Chapter 6 · Haloalkanes and Haloarenes

6.2 Nucleophilic Substitution: SN1, SN2 and Stereochemistry

A nucleophile pushes out the halide ion. It can do it in one step, from the back, turning the molecule inside out (SN2), or wait for the halide to leave first and attack a flat carbocation from either side (SN1). Rate laws tell the two apart, and chirality shows what each does to the shape of the molecule.

Many anticancer drugs of the "nitrogen mustard" family carry a $\mathrm{{-}CH_{2}CH_{2}Cl}$ arm. Inside a cell, a nitrogen atom of DNA attacks the carbon next to the chlorine, chloride ion leaves, and the drug becomes chemically bolted to the DNA of a fast-dividing tumour cell, which can no longer copy itself. The same kind of reaction, an electron-rich atom replacing a halogen, is used by chemists every day to turn alkyl halides into alcohols, ethers, amines and nitriles. It is the most important reaction of haloalkanes.

In the 1930s Edward Hughes and Christopher Ingold measured the rates of such reactions and found a puzzle. The rate of reaction of bromomethane with hydroxide ion doubles when either concentration is doubled, but the rate of reaction of 2-bromo-2-methylpropane does not depend on the hydroxide concentration at all. Two different mechanisms are at work. In this section we use the rate laws of Chapter 3 to tell them apart, and then see that they also leave different fingerprints on the three-dimensional shape of the product, a subject called stereochemistry.

Nucleophilic Substitution

📘 Definition 6.5 — Nucleophilic substitution

In a nucleophilic substitution reaction a nucleophile (an electron-pair donor, $\mathrm{Nu^{-}}$ or a neutral molecule with a lone pair) replaces the halogen of a haloalkane, which leaves as a halide ion, the leaving group:

$$\mathrm{Nu^{-}+\overset{\delta+}{R}{-}\overset{\delta-}{X}\rightarrow R{-}Nu+X^{-}}$$

The nucleophile attacks the partially positive carbon of the polar C–X bond. The reaction goes well when the new bond is stronger than the old one and the leaving group is a weak base that is stable on its own ($\mathrm{I^{-}}$, $\mathrm{Br^{-}}$, $\mathrm{Cl^{-}}$).

📐 Rule 6.5 — What a haloalkane R–X becomes
ReagentNucleophileProductClass of product
$\mathrm{NaOH}$ or $\mathrm{KOH}$ (aq)$\mathrm{HO^{-}}$$\mathrm{ROH}$alcohol
$\mathrm{H_{2}O}$$\mathrm{H_{2}O}$$\mathrm{ROH}$alcohol
$\mathrm{NaOR'}$$\mathrm{R'O^{-}}$$\mathrm{ROR'}$ether
$\mathrm{NaI}$$\mathrm{I^{-}}$$\mathrm{RI}$alkyl iodide
$\mathrm{NH_{3}}$$\mathrm{NH_{3}}$$\mathrm{RNH_{2}}$primary amine
$\mathrm{KCN}$$\mathrm{{}^{-}C{\equiv}N}$ (via C)$\mathrm{RCN}$nitrile (alkyl cyanide)
$\mathrm{AgCN}$$\mathrm{Ag{-}C{\equiv}N{:}}$ (via N)$\mathrm{RNC}$isocyanide (isonitrile)
$\mathrm{KNO_{2}}$$\mathrm{O{=}N{-}O^{-}}$ (via O)$\mathrm{R{-}O{-}N{=}O}$alkyl nitrite
$\mathrm{AgNO_{2}}$$\mathrm{Ag{-}O{-}N{=}O}$ (via N)$\mathrm{R{-}NO_{2}}$nitroalkane
$\mathrm{R'COOAg}$$\mathrm{R'COO^{-}}$$\mathrm{R'COOR}$ester
$\mathrm{LiAlH_{4}}$$\mathrm{H^{-}}$$\mathrm{RH}$alkane

With excess ammonia the main product is the primary amine; with excess haloalkane the amine reacts further to secondary and tertiary amines and a quaternary ammonium salt (Chapter 9). Reaction with cyanide adds one carbon to the chain, a useful way of lengthening it.

📘 Definition 6.6 — Ambident nucleophile

An ambident nucleophile has two different atoms that can donate an electron pair. Cyanide can attack through C or N ($\mathrm{{}^{-}C{\equiv}N{:}\leftrightarrow{:}C{=}N^{-}}$); nitrite can attack through O or N.

Which atom attacks? $\mathrm{KCN}$ is ionic and gives free cyanide ions; the carbon end attacks, because a $\mathrm{C{-}C}$ bond is more stable than a $\mathrm{C{-}N}$ bond, giving the nitrile. $\mathrm{AgCN}$ is mainly covalent: carbon is tied up in the $\mathrm{Ag{-}C}$ bond and only the nitrogen lone pair is free, giving the isocyanide. In the same way $\mathrm{KNO_{2}}$ gives alkyl nitrites $\mathrm{R{-}O{-}N{=}O}$ and $\mathrm{AgNO_{2}}$ gives nitroalkanes $\mathrm{R{-}NO_{2}}$.

✏️ Example 1 — Predicting substitution products
Give the product and its IUPAC name when 1-bromopropane is heated with: (a) aqueous $\mathrm{KOH}$; (b) sodium ethoxide in ethanol; (c) ethanolic $\mathrm{KCN}$; (d) $\mathrm{AgCN}$; (e) $\mathrm{AgNO_{2}}$; (f) excess ammonia. Which product has one more carbon in its main chain?
(a) $\mathrm{CH_{3}CH_{2}CH_{2}OH}$, propan-1-ol.
(b) $\mathrm{CH_{3}CH_{2}CH_{2}OCH_{2}CH_{3}}$, 1-ethoxypropane (an ether).
(c) $\mathrm{CH_{3}CH_{2}CH_{2}CN}$, butanenitrile: the cyanide carbon becomes C1 of a four-carbon chain.
(d) $\mathrm{CH_{3}CH_{2}CH_{2}NC}$, 1-isocyanopropane (propyl isocyanide).
(e) $\mathrm{CH_{3}CH_{2}CH_{2}NO_{2}}$, 1-nitropropane.
(f) $\mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}$, propan-1-amine.
Answer Only the nitrile, (c), has a longer carbon chain (butanenitrile). In the isocyanide (d) the extra carbon is joined through nitrogen, so the chain is not lengthened.

