Ethyl bromide is converted to succinic acid by using the following re…

The correct answer is Option (1) → C, B, A and D

Ethyl bromide has two carbon atoms, while succinic acid, \( \ce{HOOC-CH2-CH2-COOH} \), has four. The two extra carbons come from two cyanide groups, so first a halogen must be put on each of the two carbons, and then each –CN is hydrolysed to –COOH.

First (C), alcoholic KOH removes HBr (dehydrohalogenation) and gives ethene:

    \( \displaystyle \ce{CH3CH2Br} \xrightarrow{\text{KOH (alc.), }\Delta} \ce{CH2=CH2 + KBr + H2O} \)

Next (B), bromine adds across the double bond and gives 1,2-dibromoethane:

    \( \ce{CH2=CH2 + Br2 -> BrCH2-CH2Br} \)

Then (A), KCN replaces both bromine atoms by nucleophilic substitution, giving the dinitrile:

    \( \ce{BrCH2-CH2Br + 2KCN -> NC-CH2-CH2-CN + 2KBr} \)

Finally (D), acid hydrolysis turns each –CN group into –COOH (through the amide):

    \( \displaystyle \ce{NC-CH2-CH2-CN} \xrightarrow{\ce{H2O}/\ce{H+}} \ce{HOOC-CH2-CH2-COOH} \)

Shortcut: hydrolysis (D) can only be the last step, after the –CN groups are in place, so the sequence ends with D; Options (1) and (2) do, and Option (2) begins with \( \ce{Br2} \), which does not react with ethyl bromide. Only Option (1) remains.

NCERT (2026-27 edition):
From Nitriles and Amides – “Nitriles are hydrolysed to amides and then to acids in the presence of H⁺ or OH⁻ as catalyst. Mild reaction conditions are used to stop the reaction at the amide stage.”