Haloalkanes and Haloarenes • Topic 2 of 3

Reactions of Haloalkanes

The polar C–X bond makes the carbon electrophilic and the halide a good leaving group, so haloalkanes are workhorses of synthesis. Their three big reaction families are nucleophilic substitution, elimination, and reaction with metals. Which one wins, and how, depends on the substrate, the attacking species and the conditions.

Nucleophilic substitution: the two pathways

A nucleophile (Nu−) replaces X−. There are two limiting mechanisms.

SN2 (bimolecular): a one-step, concerted process. The nucleophile attacks the carbon from the side opposite the leaving group (backside attack); the C–Nu bond forms as the C–X bond breaks through a five-coordinate transition state. Rate = k[R–X][Nu] (second order). Because attack is from the back, the configuration is inverted (Walden inversion) — like an umbrella turning inside out. SN2 is fastest for methyl > 1° > 2° > 3° (less steric crowding at the carbon), favoured by strong nucleophiles and polar aprotic solvents.

SN1 (unimolecular): a two-step process. Step 1 (slow, rate-determining) is ionisation of R–X to a planar carbocation + X−. Step 2 (fast) is attack by the nucleophile. Rate = k[R–X] (first order, independent of [Nu]). Because the carbocation is planar (sp2), the nucleophile can attack either face, giving a near 1:1 mixture of enantiomers — racemisation (often with slight excess inversion). SN1 is fastest for 3° > 2° > 1° > methyl (more stable carbocation), favoured by polar protic solvents that stabilise the ions.

Factors that decide SN1 vs SN2

  • Substrate: 3° and allyl/benzyl → SN1 (stable cation); methyl/1° → SN2 (open to backside attack). 2° can do either.
  • Nucleophile: strong/charged nucleophiles (OH−, CN−, RO−) push SN2; weak nucleophiles favour SN1.
  • Solvent: polar protic (water, alcohols) stabilise the carbocation and favour SN1; polar aprotic (acetone, DMSO, DMF) leave the nucleophile “naked” and reactive, favouring SN2.
  • Leaving group: better leaving group (I− > Br− > Cl−) speeds both.

Elimination (β-elimination, E)

With a strong base (e.g. alc. KOH) the halide can lose HX to give an alkene: H is removed from the β-carbon and X from the α-carbon (dehydrohalogenation). When more than one alkene is possible, Saytzeff’s rule predicts the major product as the more highly substituted (more stable) alkene. Substitution and elimination compete: aqueous KOH favours substitution; alcoholic KOH (a stronger, less solvated base) and higher temperature favour elimination.

Reaction with metals

Grignard reagents: R–X + Mg in dry ether → R–MgX (alkylmagnesium halide). The R–Mg bond is highly polar with carbon strongly nucleophilic, so Grignard reagents build C–C bonds and react violently with any source of acidic H (water, alcohols, acids) — hence the apparatus must be perfectly dry.

Wurtz reaction: 2R–X + 2Na in dry ether → R–R + 2NaX, coupling two alkyl groups to give a symmetrical alkane of twice the carbon count.

Optical activity and chirality

A carbon bearing four different groups is a stereocentre; the molecule and its mirror image are non-superimposable enantiomers that rotate plane-polarised light equally but in opposite directions. A 1:1 mixture (racemate) is optically inactive. Stereochemistry is the clearest fingerprint of mechanism: clean inversion at a single stereocentre is SN2, while racemisation signals a planar carbocation and SN1.

SN2 backside attack and Walden inversionSₙ2: backside attack → inversion (Walden)Nu:−→CXabc(starting)transition stateNu···C···X5-coordinate→NuCabcX−(inverted)Rate = k[R-X][Nu] · one step · methyl > 1° > 2° > 3°Sₙ1 contrast: slow ionisation to planar carbocation → racemisation; 3° > 2° > 1°
Solved Examples
1
Worked Example
Identify the mechanism (SN1 or SN2) for the hydrolysis of (i) CH3Br and (ii) (CH3)3CBr, and justify.
Solution
  1. (i) CH3Br is a methyl halide — the carbon is unhindered, so backside attack is easy and there is no stabilised carbocation: it reacts by SN2.
  2. (ii) (CH3)3CBr is tertiary; backside attack is blocked by the three methyls, but it forms a stable 3° carbocation: it reacts by SN1.
  3. Consistency check: SN2 rate falls methyl > 1° > 2° > 3°, SN1 rate rises in the opposite order, matching the assignment.

Answer: (i) SN2 (unhindered methyl); (ii) SN1 (stable 3° carbocation, steric block to backside attack).

2
Worked Example
An optically active 2° alkyl halide is hydrolysed and gives a product that is almost fully inverted in configuration. What does this tell you about the mechanism and the rate law?
Solution
  1. Clean inversion at the stereocentre means the nucleophile attacked from the side opposite the leaving group in a single concerted step.
  2. That is the signature of SN2 (Walden inversion); a carbocation (SN1) would have given racemisation.
  3. An SN2 process is bimolecular, so the rate law is Rate = k[R–X][Nu] (second order).