The SN2 Mechanism: One Step, from the Back

For chloromethane and hydroxide ion, $\mathrm{CH_{3}Cl+OH^{-}\rightarrow CH_{3}OH+Cl^{-}}$, experiment gives

$$\text{rate}=k[\mathrm{CH_{3}Cl}][\mathrm{OH^{-}}],$$

first order in each reactant and second order overall (Section 3.1). The simplest mechanism that fits is a single elementary step in which both species collide: a bimolecular step, molecularity 2.

📘 Definition 6.7 — The SN2 mechanism

SN2 means substitution, nucleophilic, bimolecular. It is a one-step (concerted) reaction:

  • The nucleophile attacks the carbon from the side opposite to the leaving group (backside attack, at $180^{\circ}$ to the C–X bond), where the $\sigma^{*}$ orbital of C–X is exposed and the electron-rich halogen does not repel it.
  • The new C–Nu bond forms as the C–X bond breaks. Halfway, at the transition state, the carbon is partly bonded to both; the three other groups lie in one plane, like a flattened umbrella. There is no intermediate.
  • The three groups then flip over to the other side, like an umbrella turning inside out in the wind. The product has the inverted configuration: this is called Walden inversion.
  • Rate $=k[\mathrm{RX}][\mathrm{Nu^{-}}]$: second order.
📐 Rule 6.6 — Reactivity in SN2 reactions

Substrate. The nucleophile must reach the back of the carbon. Each alkyl group on that carbon blocks the way (steric hindrance) and crowds the five-coordinate transition state, so

$$\mathrm{CH_{3}X\gt1^{\circ}\gt2^{\circ}\gg3^{\circ}}\qquad(\text{tertiary halides practically do not react by SN2}).$$

Leaving group (for both SN1 and SN2). The weaker the C–X bond and the more stable the halide ion, the faster the reaction: $\mathrm{R{-}I\gt R{-}Br\gt R{-}Cl\gg R{-}F}$.

Nucleophile. A strong nucleophile at high concentration favours SN2: $\mathrm{HO^{-}}$, $\mathrm{RO^{-}}$, $\mathrm{CN^{-}}$, $\mathrm{I^{-}}$ react much faster than neutral $\mathrm{H_{2}O}$.

✏️ Example 2 — Reading an SN2 rate law
For $\mathrm{CH_{3}Br+OH^{-}\rightarrow CH_{3}OH+Br^{-}}$ at a certain temperature, $\text{rate}=k[\mathrm{CH_{3}Br}][\mathrm{OH^{-}}]$ with $k=3.0\times10^{-4}\ \mathrm{L\,mol^{-1}\,s^{-1}}$ (an illustrative value). (a) Find the initial rate when $[\mathrm{CH_{3}Br}]=0.050$ and $[\mathrm{OH^{-}}]=0.10\ \mathrm{mol\,L^{-1}}$. (b) What happens to the rate if $[\mathrm{OH^{-}}]$ is doubled? (c) If both concentrations are halved? (d) In $1.0\ \mathrm{mol\,L^{-1}}$ $\mathrm{NaOH}$ (a large excess) the bromomethane decays by pseudo-first-order kinetics. Find the observed rate constant and the half-life of bromomethane.
(a) $\text{rate}=3.0\times10^{-4}\times0.050\times0.10$ $=1.5\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.
(b) First order in $\mathrm{OH^{-}}$: the rate doubles, to $3.0\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.
(c) $\tfrac12\times\tfrac12=\tfrac14$: the rate falls to $3.75\times10^{-7}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ (about $3.8\times10^{-7}$).
(d) $[\mathrm{OH^{-}}]$ stays almost constant, so $\text{rate}=k'[\mathrm{CH_{3}Br}]$ with $k'=k[\mathrm{OH^{-}}]=3.0\times10^{-4}\ \mathrm{s^{-1}}$ (Section 3.2). $t_{1/2}=\dfrac{0.693}{3.0\times10^{-4}\ \mathrm{s^{-1}}}=2.31\times10^{3}\ \mathrm{s}$, about $39\ \mathrm{min}$.
✏️ Example 3 — Which reacts faster by SN2?
(a) Arrange in order of decreasing SN2 reactivity: $\mathrm{CH_{3}CH_{2}CH_{2}Br}$, $\mathrm{(CH_{3})_{3}CBr}$, $\mathrm{CH_{3}Br}$, $\mathrm{(CH_{3})_{2}CHBr}$. (b) Arrange 1-chlorobutane, 1-bromobutane, 1-iodobutane and 1-fluorobutane in order of decreasing reactivity with $\mathrm{NaOH}$. (c) Which gives a precipitate faster with $\mathrm{NaI}$ in acetone, 1-bromobutane or 2-bromobutane?
(a) Fewer alkyl groups on the carbon bearing Br means less steric hindrance: $\mathrm{CH_{3}Br\gt CH_{3}CH_{2}CH_{2}Br}$ $\mathrm{{}\gt(CH_{3})_{2}CHBr\gt(CH_{3})_{3}CBr}$ (methyl $\gt$ 1° $\gt$ 2° $\gt$ 3°).
(b) Same alkyl group, so the leaving group decides: 1-iodobutane $\gt$ 1-bromobutane $\gt$ 1-chlorobutane $\gt$ 1-fluorobutane (C–I weakest, C–F strongest).
(c) The Finkelstein reaction is SN2 ($\mathrm{I^{-}}$ is a strong nucleophile, acetone does not help ions form). The primary 1-bromobutane is less hindered and precipitates $\mathrm{NaBr}$ faster.

