Chapter 10: Q24P (page 527)

Predict the products you would expect from the reaction of NaBH4 with the following compounds. You may assume that these reactions take place in methanol as the solvent
a.CH₃(CH₂)₈-CHO
b.
c. Ph-COOH
d.
e.
f.

Short Answer

Expert verified

Answer

$C H_{3} - {(C H_{2})}_{8} - C H O + N a B H_{4} \to C H_{3} - {(C H_{2})}_{8} - C H_{2} - O H$

Step by step solution

Reaction of NaBH4 with aldehyde, ketone

The ${NaBH}_{4}$ isa strong reducing agent. It reduces aldehyde to primary alcohol and ketone to secondary alcohol but has no effect on ester and acid respectively.

Step 2: Formation of product on reaction with NaBH4

$C H_{3} - {(C H_{2})}_{8} - C H O + N a B H_{4} \to C H_{3} - {(C H_{2})}_{8} - C H_{2} - O H$

In this reactionreduces aldehyde to primary alcohol as the reactant contains the aldehyde group.

In this reaction as the reactant contains an ester group, on which the does not show any effect. Thus, no reaction takes place.

In this reaction as the reactant contains an acid group. The does not show any effect on the acid and ester group. Thus, no reaction takes place.

In this reaction the reactant contains a ketone group which is reduced to secondary alcohol by the reducing agent, i.e., NaBH₄.

In this reaction the ester group present has no effect but the aldehyde group present is reduced to primary alcohol and ketone group to secondary alcohol by the reducing agent ( ${NaBH}_{4}$ ) respectively.