Q12 V3A compound underwent two tests.In the first test, 1 mole of the compound was completely combusted, yielding 3 moles of CO2.In the second test, the compound showed a rapid reaction with bromine. The formula of the compound is likely to be:

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The problem presents a compound that, on complete combustion of 1 mole, produces 3 moles of CO2, and it also shows a rapid reaction with bromine. We will evaluate each option against these facts.
Option a: C4H10. If a C4H10 alkane combusts completely, the typical CO2 yield would be 4 moles of CO2 per mole of fuel (C4H10 + 6.5 O2 → 4 CO2 + 5 H2O). This would give 4 moles of CO2, not 3. Since the observed CO2 is 3, this option is inconsistent with the data.
Option b: C3H8. Complete combustion would produce 3 moles of CO2 (C3H8 + 5 O2 → 3 CO2 + 4 H2O), which aligns with the CO2 count. However, this compound is an alkane and typically does not react rapidly with bromine under normal conditions; alkanes require radical initiation or UV light for bromine substitution/halogenation to proceed rapidly. The rapid bromine reaction is a key clue that this is not an alkane, so this option is unlikely given the second test.
Option c: C2H4. Ethene (C2H4) would yield 2 moles of CO2 on complete combustion (C2H4 + 3 O2 → 2 CO2 + 2 H2O), which does not match the observed 3 moles of CO2. Therefore this option is incorrect.
Option d: C3H6. Propene (C3H6) combusts to produce 3 moles of CO2 (C3H6 + 4.5 O2 → 3 CO2 + 3 H2O), matching the given 3 CO2. Additionally, propene contains a C=C double bond, so it should react rapidly with bromine (Br2) in the characteristic addition reaction across the double bond. This aligns with the second test outcome. Therefore this option fits both observations.

Q12 V3A compound underwent two tests.In the first test, 1 mole of the compound was completely combusted, yielding 3 moles of CO2.In the second test, the compound showed a rapid reaction with bromine. The formula of the compound is likely to be:

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