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Q. $[P]\xrightarrow{Br_2} C_2H_4Br_2 \xrightarrow[NH_3] {NaNH_2}[Q]$
$ [Q]\xrightarrow[Hg^{2+} , \Delta ]{20 \% H_2 SO_4} [R] \xrightarrow{Zn-Hg/HCl} [S] $
The species P, Q, R and S respectively are

WBJEEWBJEE 2018Aldehydes Ketones and Carboxylic Acids

Solution:

Given reactions are
$[P] \xrightarrow{Br_2} C_2H_4Br_2 \xrightarrow[NH_3]{NaNH_2}[Q]$
$[Q] \xrightarrow[Hg^{2+},\Delta]{20\%/H_2SO_4}[R] \xrightarrow{Zn-Hg/HCl}[S]$
$[P]$ is an alkene with two carbon atoms i.e. ethene. When it reacts with $Br_2$, addition reaction occurs $1$ and $C_2H_4Br_2$ is formed.
image
$C_2H_4Br_2(1, 2$- dibromoethane) on reaction with $NaNH_2/NH_3$ gives ethyne $(C_2H_2)$.
Ethyne $[Q]$ in presence of $20\%\,H_2SO_4, Hg^{2+}$ at $333 \,K$
gives ethanal.
$CH = CH + H -OH \xrightarrow[333\,K]{Hg^{2+}/H^{+}} CH_2 = \underset{\overset{|}{O}H}{C} -H$
$\xrightarrow{\text{Tautomerisation}} \underset{\text{Ethanal[R]}}{CH_3CHO}$
Ethanal $[P]$ undergoes reduction in presence of $Zn-Hg /HCl$ to give ethane $[S]$
$\underset{\overset{\text{Ethanal}}{[R]}}{CH_3CHO} \xrightarrow[\text{Clemmensen reduction}]{Zn-Hg/HCl} \underset{\overset{\text{Ethane}}{[S]}}{CH_3 - CH_3}$