In the following reaction A is:

• Step 1: Free Radical Substitution

  When cyclohexane is treated with bromine in the presence of light ($$\text{Br}_2/h\nu$$), it undergoes photochemical free-radical bromination to form bromocyclohexane:

  $$\text{Cyclohexane} \xrightarrow{\text{Br}_2/h\nu} \text{Bromocyclohexane}$$



• Step 2: Dehydrohalogenation (Formation of A)

  When bromocyclohexane is treated with a strong, hot base like alcoholic potassium hydroxide ($$\text{KOH (alc.)}$$), it undergoes an $$\text{E}2$$ elimination reaction. The base abstracts a adjacent proton along with the loss of the bromide leaving group to introduce a carbon-carbon double bond:

  $$\text{Bromocyclohexane} \xrightarrow{\text{KOH (alc.)}} \text{Cyclohexene (A)}$$

Validation via Subsequent Steps:

We can confirm that A is cyclohexene by tracking the remaining reactions shown in the diagram:

1. Reductive Ozonolysis: Treating cyclohexene with $$\text{O}_3$$ followed by $$\text{Me}_2\text{S}$$ oxidatively cleaves the double bond down into a dicarbonyl compound, hexanedial (adipaldehyde).
2. Intramolecular Aldol Condensation: Heating hexanedial with aqueous base ($$\text{NaOH (aq.), }\Delta$$) triggers a ring-closing aldol pathway, yielding the final 5-membered cyclic conjugated product, cyclopent-1-ene-1-carbaldehyde.

Answer: Cyclohexane