The SN1 Mechanism: a Carbocation in Between

2-Bromo-2-methylpropane reacts with water or hydroxide ion in aqueous solvents, $\mathrm{(CH_{3})_{3}CBr+OH^{-}\rightarrow(CH_{3})_{3}COH+Br^{-}}$, but now

$$\text{rate}=k[\mathrm{(CH_{3})_{3}CBr}],$$

first order, and independent of $[\mathrm{OH^{-}}]$. The nucleophile cannot be involved in the slow step. The reaction must go in two steps, the first of which, the rate-determining step (Section 3.1), involves only the haloalkane.

📘 Definition 6.8 — The SN1 mechanism

SN1 means substitution, nucleophilic, unimolecular.

$$\textbf{Step 1 (slow): }\ \mathrm{(CH_{3})_{3}C{-}Br\rightarrow(CH_{3})_{3}C^{+}+Br^{-}}$$ $$\textbf{Step 2 (fast): }\ \mathrm{(CH_{3})_{3}C^{+}+OH^{-}\rightarrow(CH_{3})_{3}C{-}OH}$$
  • Step 1, the heterolysis of the C–X bond to a carbocation and a halide ion, is slow and decides the rate: $\text{rate}=k[\mathrm{RX}]$.
  • The carbocation is an intermediate: $sp^{2}$, trigonal planar, with an empty $p$ orbital above and below the plane. In step 2 the nucleophile can attack either lobe of that orbital, that is, either face.
  • SN1 reactions are favoured by polar protic solvents (water, alcohols, methanoic acid), which surround and stabilise the ions formed in step 1, and by weak or dilute nucleophiles; water or the alcohol solvent itself often acts as the nucleophile (solvolysis).
📐 Rule 6.7 — Reactivity in SN1 reactions

The rate depends on how easily the carbocation forms, that is, on its stability. Alkyl groups stabilise a carbocation by their $+I$ effect and by hyperconjugation (Class 11 Chemistry, Organic Chemistry — Some Basic Principles and Techniques), so

$$\mathrm{3^{\circ}\gt2^{\circ}\gt1^{\circ}\gt CH_{3}X}\qquad(\text{the reverse of the SN2 order}).$$

Allylic and benzylic halides react very readily by SN1, because their carbocations are stabilised by resonance: the positive charge is spread over two carbons in the allyl cation, $\mathrm{CH_{2}{=}CH{-}\overset{+}{C}H_{2}\leftrightarrow\overset{+}{C}H_{2}{-}CH{=}CH_{2}}$, and over three ring carbons as well as the $\mathrm{CH_{2}}$ in the benzyl cation $\mathrm{C_{6}H_{5}\overset{+}{C}H_{2}}$. Each extra phenyl group on the cationic carbon adds more resonance: $\mathrm{(C_{6}H_{5})_{3}C^{+}\gt(C_{6}H_{5})_{2}CH^{+}\gt C_{6}H_{5}CH_{2}^{+}}$. (Primary allylic and benzylic halides also react fast by SN2.)

The leaving-group order is the same as for SN2: $\mathrm{R{-}I\gt R{-}Br\gt R{-}Cl\gg R{-}F}$.

✏️ Example 4 — Ranking by SN1 reactivity
Arrange each set in order of decreasing reactivity in SN1 reactions, giving a reason. (a) $\mathrm{CH_{3}Br}$, $\mathrm{(CH_{3})_{3}CBr}$, $\mathrm{CH_{3}CH_{2}Br}$, $\mathrm{(CH_{3})_{2}CHBr}$. (b) $\mathrm{C_{6}H_{5}CH_{2}Cl}$, $\mathrm{(C_{6}H_{5})_{2}CHCl}$, $\mathrm{C_{6}H_{5}CH(CH_{3})Cl}$. (c) 1-chlorobutane and 3-chlorobut-1-ene.
(a) Carbocation stability decides: $\mathrm{(CH_{3})_{3}CBr\gt(CH_{3})_{2}CHBr}$ $\mathrm{{}\gt CH_{3}CH_{2}Br\gt CH_{3}Br}$ (3° $\gt$ 2° $\gt$ 1° $\gt$ methyl), exactly the reverse of Example 3(a).
(b) All three are benzylic. The diphenylmethyl cation is stabilised by two rings; the cation from $\mathrm{C_{6}H_{5}CH(CH_{3})Cl}$ is secondary benzylic (one ring plus a methyl group); the benzyl cation is primary benzylic. Order: $\mathrm{(C_{6}H_{5})_{2}CHCl\gt C_{6}H_{5}CH(CH_{3})Cl}$ $\mathrm{{}\gt C_{6}H_{5}CH_{2}Cl}$.
(c) 3-Chlorobut-1-ene, $\mathrm{CH_{2}{=}CHCHClCH_{3}}$, is a secondary allylic halide: its cation is both secondary and resonance-stabilised. 1-Chlorobutane would give an unstable primary cation and hardly reacts by SN1. 3-Chlorobut-1-ene is far faster.
Interactive 1 — SN2 or SN1? Watch the Mechanism, Read the Rate Law

Choose a mechanism and press Play. In SN2 the nucleophile hits the back of the carbon of 2-bromobutane, the three groups flatten at the single transition state and flip over (Walden inversion). In SN1 bromide leaves first from a tertiary bromide, the flat carbocation waits, and the nucleophile arrives from the left or from the right at random; the tally shows the two mirror-image products building up equally. The energy profile on the right follows the reaction. Move the concentration sliders and watch which rate law responds. The rate constants are illustrative values.

