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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 ((3^{rd})) carbon atom.)

Solution :

Given Reactants and Product

Reactants: $CH_3-CH(OH)-CH(CH_3)_2$ (3-methylbutan-2-ol) and $HBr$.
Products: $CH_3-CH_2-C(Br)(CH_3)_2$ (2-bromo-2-methylbutane).

Step 1: Protonation of the Alcohol

The lone pair of electrons on the oxygen atom of the hydroxyl group attacks the electrophilic proton ($H^+$) provided by the hydrobromic acid. This transforms the poor leaving group ($-OH$) into an excellent leaving group ($-OH_2^+$).

$CH_3-CH(OH)-CH(CH_3)_2 + H^+ \rightarrow CH_3-CH(\overset{+}{O}H_2)-CH(CH_3)_2$ [Per the Brønsted-Lowry acid-base theory]

Step 2: Formation of the Secondary Carbocation

The bond between the carbon and the oxonium ion breaks, releasing a water molecule and leaving behind a secondary carbocation ($2^\circ$ carbocation) at the $C_2$ position.

$CH_3-CH(\overset{+}{O}H_2)-CH(CH_3)_2 \rightarrow CH_3-\overset{+}{C}H-CH(CH_3)_2 + H_2O$ [Rate-determining step]

Step 3: Carbocation Rearrangement (1,2-Hydride Shift)

The secondary carbocation is less stable than a tertiary carbocation. A hydride ion ($H^-$) migrates from the adjacent $C_3$ carbon to the $C_2$ carbocation. This shift creates a more stable tertiary carbocation at $C_3$.

$CH_3-\overset{+}{C}H-CH(CH_3)_2 \xrightarrow{1,2-H \text{ shift}} CH_3-CH_2-\overset{+}{C}(CH_3)_2$ [Per the stability order of carbocations: $3^\circ > 2^\circ > 1^\circ$]

Step 4: Nucleophilic Attack by Bromide Ion

The bromide ion ($Br^-$), acting as a nucleophile, attacks the tertiary carbocation at the $C_3$ position to form the final alkyl halide product.

$CH_3-CH_2-\overset{+}{C}(CH_3)_2 + Br^- \rightarrow CH_3-CH_2-C(Br)(CH_3)_2$

Mechanism Visualization

Step 1 & 2: $CH_3CH(OH)CH(CH_3)_2 \rightarrow CH_3CH^+CH(CH_3)_2 + H_2O$ Step 3 (Rearrangement): $CH_3CH^+CH(CH_3)_2 \rightarrow CH_3CH_2C^+(CH_3)_2$ Hydride Shift Step 4: $CH_3CH_2C^+(CH_3)_2 + Br^- \rightarrow CH_3CH_2C(Br)(CH_3)_2$

Final Solution: The mechanism proceeds via the formation of a secondary carbocation followed by a 1,2-hydride shift to generate a more stable tertiary carbocation, which is then attacked by the bromide nucleophile to yield 2-bromo-2-methylbutane.


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