Carboxylic Acids and its Derivatives

\(C_3H_6O_2\) fits the general formula \(C_nH_{2n}O_2\) of saturated carboxylic acids and esters, and an acid and an ester of the same formula are functional isomers:

Ethanoic acid heated with ethanol in the presence of concentrated sulphuric acid gives ethyl ethanoate and water (the reaction is reversible):

R gives but-1-ene on dehydration, so R is a four-carbon alcohol with a straight chain. R on oxidation gives the acid Q with the same number of carbons, so R must be a primary alcohol (a secondary alcohol, butan-2-ol, would give a ketone and mainly but-2-ene). So R is butan-1-ol, \(CH_3CH_2CH_2CH_2OH\), and Q is butanoic acid, \(CH_3CH_2CH_2COOH\). P is the ester of Q and R, so P is butyl butanoate, \(CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3\), whose formula is \(C_8H_{16}O_2\) (a check: \(C_4H_8O_2+C_4H_{10}O-H_2O=C_8H_{16}O_2\)).

1. Hydrolysis of P

2. Oxidation of R with chromic acid

3. Dehydration of R

(The printed question says "C on dehydration"; the alcohol R is meant.)

Claisen condensation takes place. Two molecules of the ester condense to give a \(\beta\)-keto ester, ethyl 3-oxobutanoate (ethyl acetoacetate), and ethanol:

Alcoholysis of the anhydride takes place, and an ester and a carboxylic acid are formed:

Rosenmund reduction takes place. The acid chloride is reduced to the aldehyde ethanal. \(BaSO_4\) poisons the palladium catalyst, so the aldehyde is not reduced further to the alcohol:

Methanoic acid has an aldehyde group, so it reduces Tollens’ reagent (ammoniacal silver nitrate), and a shining silver mirror is deposited. The acid itself is oxidised to carbon dioxide and water:

Other carboxylic acids do not give this silver mirror test.

Two molecules of ethanoic acid lose one molecule of water, and ethanoic anhydride is formed (intermolecular dehydration):

In every other carboxylic acid the carboxyl carbon is bonded to an alkyl or aryl group, but in methanoic acid (\(H-COOH\)) it is bonded to a hydrogen atom. So the molecule has the aldehyde group \(H-C=O\) as well as the carboxyl group, and it shows some properties that the other acids do not show. These are called its anomalous (abnormal) properties.

Example 1: it is a reducing agent (like an aldehyde). It reduces Tollens’ reagent to a silver mirror and Fehling’s solution to a red precipitate of cuprous oxide, but acetic acid does not do either. It also decolourises pink acidified \(KMnO_4\) and reduces mercuric chloride to white \(Hg_2Cl_2\) and then to black mercury.

Example 2: action of concentrated sulphuric acid. Other acids form anhydrides with dehydrating agents, but methanoic acid is dehydrated to carbon monoxide:

It is also the strongest of the monocarboxylic acids (\(pK_a=3.75\)), because there is no alkyl group to push electrons.

In benzoic acid the \(-COOH\) group is attached to the ring through a carbon that carries a partly positive charge (\(C=O\)), so it withdraws electrons from the ring both by the \(-I\) effect and by the \(-M\) (resonance) effect. The resonating structures show a positive charge on the carbons at the ortho and para positions, while the meta carbons are not affected. So the ring is deactivated (less reactive than benzene), the meta positions are comparatively richer in electrons, and the electrophile enters at the meta position. The \(-COOH\) group is therefore a meta director, and benzoic acid does not give the Friedel-Crafts reaction. The common reactions are nitration, sulphonation and bromination:

The reagents given with \(CH_3COOH\) are:

(Acetic acid alone is not dehydrated by \(H^+\) at once; a dehydrating agent like \(P_2O_5\) is needed in practice, but the textbook gives the anhydride for C.)

Two molecules of an ester with \(\alpha\)-hydrogen condense, and a \(\beta\)-keto ester is formed.

A carboxylic acid with \(\alpha\)-hydrogen is halogenated at the \(\alpha\)-carbon in the presence of red phosphorus.

