---
title: "Nucleophilic Substitution"
book: "University Chemistry — Year 1"
subject: chemistry
language: en
chapter: 19
exercises: 12
source: https://one-course.com/books/chemistry/2/en/chapter/19-nucleophilic-substitution
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 19 — Nucleophilic Substitution

Take a bottle of (*S*)-2-bromobutane, a single enantiomer that rotates polarised light. Heated with sodium hydroxide in a polar [aprotic solvent](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes), it gives butan-2-ol that is a single enantiomer too — but the other one, (*R*): the reaction has turned the molecule inside out like an umbrella in the wind. Left instead in warm water, the same bromide gives butan-2-ol that is partly racemic. The same overall change, an OH group replacing a Br atom, follows two different paths. This chapter establishes the two mechanisms from their [rate laws](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order), explains their stereochemistry, and gives the rules that decide between them — the first full use of the tools of the kinetics and electronic-effects chapters.

**You already know.**

[Rate laws](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order), [elementary steps](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-elementary-step), the [rate-determining step](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-rds) and the [steady-state approximation](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-qssa) are in Chapters [8](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#ch-b1-rate-laws) and [9](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#ch-b1-elementary-steps); [configurations](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-isomers), Cram drawings and [optical activity](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-optical-activity) in [Chapter 16](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#ch-b1-stereochemistry-in-depth); [nucleophiles](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile), [electrophiles](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile), [leaving groups](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) and [carbocations](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) in [Chapter 18](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#ch-b1-electronic-effects); protic and [aprotic solvents](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes) in [Chapter 4](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#ch-b1-intermolecular-forces-solvents).

## 19.1 The substitution and its partners

**Definition 19.1 (Nucleophilic substitution, substrate).**

A *nucleophilic substitution* is the replacement, on a saturated carbon, of a [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) Y by a [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) Nu: $\mathrm{Nu^-} + \mathrm{R{-}Y} \to \mathrm{R{-}Nu} +
\mathrm{Y^-}$. The compound R–Y is the *substrate*; the carbon bearing Y is electrophilic because the C–Y bond is polarised towards Y.

Halogenoalkanes are the typical [substrates](#def-b1-nucleophilic-substitution-substitution), with $\ce{Cl-}$, $\ce{Br-}$ or $\ce{I-}$ as [leaving groups](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile). Hydroxide gives alcohols; alkoxides give ethers; cyanide gives nitriles (one more carbon); ammonia and amines give amines; iodide exchanges with the other halides. Water and alcohols, used as solvents, can also act as [nucleophiles](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile): the substitution is then a [solvolysis](#def-b1-nucleophilic-substitution-sn1).

## 19.2 The SN2 mechanism

When bromomethane reacts with hydroxide in water, doubling either concentration doubles the rate: $v = k[\ce{CH3Br}][\ce{OH-}]$, a reaction of order one in each reactant.

**Definition 19.2 (SN2 mechanism, Walden inversion).**

The *SN2 mechanism* (substitution, nucleophilic, bimolecular) is a single [elementary step](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-elementary-step) in which the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) attacks the carbon from the side opposite the [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) while the [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) departs. The three other substituents of the carbon flip over, like an umbrella turned inside out: the *Walden inversion*.

![The SN2 reaction of hydroxide with (S)-2-bromobutane. The nucleophile attacks the face opposite the bromine (curly arrows); in the transition state the O–C bond is forming while the C–Br bond breaks; the product is (R)-butan-2-ol: the three other groups have flipped to the other side.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-nucleophilic-substitution/fig-562d4f01988a.svg)

*The SN2 reaction of hydroxide with (*S*)-2-bromobutane. The [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) attacks the face opposite the bromine ([curly arrows](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-curly-arrow)); in the [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile) the O–C bond is forming while the C–Br bond breaks; the product is (*R*)-butan-2-ol: the three other groups have flipped to the other side.*

**Proposition 19.3 (Rate law of SN2).**

For an SN2 reaction, $v = k[\mathrm{RY}][\mathrm{Nu}]$.

**Proof.** The mechanism is a single bimolecular [elementary step](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-elementary-step); by the law of mass action for [elementary steps](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-elementary-step) ([Chapter 9](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#ch-b1-elementary-steps)) its rate is proportional to the product of the concentrations of the two partners. ∎

**Proposition 19.4 (Stereochemistry of SN2).**

An SN2 reaction at a stereogenic carbon gives a single stereoisomer, with inverted arrangement of the three unchanged groups: a reaction of this kind is stereospecific.

