HBr reacts fastest with
a. 2-Methyl-1-propanol
b. 2-Methyl-2-propanol
c. 2-propanol
d. 1-propanol
e. None of the above
The correct answer and explanation is :
The correct answer is: b. 2-Methyl-2-propanol
Explanation (300+ words)
The reaction of HBr (hydrobromic acid) with alcohols is a classic example of a nucleophilic substitution reaction, where the hydroxyl (-OH) group is replaced by a bromide (Br⁻) to form an alkyl halide.
The rate of this reaction greatly depends on the type of alcohol—whether it’s primary (1°), secondary (2°), or tertiary (3°)—and the mechanism by which the reaction proceeds:
- Tertiary alcohols typically undergo SN1 (unimolecular nucleophilic substitution) mechanisms.
- Primary alcohols undergo SN2 (bimolecular nucleophilic substitution) mechanisms.
- Secondary alcohols may undergo either mechanism, depending on the conditions.
In the SN1 mechanism, the rate-determining step is the formation of a carbocation intermediate after the alcohol is protonated and the water molecule leaves. The more stable the carbocation, the faster this step occurs.
Let’s analyze the options:
b. 2-Methyl-2-propanol (Tertiary Alcohol)
- This is a tertiary alcohol, with the structure (CH₃)₃COH.
- Upon protonation, it forms a tertiary carbocation, which is highly stable due to inductive effects and hyperconjugation from the surrounding methyl groups.
- The carbocation forms readily, making the SN1 reaction very fast.
Other Options:
- a. 2-Methyl-1-propanol – Primary alcohol, slower SN2 mechanism.
- c. 2-Propanol – Secondary alcohol, slower than tertiary alcohol.
- d. 1-Propanol – Primary alcohol, slowest via SN2.
- e. None of the above – Incorrect, because option b is a valid and correct choice.
Conclusion
2-Methyl-2-propanol reacts fastest with HBr because it forms a tertiary carbocation, which makes the SN1 mechanism highly favorable. The rate of substitution increases with the stability of the carbocation intermediate, and tertiary carbocations are the most stable among all. Therefore, option b is correct.