The Question is from neet-chemistry haloalkanes-haloarenes , and it was created by the author THIRUPATHI REDDY

Question

Predict the major product formed when 2-bromotoluene reacts with NaNH2 in liquid ammonia followed by treatment with CH3I.

THIRUPATHI REDDY · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B. N-methyl-3-methylaniline
## CONCEPT FRAMEWORK
	**Primary Concept**: Benzyne Mechanism
	**Related Topics**: Aromatic Nucleophilic Substitution, Elimination-Addition Mechanism, Alkylation
	**Prerequisites**: Understanding of leaving groups, strong bases, and nucleophiles.
## SOLUTION PATHWAY
**Given Information**:
	2-bromotoluene reacts with NaNH2 (strong base) in liquid ammonia (polar aprotic solvent) followed by CH3I (methylating agent).
**Method & Steps**:
	1. **Formation of Benzyne Intermediate:**
		Reasoning: NaNH2 is a very strong base. It abstracts the proton ortho to the bromine atom in 2-bromotoluene.  This leads to the elimination of bromide ion (Br-) and the formation of a highly reactive benzyne intermediate.
		Formula/Rule: Elimination reaction.
	2. **Nucleophilic Attack by Amide Ion:**
		Working: The amide ion (NH2-) acts as a nucleophile and attacks the benzyne intermediate.  Due to the inductive effect of the methyl group, the nucleophilic attack is more likely to occur at the less hindered carbon atom (meta to the methyl group).
		Key Point: The benzyne intermediate is symmetrical, but the attack is regioselective due to the methyl substituent.
	3. **Alkylation of Aniline:**
		Result: The resulting aniline derivative then reacts with CH3I in an SN2 reaction, leading to N-methylation. The final product is N-methyl-3-methylaniline.
		Verification: The product is consistent with the benzyne mechanism and the directing effect of the methyl group.
## OPTION ANALYSIS
[A] :x: N-methyl-2-methylaniline: This product would be formed if the nucleophilic attack occurred ortho to the methyl group, which is less favored due to steric hindrance.
[B] ✓: N-methyl-3-methylaniline: This is the correct answer as explained above. The nucleophilic attack occurs meta to the methyl group, followed by N-methylation.
[C] :x: 2-methylaniline: This product would be formed if there was no N-methylation step after the benzyne mechanism.
[D] :x: 3-methylaniline: This product would be formed if there was no N-methylation and the nucleophilic attack occurred meta to the methyl group, but without the initial presence of bromine.
## LEARNING AIDS
	**Common Mistakes**:
		1. Ignoring the steric effect of the methyl group: This can lead to incorrectly predicting N-methyl-2-methylaniline as the major product.
		2. Forgetting the N-methylation step: This can lead to incorrectly predicting 2-methylaniline or 3-methylaniline as the product.
	**Quick Tips**:
		1. Remember that NaNH2 in liquid ammonia is a strong base that favors elimination reactions leading to benzyne formation.
		2. Consider the steric and electronic effects of substituents when predicting the regioselectivity of nucleophilic attack on a benzyne intermediate.

THIRUPATHI REDDY · Answer

N-methyl-2-methylaniline

THIRUPATHI REDDY · Answer

The correct Option is "N-methyl-3-methylaniline"

THIRUPATHI REDDY · Answer

2-methylaniline

THIRUPATHI REDDY · Answer

3-methylaniline

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