**Proof.** The [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) can only reach the empty side of the carbon, opposite the [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile); the three other groups pass through a plane in the [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile) and end on the other side. The new bond points where the old one did not: each molecule of [substrate](#def-b1-nucleophilic-substitution-substitution) gives the mirror arrangement of its groups. The descriptor usually changes from (*S*) to (*R*) or back, but not always, since it depends on the ranks of the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) and of the [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile). ∎

**Definition 19.5 (Stereospecific reaction).**

A reaction is a *stereospecific reaction* when stereoisomeric [substrates](#def-b1-nucleophilic-substitution-substitution) give stereoisomerically different products, each [substrate](#def-b1-nucleophilic-substitution-substitution) giving its own product.

## 19.3 The SN1 mechanism

2-Bromo-2-methylpropane reacts with water at a rate that does not depend on the concentration of the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile): $v = k[(\ce{CH3})_3\ce{CBr}]$. A step that does not involve the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) must decide the rate.

**Definition 19.6 (SN1 mechanism, racemisation, solvolysis).**

The *SN1 mechanism* (substitution, nucleophilic, unimolecular) has two steps: the slow [heterolysis](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-cleavage) of the C–Y bond, giving a [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates), then the fast addition of the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) to the [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates). A reaction that turns a single enantiomer into a mixture of both is a *racemisation*; a substitution in which the solvent is the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) is a *solvolysis*.

![SN1 solvolysis of 2-bromo-2-methylpropane in water: slow ionisation to the tertiary carbocation, then capture by water and loss of a proton. Below: a chiral secondary carbocation (from 2-bromobutane) is planar, and water adds on either face with equal probability.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-nucleophilic-substitution/fig-f3f68e7fc171.svg)

*SN1 [solvolysis](#def-b1-nucleophilic-substitution-sn1) of 2-bromo-2-methylpropane in water: slow ionisation to the tertiary [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates), then capture by water and loss of a proton. Below: a [chiral](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-stereocentre) secondary [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) (from 2-bromobutane) is planar, and water adds on either face with equal probability.*

**Proposition 19.7 (Rate law of SN1).**

For the [SN1 mechanism](#def-b1-nucleophilic-substitution-sn1), if the [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) is captured much faster than it returns to the [substrate](#def-b1-nucleophilic-substitution-substitution), $v = k_1[\mathrm{RY}]$, independent of the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile).

**Proof.** Steps: $\mathrm{RY} \rightleftharpoons \mathrm{R^+} + \mathrm{Y^-}$ ($k_1$, $k_{-1}$), $\mathrm{R^+} + \mathrm{Nu} \to \mathrm{RNu}$ ($k_2$). The [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) is a [reactive intermediate](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates): by the [steady-state approximation](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-qssa) ([Chapter 9](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#ch-b1-elementary-steps)), $k_1[\mathrm{RY}] =
k_{-1}[\mathrm{R^+}][\mathrm{Y^-}] + k_2[\mathrm{R^+}][\mathrm{Nu}]$, so $v = k_2[\mathrm{R^+}][\mathrm{Nu}] = \dfrac{k_1k_2[\mathrm{RY}][\mathrm{Nu}]}
{k_{-1}[\mathrm{Y^-}] + k_2[\mathrm{Nu}]}$. When $k_2[\mathrm{Nu}] \gg
k_{-1}[\mathrm{Y^-}]$ this reduces to $k_1[\mathrm{RY}]$: the first step is rate-determining. ∎

**Proposition 19.8 (Stereochemistry of SN1).**

An SN1 reaction at a stereogenic carbon gives both [configurations](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-isomers): the reaction is not stereospecific, and leads to [racemisation](#def-b1-nucleophilic-substitution-sn1) when nothing distinguishes the two faces of the [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates).

