Bromoethane Reactions: SN2 vs E2 Mechanisms

Why Study Bromoethane?

Bromoethane ( $CH_3CH_2Br$ ) is the classic substrate for studying competing SN2 (nucleophilic substitution) and E2 (elimination) reactions. Because it is a primary haloalkane, the SN2 pathway is strongly favoured — but under the right conditions, E2 can dominate.

Understanding this competition is essential for predicting the products of organic reactions in exams and in the lab.

Learning Goals: By the end of this guide, you should be able to:

Draw complete curly arrow mechanisms for SN2 and E2 reactions of bromoethane.

Predict the major product based on reagent, temperature, and solvent.

Explain why primary substrates favour SN2 over SN1.

Compare the stereochemical and kinetic features of SN2 and E2.

SN2: Nucleophilic Substitution (Bimolecular)

In an SN2 reaction, the nucleophile attacks the electrophilic carbon from the back side (180° to the leaving group), and bond formation and bond breaking happen simultaneously in a single concerted step.

Reaction with NaOH (dilute, aqueous)

CH_3CH_2Br + OH^- \rightarrow CH_3CH_2OH + Br^-

Product: Ethanol

Key Features of SN2

Feature	Description
Rate law	$\text{Rate} = k[substrate][nucleophile]$ — bimolecular
Mechanism	One-step, concerted
Stereochemistry	Inversion of configuration (Walden inversion)
Substrate preference	Primary > Secondary >> Tertiary (steric hindrance)
Nucleophile	Strong nucleophile required ( $OH^-$ , $CN^-$ , $NH_3$ )

Common SN2 Reactions of Bromoethane

Nucleophile	Product	Name
$OH^-$ (dilute, warm)	$CH_3CH_2OH$	Ethanol
$CN^-$ (in ethanol)	$CH_3CH_2CN$	Propanenitrile
$NH_3$ (excess, sealed)	$CH_3CH_2NH_2$	Ethylamine
$H_2O$	$CH_3CH_2OH$	Ethanol (slow, neutral)

E2: Elimination (Bimolecular)

In an E2 reaction, a strong base abstracts a β-hydrogen while the leaving group departs simultaneously, forming a C=C double bond (alkene).

Reaction with KOH (conc., in ethanol, hot)

CH_3CH_2Br + OH^- \xrightarrow{\text{ethanol, heat}} CH_2{=}CH_2 + H_2O + Br^-

Product: Ethene

Key Features of E2

Feature	Description
Rate law	$\text{Rate} = k[substrate][base]$ — bimolecular
Mechanism	One-step, concerted (anti-periplanar geometry)
Product	Alkene (follows Zaitsev's rule for longer chains)
Substrate preference	Tertiary > Secondary > Primary
Base	Strong, bulky base favours E2 ( $KOH$ in ethanol, $t$ - $BuOK$ )

SN2 vs E2: Decision Framework

The same reagent ( $OH^-$ ) can act as either a nucleophile (SN2) or a base (E2). What determines which path wins?

Factor	Favours SN2	Favours E2
Temperature	Low/moderate	High
Solvent	Aqueous (polar protic)	Ethanol (less polar)
OH⁻ concentration	Dilute	Concentrated
Substrate	Primary, unhindered	Tertiary, hindered
Reagent	Good nucleophile, weak base	Strong, bulky base

For bromoethane specifically: SN2 dominates in most conditions because primary carbons have minimal steric hindrance. E2 only becomes significant with concentrated KOH in ethanol at reflux temperature.

Worked Examples

Example 1: Predicting the Product

Reagent: Bromoethane + NaOH (dilute, aqueous, warm)

Analysis: Dilute aqueous conditions → SN2 favoured. $OH^-$ acts as nucleophile.

Product: $CH_3CH_2OH$ (ethanol) ✅

Example 2: Conditions for Ethene

Question: What conditions convert bromoethane into ethene?

Answer: Concentrated KOH in ethanol, heated under reflux. The ethanol solvent and high temperature favour E2 elimination. $OH^-$ acts as a base, abstracting the β-hydrogen.

Example 3: Why Not SN1?

Question: Why doesn't bromoethane undergo SN1?

Answer: SN1 requires formation of a carbocation intermediate. A primary carbocation ( $CH_3CH_2^+$ ) is extremely unstable — no hyperconjugation or inductive stabilisation from bulky alkyl groups. SN2 is overwhelmingly faster for primary substrates.

Common Mistakes

Confusing nucleophile and base roles — The same species (e.g. $OH^-$ ) can be either. Context (solvent, temperature, concentration) determines competition.
Drawing SN1 for primary substrates — Primary carbocations don't form under normal conditions. Always default to SN2 for primary haloalkanes.
Forgetting the anti-periplanar requirement — In E2, the H and leaving group must be anti-periplanar (180° dihedral). For bromoethane this is easily achieved by rotation, but for cyclic molecules it matters.
Not specifying conditions clearly — Examiners want reagent and conditions: "NaOH, dilute aqueous, warm" for SN2 vs "KOH, conc., ethanol, reflux" for E2.

Exam Tips (A-Level / AP / IB)

Draw full curly arrows from lone pair/bond to destination. Half arrows are for radical mechanisms only.
For SN2, show the nucleophile attacking from the opposite side to the leaving group.
State the type of mechanism explicitly: "This is an SN2 nucleophilic substitution."
Know the three organic product types from $CH_3CH_2Br$ : alcohol (NaOH/aq), nitrile (KCN/ethanol), amine (NH₃/excess).

Frequently Asked Questions

Why does bromoethane react faster than chloroethane in SN2?

The $C-Br$ bond is weaker than the $C-Cl$ bond (276 vs 338 kJ/mol) and $Br^-$ is a better leaving group (more stable anion, more polarisable). However, $C-I$ is even weaker, so iodoethane reacts fastest of all.

Can bromoethane undergo E1 elimination?

Effectively no. E1 requires carbocation formation (like SN1), and primary carbocations are too unstable. E1 is only significant for tertiary substrates.

What happens with excess ammonia?

With excess $NH_3$ , bromoethane first forms ethylamine ( $CH_3CH_2NH_2$ ). Without excess, further alkylation can occur: diethylamine → triethylamine → tetraethylammonium salt. Excess $NH_3$ minimises this.

Curly Arrow Mechanisms — Master electron-pair movement notation for all organic reactions.
Structural Isomerism — Understand how different connectivity produces isomeric products.
Intermolecular Forces — How IMFs affect boiling points of substrates and products.