Nucleophilic Substitution and Elimination

AS · 15 min

When a halogenoalkane reacts with a nucleophile, the halogen is replaced. That much is simple. The interesting question is how: does the nucleophile push the halogen out in one smooth step, or does the halogen leave first and the nucleophile arrive afterwards? Both happen, and which one depends on whether the halogenoalkane is primary or tertiary. This note sets out the SN2 and SN1 mechanisms with every curly arrow, explains why primary halogenoalkanes use SN2 and tertiary ones SN1, and describes the competing elimination reaction. The SN1 and SN2 mechanisms are among the most frequently examined drawings in Paper 2.

Nucleophilic substitution in outline

Every nucleophilic substitution of a halogenoalkane has the same overall pattern:

\ce{Nu^- + R-X -> R-Nu + X^-}

The nucleophile (Nu) has a lone pair. At AS you need $\ce{OH-}$ , $\ce{CN-}$ , $\ce{NH3}$ and $\ce{H2O}$ .
It attacks the $\delta+$ carbon of the polar C–X bond.
The halogen leaves as a halide ion, $\ce{X-}$ , taking both electrons of the C–X bond with it (heterolytic fission). The halide ion is called the leaving group.

The name "SN" stands for Substitution, Nucleophilic. The number gives the number of species involved in the rate-determining step, the slowest step, which controls the overall rate.

The SN2 mechanism

SN2 means substitution, nucleophilic, bimolecular: two species (the halogenoalkane and the nucleophile) take part in the single, rate-determining step. Primary halogenoalkanes react by SN2.

SN2 mechanism for bromoethane and hydroxide ions. The nucleophile attacks the δ+ carbon from the side opposite the bromine while the C–Br bond breaks, through a single transition state (bonds partly formed and partly broken, shown dashed). The other three groups flip over like an umbrella in a wind.

For bromoethane and hydroxide ions:

The C–Br bond is polar: $\ce{C^{\delta+}-Br^{\delta-}}$ .
A curly arrow from a lone pair on the oxygen of $\ce{OH-}$ to the $\delta+$ carbon. The hydroxide ion approaches from the side opposite the bromine atom, where it is not repelled by the electron-rich bromine atom.
At the same time, a curly arrow from the C–Br bond to the bromine atom: the bond breaks heterolytically and bromine leaves as $\ce{Br-}$ .
Halfway through, the molecule passes through a transition state in which the C–O bond is partly formed and the C–Br bond partly broken. It carries a negative charge spread over the O and Br, and the three other groups on the carbon lie in a plane. It is drawn in square brackets with dashed partial bonds. It is not an intermediate: it cannot be isolated.
As the C–Br bond breaks completely, the three groups flip to the other side (like an umbrella in a strong wind). The products are ethanol and a bromide ion.

\ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}

Because both the halogenoalkane and the nucleophile are in the one step, the rate depends on the concentrations of both. (At A Level this is written $\text{rate} = k[\ce{RX}][\ce{OH-}]$ .)

The SN1 mechanism

SN1 means substitution, nucleophilic, unimolecular: only one species (the halogenoalkane) is involved in the rate-determining step. Tertiary halogenoalkanes react by SN1.

SN1 mechanism for 2-bromo-2-methylpropane and hydroxide ions. Step 1 (slow): the C–Br bond breaks heterolytically, both electrons going to bromine, to give a tertiary carbocation and Br⁻. Step 2 (fast): a lone pair on OH⁻ forms a bond to the positive carbon.

For 2-bromo-2-methylpropane and hydroxide ions:

Step 1 (slow, rate-determining)

The polar C–Br bond, $\ce{C^{\delta+}-Br^{\delta-}}$ , breaks heterolytically on its own. A curly arrow from the C–Br bond to the bromine atom shows both electrons going to bromine.
This forms a bromide ion and a tertiary carbocation, $\ce{(CH3)3C+}$ , the intermediate. The positive carbon has only three bonds; it is sp² hybridised and planar.

\ce{(CH3)3CBr -> (CH3)3C+ + Br-}

Step 2 (fast)

A curly arrow from a lone pair on the oxygen of $\ce{OH-}$ to the positive carbon forms the C–O bond. Because the carbocation is planar, the hydroxide ion can attack from either side.

\ce{(CH3)3C+ + OH- -> (CH3)3COH}

The product is 2-methylpropan-2-ol. The rate depends only on the concentration of the halogenoalkane, because the hydroxide ion is not involved in the slow step. (At A Level: $\text{rate} = k[\ce{RX}]$ .)

