Question

What are $\text{X}$ and $\text{Y}$ in the following reaction sequence?
$\text{C}_5\text{H}_{12}\text{O} \xrightarrow{\text{Cu}/573 \text{ K}} \text{C}_5\text{H}_{10} \xrightarrow{(\text{i}) \text{O}_3, (\text{ii}) \text{Zn}+\text{H}_2\text{O}} \text{X} + \text{Y}$

• A. Acetone + Acetaldehyde ($\text{CH}_3\text{COCH}_3 + \text{CH}_3\text{CHO}$)
• B. $\text{C}_3\text{H}_7\text{CH}_2\text{CH}_2\text{OH}$ (Incorrect formula)
• C. $\text{C}_4\text{H}_9\text{CHO}$ (Incorrect formula)
• D. $\text{C}_4\text{H}_9\text{OH}$ (Incorrect formula)

Hint:

The reaction $\text{R-OH} \xrightarrow{\text{Cu}/573 \text{ K}}$ is an oxidation for $1^\circ/2^\circ$ alcohols and dehydration for $3^\circ$ alcohols. Ozonolysis ($\text{O}_3/\text{Zn}/\text{H}_2\text{O}$) is used to cleave a $\text{C=C}$ double bond, forming a mix of aldehydes and/or ketones.

Answer 1

The Correct Option is A

Step 1: Analyze the first reaction — Oxidation/Dehydrogenation of Alcohol ($\text{C}_5\text{H}_{12}\text{O}$).
The molecular formula $\text{C}_5\text{H}_{12}\text{O}$ corresponds to a pentanol (primary, secondary, or tertiary alcohol). When vapors of alcohol are passed over hot copper at $573\,\text{K}$: \[ 1^\circ \text{alcohol} \to \text{Aldehyde}, 2^\circ \text{alcohol} \to \text{Ketone}, 3^\circ \text{alcohol} \to \text{Alkene (Dehydration)} \] Since the product has the formula $\text{C}_5\text{H}_{10}$ (alkene), the alcohol must be tertiary. Hence, the alcohol is likely 2-Methylbutan-2-ol, which undergoes dehydration to form an alkene. \[ \text{CH}_3\text{C}(\text{OH})(\text{CH}_3)\text{CH}_2\text{CH}_3 \xrightarrow[\text{573 K}]{\text{Cu}} \text{CH}_3\text{C}(\text{CH}_3)=\text{CHCH}_3 \] This product is 2-Methyl-2-butene ($\text{C}_5\text{H}_{10}$).
Step 2: Analyze the second reaction — Ozonolysis of $\text{C_5\text{H}_{10}$.}
\[ \text{CH}_3\text{C}(\text{CH}_3)=\text{CHCH}_3 \xrightarrow[\text{Zn}/\text{H}_2\text{O}]{\text{O}_3} \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{CHO} \] The products are: \[ \text{X} = \text{CH}_3\text{COCH}_3 \ (\text{Acetone}), \text{Y} = \text{CH}_3\text{CHO} \ (\text{Acetaldehyde}) \] Thus, ozonolysis gives one ketone and one aldehyde, matching the given option.
Step 3: Conclude the correct option.
\[ \text{C}_5\text{H}_{12}\text{O} \ (\text{2-Methylbutan-2-ol}) \xrightarrow{\text{Cu}/573 \text{ K}} \text{C}_5\text{H}_{10} \ (\text{2-Methyl-2-butene}) \xrightarrow{\text{O}_3/\text{Zn}/\text{H}_2\text{O}} \text{Acetone} + \text{Acetaldehyde} \] Hence, the correct answer is: \[ \boxed{\text{X} = \text{Acetone}, \ \text{Y} = \text{Acetaldehyde}} \]