stronger acid.

The total negative charge in the anion is still 1− but it is spread out more over the whole anion in the chloroethanoate ion rather than being localised on the COO− group.

Chloroethanoic acid is a stronger acid than ethanoic acidThe dissociation of ethanoic acid and chlorethanoic acid may be shown as:

The greater acidity of chloroethanoic acid is sometimes explained in terms of the stability of the acid. The Cl atom is electron-withdrawing and pulls electron density from the COOH group. This makes the H of the COOH group more positive and thus more likely to be lost as a proton.

Trichloroethanoic acid is a stronger acid than chloroethanoic acid

O–

OCl

CH

Helectron density withdrawn

from COO– group

C

Figure G28 The chloroethanoate ion.

An electron-withdrawing e� ect is sometimes called a negative inductive e� ect.

CO

O

HC

trichloroethanoic acid

Cl

Cl

Cl CO–

OC + H+

trichloroethanoate ion

Cl

Cl

Cl

Again, we will consider the stability of the anion/conjugate base formed. Three very electronegative Cl atoms have a greater electron-withdrawing e� ect than just one. More electron density is thus pulled away from the COO− group making it less negative and less likely


Figure G29 The trichloroethanoate ion.

O–

OCl

CCl

Clmore electron density

withdrawn from COO– group

C

to attract back the H+ (Figure G29). The trichloroethanoate ion is thus relatively more stable than the chloroethanoate ion and so trichloroethanoic acid dissociates more.

Chloroethanoic acid is a stronger acid than iodoethanoic acid

CO

O

HC

iodoethanoic acid

I

H

H CO–

OC + H+

iodoethanoate ion

H

I

H

CO

O

HC

X

H

H

Iodine is a less electronegative atom than chlorine and thus withdraws less electron density from the COO− group. The COO group is thus more negative in the iodoethanoate ion and the H+ ion is attracted back more strongly. The conjugate base is less stable and the acid is weaker. Iodoethanoic acid is, however, still a stronger acid than ethanoic acid because of the electron-withdrawing nature of the I atom.

In general, for compounds of the form:

if X is an electron-withdrawing group, the acid will be a stronger acid than ethanoic acid due to stabilisation of the anion/conjugate base formed when the acid dissociates. If X is an electron-releasing group the acid will be a weaker acid than ethanoic acid due to destabilisation of the anion/conjugate base formed.

An electron-releasing group will increase the electron density on the COO group in the anion and the H+ ion will thus be attracted back more strongly. The conjugate base is therefore less stable and the acid weaker.

Ethanoic acid is a weaker acid than methanoic acidEthanoic acid has an electron-releasing group attached to the COO−group in the anion. This increases the electron density on the COO group and therefore the H+ is attracted back more strongly (Figure G30). This means that the conjugate base is less stable for ethanoic acid and, it is, therefore a weaker acid.

CO

O

HC

ethanoic acid

H

H

H

CO–

OC + H+

ethanoate ion

H

H

H

HO

O

HC

methanoic acid

HO–

OC + H+

methanoate ion

The electron-releasing e� ect of an alkyl group is often called a positive inductive e� ect.

O–

OH

CH

Helectron density donated

to COO– group

C

Figure G30 The ethanoate ion.


Beyond an ethyl group, the size of the alkyl group makes little di� erence to the electron-releasing e� ect. Thus propanoic acid is slightly stronger than ethanoic acid (the electron-releasing e� ect of an ethyl group is slightly greater than that of a methyl group) but propanoic acid, butanoic acid and pentanoic acid all have very similar acid strengths.

The effect of distance of the Cl group from the COOH group on acid strength2-Chlorobutanoic acid is a stronger acid than 4-chlorobutanoic acid.

The further the Cl is from the COO− group in the anion, the smaller the electron-withdrawing e� ect it has on it.

Phenol is a signifi cantly stronger acid than aliphatic alcohols such as ethanolThe dissociation of phenol may be shown as:

C6H5OH C6H5O− + H+

or, showing structures:

The dissociation of ethanol is:

O H

2-chlorobutanoic acid

O–

OC + H+C

Cl

H

CC

H

H

H

H

H

OCC

Cl

H

CC

H

H

H

H

H

O H

4-chlorobutanoic acid

O–

OC + H+C

H

H

CC

H

H

H

Cl

H

OCC

H

H

CC

H

H

H

Cl

H

Extension

Propanoic acid, butanoic acid and pentanoic acid all have very similar acid strengths because of the mechanism by which alkyl groups donate electrons. This arises because of the overlap of electron density in a σ bond (either C–H or C–C) with a p orbital on an adjacent atom – σ-conjugation/hyperconjugation.

Phenol reacts with blue litmus and forms salts with sodium hydroxide. It is still however a weaker acid than carboxylic acids. Phenol may be distinguished from carboxylic acids by the fact that it will not liberate CO2 from sodium hydrogencarbonate solutions, i.e. it is a weaker acid than carbonic acid.

Phenol is more soluble in sodium hydroxide solution than in water. It reacts with the sodium hydroxide to form an ionic salt (C6H5O−Na+) and therefore there are stronger interactions with water molecules.

H

phenol

+ H+

O

phenoxide ion

O–

ethanol ethoxide ion

H

H

C

H

H

C

H

OH

H

H

C

H

H

C O–+ H+H

Again we will consider the stability of the anion/conjugate base. The lone pair on the O− in the phenoxide ion overlaps with the delocalised system in the benzene ring (Figure G31). This spreads out the negative charge in the ion. The O is thus less negative and the H+ is not attracted back as strongly. The anion is thus stabilised and phenol is a stronger acid than ethanol. In ethanol this delocalisation does not occur and indeed the negative charge on the O in the ethoxide ion is increased by the electron-donating alkyl group.


