Question

$2R-MgBr+CdC{l}_2\stackrel{X,Y}{\to }{R}_{2}Cd+MgB{r}_2+M{g}_C{l}_2$
What will be there in the place of $X,Y$ in the above reaction equation to convert the highly reactive Grignard and cadmium reagents to less nucleophilic species?

• A. ether , benzene
• B. ester, phenol
• C. ketone, carboxylic acid
• D. None of the above

Answer 1

The correct option is A ether , benzene

In Grignard reagent. $R-MgBr$ is highly polar. This is one which involves the reaction of an aryl halide or an alkyl halide in the presence of ether.

Ether act as a solvent for the bond between $R-Mg$

Benzene has no lone pair of electrons so does not form products.

Hence these two gives less nucleophilic species.