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

Which of the following pairs is differentiated by the iodoform test?

The reaction sequence is: $P$ $\xrightarrow{\Delta } X$ $\xrightarrow[{HCl}]{{Zn-Hg}} Y$. Here,$P$ is $2-$oxocyclohexanecarboxylic acid. Identify $Y$.

What is the product formed when a ketone reacts with $Mg-Hg$ in the presence of water?

Which reagent is suitable for the following transformation?

The reaction sequence is as follows:
$HOOC-CH_2-CH_2-CH_2-CH_2-COOH$ $\xrightarrow[BaCO_3]{\Delta} A$ $\xrightarrow{NH_2-NH_2} B$ $\xrightarrow[KOH]{\text{heat}} C$
The final product $(C)$ obtained is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module