Acidity of substituted phenolsIn general, for compounds of the form

electron densitydonated into ring

lone pair on O

Figure G31 The negative charge on O is delocalised into the ring.

OH

X

if X (in any position) is an electron-withdrawing group the compound will be a stronger acid than phenol. The electron-withdrawing group reduces the amount of electron density on the O in the conjugate base – the charge is more delocalised in the anion – meaning that the H+ is less strongly attracted and the anion is more stable. If X is an electron-donating group the compound will be a weaker acid than phenol. In this case the charge is less delocalised in the anion. With the charge more localised on the O, the H+ is attracted back more strongly and the anion is less stable.

4-methylphenol is a weaker acid than phenol

H

4-methylphenol

+ H+

O

4-methylphenoxide ion

O–

CH3

CH3

The methyl group is electron-releasing (positive inductive e� ect) and increases the amount of electron density in the aromatic ring. Delocalisation of the negative charge on O in the anion of 4-methylphenol is reduced as the ring is more negative than in phenol. There is thus more electron density on the O in the 4-methylphenoxide ion so that the H+ is attracted back more strongly and the anion is less stable.

2,4,6-trinitrophenol is a stronger acid than phenol

H

2,4,6-trinitrophenol

+ H+

NO2O2N

O

2,4,6-trinitrophenoxide ion

NO2

NO2O2N

O–

NO2

2,4,6-Trinitrophenol is such a strong acid that its trivial name is picric acid.


The NO2 group is electron-withdrawing and will therefore reduce the amount of electron density in the aromatic ring. There is thus greater delocalisation of the negative charge from the O− into the ring. With lower charge on the O the H+ is attracted back less strongly and the anion is more stable for 2,4,6-trinitrophenol.

The NO2 group is electron-withdrawing because it consists of three very electronegative atoms and there is no lone pair on the N that could be drawn into the ring. The Lewis structure for the NO2 group attached to a benzene ring is shown in Figure G32a. A dative covalent bond is formed from the N to the left hand O in this diagram and so the N is often shown with a positive charge (as it is has essentially given one electron to the O) (Figure G32b).

Oa

NO O

b

+N

–O

Figure G32 Lewis structure for (a) the NO2 group attached to a benzene ring and (b) showing the charges.

Extension

Some of the electron-withdrawing e� ect of the NO2 group also comes from a π e� ect where the p orbitals which make up the delocalised system in the benzene ring overlap with the π system of the NO2 group.

BasicityA base (Brønsted–Lowry de� nition) is a proton acceptor. Thus, when a base is put into water it accepts a proton from water:

B: + H2O HB+ + OH−

The more the position of equilibrium lies to the right, the stronger the base.

Methylamine reacts with water to form the methylammonium ion:

CH3NH2 + H2O CH3NH3+ + OH−

methylamine methylammonium ion

pKb is used to quantify the strength of bases. The lower the value of pKb, the stronger the base (Table G3).

Base Formula pKb

ammonia NH3 4.75

methylamine CH3NH2 3.36

dimethylamine (CH3)2NH 3.28

trimethylamine (CH3)3N 4.20

ethylamine CH3CH2NH2 3.27

propylamine CH3CH2CH2NH2 3.16

butylamine CH3CH2CH2CH2NH2 3.39

phenylamine C6H5NH2 9.38

Table G3 The formula and pKb of various bases.


δ–δ+NH

CH3

H3C

In all of these compounds the N is the most electronegative atom and will have a small negative charge.

H+ ions will thus be attracted to the N and the lone pair of electrons will be used to form a dative covalent bond to the H+. In general, the greater the negative charge on the N the more it will attract H+ ions and the stronger base it will be.

CH3 is an electron-releasing group (positive inductive e� ect) and so will push electron density onto the N, making it more negative and therefore more likely to attract H+. Methylamine is thus a stronger base than ammonia (Figure G33).

Similarly, dimethylamine is a stronger base than methylamine as it has two electron-releasing alkyl groups which donate electrons to the N making it even more negative (Figure G34).

It would therefore be expected that trimethylamine would be a stronger base than dimethylamine. However, although it is a stronger base than ammonia it is a weaker base than methylamine and dimethylamine. This is due to greater stabilisation of the protonated form by hydrogen bonding with water (the solvent) in the dimethylammonium ion than in the trimethylammonium ion. There are two H atoms that can hydrogen bond to water in the dimethylammonium ion but only one in the trimethylammonium ion (Figure G35). The protonated form is thus stabilised to a greater extent for dimethylamine than for trimethylamine and this outweighs the extra electron-releasing e� ect of the extra alkyl group. This means that the protonated form is more likely to be formed for dimethylamine and it is therefore ionised to a greater extent and a stronger base.

Nδ–δ+

H3C

H

δ+H

Nδ–δ+

δ+H

H

δ+H

δ–δ+NH

δ+H

H3C

Figure G33 Methylamine.

Figure G34 Dimethylamine.

δ– δ+

δ+

δ+

N+Hδ+H

CH3

a

CH3Hδ+

H

Hδ+

H

O δ–O

δ– δ+

δ+

δ+

N+H

CH3

b

CH3

CH3

H

H

O

Figure G35 Hydrogen bonding of the H atoms in the (a) dimethylammonium ion and (b) trimethylammonium ion with water.

This can also explain why methylamine is a lot stronger than ammonia but dimethylamine is only slightly stronger than methylamine.

Phenylamine is a signifi cantly weaker base than ammonia

C6H5NH2 + H2O C6H5NH3+ + OH−

phenylamine phenylammonium ion

This is because of the overlap of the lone pair on the N with the benzene delocalised system, which makes the lone pair less available for donation to H+.

Formation of salts Amines, like ammonia, form salts with acids, e.g.

