Free Nucleophilic Substitution MCQs with Answers

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

6. Reaction of bromoethane with alcoholic ammonia under pressure gives

• A. ethanol
• B. ethylamine
• C. ethanenitrile
• D. ethane

Explanation: Ammonia has a lone pair on nitrogen and acts as a nucleophile, displacing bromide to give ethylamine, though the product is itself nucleophilic so further substitution to secondary and tertiary amines readily follows. Using a large excess of ammonia limits this. Ethanenitrile would be formed with potassium cyanide instead.

7. The reaction of an alkyl halide with potassium cyanide is synthetically useful because it

• A. shortens the carbon chain by one atom
• B. produces an alkene
• C. lengthens the carbon chain by one carbon atom
• D. removes the halogen without adding anything

Explanation: The cyanide ion substitutes for the halogen, so the nitrile formed has one more carbon than the starting halide, and that nitrile can then be hydrolysed to a carboxylic acid or reduced to an amine. Building the chain one carbon at a time is a standard strategy in organic synthesis. Alcoholic conditions are used to favour substitution over elimination.

8. A Grignard reagent is formed when an alkyl halide reacts with

• A. sodium metal in dry ether
• B. magnesium in dry ether
• C. zinc in aqueous solution
• D. copper in ethanol

Explanation: Magnesium inserts itself into the carbon to halogen bond to give an alkylmagnesium halide, and the ether must be perfectly dry because even a trace of water destroys the reagent, converting it into an alkane. The carbon attached to magnesium is strongly nucleophilic, which makes Grignard reagents extremely versatile for building carbon to carbon bonds. Sodium with an alkyl halide gives the Wurtz reaction instead.

9. Which alkyl halide would react fastest by the SN1 mechanism?

• A. CH3Br
• B. CH3CH2Br
• C. (CH3)2CHBr
• D. (CH3)3CBr

Explanation: SN1 rates follow the stability of the carbocation formed, and that increases from primary through secondary to tertiary because each alkyl group releases electron density and spreads the charge. The tertiary bromide therefore ionises most readily. The order is exactly reversed for SN2, where the methyl halide reacts fastest because it is the least hindered.

10. In an SN2 reaction the configuration at the reacting carbon atom is

• A. inverted, because the nucleophile attacks from the side opposite the leaving group
• B. retained completely
• C. randomised, giving a racemic mixture
• D. unchanged because no bond is broken

Explanation: Back side attack turns the three remaining groups inside out like an umbrella in the wind, so an optically active substrate gives a product of opposite configuration, an effect called Walden inversion. An SN1 reaction gives a racemic mixture instead, because the planar carbocation can be attacked equally from either face. Stereochemistry is therefore the clearest experimental way to distinguish the two mechanisms.