Question

When CH₃CH₂CHCl₂ is treated with NaNH₂, the product formed is 

• A. CH₃ – CH = CH₂
• B. CH₃ – C ≡ CH
• C. CH₃CH₂CHNH₂NH₂
• D. CH₃CH₂CHClNH₂

Answer 1

The Correct Option is B

To determine the product formed when CH₃CH₂CHCl₂ is treated with NaNH₂, we need to understand the mechanism involved in this reaction. The compound in question is a dihalide, and the reagents suggest the possibility of a double elimination reaction.

Step-by-Step Reaction Mechanism

1. Initial Conditions: We have CH₃CH₂CHCl₂, known as 1,1-dichloropropane.

2. Reagent: Sodium amide (NaNH₂) is a strong base known for its ability to deprotonate C-H bonds and facilitate elimination reactions. In particular, NaNH₂ is used in dehydrohalogenation to convert vicinal dihalides to alkynes.

3. Elimination Reaction: The reaction proceeds through a two-step elimination process:

• First, NaNH₂ deprotonates one of the hydrogens adjacent to the chloride group, and a chloride ion is eliminated, forming a vinyl chloride intermediate (CH₃CH=CHCl).
• Next, NaNH₂ further deprotonates the vinyl chloride intermediate, and another chloride ion is eliminated, leading to the formation of an alkyne.

4. Final Product: The final product is propyne (CH₃–C≡CH).

Conclusion

The correct product of this reaction is an alkyne, CH₃–C≡CH, making it the correct choice. Here is a summary of why the other options are incorrect:

• CH₃ – CH = CH₂ (Propene) would require only one elimination step, but we are starting from a dihalide, leading to an alkyne.
• CH₃CH₂CHNH₂NH₂ and CH₃CH₂CHClNH₂ would result from substitution reactions, which are not favored under the strong basic conditions provided by NaNH₂.

Therefore, the correct answer is CH₃ – C ≡ CH (propyne).