The Question is from neet-chemistry , and it was created by the author A.RAMA NAGESWARARAO

Question

Which alkyne, upon reaction with sodamide ($NaNH_2$) followed by treatment with 1-bromopropane, would yield 4-methylhex-2-yne?

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Alkylation of terminal alkynes
	**Related Topics**: Acidity of terminal alkynes, SN2 reactions, organic synthesis
	**Prerequisites**: Understanding of alkyne reactions, acid-base chemistry, and alkyl halide reactions.
## SOLUTION PATHWAY
**Given Information**:
	Target molecule: 4-methylhex-2-yne
	Reagents: $NaNH_2$, 1-bromopropane
**Method & Steps**:
	1. **Deprotonation of the terminal alkyne:**
		Reasoning: $NaNH_2$ is a strong base that deprotonates the terminal alkyne, forming a nucleophilic acetylide anion.
		Formula/Rule: $R-C\equiv C-H + NaNH_2 ightarrow R-C\equiv C^-Na^+ + NH_3$
	2. **SN2 reaction with 1-bromopropane:**
		Working: The acetylide anion acts as a nucleophile and attacks the primary alkyl halide (1-bromopropane) in an SN2 reaction, forming a new carbon-carbon bond.
		Key Point: The reaction proceeds with inversion of configuration at the alkyl halide carbon (although this isn't relevant in this specific case with 1-bromopropane).
	3. **Formation of the product:**
		Result: The reaction yields the desired alkyne, 4-methylhex-2-yne.
		Verification: Working backward from the product, removing the propyl group leaves a 3-methylbut-1-yne anion.  Protonation of this anion gives 3-methylbut-1-yne.
## OPTION ANALYSIS
[A] :x: but-1-yne would react with $NaNH_2$ and 1-bromopropane to form hex-2-yne.
[B] ✓: 3-methylbut-1-yne would react with $NaNH_2$ and 1-bromopropane to form 4-methylhex-2-yne.
[C] :x: pent-1-yne would react with $NaNH_2$ and 1-bromopropane to form 4-methylhept-2-yne.
[D] :x: 3,3-dimethylbut-1-yne would react with $NaNH_2$ and 1-bromopropane to form 4,4-dimethylhept-2-yne.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly identifying the nucleophile: Remember that the acetylide anion is the nucleophile, not the $NaNH_2$.
		2. Forgetting the SN2 mechanism: Ensure you understand how the acetylide anion attacks the alkyl halide.
	**Quick Tips**:
		1. Work backward from the product: This can help identify the starting alkyne.
		2. Visualize the reaction mechanism: Drawing the mechanism can help understand the bond formation and product structure.

A.RAMA NAGESWARARAO · Answer

but-1-yne

A.RAMA NAGESWARARAO · Answer

The correct Option is "3-methylbut-1-yne"

A.RAMA NAGESWARARAO · Answer

pent-1-yne

A.RAMA NAGESWARARAO · Answer

3,3-dimethylbut-1-yne

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