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Question: Match the reaction intermediates formed during the reactions given in Column-I with Column-II | ...

Match the reaction intermediates formed during the reactions given in Column-I with Column-II

Column-IColumn-II
ACH3−C≡C−H→Na\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} \xrightarrow{\mathrm{Na}} P Carbocation (Non classical)
BCH3−CH=CH2→HBr\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow{\mathrm{HBr}} Q Carbocation (Classical)
CCH3−CH=CH2→PeroxideHBr\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[\text{Peroxide}]{\mathrm{HBr}} R Carbanion
DCH3−CH=CH2→/CCl4Br2S\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[\text{/CCl}_4]{\mathrm{Br}_{2}} \mathrm{S}S Alkyl free radical
A B C D
A

R S Q P

B

R Q S P

C

S R Q P

D

P Q S R

Answer

2

Explanation

Solution

The problem asks us to match the given reactions with the reaction intermediates formed during them. We will analyze each reaction individually.

A. CH3−C≡C−H→Na\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} \xrightarrow{\mathrm{Na}}

This is the reaction of a terminal alkyne with sodium metal. Terminal alkynes have an acidic hydrogen. Sodium is a strong base and will deprotonate the alkyne, forming an acetylide anion.
CH3−C≡C−H+Na→CH3−C≡C−Na++12H2\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H} + \mathrm{Na} \rightarrow \mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}^- \mathrm{Na}^+ + \frac{1}{2} \mathrm{H}_2
The intermediate formed is CH3−C≡C−\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}^-, which is a Carbanion.
Thus, A matches with R.

B. CH3−CH=CH2→HBr\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow{\mathrm{HBr}}

This is an electrophilic addition reaction of HBr to propene, following Markovnikov's rule. The first step involves the addition of a proton (H+\mathrm{H}^+) to the double bond, forming the more stable carbocation.
CH3−CH=CH2+H+→CH3−CH+−CH3\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} + \mathrm{H}^+ \rightarrow \mathrm{CH}_{3}-\mathrm{CH}^+-\mathrm{CH}_{3} (secondary carbocation)
This is a carbon atom with a positive charge, which is a Classical Carbocation.
Thus, B matches with Q.

C. CH3−CH=CH2→PeroxideHBr\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[\text{Peroxide}]{\mathrm{HBr}}

This is the addition of HBr to propene in the presence of peroxide, which is an anti-Markovnikov addition (Kharasch effect) via a free radical mechanism. The propagation step involves the addition of a bromine radical (Br⋅\mathrm{Br} \cdot) to the alkene, forming the more stable alkyl free radical.
CH3−CH=CH2+Br⋅→CH3−CH⋅−CH2Br\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} + \mathrm{Br} \cdot \rightarrow \mathrm{CH}_{3}-\mathrm{CH} \cdot -\mathrm{CH}_{2}\mathrm{Br} (secondary alkyl radical)
The intermediate formed is an Alkyl free radical.
Thus, C matches with S.

D. CH3−CH=CH2→/CCl4Br2\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[\text{/CCl}_4]{\mathrm{Br}_{2}}

This is an electrophilic addition reaction of Br2\mathrm{Br}_2 to propene. The mechanism involves the formation of a cyclic bromonium ion. The Br2\mathrm{Br}_2 molecule is polarized, and one bromine atom attacks the double bond, forming a three-membered ring intermediate with a positive charge on the bromine atom. This bridged ion is considered a Non-classical Carbocation.
Thus, D matches with P.

Summary of Matches:
A - R
B - Q
C - S
D - P

Comparing this with the given options:
(1) R S Q P
(2) R Q S P
(3) S R Q P
(4) P Q S R

The correct sequence is R Q S P, which corresponds to option (2).