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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: 33seven on November 09, 2010, 04:21:23 PM

Title: Free Radical Bromination of Ethane question, need assistance, lol wtf?
Post by: 33seven on November 09, 2010, 04:21:23 PM

Ok, question from my organic exam, copied word for word w/ my attempt:

1. During the free radical bromination or ethane, which of the following reactions has the lowest collision frequency? (i.e. Which reaction is the least likely to happen?)

a) CH₃CH₂. +Br₂ :rarrow: CH₃CH₂Br + Br.
b) Br. + CH₃CH₃ :rarrow: HBr + CH₃CH₂.
c) Br. + Br₂  :rarrow: Br₂ + Br.
d) CH₃CH₂. + Br. :rarrow: CH₃CH₂Br
e) CH₃CH₂. + CH₃CH₃ :rarrow: CH₃CH₃ + CH₃CH₂.

So we're given a sheet with Bond-dissociation enthalpies for homolytic bond cleavage(relevant energies):
Br-Br     46 Kcal/mol
CH₃CH₂-Br     68
CH₃CH₂-H     98
H-Br     88
    


My attempt:
a) 46 - 68 = -22
b) 98 - 88 = +10
c) 46 - 46 = 0
d) 0 - 68 = -68
e) 98- 98 = 0

So, obvious answer is B?

Title: Re: Free Radical Bromination of Ethane question, need assistance, lol wtf?
Post by: 33seven on November 15, 2010, 02:03:17 PM

Nevermind, i figured it out on my own. The question had little or nothing to do with the bond energies, b/c 4 of the 5 choices are propagation steps and only 1 is a termination step. B/c the amount of free radicals available to react is significantly lower in a termination step to a free radical reaction, the probability that a termination reaction will occur is much less than the probability that a propagation reaction will occur. So obvious answer D. You guys were so helpful!! Thanks, I couldn't have done it without your h elp!! What a joke.