Compound 'X' with molecular formula C₄H₉Br reacts with aqueous KOH to give an alcohol. The rate of this reaction depends only on the concentration of the compound 'X'. When an optically active isomer 'Y' of the compound 'X' was treated with aqueous KOH solution, the rate of reaction was found to be dependent on concentration of compound 'Y' and aqueous KOH both. (3 × 1)

(a) Write down the structural formula of both 'X' and 'Y'.

(b) Out of 'X' and 'Y', which one will undergo racemisation and why?

(c) Out of 'X' and 'Y', which one will form product with inversion of configuration and why?

Q: Compound 'X' with molecular formula C₄H₉Br reacts with aqueous KOH to give an alcohol. The rate of this reaction depends only on the concentration of the compound 'X'. When an optically active isomer 'Y' of the compound 'X' was treated with aqueous KOH solution, the rate of reaction was found to be dependent on concentration of compound 'Y' and aqueous KOH both. (3 × 1) (a) Write down the structural formula of both 'X' and 'Y'. (b) Out of 'X' and 'Y', which one will undergo racemisation and why? (c) Out of 'X' and 'Y', which one will form product with inversion of configuration and why?

**Assigning the isomers (both C₄H₉Br):** - **X** — rate depends only on [X] → first order → **SN1** → a tertiary halide → **X = 2-bromo-2-methylpropane, (CH₃)₃C–Br** (tert-butyl bromide). - **Y** — optically active and rate depends on [Y] and [KOH] → second order → **SN2** → the chiral secondary halide → **Y = 2-bromobutane, CH₃–CHBr–CH₂–CH₃** (sec-butyl bromide). **(a) Structural formulae:** - X: (CH₃)₃C–Br - Y: CH₃CH(Br)CH₂CH₃ **(b) Which undergoes racemisation and why:** - **X** proceeds by **SN1** through a **planar (sp²) carbocation** intermediate. The nucleophile (OH⁻) can attack this flat carbocation equally from **both faces**, so an SN1 reaction is accompanied by **racemisation**. **(c) Which gives inversion of configuration and why:** - **Y** proceeds by **SN2**: OH⁻ attacks from the side **opposite** to the leaving Br⁻ (backside attack) in a single step, so the configuration is **inverted (Walden inversion)**.

Haloalkanes and Haloarenes

SN1 vs SN2 Mechanism of C₄H₉Br

Official Answer

Assigning the isomers (both C₄H₉Br):

X — rate depends only on [X] → first order → SN1 → a tertiary halide → X = 2-bromo-2-methylpropane, (CH₃)₃C–Br (tert-butyl bromide).
Y — optically active and rate depends on [Y] and [KOH] → second order → SN2 → the chiral secondary halide → Y = 2-bromobutane, CH₃–CHBr–CH₂–CH₃ (sec-butyl bromide).

(a) Structural formulae:

X: (CH₃)₃C–Br
Y: CH₃CH(Br)CH₂CH₃

(b) Which undergoes racemisation and why:

X proceeds by SN1 through a planar (sp²) carbocation intermediate. The nucleophile (OH⁻) can attack this flat carbocation equally from both faces, so an SN1 reaction is accompanied by racemisation.

(c) Which gives inversion of configuration and why:

Y proceeds by SN2: OH⁻ attacks from the side opposite to the leaving Br⁻ (backside attack) in a single step, so the configuration is inverted (Walden inversion).

SN1 mechanismSN2 mechanismtert-butyl bromide2-bromobutanecarbocation planarracemisationinversion of configurationWalden inversion

Marking Scheme

11 mark (a): X = (CH₃)₃CBr (2-bromo-2-methylpropane) AND Y = CH₃CHBrCH₂CH₃ (2-bromobutane); ½ each.
21 mark (b): X undergoes racemisation because it reacts by SN1 via a planar carbocation attacked from both faces.
31 mark (c): Y gives product with inversion of configuration because it reacts by SN2 (backside/Walden inversion).

Hint

Order of reaction fixes the mechanism: only-[X] first order = SN1 = tertiary (tert-butyl bromide) = racemisation; both-concentrations second order = SN2 = chiral 2-bromobutane = inversion.

Quick Oral Answer

Because X's rate depends only on X it reacts by SN1 and is tertiary tert-butyl bromide, going through a planar carbocation attacked from both sides, so it racemises; Y's rate depends on both reactants so it reacts by SN2 and, being the chiral 2-bromobutane, undergoes backside attack giving inversion of configuration.

Analysis & Explanation

This is a mechanism-identification problem: the kinetics (order of reaction) tells you the mechanism, the mechanism tells you the structure, and the mechanism also fixes the stereochemical outcome.

Reading the clues:

'Rate depends only on [X]' = first order = SN1. SN1 is favoured by the most stable carbocation, i.e. a tertiary halide → tert-butyl bromide.
'Optically active' and 'rate depends on both concentrations' = second order = SN2, and to be optically active the molecule must be chiral. Among C₄H₉Br isomers only 2-bromobutane has a stereocentre, so Y = sec-butyl bromide.

Stereochemistry — the heart of the question:

SN1 → racemisation: the intermediate carbocation is planar (sp²); OH⁻ attacks from either face with (nearly) equal probability, generating both configurations.
SN2 → inversion: concerted backside attack flips the molecule inside-out like an umbrella in the wind (Walden inversion), so the optically active Y gives a product of opposite configuration.

Exam trap / subtle point: tert-butyl bromide (X) is itself achiral, so you cannot 'see' optical rotation change in its own product — but the mechanistic answer expected is that the SN1 pathway is the one associated with racemisation, while the SN2 substrate (Y), being optically active, is the one that visibly shows inversion. Answer strictly by mechanism: SN1 = racemisation (X), SN2 = inversion (Y).

Common Mistakes

1Assigning X as the optically active/secondary halide — X follows SN1 (rate depends only on X), so it must be the tertiary tert-butyl bromide; Y (optically active, SN2) is 2-bromobutane.
2Swapping the stereochemistry: writing that SN1 gives inversion and SN2 gives racemisation. SN1 → racemisation, SN2 → inversion.
3Choosing 1-bromobutane or isobutyl bromide for Y — these are primary and not chiral, so they cannot be the optically active isomer.

Interesting Facts

The terms SN1 and SN2 were coined by Edward Hughes and Christopher Ingold in the 1930s, based precisely on this kind of reaction-order (kinetics) evidence.

The stereochemical inversion in SN2 is called 'Walden inversion' after Paul Walden, who demonstrated configuration reversal in 1896 through a cycle of substitution reactions.

Tertiary halides essentially never react by SN2 because the three bulky groups block backside attack, which is exactly why tert-butyl bromide is forced down the SN1 (carbocation) path.

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Frequently Asked Questions

How do you decide which isomer is SN1 and which is SN2 from the given data?

Use the rate law. If the rate depends only on the concentration of the halide (first order), the mechanism is SN1, favoured by a tertiary halide (X = tert-butyl bromide). If the rate depends on both the halide and the nucleophile (second order), the mechanism is SN2 (Y = 2-bromobutane, the optically active secondary halide).

Why does SN1 lead to racemisation while SN2 gives inversion?

SN1 goes through a flat sp² carbocation intermediate, and the nucleophile can attack it from either face with about equal probability, producing both configurations (racemisation). SN2 is a one-step reaction in which the nucleophile attacks from the side opposite the leaving group, flipping the configuration (Walden inversion).