JEE MAIN - Chemistry (2013 (Offline) - No. 4)
An organic compound A upon reacting with $$N{H_3}$$ gives $$B.$$ On heating $$B$$ gives $$C.$$ $$C$$ in presence of $$KOH$$ reacts with $$B{r_2}$$ to given $$C{H_3}C{H_2}N{H_2}2.$$ $$A$$ is :
$$C{H_3}COOH$$
$$C{H_3}C{H_2}C{H_2}COOH$$
_en_4_1.png)
$$C{H_3}C{H_2}COOH$$
Explanation
_en_4_2.png)
Reaction $$\left( {{\rm I}{\rm I}{\rm I}} \right)$$ is a Hofmann bromamide reaction formation of $$C{H_3}C{H_2}N{H_2}$$ is possible only from a compound $$C{H_3}C{H_2}CON{H_2}$$ which can be obtained from the compound $$C{H_3}C{H_2}CO{O^ - }\,NH_4^ + $$ $$(B)$$ in $$\left( {{\rm I}{\rm I}} \right)$$ reaction further propanic acid $$\left( {C{H_3}C{H_2}COOH} \right)$$ on reaction with $$N{H_3}$$ produce $$C{H_3}C{H_2}CO{O^ - }NH_4^ - $$ (reaction $${\rm I}$$) hence the reaction will be
_en_4_3.png)
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