Tip

Extension (A Level): because SN2 attack comes from the back, an SN2 reaction at a chiral centre inverts its configuration, giving a single optical isomer. Because SN1 attack on a planar carbocation is equally likely from either face, an SN1 reaction at a chiral centre gives a racemic mixture (equal amounts of both enantiomers). This is used at A Level as evidence for the mechanisms.

Why primary goes SN2 and tertiary goes SN1

Two factors decide the mechanism, and they point the same way.

Carbocation stability: the inductive effect

SN1 needs a carbocation to form on its own. Alkyl groups are electron-donating (positive inductive effect): each one pushes electron density towards the positive carbon, spreading out the charge and stabilising the ion.

A tertiary carbocation has three alkyl groups on the positive carbon. It is stable enough to form at a reasonable rate, so tertiary halogenoalkanes can ionise: SN1.
A primary carbocation has only one alkyl group. It is too unstable to form, so SN1 does not happen for primary halogenoalkanes.

Steric hindrance

SN2 needs the nucleophile to reach the carbon from the back, through the gap between the three other groups.

In a primary halogenoalkane, two of those groups are small hydrogen atoms, so the back of the carbon is open: SN2 is easy.
In a tertiary halogenoalkane, three bulky alkyl groups crowd round the carbon and block the approach of the nucleophile (steric hindrance): SN2 is effectively impossible.

Key result

halogenoalkane	main mechanism	why
primary	SN2	carbon is unhindered; primary carbocation too unstable for SN1
secondary	both SN1 and SN2	intermediate hindrance and carbocation stability
tertiary	SN1	stable tertiary carbocation (three electron-donating alkyl groups); carbon too hindered for SN2

feature	SN2	SN1
number of steps	one	two
species in the rate-determining step	two (RX and $\ce{Nu-}$ )	one (RX)
intermediate	none (a transition state only)	carbocation
direction of attack	from the side opposite the halogen	either face of the planar carbocation
typical substrate	primary	tertiary
effect of doubling $[\ce{Nu-}]$	rate doubles	no effect on rate

Mechanisms with other nucleophiles

The same two mechanisms work with every nucleophile on the syllabus. Only the attacking species changes.

Cyanide ions

For bromoethane (primary, SN2) with cyanide in ethanol:

A curly arrow from the lone pair on the carbon atom of $\ce{CN-}$ to the $\delta+$ carbon of C–Br.
A curly arrow from the C–Br bond to Br; $\ce{Br-}$ leaves.
Product: propanenitrile, $\ce{CH3CH2CN}$ , with a new C–C bond.

The arrow must start from the carbon end of cyanide, because the new bond is C–C.

Ammonia

For bromoethane (SN2) with ammonia:

A curly arrow from the lone pair on the nitrogen of $\ce{NH3}$ to the $\delta+$ carbon.
A curly arrow from the C–Br bond to Br; $\ce{Br-}$ leaves.
The product at this point is an ion, $\ce{CH3CH2NH3+}$ , because nitrogen now has four bonds.
A second ammonia molecule acts as a base: a curly arrow from its nitrogen lone pair to one of the H atoms on the $\ce{N+}$ , and a curly arrow from that N–H bond to the nitrogen. This removes $\ce{H+}$ , giving ethylamine, $\ce{CH3CH2NH2}$ , and $\ce{NH4+}$ .

Water

Water is a much weaker nucleophile than hydroxide (it is neutral, so less strongly attracted to the $\delta+$ carbon). This is why hydrolysis by water in the silver nitrate test is slow, and why it is a useful way to compare rates.

Elimination

The hydroxide ion has two possible roles. In aqueous solution it acts mainly as a nucleophile, attacking carbon (substitution). In ethanol it acts mainly as a base, removing a proton (elimination).

Key result

Elimination: halogenoalkane heated under reflux with NaOH (or KOH) in ethanol gives an alkene. An H atom from a carbon next to the C–X carbon and the X atom are removed (as $\ce{H2O}$ and $\ce{X-}$ , effectively HX).

\ce{CH3CHBrCH3 + OH- -> CH2=CHCH3 + H2O + Br-}

Elimination is favoured by:

ethanol as solvent (rather than water);
higher temperature;
a concentrated, strong base;
tertiary (and secondary) halogenoalkanes, where the carbon is too hindered for substitution but the hydrogens on neighbouring carbons are exposed.

