Question 36: (+) tartaric acid has a specific rotation of +12.0°. Then the specific rotation…

Question 36: (+) tartaric acid has a specific rotation of +12.0°. Then the specific rotation of a mixture of 68% (+) tartaric acid and 32% (−) tartaric acid is x × 10−1. The value of x is (nearest integer)

The correct answer is: $43$

Explanation

The observed specific rotation is proportional to the enantiomeric excess (difference in fractions) because the two enantiomers have equal and opposite rotations. Here the enantiomeric excess is $68\% – 32\% = 36\%$, so the net rotation is $0.36$ times $+12.0^\circ$.

Steps:

  1. Enantiomeric excess: $$ee = 0.68 – 0.32 = 0.36$$
  2. Net specific rotation: $$[α]_{\text{net}} = ee \times (+12.0^\circ) = 0.36 \times 12.0^\circ = 4.32^\circ$$
  3. Write as $x\times 10^{-1}$: $$4.32 = x\times 10^{-1}\Rightarrow x = 43.2 \approx 43$$

Therefore the nearest integer value of $x$ is 43.

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