CH_3CO_2Et + (CH_2)_5(MgBr)_2 xrightarrow[(2) | vedclass

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Question

(A) The reaction involves the reaction of an ester with a Grignard reagent. $CH_3CO_2Et$ reacts with the bifunctional Grignard reagent $(CH_2)_5(MgBr)

Vedclass · Accepted Answer

$1-$methylcyclohexanol

Vedclass · Answer

(A) The reaction involves the reaction of an ester with a Grignard reagent. $CH_3CO_2Et$ reacts with the bifunctional Grignard reagent $(CH_2)_5(MgBr)_2$. First,one end of the Grignard reagent attacks the carbonyl carbon of the ester,displacing the ethoxide group to form a ketone intermediate: $CH_3-C(=O)-(CH_2)_5-MgBr$. Next,the second Grignard end (intramolecularly) attacks the newly formed ketone carbonyl group. This forms a cyclic alkoxide intermediate,which upon acidic workup $(H^+)$ yields the cyclic tertiary alcohol,$1-methylcyclohexanol$.

$CH_3CO_2Et + (CH_2)_5(MgBr)_2 \xrightarrow[(2) H^{+}]{(1)} \mathop {C_7H_{14}O}\limits_{(A)}$ ; compound $(A)$ will be

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