An organic compound ‘A’ on treatment with NH3, gives ‘B’ which on heating gives ‘C’, ‘C’ when treated with Br2, in the presence of KOH produces ethylamine. Compound ‘A’ is:

A
${CH}_{3} COOH$
B
${CH}_{3} {CH}_{2} {CH}_{2} COOH$
C
D
${CH}_{3} {CH}_{2} COOH$

Solution:

Since, C when heated with Br₂, in presence of KOH produces ethylamine, hence it must be propanamide and hence the organic compound (A) will be propanoic acid. The reactions follows.
$\underset{(A)}{{CH}_{3} - {CH}_{2} - COOH \overset{{NH}_{3}}{⟶}} \underset{(B)}{{CH}_{3} - {CH}_{2} - {COONH}_{4} \overset{Δ}{⟶}}$ $\underset{(C)}{{CH}_{3} - {CH}_{2} - {CONH}_{2}} \to_{Hoffmann (C) bromamide reaction}^{KOH + Br 2}$ $\underset{(Ethylamine)}{{CH}_{3} - {CH}_{2} - {NH}_{2}}$

An organic compound ‘A’ on treatment with NH₃, gives ‘B’ which on heating gives ‘C’, ‘C’ when treated with Br₂, in the presence of KOH produces ethylamine. Compound ‘A’ is:

Solution:

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