Acetyl bromide reacts with excess of CH_3MgI | vedclass

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(A) Acetyl bromide $(CH_3COBr)$ reacts with the first equivalent of $CH_3MgI$ to form acetone $(CH_3COCH_3)$.Since $CH_3MgI$ is in excess,it further r

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$2$-methyl-$2$-propanol

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(A) Acetyl bromide $(CH_3COBr)$ reacts with the first equivalent of $CH_3MgI$ to form acetone $(CH_3COCH_3)$.Since $CH_3MgI$ is in excess,it further reacts with the formed acetone to produce a tertiary alkoxide intermediate $(CH_3-C(CH_3)_2-OMgI)$.Finally,hydrolysis of this intermediate with a saturated solution of $NH_4Cl$ yields $2$-methyl-$2$-propanol $(CH_3-C(CH_3)_2-OH)$.Reaction: $CH_3COBr + 2CH_3MgI$ $\rightarrow CH_3-C(CH_3)_2-OMgI$ $\xrightarrow{NH_4Cl} CH_3-C(CH_3)_2-OH$.

Acetyl bromide reacts with excess of $CH_3MgI$ followed by treatment with a saturated solution of $NH_4Cl$ gives:

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