Number of moles of CH_3MgBr consumed in the a | vedclass

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

Question

(B) The reactant is phthaloyl chloride $(C_6H_4(COCl)_2)$.Each acid chloride group $(-COCl)$ reacts with $2$ moles of Grignard reagent $(CH_3MgBr)$ to

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The reactant is phthaloyl chloride $(C_6H_4(COCl)_2)$.Each acid chloride group $(-COCl)$ reacts with $2$ moles of Grignard reagent $(CH_3MgBr)$ to form a tertiary alcohol.Specifically,the first mole of $CH_3MgBr$ attacks the carbonyl carbon to form a ketone intermediate,and the second mole attacks the ketone to form the tertiary alcohol after acidic workup $(H^+)$.Since there are two $-COCl$ groups in the molecule,a total of $2 	imes 2 = 4$ moles of $CH_3MgBr$ are consumed.Therefore,the correct option is $B$.

Number of moles of $CH_3MgBr$ consumed in the above reaction is:

Similar Questions

Consider the following reaction. Which of the following products is not expected to be formed?
$CH_3CH_2CH(Br)CH_3 + HC \equiv C^- Na^+ \rightarrow ?$

Assertion $(A)$: $1-$Bromopentane reacts with $AgCN$ to give pentylisocyanide.
Reason $(R)$: $AgCN$ is mainly ionic in nature.

Chlorination of toluene in the presence of light and heat followed by reaction with aqueous $NaOH$ gives ...................

The reaction for the preparation of ethene from ethyl chloride is known as......

$Me_2C = CH - CH_2 - CH_2 - Cl$ $\xrightarrow[CaCO_3]{H_2O} (X);$ What is the major product $(X)$ of the reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module