Question

Identify the compound (Z) in the following reaction sequence

• A. Propanal
• B. Propanone
• C. Propanoic acid
• D. Propanamide

Hint:

Hydration of alkynes (adding \(H_2O\) with \(Hg^{2+}/H^+\)) is a key reaction. - For terminal alkynes (except acetylene), it follows Markovnikov's rule and always produces a methyl ketone. - For symmetrical internal alkynes, it gives a single ketone. - For unsymmetrical internal alkynes, it gives a mixture of two ketones.

Answer 1

The Correct Option is B

Let's trace the reaction sequence step by step.
The starting material is 1,1-dibromopropane.
Step 1: Formation of X.
The starting material reacts with (i) alcoholic KOH and (ii) sodium amide (NaNH\(_2\)). This is a double dehydrohalogenation reaction. Geminal dihalides (halogens on the same carbon) are converted to alkynes.
First, alcoholic KOH eliminates one molecule of HBr to form an alkenyl halide.
\( \text{CH}_3\text{CH}_2\text{CHBr}_2 \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH=CBr} + \text{KBr} + \text{H}_2\text{O} \) (major product).
Then, the stronger base NaNH\(_2\) is used to eliminate the second molecule of HBr from the less reactive alkenyl halide to form an alkyne.
\( \text{CH}_3\text{CH=CBr} \xrightarrow{\text{NaNH}_2} \text{CH}_3\text{C}\equiv\text{CH} + \text{NaBr} + \text{NH}_3 \).
So, compound X is propyne.
Step 2: Formation of Y.
Propyne (X) reacts with water in the presence of an acid catalyst (H\(^+\)) and mercuric ions (Hg\(^{2+}\)). This is the hydration of an alkyne (Kucherov's reaction).
The reaction follows Markovnikov's rule. The -OH group adds to the more substituted carbon of the triple bond, and the H adds to the less substituted carbon.
\( \text{CH}_3\text{C}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}, \text{H}^+} [\text{CH}_3\text{C(OH)=CH}_2] \).
This initially forms an enol (prop-1-en-2-ol). This enol is unstable and immediately tautomerizes to its more stable keto form.
\( [\text{CH}_3\text{C(OH)=CH}_2] \rightleftharpoons \text{CH}_3\text{C(=O)CH}_3 \).
The product Y is propanone (acetone).
Step 3: Formation of Z.
The reaction shown is "Isomerization". However, propanone is already the stable product of the hydration. The label "Isomerization" likely refers to the tautomerization step that converts the enol to the ketone (Y). In this context, Y and Z are the same compound, or Z is simply the final stable product of the sequence.
Therefore, Z is propanone.