Why a Crowded Carbon Changes How a Haloalkane Reacts

Swap a hydrogen in an alkane for a halogen and the molecule suddenly becomes reactive: the carbon-halogen bond is polar, and the halide leaves easily. That makes haloalkanes a gateway to alcohols, ethers, amines and nitriles — and the route they take depends on how crowded the carbon is.

This lesson covers naming and preparing haloalkanes, and the SN1 and SN2 mechanisms of nucleophilic substitution with the factors that control them.

Haloalkanes are named as halo-substituted alkanes and classified as primary, secondary or tertiary by the carbon carrying the halogen, and they are prepared from alcohols, from alkenes, or by halogen exchange.

Naming and classification:

- $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}$ is chloroethane, $\mathrm{C}{\mathrm{H}}_3\mathrm{CHBrC}{\mathrm{H}}_3$ is 2-bromopropane, and $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{CCl}$ is 2-chloro-2-methylpropane
- A haloalkane is primary, secondary or tertiary according to how many carbons are attached to the carbon carrying the halogen
- The C-X bond is polar, with a partial positive charge on carbon

From alcohols:

- With hydrogen halides: $\mathrm{ROH}+\mathrm{HCl}\stackrel{\mathrm{ZnC}{\mathrm{l}}_2}{\to }\mathrm{RCl}+{\mathrm{H}}_2\mathrm{O}$
- With phosphorus halides: $3ROH+\mathrm{PC}{\mathrm{l}}_3\to 3RCl+{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3$
- With thionyl chloride, the cleanest method because both by-products are gases: $\mathrm{ROH}+\mathrm{SOC}{\mathrm{l}}_2\to \mathrm{RCl}+\mathrm{S}{\mathrm{O}}_2+\mathrm{HCl}$

From alkenes. Adding HX follows Markovnikov's rule, so propene with HBr gives 2-bromopropane; adding bromine gives a vicinal dihalide, $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$.

Halogen exchange:

- Finkelstein reaction — $\mathrm{RCl}+\mathrm{NaI}\to \mathrm{RI}+\mathrm{NaCl}$ in dry acetone, where sodium chloride precipitates and drives the reaction forward
- Swarts reaction — alkyl chlorides or bromides heated with metal fluorides such as AgF give alkyl fluorides

Worked example. To make 1-chlorobutane from butan-1-ol with the simplest clean-up, use thionyl chloride: sulphur dioxide and hydrogen chloride escape as gases, leaving the product almost pure.

An everyday example. Chloroethane sprays used by sports physiotherapists numb an injured area quickly, because the liquid evaporates rapidly and cools the skin.

The substance. Tertiary alcohols react with hydrogen halides fastest — their reaction passes through a stable carbocation, the same idea behind the Lucas test.

In SN2, the nucleophile attacks carbon from the side opposite the leaving group in a single step, while in SN1 the halide leaves first to form a carbocation that the nucleophile then attacks; the haloalkane's structure, the nucleophile, the solvent and the leaving group decide which path wins.

SN2 — bimolecular substitution:

- One step: the nucleophile bonds as the halide leaves, through a transition state with five groups around carbon
- Rate $=k[\mathrm{RX}][\mathrm{N}{\mathrm{u}}^{-}]$, so the reaction is second order
- Attack from the back causes inversion of configuration at carbon
- Reactivity: methyl > primary > secondary > tertiary, because bulky groups block the nucleophile

SN1 — unimolecular substitution:

- Slow step: $\mathrm{R}-\mathrm{X}\to {\mathrm{R}}^{+}+{\mathrm{X}}^{-}$, forming a planar carbocation
- Fast step: the nucleophile attacks the carbocation from either face
- Rate $=k[\mathrm{RX}]$, so the reaction is first order
- Reactivity: tertiary > secondary > primary > methyl, following carbocation stability
- An optically active halide gives a largely racemic product

Factors that decide the mechanism:

- Structure — tertiary, benzylic and allylic halides favour SN1; methyl and primary halides favour SN2
- Nucleophile — a strong nucleophile at high concentration favours SN2
- Solvent — polar protic solvents such as water stabilise carbocations and favour SN1; polar aprotic solvents such as acetone favour SN2
- Leaving group — $\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl}>\mathrm{R}-\mathrm{F}$ for both, because the C-I bond is weakest

Worked example. $\mathrm{C}{\mathrm{H}}_3\mathrm{Br}$ with hydroxide in acetone reacts by SN2, and its rate doubles if either concentration doubles. $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{CBr}$ in water reacts by SN1 to give 2-methylpropan-2-ol, at a rate that does not depend on the nucleophile at all.

An everyday example. Drug makers in India's pharmaceutical hubs must control which mirror-image form of a molecule they produce, and an SN2 step, with its predictable inversion, gives a single form where an SN1 step would give a racemic mixture.

The substance. Benzylic and allylic halides can react by either mechanism — resonance stabilises their carbocation for SN1, while their unhindered carbon still allows SN2 attack.