Mechanisms (SN1, SN2, stereochemistry) · Haloalkanes and Haloarenes

CHEReactions of Haloalkanes

Mechanisms (S_N1, S_N2, stereochemistry)

Nucleophilic substitution runs by two limiting mechanisms — $S_N1$ (two steps, via a carbocation) and $S_N2$ (one concerted step). Which one a substrate follows controls its rate law, its stereochemistry, and how it responds to solvent and structure.

S_N2

rate law (second order)

\text{rate} = k\,[R\text{-}X]\,[Nu^-]

single concerted step; rate depends on both substrate and nucleophile

S_N1

rate law (first order)

\text{rate} = k\,[R\text{-}X]

slow step is carbocation formation; rate is independent of

[Nu^-]

Structural reactivity orders

S_N2:\; CH_3X > 1^\circ > 2^\circ > 3^\circ \qquad S_N1:\; 3^\circ > 2^\circ > 1^\circ > CH_3X

S_N2

set by steric crowding;

S_N1

set by carbocation stability

$S_N2$ : single step, $\text{rate} = k[R\text{-}X][Nu^-]$ (second order); backside attack inverts configuration (Walden inversion).
$S_N1$ : two steps through a planar carbocation, $\text{rate} = k[R\text{-}X]$ (first order); gives racemisation (with a slight excess of inversion).
$S_N2$ reactivity: $CH_3X > 1^\circ > 2^\circ > 3^\circ$ (steric crowding); $S_N1$ reactivity: $3^\circ > 2^\circ > 1^\circ > CH_3X$ (carbocation stability).
Polar protic solvents favour $S_N1$ (they stabilise the carbocation); polar aprotic solvents favour $S_N2$ .
Allylic and benzylic halides are unusually reactive — their carbocations are resonance-stabilised (fast $S_N1$ ).
Quick read: strong/concentrated nucleophile + aprotic solvent → $S_N2$ ; weak nucleophile + protic solvent + stable cation → $S_N1$ .
$S_N2$ has a single bimolecular transition state in which the nucleophile bonds to carbon as the leaving group departs; the carbon is $sp^2$ -like with the five groups in a trigonal-bipyramidal arrangement (umbrella inversion).
$S_N1$ stereochemistry: the planar carbocation can be attacked from either face, so a single enantiomer gives a nearly racemic product — proof that a free carbocation forms.
Observing inversion (e.g. $(-)$ -2-bromooctane $\rightarrow$ $(+)$ -octan-2-ol) is evidence of an $S_N2$ pathway, since backside attack inverts every molecule.
A strong nucleophile (high charge, good base, e.g. $OH^-$ , $CN^-$ , $OR^-$ ) favours $S_N2$ ; a weak nucleophile / solvolysis favours $S_N1$ .
$2^\circ$ halides are borderline and can follow either path; conditions (solvent, nucleophile strength) decide which dominates.
Better leaving groups (weaker bases: $I^- > Br^- > Cl^-$ ) speed up both mechanisms because C–X cleavage is involved in the rate-determining step of each.

	$S_N1$	$S_N2$
Rate law	Rate $= k[\mathrm{RX}]$ — first order	Rate $= k[\mathrm{RX}][\mathrm{Nu^{-}}]$ — second order
Steps	Two, via a carbocation	One, concerted
Favoured by	$3^{\circ} > 2^{\circ} > 1^{\circ}$	$1^{\circ} > 2^{\circ} > 3^{\circ}$
Why	Carbocation stability	Less steric hindrance to backside attack
Stereochemistry	Racemisation (planar intermediate)	Inversion — the Walden inversion
Solvent	Polar protic	Polar aprotic

Every reasoning question in this subtopic is answered from one of these rows. The rate law is the clincher: it is the one piece of evidence that distinguishes the two mechanisms experimentally.

\mathrm{HO^{-}} + \mathrm{CH_3\text{–}Br} \longrightarrow \left[\mathrm{HO\cdots CH_3\cdots Br}\right]^{\ddagger} \longrightarrow \mathrm{HO\text{–}CH_3} + \mathrm{Br^{-}}

Conditions

S_N2

— one step, through a five-coordinate transition state

The nucleophile attacks from the side opposite the leaving group, so the three remaining bonds turn inside out like an umbrella in the wind. That is why the product is inverted.

Worked example · 2 marks

A compound [A] of formula

\mathrm{C_4H_9Br}

reacts with aqueous

\mathrm{KOH}

at a rate that depends only on

[\mathrm{A}]

. Its structural isomer [B] reacts at a rate depending on both

[\mathrm{B}]

and

[\mathrm{KOH}]

. Identify [A] and [B].

A rate depending only on the halide is first order, so [A] reacts by $S_N1$ — which needs the most stable carbocation, so [A] is tertiary.
A rate depending on both is second order, so [B] reacts by $S_N2$ — which needs the least hindered carbon, so [B] is primary.
Write the $\mathrm{C_4H_9Br}$ isomers and pick the tertiary and the primary one.

Answer[A] is _tert_-butyl bromide,

\mathrm{(CH_3)_3CBr}

(2-bromo-2-methylpropane); [B] is _n_-butyl bromide,

\mathrm{CH_3CH_2CH_2CH_2Br}

(1-bromobutane).

Where the marks go

Writing "inversion" for $S_N1$ — it gives racemisation (a racemic or near-racemic product), not clean inversion.
Reversing the reactivity orders: $3^\circ$ is fastest for $S_N1$ but slowest for $S_N2$ ; do not use one order for both.
Claiming the nucleophile concentration affects the $S_N1$ rate — it does not, because the nucleophile enters after the rate-determining step.
Confusing solvent effects: polar protic favours $S_N1$ , polar aprotic favours $S_N2$ ; swapping these is a common slip.
Saying a racemic product means 50% inversion plus 50% retention without noting $S_N1$ usually shows a slight excess of inversion due to ion-pair shielding.

How the board asks it

Distinguish $S_N1$ vs $S_N2$ mechanisms2 mkAsked 2025
Distinguish between $S_N1$ and $S_N2$ reactions with respect to their molecularity, order (rate law) and the stereochemistry of the product formed.
Mechanism $S_N2$ backside attack and walden inversion
Explain the mechanism of the alkaline hydrolysis of $(-)$ -2-bromooctane by the $S_N2$ pathway, showing the transition state and explaining why the product $(+)$ -octan-2-ol shows inverted configuration.
Give reasons $S_N1$ racemisation as proof of a carbocation3 mk
Give reasons: when optically active 2-bromooctane is hydrolysed under $S_N1$ conditions, the product is almost completely racemised.
Give reasonsstructural reactivity orders3 mk
Account for the fact that tertiary alkyl halides undergo $S_N1$ substitution much faster than primary halides, whereas the reverse order ( $CH_3X > 1^\circ > 2^\circ > 3^\circ$ ) holds for $S_N2$ reactions.
Give reasonssolvent effects on mechanism
Why does a polar protic solvent favour an $S_N1$ reaction while a polar aprotic solvent favours an $S_N2$ reaction?
Assertion–Reasonnucleophile concentration and the rate law2 mkAsked 2025
Assertion: the rate of an $S_N1$ reaction is independent of the concentration of the nucleophile. Reason: the nucleophile attacks only after the rate-determining ionisation step. Choose the correct option.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.