Answer: The reaction is SN2 with inversion of configuration; Rate = k[R–X][Nu] (second order).

3
Worked Example
Predict the major alkene when 2-bromobutane is treated with alcoholic KOH and name the rule used.
Solution
  1. Alcoholic KOH is a strong base → β-elimination (dehydrohalogenation): loss of H from a β-carbon and Br from C2.
  2. Two alkenes are possible: but-1-ene (terminal) and but-2-ene (internal).
  3. By Saytzeff’s rule the more substituted, more stable alkene predominates: but-2-ene (CH3CH=CHCH3) is the major product.

Answer: But-2-ene is the major product (Saytzeff’s rule — the more highly substituted alkene).

4
Worked Example
Why must the apparatus be perfectly dry when preparing and using a Grignard reagent?
Solution
  1. In R–MgX the C–Mg bond is strongly polarised with carbon highly nucleophilic (essentially carbanion-like).
  2. Such a carbon reacts immediately with any acidic hydrogen; water supplies it: R–MgX + H2O → R–H + Mg(OH)X.
  3. Even traces of moisture destroy the reagent (converting it to an alkane) before it can do the intended C–C bond reaction, so everything must be anhydrous.

Answer: The nucleophilic carbon of R–MgX reacts instantly with the acidic H of water to give the alkane (R–H), destroying the reagent; hence dry conditions are essential.

5
Worked Example
How does changing the solvent from acetone (aprotic) to water (protic) affect the substitution of a secondary alkyl bromide?
Solution
  1. A 2° halide can react by either SN1 or SN2; the solvent tips the balance.
  2. Acetone (polar aprotic) does not solvate the nucleophile’s lone pair, leaving it reactive → favours SN2.
  3. Water (polar protic) hydrogen-bonds to and stabilises both the nucleophile (slowing it) and especially the carbocation, so it favours SN1.

Answer: Acetone favours SN2 (naked, reactive nucleophile); water favours SN1 (stabilises the carbocation).

6
Worked Example
Write the product of the Wurtz reaction of chloroethane with sodium in dry ether, and explain why it is not a good method for making unsymmetrical alkanes.
Solution
  1. Wurtz: 2CH3CH2Cl + 2Na →(dry ether) CH3CH2–CH2CH3 + 2NaCl, i.e. n-butane.
  2. The reaction couples two alkyl groups; with two different halides (R–X + R′–X) three products form: R–R, R′–R′ and R–R′.
  3. That mixture is hard to separate and lowers the yield of the desired unsymmetrical alkane, so Wurtz is useful mainly for symmetrical alkanes.

Answer: Product is n-butane; for unsymmetrical alkanes a mixture of three coupling products forms, making Wurtz impractical there.

Key Points

  • SN2: one-step, backside attack, inversion (Walden), Rate = k[R–X][Nu]; fastest for methyl > 1° > 2° > 3°; favoured by strong nucleophiles and polar aprotic solvents.
  • SN1: two-step via planar carbocation, racemisation, Rate = k[R–X]; fastest for 3° > 2° > 1° > methyl; favoured by polar protic solvents.
  • β-elimination (alc. KOH) gives alkenes; Saytzeff’s rule → the more substituted (more stable) alkene is major; aq. KOH favours substitution, alc. KOH favours elimination.
  • Metals: Mg/dry ether → Grignard R–MgX (nucleophilic C, must be anhydrous); 2R–X + 2Na/dry ether → R–R (Wurtz, symmetrical alkanes).
  • A C with four different groups is chiral; enantiomers rotate light oppositely; inversion signals SN2, racemisation signals SN1.
Quick Check — 4 MCQs
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Q1.The hydrolysis of (CH3)3C–Br proceeds by SN1. Its rate law is:
Explanation: SN1 is unimolecular; the slow step is ionisation of the substrate only, so the rate depends solely on [R–X] and is independent of the nucleophile concentration.
Q2.Inversion of configuration (Walden inversion) at the reacting carbon is characteristic of:
Explanation: SN2 involves backside attack opposite the leaving group, so the configuration is inverted. SN1 goes through a planar carbocation and gives racemisation.
Q3.When 2-bromo-2-methylbutane is heated with alcoholic KOH, the major alkene (by Saytzeff’s rule) is:
Explanation: Saytzeff’s rule states that in β-elimination the more highly substituted, more stable alkene is formed as the major product.
Q4.Polar aprotic solvents such as DMSO or acetone speed up substitution by favouring:
Explanation: Polar aprotic solvents solvate cations but not anions well, so the nucleophile stays "naked" and highly reactive, accelerating the bimolecular SN2 pathway.
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