NH3 + HCl → NH4+Cl−

ammonium chloride

Amines are more soluble in hydrochloric acid than in water due to the formation of the ionic salt and, therefore, stronger interactions with water molecules.


CH3NH2 + HCl → CH3NH3+Cl−

methylammonium chloride

An alkylammonium salt can be converted back to an amine by reaction with a strong base such as sodium hydroxide:

CH3NH3+Cl− + NaOH → CH3NH2 + H2O + NaCl

This can be shown as an ionic equation:

CH3NH3+ + OH− → CH3NH2 + H2O

G9 Addition–elimination reactions

Acyl chlorides (acid chlorides)The basic structure of an acyl chloride is:

The alkylammonium ion is acting as a Brønsted–Lowry acid. NaOH is a stronger base than the amine and therefore can take the proton away from it. Test yourself

13 Arrange the following in order of acid strength (weakest � rst): phenol propanoic acid 2-chloropropanoic acid 2,2-dichloropropanoic acid ethanol 2,2-di� uoropropanoic acid 3-chloropropanoic acid 2-methylphenol

HLLearning objectives

• Describe the reactions of acyl chlorides with nucleophiles

• Explain the mechanism for the reaction of acyl chlorides with nucleophiles

• Describe the reactions of acid anhydrides with nucleophiles

R CO

Cl

O

Cl

H

CH

Hethanoyl chloride

CO

Cl

H

CC

Hbutanoyl chloride

H

C

H

H

H

H

CO

Cl

H

CC

H3-methylbutanoyl chloride

CH3

C

H

H

H

H

C

Acyl chlorides may be prepared by reacting carboxylic acids with SOCl2 or PCl5.

They are named as alkanoyl chloride, e.g.

These are extremely reactive and react to add an acyl group to nucleophiles with the elimination of HCl.

OR

acyl group

C

O

Cl

H

CH

Hethanoyl chloride

CO

H

CH

Hethanoic acid

ClHC+ +HH

OO H

1 Reaction with water Acyl chlorides react violently with water to from carboxylic acids.


In all the reactions we will meet the acyl group will be added to the most electronegative atom of the nucleophile (N or O) and H (from the nucleophile) and Cl (from the acyl chloride) will be eliminated (Figure G36).

2 Reaction with alcohols Acyl chlorides react with alcohols to form esters.

Figure G36 Reaction of an acyl chloride with water.

ClH+ +O

Cl

H

CC

Hbutanoyl chloride

H

C

H

H

H

H

H

C HC

H

H

O

H

H

C

H

CC

Hethyl butanoate

H

C

H

H

H H

H

H

CC

H

H

O

H

O

C

HL

Cl

H O

CH

H

CH

acyl group added to O

HCl eliminated

O H

This reaction is usually carried out in the presence of a base such as pyridine (C5H5N).An alkaline solution of phenol reacts with benzoyl chloride to form

phenyl benzoate:

ClH+ +

Cl O

O

COH O

C

phenyl benzoate

ClH++

H

H

C

H

H

C

H

CHO

ClN

HH

propanamide

H

H

C

H

H

C CHO

NH2

A more accurate way of writing this reaction would be as:

CH3CH2COCl + 2NH3 → CH3CH2CONH2 + NH4Cl

HCl would not be formed in the presence of a base such as ammonia but rather the salt ammonium chloride.

3 Reactions with ammoniaWhen concentrated ammonia solution is added to an acyl chloride at 0 °C a primary amide is formed.

4 Reactions with amines When acyl chlorides are reacted with amines (again at low

temperature) N-substituted amides are formed.

ClH++

H

H

C

H

H

C CHO

Cl H

HH

H

C

H

H

C HN

H

H

H

C

H

H

C HN

N-ethylpropanamide

ethylamine

H

H

C

H

H

C CHO

Again, the balanced equation should, more correctly, be written as:

CH3CH2COCl + 2CH3CH2NH2 → CH3CH2CONHCH2CH3 + CH3CH2NH3Cl


The addition–elimination mechanismThe mechanism of these reactions involves two steps:1 initial addition of the nucleophile 2 elimination of the elements of HCl.

The basic mechanism can be illustrated by representing the nucleophile as X−:

C

H

H X–

H

O

CCl

δ–

δ+ C

H

H

H

O

C + Cl–

XC

H

H X

CH

O–

Cl

The nucleophile attacks the δ+ C of the C=O group. The π component of the C=O breaks with the pair of electrons going to the O to generate O−. A lone pair from the O then is used to re-form the double bond and at the same time the C–Cl bond breaks.

Now, if we look at the reaction with water:

C

H

H

H

O

CCl

δ–

δ+

δ+

δ+

+ Cl–

HH

C

H

H O+

HHO

CH

O–

Cl

H

HO

C

H

H

H

O

CO H

H

C

H

H O

HHO+

CH

O–

H

Cl

O + H+

HH

C

H

H O+

CH

O–

Cl

H

C

H

H

CH

O–

Cl

C

H

HH

H

O

CCl

δ–

δ–

δ+

δ+

δ+

δ+

+ Cl–

HH

C

H

H N+

HHN

HHN

C

H

H

O–

HN H

Cl

H

H

C

H

HH

H

O

CN H

+

C

H

H

HHN+

CH

O–

H

Cl

The second stage could also have been shown more simply as loss of a proton (H+):

In the second stage H+ is removed by interaction with a base. In the reaction of an alcohol with an acyl chloride the reaction is conducted in the presence of a base. In the reaction with ammonia the proton is removed by the ammonia/amine.

HL

Acid anhydridesThe basic structure of an acid anhydride is:

R C

R C

O

O

O


This can be regarded as being formed from two molecules of carboxylic acid with water removed (Figure G37).

Acid anhydrides are named according to the carboxylic acid from which they are derived, e.g.

