C341/Spring 2010
Chapter 8 Discussion Worksheet
1. For the following hydrocarbon, NAME each type of proton (e.g. 2o alkyl, vinylic, etc.) in the spaces provided below. Secondly, put the protons in order of their ease of abstraction/removal (i.e. lowest bond dissociation energy)? Ha
Ha = 2o alkyl Hb = 2o allyl H3C
Hb Hd = 3o allyl Hc = vinyl Hd
Hc Ease of abstraction: Hd > Hb > Ha > Hc Bond energy: Hc > Ha > Hb > Hd 2. The following reactions are in a random order. Using the spaces provided to the right, identify the following steps as either initiation(I), propagation(P), or termination (T) steps in the following free radical mechanism. Br
A.
Br
Br
Br
H
Br
Br
D.
h
Br
E.
HBr
Br
C.
Br
Br
Br
Br
Br
P
T
Br
B.
Br
P
T
I
or heat
T
F.
3. Using propane and Br2, name and demonstrate the three steps in a radical halogenation mechanism below using appropriate arrows. Remember to use half‐arrows for free radical reactions. Page 1 of 2
C341/Spring 2010
Chapter 8 Discussion Worksheet
4. Arrange the following radical intermediates in order of stability. Ph = phenyl and is just a benzene ring. C > D = A > B > E 5. Provide the major product(s) for the reaction conditions provided below (showing stereochemistry if necessary). 6. Draw all the expected products from the reaction of 2‐methyl‐2‐pentene with Br2, and indicate the relative amounts of products that you might expect to be formed using >, < and = signs. This is a difficult question to provide a definite MAJOR product for but discuss which two might be the most stable and why. Discuss if the methyl protons substitute at different rates and why. B > A > C > E > D (trans > cis and because they are allylic methyls) > F Compounds A and B might be equal in stability and show equal product distributions (see pg. 307 for further explanation on this). Product A is tertiary allylic, but it has a less stable double bond than B. Product B is secondary, but it has a more stable double bond. It would be hard to make a call here which one is truly more stable. If I had to make a call, I guess I would say B is more stable, but I would not expect you to know the difference although I would expect you to be able to have an opinion and explain why these might be close. Compounds B and C are chiral, and therefore, have enantiomers or equal stability, respectively. Compound B is more stable because its double bond is more substituted than C’s even though they were both formed from secondary radicals. Compounds D and E are both primary allylic structures, but D is more stable because it is trans > cis (E). Finally, compound F is least stable because it is primary and most likely would not be formed at all although I expect you to be able to have skill in drawing all products. Page 2 of 2