This question is about the alkenes and their reactions. Fig. 15.1 shows the displayed formula of a propene molecule. Fig. 15.1 (a) State two ways in which t...

This question covers how alkenes differ from alkanes, testing propene, drawing its reaction product with bromine, a cracking equation, boiling-point trends and polymerisation.

(a) Two ways alkenes differ from alkanes, any two of [2]: alkenes are unsaturated (contain a C=C double bond) whereas alkanes are saturated; alkenes have the general formula C_nH_2n whereas alkanes are C_nH_2n+2; alkenes are more reactive and decolourise bromine water while alkanes do not.

(b) Propene has three carbons, so its molecular formula is C₃H₆ [1], and the general formula of the alkenes is C_nH_2n [1].

(c) To distinguish propene from propane [3]: add bromine water to each [1]; with propene the bromine water turns from orange to colourless (decolourised) [1]; with propane there is no change (stays orange) [1]. Only the unsaturated propene reacts.

(d) Propene adds bromine across its double bond to give 1,2-dibromopropane; the type of reaction is addition [1]. The displayed formula shows three carbons joined by single bonds, a bromine atom on each of the first two carbons, and enough hydrogen atoms so every carbon has four bonds [2]:

(e) The equation must balance. C₉H₂₀ gives two C₃H₆ (6 C, 12 H), leaving 9 - 6 = 3 carbons and 20 - 12 = 8 hydrogens, so the missing product is C₃H₈ (propane) [1].

(f)(i) The boiling point increases as the number of carbon atoms increases [1], because larger molecules have stronger intermolecular forces.

(f)(ii) Pent-1-ene is a liquid at 20 °C [1], because it is a larger molecule than but-1-ene with even stronger intermolecular forces, so its boiling point is above 20 °C [1]. (The pattern from -104, -47, -6 °C is rising past room temperature.)

(g) The monomer is propene [1]; during polymerisation the carbon-carbon double bonds open [1] and the molecules join together to form one long chain molecule [1].