Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Two haloalkanes are heated with alcoholic potassium hydroxide: 2-bromo-3-methylpentane, , and 3-bromo-2,3-dimethylpentane, . Count the different alkenes that each can give, counting cis and trans forms as different compounds. Then count the beta hydrogen atoms of 3-bromo-2,3-dimethylpentane. Say which alkene is the main product from each halide by Saytzeff's rule.
Elimination takes a beta hydrogen and the halogen away and leaves a double bond. The more substituted alkene is the main product (Saytzeff). Each double bond with different groups on both ends counts as cis and trans. Alcoholic KOH and high temperature favour elimination. The ratio of two rates is .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. 2-Bromo-3-methylpentane. The beta carbons are carbon 1 (a methyl group) and carbon 3 (a CH group). Towards carbon 1 the product is 3-methylpent-1-ene. Towards carbon 3 it is 3-methylpent-2-ene, which has cis and trans forms. That is alkenes. The main product is 3-methylpent-2-ene, with three alkyl groups on the double bond.
Step 2. 3-Bromo-2,3-dimethylpentane. The beta carbons are carbon 2, the methyl on carbon 3, and carbon 4. Towards carbon 2 the product is 2,3-dimethylpent-2-ene, with no cis and trans forms. Towards the methyl it is 2-ethyl-3-methylbut-1-ene. Towards carbon 4 it is 3,4-dimethylpent-2-ene, with cis and trans forms. That is alkenes. The main product is 2,3-dimethylpent-2-ene, with four alkyl groups.
Step 3. Beta hydrogens. Carbon 2 has 1, the methyl on carbon 3 has 3, and carbon 4 has 2. The total is .
Answer: 3 alkenes, 4 alkenes and 6 beta hydrogen atoms.
Check. Each beta carbon with hydrogen gives a direction, and only the middle direction gives cis and trans forms in each case.
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