An organic compound A reacts with NH_3 on hea | vedclass

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Question

(D) The reaction of $B$ with $Br_2$ and $KOH$ is the Hofmann bromamide degradation reaction,which converts an amide $(RCONH_2)$ into a primary amine $

Vedclass · Accepted Answer

$CH_3CH_2COOH$

Vedclass · Answer

(D) The reaction of $B$ with $Br_2$ and $KOH$ is the Hofmann bromamide degradation reaction,which converts an amide $(RCONH_2)$ into a primary amine $(RNH_2)$ with one carbon atom less.Given that the product is $CH_3CH_2NH_2$ (ethanamine),the amide $B$ must be $CH_3CH_2CONH_2$ (propanamide).Since $A$ reacts with $NH_3$ on heating to form propanamide $(B)$,$A$ must be propanoic acid $(CH_3CH_2COOH)$.The reaction sequence is: $CH_3CH_2COOH + NH_3$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (B)$ $\xrightarrow{Br_2/KOH} CH_3CH_2NH_2$.

List-$I$ (Compounds)	List-$II$ (Reactions)
$A. \ CH_3(CH_2)_3NH_2$	$(i)$ Alkaline hydrolysis
$B. \ CH_3C\equiv CH$	$(ii)$ With $KOH$ (alc.) and $CHCl_3$ produces bad smell
$C. \ CH_3CH_2COOCH_3$	$(iii)$ Gives white ppt. with ammoniacal $AgNO_3$
$D. \ CH_3CH(OH)CH_3$	$(iv)$ With Lucas reagent cloudiness appears after $5 \ minutes$

List-$I$	List-$II$
$1.$ Aniline	$a.$ Used in the preparation of azo dyes
$2.$ Nitrobenzene	$b.$ Sulpha drug
$3.$ Sulphanilamide	$c.$ Solvent in Friedel-Crafts reaction
$4.$ Trinitrotoluene	$d.$ Used as an explosive

An organic compound $A$ reacts with $NH_3$ on heating to give $B$. Compound $B$ reacts with $Br_2$ in the presence of $KOH$ to give $CH_3CH_2NH_2$. What is $A$?

Similar Questions

Match List-$I$ with List-$II$ and select the correct answer.

List-$I$ List-$II$

$1.$ Aniline $a.$ Used in the preparation of azo dyes

$2.$ Nitrobenzene $b.$ Sulpha drug

$3.$ Sulphanilamide $c.$ Solvent in Friedel-Crafts reaction

$4.$ Trinitrotoluene $d.$ Used as an explosive

What are $X$ and $Y$ respectively in the following set of reactions?

The product $(Y)$ of the following reaction is:

The correct order of basic strengths of the given amines is:

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