O

O

H

CH

H

CH

O

O

H

CH

H

CH

Figure G37 Formation of an acid anhydride.

H3C

H3C

C

C

O

O

O

ethanoic anhydride

CH3CH2

CH3CH2 C

C

O

O

O

propanoic anhydride

Mixed anhydrides are also possible and they would be named as, e.g. ethanoic propanoic anhydride.

Like acyl chlorides, acid anhydrides are acylating agents – they react by adding the acyl group to other species. This time it is a molecule of carboxylic acid that is eliminated.

The reactions of acid anhydrides are basically the same as those of acyl chlorides.1 Reaction with water

Acid anhydrides react when warmed with water to form carboxylic acids:

Acid anhydrides are actually made by the reaction between an acyl chloride and the sodium salt of a carboxylic acid.

Note that acid anhydrides are less reactive than acyl chlorides.

The systematic name of aspirin is 2-ethanoyloxybenzenecarboxylic acid!

Aspirin contains an ester linkage.

H3C

H3C

C

C

O

OO

O

ethanoic anhydride ethanoic acid

+ +

H

H

H

C CO

OO

HH H

ethanoic acid

H

H

H

C CO

O H

O

propanoic acidethyl propanoate

+ + C

H

H

C

H

H

CH

O

C O

H

H

C

H

H

C H

H

H

C

H

H

CH

H

H

C H

H

H

C

H

OO

O HCH3CH2

CH3CH2 C

C

O

O

propanoic anhydride

2 Reaction with alcoholsAcid anhydrides react when heated with alcohols to form esters:

HL

ethanoic acid

+ + C

H

H

CHO

O H

H

H

H

CCO

OH3C

H3C

C

C

O

O

O

ethanoicanhydride

COO

H

O H

2-hydroxybenzoic acid

COO

H

aspirin

Ethanoic anhydride can be heated with 2-hydroxybenzoic acid to make aspirin:


4 Reactions with amines When acid anhydrides are reacted with amines N-substituted amides

are formed:

O

propanoic acidN-ethylpropanamideethylamineethanamine

+ + C

H

H

C

H

H

CHO

OC

H

H H

C

H

H

CHO

N H

H

H

C

H

H

C

H

HN H

H

H

C

H

H

CHCH

3CH

2

CH3CH

2C

C

O

O

propanoic anhydride

N

ethanoic acid

+ + C

H

H

CHO

O H

H3C

H3C

C

C

O

O

O

ethanoicanhydride4-aminophenol

HCH3C

OH

NH2

OH

paracetamolO

Ethanoic anhydride reacts with 4-aminophenol at room temperature to form paracetamol:

The systematic name of paracetamol is N-(4-hydroxyphenyl)ethanamide!

3 Reactions with ammonia When excess concentrated aqueous ammonia solution is reacted with

an acid anhydride a primary amide is formed:

O

propanoic acidpropanamide

+ + C

H

H

C

H

H

CHO

OC

H

H

C

H

H

CHO

NH2H

HHN

HCH3CH2

CH3CH2 C

C

O

O

propanoic anhydride

HL

Test yourself 14 All the following compounds can be made in a single step

using either an acyl chloride or an acid anhydride. Give the names of two compounds from which each of the following compounds could be made:

a butyl ethanoate b pentanamide c N-ethylbutanamide


In the � rst step a pair of electrons from the benzene π delocalised system is used to form a bond to the electrophile. In order to form a bond the C atom must use one of the orbitals that was previously forming part of the π delocalised system. The delocalised system can then only extend over the other � ve C atoms (Figure G38). This is shown by the partial ring. There were originally six electrons in the π delocalised system (one from each C atom) but two are now used to form the C–X bond, which leaves only four electrons shared over � ve C atoms; the ring thus has a positive charge.

X

H

In the second step the C–H bond breaks. This releases an orbital and a pair of electrons to complete the π delocalised system again.

The nitration of benzeneWhen benzene is heated with a mixture of concentrated nitric and sulfuric acids, nitrobenzene is formed:

Figure G38 Partial delocalisation in the intermediate in an electrophilic substitution reaction.

Another way of thinking about this is that the initial reaction involved a neutral species reacting with a positively charged species, therefore the intermediate must also have a positive charge.

conc. H2SO4

heat under reflux60°C

NO2

+ HNO3

nitrobenzene

+ H2O

conc. HNO3

C6H6 + HNO3 → C6H5NO2 + H2O

Formation of the electrophileThe electrophile in this reaction is the NO2

+ ion, which is formed when the concentrated sulfuric acid reacts with the concentrated nitric acid:

HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4

−

Learning objectives

• Describe the chlorination, alkylation, acylation and nitration reactions of benzene

• Explain the mechanism for electrophilic substitution

• Describe the chlorination, alkylation, acylation and nitration reactions of methylbenzene

• Explain the relative rate of electrophilic substitution in benzene and substituted benzene rings

• Explain the position of substitution in substituted benzene rings

G10 Electrophilic substitution reactionsBenzene reacts with electrophiles via an electrophilic substitution reaction. The general mechanism is:

X+ +

electrophile

X + H+X

H

HL

A simpler version of this equation is:

HNO3 + H2SO4 → NO2+ + H2O + HSO4

−


The mechanism is basically the same as shown in the general mechanism on page 33:

The second stage may also be shown slightly di� erently with the HSO4

− ion (formed when the electrophile was formed) removing the H+ ion. The H2SO4 is thus re-formed and can be regarded as a catalyst in this reaction.

The reaction must be carried out in the absence of direct sunlight; otherwise a free-radical reaction can occur to produce C6H6Cl6.

AlCl3 acts as a Lewis acid.