Substitution and elimination always compete. In practice a mixture forms; the conditions shift the balance.

Tip

The mechanism of elimination is not required at AS. In outline (E2): the hydroxide ion uses a lone pair to remove an H+ from the carbon next to C–X; the C–H bonding pair moves in to form the C=C $\pi$ bond; and the C–X bond breaks, releasing $\ce{X-}$ .

Worked examples

Routine: describing SN2

Describe the mechanism of the reaction between 1-bromopropane and aqueous hydroxide ions. Name the mechanism and the product.

Solution

SN2 (1-bromopropane is primary).

Dipole on C–Br: $\ce{C^{\delta+}-Br^{\delta-}}$ .
Curly arrow from a lone pair on O of $\ce{OH-}$ to the $\delta+$ carbon, approaching from the side opposite the Br.
Curly arrow from the C–Br bond to Br.
Transition state: $\ce{[HO\cdots C\cdots Br]-}$ (partial bonds, negative charge), with the carbon's three other groups ( $\ce{C2H5}$ , H, H) in a plane.
Products: propan-1-ol, $\ce{CH3CH2CH2OH}$ , and $\ce{Br-}$ .

Routine: describing SN1

Describe the mechanism of the hydrolysis of 2-chloro-2-methylpropane by hydroxide ions.

Solution

SN1 (tertiary halogenoalkane).

Step 1 (slow): curly arrow from the C–Cl bond to Cl; heterolytic fission gives $\ce{Cl-}$ and the tertiary carbocation $\ce{(CH3)3C+}$ .

Step 2 (fast): curly arrow from a lone pair on O of $\ce{OH-}$ to the positive carbon.

Product: 2-methylpropan-2-ol, $\ce{(CH3)3COH}$ .

Standard: predicting the mechanism

For each halogenoalkane, predict whether it reacts with aqueous hydroxide mainly by SN1 or SN2, and explain: (a) 1-iodobutane, (b) 2-bromo-2-methylbutane, (c) 2-chlorobutane.

Solution

(a) Primary: SN2. The back of the C–I carbon is unhindered (it carries two H atoms), and a primary carbocation would be too unstable to form.

(b) Tertiary: SN1. The C–Br carbon carries three alkyl groups, which block backside attack (steric hindrance) and, by their electron-donating inductive effect, stabilise the tertiary carbocation.

(c) Secondary: both mechanisms occur. The secondary carbocation is moderately stable and the carbon moderately hindered.

Standard: the cyanide mechanism

1-chloropropane is heated with potassium cyanide in ethanol. (a) Name and describe the mechanism. (b) Name the product and state why this reaction is useful in synthesis.

Solution

(a) SN2 (primary). Curly arrow from the lone pair on the carbon of $\ce{CN-}$ to the $\delta+$ carbon of C–Cl (attack from the opposite side to Cl); curly arrow from the C–Cl bond to Cl. Transition state $\ce{[NC\cdots C\cdots Cl]-}$ ; products $\ce{CH3CH2CH2CN}$ and $\ce{Cl-}$ .

(b) Butanenitrile. The reaction forms a new carbon–carbon bond, lengthening the chain by one carbon; the nitrile can then be hydrolysed to a carboxylic acid (butanoic acid) or reduced to an amine.

Exam-hard: rate evidence

When the concentration of hydroxide ions is doubled, the rate of hydrolysis of 1-bromobutane doubles, but the rate of hydrolysis of 2-bromo-2-methylpropane does not change. Explain these observations in terms of the mechanisms.

Solution

1-bromobutane is primary and reacts by SN2: the hydroxide ion and the halogenoalkane collide in a single step, which is the rate-determining step. The rate depends on how often they collide, so it is proportional to $[\ce{OH-}]$ : doubling $[\ce{OH-}]$ doubles the rate.

2-bromo-2-methylpropane is tertiary and reacts by SN1. The slow, rate-determining step is the ionisation of the halogenoalkane, $\ce{(CH3)3CBr -> (CH3)3C+ + Br-}$ , which does not involve hydroxide. Hydroxide only reacts in the fast second step, so changing its concentration does not change the overall rate.

Exam-hard: substitution vs elimination

2-bromo-2-methylbutane, $\ce{CH3CBr(CH3)CH2CH3}$ , is heated with potassium hydroxide.

