Q.Which of the following compounds will give racemic mixture on nucleophilic substitution by $\mathrm{OH^-}$ ion?

The key idea is that a racemic mixture forms when the reaction proceeds through a planar carbocation intermediate — that is, via an SN1 mechanism. A chiral centre that becomes planar during the reaction will be attacked from either face with equal probability, giving a racemate.

Step 1: Identify the substrate class and check which carbon is actually a stereocentre.

• (a) $CH_3{-}CH(Br){-}C_2H_5$ is a secondary alkyl halide ($2^\circ$). The Br-bearing carbon is attached to $CH_3$, $C_2H_5$, $H$, and $Br$ — four different groups, so it IS a genuine stereocentre.
• (b) $CH_3{-}C(Br)(C_2H_5){-}CH_3$ is a tertiary alkyl halide ($3^\circ$). But look at all four groups on the Br-bearing carbon: $CH_3$, $CH_3$, $C_2H_5$, $Br$ — two of them are identical methyl groups, so this carbon is not a stereocentre at all, regardless of how easily it ionises.
• (c) $CH_3{-}CH(C_2H_5){-}CH_2Br$ does have a genuine stereocentre (the $CH$ carbon bonded to $CH_3$, $C_2H_5$, $H$, $CH_2Br$), but the leaving group $Br$ sits on the adjacent primary $CH_2Br$ carbon, not on the stereocentre itself.

Step 2: Determine whether racemisation is actually possible in each case.

• (a): the stereocentre IS the carbon that loses Br. Under conditions with some $S_N1$ character, the resulting planar carbocation can be attacked by $\mathrm{OH^-}$ from either face, giving both enantiomers — a genuine racemic mixture is possible here. …

What "gives a racemic mixture" actually requires

A racemic mixture means two enantiomers of a chiral product forming in equal amounts. That is only possible when the carbon losing the leaving group is itself a genuine stereocentre (four different groups attached) both before the reaction AND in the product — a mechanism with carbocation character (planar intermediate, attackable from either face) is what lets both mirror-image outcomes form.

(a) $CH_3{-}CH(Br){-}C_2H_5$ (2-bromobutane)

The bromine-bearing carbon is bonded to $CH_3$, $C_2H_5$, $H$, and $Br$ — four different groups, a genuine stereocentre, and this is exactly the carbon where substitution happens. If the mechanism has carbocation character, the resulting planar cation can be attacked from either face, giving both enantiomers of the alcohol product — a real racemic mixture is possible here.

(b) $CH_3{-}C(Br)(C_2H_5){-}CH_3$ (2-bromo-2-methylbutane)

Look carefully at the reacting carbon's four substituents: $CH_3$, $CH_3$, $C_2H_5$, $Br$ — two of them are identical methyl groups. This carbon is not a stereocentre at all, either before the reaction or in the product (replacing Br with OH still leaves two identical methyls). Since there is no stereocentre to begin with, "racemic mixture" does not even apply here — attacking the (achiral) carbocation from either face gives the exact same molecule both times, not two enantiomers. …

Method: SN1 Reactivity & Racemisation Check

Step 1 – Recall the condition for racemisation

A racemic mixture forms when the reaction proceeds via an SN1 mechanism — that is, through a planar carbocation intermediate. The nucleophile can attack from either face, giving equal amounts of both enantiomers.

Step 2 – Identify the substrate type

SN1 requires a stable carbocation (tertiary > secondary > primary).

Also, the leaving group must be on a chiral carbon for racemisation to be observed.

Let’s examine each compound:

• (a) $\mathrm{CH_3-\underset{\underset{\displaystyle C_2H_5}{|}}{CH}-Br}$

  → Secondary alkyl halide. Can form a moderately stable carbocation.

  → The carbon bearing Br is chiral (4 different groups: H, CH₃, C₂H₅, Br).

  → SN1 possible → racemic mixture.

• (b) $\mathrm{CH_3-\overset{\overset{\displaystyle Br}{|}}{\underset{\underset{\displaystyle C_2H_5}{|}}{C}}-CH_3}$

  → Tertiary alkyl halide. Forms a very stable carbocation.

  → The Br-bearing carbon carries CH₃, CH₃, C₂H₅ and Br — two identical methyl groups, so it is not chiral. …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing SN1 with SN2 reactivity

The error: Students assume any alkyl halide with a chiral centre will give a racemic mixture, forgetting that racemic mixture formation is a hallmark of SN1, not SN2.

How to avoid:

• SN1 proceeds via a planar carbocation intermediate → nucleophile attacks from either face → racemic mixture (if the carbon is a stereocentre).
• SN2 proceeds via backside attack → inversion of configuration (no racemisation).

Key rule: Racemic mixture → only if the reacting carbon is a genuine stereocentre AND the mechanism is SN1.

Mistake 2: Assuming a tertiary carbon with a bromine is automatically chiral

The error: Students see compound (b), $\mathrm{CH_3-\overset{\overset{\displaystyle Br}{|}}{\underset{\underset{\displaystyle C_2H_5}{|}}{C}}-CH_3}$, and reason "tertiary carbon, undergoes SN1 easily, so it must give a racemic mixture."

Why it's wrong: Look at ALL FOUR groups on that carbon: $CH_3$, $Br$, $C_2H_5$, and... another $CH_3$. Two of the four groups are identical methyls. A carbon needs four genuinely DIFFERENT groups to be a stereocentre — (b)'s Br-bearing carbon is NOT chiral at all, even though it's tertiary. With no stereocentre to begin with, there's nothing to racemise: the product (2-methylbutan-2-ol) is a single, achiral compound either way, not a "racemic mixture" of two enantiomers.

How to avoid: Before asking "SN1 or SN2," first check: does this carbon actually have four different substituents? Count them explicitly, including any that repeat. If two match, it's not a stereocentre, and the racemisation question doesn't even apply.

Mistake 3: Ignoring the leaving group's position relative to the stereocentre

The error: Students see a chiral carbon elsewhere in the molecule and assume the reaction will affect it, without checking if the leaving group is actually attached to that carbon.

Example from compound (c): $\mathrm{CH_3-\underset{\underset{\displaystyle C_2H_5}{|}}{CH}-CH_2Br}$ — the chiral carbon is the second carbon (bonded to $CH_3$, $C_2H_5$, $H$, and $CH_2Br$ — four different groups), but Br sits on the adjacent primary carbon, not on the stereocentre itself.

Why it's wrong: Substitution happens at the primary $CH_2Br$ carbon via clean $S_N2$ (fast, no carbocation, and that carbon isn't a stereocentre in the first place). The actual stereocentre is never touched by the reaction — it keeps its original configuration throughout. The product is optically ACTIVE (same single enantiomer as the reactant), not racemic.

How to avoid: Always locate the carbon bearing the leaving group first, and check whether it's the SAME carbon as any stereocentre in the molecule. If the leaving group and the stereocentre are on different carbons, the stereocentre is usually untouched.

Mistake 4: Forgetting to check carbocation stability for compound (a) …