NO2+ +

NO2

electrophile

NO2 + H+

H

NO2NO2 + H2SO4H O

O

O–

OS

H

+

Chlorination of benzeneThe overall equation for the reaction between benzene and chlorine is:

C6H6 + Cl2 → C6H5Cl + HCl

Benzene does not react with halogens in the dark because the halogen molecule is non-polar and not a su� ciently strong electrophile. Therefore a catalyst is required which polarises the molecule. The catalysts used are AlCl3 or iron (forms FeCl3); these are known as halogen carriers.

Formation of the electrophileAluminium chloride is electron de� cient (only six electrons in the outer shell of the Al) and accepts a pair of electrons from the Cl2. This is often shown as:

Cl2 + AlCl3 → [AlCl4]− + Cl+

For bromination, iron or pyridine could be used.

All reactions involving AlCl3 must be carried out in the absence of water as AlCl3 reacts with this.

Examiner’s tipThis is the equation you need for the examination.

Cl+ is the electrophile.

HL

Extension

Actually it is unlikely that Cl+ is actually formed in the reaction mixture and it is more likely that the Cl2 is just very polarised by coordination to the AlCl3.

Cl

Cl

Cl Cl Al Clδ+


The mechanism is basically the same as the general mechanism given on page 33:

+Cl+

electrophile

Cl + H+ClH

+Cl

Cl + HCl + AlCl3

catalyst re-formedH

Al–Cl Cl

Cl

Cl

+ HCl+ CH3

CH3

AlCl35 °C

Cl

CH3C

CH3

CH3

CH3C

(1,1-dimethylethyl)benzene

The second step can also be shown with the [AlCl4]− removing the proton to reform the catalyst.

Alkylation (also known as Friedel–Crafts alkylation)Halogenoalkanes react with benzene in the presence of a halogen carrier catalyst. Consider the reaction between 2-chloro-2-methylpropane and benzene:

(CH3)3CCl + C6H6 → C6H5C(CH3)3 + HCl

Formation of the electrophile

(CH3)3CCl + AlCl3 → (CH3)3C+ + [AlCl4]−

For tertiary halogenoalkanes it is likely that the electrophile is the carbocation, but for secondary and primary halogenoalkanes it is probably a highly polarised complex. For the examinations, however, the electrophile can be regarded as R+ where R is the alkyl group of the halogenoalkane.

The mechanism is the same as for halogenation.

HL The � rst stage in this mechanism can be regarded as a Lewis acid–base reaction: the benzene ring is the Lewis base (electron pair donor) and the Cl+ the Lewis acid (elecron pair acceptor).

CH3

CH3

CH3

C CH3

CH3

CH3

CCH3

CH3

CH3

C

electrophile

+ H+

H++

CH3

CH3

CH3

CCH3

CH3

CH3

C+ HCl + AlCl3H

catalyst re-formed

Al–Cl Cl

Cl

Cl

+

Again, the second step could be shown with [AlCl4]− removing the H+ to reform the catalyst.


+ HClAlCl350 °C

C

C

phenylethanone

O

Cl

H

C+ H

H

OH

CH

H

O

CH3

C

CH3

CCH3

C

electrophile

+ H+

H

O

O++

Acylation (also known as Friedel–Crafts acylation)Acyl halides react with benzene in the presence of a halogen carrier catalyst. Consider the reaction of ethanoyl chloride with benzene:

CH3COCl + C6H6 → C6H5COCH3 + HCl

Formation of the electrophileThis is a similar reaction to that already seen on pages 34–35 for the formation of the electrophiles for chlorination and alkylation.

CH3COCl + AlCl3 → CH3CO+ + [AlCl4]−

The mechanism is the same as for alkylation.

The benzene ring here is named as a phenyl substituent.

Again, the second step could be shown with [AlCl4]− removing the H+ to reform the catalyst.

Reactions of methylbenzeneMethylbenzene reacts in the same way as benzene, via an electrophilic-substitution mechanism. The conditions for the reactions of methylbenzene are slightly milder than those for the reactions of benzene as the methyl group is an activating group. The methyl group donates electron density into the benzene ring (positive inductive e� ect). This increases the amount of electron density in the ring so that it is more attractive to electrophiles and reacts more readily.

The methyl group is a 2,4-directing group and so the major products of substitution are:

HL

CH3

X

CH3

X

Some of the electrophilic substitution reactions of methylbenzene are shown in Figure G39.


CH3

Cl

O

CO

H

H

C

H

C CH3

O

CH3

NO2

CH3

Cl

Cl

CH3

NO2

CH3

CH3

CH2CH3

CH2CH3

CH3

C

CH3

CH3

H3C

conc. HNO3

conc. H2SO4

heat under reflux

AlCl3

AlCl3 C2H5Cl

AlCl3Cl2

Figure G39 Electrophilic substitution reactions of methylbenzene.

HL

Chlorination of methylbenzeneIf methylbenzene is reacted with chlorine in the presence of a halogen carrier catalyst (AlCl3) at room temperature 2-chloromethylbenzene and 4-chloromethylbenzene are formed. This is chlorination in the ring.

CH3 CH3 CH3

Cl

Cl2

AlCl3

2-chloromethylbenzene

4-chloromethylbenzene

+ HCl+

Cl

C

Cl2

UV

(chloromethyl)benzene

+ HCl

H

H H C

H

H Cl

If methylbenzene is reacted with chlorine in the presence of UV light then side-chain substitution occurs where a hydrogen atom in the methyl group is substituted by a Cl atom. This involves a free radical substitution mechanism as for alkanes.

Examiner’s tipThe mechanism here is as for alkanes (see Chapter 10, page 434 of the Coursebook) and can be summarised as UVCl2 ⎯→ 2Cl•

RCH3 + Cl• → RCH2• + HCl

RCH2• + Cl2 → RCH2Cl + Cl•

plus a termination step.


