Carboxylic Acids and Derivatives: Question 3

Worked solution

Part (a): Acid hydrolysis of the nitrile

Nitriles are hydrolysed all the way to carboxylic acids when heated under reflux with an excess of dilute acid (or with dilute alkali followed by acidification). Here, dilute hydrochloric acid both hydrolyses the nitrile and supplies the acidic conditions in one step, so no separate acidification is needed:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH} + \text{NH}_4\text{Cl}$

Type of reaction: hydrolysis. Products: pentanoic acid and ammonium chloride (the nitrile nitrogen ends up as the ammonium ion, $\text{NH}_4^+$, paired with the chloride ion from the hydrochloric acid). The nitrile carbon becomes the carboxyl carbon, so the carbon chain length is unchanged: pentanenitrile (5 carbons, including the nitrile carbon) gives pentanoic acid (5 carbons, including the carboxyl carbon).

Part (b): Reduction with LiAlH4

$\text{LiAlH}_4$ is a strong, reactive source of hydride ($\text{H}^-$) ions and reduces a carboxylic acid all the way to a primary alcohol (via an aldehyde intermediate that is not isolated):

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{H}_2\text{O}$

Reducing the $-\text{COOH}$ carbon (which is bonded to only one carbon chain and to oxygen) always gives a $-\text{CH}_2\text{OH}$ group, i.e. a primary alcohol, here, pentan-1-ol. The reaction must be carried out in dry (anhydrous) conditions, typically in dry ether/ethoxyethane, because $\text{LiAlH}_4$ reacts violently with water/protic solvents; a dilute acid is only added afterwards, on work-up, to protonate the intermediate alkoxide and liberate the alcohol.

Part (c): Why NaBH4 cannot perform this reduction

$\text{NaBH}_4$ is a considerably weaker (less reactive, less ionic) source of hydride ions than $\text{LiAlH}_4$. It is only reactive enough to attack the strongly electrophilic carbonyl carbon of an aldehyde or ketone. In a carboxylic acid, the carbonyl carbon is already flanked by an $-\text{OH}$ oxygen that donates electron density into the carbonyl by resonance/lone-pair donation, which reduces the positive character (electrophilicity) of the carbonyl carbon. This makes the carboxylic acid’s carbonyl carbon far less susceptible to nucleophilic hydride attack than an aldehyde or ketone’s, so the mild $\text{NaBH}_4$ hydride cannot react with it, whereas the much more powerful $\text{LiAlH}_4$ hydride can.

Final answers

• (a) Hydrolysis; products are pentanoic acid and ammonium chloride; $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COOH} + \text{NH}_4\text{Cl}$.
• (b) Primary alcohol, pentan-1-ol; reaction needs anhydrous/dry conditions (e.g. dry ether).
• (c) NaBH4 is too weak/unreactive to attack the less electrophilic carbonyl carbon of a carboxylic acid, unlike the stronger, more reactive LiAlH4.