Question

In the conversion of Grignard reagent into higher aldehydes, the other compound used is

• A. Ethyl formate
• B. Ethyl acetate
• C. HCN
• D. Both A and C

Answer 1

The correct option is D

Both A and C

The explanation for correct answers:

Option (D)

• A Grignard reagent is an organomagnesium compound and represented by the general formula $\mathrm{R}-\mathrm{Mg}-\mathrm{X}$ where $\mathrm{R}$ is an alkyl or aryl group and $\mathrm{X}$ is a halogen group.
• Ethyl formate$\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{COOH}\right)$ reacts with Methyl magnesium bromide and produces Acetaldehyde.

$\underset{\mathrm{Ethyl}\mathrm{formate}}{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{COOH}}+\underset{\mathrm{Methyl}\mathrm{magnesium}\mathrm{bromide}}{{\mathrm{CH}}_3\mathrm{MgBr}}\to \underset{\mathrm{Acetaldehyde}}{{\mathrm{CH}}_3\mathrm{CHO}}$

• Hydrogen cyanide$\left(\mathrm{HCN}\right)$ reacts with Methyl magnesium bromide and produces Acetaldehyde.

$\underset{\mathrm{Hydrogen}\mathrm{cyanide}}{\mathrm{HCN}}+\underset{\mathrm{Methyl}\mathrm{magnesium}\mathrm{bromide}}{{\mathrm{CH}}_3\mathrm{MgBr}}\to \underset{\mathrm{Acetaldehyde}}{{\mathrm{CH}}_3\mathrm{CHO}}$

• Hence option (D) Both A and C is correct.

The explanation for incorrect answers:

Option (B) Ethyl acetate

• Ethyl acetate $\left({\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{COOCH}}_3\right)$ reacts with Grignard reagent and produces Ketone.

$\underset{\mathrm{Ethyl}\mathrm{acetate}}{{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{COOCH}}_3}+\underset{\mathrm{Methylmagnesium}\mathrm{bromide}}{{\mathrm{CH}}_3\mathrm{MgBr}}\to \underset{\mathrm{Acetone}}{{\mathrm{CH}}_3{\mathrm{COCH}}_3}$

• Hence option (B) Ethyl acetate is incorrect

Therefore, option (D) is the correct answer.