C

Cl2

UV

(chloromethyl)benzene

H

H H C

H

H Cl

Cl2

UV

(dichloromethyl)benzene

C

Cl

H Cl

Cl2

UV

(trichloromethyl)benzene

C

Cl

Cl Cl

C

Cl2

UV

C

H H

H

H H C

H

H H C

H

H Cl

CH Cl

+

CH H

Multiple substitution can occur to form (dichloromethyl)benzene and (trichloromethyl)benzene.

This reaction occurs more generally when an alkyl group attached to a benzene ring reacts with a halogen in the presence of UV light. Thus ethylbenzene would react with chlorine in the presence of UV light to form a mixture of monochlorinated products:

Naming multi-substituted benzene ringsNames of some simple molecules, where both groups are the same, are shown here.

CH3

CH3

CH3

CH31,2-dimethylbenzene

NO2

NO21,3-dinitrobenzene

1,4-dimethylbenzene

When the groups are di� erent, various ways of naming the compound are usually accepted. The IUPAC system involves giving the substituent that comes � rst in the alphabet the number 1 and numbering other substituents relative to that. The numbers are chosen to give the lowest possible numbers. Some examples are shown in Table G4. Commonly used alternative names are also given in the table. Derivatives of phenol are named so that the OH group of the phenol is counted as position 1.

IUPAC name 1-chloro-2-methylbenzene 1-methyl-2-nitrobenzene 4-chlorophenol

Alternative names

2-chloromethylbenzene 2-nitromethylbenzene

2-methylchlorobenzene 2-methylnitrobenzene

Table G4 Naming multi-substituted benzene rings.

Cl

CH312

CH3

NO212

OH

Cl

HL


Reactions of substituted benzene ringsSubstituted benzene rings undergo basically the same reactions as a benzene ring, i.e. electrophilic substitution. The nature of the substituent determines the position of further substitution and the rate of the reaction relative to unsubstituted benzene.

Substituents on a benzene ring may be divided into two groups: those which cause substitution predominantly at positions 2 and 4 (and 6) (ortho and para positions) and those that cause substitution at position 3 (and 5) (the meta position).

Substituents that cause substitution faster than with benzene are called activating groups and those that cause substitution to occur more slowly than with benzene are called deactivating groups (Table G5).

Substituent Main product of chlorination Rate of substitution

relative to benzene

Nature of effect on electrophilic

substitution

–CH3

and other alkyl group

faster activating and 2,4-directing

–OH faster activating and 2,4-directing

–NO2 slower deactivating and 3-directing

Table G5 Reactions of substituted benzene rings.

CH3 CH3

Cl

Cl

OH

Cl Cl

OH

Cl

OH

Cl

Cl

Cl

NO2

2,4-directing groups usually cause substitution faster than benzene and 3-directing groups normally cause electrophilic substitution to occur more slowly than benzene (chlorine as a substituent is an exception to this – it is a 2,4-directing group and chlorobenzene reacts more slowly than benzene).

Rate of substitutionThe rate-determining step in the reaction is the attack of the electrophile on the ring. When an activating group is present this step occurs more quickly as there is more electron density in the ring so that an electrophile is attracted more strongly. When the ring is deactivated by the withdrawal of electron density, the electrophile is attracted less strongly and the reaction occurs more slowly.

Thus methylbenzene reacts more readily than benzene due to the electron-releasing e� ect of the –CH3 group (positive inductive e� ect). The methyl group activates the ring towards electrophilic substitution by donating electron density into the ring. This makes the

CH3

HL


ring more negative, i.e. more attractive towards electrophiles and the reaction occurs more quickly than with benzene.

Why does phenol react more readily than benzene?At � rst sight we might expect the –OH group to be electron-withdrawing due to the high electronegativity of O. However, the –OH group also possesses a lone pair of electrons and overlap of this lone pair into the ring activates the benzene ring (Figure G40).

This π donation into the ring is a bigger e� ect than the σ-withdrawing e� ect (due to the electronegativity of O). Therefore there is net donation of electron density into the ring and the ring will attract electrophiles more strongly.

Reaction of phenol with chlorineOne particularly good example of the ease of reactivity of phenol is in its reaction with chlorine.

Phenol reacts with chlorine water – Cl2(aq) – to form 2,4,6-trichlorophenol (no halogen carrier required):

Nitrobenzene reacts more slowly than benzeneThe NO2 group is electron-withdrawing and deactivates the benzene ring. With less electron density in the ring it attracts electrophiles less strongly, therefore nitrobenzene reacts more slowly than benzene. NO2 is electron-withdrawing because of the high electronegativity of the atoms in the group and, unlike the O in phenol, it does not possess a lone pair of electrons on the N that can be donated into the ring. There is also delocalisation of the π electrons from the ring onto the NO2 group.

Position of substitution in arenesThe position of substitution is determined by the charge distribution in the intermediate. The more stable the intermediate the more likely it is to be formed.

This is most usually explained using the Kekulé structure of benzene but to understand the explanation we must � rst understand something about resonance.

Two possible Lewis structures may be drawn for benzene, with the double bonds between di� erent atoms (Figure G41). These are called resonance structures. The actual structure of benzene is regarded as a resonance hybrid of these two structures – not one or the other but somewhere in between.

The position of substitution in nitrobenzeneThe position of substitution in nitrobenzene can be explained in terms of the stability of the intermediate carbocation formed.

CC C

C O H

lone pair

π donationinto ring

CC

Figure G40 Activation of the benzene ring.

OH

Cl Cl

OH

Cl

2,4,6-trichlorophenol

+ 3Cl2 + 3HCl

Phenol is also made more reactive in water since it is acidic and produces the phenoxide ion C6H5O−. The negative charge on the O means σ withdrawal is reduced and π donation is increased, therefore the ring is more attractive to electrophiles.

