Q. $CH \equiv CH\underset{dil.H_{2}SO_{4}}{\overset{\ \ \ \ \ \ \ HgSO_{4}\ \ \ \ }{\rightarrow}}A\underset{NaOH}{\overset{\ \ \ \ Dilute\ \ \ }{\rightarrow}}B$ The compound B is
A) $CH_{3} - CH - CH_{2} - CHO$
|
OH
B) $CH_{3} - CH - CH_{2} - COONa$
|
OH
C) O
||
$CH_{3} - CH - CH_{2} - C - CH_{3}$
|
OH
D) OH O
| ||
$CH_{3} - C - CH_{2} - C - CH_{3}$
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$\ \ \ \ \ \ \ \ \ \ \ \ \ CH_{3}$
Explanation
(a)
$CH \equiv CH\underset{Dil.H_{2}SO_{4}}{\overset{\ \ \ HgSO_{4}\ \ \ \ \ \ \ \ }{\rightarrow}}CH_{3}CHO$
acetaldehyde
[A]
$\underset{\begin{array}{r}
NaOH \\
\lbrack aldol\ condensation\rbrack
\end{array}}{\overset{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Dil\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }{\rightarrow}}CH_{3} - CH - CH_{2} - CHO$
|
OH
aldol
[B]
This reaction is followed by acidic oxidation and aldol condensation respectively.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.