JEE Advanced 2018 · Paper 2 · Chemistry Q10

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ChemistryNumerical+3Question 28 of 54

For the given compound X, the total number of optically active stereoisomers is ____.

Compound X: a cyclopentane bearing two OH groups (both wedged) at C1 and C3 and two side chains at C4 and C5. The C5 chain (wedged bond) is CH=CH-CH(OH)-(CH₂)₄CH₃ with a fixed E double bond and a wedged OH. The C4 chain is attached by a wavy bond and is CH=CH-CH(OH)-(CH₂)₄CH₃ with a wavy double bond and a wavy OH. Legend: a wedge means the configuration at that carbon and the double-bond geometry are fixed; a wavy bond means they are NOT fixed.

Your answer

Answer and worked solution

Official answer: 7

Count the stereo elements whose configuration is not fixed (wavy bonds):

The ring carbon carrying the lower side chain.
The C=C geometry in the lower chain.
The carbinol carbon in the lower chain.

Everything else is fixed. That gives $2^3=8$ stereoisomers.

Meso form. The ring is $\ce{CH(OH)-CH(R)-CH(R)-CH(OH)-CH2}$ , so it has symmetric constitution about a plane through the $\ce{CH2}$ . The two chains are constitutionally identical. When all three free elements are chosen as the mirror image of the fixed upper half, the molecule has a mirror plane and is achiral (meso).

Optically active: $8-1=7$ .

Answer: 7

Topics:Optical isomerism and chirality Geometrical isomerism

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JEE Advanced 2018 · Paper 2, Chemistry question 10