Q.896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation of 'A' gives 2-methylpentane. Also 'A' on hydration in the presence of H2SO4 and HgSO4 gives a ketone 'B' having molecular formula C6H12O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reactions involved.

The key idea is to first determine the molecular formula of the hydrocarbon using the vapour density data and percentage composition, then use the chemical reactions to deduce the structure.

Step 1: Find the molecular formula of A.

At STP, 22400 mL of vapour weighs $\frac{3.28}{896} \times 22400 = 82 \, \text{g}$, so molar mass = 82 g/mol.

Carbon: $\frac{87.80}{12} = 7.32$, Hydrogen: $\frac{12.19}{1} = 12.19$. Ratio $\frac{7.32}{7.32} : \frac{12.19}{7.32} \approx 1 : 1.67 \approx 3 : 5$, so empirical formula = $\text{C}_3\text{H}_5$ (mass 41).

$n = \frac{82}{41} = 2$, hence molecular formula = $\text{C}_6\text{H}_{10}$.

Step 2: Use hydrogenation result.

Hydrogenation of A gives 2-methylpentane, so the carbon skeleton of A is the same: a straight chain of 5 carbons with a methyl branch at C-2. A has two degrees of unsaturation (from $\text{C}_6\text{H}_{10}$ vs $\text{C}_6\text{H}_{14}$).

Step 3: Use hydration and iodoform test. …

1. Finding the molecular formula of A

First, the vapour data: 896 mL at STP corresponds to

$\frac{896}{22400} = 0.04$ moles.

Mass of this sample = 3.28 g, so the molar mass is

$M = \frac{3.28}{0.04} = 82\ \text{g mol}^{-1}$.

Now the percentage composition:

Carbon: 87.80 % → in 100 g, mass of C = 87.80 g → moles of C = $\frac{87.80}{12} = 7.317$

Hydrogen: 12.19 % → in 100 g, mass of H = 12.19 g → moles of H = $\frac{12.19}{1} = 12.19$

Divide by the smaller number (7.317) to get the simplest ratio:

C : H = $1 : 1.666$ → multiply by 3 → C₃H₅ as the empirical formula.

Empirical formula mass = $3 \times 12 + 5 \times 1 = 41$ g mol⁻¹.

Since molar mass = 82 g mol⁻¹, the molecular formula is twice the empirical: C₆H₁₀.

2. Hydrogenation gives 2‑methylpentane

Hydrogenation of A with H₂/Ni (or Pd/C) adds H₂ across all multiple bonds, yielding a saturated alkane. The product is 2‑methylpentane, whose carbon skeleton is:

   C
   |
C—C—C—C

That is, a five‑carbon straight chain with a methyl branch on carbon 2. So the carbon skeleton of A must be exactly this: 2‑methylpentane skeleton.

Since A has two degrees of unsaturation and the skeleton is fixed, A must be an alkyne (one triple bond) or a diene. The hydration reaction will decide.

3. Hydration gives a ketone B (C₆H₁₂O) that gives a positive iodoform test

Hydration of an alkyne in the presence of H₂SO₄ and HgSO₄ follows Markovnikov’s rule: the OH ends up on the more substituted carbon of the triple bond, and the product tautomerises to a carbonyl compound.

• If the triple bond is terminal (‑C≡CH), hydration gives a methyl ketone (‑COCH₃).
• If the triple bond is internal, hydration gives a mixture or a different ketone.

The ketone B has formula C₆H₁₂O (one oxygen, so it’s a saturated ketone). A positive iodoform test means B contains a CH₃‑CO‑ group (methyl ketone). Therefore, A must have a terminal triple bond.

So A is a terminal alkyne with the 2‑methylpentane skeleton. The only place to put a terminal triple bond is at the end of the chain: