Question

The correct statement regarding X and Y formed in the following reaction is: 

\[ (\mathrm{CH}_3)_3\mathrm{COC}_2\mathrm{H}_5 \xrightarrow{\text{HI},\ \Delta} \text{Halide (X) + Alcohol (Y)} \]

• A. X undergoes substitution by \(\mathrm{S}_\text{N}2\) mechanism
• B. X undergoes substitution with water in two steps
• C. Y gets converted to corresponding chloride with conc.HCl at room temperature
• D. Reaction of Y with Cu / 573 K gives ketone

Hint:

Remember the reactivity patterns for ethers and alcohols: Ether cleavage with HX:} For unsymmetrical ethers, if one alkyl group is primary/secondary and the other is tertiary, the tertiary group forms the halide via \(\operatorname{S}_{N}}1\). If both are primary/secondary, the reaction typically proceeds via \(\operatorname{S}_{N}}2\) with the halide attacking the less hindered carbon.

Answer 1

The Correct Option is B

Step 1: Determine products X (halide) and Y (alcohol) from the ether cleavage. 
The reactant is tert-butyl ethyl ether, $(\operatorname{CH}_3)_3\operatorname{COC}_2\operatorname{H}_5$. This is an unsymmetrical ether containing a primary alkyl group (ethyl) and a tertiary alkyl group (tert-butyl).
When an unsymmetrical ether reacts with a strong acid like HI, the cleavage mechanism depends on the nature of the alkyl groups. If one of the alkyl groups is tertiary, the cleavage proceeds via an $\operatorname{S}_{N}1$ mechanism, leading to the formation of the more stable tertiary carbocation. The halide ion then attacks this carbocation, while the less hindered group forms the alcohol. 
1. Protonation of the ether:
$(\operatorname{CH}_3)_3\operatorname{COC}_2\operatorname{H}_5 + \operatorname{HI} \rightleftharpoons (\operatorname{CH}_3)_3\operatorname{C}-\operatorname{O}^+\operatorname{H}-\operatorname{C}_2\operatorname{H}_5 + \operatorname{I}^-$ 
2. Heterolytic cleavage to form carbocation (rate-determining $\operatorname{S}_{N}1$ step):
The bond between the oxygen and the tertiary carbon breaks to form a stable tertiary carbocation and an alcohol. $(\operatorname{CH}_3)_3\operatorname{C}-\operatorname{O}^+\operatorname{H}-\operatorname{C}_2\operatorname{H}_5 \rightarrow (\operatorname{CH}_3)_3\operatorname{C}^+ + \operatorname{C}_2\operatorname{H}_5\operatorname{OH}$ So, Y is ethyl alcohol ($\operatorname{C}_2\operatorname{H}_5\operatorname{OH}$). 
3. Nucleophilic attack by iodide ion:
The highly reactive iodide ion attacks the tertiary carbocation. $(\operatorname{CH}_3)_3\operatorname{C}^+ + \operatorname{I}^- \rightarrow (\operatorname{CH}_3)_3\operatorname{CI}$ So, X is tert-butyl iodide ($(\operatorname{CH}_3)_3\operatorname{CI}$). 
Now let's evaluate each option based on the identified products X and Y. 
Option (1) X undergoes substitution by $\operatorname{S}_{N}2$ mechanism:
X is tert-butyl iodide, which is a tertiary alkyl halide. Tertiary alkyl halides are highly hindered and undergo substitution reactions predominantly via an $\operatorname{S}_{N}1$ mechanism, not $\operatorname{S}_{N}2$. Therefore, this statement is incorrect. 
Option (2) X undergoes substitution with water in two steps:
X is tert-butyl iodide. Tertiary alkyl halides react with weak nucleophiles like water via an $\operatorname{S}_{N}1$ mechanism. The $\operatorname{S}_{N}1$ mechanism is a two-step process: (i) ionization of the alkyl halide to form a carbocation and a leaving group, and (ii) nucleophilic attack on the carbocation. Therefore, this statement is correct. 
Option (3) Y gets converted to corresponding chloride with conc. HCl at room temperature:
Y is ethyl alcohol ($\operatorname{C}_2\operatorname{H}_5\operatorname{OH}$), which is a primary alcohol. Primary alcohols react very slowly or not at all with concentrated HCl alone at room temperature. The reaction requires a Lewis acid catalyst like anhydrous $\operatorname{ZnCl}_2$ (Lucas reagent) or heating. Therefore, this statement is incorrect. 
Option (4) Reaction of Y with Cu / 573 K gives ketone:
Y is ethyl alcohol ($\operatorname{C}_2\operatorname{H}_5\operatorname{OH}$), a primary alcohol. When primary alcohols are passed over hot copper at 573 K, they undergo catalytic dehydrogenation to form aldehydes. Ketones are formed from secondary alcohols under these conditions. \[ \operatorname{CH}_3\operatorname{CH}_2\operatorname{OH} \xrightarrow{\text{Cu}, 573 \text{ K}} \operatorname{CH}_3\operatorname{CHO} \ (\text{acetaldehyde}) + \operatorname{H}_2 \] Therefore, this statement is incorrect. 
Based on the analysis, only Option (2) is the correct statement. 
The final answer is: $\boxed{\text{X undergoes substitution with water in two steps}}$.