Single Choice

An organic compound 'A' on treatment with $$NH_{3}$$ gives 'B', which on heating gives 'C'. 'C', when treated with $$Br_{2}$$ in the presence of KOH, produces ethylamine. Compound 'A' is:

A$$CH_{3}CH_{2}COOH$$
Correct Answer
B$$CH_{3}COOH$$
C$$CH_{3}CH_{2}CH_{2}COOH$$
D$$CH_3-CH(CH_3)-COOH$$

Solution

When propanoic acid is reacted with liq.ammonia it gives 1-amino propanone and water as a byproduct
1-amino propanone is reacted with bromine in presence of potassium hydroxide, Hofmann bromamide reaction takes place which gives ethylamine as a product.
Thus Propanoic acid ($$CH_3CH_2COOH$$) is the correct answer.


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