Chapter 17: Q23E (page 567)

Predict the product from reaction of the following substance with:
${NaBH}_{4}$ ; then $H_{3} O^{+}$
${LiAlH}_{4}$ ; then $H_{3} O^{+}$
3. $2 {CH}_{3} {CH}_{2} MgBr$ ; then $H_{3} O^{+}$

Short Answer

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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 ${NaBH}_{4}$ , ${LiAlH}_{4}$ ,and ${CH}_{3} {CH}_{2} MgBr$ . The ${NaBH}_{4}$ 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 ${NaBH}_{4}$ . This is because ${NaBH}_{4}$ does not react with esters. ${NaBH}_{4}$ only reduces ketones and aldehydes to alcohol products.

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

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

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

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

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

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