An organic compound A upon reacting with \(\mathrm{NH_3}\) gives B. On heating B gives C. C in presence of KOH reacts with \(\mathrm{Br_2}\) to given \(\mathrm{CH_3CH_2NH_2}\)2. A is :
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- \(\mathrm{CH_3COOH}\)
- \(\mathrm{CH_3CH_2CH_2COOH}\)
- \(\mathrm{CH_3-CH(CH_3)-COOH}\)
- \(\mathrm{CH_3CH_2COOH}\)
The Correct Option is D
Solution and Explanation
So C is the amide with one more carbon: \(\mathrm{CH_3CH_2CONH_2}\) (propanamide).
C is obtained by heating the ammonium salt B: \(\mathrm{CH_3CH_2COONH_4 \xrightarrow{\Delta} CH_3CH_2CONH_2 + H_2O}\).
B is formed from the acid A and \(\mathrm{NH_3}\): \(\mathrm{CH_3CH_2COOH + NH_3 \to CH_3CH_2COONH_4}\).
Hence A is propanoic acid.
Answer: (4) \(\mathrm{CH_3CH_2COOH}\)
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