Q.Propyne on treatment with water in the presence of $H_2SO_4$ and $HgSO_4$ first forms an unstable intermediate 'A' (an enol), which then rearranges to the final product (propan-2-one). The structure of 'A' and the type of isomerism (between 'A' and the product) are respectively:

The key idea is IUPAC Nomenclature combined with tautomerism (keto-enol equilibrium).

Step 1 -- Hydration of propyne.

Propyne (CH3-C=CH) adds water across the triple bond following Markovnikov's rule. The OH attaches to the more substituted carbon, giving an enol.

Step 2 -- Identify the enol.

The enol has the OH on the middle carbon: CH3-C(OH)=CH2. Its IUPAC name is prop-1-en-2-ol.

Step 3 -- Rearrangement and isomerism. …

1. The reaction: Hydration of an alkyne. Propyne (CH3C=CH) is a terminal alkyne. In the presence of dilute H2SO4 and HgSO4, water adds across the triple bond following Markovnikov's rule: H adds to the terminal carbon, OH to the internal carbon, giving an enol.

2. Identifying the enol intermediate 'A'. Adding H2O to CH3C=CH: H+ adds to C1 (terminal), OH- adds to C2. Result: CH3C(OH)=CH2. IUPAC name: prop-1-en-2-ol.

3. The rearrangement: Keto-enol tautomerism. The enol is unstable and rearranges to propan-2-one (acetone): CH3C(OH)=CH2 -> CH3C(=O)CH3. This is keto-enol tautomerism -- the two isomers differ in the position of a hydrogen atom and a double bond and exist in dynamic equilibrium.

4. Why the other options are wrong. …

Method: Hydration of Alkynes (Kucherov Reaction) — Followed by Keto-Enol Tautomerism

Step 1: Identify the reaction type

Propyne ($CH_3-C \equiv CH$) undergoes acid-catalyzed hydration in the presence of $HgSO_4$ and $H_2SO_4$. This is the Kucherov reaction.

Step 2: Apply Markovnikov’s rule for addition

• Water adds across the triple bond.
• The OH group attaches to the more substituted carbon (Markovnikov addition).
• For propyne: $CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}/H^+} CH_3-C(OH)=CH_2$

This gives prop-1-en-2-ol (the enol form).

Step 3: Identify the unstable intermediate 'A'

• The enol formed is prop-1-en-2-ol.
• Structure: $CH_3-C(OH)=CH_2$

Step 4: Recognize the rearrangement

• The enol is unstable and tautomerizes to the more stable keto form.
• Keto-enol tautomerism occurs: …

Let’s break this down step-by-step — first the chemistry, then the common mistakes.

Step 1 — The reaction

Propyne ($CH_3C \equiv CH$) reacts with water in the presence of $H_2SO_4$ and $HgSO_4$ (Markovnikov hydration of alkynes).

The initial product is an enol (unstable intermediate ‘A’):

This enol is prop-1-en-2-ol.

It then rearranges to the keto form:

The isomerism between the enol and the keto form is tautomerism (specifically keto-enol tautomerism).

So the correct pair is:

Prop-1-en-2-ol, tautomerism → Option (D).

Common mistakes students make

✗ Mistake 1: Confusing the enol structure

• Students often write prop-1-en-1-ol (double bond between C1 and C2, OH on C1). But the correct enol from Markovnikov addition has the OH on the more substituted carbon (C2), giving prop-1-en-2-ol.

How to avoid:

Always apply Markovnikov’s rule: in hydration of an unsymmetrical alkyne, the OH goes to the more substituted carbon of the triple bond.

✗ Mistake 2: Confusing tautomerism with other isomerisms

• Metamerism (different alkyl groups on either side of a functional group) — not applicable here.
• Geometrical isomerism (cis/trans) — requires a double bond with restricted rotation and two different groups on each carbon; prop-1-en-2-ol has two identical H’s on one carbon, so no geometrical isomers.

How to avoid:

Remember: tautomerism is a special case of functional group isomerism where the isomers (enol and keto) are in dynamic equilibrium and differ in the position of a proton and a double bond.

✗ Mistake 3: Forgetting that the enol is unstable

• Some students think the enol is the final product. The question explicitly says ‘A’ is unstable and rearranges. …