(a) In aqueous solution the main product is an alcohol. Name it and name the mechanism. (b) In ethanol, two alkenes form. Name both and explain why there are two. (c) Explain the role of the hydroxide ion in each case.

Solution

(a) 2-methylbutan-2-ol, $\ce{CH3C(OH)(CH3)CH2CH3}$ , by SN1 (tertiary halogenoalkane).

(b) The Br is on C2. An H can be removed from C1 (or the equivalent methyl branch) giving 2-methylbut-1-ene, $\ce{CH2=C(CH3)CH2CH3}$ , or from C3 giving 2-methylbut-2-ene, $\ce{(CH3)2C=CHCH3}$ . Two different sets of neighbouring hydrogens give two alkenes. (Neither shows cis/trans isomerism: each has a carbon of the C=C with two identical groups.)

(c) In water, $\ce{OH-}$ acts as a nucleophile, donating a lone pair to the positive carbon of the carbocation. In ethanol, $\ce{OH-}$ acts as a base, accepting a proton from a carbon next to the C–Br carbon.

Watch out

Arrow from the minus sign. The nucleophile's arrow starts at a lone pair, which you must draw, on the correct atom (O of $\ce{OH-}$ , C of $\ce{CN-}$ , N of $\ce{NH3}$ ).
SN2 with a carbocation. SN2 has no intermediate. If you draw a carbocation, you have drawn SN1.
Missing the slow step. For SN1, label step 1 as slow (rate-determining). Mixing the two steps into one loses the mechanism.
Transition state without charge or brackets. The SN2 transition state is in square brackets with a negative charge (for an anionic nucleophile) and dashed partial bonds to both the nucleophile and the leaving group.
"Tertiary carbocations are stable because they are bigger". The reason is the electron-donating (positive inductive) effect of three alkyl groups.
Mixing up the role of $\ce{OH-}$ . Nucleophile in substitution (attacks carbon); base in elimination (removes $\ce{H+}$ ).

Exam tip

Mechanism marks for SN2: dipole on C–X; arrow from lone pair on Nu to $\delta+$ C; arrow from C–X bond to X; transition state (brackets, charge, partial bonds); products. For SN1: dipole and arrow from C–X to X; carbocation with + charge; arrow from lone pair on Nu to $\ce{C+}$ ; product.
When asked "explain why tertiary halogenoalkanes react by SN1", give both points: carbocation stabilised by three electron-donating alkyl groups, and steric hindrance preventing SN2.
Be careful which halogenoalkane is named: "2-bromo-2-methylpropane" is tertiary, "1-bromo-2-methylpropane" is primary, even though both are $\ce{C4H9Br}$ .
"State the role of the hydroxide ion" has a one-word answer: nucleophile (substitution) or base (elimination).

Summary

Nucleophilic substitution: Nu (lone pair) attacks $\delta+$ carbon; $\ce{X-}$ leaves (heterolytic fission).
SN2: one step, two species in the rate-determining step, backside attack, transition state with partial bonds, no intermediate. Primary halogenoalkanes.
SN1: two steps; slow ionisation to a carbocation (one species in the rate-determining step), then fast attack by Nu. Tertiary halogenoalkanes.
Tertiary → SN1 because three alkyl groups stabilise the carbocation (inductive effect) and block backside attack (steric hindrance). Primary → SN2. Secondary → both.
The same mechanisms apply with $\ce{CN-}$ (attacks through C) and $\ce{NH3}$ (attacks through N, then loses $\ce{H+}$ to a second $\ce{NH3}$ ).
Hydroxide in ethanol, heat: elimination to an alkene ( $\ce{OH-}$ acts as a base). Favoured by ethanol, heat and tertiary substrates.

Practice

Question

What do "S", "N" and "2" stand for in SN2?
State the mechanism by which each reacts with aqueous hydroxide: (a) 1-chlorobutane, (b) 2-chloro-2-methylpropane, (c) 2-chlorobutane.
Describe, with numbered steps, the mechanism of the reaction of iodomethane with hydroxide ions.
Explain why tertiary halogenoalkanes do not react by the SN2 mechanism, and why primary halogenoalkanes do not react by SN1.
Explain the difference between a transition state and an intermediate, using the SN2 and SN1 mechanisms as examples.
Describe the mechanism of the reaction of bromoethane with cyanide ions. From which atom of the cyanide ion does the curly arrow start, and why?
State three conditions that favour elimination over substitution when a halogenoalkane is treated with hydroxide ions.
Write an equation for the reaction of 2-bromo-2-methylpropane with potassium hydroxide in ethanol. Name the organic product and give the role of the hydroxide ion.
Describe the full mechanism of the reaction between 1-bromopropane and ammonia, including the step that forms the amine.
1-bromo-2,2-dimethylpropane, $\ce{(CH3)3CCH2Br}$ , is a primary halogenoalkane, but it reacts extremely slowly with hydroxide ions by both SN1 and SN2. Explain why each mechanism is slow for this compound.

