Final answer:
Different alkenes are produced through elimination reactions when chlorinated alkanes are treated with a strong base. The specific alkenes formed depend on the structure of the starting material and its substituents.
Step-by-step explanation:
Treating these substrates with a strong base usually results in an elimination reaction, forming alkenes. The specific number of alkenes formed depends on the structure of the starting material and the location of substituents.
1-Chloropentane
This substrate will yield 1-pentene and 2-pentene due to dehydrohalogenation. Because hydrogen atoms can be removed from either carbon-2 or carbon-3 in the chain, leading to the formation of the two different alkenes.
3-Chloropentane
When treated with a strong base, this substrate will give 2-pentene and 3-pentene as products. Again, this is because hydrogens are available for elimination on two different carbons adjacent to the one bearing the chlorine atom.
2-Chloro-2-methylpentane
The treatment of this substrate will only produce one alkene: 2,3-dimethyl-2-butene. Due to the steric hindrance caused by the methyl group on carbon-2, this substrate has only one set of β-hydrogens available for elimination.