Concept:
We must identify compound A using all clues:
- Organic liquid containing C, H, O
- p$K_a = 15.9$
- Used in polishing woods
- On heating with conc. H$_2$SO$_4$ gives gaseous B
- B has empirical formula CH$_2$
- B decolourises bromine water and KMnO$_4$
These clues indicate alcohol dehydration producing an alkene.
Step 1: Identify compound A from p$K_a$.
A p$K_a$ around $15.9$ is characteristic of:
$$\text{ethanol}$$
because alcohols are weak acids with p$K_a \approx 16$.
Also ethanol is commonly used in:
- polishes
- varnishes
- solvents
So A is likely ethanol.
Step 2: Reaction with concentrated sulphuric acid.
When ethanol is heated with conc. H$_2$SO$_4$ at higher temperature, dehydration occurs:
$$CH_3CH_2OH \xrightarrow[\Delta]{conc.\ H_2SO_4} CH_2=CH_2 + H_2O$$
So gaseous product B is ethene.
Step 3: Check empirical formula of B.
Ethene molecular formula:
$$C_2H_4$$
Empirical formula:
$$CH_2$$
Matches the question.
Step 4: Check test with bromine water and KMnO$_4$.
Ethene contains double bond.
Therefore it:
- Decolourises bromine water
- Decolourises dilute alkaline KMnO$_4$ (Baeyer's test)
So this clue also confirms alkene.
Step 5: Reject other options.
- Methane / ethane do not decolourise bromine water easily.
- Phenol does not dehydrate to ethanol.
- Methanol gives different products.
Step 6: Final answer.
Thus:
$$A=\text{ethanol}, \qquad B=\text{ethene}$$
Hence correct option is:
$$\boxed{\text{(D)}}$$
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