**Proof.** The [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) is planar ([Chapter 18](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#ch-b1-electronic-effects)): its two faces are mirror images. If the [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) has gone far before the capture, the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) adds to either face at the same rate, giving the two [enantiomers](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-enantiomers) in equal amounts. In practice the leaving ion still shields one face for a while, and a slight excess of inversion is often observed. ∎

![Schematic energy profiles. SN2: a single transition state, five groups around carbon. SN1: a first, higher barrier (ionisation, the rate-determining step), the carbocation in a shallow well, then a small barrier to its capture.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-nucleophilic-substitution/fig-953a01bf5cc6.svg)

*Schematic [energy profiles](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile). SN2: a single [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile), five groups around carbon. SN1: a first, higher barrier (ionisation, the [rate-determining step](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-rds)), the [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) in a shallow well, then a small barrier to its capture.*

## 19.4 What decides the mechanism

**Proposition 19.9 (Substrate).**

Methyl and primary [substrates](#def-b1-nucleophilic-substitution-substitution) react by SN2; tertiary [substrates](#def-b1-nucleophilic-substitution-substitution) by SN1; secondary [substrates](#def-b1-nucleophilic-substitution-substitution) by either, depending on the other factors. Allylic and benzylic [substrates](#def-b1-nucleophilic-substitution-substitution) react fast by both.

**Proof.** SN2 needs room behind the carbon: each alkyl group in place of a hydrogen crowds the [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile), which carries five groups, and slows the reaction; three groups block it. SN1 needs a stable [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates): the order of stability ([Proposition 18.8](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#prop-b1-electronic-effects-carbocation-order)) makes ionisation fast for tertiary, possible for secondary, and too slow for primary and methyl [carbocations](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates). Allylic and benzylic [carbocations](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) are stabilised by resonance, and the same $\pi$ system stabilises the SN2 [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile). ∎

**Proposition 19.10 (Leaving group).**

Both mechanisms are faster the better the [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile), that is the weaker the base it becomes: $\ce{I-} > \ce{Br-} > \ce{Cl-} \gg \ce{F-}$; $\ce{OH-}$ and $\ce{RO-}$ do not leave.

**Proof.** In both, the C–Y bond is broken in or before the [rate-determining step](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-rds), and the [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile) carries part of the negative charge on Y. A group that holds a negative charge well — the conjugate base of a [strong acid](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) — lowers that [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile). $\ce{HI}$, $\ce{HBr}$ and $\ce{HCl}$ are [strong acids](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid), $\ce{HF}$ a weak one ($\mathrm{p}K_a$ 3.2), water and alcohols very weak ones: $\ce{OH-}$ and $\ce{RO-}$ are [strong bases](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) and poor [leaving groups](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile). ∎

**Proposition 19.11 (Solvent).**

Polar [protic solvents](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes) (water, alcohols) favour SN1; polar [aprotic solvents](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes) (propanone, dimethyl sulfoxide, dimethylformamide) favour SN2 with anionic [nucleophiles](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile).

**Proof.** SN1 creates two ions from a neutral [substrate](#def-b1-nucleophilic-substitution-substitution): a polar [protic solvent](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes) stabilises both, the cation through its [lone pairs](https://one-course.com/books/chemistry/2/en/chapter/3-lewis-structures-resonance-and-vsepr#def-b1-lewis-resonance-vsepr-lewis) and the anion through [hydrogen bonds](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-hbond), and lowers the ionisation barrier. In SN2 the charge of an anionic [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) is spread in the [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile); a [protic solvent](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes), which binds the small anion strongly by [hydrogen bonds](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-hbond), slows it, whereas an [aprotic solvent](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes) leaves the anion free and reactive ([Chapter 4](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#ch-b1-intermolecular-forces-solvents)). ∎

**Method 19.12 (Choosing the mechanism).**

1. [Substrate](#def-b1-nucleophilic-substitution-substitution) : methyl or primary $\to$ SN2; tertiary $\to$ SN1; secondary $\to$ go on.
2. [Nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) : strong and anionic ( $\ce{OH-}$ , $\ce{RO-}$ , $\ce{CN-}$ , $\ce{I-}$ ) $\to$ SN2; weak and neutral (water, alcohol) $\to$ SN1.
3. Solvent: aprotic $\to$ SN2; protic $\to$ SN1.
4. Predict the stereochemistry (inversion or [racemisation](#def-b1-nucleophilic-substitution-sn1) ) and check that a [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) does not rather cause an elimination ( [Chapter 20](https://one-course.com/books/chemistry/2/en/chapter/20-elimination#ch-b1-elimination) ).

