Question

Predict the product from reaction of the following substance with:

1. ${\mathbf{NaBH}}_4$; then ${\mathbf{H}}_3{\mathbf{O}}^{+}$
2. ${\mathbf{LiAlH}}_4$; then ${\mathbf{H}}_3{\mathbf{O}}^{+}$

3. $\mathbf{2}{\mathbf{CH}}_3{\mathbf{CH}}_2\mathbf{MgBr}$; then ${\mathbf{H}}_3{\mathbf{O}}^{+}$

Short answer

a.

b.

c.

Step-by-step solution

Step-by-Step SolutionStep 1: Reduction of carbonyl compounds

The synthesis of alcohols can be done by reducing aldehydes, ketones, esters, and carboxylic compounds with various reagents like ${\mathbf{NaBH}}_4$ ,${\mathbf{LiAlH}}_4$,and ${\mathbf{CH}}_3{\mathbf{CH}}_2\mathbf{MgBr}$. The${\mathbf{NaBH}}_4$reagent only reduces aldehydes and ketones to alcohol, not esters and carboxylic acids.

The product of the reaction with NaBH4 and then with H3O+  

a. No reaction is observed when an ester-containing compound reacts with ${\text{NaBH}}_{\text{4}}$. This is because ${\text{NaBH}}_{\text{4}}$does not react with esters. ${\text{NaBH}}_{\text{4}}$only reduces ketones and aldehydes to alcohol products.

Product of the reaction with${\text{NaBH}}_{\text{4}}$

The product of the reaction with LiAlH4 and then with  H3O+

b. When an ester-containing compound reacts with ${\text{LiAlH}}_{\text{4}}$, it gives primary alcohol product. ${\text{LiAlH}}_{\text{4}}$reduces esters to primary alcohols.

Product of the reaction with${\text{LiAlH}}_{\text{4}}$

The product of the reaction with 2 CH3CH2MgBr  and then with  H3O+ 

c.When an ester undergoes a reaction with Grignard reagent (${\text{2\hspace{0.17em}CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$), a tertiary alcohol product is formed. So, we get tertiary alcohol and methanol as a product in this reaction.

Product of the reaction with${\text{2\hspace{0.17em}CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$