When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

$\mathrm{CH_3-\underset{\underset{\displaystyle CH_3}{|}}{CH}-\underset{\underset{\displaystyle OH}{|}}{CH}-CH_3\ \xrightarrow{HBr}\ CH_3-\overset{\overset{\displaystyle Br}{|}}{\underset{\underset{

Question

When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

$\mathrm{CH_3-\underset{\underset{\displaystyle CH_3}{|}}{CH}-\underset{\underset{\displaystyle OH}{|}}{CH}-CH_3\ \xrightarrow{HBr}\ CH_3-\overset{\overset{\displaystyle Br}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-CH_2-CH_3}$

Give a mechanism for this reaction.

(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.

Accepted Answer

> [!TLDR]
> The reaction proceeds via an **Sₙ1 mechanism** where the initial secondary carbocation undergoes a **1,2-hydride shift** to form a more stable tertiary carbocation, which is then attacked by Br⁻ to give the rearranged product.

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### Why this happens — the concept

The hint tells you the key: a hydride shift. But why does the carbocation bother to rearrange at all? The answer is **carbocation stability**. Tertiary carbocations are more stable than secondary ones because the three alkyl groups donate electron density via hyperconjugation and inductive effects, spreading the positive charge. The reaction is driven by the thermodynamic urge to form the most stable intermediate possible.

In Williamson ether synthesis, you'd use an alkoxide and an alkyl halide — but here we have an alcohol with HBr, which is a classic **Sₙ1** or **Sₙ2** situation. Because the alcohol is secondary and the acid provides a good leaving group (water), the reaction favours an Sₙ1 pathway, especially when rearrangement can lead to a more stable carbocation.

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### Step-by-step mechanism

**1. Protonation of the hydroxyl group**  
The lone pair on oxygen of 3-methylbutan-2-ol attacks a proton from HBr, forming an oxonium ion. This makes the OH group a much better leaving group (water instead of hydroxide).

$$
\ce{CH3-CH(CH3)-CH(OH)-CH3 + HBr -> CH3-CH(CH3)-CH(OH2+)-CH3 + Br-}
$$

**2. Loss of water to form a secondary carbocation**  
The C–O bond breaks heterolytically, ejecting a water molecule and leaving behind a secondary carbocation at the 2nd carbon.

$$
\ce{CH3-CH(CH3)-CH(OH2+)-CH3 -> CH3-CH(CH3)-CH+ -CH3 + H2O}
$$

This carbocation is secondary — it has two alkyl groups attached to the positive carbon. It's reasonably stable, but not as stable as it could be.

> [!WARNING]
> A common mistake is to stop here and attack Br⁻ directly. But if you do that, you'd get the *unrearranged* product (2-bromo-3-methylbutane), which is **not** what the question shows. The product

Alkyl halide	Works?	Why
Methyl ( $CH_3X$ )	Yes	Least hindered, fastest SN2
Primary (1°)	Yes	Clean SN2
Secondary (2°)	Sometimes	Works if not too bulky; elimination competes
Tertiary (3°)	No	Elimination dominates
Aryl (e.g., bromobenzene)	No	SN2 impossible on $sp^2$ carbon

Alkyl Halide Type	Steric Hindrance	$\text{S}_\text{N}2$ Reactivity
Methyl ( $\text{CH}_3\text{X}$ )	Minimal	Very fast
Primary ( $\text{RCH}_2\text{X}$ )	Low	Fast
Secondary ( $\text{R}_2\text{CHX}$ )	Moderate	Slow
Tertiary ( $\text{R}_3\text{CX}$ )	High	Does not occur

Halide	Leaving Group Ability	Reason
$\text{I}^-$	Excellent	Large, polarizable, weak base
$\text{Br}^-$	Good	Moderate size, weak base
$\text{Cl}^-$	Fair	Smaller, stronger base
$\text{F}^-$	Poor	Small, strong base, holds tightly

Why this happens — the concept

Step-by-step mechanism

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