|  | SN2 | SN1 |
| --- | --- | --- |
| steps | one | two, via a [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) |
| [rate law](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) | $k[\mathrm{RY}][\mathrm{Nu}]$ | $k[\mathrm{RY}]$ |
| [substrate](#def-b1-nucleophilic-substitution-substitution) | methyl $>$ primary $>$ secondary | tertiary $>$ secondary |
| [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) | strong, anionic | weak, neutral (often the solvent) |
| solvent | polar aprotic | polar protic |
| stereochemistry | inversion (stereospecific) | [racemisation](#def-b1-nucleophilic-substitution-sn1) (mostly) |

*The two mechanisms side by side.*

## 19.5 Exercises

**Exercise 19.1 ★.**

Write the products of: bromoethane with sodium hydroxide; iodomethane with sodium ethoxide; 1-chloropropane with potassium cyanide; bromomethane with ammonia (first step). Identify [substrate](#def-b1-nucleophilic-substitution-substitution), [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) and [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile).

**Solution of Exercise 19.1.**

$\ce{CH3CH2OH + Br-}$; $\ce{CH3OCH2CH3 + I-}$; $\ce{CH3CH2CH2CN + Cl-}$; $\ce{CH3NH3+ + Br-}$. [Substrates](#def-b1-nucleophilic-substitution-substitution): the halogenoalkanes; [nucleophiles](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile): $\ce{OH-}$, $\ce{CH3CH2O-}$, $\ce{CN-}$, $\ce{NH3}$; [leaving groups](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile): the halide ions.

**Exercise 19.2 ★.**

For a reaction $\mathrm{RY} + \mathrm{Nu}$ with $v = k[\mathrm{RY}][\mathrm{Nu}]$, what happens to the [initial rate](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-initial-rate) if $[\mathrm{Nu}]$ is tripled? If both concentrations are halved? Same questions if $v = k[\mathrm{RY}]$.

**Solution of Exercise 19.2.**

Second order: tripled; both halved, divided by 4. First order in RY only: unchanged; halved.

**Exercise 19.3 ★.**

Predict the mechanism (SN1 or SN2): 1-bromobutane with sodium iodide in propanone; 2-bromo-2-methylpropane in water; bromomethane with hydroxide; 2-bromobutane in ethanol without added base.

**Solution of Exercise 19.3.**

SN2 (primary, good [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile), aprotic); SN1 (tertiary, water); SN2 (methyl, strong [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile)); SN1 [solvolysis](#def-b1-nucleophilic-substitution-sn1) (secondary, weak neutral [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile), [protic solvent](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes)), possibly with some SN2.

**Exercise 19.4 ★.**

Draw the product of the SN2 reaction of cyanide with (*R*)-2-bromobutane in [Cram representation](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-representations) and give its descriptor.

**Solution of Exercise 19.4.**

The cyanide takes the place opposite the bromine and the three other groups flip. In the product the CN group ranks first, as Br did, the other ranks unchanged: the inverted arrangement has the opposite descriptor, (*S*)-2-methylbutanenitrile.

**Exercise 19.5 ★★.**

Write the full [SN1 mechanism](#def-b1-nucleophilic-substitution-sn1) of the hydrolysis of 2-bromo-2-methylpropane, with [curly arrows](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-curly-arrow), including the loss of the proton. Why is the [rate law](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) first order although three species take part?

**Solution of Exercise 19.5.**

1. [Heterolysis](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-cleavage): arrow from the C–Br bond to Br, giving $\ce{(CH3)3C+}$ and $\ce{Br-}$ (slow). 2. A [lone pair](https://one-course.com/books/chemistry/2/en/chapter/3-lewis-structures-resonance-and-vsepr#def-b1-lewis-resonance-vsepr-lewis) of water to the cationic carbon: $\ce{(CH3)3C-OH2+}$. 3. A [lone pair](https://one-course.com/books/chemistry/2/en/chapter/3-lewis-structures-resonance-and-vsepr#def-b1-lewis-resonance-vsepr-lewis) of another water molecule takes a proton of the $\ce{OH2+}$ group, the O–H pair returning to oxygen. Only the first, slow step decides the rate; water, the solvent, is in such excess that its concentration does not change.

**Exercise 19.6 ★★.**

Explain why 1-bromo-2,2-dimethylpropane, a primary bromide, reacts very slowly by SN2 and does not react by SN1.

**Solution of Exercise 19.6.**

The carbon bearing Br is primary, but next to it a tert-butyl group fills the space behind the C–Br bond: the SN2 [transition state](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile) is very crowded. SN1 would need a primary [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates), too unstable.

