What is the product of the following reaction | vedclass

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(C) The reaction involves the alkylation of cyclohexanone using excess $CH_3I$ and $KH$. $KH$ is a strong base that deprotonates the $\alpha$-carbons

Vedclass · Accepted Answer

$2,2,6,6-$tetramethylcyclohexanone

Vedclass · Answer

(C) The reaction involves the alkylation of cyclohexanone using excess $CH_3I$ and $KH$. $KH$ is a strong base that deprotonates the $\alpha$-carbons of the ketone to form enolate ions. Since $KH$ and $CH_3I$ are in excess,all four $\alpha$-hydrogens of the cyclohexanone are replaced by methyl groups. First,the two $\alpha$-carbons are deprotonated to form a dienolate,which then reacts with $CH_3I$ to form $2,6-$dimethylcyclohexanone. Further treatment with excess $KH$ and $CH_3I$ leads to the substitution of the remaining two $\alpha$-hydrogens,resulting in $2,2,6,6-$tetramethylcyclohexanone.

$A$. Lucas reagent	$I$. $SnCl_2 + HCl$
$B$. Clemmensen reagent	$II$. $[Ag(NH_3)_2]^+$
$C$. Tollen's reagent	$III$. Anhydrous $ZnCl_2 + conc. HCl$
$D$. Stephen reagent	$IV$. $Zn-Hg/HCl$

What is the product of the following reaction?

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