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NVAMediumJEE 2023Reaction Mechanisms (Substitution, Addition, Elimination)

JEE Chemistry 2023 Question with Solution

The number of possible isomeric products formed when 3-chloro-1-butene reacts with HCl through carbocation formation is _____.

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: 3-chloro-1-butene reacts with HCl through carbocation formation.

Find: The total number of possible isomeric products.

Step 1: Protonation of the alkene The double bond is protonated by HCl. Following Markovnikov addition, a secondary carbocation is formed.

Reaction scheme showing protonation of 3-chloro-1-butene by H+ to form a secondary carbocation intermediate with chlorine substituent retained.

Step 2: Carbocation rearrangement The initially formed secondary carbocation undergoes a 1,2-hydride shift to generate a more stable tertiary carbocation.

Reaction scheme showing 1,2-hydride shift from the secondary carbocation to form a more stable tertiary carbocation intermediate.

Step 3: Attack by chloride ion Chloride ion can attack both carbocation intermediates.

From attack on the secondary carbocation, 1 product is formed:

CH3CHClCH2CH2Cl\text{CH}_3\text{CHClCH}_2\text{CH}_2\text{Cl}
Reaction scheme showing chloride ion attack on the secondary carbocation to give one dichlorobutane product.

From attack on the tertiary carbocation, 3 products are formed for

CH3CHClCHClCH3\text{CH}_3\text{CHClCHClCH}_3

namely one meso compound and one pair of enantiomers.

Reaction scheme showing chloride ion attack on the tertiary carbocation leading to three stereoisomeric products: one meso form and one enantiomeric pair.

Therefore, total possible isomeric products are

1+3=41 + 3 = 4

So, the total number of possible isomeric products is 44.

Product Counting by Intermediates

Given: Addition of HCl to 3-chloro-1-butene proceeds through carbocation formation.

Find: Count all distinct isomeric products formed after direct formation and rearrangement.

The key idea is that the reaction does not stop at only one carbocation. After protonation, the initially formed carbocation can either be trapped directly by Cl−\text{Cl}^- or rearrange before capture.

  1. Initial secondary carbocation gives one constitutional product after chloride attack.
  2. Rearranged tertiary carbocation gives a product with two stereogenic centers, so stereoisomerism must be counted.
  3. That stereochemical set contains one meso form and two enantiomers.

Thus, the total count is

1+3=41 + 3 = 4

Hence, the numerical value of the answer is 44.

Common mistakes

  • Counting only the product from the initially formed carbocation is incorrect because the solution shows a 1,2-hydride shift to a more stable tertiary carbocation. Both unrearranged and rearranged pathways must be considered.

  • Treating the tertiary-carbocation product as a single product is wrong because CH3CHClCHClCH3\text{CH}_3\text{CHClCHClCH}_3 shows stereoisomerism. You must count one meso compound and one pair of enantiomers separately.

  • Ignoring rearrangement stability is a conceptual error. Carbocations often rearrange to more stable intermediates before nucleophilic attack, so product counting must include rearranged intermediates whenever the mechanism permits.

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