Acetyl bromide reacts with excess of CH_3MgI | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$-methyl-$2$-propanol

Vedclass · Answer

(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:

Similar Questions

How will you convert butan-$2$-one to propanoic acid?
$CH_3-CH_2-CO-CH_3 + I_2 + NaOH$ $\rightarrow C_2H_5CO_2^-Na^+ + CHI_3$ $\xrightarrow{H^+} C_2H_5COOH$

Acetals are

Which of the following compounds exhibits the Schiff test?

Clemmensen reduction of ketones is carried out in

Match the following lists:

Vedclass Test Series

Exam Paper Generator

Online Exam Module