3-methylpentan-3-ol will be prepared from | vedclass

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(B) To prepare $3-$methylpentan$-3-$ol,we need a Grignard reagent to attack an ester. The structure of $3-$methylpentan$-3-$ol is $CH_3CH_2-C(OH)(CH_3

Vedclass · Accepted Answer

ethyl ethanoate and ethyl magnesium bromide

Vedclass · Answer

(B) To prepare $3-$methylpentan$-3-$ol,we need a Grignard reagent to attack an ester. The structure of $3-$methylpentan$-3-$ol is $CH_3CH_2-C(OH)(CH_3)-CH_2CH_3$.This tertiary alcohol can be formed by the reaction of an ester with two equivalents of a Grignard reagent.Specifically,reacting ethyl propanoate $(CH_3CH_2COOC_2H_5)$ with two equivalents of methyl magnesium bromide $(CH_3MgBr)$ yields $3-$methylpentan$-3-$ol.The reaction proceeds as follows:$CH_3CH_2COOC_2H_5 + CH_3MgBr \rightarrow CH_3CH_2COCH_3 + C_2H_5OMgBr$$CH_3CH_2COCH_3 + CH_3MgBr \rightarrow CH_3CH_2C(OMgBr)(CH_3)_2$After hydrolysis: $CH_3CH_2C(OH)(CH_3)_2$ (which is $2-$methylbutan$-2-$ol). Wait,let us re-evaluate.To get $3-$methylpentan$-3-$ol $(CH_3CH_2-C(OH)(CH_3)-CH_2CH_3)$,we need to react ethyl ethanoate $(CH_3COOC_2H_5)$ with two equivalents of ethyl magnesium bromide $(C_2H_5MgBr)$.$CH_3COOC_2H_5 + 2C_2H_5MgBr \rightarrow CH_3C(OH)(C_2H_5)_2$.Thus,the correct option is $B$.

$3-$methylpentan$-3-$ol will be prepared from

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