Haloalkanes Class 12: SN1 and SN2 Worked Questions

To solve a Haloalkanes and Haloarenes question, first locate the carbon bonded to the halogen. Then read the reagent, solvent and conditions before choosing a mechanism. The same substrate can behave differently when those conditions change. Memorising "tertiary is fastest" without naming the reaction is an easy way to reverse an answer.

This Class 12 guide covers selected substitution and elimination questions within CBSE's 2026-27 Unit 6. It is not complete chapter coverage or an official answer key. The examples below are original paper exercises, not instructions for carrying out chemical reactions at home.

Start at the carbon carrying the halogen

In a haloalkane, the halogen is attached to an sp³ carbon. A haloarene has the halogen directly attached to an aromatic-ring carbon. That word "directly" matters.

Compare C₆H₅CH₂Cl with C₆H₅Cl. The first is benzyl chloride: chlorine is attached to the side-chain carbon, not the ring. The second is chlorobenzene, an aryl halide. Do not classify both as haloarenes simply because both contain benzene rings.

For an alkyl halide, count the carbon neighbours of the carbon bearing X. One makes it primary, two secondary and three tertiary. Count carbon neighbours, not the total number of carbons in the molecule.

SN2: leave room for the incoming nucleophile

An SN2 reaction forms the new bond while the carbon-halogen bond breaks in a single step. There is no carbocation intermediate. Backside approach explains why crowding around the reacting carbon slows the process.

For comparable simple alkyl halides with the same leaving group under the same SN2 conditions, the familiar trend is methyl, then primary, then secondary; tertiary substrates are strongly hindered. Nearby branching also matters, so "primary" alone does not settle every comparison.

Worked choice: compare chloroethane with 2-chloro-2-methylbutane under conditions specified to favour SN2. Choose chloroethane. Its reacting carbon is less crowded. An answer about a more stable tertiary carbocation would explain the wrong mechanism.

At a stereogenic reacting carbon, backside attack gives inversion of configuration. Inversion describes spatial arrangement; it does not automatically mean that the sign of optical rotation changes from positive to negative.

SN1: consider the carbocation instead

In the usual SN1 model, slow ionisation forms a carbocation and a halide ion. The nucleophile then attacks. For comparable simple alkyl halides, easier formation of a more stable carbocation favours the tertiary substrate over secondary and primary ones. Benzylic and allylic systems need separate attention because resonance can stabilise their carbocations.

Worked choice: compare 2-bromo-2-methylpentane with 1-bromohexane under comparable conditions specified to favour SN1. The tertiary halide is favoured because its ionisation produces the more stable carbocation. Both contain six carbons; total chain length is not the deciding feature.

A planar carbocation allows attack from either face. At school level this explains racemisation from an optically active substrate. Do not use the SN2 inversion argument for an SN1 question.

Use the rate law to test your choice

For the simple models, SN2 has rate = k[RX][Nu], whereas SN1 has rate = k[RX]. Brackets represent concentrations. The nucleophile still takes part in SN1 chemistry even though it is absent from this rate law.

Original calculation: an SN2 experiment starts at rate r. If [RX] becomes twice its original value and [Nu] becomes three times its original value, the new rate is 2 × 3 × r = 6r, assuming the same temperature and rate constant.

For an SN1 experiment, doubling [RX] doubles the rate. Tripling nucleophile concentration alone leaves the predicted rate unchanged in this model. Write the rate law before multiplying concentration factors.

Reagent conditions can change the product

In the standard school comparison, aqueous KOH favours substitution to an alcohol, while heating a suitable haloalkane with alcoholic KOH favours elimination to an alkene. Elimination needs a hydrogen on a beta-carbon adjacent to the carbon bearing the halogen.

For 1-bromopropane, the corresponding paper products are propan-1-ol in the aqueous substitution comparison and propene in the heated alcoholic elimination comparison. Write the conditions beside the arrow. These are useful textbook patterns, not a claim that every substrate and solvent combination behaves identically.

Keep chlorobenzene outside the simple alkyl-halide ranking

Chlorobenzene resists ordinary nucleophilic substitution much more than a typical alkyl chloride. Resonance gives its C-Cl bond partial double-bond character, and the bond to an sp² carbon is shorter and stronger. A phenyl carbocation is not the resonance-stabilised benzyl carbocation.

Do not write "chlorobenzene never reacts". NCERT discusses substitution under demanding conditions and activation by suitable electron-withdrawing groups. The useful distinction is resistance under ordinary conditions, not absolute impossibility.

Finish with an explained answer

Try a fresh rate question: [RX] is halved and [Nu] is doubled. SN2 predicts unchanged rate because 0.5 × 2 = 1; SN1 predicts half the original rate. State that temperature and k remain unchanged. If you chose the wrong model, record the missing clue in an exam mistake log and retry without looking at the solution.

Sources and syllabus check

Scope checked against the CBSE Chemistry syllabus for 2026-27. Mechanisms and qualifications follow NCERT Chemistry Part II, Haloalkanes and Haloarenes. Examples here are independently written, not copied exercise solutions.

Frequently asked questions

Does tertiary always mean faster?

No. It commonly favours SN1 among comparable simple alkyl halides, but steric crowding strongly hinders SN2.

Is benzyl chloride a haloarene?

No. Chlorine is bonded to a side-chain sp³ carbon, not directly to the aromatic ring.

Does SN2 form a carbocation?

No. Bond formation and bond breaking occur together through a transition state, without a carbocation intermediate.

Should every substitution answer include conditions?

Include the stated reagent and conditions. They help distinguish substitution from elimination and make a product prediction meaningful.