0.100.10
rate law—
rate (mol L⁻¹ s⁻¹)—
if [OH⁻] is doubled—
stage—
stereochemistry—
🧭 Problem-solving strategy — SN1 or SN2?
Factorfavours SN2favours SN1
substrate$\mathrm{CH_{3}X}$, 1° (also 2°)3°, allylic, benzylic (also 2°)
nucleophilestrong, concentrated ($\mathrm{HO^{-}}$, $\mathrm{RO^{-}}$, $\mathrm{CN^{-}}$, $\mathrm{I^{-}}$)weak or dilute ($\mathrm{H_{2}O}$, $\mathrm{ROH}$)
solventless polar or aprotic (acetone)polar protic (water, alcohols)
rate law$k[\mathrm{RX}][\mathrm{Nu^{-}}]$$k[\mathrm{RX}]$
stereochemistryinversionracemisation (mostly)

Look at the substrate first: methyl and primary halides go SN2, tertiary halides go SN1; secondary halides can go either way, and the nucleophile and solvent decide.

Stereochemistry: Chirality, Enantiomers and Optical Activity

Ordinary light vibrates in every plane containing its direction of travel. After passing through a polarising filter (a Polaroid sheet or a Nicol prism) it vibrates in one plane only: it is plane-polarised light. Some substances, in solution or as pure liquids, rotate that plane as the light passes through them. They are optically active. The angle of rotation is measured in a polarimeter. A substance that rotates the plane clockwise (as seen by an observer looking towards the light) is dextrorotatory, written $d$ or $(+)$; one that rotates it anticlockwise is laevorotatory, $l$ or $(-)$.

📘 Definition 6.9 — Chirality and the stereocentre

An object (or molecule) is chiral (Greek cheir, hand) if it cannot be superimposed on its mirror image, as a left hand cannot be fitted exactly onto a right hand. An object that can be superimposed on its mirror image is achiral.

The commonest cause of chirality is a carbon atom bonded to four different groups, called an asymmetric carbon or stereocentre (often marked with an asterisk, $\mathrm{C^{*}}$). In 2-bromobutane, $\mathrm{CH_{3}{-}\overset{*}{C}HBr{-}CH_{2}CH_{3}}$, C2 carries $\mathrm{H}$, $\mathrm{Br}$, $\mathrm{CH_{3}}$ and $\mathrm{C_{2}H_{5}}$. If any two groups on a carbon are the same, as in 2-bromopropane, the molecule has a plane of symmetry and is achiral.

📘 Definition 6.10 — Enantiomers, racemic mixture and racemisation
  • Enantiomers are stereoisomers that are non-superimposable mirror images of each other. They have identical melting points, boiling points, densities, solubilities and reactions with achiral reagents, but rotate plane-polarised light by equal angles in opposite directions. They can differ greatly in their effects in the body, whose receptors and enzymes are themselves chiral.
  • A racemic mixture (racemate) contains equal amounts of two enantiomers. The rotations cancel, so it is optically inactive; it is written $(\pm)$ or $dl$, e.g. $(\pm)$-butan-2-ol.
  • Racemisation is the conversion of one enantiomer into a racemic mixture.

If a sample contains a fraction $f$ of the $(+)$ enantiomer and $1-f$ of the $(-)$, its observed rotation is $\alpha=(2f-1)\,\alpha_{\text{pure}}$, where $\alpha_{\text{pure}}$ is the rotation of the pure $(+)$ enantiomer measured under the same conditions.

📐 Rule 6.8 — Stereochemistry of nucleophilic substitution

The spatial arrangement of the groups around a stereocentre is its configuration. When a reaction at a stereocentre gives a product whose groups keep the same relative arrangement, the reaction proceeds with retention of configuration; when the arrangement becomes the mirror-image one, with inversion of configuration.

  • SN2: backside attack always gives inversion. An optically active haloalkane gives a single enantiomer of the product, of inverted configuration.
  • SN1: the planar carbocation is attacked equally from both faces, so an optically active haloalkane gives a racemic (or nearly racemic) product: half retention, half inversion. In practice the departing halide ion briefly shields one face, so slightly more inversion than retention is often found (partial racemisation).
  • The sign of rotation is not a guide to configuration: the product is a different compound, and inversion of configuration may leave the sign of rotation unchanged or change it.
✏️ Example 5 — Finding stereocentres
Which of these molecules are chiral? Mark each stereocentre. (a) butan-2-ol; (b) 1-chlorobutane; (c) 2-chloropropane; (d) 3-bromopentane; (e) 2-bromopentane; (f) 1-bromo-2-methylbutane; (g) 2-bromo-2-methylbutane; (h) 2-chloro-3-methylpentane.
Method For every $sp^{3}$ carbon, list its four groups. A carbon with two hydrogens (or two methyls, or two identical chains) cannot be a stereocentre.
Chiral (a) $\mathrm{CH_{3}\overset{*}{C}H(OH)CH_{2}CH_{3}}$: H, OH, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$. (e) $\mathrm{CH_{3}\overset{*}{C}HBrCH_{2}CH_{2}CH_{3}}$: H, Br, $\mathrm{CH_{3}}$, $\mathrm{C_{3}H_{7}}$. (f) $\mathrm{BrCH_{2}\overset{*}{C}H(CH_{3})CH_{2}CH_{3}}$: C2 carries H, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{CH_{2}Br}$. (h) $\mathrm{CH_{3}\overset{*}{C}HCl\overset{*}{C}H(CH_{3})CH_{2}CH_{3}}$ has two stereocentres, C2 and C3.
Achiral (b) no carbon has four different groups; (c) C2 carries two $\mathrm{CH_{3}}$; (d) C3 carries two $\mathrm{C_{2}H_{5}}$; (g) C2 carries two $\mathrm{CH_{3}}$.
Answer Chiral: (a), (e), (f), (h). Achiral: (b), (c), (d), (g).
✏️ Example 6 — What the polarimeter shows
A pure enantiomer of butan-2-ol rotates plane-polarised light by $13.5^{\circ}$ under the conditions used (a literature value, rounded). (a) What rotation does its enantiomer show, and a racemic mixture? (b) An optically pure sample of one enantiomer of 2-bromobutane is converted into butan-2-ol by $\mathrm{OH^{-}}$ under SN2 conditions. Is the product optically active? (c) The same bromide is heated with water under conditions where SN1 dominates; the alcohol shows almost no rotation. Explain. (d) A sample of butan-2-ol shows $+5.4^{\circ}$, the pure $(+)$ enantiomer showing $+13.5^{\circ}$. Find the percentage of each enantiomer.
(a) The enantiomer rotates by the same angle in the opposite direction ($-13.5^{\circ}$ if the first is $+13.5^{\circ}$). The racemic mixture shows $0^{\circ}$.
(b) Yes. SN2 inverts every molecule, so a single enantiomer of butan-2-ol (of inverted configuration) is formed, with the full rotation of $13.5^{\circ}$. Whether it is $+$ or $-$ cannot be predicted from the sign of the bromide.
(c) Water is a weak nucleophile and a polar protic solvent, so the reaction goes through the planar 2° carbocation, which is attacked from both faces almost equally: the product is (nearly) racemic.
(d) $\alpha=(2f-1)\alpha_{\text{pure}}$: $5.4=(2f-1)\times13.5$, so $2f-1=0.40$ and $f=0.70$. The sample is $70\%$ $(+)$ and $30\%$ $(-)$ butan-2-ol.
Check $0.70\times13.5+0.30\times(-13.5)$ $=9.45-4.05=5.40^{\circ}$ ✓.
Interactive 2 — Chirality Explorer: Mirror Images and the Polarimeter

