Question

What is the final product obtained in the reaction between ethane nitrile and ethyl magnesium bromide?
A.Methoxy propane
B.Butanamine
C.Butan-2-one
D.Pent-2-one

Answer 1

Hint: The chemical compounds which have the general formula ${\text{R - Mg - X}}$ where X represents a halogen and R represents an organic group, which may be alkyl or aryl, are termed as ‘Grignard reagents’.
Aliphatic ketones can be prepared by the action of a suitable Grignard reagent on an alkyl nitrile. This method of preparation is useful for preparing ketones only.

Complete step by step answer:
We need to determine what the final product is in the reaction between ethane nitrile and ethyl magnesium bromide.
Ethane nitrile, also known as methyl cyanide or acetonitrile is the simplest organic nitrile and has the molecular formula ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CN}}$ .
Ethyl magnesium bromide is a Grignard reagent having the formula ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{MgBr}}$ .
Thus, when ethane nitrile is acted upon by ethyl magnesium bromide, the strongly organometallic reagent adds to the carbon-nitrogen triple bond in a way similar to aldehydes and ketones. The reaction first proceeds through an imine salt intermediate which is then hydrolysed to give butan-2-one.
 butan-2-one.

Therefore, option C is correct.

Methoxy propane or methyl propyl ether is an ether, not a ketone. So it cannot be the product of the reaction between a nitrile and a Grignard reagent. So, A is incorrect.

Butanamine is an amine of butane and so it cannot be the product of the reaction between a nitrile and a Grignard reagent. So, B is incorrect.

Pent-2-one is methyl propyl ketone and it will be formed when propyl magnesium bromide reacts with ethane nitrile. So. D is also incorrect.

Hence option C is correct.

Note:
If hydrocyanic acid is used instead of nitrile, then aldehyde is formed.