Question

Predict the alkene that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane.

Hint:

Internal double bonds are generally more stable than terminal/exocyclic ones due to more hyperconjugative stabilization.

Answer 1

Concept: 
• Dehydrohalogenation follows Zaitsev's (Saytzeff's) rule.
• The rule states that during elimination, the preferred product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbon atoms (the more substituted alkene).

Step 1: Analyze the substrate structure
1-Bromo-1-methylcyclohexane has a bromine atom and a methyl group on the same carbon (C1) of the cyclohexane ring. 
There are two types of beta-hydrogens available for elimination:
1. Hydrogens on the methyl group (\( CH_3 \)).
2. Hydrogens on the C2/C6 positions of the ring (\( CH_2 \)). 

Step 2: Evaluate potential products
Path 1: Elimination using methyl hydrogens forms methylenecyclohexane (exocyclic double bond).
Path 2: Elimination using ring hydrogens forms 1-methylcyclohexene (endocyclic double bond). 

Step 3: Apply Zaitsev's Rule
1-methylcyclohexene has a trisubstituted double bond. Methylenecyclohexane has a disubstituted double bond. 
Therefore, 1-methylcyclohexene is the more stable, major product. The final answer is 1-methylcyclohexene.