Q17 / 75 · JEE Main 2026 (06 April Shift 2) · Chemistry
Consider the following organic reaction sequence. Choose the final product $(X)$ from the following (consider the major product in all intermediate reactions)


- Correct answerABenzene
- BPhenol
- CPropanol
- DChlorobenzene
Solution
The starting material is propanoic acid ($CH_3CH_2COOH$).
In step (i), propanoic acid reacts with $NH_3$ upon heating to form propanamide.
$CH_3CH_2COOH \xrightarrow{NH_3, \Delta} CH_3CH_2CONH_2$
In step (ii), propanamide undergoes Hoffmann bromamide degradation with $NaOH$ and $Br_2$ to form ethylamine.
$CH_3CH_2CONH_2 \xrightarrow{NaOH, Br_2} CH_3CH_2NH_2$
In step (iii), ethylamine reacts with nitrous acid ($HNO_2$) to form an unstable diazonium salt, which hydrolyzes to give ethanol.
$CH_3CH_2NH_2 \xrightarrow{HNO_2, H_2O} CH_3CH_2OH$
In step (iv), ethanol reacts with benzenediazonium chloride ($C_6H_5N_2Cl$). Ethanol reduces the diazonium salt to benzene, while getting oxidized to ethanal.
$CH_3CH_2OH + C_6H_5N_2Cl \rightarrow C_6H_6 + CH_3CHO + N_2 + HCl$
The final major product (X) is benzene.
Answer: Benzene
In step (i), propanoic acid reacts with $NH_3$ upon heating to form propanamide.
$CH_3CH_2COOH \xrightarrow{NH_3, \Delta} CH_3CH_2CONH_2$
In step (ii), propanamide undergoes Hoffmann bromamide degradation with $NaOH$ and $Br_2$ to form ethylamine.
$CH_3CH_2CONH_2 \xrightarrow{NaOH, Br_2} CH_3CH_2NH_2$
In step (iii), ethylamine reacts with nitrous acid ($HNO_2$) to form an unstable diazonium salt, which hydrolyzes to give ethanol.
$CH_3CH_2NH_2 \xrightarrow{HNO_2, H_2O} CH_3CH_2OH$
In step (iv), ethanol reacts with benzenediazonium chloride ($C_6H_5N_2Cl$). Ethanol reduces the diazonium salt to benzene, while getting oxidized to ethanal.
$CH_3CH_2OH + C_6H_5N_2Cl \rightarrow C_6H_6 + CH_3CHO + N_2 + HCl$
The final major product (X) is benzene.
Answer: Benzene
Hence, the correct answer is option: 'A'.