ChemistryCBSE Class 12 · Haloalkanes and Haloarenes

Stereochemistry of Nucleophilic Substitution: Chirality, Inversion and Racemisation

Updated 15 October 2026 · Fact-checked

A carbon with four different groups is chiral, and the molecule exists as two non-superimposable mirror images (enantiomers) that rotate plane-polarised light in opposite directions. In SN2, backside attack inverts the configuration. In SN1, the flat carbocation is attacked from both faces, so an optically active haloalkane gives a racemic, optically inactive product.

Understand Stereochemistry of Nucleophilic Substitution

Plane-polarised light and optical activity. Ordinary light vibrates in all planes perpendicular to its direction. Passing it through a Nicol prism lets through only light vibrating in one plane: plane-polarised light. Some compounds rotate this plane when the light passes through their solution, measured with a polarimeter. Such compounds are optically active. If the rotation is clockwise (to the right) the compound is dextrorotatory, written d or (+); if anticlockwise (to the left), laevorotatory, written l or (−). The two forms are optical isomers.

Chirality. Van't Hoff and Le Bel linked optical activity to three-dimensional shape. A molecule that cannot be superimposed on its mirror image is chiral, like your left and right hands. The most common cause is an asymmetric carbon (stereocentre): a carbon bonded to four different groups. Butan-2-ol has one (C2 carries $H$ , $O H$ , $C H_{3}$ and $C_{2} H_{5}$ ), so it is chiral; propan-2-ol has two identical $C H_{3}$ groups on C2, so it is achiral. A molecule that is superimposable on its mirror image is achiral and optically inactive.

Enantiomers and racemic mixtures. The two non-superimposable mirror-image forms of a chiral molecule are enantiomers. They have identical physical properties (melting point, boiling point, solubility, refractive index) except that they rotate plane-polarised light by equal amounts in opposite directions. A 1 : 1 mixture of the two is a racemic mixture, written dl or (±). It shows zero rotation because the rotation by one enantiomer cancels that by the other (external compensation). Converting an enantiomer into a racemic mixture is racemisation.

Retention and inversion. Configuration means the spatial arrangement of groups around the stereocentre. If a reaction keeps the same arrangement, it shows retention; this is expected when no bond to the stereocentre is broken. If the new group ends up on the opposite side to where the old group was, the reaction shows inversion.

SN2 gives inversion. The nucleophile attacks from the side opposite the leaving group. As the leaving group goes, the other three groups flip through a planar arrangement in the transition state, like an umbrella turning inside out. So an optically active haloalkane gives a product with inverted configuration. NCERT's example: (−)-2-bromooctane with hydroxide gives (+)-octan-2-ol, with the $O H$ in the position opposite to where Br was. This is also why SN2 at a chiral carbon gives a single enantiomer of the product, not a mixture.

SN1 gives racemisation. In SN1, the leaving group departs first, forming a planar $s p^{2}$ carbocation. The nucleophile can attack this flat ion equally from either face. Attack from the side where the halogen was gives the product with retention; attack from the opposite side gives inversion. Equal amounts of both mean a racemic mixture: optically active 2-bromobutane, hydrolysed by the SN1 route, gives (±)-butan-2-ol.

These results are exam evidence for the mechanisms: inversion points to SN2; racemisation points to SN1.

Stereochemistry of Nucleophilic Substitution: formulas and key facts

Condition for a chiral carbon: $C (a) (b) (c) (d) : four different groups \Rightarrow stereocentre$; Two identical groups on the carbon make it achiral at that centre.
Racemic mixture: $(+) : (-) = 1 : 1 \Rightarrow observed rotation = 0$; Written (±) or dl; optically inactive by external compensation.
SN2 stereochemistry: $(-) -2-bromooctane O H^{-} (+) -octan-2-ol (inversion)$; Backside attack; one enantiomer of product.
SN1 stereochemistry: $optically active R - X \to R^{+} (planar) \to (\pm) -product$; Attack from both faces: racemisation.

How to solve Stereochemistry of Nucleophilic Substitution questions

Board questions here ask you to identify chiral molecules, define terms, and predict whether a substitution gives inversion or racemisation. Use this order.

