Number of moles of CH_3MgBr consumed in the a | vedclass

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(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:

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