Ochem E2 or E1 reaction?

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bchang57

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2-iodobutane reacts with NaOCH3 with CH3OH as the solvent

(this is the iodo group, not a carbon chain)
I
| NaOCH3 C-C=C-C
/\/ CH3OH


according to destroyer this is an E2 reaction (with some competition from SN2) but i'm having trouble seeing this.

My thinking is:
I is a good leaving group, NaOCH3 is a strong base and strong, non-bulky nucleophile, but because CH3OH is a protic solvent, this would lean toward E1?
 
Its def E2. You have a secondary halide in a protic solvent with a strong nuc./base. Remember, E2 and SN2 favor aprotic solvent's, but they can also occur in protic solvent's as well. If something is in an aprotic solvent, you can immediately eliminate SN1 and E1, but this is not the case with protic solvent's.
 
ohh thank you! that 2nd sentence made me expand my thinking and clear things up =)
 
Its def E2. You have a secondary halide in a protic solvent with a strong nuc./base. Remember, E2 and SN2 favor aprotic solvent's, but they can also occur in protic solvent's as well. If something is in an aprotic solvent, you can immediately eliminate SN1 and E1, but this is not the case with protic solvent's.

What he said. Totally true.

E1, SN1 = Protic ONLY
........meaning if you have a aprotic solvent (ie. DMSO, acetone ethers, etc.) it is automatically E2 or SN2.

E2, SN2 = FAVOR Aprotic (but can still undergo e2 or sn2 even with Protic solvents).
 
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