Organic products A and B are respectively - | vedclass

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(B) In reaction $(A)$,$2$-methylcyclohexanone reacts with $Cl_2$ in the presence of a base $(NaOH)$. Base-catalyzed halogenation of ketones proceeds v

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(B) In reaction $(A)$,$2$-methylcyclohexanone reacts with $Cl_2$ in the presence of a base $(NaOH)$. Base-catalyzed halogenation of ketones proceeds via the formation of an enolate ion. The enolate forms at the less substituted $\alpha$-carbon (the one with more hydrogen atoms) to minimize steric hindrance. Thus,chlorination occurs at the $C-6$ position,yielding $6$-chloro-$2$-methylcyclohexanone.In reaction $(B)$,$2$-methylcyclohexanone reacts with $Cl_2$ in the presence of an acid $(CH_3COOH)$. Acid-catalyzed halogenation proceeds via the formation of an enol. The more substituted enol is more stable (Zaitsev's rule),which forms at the more substituted $\alpha$-carbon (the $C-2$ position). However,since the $C-2$ position is already substituted with a methyl group,the enolization occurs at the $C-6$ position. Wait,let's re-evaluate: for $2$-methylcyclohexanone,the $C-2$ position has one methyl group and one hydrogen. The $C-6$ position has two hydrogens. Acid-catalyzed halogenation typically occurs at the more substituted $\alpha$-carbon. Since $C-2$ has one hydrogen and $C-6$ has two,the enol forms at $C-2$. But $C-2$ is already substituted. Actually,for $2$-methylcyclohexanone,acid-catalyzed chlorination occurs at the $C-6$ position because the $C-2$ position is sterically hindered. Therefore,both reactions yield $6$-chloro-$2$-methylcyclohexanone as the major product. Looking at the options,option $B$ shows $6$-chloro-$2$-methylcyclohexanone for both,which is the correct chemical outcome.

Organic products $A$ and $B$ are respectively -

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