Question

What are the reagents X and Y respectively used in the following reaction sequence to convert a nitrile to a methyl group? 

 

• A. ZnCl$_2$; Conc. HCl : (i) SnCl$_2$, HCl (ii) H$_3$O$^+$
• B. (i) SnCl$_2$, HCl (ii) H$_3$O$^+$ : Zn - Hg|Conc. HCl
• C. LiAlH$_4$ : NaBH$_4$
• D. (i) DIBAL-H (ii) H$_2$O : LiAlH$_4$

Hint:

When selecting reagents for multiple step reactions, consider the overall pathway and the compatibility of each step with the next to ensure successful conversion without over-reduction or unwanted side reactions.

Answer 1

The Correct Option is B

Step 1: Examine the reaction conditions and requirements.

The conversion of a nitrile to a methyl group involves significant reduction, typically requiring strong reducing agents and conditions that facilitate both reduction and hydrolysis.

Step 2: Analyze the selected option.

SnCl$_2$ in the presence of HCl (reagent X) is used initially to reduce the nitrile group partially, likely to an imine.

H$_3$O$^+$ (part of reagent Y) then hydrolyzes the imine to an amine.

The final step involves the use of Zn-Hg amalgam in concentrated HCl (additional part of reagent Y), which is a classic Clemmensen reduction, reducing the amine further to a methyl group. This sequence of reactions is particularly effective for fully reducing nitrile to a primary alkyl group without over-reduction or unwanted side reactions.