Pick a molecule. The lab finds every $sp^{3}$ carbon with four different groups (red asterisk) and draws the selected stereocentre in 3-D with its mirror image. Press Play (or drag turn) to rotate the mirror image about its vertical axis and try to lay it on the original: for a chiral centre two groups always end up swapped; for an achiral carbon the two match. Then mix enantiomers with the composition slider and read the rotation a polarimeter would show.

100%
stereocentres—
groups on the drawn carbon—
superimposable on mirror image?—
observed rotation—
⚠️ Common mistakes with SN1, SN2 and stereochemistry

1. Reversed reactivity orders. SN2: $\mathrm{CH_{3}\gt1^{\circ}\gt2^{\circ}\gt3^{\circ}}$ (steric). SN1: $\mathrm{3^{\circ}\gt2^{\circ}\gt1^{\circ}\gt CH_{3}}$ (carbocation stability). Learn the reason, and the order follows.

2. "Unimolecular" does not mean only one molecule reacts. The 1 in SN1 refers to the molecularity of the slow step; the nucleophile still takes part, but only in the fast second step.

3. $d/l$ and $(+)/(-)$ describe rotation, not shape. They are measured, not predicted from a drawing.

4. Every SN2 is an inversion, but you only see it when the carbon attacked is a stereocentre.

5. KCN vs AgCN. KCN gives nitriles R–CN (via C); AgCN gives isocyanides R–NC (via N). Likewise $\mathrm{KNO_{2}}$ gives nitrites R–ONO and $\mathrm{AgNO_{2}}$ nitroalkanes R–NO₂.

🚀 Beyond the board syllabus (JEE Main · NEET) — naming configurations R and S

In JEE Main and NEET 2026 (stereoisomerism and its nomenclature); not in the CBSE syllabus, and JEE Advanced explicitly excludes R,S configurations.

The Cahn–Ingold–Prelog rules label a stereocentre. Rank the four groups by the atomic number of the atom directly attached (higher first; if tied, compare the next atoms out). Point the lowest-ranked group away from you. If the path $1\rightarrow2\rightarrow3$ runs clockwise the centre is R (rectus); anticlockwise, S (sinister). For 2-bromobutane: $\mathrm{Br\gt C_{2}H_{5}\gt CH_{3}\gt H}$. In SN2 hydrolysis $\mathrm{OH}$ takes the place of Br and is also the top-ranked group, so the geometric inversion turns (R)-2-bromobutane into (S)-butan-2-ol. R/S describes the arrangement in space and has no fixed connection with the sign of rotation, which can only be measured.

