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Basics of Organic Chemistry question

2022 · 27 Jul · Shift 1 · Q22
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Basics of Organic Chemistry question

2022 · 27 Jul · Shift 1 · Q22

JEE MainChemistryBasics of Organic ChemistryNumerical+4 / −1
Optical activity of an enantiomeric mixture is +12.6∘+12.6^{\circ}+12.6∘ and the specific rotation of (+)(+)(+) isomer is +30∘+30^{\circ}+30∘. The optical purity is ‾\underline{\hspace{2cm}}​%\%%.
Numerical answer
View written solutionFree

Correct answer: 42

  1. Use the relation between observed rotation and optical purity

For an enantiomeric mixture,

optical purity=observed rotationrotation of pure enantiomer×100\text{optical purity} = \frac{\text{observed rotation}}{\text{rotation of pure enantiomer}} \times 100optical purity=rotation of pure enantiomerobserved rotation​×100

Given:

  • Observed optical activity =+12.6∘= +12.6^\circ=+12.6∘
  • Specific rotation of pure (+)(+)(+) isomer =+30∘= +30^\circ=+30∘

So,

optical purity=12.630×100\text{optical purity} = \frac{12.6}{30} \times 100optical purity=3012.6​×100
  1. Calculate
12.630=0.42\frac{12.6}{30} = 0.423012.6​=0.42

Thus,

optical purity=0.42×100=42%\text{optical purity} = 0.42 \times 100 = 42\%optical purity=0.42×100=42%
  1. Final answer
42%\boxed{42\%}42%​

This also equals the enantiomeric excess.

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