Q.Addition of water to alkynes occurs in acidic medium and in the presence of $Hg^{2+}$ ions as a catalyst. Which one of the following products will be formed on addition of water to but-1-yne under these conditions?

The key idea is Markovnikov hydration of alkynes via an enol intermediate that tautomerizes to a carbonyl compound.

1. But-1-yne is a terminal alkyne: $CH_3CH_2C \equiv CH$.
2. In acidic $Hg^{2+}$-catalysed hydration, water adds according to Markovnikov’s rule — the $OH$ goes to the more substituted carbon of the triple bond. This gives the enol $CH_3CH_2C(OH)=CH_2$.
3. This enol is unstable and tautomerises (keto-enol tautomerism) to the more stable ketone, not an aldehyde. The double bond shifts to give $CH_3CH_2COCH_3$.

The reaction you’re looking at is acid-catalysed hydration of alkynes — a classic way to make carbonyl compounds from alkynes. The key is that the addition of water follows Markovnikov’s rule, and the initial product is an enol, which immediately rearranges to a more stable keto form (keto-enol tautomerism).

For a terminal alkyne like but-1-yne, the triple bond is between C1 and C2. The $Hg^{2+}$ catalyst coordinates to the triple bond, making it more electrophilic. Water attacks the more substituted carbon of the triple bond — that’s Markovnikov addition. Let’s trace it step by step.

1. Identify the structure of but-1-yne

   But-1-yne is $CH_3-CH_2-C \equiv CH$. The triple bond is between carbon 1 (terminal) and carbon 2.

2. Markovnikov addition of water

   In the presence of $Hg^{2+}$ and $H^+$, water adds such that the $OH$ group ends up on the more substituted carbon of the triple bond.

   The two carbons of the triple bond:

   • C1 (terminal, less substituted — attached to one H)
   • C2 (internal, more substituted — attached to an ethyl group) So the $OH$ goes to C2, and the H goes to C1. This gives an enol:

3. Keto-enol tautomerism That enol is unstable. It tautomerises: the $OH$ hydrogen shifts to the terminal carbon, and the double bond moves to become a $C=O$ at C2. The result is:

That’s butan-2-one (also called methyl ethyl ketone).

4. Check the options
   • (i) Butanal — that would come from anti-Markovnikov addition, not happening here.
   • (ii) Butan-2-one — matches our product.
   • (iii) A hydroxy-aldehyde — not formed; tautomerism gives a ketone, not an aldehyde.
   • (iv) Butan-2-ol — that’s an alcohol, not a carbonyl; hydration of alkynes gives carbonyls, not alcohols.

Method: Hydration of Alkynes (Markovnikov Addition via Enol–Keto Tautomerism)

Concept Summary

Addition of water to an alkyne in acidic medium with $\ce{Hg^{2+}}$ catalyst follows Markovnikov’s rule — the $\ce{-OH}$ group attaches to the more substituted carbon of the triple bond. The initial product is an enol, which rapidly tautomerizes to the more stable keto form.

Step-by-Step Reasoning

1. Identify the substrate

   But-1-yne: $\ce{CH3-CH2-C#CH}$

   Triple bond is between C1 and C2.

2. Apply Markovnikov addition

   Water adds such that $\ce{-OH}$ goes to the more substituted carbon of the triple bond.

   • C1 (terminal) is less substituted (1 H).
   • C2 (internal) is more substituted (0 H). So $\ce{-OH}$ attaches to C2, and $\ce{-H}$ attaches to C1.

3. Write the enol formed

   After addition:

   $\ce{CH3-CH2-C(OH)=CH2}$

   This is an enol (alkene + alcohol).

4. Tautomerization

   The enol is unstable and rearranges to a ketone:

   $\ce{CH3-CH2-C(OH)=CH2 -> CH3-CH2-CO-CH3}$

   This is butan-2-one.

Final Answer

✓ Option (B): $\ce{CH3-CH2-CO-CH3}$ (butan-2-one)

Key exam point: Terminal alkynes give methyl ketones upon hydration, not aldehydes. Option (A) would require anti-Markovnikov addition, which does not occur under these conditions.

🧪 The Reaction Context

Reaction: Hydration of but-1-yne

Conditions: Acidic medium, $\ce{Hg^{2+}}$ catalyst (oxymercuration)

Key rule: Follows Markovnikov’s rule — the $\ce{-OH}$ ends up on the more substituted carbon of the double bond (after tautomerization).

But-1-yne:

$\ce{CH3-CH2-C#CH}$

✗ Mistake 1: Forgetting that terminal alkynes give methyl ketones

Many students think the product is an aldehyde (option A).

Why they slip: They remember that hydration of ethene gives ethanol, so they assume a similar pattern.

Correct logic:

• Water adds across the triple bond.
• The $\ce{-OH}$ goes to the more substituted carbon (Markovnikov).
• The enol formed tautomerizes to a ketone, not an aldehyde.

Result:

$\ce{CH3-CH2-C#CH ->[H2O/Hg^{2+}] CH3-CH2-C(OH)=CH2 ->[tautomerize] CH3-CH2-CO-CH3}$

✓ Option B: butan-2-one

How to avoid:

• Memorise: Terminal alkyne → methyl ketone (except with special reagents like disiamylborane).
• Always check: if the triple bond is at the end, the carbonyl ends up at carbon-2.

✗ Mistake 2: Confusing hydration of alkynes with hydration of alkenes

Some students pick option D (butan-2-ol), thinking it’s like alkene hydration.

Why they slip: They remember that alkenes give alcohols, so they assume alkynes do too.

Correct logic:

• Alkynes first form an enol, which is unstable.
• Enol immediately tautomerizes to a carbonyl compound (ketone or aldehyde).
• You never get an alcohol as the final product under these conditions.

How to avoid:

• Draw the enol intermediate every time.
• Remember: enol → keto tautomerism is spontaneous under acidic conditions.

✗ Mistake 3: Misapplying anti-Markovnikov addition

Some students pick option A (butanal), thinking water adds anti-Markovnikov.

Why they slip: They confuse this with hydroboration-oxidation (which gives anti-Markovnikov products).

Correct logic:

• $\ce{Hg^{2+}}$ catalyzed hydration is Markovnikov.
• Anti-Markovnikov requires different reagents (e.g., $\ce{B2H6}$ then $\ce{H2O2/OH-}$).

How to avoid:

• Make a mental checklist:
  • $\ce{Hg^{2+}/H+}$ → Markovnikov → ketone
  • $\ce{B2H6}$ then $\ce{H2O2/OH-}$ → anti-Markovnikov → aldehyde (for terminal alkynes)

✗ Mistake 4: Not checking the carbon skeleton

Option C ($\ce{CH3-CH(OH)-CH2-CHO}$) has a 4-carbon chain with both an alcohol and an aldehyde — that’s not possible from a simple hydration of but-1-yne.

Why they slip: They try to “force” a product without counting carbons or functional groups.

Correct logic:

• But-1-yne has 4 carbons, one triple bond.
• Hydration adds one water molecule — you get one carbonyl group, not two.
• Option C has two oxygen-containing functional groups — impossible here.

How to avoid:

• Count carbons and oxygens in the product.
• If the reactant has one triple bond, the product has one carbonyl (unless further reaction occurs).

✓ Final Answer

Correct option: (B) butan-2-one

$\ce{CH3-CH2-CO-CH3}$

📝 Quick Revision Table

Final tip: In IUPAC nomenclature + reaction questions, always draw the full mechanism step-by-step — even if rough. That single habit eliminates 90% of these errors.