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MCQEasyJEE 2024Alkynes (Acidic Nature, Reactions)

JEE Chemistry 2024 Question with Solution

The final product A, formed in the following multistep reaction sequence is: CH₃ - C≡CH + Na → A B → CH₃ - C≡C - CH₂ - CH₂ - CH₃ + NaBr

Choose the correct answer from the options below:

  • A

    A = CH₃–C≡CNa, B = CH₃–CH₂–CH₂–Br

  • B

    A = CH₃–CH₂–CH₂Br, B = CH₃–C≡C–CH₃

  • C

    A = CH₃–C≡CNa, B = CH₃–C≡CH

  • D

    A = CH₃–C≡CNa, B = CH₃–CH₂–CH₃

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The starting compound is CH₃ - C≡CH and it reacts with Na to form A. Then B reacts to give CH₃ - C≡C - CH₂ - CH₂ - CH₃ + NaBr.

Find: The correct identities of A and B.

A terminal alkyne reacts with sodium metal to form its sodium salt. Therefore,

CH3−C≡CH+Na→CH3−C≡CNa\text{CH}_3-\text{C}\equiv\text{CH} + \text{Na} \rightarrow \text{CH}_3-\text{C}\equiv\text{CNa}

So, A = CH₃–C≡CNa.

Now the product formed is CH₃ - C≡C - CH₂ - CH₂ - CH₃ with NaBr as a by-product. This indicates an alkylation of the sodium acetylide with an alkyl bromide. To attach the CH₂ - CH₂ - CH₃ group, B must be CH₃–CH₂–CH₂–Br.

Thus,

CH3−C≡CNa+CH3CH2CH2Br→CH3−C≡C−CH2−CH2−CH3+NaBr\text{CH}_3-\text{C}\equiv\text{CNa} + \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \rightarrow \text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_3 + \text{NaBr}

Therefore, the correct option is A.

Source solution discrepancy noted

Given: The source the solution states that the correct option is A / Option (1).

Find: Whether that matches the reaction sequence shown in the question.

The explanatory text in the solution discusses hydration, oxidation with CrO₃, and Wolff-Kishner reduction. That chemistry is unrelated to the given alkyne alkylation question, so that working is for a different question.

Using the actual reaction shown in the question:

  1. CH₃ - C≡CH is a terminal alkyne.
  2. Reaction with Na forms the sodium acetylide CH₃–C≡CNa.
  3. To obtain CH₃ - C≡C - CH₂ - CH₂ - CH₃, the reagent B must provide a propyl group through substitution, so B = CH₃–CH₂–CH₂–Br.

Hence the option consistent with the question is Option (1), i.e. A.

Common mistakes

  • Mistake: Treating Na as if it adds across the triple bond. Why wrong: sodium metal here removes the acidic terminal hydrogen of the alkyne to form a sodium acetylide. What to do instead: first identify CH₃–C≡CH as a terminal alkyne and convert it to CH₃–C≡CNa.

  • Mistake: Choosing an alkene, alkane, or alkyne as B instead of an alkyl bromide. Why wrong: the final step produces NaBr, which signals nucleophilic substitution with an alkyl bromide. What to do instead: infer that B must be CH₃–CH₂–CH₂–Br.

  • Mistake: Using a secondary or mismatched carbon skeleton for B. Why wrong: the final product requires addition of a straight CH₂–CH₂–CH₃ fragment to the acetylide carbon. What to do instead: match the exact fragment being introduced before selecting the alkyl halide.

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