Three alkyl halides are heated separately under reflux with a solution of potassium hydroxide in…
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Three alkyl halides are heated separately under reflux with a solution of potassium hydroxide in ethanol. In each case a hydrogen halide is removed from two neighbouring carbon atoms. The gas or liquid that distils off is a mixture of alkenes, which are separated by gas chromatography and identified one by one. Every alkene that can form is found in at least a trace. The question asks for the number of different alkenes in each mixture, counting cis and trans isomers as separate compounds. She notes that the three halides differ in the number of carbon atoms and in the position of the bromine atom. So the carbon atoms next to the one that bears bromine differ from one halide to the next. Hydrogen atoms that are alike by symmetry lead to the same alkene. Use the rule that decides which alkene is found in the largest amount.
- 2-Bromopentane.
- 3-Bromopentane.
- 2-Bromo-2-methylbutane.
Also say which alkene is the main product of (a) and (c), and why.
Need a reminder? The key ideaShow
Geometrical isomers need two different groups on each carbon of the double bond. Alcoholic KOH removes HX from an alkyl halide, and the more substituted alkene is the main product (Saytzeff). The more alkyl groups on the double bond, the more stable the alkene and the less heat it gives out on hydrogenation.
The worked solution
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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. (a) 2-Bromopentane has neighbours carbon 1 (a methyl group) and carbon 3 (a group). Hydrogen from carbon 1 gives pent-1-ene. Hydrogen from carbon 3 gives pent-2-ene, as the cis and the trans isomer. That is alkenes. The main product is pent-2-ene, which has two alkyl groups on the double bond (Saytzeff's rule).
Step 2. (b) 3-Bromopentane is symmetrical. Both neighbours are groups, and both give pent-2-ene. The product is the cis and the trans isomer: alkenes.
Step 3. (c) 2-Bromo-2-methylbutane has neighbours: two methyl groups on carbon 2 (alike) and the group of carbon 3. The methyl route gives 2-methylbut-1-ene. The route gives 2-methylbut-2-ene. Neither has cis-trans isomers: alkenes. The main product is 2-methylbut-2-ene, which has three alkyl groups on the double bond.
Answer: , and .
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