Figure G41 Resonance structures of benzene.

HL


The structure of nitrobenzene showing the bonds involved is shown in Figure G42. The nitrogen forms a dative covalent bond to one of the O atoms, therefore there is a positive charge on the N.

If substitution were to occur at position 2 then one of the resonance forms that contributes to the intermediate has two positive charges next to each other, which makes the intermediate less stable.

O+N

–O

Figure G42 The structure of nitrobenzene showing charge on N.

HL

X+

substitutionat position 2

X+

+ +H

2 positive chargesnext to each other

X

H

curly arrows represent delocalisation of electrons

some resonance forms of the intermediate

X

H

ON

–O+

ON

–O+

ON

–O+

ON

–O+

X+


X H X H X H2 positive chargesnext to each other

+ +

+

ON

–O+

ON

–O+

ON

–O+

ON

–O+

A similar situation occurs if substitution occurs at position 4.

However, when substitution occurs at position 3 it is not possible for the two positive charges to be next to each other and the intermediate is more stable and more likely to be formed.

X+


X

H

X

H

X

H+

+ +

ON

–O+

ON

–O+

ON

–O+

ON

–O+

O

H


O

H

X+X

H


intermediate stabilisedby donation of lone pairinto ring

O

H

X

H

+

+

Position of substitution in phenolWhen substitution occurs at position 2 the intermediate is stabilised by donation of the lone pair into the ring.


HL

O

H


O

H


intermediate stabilisedby donation of lone pairinto ring

O

H

X+ X H X H

+

+

O

H

O

H

X+


delocalisation not possibleas C has 4 bonds

X

H+

The same stabilisation is possible when substitution occurs at position 4.

If substitution were to occur at position 3, stabilisation of the positively charged intermediate is not possible by donation of the lone pair on the O into the ring.

Position of substitution in methyl benzeneThe methyl group has an electron-releasing e� ect and this also stabilises the intermediate when substitution occurs at positions 2 and 4.

The electron-releasing e� ect of the methyl group comes from the overlap of the electron density in the C–H σ bond with a p orbital on an adjacent atom (Figure G43).

The stabilising e� ect of the methyl group on the intermediate can be shown as:

CC C

C C H

Hdonationinto ring

CC H

Figure G43 The electron-releasing effect of the methyl group.


X+X

H

intermediate stabilised bydonation of electrons into ring

C

HHH

C

+

HHH

Examiner’s tipA summary of the explanations is that an electron-releasing group stabilises the intermediate by electron donation into the ring. This stabilisation is only possible when substitution occurs at positions 2, 4 and 6. An electron-withdrawing group destabilises the intermediate by withdrawing electron density from the ring. This destabilisation is greatest when substitution occurs at positions 2, 4 and 6, therefore substitution at position 3 is preferred.

As for phenol, this stabilisation is only possible when substitution occurs at positions 2 or 4.


Steric effectsSteric e� ects can also in� uence the position of substitution in a benzene ring. On a purely statistical basis, for a 2,4-directing group, substitution at position 2 should be twice as likely as substitution at position 4 as there are two position 2s but only one position 4. For nitration of methyl benzene there is indeed greater substitution at position 2, but when the much bulkier –C(CH3)3 group is attached to the benzene ring, nitration results in over four times more of the 4-substituted product. The bulky –C(CH3)3 group blocks the approach of the electrophile to position 2 (or 6) more e� ectively than the –CH3 group.

Acylation of methylbenzene usually results in a very high proportion of the 4-substituted product. The can be also explained by steric e� ects, where the methyl group hinders the approach of the electrophile to positions 2 or 6.

An alternative way of explaining the directing e� ect of substituent groups (not for examinations) is in terms of the charge distribution in the initial compound. An electron-releasing group will increase the electron density at positions 2, 4 and 6 and therefore the electrophile is more attracted to these positions. An electron-withdrawing group withdraws electron density from positions 2, 4 and 6 and, therefore, there is more electron density at positions 3 and 5 and an electrophile is attracted to these positions.

This can also be shown using resonance forms.

electron-releasinggroup

electrophilemore attracted

X

δ–δ–

δ–

electron-withdrawinggroup

electrophilemore attracted

X

δ+δ+

δ+

electron density reducedat position 2

electron density increasedat position 2

O

H

O

HO

N

–O+

O–

N

–O+

– +

Test yourself 15 Name the following molecules: 16 The molecule in Question 15a can react with

chlorine under suitable conditions to form � ve possible monochlorinated products. Draw the structures of the possible products and suggest which is least likely to be formed when the molecule reacts with chlorine.

HC C

H

H

H

H

OH

Cl

NO2a b c

CH3


O

Cl

H

CH

H

C

O

Cl

H

CHAlCl3

NH3

AlCl3

H2O

AlCl3

CH3Cl

NaOH(aq)heat

H

C

O

N

H

CH

H

C

O

NH2

H

CH

H

C

OH

CH

H

C

H

CH

H

C

H

Hcold

base

coldconc. NH3(aq)

NH2

H

CH

H

C

H

H

OH

H

C

HH

C

H

H

H

O

O

H

CH

H

CH

C

H

C

H

H

H

O

O

H

CH

H

CH

OCO

H

H

C H

Cl2

AlCl3

Cl2

AlCl3

NaOH(aq)

heatno reaction

Cl2

UV light

Mg

dry ethoxyethane

(i) CO2

(ii) H+ / H2O

conc. H2SO4

heat under reflux

conc. HNO3

CH3

CH3

ClCH3 CH2MgClCH2Cl

CH2OH

ClCH3Cl AlCl3

NO2NO2

Cl

CH3

Figure G44 Summary of reactions of aromatic compounds and acyl chlorides.

