JEE Advanced
Chemistry
Haloalkanes and Haloarenes
2023
JEE Advanced 2023 (Paper 1)
JEE Advanced Chemistry Question (2023) — Solution
Question
Match the reactions in List-I with the features of their products in List-II and choose the correct option. List I List II ( P ) - - 1 - Bromo - 2 - ethylpentane ( Single   enantiomer )     → S N   2     reaction aq .  NaOH 1 Inversion of configuration ( Q ) - - 2 - Bromopentane Single   Enantiomer     → S N   2   reaction aq .  NaOH         2 Retention of configuration ( R ) - - 3 - Bromo - 3 -   methylhexane   Single   Enatiomer       → S N 1 aq . NaOH 3 Mixture of enantiomers ( S ) 4 Mixture of structural isomers 5 Mixture of diastereomers
Options
- A. P   →   1 ;   Q   →   2 ;   R   →   5 ;   S   →   3
- B. P   →   2 ;   Q   →   1 ;   R   →   3 ;   S   →   5
- C. P → 1 ;   Q → 2 ;   R → 5 ;   S → 4
- D. P   →   2 ;   Q   →   4 ;   R   →   3 ;   S   →   5
Answer
B. P   →   2 ;   Q   →   1 ;   R   →   3 ;   S   →   5
Step-by-step solution
1 - Bromo - 2 - ethylpentane ( Single   enantiomer )     → S N   2     reaction aq .  NaOH 2 - ethylpentanol In the above reaction the chiral carbon is not involving in the reaction as S N 2 reaction takes place via formation of transition state but not through formation of intermediate. In the above reaction, bromine attached to chiral carbon, and it is involving in S N 2 reaction. Hence, the alcohol formed in inversion configuration. The substrate is undergoing S N 1 reaction. It takes place via formation of carbocation. Carbocation as planar, hence, the alcohol product formed is racemic mixture as nucleophile can attack from both the planes. The substrate is undergoing S N 1 reaction. It takes place via formation of carbocation. Carbocation as planar, hence, the two alcohol products are formed as nucleophile can attack from both the planes but the substrate already having the chiral centre, the mixture formed is diastereomeric mixture.
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Related: Chemistry — Haloalkanes and Haloarenes · All PYQ Banks