A pure sample of 2-chlorobutane shows an observed rotation of under standard conditions. If this sam — Isomerism and Stereochemistry Chemistry Question
Question
A pure sample of 2-chlorobutane shows an observed rotation of $+30^\circ$ under standard conditions. If this sample is contaminated by its enantiomer such that the new observed rotation becomes $+22.5^\circ$, what is the percentage of the dextrorotatory ($d$ -form) in the contaminated mixture?
Answer: 87.5
💡 Solution & Explanation
The optical purity (enantiomeric excess) is given by $ee = (\alpha_{obs} / \alpha_{pure}) \times 100 = (22.5 / 30) \times 100 = 75\%$. Let $x$ be the fraction of the $d$ -form. Then $x - (1-x) = 0.75 \implies 2x = 1.75 \implies x = 0.875$. Thus, the mixture contains 87.5% of the $d$ -form.
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