Answers

Substitution, nucleophilic, bimolecular (two species are involved in the rate-determining step).
(a) SN2 (primary). (b) SN1 (tertiary). (c) Both SN1 and SN2 (secondary).
SN2. (1) C–I is polar, $\ce{C^{\delta+}-I^{\delta-}}$ . (2) Curly arrow from a lone pair on the O of $\ce{OH-}$ to the $\delta+$ carbon, from the side opposite the iodine. (3) Curly arrow from the C–I bond to the I atom. (4) Transition state $\ce{[HO\cdots CH3\cdots I]-}$ with partial bonds and the three H atoms in a plane. (5) Products: methanol, $\ce{CH3OH}$ , and $\ce{I-}$ .
Tertiary: the three bulky alkyl groups on the C–X carbon block the nucleophile's approach from the back (steric hindrance), so SN2 cannot occur. Primary: SN1 would need a primary carbocation, which has only one electron-donating alkyl group to stabilise the positive charge; it is too unstable to form at a significant rate.
A transition state is the highest-energy arrangement during a step, with bonds partly made and partly broken; it exists only momentarily and cannot be isolated (for example $\ce{[HO\cdots C\cdots Br]-}$ in SN2). An intermediate is a species formed in one step and used up in a later step; it is a real (if short-lived) species with complete bonds, such as the carbocation in SN1. SN2 has one step and so a transition state but no intermediate; SN1 has two steps and a carbocation intermediate.
SN2. Curly arrow from the lone pair on the carbon atom of $\ce{:C#N-}$ to the $\delta+$ carbon of C–Br (from the side opposite Br); curly arrow from the C–Br bond to Br; transition state $\ce{[NC\cdots C\cdots Br]-}$ ; products propanenitrile, $\ce{CH3CH2CN}$ , and $\ce{Br-}$ . The arrow starts from carbon because the product is a nitrile, with a new C–C bond; the carbon of cyanide carries the lone pair used.
Ethanol as the solvent rather than water; a high temperature (heat under reflux); a concentrated strong base. (Also: a tertiary rather than primary halogenoalkane.)
$\ce{(CH3)3CBr + KOH -> CH2=C(CH3)2 + KBr + H2O}$ . Product: 2-methylpropene (methylpropene). The hydroxide ion acts as a base (removes a proton from a carbon next to the C–Br carbon).
SN2. (1) Dipole on C–Br. (2) Curly arrow from the lone pair on N of $\ce{NH3}$ to the $\delta+$ carbon. (3) Curly arrow from the C–Br bond to Br, releasing $\ce{Br-}$ . (4) Intermediate ion $\ce{CH3CH2CH2NH3+}$ . (5) A second $\ce{NH3}$ uses its lone pair to remove an $\ce{H+}$ from the $\ce{N+}$ : curly arrow from the N lone pair of $\ce{NH3}$ to an H on $\ce{N+}$ , and curly arrow from that N–H bond to the N. (6) Products propylamine, $\ce{CH3CH2CH2NH2}$ , and $\ce{NH4+}$ (with $\ce{Br-}$ , overall $\ce{NH4Br}$ ).
SN2: although the C–Br carbon is primary, the carbon next to it carries three methyl groups. This bulky $\ce{C(CH3)3}$ group crowds the back of the C–Br carbon and blocks the nucleophile's approach (steric hindrance), so backside attack is very slow. SN1: ionisation would give a primary carbocation, $\ce{(CH3)3CCH2+}$ , with only one alkyl group bonded to the positive carbon; it is too unstable to form readily. With both routes hindered, the reaction is very slow.

Nucleophilic substitution in outline

The SN2 mechanism

The SN1 mechanism

Why primary goes SN2 and tertiary goes SN1

Carbocation stability: the inductive effect

Steric hindrance

Mechanisms with other nucleophiles

Cyanide ions

Ammonia

Water

Elimination

Worked examples

Practice

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