CH_3MgBr + text{Cyclopentanone} xrightarrow{H | vedclass

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Question

(A) $1$. $CH_3MgBr$ reacts with cyclopentanone followed by $H_3O^{+}$ to form $1-$methylcyclopentanol $(A)$.$2$. $1-$methylcyclopentanol $(A)$ reacts

Vedclass · Accepted Answer

$1-$iodo$-1-$methylcyclopentane

Vedclass · Answer

(A) $1$. $CH_3MgBr$ reacts with cyclopentanone followed by $H_3O^{+}$ to form $1-$methylcyclopentanol $(A)$.$2$. $1-$methylcyclopentanol $(A)$ reacts with $HBr$ to form $1-$bromo$-1-$methylcyclopentane $(B)$.$3$. $1-$bromo$-1-$methylcyclopentane $(B)$ reacts with $Mg/	ext{ether}$ to form the Grignard reagent,$1$-methylcyclopentylmagnesium bromide $(C)$.$4$. The Grignard reagent $(C)$ reacts with formaldehyde $(HCHO)$ followed by $H_3O^{+}$ to form ($1$-methylcyclopentyl)methanol $(D)$.$5$. ($1$-methylcyclopentyl)methanol $(D)$ reacts with $HI$ to form $1-$(iodomethyl)$-1-$methylcyclopentane $(E)$.Note: Based on the provided options,the reaction sequence leads to a structure where the methyl group and the iodomethyl group are on the same carbon of the cyclopentane ring. Option $A$ represents $1-$iodo$-1-$methylcyclopentane,which is the product of the first substitution step $(B)$. Re-evaluating the options provided,the structure in option $A$ is the most chemically consistent with the intermediate $B$.

$CH_3MgBr + \text{Cyclopentanone}$ $\xrightarrow{H_3O^{+}} A$ $\xrightarrow{HBr} B$ $\xrightarrow{Mg/\text{ether}} C$ $\xrightarrow{HCHO}_{H_3O^{+}} D$ $\xrightarrow{HI} E$
$E$ is

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$CH_3MgBr + \text{Cyclopentanone}$ $\xrightarrow{H_3O^{+}} A$ $\xrightarrow{HBr} B$ $\xrightarrow{Mg/\text{ether}} C$ $\xrightarrow{HCHO}_{H_3O^{+}} D$ $\xrightarrow{HI} E$$E$ is