Exercises

I
Substitution products.
  • 1 Write the product formed when bromoethane reacts with: (a) aqueous $\mathrm{NaOH}$; (b) sodium methoxide; (c) $\mathrm{KCN}$; (d) $\mathrm{AgCN}$; (e) $\mathrm{KNO_{2}}$; (f) $\mathrm{AgNO_{2}}$; (g) silver ethanoate; (h) $\mathrm{LiAlH_{4}}$. Name each product.
  • 2 Explain why $\mathrm{KCN}$ gives mainly nitriles but $\mathrm{AgCN}$ gives mainly isocyanides with haloalkanes.
  • 3 How would you prepare propan-1-amine from 1-bromopropane? Why is a large excess of ammonia used?
  • 4 Write the equation for 2-bromopropane with sodium iodide in acetone, and name the product.
  • 5 What is an ambident nucleophile? Name the two donor atoms of the nitrite ion and the product type each gives.
  • 6 Bromomethane reacts much faster with $\mathrm{OH^{-}}$ than with water, although both give methanol. Why?
1 (a) $\mathrm{C_{2}H_{5}OH}$ ethanol; (b) $\mathrm{C_{2}H_{5}OCH_{3}}$ methoxyethane; (c) $\mathrm{C_{2}H_{5}CN}$ propanenitrile; (d) $\mathrm{C_{2}H_{5}NC}$ isocyanoethane (ethyl isocyanide); (e) $\mathrm{C_{2}H_{5}{-}O{-}N{=}O}$ ethyl nitrite; (f) $\mathrm{C_{2}H_{5}NO_{2}}$ nitroethane; (g) $\mathrm{CH_{3}COOC_{2}H_{5}}$ ethyl ethanoate; (h) $\mathrm{C_{2}H_{6}}$ ethane.
2 $\mathrm{KCN}$ is ionic: free $\mathrm{CN^{-}}$ attacks through carbon, forming the more stable C–C bond. $\mathrm{AgCN}$ is covalent (Ag–C bond); only the nitrogen lone pair is available, so N attacks and gives R–NC.
3 Heat 1-bromopropane with excess ethanolic ammonia in a sealed tube: $\mathrm{CH_{3}CH_{2}CH_{2}Br+NH_{3}}$ $\mathrm{\rightarrow CH_{3}CH_{2}CH_{2}NH_{2}+HBr}$ (the HBr is taken up by more ammonia). Excess ammonia makes it more likely that the haloalkane meets ammonia rather than the amine already formed, which would give secondary and tertiary amines.
4 $\mathrm{(CH_{3})_{2}CHBr+NaI\rightarrow(CH_{3})_{2}CHI+NaBr\downarrow}$; 2-iodopropane.
5 A nucleophile with two different atoms that can donate an electron pair. Nitrite, $\mathrm{O{=}N{-}O^{-}}$: attack through O gives an alkyl nitrite R–O–N=O (with $\mathrm{KNO_{2}}$); through N gives a nitroalkane R–NO₂ (with $\mathrm{AgNO_{2}}$).
6 Methyl halides react by SN2, whose rate is proportional to the nucleophile's strength and concentration. The negative $\mathrm{OH^{-}}$ is a much stronger nucleophile than neutral water.
II
Mechanisms and rate laws. Rate constants here are illustrative values.
  • 1 Write the rate law for $\mathrm{CH_{3}Cl+OH^{-}\rightarrow CH_{3}OH+Cl^{-}}$ and state its order and molecularity. What happens to the rate if $[\mathrm{OH^{-}}]$ is doubled?
  • 2 The hydrolysis of 2-chloro-2-methylpropane follows $\text{rate}=k[\mathrm{(CH_{3})_{3}CCl}]$. By what factor does the rate change if (a) $[\mathrm{OH^{-}}]$ is tripled; (b) $[\mathrm{(CH_{3})_{3}CCl}]$ is tripled?
  • 3 An SN2 reaction has $k=2.0\times10^{-3}\ \mathrm{L\,mol^{-1}\,s^{-1}}$. Find the rate when $[\mathrm{RX}]=0.040$ and $[\mathrm{Nu^{-}}]=0.050\ \mathrm{mol\,L^{-1}}$.
  • 4 An SN1 hydrolysis has $k=1.2\times10^{-4}\ \mathrm{s^{-1}}$. Find the half-life of the haloalkane, and the fraction left after $3$ half-lives.
  • 5 Sketch and label the energy profiles of an SN2 and an SN1 reaction. How many transition states and intermediates does each have, and which step of SN1 is rate-determining?
  • 6 Why do tertiary halides not react by SN2, and why do primary halides (other than allylic and benzylic ones) not react by SN1?
  • 7 Why does an SN1 reaction go faster in water than in a non-polar solvent?
  • 8 For $\mathrm{RX+OH^{-}}$: with $[\mathrm{RX}]=0.10$, $[\mathrm{OH^{-}}]=0.10$ the initial rate is $2.0\times10^{-5}$; with $[\mathrm{RX}]=0.20$, $[\mathrm{OH^{-}}]=0.10$ it is $4.0\times10^{-5}$; with $[\mathrm{RX}]=0.10$, $[\mathrm{OH^{-}}]=0.20$ it is $2.0\times10^{-5}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. Find the rate law and $k$, and say which mechanism is operating.
1 $\text{rate}=k[\mathrm{CH_{3}Cl}][\mathrm{OH^{-}}]$: second order overall, a bimolecular elementary step (SN2). Doubling $[\mathrm{OH^{-}}]$ doubles the rate.
2 (a) No change: $\mathrm{OH^{-}}$ is not in the rate law. (b) The rate triples.
3 $\text{rate}=2.0\times10^{-3}\times0.040\times0.050$ $=4.0\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.
4 $t_{1/2}=\dfrac{0.693}{1.2\times10^{-4}}=5.78\times10^{3}\ \mathrm{s}$ (about $96\ \mathrm{min}$). After 3 half-lives, $\left(\tfrac12\right)^{3}=\tfrac18$ is left.
5 SN2: one hump (one transition state), no intermediate. SN1: two humps (two transition states) with a dip between them, the carbocation intermediate; the first hump (ionisation) is the higher one, so step 1 is rate-determining.
6 In a tertiary halide three alkyl groups block the backside approach, so SN2 is impossible. A primary carbocation is too unstable to form, so primary halides cannot ionise (allylic and benzylic cations are the exception, being resonance-stabilised).