1Draw or write each carbon that bears a heteroatom or a branch, and list its four attached groups.
2If all four are different, mark it with an asterisk as a chiral (asymmetric) carbon; if any two match, it is not.
3A molecule with one chiral carbon is chiral and shows optical isomerism (two enantiomers).
4For substitution at a chiral carbon, decide the mechanism first (primary/strong nucleophile → SN2; tertiary/weak nucleophile, protic solvent → SN1).
5SN2: write 'inversion of configuration', and say the product is optically active.
6SN1: write 'racemisation', and say the product is a (±) mixture with zero rotation (possibly with slight excess of inversion, but board answers expect racemic).
7When defining terms, give one example each: enantiomers (d- and l-butan-2-ol), racemic mixture ((±)-butan-2-ol).

Stereochemistry of Nucleophilic Substitution in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A carbocation route gives a racemic product

Suppose a company wants to make a single enantiomer of a medicine whose molecule has one chiral carbon, because only that enantiomer is active. A chemist proposes making it by substitution on a tertiary halide that already has the correct configuration, using water as the nucleophile.

Predict the stereochemical outcome and explain why the plan would waste about half of the product. What kind of substrate and conditions would be better?

Try it first, then show the solutionSolution

A tertiary halide with a weak nucleophile (water) reacts by SN1.
Ionisation gives a planar carbocation, which loses the original arrangement around the carbon.
Water attacks both faces equally, giving a racemic (±) mixture: about 50% of the wanted enantiomer and 50% of its mirror image.
A substrate that reacts by SN2 (for example a less hindered secondary or primary halide) with a strong nucleophile gives clean inversion, so starting from the opposite configuration gives a single product enantiomer.

Answer: SN1 gives a racemic mixture, so only about half is the active enantiomer; an SN2 route with inversion gives a single enantiomer.

Memory hook: Flat carbocations forget which side they came from, so SN1 gives both hands; SN2 flips the hand.

Tricks to remember Stereochemistry of Nucleophilic Substitution

Short memory aids. Say each one aloud once, then use it on a question the same day.

Four different friends
A chiral carbon must hold hands with four different friends; two identical friends make it ordinary (achiral).
Use it when: Spotting stereocentres fast.
SN2 = umbrella flip, SN1 = coin toss
SN2 turns the molecule inside out like an umbrella (inversion); SN1 lets the nucleophile choose either face like a coin toss (racemisation).
Use it when: Stereochemical outcome questions.
Equal and opposite cancel
(+) and (−) in equal amounts cancel to zero, just like +5 and −5.
Use it when: Explaining why a racemic mixture is optically inactive.

Stereochemistry of Nucleophilic Substitution: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Spot the chiral molecule · 1 mark · MCQBoard level

Which of the following compounds is chiral?
(a) Propan-2-ol (b) 1-Chlorobutane (c) Butan-2-ol (d) 2-Chloropropane

Try it first, then show the solutionSolution

Butan-2-ol: C2 carries H, OH, $C H_{3}$ and $C_{2} H_{5}$ , four different groups.
The others have two identical groups on the relevant carbon.

Answer: (c) Butan-2-ol.

Example 2: Enantiomers and racemic mixtures · 2 marks · Very short answerBoard level

Define enantiomers and racemic mixture. Why is a racemic mixture optically inactive?

Try it first, then show the solutionSolution

Enantiomers: stereoisomers that are non-superimposable mirror images of each other, e.g. (+)- and (−)-butan-2-ol (½ mark).
Racemic mixture: an equimolar mixture of the two enantiomers, written (±) or dl (½ mark).
The two enantiomers rotate plane-polarised light equally but in opposite directions, so their rotations cancel and the net rotation is zero (1 mark).

Answer: Non-superimposable mirror images; 1 : 1 mixture of them; rotations cancel (external compensation).

Example 3: Finding stereocentres · 3 marks · Short answerHarder board level

Identify which of the following have a chiral carbon, and mark it: (a) 2-chlorobutane (b) 2-bromo-2-methylbutane (c) 3-bromopentane (d) 2-bromopentane (e) 1-chlorobutane

Try it first, then show the solutionSolution

(a) $C H_{3} - C^{*} H (C l) - C H_{2} C H_{3}$ : C2 has H, Cl, $C H_{3}$ , $C_{2} H_{5}$ : chiral (½ mark).
(b) $C H_{3} C H_{2} - C (B r) (C H_{3}) - C H_{3}$ : C2 has two $C H_{3}$ : not chiral (½ mark).
(c) $C H_{3} C H_{2} - C H (B r) - C H_{2} C H_{3}$ : C3 has two $C_{2} H_{5}$ : not chiral (½ mark).
(d) $C H_{3} - C^{*} H (B r) - C H_{2} C H_{2} C H_{3}$ : H, Br, $C H_{3}$ , $C_{3} H_{7}$ : chiral (½ mark).
(e) $C H_{3} C H_{2} C H_{2} C H_{2} C l$ : no carbon has four different groups: not chiral (1 mark).

