Q.Butan-1-ol is oxidised completely with hot acidified $\text{KMnO}_4$. Identify the product and state which functional group is formed first as an isolable intermediate under milder oxidation.

Since butanoic acid has the same number of carbons (4) as 1-chlorobutane, the conversion must NOT go through cyanide (which would add a carbon); instead, hydrolyse to the alcohol and then oxidise it fully to the acid.

1-Chlorobutane, $CH_3CH_2CH_2CH_2Cl$, already has four carbons — exactly the number needed for butanoic acid, $CH_3CH_2CH_2COOH$. So the terminal $-CH_2Cl$ carbon simply needs to become a $-COOH$ carbon, with no chain extension required (ruling out the usual $KCN$ chain-extension route used for acids with one extra carbon).

Step 1 — Hydrolysis (nucleophilic substitution by hydroxide):

$CH_3CH_2CH_2CH_2Cl + KOH \xrightarrow{aq.} CH_3CH_2CH_2CH_2OH\;(\text{butan-1-ol}) + KCl$

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Hydrolysis of benzamide ($C_6H_5CONH_2$) breaks the C–N bond to give benzoic acid ($C_6H_5COOH$) and ammonia.

Concept — hydrolysis of an amide. An amide is a carboxylic-acid derivative; on hydrolysis (acidic or basic) the $-CONH_2$ group is converted back to the parent carboxylic acid, releasing the nitrogen as ammonia (or ammonium salt).

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Complete acidic hydrolysis of a nitrile gives a carboxylic acid; CH3CN gives CH3COOH.

Nitriles undergo hydrolysis. On complete hydrolysis with aqueous acid (H3O+), the –CN group is converted, via the amide, all the way to the carboxylic acid:

CH3CN --H3O+--> CH3CONH2 --H3O+--> CH3COOH + NH4+

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