Free Alkyl Halides MCQs with Answers

1. The carbon atom bonded to the halogen in an alkyl halide is attacked by nucleophiles because it carries

• A. a partial positive charge, since the halogen is more electronegative
• B. a partial negative charge
• C. a full negative charge
• D. no charge at all

Explanation: The carbon to halogen bond is polar, with electron density drawn towards the halogen, so the carbon is electron deficient and attracts species with a lone pair. This polarity is the reason alkyl halides are far more reactive than alkanes despite looking similar. The halogen leaves as a stable halide ion, which is why it is described as a good leaving group.

2. The compound (CH3)3CBr is classified as

• A. a primary alkyl halide
• B. a secondary alkyl halide
• C. a tertiary alkyl halide
• D. an aryl halide

Explanation: The carbon carrying the bromine is attached to three other carbon atoms, which makes it tertiary, whereas a primary carbon is attached to one and a secondary to two. This classification matters because tertiary halides react overwhelmingly by the SN1 route while primary halides prefer SN2. An aryl halide would have the halogen attached directly to a benzene ring.

3. The order of reactivity of alkyl halides towards nucleophilic substitution, for the same alkyl group, is

• A. R-F greater than R-Cl greater than R-Br greater than R-I
• B. R-I greater than R-Br greater than R-Cl greater than R-F
• C. all four react at the same rate
• D. R-Cl greater than R-I greater than R-Br greater than R-F

Explanation: Reactivity is governed by the strength of the carbon to halogen bond, and that bond weakens down the group as the halogen gets larger, so the iodide breaks most easily and the fluoride hardly reacts at all. This outweighs the fact that fluorine creates the largest dipole. Iodide is also the most stable leaving group of the four.

4. The reaction of bromoethane with aqueous potassium hydroxide gives

• A. ethene
• B. ethanoic acid
• C. ethane
• D. ethanol

Explanation: In aqueous solution the hydroxide ion acts as a nucleophile, replacing the bromine to give ethanol in a substitution reaction. The same reagent in hot ethanolic solution behaves as a base instead, removing a hydrogen and giving ethene by elimination, so the solvent decides the outcome. This pair of reactions is examined constantly and turns entirely on the conditions.

5. An SN2 reaction is characterised by

• A. a two step mechanism through a carbocation
• B. first order kinetics involving only the alkyl halide
• C. a single step in which the nucleophile attacks as the leaving group departs, with second order kinetics
• D. the formation of a free radical intermediate

Explanation: The nucleophile approaches from the side opposite the leaving group and bond forming and bond breaking occur together through a five coordinate transition state, so the rate depends on the concentrations of both the halide and the nucleophile. Because attack is from the back, the configuration at that carbon is inverted, an effect known as Walden inversion. Primary halides follow this route because they are least hindered.

6. A tertiary alkyl halide usually undergoes nucleophilic substitution by the SN1 mechanism because

• A. the tertiary carbocation formed is relatively stable and the carbon is too hindered for back side attack
• B. tertiary halides have the weakest carbon to halogen bond
• C. tertiary carbon atoms carry a full positive charge already
• D. the nucleophile is always weak

Explanation: Three electron releasing alkyl groups spread the positive charge and stabilise the carbocation, so the slow ionisation step becomes feasible, and at the same time those bulky groups block the approach a concerted SN2 attack would need. The rate therefore depends only on the halide concentration, giving first order kinetics. Because the planar carbocation can be attacked from either face, a racemic mixture results.

7. The rate of an SN1 reaction depends on the concentration of

• A. both the alkyl halide and the nucleophile
• B. the alkyl halide only
• C. the nucleophile only
• D. the solvent only

Explanation: The slow, rate determining step is the ionisation of the alkyl halide into a carbocation and a halide ion, and the nucleophile is not involved until the fast second step, so it does not appear in the rate equation. This is why the reaction is called unimolecular. Detecting first order kinetics is the standard experimental evidence for the SN1 pathway.

8. 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.

9. 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.

10. 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.