7 Water is polar and protic: it solvates the carbocation through its oxygen lone pairs and the halide ion through hydrogen bonds, lowering the energy of the ions and of the transition state leading to them.
8 Doubling $[\mathrm{RX}]$ doubles the rate (first order in RX); doubling $[\mathrm{OH^{-}}]$ has no effect (zero order). $\text{rate}=k[\mathrm{RX}]$, $k=\dfrac{2.0\times10^{-5}}{0.10}=2.0\times10^{-4}\ \mathrm{s^{-1}}$. SN1.
III
Reactivity and choice of mechanism.
  • 1 Arrange 1-bromopentane, 2-bromopentane and 2-bromo-2-methylbutane in order of increasing SN2 reactivity.
  • 2 Arrange the same three compounds in order of increasing SN1 reactivity.
  • 3 Which reacts faster with $\mathrm{NaOH}$, 1-chlorobutane or 1-iodobutane? Why?
  • 4 Arrange $\mathrm{C_{6}H_{5}CH_{2}Br}$, $\mathrm{C_{6}H_{5}CH(CH_{3})Br}$ and $\mathrm{C_{6}H_{5}C(CH_{3})_{2}Br}$ in order of increasing SN1 reactivity.
  • 5 In each pair, which reacts faster by SN1? (a) $\mathrm{(CH_{3})_{3}CCl}$ or $\mathrm{(CH_{3})_{3}CBr}$; (b) $\mathrm{CH_{2}{=}CHCH_{2}Cl}$ or $\mathrm{CH_{3}CH_{2}CH_{2}Cl}$.
  • 6 Which reacts faster by SN2 with $\mathrm{I^{-}}$ in acetone: 1-chlorobutane or 2-chlorobutane?
  • 7 Predict the main mechanism: (a) bromoethane with concentrated aqueous $\mathrm{NaOH}$; (b) 2-bromo-2-methylpropane warmed in aqueous ethanol; (c) iodomethane with $\mathrm{NaCN}$ in acetone.
1 2-Bromo-2-methylbutane (3°) $\lt$ 2-bromopentane (2°) $\lt$ 1-bromopentane (1°).
2 1-Bromopentane $\lt$ 2-bromopentane $\lt$ 2-bromo-2-methylbutane.
3 1-Iodobutane: the C–I bond is weaker ($234$ against $351\ \mathrm{kJ\,mol^{-1}}$ for C–Cl in $\mathrm{CH_{3}X}$) and $\mathrm{I^{-}}$ is a better leaving group.
4 $\mathrm{C_{6}H_{5}CH_{2}Br\lt C_{6}H_{5}CH(CH_{3})Br}$ $\mathrm{{}\lt C_{6}H_{5}C(CH_{3})_{2}Br}$ (1°, 2°, 3° benzylic cations).
5 (a) $\mathrm{(CH_{3})_{3}CBr}$: same cation, better leaving group. (b) $\mathrm{CH_{2}{=}CHCH_{2}Cl}$: its allyl cation is resonance-stabilised.
6 1-Chlorobutane (primary, less hindered).
7 (a) SN2 (primary, strong concentrated nucleophile); (b) SN1 (tertiary, weak nucleophile, polar protic solvent); (c) SN2 (methyl, strong nucleophile, acetone).
IV
Stereochemistry.
  • 1 Define: chiral molecule, enantiomers, racemic mixture. Why is a racemic mixture optically inactive?
  • 2 Mark the stereocentres (if any) in: (a) 2-chlorobutane; (b) 1-chloro-2-methylpropane; (c) 3-chloro-3-methylhexane; (d) 2,3-dichlorobutane; (e) 1,2-dichloropropane.
  • 3 How many of the structural isomers of $\mathrm{C_{4}H_{9}Cl}$ are chiral?
  • 4 Name the chiral structural isomers of $\mathrm{C_{5}H_{11}Cl}$.
  • 5 A pure enantiomer of a compound gives a rotation of $+40.0^{\circ}$ in a polarimeter. A sample of the same compound, measured in the same way, gives $+10.0^{\circ}$. Find the percentage of each enantiomer.
  • 6 Optically active 2-bromobutane slowly loses its optical activity when it is kept dissolved in acetone with sodium bromide. Explain.
  • 7 Distinguish between retention and inversion of configuration, and state which of SN1 and SN2 gives each.
  • 8 Why is the alcohol obtained by hydrolysis of 2-bromo-2-methylbutane optically inactive, whatever the mechanism?
1 Chiral: not superimposable on its mirror image. Enantiomers: a pair of non-superimposable mirror-image stereoisomers. Racemic mixture: equal amounts of two enantiomers; the equal and opposite rotations cancel.
2 (a) one, C2; (b) none (C2 carries two $\mathrm{CH_{3}}$); (c) one, C3 (Cl, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{C_{3}H_{7}}$); (d) two, C2 and C3; (e) one, C2 (H, Cl, $\mathrm{CH_{3}}$, $\mathrm{CH_{2}Cl}$).
3 One: 2-chlorobutane (the others are 1-chlorobutane, 1-chloro-2-methylpropane, 2-chloro-2-methylpropane).
4 Three: 2-chloropentane, 1-chloro-2-methylbutane and 2-chloro-3-methylbutane.
5 $10.0=(2f-1)\times40.0$ gives $f=0.625$: $62.5\%$ of the $(+)$ and $37.5\%$ of the $(-)$ enantiomer.
6 $\mathrm{Br^{-}}$ is a good nucleophile in acetone and replaces Br by SN2. Each substitution inverts the configuration of that molecule, so molecules of the original enantiomer are steadily converted into the other one until the two are present in equal amounts: racemisation.
7 Retention: the product has the same spatial arrangement as the reactant. Inversion: the mirror-image arrangement. SN2 gives inversion; SN1 gives both (racemisation).
8 2-Methylbutan-2-ol, $\mathrm{(CH_{3})_{2}C(OH)CH_{2}CH_{3}}$, has two $\mathrm{CH_{3}}$ groups on C2 and no stereocentre; it is achiral (and so was the bromide).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Describe the SN2 mechanism for the reaction of chloromethane with hydroxide ion. Why does it lead to inversion of configuration, and why is it called bimolecular?

