Destroyer Ochem #50.

[CANgnome]

Advertisement - Members don't see this ad

I am wondering why e) does not form an alkene.

2-iodo-2-methyl-propane ----methanol (aq)----> does not produce alkene, only the SN1 product.

Polar protic solvent.
weak nucleophile/base

commonly you would see both SN1 and E1 reactions competing in such an environment. On chad's notes there is also the EXACT same example written down on the E1 reactons section showing the formation of an alkene. The leaving group is good, also, as it is an iodine.

Is this an error by destroyer or what?

 

[UndergradGuy7]

Sn1 because methanol is more nucleophilic than basic?

Chad does reaction of a tertiary alkane + CH3OH and says it does Sn1...
It would favor E1 more if it said you add heat.

 

[CANgnome]

yes.... It would favor E1 if heat was added. But this doesn't say. I want to put into my mind that if there is no heat, ordinarily SN1 would be favored, but do not want to commit to something like that unless i get firm proof 😕

 

[jirotrom]

i did this just last night... "e" is solvolysis... Sn1 ... i believe this was stated in chad's if im correct.

just remember if the solvent also acts as the nucleophile its sN1, unless there is heat.

 

[CANgnome]

in question 50, it is not solvolysis. The ethanol is in an aqueous solution, so the solvent is water.

 

[jirotrom]

CANgnome... trust me, I know how you feel... I spent a few hours debating Sn1 and E1 with friends and came to the conclusion that its 50/50 unless there is heat. but if I see ethanol im picking Sn1 until I see a rxn that shows otherwise