Answer: Chiral: (a) and (d). Not chiral: (b), (c), (e).

Example 4: Inversion in SN2 · 3 marks · Short answerHarder board level

(−)-2-Bromooctane reacts with aqueous NaOH under SN2 conditions. (a) What is the product, and what is the relation between its configuration and that of the starting material? (b) Explain why, using the mechanism. (c) Will the product be optically active?

Try it first, then show the solutionSolution

(a) (+)-Octan-2-ol, with inverted configuration: the OH takes the position opposite to that of the Br (1 mark).
(b) $O H^{-}$ attacks C2 from the side opposite Br. In the transition state the three other groups lie in a plane; as Br leaves they flip to the other side (umbrella inversion) (1 mark).
(c) Yes. Only one enantiomer forms, so the product rotates plane-polarised light (1 mark).

Answer: (a) (+)-Octan-2-ol with inverted configuration. (b) Backside attack flips the groups. (c) Yes, optically active.

Example 5: Using stereochemistry to identify the mechanism · 3 marks · Short answerHOTS / competency

An optically active sample of 2-bromobutane is hydrolysed under two different conditions. Under condition X the butan-2-ol formed is optically active with fully inverted configuration. Under condition Y the butan-2-ol formed shows almost no optical rotation. (a) Identify the mechanism in each case and explain the result. (b) Which condition likely used a high concentration of $O H^{-}$ in a polar aprotic solvent, and which water as solvent and nucleophile?

Try it first, then show the solutionSolution

X: inversion shows SN2 (backside attack in one step) (1 mark).
Y: loss of rotation shows SN1: the planar carbocation is attacked from both faces, giving a racemic mixture (1 mark).
(b) X: strong nucleophile at high concentration, aprotic solvent (favours SN2). Y: water as weak nucleophile in a polar protic medium (favours ionisation, SN1) (1 mark).

Answer: (a) X: SN2, inversion; Y: SN1, racemisation. (b) X: concentrated $O H^{-}$ , aprotic; Y: water, protic.

Where students lose marks in Stereochemistry of Nucleophilic Substitution

The slips examiners see most often, and what to write instead.

Mistake: Calling a carbon chiral when two groups are the same alkyl group
Wrong working
3-Bromopentane: C3 is chiral
Right working
C3 carries two ethyl groups, so it is not chiral
Do this: Write out all four groups in full before deciding.
Usually costs: 1 mark.
Mistake: Saying SN2 gives racemisation
Do this: SN2 → inversion (one enantiomer); SN1 → racemisation.
Usually costs: 1 mark.
Mistake: Defining a racemic mixture as 'a mixture of isomers'
Do this: It is an equimolar (1 : 1) mixture of the two enantiomers.
Usually costs: ½ mark.
Mistake: Confusing (+)/(−) with the configuration
Students think inversion always changes (+) to (−).
Do this: The sign of rotation is measured experimentally; inversion refers to the spatial arrangement. In NCERT's example (−)-2-bromooctane gives (+)-octan-2-ol, but the product is a different compound.
Usually costs: ½ mark in reasoning answers.

Exam tips

Mark chiral carbons with an asterisk (*) in structural formulas.
Use the exact words 'non-superimposable mirror images' when defining enantiomers.
Link each mechanism to its stereochemical result in one sentence: 'SN2: inversion; SN1: racemisation'.
Quote NCERT's examples: (−)-2-bromooctane → (+)-octan-2-ol (SN2); optically active 2-bromobutane → (±)-butan-2-ol (SN1).
For 'why optically inactive' questions, say 'external compensation' for racemic mixtures.