**Exercise 19.7 ★★.**

(*S*)-2-iodooctane loses its [optical activity](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-optical-activity) when kept in propanone with sodium iodide, although no new compound forms. Explain, and say why the rate of loss of activity is twice the rate of substitution.

**Solution of Exercise 19.7.**

Iodide replaces iodide by SN2, each time with inversion: (*S*) becomes (*R*) until both are equally present, a [racemic mixture](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-enantiomers). Each substitution turns one (*S*) molecule into one (*R*): the rotation falls by two molecules’ worth, so the rotation decays twice as fast as the [substrate](#def-b1-nucleophilic-substitution-substitution) is substituted.

**Exercise 19.8 ★★.**

Rank by rate of SN2 with iodide: 2-bromo-2-methylpropane, bromomethane, 2-bromopropane, 1-bromopropane. Rank the same compounds by rate of SN1 [solvolysis](#def-b1-nucleophilic-substitution-sn1).

**Solution of Exercise 19.8.**

SN2: bromomethane $>$ 1-bromopropane $>$ 2-bromopropane $>$ 2-bromo-2-methylpropane (crowding). SN1: the reverse order ([carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) stability), bromomethane and 1-bromopropane hardly reacting.

**Exercise 19.9 ★★.**

An optically pure secondary bromide undergoes [solvolysis](#def-b1-nucleophilic-substitution-sn1); the product rotates light at $12\,\%$ of the value expected for pure inversion. Give the proportions of the two [enantiomers](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-enantiomers) of the product and interpret.

**Solution of Exercise 19.9.**

$x_{\text{inv}} - x_{\text{ret}} = 0.12$ and $x_{\text{inv}} + x_{\text{ret}} = 1$: $56\,\%$ inverted, $44\,\%$ retained. Mostly SN1 [racemisation](#def-b1-nucleophilic-substitution-sn1), with a slight excess of inversion: the departing bromide still shields its side when water arrives.

**Exercise 19.10 ★★★.**

Derive the general SN1 [rate law](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) with the [steady-state approximation](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-qssa), and show that adding the [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) ion $\ce{Y-}$ at the start slows the reaction (the [common-ion effect](https://one-course.com/books/chemistry/2/en/chapter/11-precipitation-and-solubility#def-b1-precipitation-common-ion) of kinetics). Under which condition is this effect negligible?

**Solution of Exercise 19.10.**

$v = k_1k_2[\mathrm{RY}][\mathrm{Nu}]/(k_{-1}[\mathrm{Y^-}] + k_2[\mathrm{Nu}])$ ([Proposition 19.7](#prop-b1-nucleophilic-substitution-sn1-law)). Adding $\ce{Y-}$ increases the denominator: more [carbocations](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) return to the [substrate](#def-b1-nucleophilic-substitution-substitution) instead of being captured, and the rate falls. Negligible when $k_2[\mathrm{Nu}] \gg k_{-1}[\mathrm{Y^-}]$, as when the [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) is the solvent.

**Exercise 19.11 ★★★.**

Two half-lives for the reaction of bromomethane with hydroxide: with $[\ce{OH-}]_0 = [\ce{CH3Br}]_0 = C_0$, show that $1/[\ce{CH3Br}] = 1/C_0 +
kt$ and give $t_{1/2}$. With $[\ce{OH-}]_0 = 50\,C_0$, show that the decay is exponential and give $t_{1/2}$. Data of the exercise: $k =
2.0 \times 10^{-4}\,\mathrm{L}/(\mathrm{mol}\,\mathrm{s})$, $C_0 = 0.010\,\mathrm{mol}/\mathrm{L}$.

**Solution of Exercise 19.11.**

Equal concentrations: $-\mathrm{d}c/\mathrm{d}t = kc^2$, $1/c = 1/C_0 + kt$, $t_{1/2} = 1/(kC_0) = 5.0 \times 10^{5}\,\mathrm{s}$ (about six days). With a fifty-fold excess of hydroxide, $[\ce{OH-}] \approx 50C_0$ is constant: $c = C_0
e^{-k't}$ with $k' = 50kC_0 = 1.0 \times 10^{-4}\,\mathrm{s}^{-1}$, $t_{1/2} = \ln 2/k' =
6.9 \times 10^{3}\,\mathrm{s}$, about two hours.

