JEE Advanced 2018 · Paper 1 · Chemistry Q4

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ChemistryMultiple correct+4 / -2Question 22 of 54

In the following reaction sequence, the correct structure(s) of X is (are)

X treated with (1) PBr₃, Et₂O, (2) NaI, Me₂CO, (3) NaN₃, HCONMe₂ gives enantiomerically pure trans-1-azido-3-methylcyclopentane: Me wedged on the upper-left ring carbon and N₃ hashed on the upper-right carbon; (1R,3R).

trans-3-Methylcyclopentan-1-ol drawn with the ring apex up: Me wedged on the upper-left carbon, OH hashed on the upper-right carbon; (1R,3R).

cis-3-Methylcyclopentan-1-ol drawn with the ring apex up: Me wedged on the upper-left carbon, OH wedged on the upper-right carbon; (1S,3R).

trans-3-Methylcyclopentan-1-ol drawn with the ring apex down (mirror image of A): Me wedged, OH hashed; (1S,3S).

cis-3-Methylcyclopentan-1-ol drawn with the ring apex down (mirror image of B): Me wedged, OH wedged; (1R,3S).

Answer and worked solution

Official answer: B

Count the inversions.

$\ce{PBr3}$ converts the secondary alcohol to a bromide by an $S_N2$ step: inversion.
$\ce{NaI}$ in acetone (Finkelstein, $S_N2$ ): inversion.
$\ce{NaN3}$ in DMF ( $S_N2$ ): inversion.

Three inversions give an overall inversion at C-1. The methyl-bearing C-3 is untouched.

Work backwards. The product is trans-1-azido-3-methylcyclopentane, (1R,3R). So X has the opposite configuration at C-1 and the same at C-3: (1S,3R)-3-methylcyclopentan-1-ol. That is the cis alcohol.

Structure (B) is (1S,3R). (D) is its enantiomer (1R,3S), which would give the enantiomeric azide.

(B)

Topics:SN1 and SN2 Optical isomerism and chirality

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JEE Advanced 2018 · Paper 1, Chemistry question 4