Question

What is 'Z' in the following reaction sequence? (Alcohol = colorless) $$ \mathrm{C_3H_6} \xrightarrow[\text{CCl}_4]{\mathrm{Br_2}} \mathrm{X} \xrightarrow[\Delta]{\text{i) KOH/alcohol} \\ \text{ii) NaNH}_2} \mathrm{Y} \xrightarrow[\mathrm{Hg}^{2+}, \mathrm{H}^+]{333\,K} \mathrm{Z} $$

• A. Acetone
• B. Propanal
• C. Propanol-2
• D. Methoxy ethane

Hint:

Bromination followed by dehydrohalogenation and hydration of alkynes typically leads to ketone formation (like acetone).

Answer 1

The Correct Option is A

Step 1: The reaction with \(\mathrm{Br_2}\) in \(\mathrm{CCl_4}\) adds bromine across the double bond of propene (\(\mathrm{C_3H_6}\)), forming a vicinal dibromide (X).
Step 2: Treatment with alcoholic KOH causes dehydrohalogenation, forming an alkyne (Y).
Step 3: Reaction with sodium amide (\(\mathrm{NaNH_2}\)) ensures formation of the alkyne, which on hydration with \(\mathrm{Hg^{2+}}\), \(\mathrm{H^+}\) (acidic medium) undergoes keto-enol tautomerism forming the ketone acetone (Z). Thus, Z is acetone.