[susdujcrd]

who's going to tend more to E1 and why?

[discodermolide]

who's going to tend more to E1 and why?

You got any ideas?

[susdujcrd]

well, the only difference is the leaving group. I know iodine is a better leaving
group than chlorine, but a good leaving group affects both E1 and E2, so that's pretty much it.

[discodermolide]

well, the only difference is the leaving group. I know iodine is a better leaving
group than chlorine, but a good leaving group affects both E1 and E2, so that's pretty much it.

What sort of reaction conditions would you expect?

[susdujcrd]

for E1? a weak base and a protic polar solvent, but I know nothing on both.

[discodermolide]

for E1? a weak base and a protic polar solvent, but I know nothing on both.

Read this page;
http://en.wikipedia.org/wiki/Elimination_reaction

[susdujcrd]

read it.
I believe 2-iodobutane will tend more to E1 because iodine will leave "faster"
making the molecule more favorable of E1 than that with chlorine.

[discodermolide]

read it.
I believe 2-iodobutane will tend more to E1 because iodine will leave "faster"
making the molecule more favorable of E1 than that with chlorine.

That seems reasonable. Remember that the mechanism will depend on the base and the solvent used as well.

[PhDoc]

In regards to the E1 vs. E2 of chloro- vs. iodobutane, there is more to the outcome than just the leaving group. It's important to consider the bond dissociation energies. The results from the numbers reflect common sense insofar that the C-I bond is weaker than the C-Cl bond. Please consult your textbook or the following link for information:

http://www.colorado.edu/chem/ellison/papers/Blanksby_Acct_Chem_Res_2003.pdf

Just as important is the base/nucleophile and solvent. When in a dipolar aprotic solvent such as DMF or DMSO, the base/nucleophile reacts with the electrophile as if it were in the gas phase. This occurs from a solvent cage forming preferentially around the metal cation of the ion pair, resulting in an exposed anion being left to react freely without solvation. If the reaction is conducted in a polar protic solvent such as ethanol, then the ability of the solvent to aid leaving group departure through stabilization of the incipient carbocation becomes a predominant factor. Additionally, the polar protic solvent reduces the activity of the base/nucleophile through solvation.

If the base/nucleophile is strong, you will get an E2 regardless. This is especially so in the case of a tert-butoxide/solvent when heated. If the base/nucleophile is weak, you will get SN1/E1 in the polar protic solvent, and SN2 in the dipolar aprotic solvent.

Far too often students of organic chemistry fall into the trap of considering only one aspect of the reaction. This results in frustration from not understanding the consequences of "tunnel vision." SN1, SN2, E1, E2 will continue to be the student's introduction to mechanistic organic chemistry because it requires seeing the "big picture" for mastery of the subject.

These articles should help you with your study of the subject matter.

SN1
http://lennoxtutoring.com/2011/10/01/the-sn1-reaction/

SN2
http://lennoxtutoring.com/2011/09/30/the-sn2-reaction/

E1
http://lennoxtutoring.com/2011/10/04/the-e1-reaction/

E2
http://lennoxtutoring.com/2011/10/03/the-e2-reaction/

Basicity vs. Nucleophilicity
http://lennoxtutoring.com/2011/09/30/basicity-vs-nucleophilicity/

Good luck!

Kindest Regards,
Dr. Lennox

[PhDoc]

Here's an interesting problem for you. What happens when you react 2-bromo-1,1,3,4,4-pentamethyl-cyclohexane with KOt-Bu in HOt-Bu at 55C? In the case of bromobutane, E2 would be favored. What mechanism is favored here? What product is favored here?