Practice questions on Stereochemistry of Nucleophilic Substitution

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The optical rotation of a racemic mixture is:
(a) positive (b) negative (c) zero (d) twice that of one enantiomer
Hint
Equal and opposite rotations.
Answer
(c) zero.
Q2 · 2 marks · Very short answerBoard level
What is meant by chirality? Give one example each of a chiral and an achiral alcohol.
Hint
Mirror image test.
Answer
Chirality is the property of a molecule of being non-superimposable on its mirror image. Chiral: butan-2-ol. Achiral: propan-2-ol (or ethanol).
Q3 · 2 marks · Very short answerHarder board level
Mark the chiral carbon atoms, if any, in: (a) $C H_{3} C H (O H) C O O H$ (b) $H O C H_{2} C H (O H) C H_{2} O H$ (c) $C H_{3} C H C l C H_{2} C H_{3}$ (d) $(C H_{3})_{2} C H C l$
Hint
Check the four groups on each carbon.
Answer
(a) C2 is chiral (H, OH, $C H_{3}$ , COOH). (b) None (C2 has two identical $C H_{2} O H$ groups). (c) C2 is chiral (H, Cl, $C H_{3}$ , $C_{2} H_{5}$ ). (d) None.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Hydrolysis of an optically active tertiary haloalkane such as 3-bromo-3-methylhexane by the SN1 route gives an optically inactive (racemic) alcohol.
Reason (R): The planar carbocation intermediate is attacked by the nucleophile equally from both faces.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Racemisation.
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
Explain retention, inversion and racemisation with reference to reactions at a chiral carbon.
Hint
Compare the spatial arrangement before and after.
Answer
Retention: the product keeps the same spatial arrangement of groups around the stereocentre (no bond to it is broken). Inversion: the new group takes the position opposite to the old one, giving the opposite configuration (SN2). Racemisation: an optically active compound is converted into an equimolar mixture of both enantiomers (SN1 through a planar carbocation).
Q6 · 1 mark · Very short answerHarder board level
In each pair, identify the chiral molecule: (a) 1-bromobutane / 2-bromobutane (b) propan-2-ol / butan-2-ol
Hint
Four different groups.
Answer
(a) 2-Bromobutane. (b) Butan-2-ol.
Q7 · 3 marks · Short answerHOTS / competency
There are four structural isomers of $C_{4} H_{9} B r$ . Name them and identify which can exist as enantiomers. If the chiral isomer reacts with aqueous KOH by SN2, what will be the stereochemical result?
Hint
Two straight-chain and two branched isomers.
Answer
1-Bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane. Only 2-bromobutane has a chiral carbon (H, Br, $C H_{3}$ , $C_{2} H_{5}$ ), so only it has enantiomers. By SN2 it gives butan-2-ol with inverted configuration (optically active).
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student uses a polarimeter on three samples: (P) pure (+)-butan-2-ol, which rotates the plane of polarised light to the right; (Q) a 1 : 1 mixture of (+)- and (−)-butan-2-ol; (R) the butan-2-ol obtained by SN1 hydrolysis of optically active 2-chlorobutane.
(a) What is meant by dextrorotatory? Which sample is dextrorotatory? (1 mark)
(b) What rotation will Q show, and what is such a mixture called? (1 mark)
(c) Predict the rotation of R and explain. (1 mark)
(d) Would propan-2-ol rotate the plane of polarised light? Why? (1 mark)
Hint
Equal amounts of enantiomers cancel.
Answer
(a) Rotates the plane of polarised light clockwise (to the right); P. (b) Zero; a racemic mixture. (c) About zero: SN1 goes through a planar carbocation attacked from both faces, giving a racemic mixture. (d) No; it has no chiral carbon (C2 carries two identical $C H_{3}$ groups), so it is achiral.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a chiral carbon?Show answer
A carbon atom bonded to four different groups.
2.Symbol for a racemic mixture?Show answer
(±) or dl.
3.What stereochemical change occurs in SN2?Show answer
Inversion of configuration.
4.Why does SN1 give racemisation?Show answer
The planar carbocation is attacked equally from both faces.
5.Instrument used to measure optical rotation?Show answer
Polarimeter.
6.Is propan-2-ol chiral?Show answer
No.

Stereochemistry of Nucleophilic Substitution: frequently asked questions

What is chirality in Class 12 Chemistry?

The property of a molecule that makes it non-superimposable on its mirror image, usually because it has a carbon bonded to four different groups.

Why is a racemic mixture optically inactive?

It contains equal amounts of the two enantiomers, which rotate plane-polarised light by equal amounts in opposite directions, so the rotations cancel.

Why does SN2 cause inversion of configuration?

The nucleophile attacks from the side opposite the leaving group, and the other three groups flip over as the leaving group departs.

Why does SN1 lead to racemisation?

The carbocation intermediate is planar, so the nucleophile can attack from either face with equal chance, giving both enantiomers equally.

Do enantiomers have different boiling points?

No. Enantiomers have identical physical properties except the direction in which they rotate plane-polarised light.