**Exercise 19.12 ★★★.**

Explain why an alcohol, $\ce{ROH}$, does not react with sodium bromide, but reacts with concentrated hydrobromic acid to give $\ce{RBr}$. Write the mechanism for a tertiary and for a primary alcohol.

**Solution of Exercise 19.12.**

Bromide would have to expel $\ce{OH-}$, a [strong base](https://one-course.com/books/chemistry/2/en/chapter/10-acidbase-equilibria-and-the-predominant-reaction-method#def-b1-predominant-reaction-strong-acid) and a poor leaving group: no reaction. In concentrated HBr the OH is protonated to $\ce{-OH2+}$, whose [leaving group](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) is water. Tertiary alcohol: water leaves, giving the [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates), captured by $\ce{Br-}$ (SN1). Primary alcohol: $\ce{Br-}$ attacks the carbon of $\ce{R-OH2+}$ from behind as water departs (SN2).

## 19.6 Problem: Solvolysis of tert-Butyl Chloride

**Problem 19.1.**

Weekend problem — order and rate constant from conductivity data, the half-life, the mechanism and its stereochemistry, and the activation energy of the solvolysis

2-Chloro-2-methylpropane (tert-butyl chloride) is dissolved, at a small concentration, in a water–propanone mixture. Its [solvolysis](#def-b1-nucleophilic-substitution-sn1) $\ce{(CH3)3CCl + H2O -> (CH3)3COH + H+ + Cl-}$ produces ions, and the conductivity $\kappa$ of the solution, which is proportional to the amount of ions formed, is recorded. After a very long time it reaches $\kappa_\infty = 2.000\,\mathrm{mS}/\mathrm{cm}$ at both temperatures. Measurements ($\mathrm{mS}/\mathrm{cm}$):

| $t$ (min) | 0 | 5 | 10 | 15 | 20 | 30 | 45 | 60 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- |
| $\kappa$, $25.0\,{}^{\circ}\mathrm{C}$ | 0 | 0.172 | 0.329 | 0.473 | 0.605 | 0.835 | 1.110 | 1.321 |
| $\kappa$, $35.0\,{}^{\circ}\mathrm{C}$ | 0 | 0.507 | 0.885 | 1.168 | 1.379 | 1.654 | 1.856 | 1.940 |

**Part I — The order.**

1. Why is the conductivity proportional to the amount of [substrate](#def-b1-nucleophilic-substitution-substitution) that has reacted?
2. Express $[\mathrm{RCl}]/[\mathrm{RCl}]_0$ in terms of $\kappa$ and $\kappa_\infty$ .
3. Write the integrated law of a first-order reaction and the quantity to plot against $t$ to test it.
4. Compute $\ln(\kappa_\infty - \kappa)$ at $25.0\,{}^{\circ}\mathrm{C}$ for each time.
5. Is the plot a straight line? Conclude on the order.
6. Water is in large excess. What does that hide in the order?

**Part II — [Rate constant](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) and [half-life](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-half-life).**

7. Find the [rate constant](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) at $25.0\,{}^{\circ}\mathrm{C}$ , in $\mathrm{s}^{-1}$ .
8. Compute the [half-life](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-half-life) .
9. At what time is $90\,\%$ of the [substrate](#def-b1-nucleophilic-substitution-substitution) consumed?
10. Find the [rate constant](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) at $35.0\,{}^{\circ}\mathrm{C}$ .
11. By what factor has the rate increased for ten degrees?
12. Check one value of the $35\,{}^{\circ}\mathrm{C}$ series against the law.

**Part III — The mechanism.**

13. Which mechanism do the [substrate](#def-b1-nucleophilic-substitution-substitution) and the solvent suggest?
14. Write it with [curly arrows](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-curly-arrow) .
15. Show that it agrees with the order found.
16. Adding sodium hydroxide at $0.01\,\mathrm{mol}/\mathrm{L}$ hardly changes the [rate of disappearance](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-rate) of the chloride. Why?
17. The same experiment with the [chiral](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-stereocentre) tertiary chloride 3-chloro-3-methylhexane, optically pure, gives an alcohol of almost zero [optical activity](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-optical-activity) . Explain.
18. Draw an [energy profile](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-energy-profile) of the reaction and mark the [rate-determining step](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-rds) .
19. Why would 1-chlorobutane hardly react in the same conditions?

