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Q7.33:

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

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.

Solution :

1. Reactant Structure and Initial Protonation

The starting material is $3\text{-methylbutan-2-ol}$, which possesses the structural formula $(CH_3)_2CH-CH(OH)-CH_3$. In the presence of $HBr$, the lone pair on the oxygen atom of the hydroxyl group undergoes protonation [Per the Bronsted-Lowry acid-base theory].

Mechanism Step I: Protonation

$(CH_3)_2CH-CH(OH)-CH_3 + H^+ \rightleftharpoons (CH_3)_2CH-CH(OH_2^+)-CH_3$

2. Formation of Secondary Carbocation

The protonated alcohol (alkyloxonium ion) is a good leaving group. Loss of a water molecule ($H_2O$) results in the formation of a secondary carbocation at the $C_2$ position [By unimolecular nucleophilic substitution kinetics, $S_N1$].

Mechanism Step II: Dehydration

$(CH_3)_2CH-CH(OH_2^+)-CH_3 \rightarrow (CH_3)_2CH-CH^+-CH_3 + H_2O$

3. 1,2-Hydride Shift Rearrangement

The $C_2$ secondary carbocation is adjacent to a $C_3$ carbon which is tertiary. A hydride ion ($H^-$) shifts from $C_3$ to $C_2$ [According to the principle of thermodynamic stability; tertiary carbocations are more stable than secondary due to hyperconjugation and inductive effects].

Mechanism Step III: Rearrangement

$(CH_3)_2CH-CH^+-CH_3 \xrightarrow{1,2-H^- \text{ shift}} (CH_3)_2C^+-CH_2-CH_3$

This generates a more stable tertiary carbocation ($2\text{-methylbutan-2-yl cation}$).

4. Nucleophilic Attack by Bromide

The bromide ion ($Br^-$), acting as a nucleophile, attacks the electrophilic tertiary carbocation to form the final product, $2\text{-bromo-2-methylbutane}$.

Mechanism Step IV: Nucleophilic Addition

$(CH_3)_2C^+-CH_2-CH_3 + Br^- \rightarrow (CH_3)_2C(Br)-CH_2-CH_3$

Visual Representation of Rearrangement

Secondary C+ (C2) Hydride Shift Tertiary C+ (C2) CH(CH3)2-CH+-CH3 (CH3)2C+-CH2-CH3

Summary of Reaction Steps

Step Process Energetic Significance
I Protonation of $-OH$ Converts poor leaving group to excellent leaving group
II Carbocation Formation Creation of reactive intermediate
III $1,2-Hydride$ Shift Increased stability (Secondary $\rightarrow$ Tertiary)
IV Nucleophilic Attack Completion of substitution

Final Solution: The reaction proceeds via the formation of a secondary carbocation followed by a 1,2-hydride shift to form a stable tertiary carbocation, which is subsequently attacked by the bromide ion to yield 2-bromo-2-methylbutane.


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