Hex-4-ene-2-ol on treatment with PCC gives 'A'. 'A' on reaction with sodium hypoiodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is

We need to follow the sequence of reactions starting from hex-4-en-2-ol, which has the structure: $$CH_3-CH(OH)-CH_2-CH=CH-CH_3$$.

PCC (Pyridinium Chlorochromate) is a mild oxidizing agent that converts a secondary alcohol to a ketone without further oxidation. $$CH_3-CH(OH)-CH_2-CH=CH-CH_3 \xrightarrow{PCC} CH_3-CO-CH_2-CH=CH-CH_3$$ Compound A is hex-4-en-2-one.

Sodium hypoiodite is the iodoform reagent. It reacts with methyl ketones ($$R-CO-CH_3$$) to give a carboxylate salt and iodoform ($$CHI_3$$): $$CH_3-CO-CH_2-CH=CH-CH_3 + 3NaOI \to CHI_3 + CH_3-CH=CH-CH_2-COONa$$. Compound B is sodium pent-3-enoate ($$CH_3CH=CHCH_2COONa$$).

Heating sodium pent-3-enoate with soda lime ($$NaOH + CaO$$) causes decarboxylation, removing $$CO_2$$ and replacing the $$-COONa$$ group with $$-H$$: $$CH_3-CH=CH-CH_2-COONa \xrightarrow{NaOH/CaO} CH_3-CH=CH-CH_3 + Na_2CO_3$$. Compound C is 2-butene ($$CH_3CH=CHCH_3$$).

The correct answer is Option C: 2-butene.