Test: Solved Practice Questions on LiAlH4, NaBh4 & grignard reagent - Chemistry MCQ

Reaction (b) with two equivalents of the Grignard reagent and the ester followed by an aqueous workup gives a tertiary alcohol, but it is 1,1- diphenylethanol (CH₃CPh₂OH).

In reaction (d), the reduction of the amide with UAIH4 leads to the amine, MeOCH₂CH₂NMe₂.

Esters (other than formates) and ketones generally react with Grignard reagents to give tertiary alcohols, but aldehydes give secondary alcohols.

In reaction (a), the ester cannot be reduced with NaBH₄.

Draw structures of the carbonyl compounds first, then identify the products of their reactions with the Grignard reagent.

In this reaction, the B-H bond provides the electron pair which forms the new bond, so this is where the curly arrow should start; the formal negative charge on B does not represent a lone pair. 

LiAIH₄ is sufficiently reactive to reduce both ketone and ester groups: the ester part initially becomes the conjugate base of a hemiacetal which forms a ring-opened aldehydic alkoxide in the reaction mixture, and the aldehyde residue is then reduced further; acidification gives the triol shown.

The total number of pets sold in March and April combined is 500. So, the equation is:

(Dogs sold in March + Cats sold in March) + (Dogs sold in April + Cats sold in April) = 500

Substitute the values:

(2x + x) + (4x + 3x) = 500

Simplifying:

3x + 7x = 500

10x = 500

Solve for x:

x = 50

Now, the number of dogs sold in March is 2x = 2 * 50 = 100.

So, the correct answer is b) 100.

LiAIH₄ is a powerful hydride donor which generally reduces all types of carbonyl functional groups.

NaBH₄ is a mild hydride donor which, in general, reduces aldehydes and ketones but not esters or amides.