Learning objectives

• Deduce reaction pathways using the reactions in the sections above.

G11 Reaction pathways

Reaction pathwaysSome reactions of aromatic compounds and acyl chlorides are summarised in Figure G44.

HL


Worked examplesDesign a reaction pathway showing all reagents and conditions for the conversion of benzene to 1-methyl-2-nitrobenzene.

The molecule has one group substituted at position 2 relative to the other, therefore the methyl group must be put on the ring � rst as it is a 2,4-directing group.

Design a reaction pathway showing all reagents and conditions for the conversion of benzene to 3-nitromethylbenzene.

This time the groups are 1,3 relative to each other and therefore the nitro group must go on � rst as it is a 3-directing group.

Design a reaction pathway showing all reagents and conditions for the conversion of ethane to ethylbenzene.

Ethane is not very reactive and must be converted to something else before it can be reacted with benzene. The only reaction that allows us to convert it to something more reactive is free-radical substitution with chlorine in the presence of UV light. The chloroethane produced can then be reacted with benzene in the presence of an AlCl3 catalyst.

CH3

NO2

CH3

CH3Cl

AlCl3

conc. HNO3

conc. H2SO4

heat under reflux

NO2

CH3

NO2

CH3Cl

AlCl3

conc. HNO3

conc. H2SO4

heat under reflux

Cl

H

H

H

C C

H

H

AlCl3

Cl2

UV lightH

H

H

H

C C

H

H

H

H

H

H

C C

H

1-methyl-2-nitrobenzene is also called 2-nitromethylbenzene.


Exam-style questions

1 a Draw the full structural formula of 2-methylpent-2-ene. [1]

b i Draw the full structural formula of the major product formed when 2-methylpent-2-ene reacts with hydrogen bromide in the dark. [1]

ii Write the mechanism for this reaction and explain the formation of the major product. [5]

c 2-methylpent-2-ene can be formed when an alcohol is heated with concentrated phosphoric acid. i Name the type of reaction that occurs. [1] ii Give the structures of two possible alcohols that would produce 2-methylpent-2-ene via this reaction. [2] iii In each case another alkene will also be formed in the reaction. For each of the alcohols in part ii

draw the structure of the other alkene formed. [2] iv Describe using curly arrows the mechanism for the formation of 2-methylpent-2-ene from one of the

alcohols in part ii. [4] v One of the alcohols in part ii can be formed from the reaction between a Grignard reagent and

a ketone. Draw the structure of the Grignard reagent and the ketone that could be reacted together. [2]

2 A laboratory test for the presence of a carbonyl group involves reaction with 2,4-dinitrophenylhydrazine.

a Draw the structure of 2,4-dinitrophenylhydrazine. [1]

b Write an equation using structural formulae for the reaction between 2,4-dinitrophenylhydrazine and ethanal. [2]

c State the type of reaction between 2,4-dinitrophenylhydrazine and ethanal. [1]

d Ethanal reacts with hydrogen cyanide. i Write the mechanism for this reaction using curly arrows. [4] ii Write a balanced equation for the reaction that occurs when the organic product of this reaction

is boiled with aqueous acid. [2]

3 a Draw the isomers of C6H4Cl2 and explain how the number of isomers provides evidence for a delocalised structure of benzene. [4]

b There are several isomers of C7H7Cl that contain a benzene ring. i Draw the structure of one isomer that will react with aqueous sodium hydroxide to form a

compound with molecular formula C7H8O. [1] ii Draw the structure of one isomer which will not react with aqueous sodium hydroxide and explain

why it does not react. [3]

4 a Explain why chloroethanoic acid is a stronger acid than ethanoic acid. [2]

b i Write an equation for the dissociation of phenol in aqueous solution. [2] ii Predict and explain the relative acidities of 2-methylphenol and phenol. [3]

c Explain why dimethylamine is a stronger base than methylamine. [2]

5 2-chloropropane can form a Grignard reagent.

a State the reagents and conditions for the conversion of 2-chloropropane to a Grignard reagent. [2]

b Draw the full structural formula of the Grignard reagent formed. [1]

c The Grignard reagent formed from 2-chloropropane reacts under appropriate conditions with carbon dioxide. Draw the full structural formula of the � nal product of this reaction. [1]


d Complete the following equation by drawing the structure of the organic product formed. [1]

O

(ii) H+ / H2O

C

H

H

C

H

H

C H

H

H

C(i) H

C2H5Mgl

C

H

H

H

C C

H

H O

C

X

H3C

CH3

NO2

CH3

6 Ethanoyl chloride and ethanoic anhydride react in the same way with nucleophiles.

a i Draw the structure of ethanoyl chloride. [1] ii Write an equation showing full structural formulas for the reaction between ethanoyl chloride

and propan-2-ol. [2] iii To what class of compounds does the organic product in part ii belong? [1] iv Draw out the mechanism for the reaction between ethanoyl chloride and water. [4]

b i Draw the structure of propanoic anhydride. [1] ii Draw the full structural formulas of both organic products of the reaction between

propanoic anhydride and ammonia. [2]

7 a The compound shown can be formed from benzene in a single step.

i Write an equation, showing all reagents and conditions for the formation of this compound from benzene. [2]

ii Draw the mechanism for the reaction in part i. [3]

b Explain why phenol reacts more rapidly with chlorine than benzene does. [2]

c Nitrobenzene reacts with a mixture of concentrated sulfuric and concentrated nitric acid to form a compound with the molecular formula C6H4N2O4. Draw the structural formula of the product. [1]

d Compound X, shown below, can be formed from benzene in a two-step reaction sequence. Design a reaction pathway showing all reagents and conditions and the intermediate compound for the conversion of benzene to X. [4]

HL

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