Ethanoyl chloride reacts with ethanol (alcoholysis) to give an ester and hydrogen chloride.

Nitration of benzoic acid with a mixture of concentrated nitric and sulphuric acids gives 3-nitrobenzoic acid, because \(-COOH\) is a meta director.

This question appears a second time in the textbook, with the same five parts. The answers are given again here.

Claisen condensation takes place. Two molecules of the ester condense to give a \(\beta\)-keto ester, ethyl 3-oxobutanoate (ethyl acetoacetate), and ethanol:

Alcoholysis of the anhydride takes place, and an ester and a carboxylic acid are formed:

Rosenmund reduction takes place. The acid chloride is reduced to the aldehyde ethanal. \(BaSO_4\) poisons the palladium catalyst, so the aldehyde is not reduced further to the alcohol:

Methanoic acid has an aldehyde group, so it reduces Tollens’ reagent (ammoniacal silver nitrate), and a shining silver mirror is deposited. The acid itself is oxidised to carbon dioxide and water:

Other carboxylic acids do not give this silver mirror test.

Two molecules of ethanoic acid lose one molecule of water, and ethanoic anhydride is formed (intermolecular dehydration):

Work backwards from Z.

Z. A primary alcohol that gives a positive iodoform test must have the \(CH_3CH_2OH\) unit, and ethanol is the only primary alcohol that does. So Z is ethanol.

Y. An aliphatic primary amine reacts with nitrous acid (\(NaNO_2+HCl\)) to give an alcohol and nitrogen, so Y is ethanamine, \(CH_3CH_2NH_2\).

X. Y is formed from an amide by Hofmann bromamide degradation, which removes one carbon atom. So X has three carbons: propanamide, \(CH_3CH_2CONH_2\).

W. An amide is made from an acid chloride and ammonia, so W is propanoyl chloride, \(CH_3CH_2COCl\).

V. \(SOCl_2\) changes a carboxylic acid into the acid chloride, so V is propanoic acid, \(CH_3CH_2COOH\).

The last equation is the iodoform test of Z (yellow precipitate of iodoform).

The only secondary alcohol with three carbon atoms is propan-2-ol (P), \(CH_3CH(OH)CH_3\). A secondary alcohol is oxidised to a ketone, so Q is propanone (acetone), \(CH_3COCH_3\). Propanone has \(\alpha\)-hydrogen atoms, so with cold dilute \(NaOH\) two molecules undergo aldol addition (the ketone version is also called ketol condensation). R is 4-hydroxy-4-methylpentan-2-one (diacetone alcohol).

The enolate ion of one molecule attacks the carbonyl carbon of the second molecule, and the alkoxide formed takes a proton from water.

Q is obtained by the reduction of chlorobenzene, so Q is benzene. P loses \(CO_2\) with soda lime to give benzene, so P is the sodium salt of benzoic acid, sodium benzoate. Benzene with \(CH_3Cl/AlCl_3\) undergoes Friedel-Crafts alkylation, so R is toluene. Chromyl chloride (\(CrO_2Cl_2\)) oxidises the methyl group of toluene only as far as the aldehyde (Etard reaction), so S is benzaldehyde. Benzaldehyde heated with alcoholic \(KCN\) undergoes benzoin condensation, so T is benzoin.

1. Oxidation of ethanol (a primary alcohol is oxidised first to the aldehyde and then to the acid):

2. Hydrolysis of ethanenitrile (the nitrile is changed to the amide, and the amide to the acid):

3. From sodium methoxide and carbon monoxide (carbon monoxide is inserted between O and Na, and acidification gives the acid):

Other methods are the carbonation of methylmagnesium bromide (\(CH_3MgBr+CO_2\), then \(H_3O^+\)) and the heating of malonic acid (\(CH_2(COOH)_2\xrightarrow{\Delta}CH_3COOH+CO_2\)).

D. Methanol is obtained from a primary amine with \(NaNO_2/HCl\), so D is methanamine, \(CH_3NH_2\).

C. The \(CO\) group lost in going from C to D is the loss of one carbon in Hofmann bromamide degradation (the carbonyl carbon leaves as carbonate), so C is the amide with one more carbon, ethanamide, \(CH_3CONH_2\).

