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MCQEasyJEE 2024Alkynes (Acidic Nature, Reactions)

JEE Chemistry 2024 Question with Solution

Compound A formed in the following reaction reacts with B, giving the product C. CH3−C≡CH+Na→A\mathrm{CH_3-C\equiv CH + Na \rightarrow A} B→CH3−C≡C−CH2−CH2−CH3+NaBr\mathrm{B \rightarrow CH_3-C\equiv C-CH_2-CH_2-CH_3 + NaBr} Find out A and B.

  • A

    A = CH3−C≡CNa\mathrm{CH_3-C\equiv CNa}, B = CH3−CH2−CH2−Br\mathrm{CH_3-CH_2-CH_2-Br}

  • B

    A = CH3−CH2−CH2Br\mathrm{CH_3-CH_2-CH_2Br}, B = CH3−C≡C−CH3\mathrm{CH_3-C\equiv C-CH_3}

  • C

    A = CH3−C≡CNa\mathrm{CH_3-C\equiv CNa}, B = CH3−C≡CH\mathrm{CH_3-C\equiv CH}

  • D

    A = CH3−C≡CNa\mathrm{CH_3-C\equiv CNa}, B = CH3−CH2−CH3\mathrm{CH_3-CH_2-CH_3}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Propyne reacts with sodium to form intermediate A, and then A reacts with B to give CH3−C≡C−CH2−CH2−CH3\mathrm{CH_3-C\equiv C-CH_2-CH_2-CH_3} and NaBr\mathrm{NaBr}.

Find: The identities of A and B.

A terminal alkyne forms a sodium acetylide with sodium metal. Thus, from propyne:

CH3−C≡CH+Na→CH3−C≡C−Na++12H2\mathrm{CH_3-C\equiv CH + Na \rightarrow CH_3-C\equiv C^-Na^+ + \tfrac{1}{2}H_2}

So, A is CH3−C≡CNa\mathrm{CH_3-C\equiv CNa}.

Now the acetylide ion acts as a nucleophile and undergoes SN2\mathrm{S_N2} substitution with a primary alkyl bromide to form the longer alkyne:

CH3−C≡C−Na++CH3CH2CH2Br→CH3−C≡C−CH2−CH2−CH3+NaBr\mathrm{CH_3-C\equiv C^-Na^+ + CH_3CH_2CH_2Br \rightarrow CH_3-C\equiv C-CH_2-CH_2-CH_3 + NaBr}

Therefore, B is CH3−CH2−CH2−Br\mathrm{CH_3-CH_2-CH_2-Br}.

The solution contains a contradictory header stating Option C, but the worked chemistry clearly identifies A = CH3−C≡CNa\mathrm{CH_3-C\equiv CNa} and B = CH3−CH2−CH2−Br\mathrm{CH_3-CH_2-CH_2-Br}, which matches option A in the extracted options.

Therefore, the correct option is A.

Reaction Logic

Given: CH3−C≡CH\mathrm{CH_3-C\equiv CH} is a terminal alkyne.

Find: Which sodium salt forms first, and which bromide must react next to produce the final internal alkyne.

Terminal alkynes have an acidic hydrogen on the carbon of the triple bond. Sodium removes this hydrogen and forms the sodium salt of the alkyne. Hence intermediate A must be sodium propynide, not an alkane or alkene derivative.

The final product has the skeleton:

CH3−C≡C−CH2−CH2−CH3\mathrm{CH_3-C\equiv C-CH_2-CH_2-CH_3}

Compared with CH3−C≡CH\mathrm{CH_3-C\equiv CH}, the hydrogen of the terminal alkyne has been replaced by a propyl group (−CH2−CH2−CH3)\mathrm{(-CH_2-CH_2-CH_3)}.

So the required electrophile B must be the corresponding primary alkyl bromide:

CH3−CH2−CH2−Br\mathrm{CH_3-CH_2-CH_2-Br}

Thus, A = CH3−C≡CNa\mathrm{CH_3-C\equiv CNa} and B = CH3−CH2−CH2−Br\mathrm{CH_3-CH_2-CH_2-Br}. This corresponds to option A.

Common mistakes

  • Choosing propyne again as B is incorrect because an acetylide ion forms a new carbon-carbon bond by reacting with a suitable alkyl halide, not with another terminal alkyne under these conditions. Look for the group added to the terminal carbon and identify the matching alkyl bromide.

  • Ignoring the formation of NaBr\mathrm{NaBr} is a mistake because NaBr\mathrm{NaBr} directly indicates that B must contain bromine. Use the by-product to infer that the second reactant is an alkyl bromide.

  • Selecting propane as B is wrong because propane is not an electrophile for nucleophilic substitution. The acetylide ion requires a primary alkyl halide such as CH3CH2CH2Br\mathrm{CH_3CH_2CH_2Br} to extend the carbon chain.

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