Question

Ethyl formate ester reacts with \[{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{MgBr}}\] to give secondary alcohol. The alcohol formed is

A)

B)

C)

D)

Answer 1

Hint: Formic acid contains one carbon atom. Propyl magnesium bromide contains three carbon atoms. The product will have seven carbon atoms.

Complete answer:
The structures of the starting materials ethyl formate and propyl magnesium bromide are as shown below:

Write the chemical reaction

Two moles of propyl magnesium bromide react with one mole of ethyl formate to form heptan-4-ol. Propyl magnesium bromide is a Grignard reagent. Ethyl formate is an ester. Heptan-2-ol is a secondary alcohol. Thus, one mole of ester reacts with two moles of grignard reagent to form one mole of heptan-2-ol. In the reaction, two moles of \[{\text{Mg}}\left( {{\text{OH}}} \right){\text{Br}}\] and one mole of ethanol are the side product.

Hence, the option B ) heptan-4-ol is the correct option.

Additional Information: An ester of formic acid reacts with two equivalents of a grignard reagent to form a secondary alcohol. An ester reacts with two equivalents of a grignard reagent to form a tertiary alcohol. An aldehyde reacts with one equivalent of grignard reagent to form secondary alcohol. A ketone reacts with one equivalent of grignard reagent to form tertiary alcohol.

Note: If only one equivalent of propyl magnesium bromide reacts, then ethyl formate will give butane aldehyde. Since in the question, it is mentioned that the product is a secondary alcohol, two equivalents of propyl magnesium bromide will react with ethyl formate to form heptan-4-ol. In the secondary alcohol, the carbon atom bearing hydroxyl group has one hydrogen atom and two other carbon atoms.