Match List-I with List-II
| List-I | List-II |
| (Chemical Reactions) | (Characteristic Feature) |
| (A) Substitution nucleophilic unimolecular | (I) Anti-Markovnikov’s product |
| (B) Electrophilic addition of HX to unsymmetrical alkene | (II) Inversion |
| (C) Substitution nucleophilic bimolecular | (III) Racemization |
| (D) Hydroboration oxidation | (IV) Markovnikov’s product |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Each reaction leaves its own mark on the product. In a substitution the mark is the stereochemistry (from which side the new group enters); in an addition to an unsymmetrical alkene it is the regiochemistry (to which carbon the new group goes).
(A) Substitution nucleophilic unimolecular — This matches (III) Racemization. In SN1 the leaving group departs first and gives a planar carbocation. The nucleophile can attack from either face, so an optically active substrate gives a mixture of both enantiomers.
(B) Electrophilic addition of HX to unsymmetrical alkene — This matches (IV) Markovnikov’s product. \( \ce{H^{+}} \) adds so that the more stable carbocation forms, and \( \ce{X^{-}} \) then goes to that carbon.
\( \displaystyle \ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3} \)
(C) Substitution nucleophilic bimolecular — This matches (II) Inversion. In SN2 the nucleophile attacks from the side opposite the leaving group in a single step, and the configuration turns over like an umbrella in the wind.
(D) Hydroboration oxidation — This matches (I) Anti-Markovnikov’s product. Boron adds to the less substituted carbon, and \( \ce{H2O2} \)/\( \ce{OH^{-}} \) then replaces boron by –OH, so the –OH ends up on the less substituted carbon.
\( \displaystyle \ce{CH3-CH=CH2} \xrightarrow[\text{(ii) }\ce{H2O2/OH^-}]{\text{(i) }\ce{B2H6}} \ce{CH3-CH2-CH2-OH} \)
Shortcut: The pair (A)–(III) appears in Options (3) and (4), and (C)–(II) then leaves only Option (3).