Q23E - Organic Chemistry (Mcmurry)

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  ${\mathbf{H}}_3{\mathbf{O}}^{+}$

                                                                                           

Step-by-Step Solution

Answer

a. 

b.

c.

Step-by-Step Solution Step 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.

Step 2: The product of the reaction with NaBH 4 and then with H 3 O +

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 ${\text{NaBH}}_{\text{4}}$

Step 3: The product of the reaction with LiAlH 4 and then with H 3 O +

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 ${\text{LiAlH}}_{\text{4}}$

Step 4: The product of the reaction with 2   CH 3 CH 2 MgBr and then with H 3 O +

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 ${\text{2\hspace{0.17em}CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$