Mechanism $\mathrm{HO^{-}}$ attacks the carbon from the side opposite Cl; in one step the C–O bond forms as the C–Cl bond breaks, through a transition state $\mathrm{[HO\cdots CH_{3}\cdots Cl]^{-}}$ in which the three H atoms lie in a plane. (1 mark)
Inversion Because the nucleophile enters from the back, the three H atoms are pushed over to the other side, like an umbrella turned inside out (Walden inversion). (1 mark)
Bimolecular The single (rate-determining) step involves two species, so $\text{rate}=k[\mathrm{CH_{3}Cl}][\mathrm{OH^{-}}]$. (1 mark)

E2. NEET Which of the following reacts fastest by the SN1 mechanism? (A) $\mathrm{CH_{3}CH_{2}Br}$  (B) $\mathrm{(CH_{3})_{2}CHBr}$  (C) $\mathrm{(C_{6}H_{5})_{2}CHBr}$  (D) $\mathrm{C_{6}H_{5}CH_{2}Br}$

Idea The most stable carbocation forms fastest. $\mathrm{(C_{6}H_{5})_{2}CH^{+}}$ has its charge delocalised over two benzene rings.
Answer (C).

E3. JEE Main 1-Bromopropane is heated separately with (i) alcoholic $\mathrm{KCN}$ and (ii) $\mathrm{AgCN}$. The major organic products are, respectively: (A) $\mathrm{C_{3}H_{7}CN}$, $\mathrm{C_{3}H_{7}NC}$  (B) $\mathrm{C_{3}H_{7}NC}$, $\mathrm{C_{3}H_{7}CN}$  (C) $\mathrm{C_{3}H_{7}CN}$ in both  (D) $\mathrm{C_{3}H_{7}OH}$ in both

Idea Ionic $\mathrm{KCN}$: attack through C (nitrile). Covalent $\mathrm{AgCN}$: attack through N (isocyanide).
Answer (A): butanenitrile and 1-isocyanopropane.

E4. JEE Main Numerical value. The number of stereocentres in 3-bromo-2-chloro-4-methylhexane is ______.

Structure $\mathrm{CH_{3}{-}CHCl{-}CHBr{-}CH(CH_{3}){-}CH_{2}CH_{3}}$.
Check each carbon C2: H, Cl, $\mathrm{CH_{3}}$, rest of chain ✓. C3: H, Br, $\mathrm{CHClCH_{3}}$, $\mathrm{CH(CH_{3})C_{2}H_{5}}$ ✓. C4: H, $\mathrm{CH_{3}}$, $\mathrm{C_{2}H_{5}}$, $\mathrm{CHBr}$… ✓. C1, C5, C6 and the methyl carry two or three H.
Answer $\mathbf{3}$.

E5. JEE Advanced One or more correct. An optically pure sample of 2-bromobutane is treated with a high concentration of $\mathrm{NaOH}$ in aqueous acetone, and the reaction follows second-order kinetics. Which statements are correct? (A) Doubling $[\mathrm{OH^{-}}]$ doubles the initial rate. (B) The butan-2-ol formed has the inverted configuration. (C) The butan-2-ol formed is racemic. (D) The reaction passes through a planar carbocation intermediate.

Mechanism Second-order kinetics with a strong, concentrated nucleophile on a secondary substrate: SN2, $\text{rate}=k[\mathrm{RBr}][\mathrm{OH^{-}}]$.
(A), (B) True: first order in $\mathrm{OH^{-}}$; backside attack inverts the configuration, giving a single enantiomer.
(C), (D) False: racemisation and a carbocation belong to SN1.
Answer (A) and (B).

E6. Board · Case study A student follows two hydrolyses at $298\ \mathrm{K}$. For bromoethane, doubling either $[\mathrm{C_{2}H_{5}Br}]$ or $[\mathrm{OH^{-}}]$ doubles the initial rate. For 2-bromo-2-methylpropane, doubling $[\mathrm{(CH_{3})_{3}CBr}]$ doubles the rate, but doubling $[\mathrm{OH^{-}}]$ has no effect. (a) Write the rate law for each reaction and give its overall order. (2 marks) (b) Name the mechanism of each. (1 mark) (c) If an optically active tertiary bromide reacted like 2-bromo-2-methylpropane, what would you expect for the optical activity of the product? (1 mark)

(a) Bromoethane: $\text{rate}=k[\mathrm{C_{2}H_{5}Br}][\mathrm{OH^{-}}]$, second order. 2-Bromo-2-methylpropane: $\text{rate}=k[\mathrm{(CH_{3})_{3}CBr}]$, first order.
(b) Bromoethane: SN2. 2-Bromo-2-methylpropane: SN1.
(c) SN1 goes through a planar carbocation attacked from both faces, so the product would be a racemic mixture: optically inactive (or nearly so).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How pure is a single-enantiomer medicine, and how could a quality-control laboratory check it with a polarimeter?

Why it matters. The receptors and enzymes of the body are chiral, so the two enantiomers of a drug can act very differently. The anti-inflammatory drug naproxen is sold as a single enantiomer; the painkiller ibuprofen is usually sold as the racemate, although one enantiomer does most of the work. Regulators around the world require manufacturers to state and control the enantiomeric composition of chiral medicines.

What to do.

  1. (Illustrative data, constructed for this exercise.) A pure sample of the active enantiomer of a drug gives $+12.0^{\circ}$ in a polarimeter. Three production batches, measured in the same way, give $+12.0^{\circ}$, $+6.0^{\circ}$ and $+0.6^{\circ}$. Use $\alpha=(2f-1)\alpha_{\text{pure}}$ to find the percentage of each enantiomer in each batch.
  2. The "enantiomeric excess" is $\mathrm{ee}=|2f-1|\times100\%$. Find it for each batch.
  3. One batch was made by an SN1 step on a chiral intermediate. Which batch is it most likely to be, and why?
  4. Find out why the drug thalidomide could not have been made safe simply by selling one enantiomer.

Tools: Interactive 2 (composition slider); a calculator; reliable web sources on chiral drugs.

Hand in: A table of composition and ee for the three batches, your answer to step 3, and a short paragraph on step 4.

Going further. Polarimetry cannot detect a small amount of a different chiral impurity. Find out how chiral chromatography separates enantiomers and why it is now the standard method.

Instructor note (the rotations are illustrative data): $f=\tfrac12\left(1+\tfrac{\alpha}{\alpha_{\text{pure}}}\right)$. Batches: $100\%$ / $0\%$ (ee $100\%$); $75\%$ / $25\%$ (ee $50\%$); $52.5\%$ / $47.5\%$ (ee $5\%$). The nearly racemic third batch fits an SN1 step (planar carbocation attacked from both faces). Thalidomide's enantiomers interconvert in the body, so even a pure enantiomer gives a mixture after it is taken.

🧪 Self-check quiz

0/5 correct

Q1. The rate law for an SN2 reaction of R–X with $\mathrm{Nu^{-}}$ is

Q2. Which bromide reacts fastest by SN1?

Q3. Iodoethane heated with $\mathrm{AgCN}$ gives mainly

Q4. Which molecule is chiral?

Q5. An optically active haloalkane undergoes hydrolysis by a pure SN1 mechanism. The product is

← 6.1 Classification, Nomenclature and Preparation6.3 Elimination, Reactions of Haloarenes and Polyhalogen Compounds →