Aldehydes and Ketones - Result Question 56
####66. (a) Compound $A\left(C _8 H _8 O\right)$ on treatment with $NH _2 OH$. $HCl$ given $B$ and $C . B$ and $C$ rearrange to give $D$ and $E$, respectively, on treatment with acid. $B, C, D$ and $E$ are all isomers of molecular formula $\left(C _8 H _9 NO\right)$. When $D$ is boiled with alcoholic $KOH$, an oil $F\left(C _6 H _7 N\right)$ separates out. $F$ reacts rapidly with $CH _3 COCl$ to give back $D$. On the other hand, $E$ on boiling with alkali followed by acidification gives a white solid $G\left(C _7 H _6 O _2\right)$. Identify $A$ - $G$.
(b) Carry out the following transformation in not more than three steps.
1-butyne $\longrightarrow 2$-pentanone
(1999, 3M)
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Solution:
- (a) $G$ is benzoic acid $C _6 H _5-COOH, B$ and $C$ are two stereomeric oximes which undergo Beckmann’s rearrangement on treatment with acid to give amides $D$ and $E$.
$C _6 H _5-\underset{E}{\stackrel{O}{C}-NHCH _3} \xrightarrow[H _2 O]{OH^{-}} C _6 H _5 COOH+CH _3 NH _2$
B $\xrightarrow[H _2 O]{KOH} CH _3 COOH+\underset{F}{C _6 H _5 NH _2}$
$F+CH _3 COCl \longrightarrow C _6 H _5 NHCOCH _3$
$\Rightarrow \quad A=C _6 H _5-\stackrel{O}{C}-CH _3$
(b) $CH _3-CH _2-C \equiv CH$
$\xrightarrow[H _2 O _2 / OH^{-}]{B _2 H _6} CH _3 CH _2 CH _2-\stackrel{O}{O}-H \xrightarrow[\text { (ii) } H _3 O^{+}]{\text {(i) } CH _3 MgBr}$
