JEE Advanced 2020 · Paper 2 · Chemistry Q12

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ChemistryMultiple correct+4 / -2Question 30 of 54

Consider the following transformations of a compound P.

Choose the correct option(s).

P (C₉H₁₂) with Pt/H₂ gives propylcyclohexane. P with (i) NaNH₂, (ii) C₆H₅COCH₃, (iii) H₃O⁺/Δ gives R (optically active). P with (i) reagent X, (ii) KMnO₄/H₂SO₄/Δ gives Q (C₈H₁₂O₆, an optically active acid).

P is

Cyclohexene carrying a prop-2-ynyl (CH₂C≡CH) group at C4, opposite the ring double bond.

X is Pd-C/quinoline/

\ce{H2}

P is

Cyclohexene carrying a prop-2-ynyl (CH₂C≡CH) group at C3, next to the ring double bond.

R is

Ph-C(=CH₂)-C≡C-CH₂- attached to C4 of a cyclohexene ring (opposite the ring double bond).

Answer and worked solution

Official answer: B, C

What P must be. P ( $\ce{C9H12}$ , 4 degrees of unsaturation) hydrogenates to propylcyclohexane. With $\ce{NaNH2}$ it forms an acetylide, so it is a cyclohexene with a $\ce{CH2C#CH}$ side chain.

Which cyclohexene. Partial reduction of the alkyne with X, then hot $\ce{KMnO4}$ , cleaves both C=C bonds; the terminal $\ce{=CH2}$ is lost as $\ce{CO2}$ .

If the side chain is at C3, next to the ring C=C, the product is $\ce{HO2C-CH(CH2CO2H)-CH2CH2CH2-CO2H}$ . That is $\ce{C8H12O6}$ , and its CH is a stereocentre, so it is optically active. ✓
If the side chain is at C4, the product is $\ce{HO2C-CH2-CH(CH2CO2H)-CH2-CO2H}$ . Two identical arms make it achiral. ✗

So P is the C3 isomer, (C), not (A). X is the Lindlar-type catalyst Pd-C/quinoline/ $\ce{H2}$ , (B).

R. The acetylide adds to acetophenone, and acid/heat dehydrates the tertiary alcohol to $\ce{Ph-C(=CH2)-C#C-CH2-}$ (cyclohex-2-enyl). R is optically active because of the ring stereocentre. Option (D) draws the substituent at C4 of the ring, which would be achiral, so (D) is wrong.

(B), (C)

Topics:Alkynes Alkenes Optical isomerism and chirality

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JEE Advanced 2020 · Paper 2, Chemistry question 12