The reaction is given as follows: (CH_3)_2CHC | vedclass

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(C) The reaction of isobutyraldehyde $((CH_3)_2CHCHO)$ with methylmagnesium bromide $(CH_3MgBr)$ in the presence of ether follows a nucleophilic addit

Vedclass · Accepted Answer

$3-$methylbutan$-2-$ol

Vedclass · Answer

(C) The reaction of isobutyraldehyde $((CH_3)_2CHCHO)$ with methylmagnesium bromide $(CH_3MgBr)$ in the presence of ether follows a nucleophilic addition mechanism. $1$. The Grignard reagent $(CH_3^-)$ attacks the carbonyl carbon of the aldehyde to form an intermediate alkoxide $(A)$,which is $(CH_3)_2CH-CH(OMgBr)-CH_3$. $2$. Subsequent acid hydrolysis $(H_3O^+)$ converts the intermediate into the corresponding alcohol $(B)$,which is $(CH_3)_2CH-CH(OH)-CH_3$. $3$. The structure of $B$ is $CH_3-CH(CH_3)-CH(OH)-CH_3$. $4$. According to $IUPAC$ nomenclature,the longest chain containing the hydroxyl group has $4$ carbons. Numbering from the right gives the hydroxyl group at position $2$ and the methyl group at position $3$. Thus,the $IUPAC$ name is $3-$methylbutan$-2-$ol.

The reaction is given as follows: $(CH_3)_2CHCHO + CH_3MgBr$ $\xrightarrow{Ether} A$ $\xrightarrow{H_3O^+} B$. The $IUPAC$ name of $B$ is:

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