Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
1-Bromo-1-methylcyclohexane
1-Methylcyclohexene is the only product formed. Since the β-hydrogens on adjacent carbons (C2 and C6) are equivalent in the symmetric ring, elimination gives the same alkene. This is the major product.
2-Chloro-2-methylbutane
Two alkenes are formed: 2-Methylbut-2-ene (major) and 2-Methylbut-1-ene (minor). According to Zaitsev's rule, the alkene with greater number of alkyl groups attached to doubly bonded carbon atoms is the major product. 2-Methylbut-2-ene has three alkyl groups attached compared to two in 2-Methylbut-1-ene.
2,2,3-Trimethyl-3-bromopentane
2,2,3-Trimethylpent-2-ene is the only product formed. The β-carbon C2 has no β-hydrogen atoms (both positions occupied by methyl groups), so elimination can only occur from C4, yielding a single alkene as the major product.
Explanation
This question tests the application of Zaitsev's rule in dehydrohalogenation reactions. The rule states that 'in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.' For each substrate, students must identify all available β-hydrogen atoms and apply Zaitsev's rule to determine the major product. In cases where β-hydrogens are unavailable or equivalent, only one product forms.