B. The ammonolysis of an acid anhydride gives an amide and a carboxylic acid. Here the amide is ethanamide and the acid is ethanoic acid, so B is ethanoic anhydride, \((CH_3CO)_2O\).

A. A compound that gives effervescence with \(NaHCO_3\) is a carboxylic acid. Dehydration with \(P_2O_5\) gives the anhydride, so A is ethanoic acid, \(CH_3COOH\).

\(SOCl_2\) changes the acid into the acid chloride: P is ethanoyl chloride, \(CH_3COCl\) (\(PCl_5\) or \(PCl_3\) would give the same product). Rosenmund reduction gives the aldehyde: Q is ethanal, \(CH_3CHO\). Hydrogen cyanide adds to the carbonyl group to form a cyanohydrin: R is 2-hydroxypropanenitrile (ethanal cyanohydrin), \(CH_3CH(OH)CN\). Hydrolysis of the nitrile group gives the acid: S is 2-hydroxypropanoic acid (lactic acid), \(CH_3CH(OH)COOH\).

Work backwards from Z.

Z. A primary alcohol that gives a positive iodoform test must have the \(CH_3CH_2OH\) unit, and ethanol is the only primary alcohol that does. So Z is ethanol.

Y. An aliphatic primary amine reacts with nitrous acid (\(NaNO_2+HCl\)) to give an alcohol and nitrogen, so Y is ethanamine, \(CH_3CH_2NH_2\).

X. Y is formed from an amide by Hofmann bromamide degradation, which removes one carbon atom. So X has three carbons: propanamide, \(CH_3CH_2CONH_2\).

W. An amide is made from an acid chloride and ammonia, so W is propanoyl chloride, \(CH_3CH_2COCl\).

V. \(SOCl_2\) changes a carboxylic acid into the acid chloride, so V is propanoic acid, \(CH_3CH_2COOH\).

The last equation is the iodoform test of Z (yellow precipitate of iodoform).

The primary alcohol with two carbon atoms is ethanol (P). Oxidation with a mild oxidising agent (such as \(PCC\), or \(K_2Cr_2O_7/H_2SO_4\) with distillation of the aldehyde as it forms) gives ethanal (Q), \(CH_3CHO\). Ethanal has \(\alpha\)-hydrogens, so with cold dilute \(NaOH\) two molecules undergo aldol condensation, and R is 3-hydroxybutanal (aldol).

On heating, the aldol loses water to give but-2-enal (crotonaldehyde), \(CH_3CH=CHCHO\).

C. A primary aliphatic amine of molecular mass \(45\) is \(C_2H_7N\) (\(2\times12+7\times1+14=45\)), which is ethanamine, \(CH_3CH_2NH_2\).

B. C is formed by Hofmann bromamide degradation (\(Br_2/aq.KOH\)) of an amide with one more carbon atom, so B is propanamide, \(CH_3CH_2CONH_2\).

A. The amide is made from ammonia and a carboxylic acid (through the ammonium salt, which is heated), so A is propanoic acid, \(CH_3CH_2COOH\). (Propanoyl chloride, propanoic anhydride or an ester of propanoic acid would also give propanamide with ammonia.)

D. A primary aliphatic amine with nitrous acid gives an alcohol, so D is ethanol, \(CH_3CH_2OH\).

E. Ethanol gives the iodoform test with \(I_2/NaOH\), so E is iodoform, \(CHI_3\) (yellow precipitate), along with sodium methanoate.

Two molecules of an ester with \(\alpha\)-hydrogen condense, and a \(\beta\)-keto ester is formed.

ii. Hell-Volhard-Zelinsky reaction

A carboxylic acid with \(\alpha\)-hydrogen is halogenated at the \(\alpha\)-carbon in the presence of red phosphorus.

iii. Ethanolysis of ethanoyl chloride

Ethanoyl chloride reacts with ethanol (alcoholysis) to give an ester and hydrogen chloride.