Question

Which of the following will react most readily with HI?
a.) $\text{C}{{\text{H}}_{\text{3}}}\text{OH}$
b.) ${{\left( \text{C}{{\text{H}}_{\text{3}}} \right)}_{\text{3}}}\text{COH}$
c.) $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}$
d.) ${{\left( \text{C}{{\text{H}}_{\text{3}}} \right)}_{\text{2}}}\text{CHOH}$

Answer 1

Hint: To know which among the following options is correct, we need to have few concepts clear. Firstly, we need to be clear about the fact that which kind of alcohol readily reacts with HI. Then, we need to take into consideration the concept of +I effect.

Complete step-by-step answer:
We should know that a tertiary alcohol reacts more readily with HI. This is because there is an increased number of methyl alkyl groups. This increased number of methyl alkyl groups increases the +I effect. So, the charge density on the C atom increases and hence cleavage of the C - O bond becomes easier.
The above process results in the formation of a $3{}^\circ$ carbocation is formed. The $3{}^\circ$ carbocation, which is formed, is more stable than the $1{}^\circ$ and $2{}^\circ$ carbocation.
Out of the mentioned options, we need to find out what is the degree of the alcohols that are mentioned. Once, we know the kind, we can easily say which will react with HI more readily.

In the first option we have $\text{C}{{\text{H}}_{\text{3}}}\text{OH}$, which is also known as Methanol. Methanol is an example of primary alcohol, even though there are no alkyl groups attached to the carbon which contains the –OH.

In the second option we have ${{\left( \text{C}{{\text{H}}_{\text{3}}} \right)}_{\text{3}}}\text{COH}$, which is the simplest tertiary alcohol. This compound is also known as the Tert- Butanol. This reacts with HI to form Tert-butyl chloride.

In the third option we have $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}$, which is also known as Ethanol. This is an example of primary alcohol.

The fourth option is ${{\left( \text{C}{{\text{H}}_{\text{3}}} \right)}_{\text{2}}}\text{CHOH}$, which is also known as Isopropyl alcohol. The compound is $3{}^\circ$ alcohol, which is the most stable among the mentioned alcohols. The compound that is formed in reaction with Isopropyl Iodide.

So, as we can see that iodide is more stable than chloride (due to the larger size) and also taking into the consideration of being tertiary alcohol, the most stable option is ${{\left( \text{C}{{\text{H}}_{\text{3}}} \right)}_{\text{2}}}\text{CHOH}$.

Therefore, the correct option is Option D.

Note: We should know that reason why HI is most reactive towards alcohol. The reason is that; the acid protonated the hydroxyl group. We know that Hydrogen Iodide is a stronger acid than hydrogen chloride, and hydrogen bromide. So that this protonolysis should be more fragile, the halide counter ion acts as a nucleophile. This results in the displacement of water, which is considered as a good leaving group.