Question

Compound A formed in the following reaction reacts with B gives the product C. Find out A and B.\[ \text{CH}_3 - \text{C} \equiv \text{CH} + \text{Na} \xrightarrow{} \text{A}\xrightarrow{B} \text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_2 - \text{CH}_2 + \text{NaBr}(C)\]

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is C

The problem presents a two-step reaction sequence and asks to identify the intermediate compound A and the reactant B. The starting material is propyne (\(CH_3-C \equiv CH\)), and the final products are a substituted alkyne (C) and sodium bromide (\(NaBr\)).

Concept Used:

The solution involves two key reactions in alkyne chemistry:

1. Formation of an Acetylide Anion: Terminal alkynes (alkynes with a hydrogen atom attached to a triply-bonded carbon) are weakly acidic. They can be deprotonated by a strong base, such as sodium metal (\(Na\)) or sodamide (\(NaNH_2\)), to form a sodium acetylide salt. In this reaction, the sodium acts as a reducing agent, producing the sodium salt and hydrogen gas.

\[ 2 R-C \equiv CH + 2Na \longrightarrow 2 R-C \equiv C^-Na^+ + H_2(g) \]

2. Nucleophilic Substitution by an Acetylide Anion: The acetylide anion (\(R-C \equiv C^-\)) is a strong nucleophile. It can react with a primary alkyl halide in a nucleophilic substitution reaction (specifically, an \(S_N2\) reaction) to form a new carbon-carbon bond, resulting in a longer, internal alkyne.

\[ R-C \equiv C^-Na^+ + R'-X \longrightarrow R-C \equiv C-R' + NaX \]

Step-by-Step Solution:

Step 1: Determine the structure of compound A.

The first reaction is the reaction of propyne (\(CH_3-C \equiv CH\)) with sodium metal (\(Na\)). Propyne is a terminal alkyne, and the hydrogen atom on the terminal carbon is acidic. Sodium metal will deprotonate the alkyne to form a sodium propynide salt.

\[ CH_3 - C \equiv CH + Na \longrightarrow CH_3 - C \equiv C^-Na^+ + \frac{1}{2}H_2(g) \]

Therefore, the intermediate compound A is sodium propynide, \(CH_3 - C \equiv C^-Na^+\).

Step 2: Determine the structure of compound B.

 \(CH_3-C \equiv C-CH_2-CH_2-CH_3\) with a \(CH_3\) branch on the second carbon of the added chain. This is likely a typographical error in the problem's depiction of product C. Let's assume the product is the straight-chain alkyne \(CH_3-C \equiv C-CH_2-CH_2-CH_3\) (hex-2-yne), as would be formed from 1-bromopropane, which is a more common reactant in textbook examples and aligns with the options.

Assuming the product is hex-2-yne, the added group is \( -CH_2-CH_2-CH_3 \) (a propyl group). This would mean B is 1-bromopropane, \(CH_3-CH_2-CH_2-Br\). This matches the structure of B given in option (1).

Let's proceed by assuming the product shown has a typo and the intended product is \(CH_3-C \equiv C-CH_2CH_2CH_3\). Reaction: \[ CH_3 - C \equiv C^-Na^+ + CH_3CH_2CH_2-Br \longrightarrow CH_3-C \equiv C-CH_2CH_2CH_3 + NaBr \]

Final Computation & Result:

Step 3: Match the identified structures of A and B with the given options.

• Compound A is sodium propynide: \(A = CH_3 - C \equiv C^-Na^+\)
• Compound B  is 1-bromopropane: \(B = CH_3 - CH_2 - CH_2 - Br\)

Let's evaluate the other options:

• Option (2): A is incorrect (propene), B is incorrect.
• Option (3): A is incorrect (propane), B is incorrect.
• Option (4): A is correct, but B is incorrect (propane).

The correct choice is (1) \(A = CH_3 - C \equiv C^-Na^+, B = CH_3 - CH_2 - CH_2 - Br\).

Answer 2

The given reaction suggests that sodium acetylide reacts with an alkyl halide to yield the final product. Here, compound A is formed as sodium acetylide:

CH₃ – C ≡ CH + Na → CH₃ – C ≡ CNa

Then, it reacts with compound B (1-bromopropane):

CH₃ – C ≡ CNa + CH₃CH₂CH₂Br → CH₃ – C ≡ C – CH₂CH₂CH₃ + NaBr