Question: The hydrocarbon propene (CH3-CH=CH2) can react in two different ways with bromine (Chapters 12 and 14). (i) (ii) (a) Using the bond strengths (kcal mol-1)

The hydrocarbon propene (CH3-CH=CH2) can react in two different ways with bromine (Chapters 12 and 14).
(i)
The hydrocarbon propene (CH3-CH=CH2) can react in two different ways

(ii)

The hydrocarbon propene (CH3-CH=CH2) can react in two different ways

(a) Using the bond strengths (kcal mol-1) given in the margin, calculate –³H° for each of these reactions, (b) –³So ‰ˆ 0 cal K-l mol-1 for one of these reactions and -35 cal K-l mol-1 for the other. Which reaction has which –³So? Briefly explain your answer, (c) Calculate –³G° for each reaction at 25°C and at 600°C. Are both of these reactions thermodynamically favorable at 25°C? At 600°C?

Br B CH,-CH-CH2 + Br2 CH,-CH-CH, + HBr Br

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a Propene reacts with bromine molecule to give 12dibromopropane Formation of 12dibromopropane by the cleavage of CC double bond and BrBr bond and form... View full answer

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