Question

In the following reaction, 13.4 grams of aldehyde P gave a diastereomeric mixture of alcohols Q and R in a ratio of 2:1. If the yield of the reaction is 80 percent, then the amount of Q (in grams) obtained is ___________ (in integer). 

Hint:

In reactions involving chiral centers, nucleophilic addition to a carbonyl group can lead to diastereomeric products. The ratio of these diastereomers depends on the stereochemical environment of the existing chiral center. The yield of the reaction must be taken into account when calculating the actual amount of product obtained.

Answer 1

The reaction involves the addition of methyl lithium (MeLi) to a chiral aldehyde (P). This nucleophilic addition to a chiral carbonyl center creates a new chiral center, leading to a mixture of two diastereomeric alcohols, Q and R.

Step 1: Molecular weight of aldehyde P (C₁₀H₁₂O):
MW(P) = (10 × 12.01) + (12 × 1.01) + 16.00 = 120.1 + 12.12 + 16.00 = 148.22 g/mol

Step 2: Moles of aldehyde P used:
Moles = 13.4 g / 148.22 g/mol ≈ 0.0904 mol

The reaction with MeLi will produce a 1:1 mixture of alcohols if the reaction went to completion without stereochemical influence. However, the problem states that a diastereomeric mixture of Q and R is formed in a 2:1 ratio, indicating that the existing chiral center affects the stereochemical outcome.

Step 3: Molecular weight of alcohols Q and R (C₁₁H₁₆O):
MW(Q) = MW(R) = (11 × 12.01) + (16 × 1.01) + 16.00 = 132.11 + 16.16 + 16.00 = 164.27 g/mol

Step 4: Theoretical yield of Q + R (100% yield):
= 0.0904 mol × 164.27 g/mol ≈ 14.85 g

Step 5: Actual yield (80% of theoretical):
= 0.80 × 14.85 g ≈ 11.88 g

Step 6: Distribution in 2:1 ratio:
Q = \( \frac{2}{3} \) × 11.88 g ≈ 7.92 g
R = \( \frac{1}{3} \) × 11.88 g ≈ 3.96 g

Final Answer:
Rounding to the nearest integer, the amount of Q obtained is 8 grams.