**Part IV — [Activation energy](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-activation-energy).**

20. Write the [Arrhenius law](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#thm-b1-rate-laws-arrhenius) .
21. Express $E_a$ from the two [rate constants](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) .
22. Compute $E_a$ ( $R = 8.314\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ).
23. What does this energy measure, in the mechanism?
24. Predict the [rate constant](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-order) at $45.0\,{}^{\circ}\mathrm{C}$ .
25. State the [activation energy](https://one-course.com/books/chemistry/2/en/chapter/8-chemical-kinetics-rate-laws#def-b1-rate-laws-activation-energy) of the [solvolysis](#def-b1-nucleophilic-substitution-sn1) , to two significant figures.

**Solution of Problem 19.1.**

**1.** Each molecule that reacts gives one $\ce{H+}$ and one $\ce{Cl-}$; the conductivity is the sum of the ionic contributions, proportional to the amount of ions. **2.** $[\mathrm{RCl}]/[\mathrm{RCl}]_0 = (\kappa_\infty -
\kappa)/\kappa_\infty$. **3.** $\ln[\mathrm{RCl}] = \ln[\mathrm{RCl}]_0 - kt$: plot $\ln(\kappa_\infty - \kappa)$ against $t$. **4.** 0.693, 0.603, 0.513, 0.423, 0.333, 0.153, $-0.117$, $-0.387$. **5.** Yes, slope $-0.0180$ per minute throughout: first order. **6.** Any order in water: its concentration is constant (degeneracy of the order). **7.** $k = 0.0180/60 = 3.0 \times 10^{-4}\,\mathrm{s}^{-1}$. **8.** $t_{1/2} = \ln 2/k = 2.3 \times 10^{3}\,\mathrm{s}$, $38.5\,\mathrm{min}$. **9.** $\ln 10/k = 7.7 \times 10^{3}\,\mathrm{s}$, $128\,\mathrm{min}$. **10.** $\ln(\kappa_\infty - \kappa)$: 0.693 at $t = 0$, 0.109 at $10\,\mathrm{min}$, and so on: slope $-0.0584$ per minute, $k = 9.7 \times 10^{-4}\,\mathrm{s}^{-1}$. **11.** $9.7/3.0 = 3.2$. **12.** At $30\,\mathrm{min}$: $2.000(1 - e^{-0.0584 \times 30}) = 1.653$, measured 1.654. **13.** SN1: tertiary [substrate](#def-b1-nucleophilic-substitution-substitution), [protic solvent](https://one-course.com/books/chemistry/2/en/chapter/4-intermolecular-forces-and-solvents#def-b1-intermolecular-forces-solvents-solvent-classes), neutral [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile). **14.** As in [Exercise 19.5](#exo-b1-nucleophilic-substitution-5), with Cl. **15.** The slow ionisation involves only the [substrate](#def-b1-nucleophilic-substitution-substitution): $v =
k[\mathrm{RCl}]$. **16.** Hydroxide could only act in the fast capture step, after the rate-determining ionisation; the rate does not depend on it. **17.** The [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) is planar and [achiral](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#def-b1-stereochemistry-in-depth-stereocentre); water adds on both faces equally: racemic alcohol. **18.** A high first barrier (ionisation) to the [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) in a shallow well, then a low barrier to capture: the first step is rate-determining. **19.** It would form a primary [carbocation](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-intermediates) (too unstable), and water is too weak a [nucleophile](https://one-course.com/books/chemistry/2/en/chapter/18-electronic-effects-and-reactive-intermediates#def-b1-electronic-effects-nucleophile) for a fast SN2. **20.** $k = A\,e^{-E_a/RT}$. **21.** $E_a = R\ln(k_2/k_1)/(1/T_1 - 1/T_2)$. **22.** $E_a = 8.314 \times \ln(9.74/3.00)/(1/298.15 - 1/308.15) =
9.0 \times 10^{4}\,\mathrm{J}/\mathrm{mol}$. **23.** The height of the barrier of the ionisation, the [rate-determining step](https://one-course.com/books/chemistry/2/en/chapter/9-reaction-mechanisms-elementary-steps-and-approximations#def-b1-elementary-steps-rds). **24.** $k = 3.0 \times 10^{-4} \exp\bigl(\frac{90\,000}{8.314}
(\frac{1}{298.15} - \frac{1}{318.15})\bigr) = 2.9 \times 10^{-3}\,\mathrm{s}^{-1}$. **25.** **$E_a = 90\,\mathrm{kJ}/\mathrm{mol}$**.