Free Elimination Reactions MCQs with Answers

1. Reaction of an alkyl halide with alcoholic potassium hydroxide gives mainly

• A. an alcohol
• B. an alkane
• C. a carboxylic acid
• D. an alkene, by elimination of hydrogen halide

Explanation: In ethanol the hydroxide ion acts as a base rather than as a nucleophile, removing a hydrogen from the carbon next to the one bearing the halogen, so a double bond forms and the hydrogen halide is eliminated. Heat favours this route as well. The contrast with aqueous conditions, which give the alcohol, is the essential comparison in this topic.

2. According to Saytzeff's rule, the major product of an elimination reaction is

• A. the less substituted, less stable alkene
• B. always the cis isomer
• C. the more highly substituted and therefore more stable alkene
• D. an alkane

Explanation: When more than one alkene can form, the preferred product is the one with the greater number of alkyl groups attached to the doubly bonded carbons, because such alkenes are thermodynamically more stable. So 2-bromobutane gives mainly but-2-ene rather than but-1-ene. Using a bulky base can override this preference and favour the less substituted product instead.

3. Elimination reactions of alkyl halides are favoured over substitution by

• A. a high concentration of a strong base, a non aqueous solvent and a higher temperature
• B. dilute aqueous conditions at room temperature
• C. the use of a weak nucleophile in water
• D. cooling the mixture in ice

Explanation: Elimination has the higher activation energy and produces more particles, so heat and a strong base in ethanol push the reaction that way, while water and mild conditions favour substitution. The two pathways always compete, and a given reaction usually produces some of both products. Tertiary halides eliminate most readily because the carbon is too crowded for substitution.

4. Conversion of dihaloalkane into alkyne does not involve

• A. Addition
• B. Elimination
• C. Base
• D. Heat

Explanation: A vicinal dihalide is converted to an alkyne by removing two molecules of hydrogen halide with hot alcoholic potassium hydroxide, which is a double dehydrohalogenation and therefore an elimination. Addition is the opposite process and would move back towards the saturated compound. The base and the heat are both required conditions for the elimination.