Question

An alcohol X ($\text{C}_5\text{H}_{12}\text{O}$) produces turbidity instantly with conc. $\text{HCl}/\text{ZnCl}_2$. Isomer (Y) of $\text{X}$ undergoes dehydration with conc. $\text{H}_2\text{SO}_4$ at $443 \text{ K}$. $\text{X}$ and $\text{Y}$ respectively are

• A. $\text{Structure 1}, \text{Structure 2}$
• B. $\text{Structure 2}, \text{Structure 1}$
• C. $\text{Structure 3}, \text{Structure 4}$
• D. $\text{Structure 4}, \text{Structure 3}$

Hint:

The Lucas test ($\text{HCl}/\text{ZnCl}_2$) is used to distinguish between primary, secondary, and tertiary alcohols. The reactivity order is $3^\circ$ (instant turbidity) $> 2^\circ$ (turbidity in 5-10 min) $>1^\circ$ (no turbidity). Dehydration temperature follows the reverse order ($1^\circ$ highest, $3^\circ$ lowest).

Answer 1

The Correct Option is B

Step 1: Determine the nature of Alcohol X.
The molecular formula is $\text{C}_5\text{H}_{12}\text{O}$. The alcohol X produces turbidity instantly with Lucas reagent (conc. $\text{HCl}/\text{ZnCl}_2$).
The reaction is: $\text{R-OH} + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{R-Cl} + \text{H}_2\text{O}$.
The order of reactivity for Lucas test is $3^\circ \text{ alcohol}>2^\circ \text{ alcohol}>1^\circ \text{ alcohol}$.
Instantaneous turbidity indicates that $\text{X}$ must be a tertiary ($3^\circ$) alcohol.

Step 2: Identify the tertiary alcohol X with formula $\text{C_5\text{H}_{12}\text{O}$.}
The only tertiary alcohol with $\text{C}_5\text{H}_{12}\text{O}$ is 2-Methylbutan-2-ol. $\text{Structure 2}$ is 2-Methylbutan-2-ol ($\text{CH}_3-\text{C}(\text{OH})(\text{CH}_3)-\text{CH}_2-\text{CH}_3$).
So, $\text{X}$ is $\text{Structure 2}$.

Step 3: Determine the nature of Alcohol Y.
Alcohol $\text{Y}$ is an isomer of $\text{X}$ ($\text{C}_5\text{H}_{12}\text{O}$). $\text{Y}$ undergoes dehydration with conc. $\text{H}_2\text{SO}_4$ at $443 \text{ K}$ (a high temperature). $3^\circ$ alcohols dehydrate the easiest, and $1^\circ$ alcohols the hardest (highest temperature). The high temperature suggests $\text{Y}$ is a primary ($1^\circ$) or secondary ($2^\circ$) alcohol.
If $\text{X}$ is $3^\circ$, then $\text{Y}$ must be $1^\circ$ or $2^\circ$. Let's check the options. $\text{Structure 1}$ is Pentan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$), a $1^\circ$ alcohol. $1^\circ$ alcohols require high temperature for dehydration.

Step 4: Conclude the final correct option.
$\text{X}$ must be $\text{Structure 2}$ (2-Methylbutan-2-ol, $3^\circ$ alcohol).
$\text{Y}$ must be $\text{Structure 1}$ (Pentan-1-ol, $1^\circ$ alcohol, which is an isomer of X).
The pair ($\text{X}$, $\text{Y}$) is ($\text{Structure 2}$, $\text{Structure 1}$).