I have this.



The red writing is the professor's. So I know the answer is correct.

The problem I'm having is I can't figure out why the hell H+ is part of the answer.

When I looked at it myself I jumped to EtOH as the thing in the middle.

The H would get deprotonated by the pi bond, then EtO- would attack the carbocation and bam you did it.

So why is the H+ part of it? I can't figure out what I'm missing

its been like 8 years for me, but hopefully this is correct:

you will get an intermediate form in which the H+ will bind with the terminal carbon (right carbon of the double bond), leaving a slightly positive charge at the left carbon of the double bond to which the oxygen of ethanol binds. When that bond forms, a hydrogen bound to the oxygen molecule will then fall off.

https://en.wikipedia.org/wiki/Hydration_reaction

similar to how the reaction of hydration of 1-methylcyclohexanol proceeds, but using proton and ethanol instead of h2o / h3o+


The problem with your thinking is the resulting EtO- is very unstable and highly basic, so it wouldn't want to give up the hydrogen atom to form in the first place

This post was edited by cialda on Nov 3 2017 06:40pm

I concur in that it is required to form an intermediate. If this has not been part of your course, it may be worthwhile to ask fellow students or a professor.

the H+ implies you are under acidic conditions, I'll write out the mechanism for you in like 10-15 minutes

Before the reaction occurs, you are under acidic conditions. This implies the ethanol will be protonated ahead of time.

1. Double bond abstracts proton from the protonated ethanol, which leaves a secondary carbocation behind (remember your carbocation stability rules and this will explain why the hydrogen abstracted from the protonated ethanol migrates to the primary carbon (least substituted carbon).
2. Nucleophilic attack of the carbocation by ethanol. The electrons here attack the carbocation forming a bond and positively charged oxygen. The charge on the oxygen needs to be dealt with in step 3.
3. Deprotonation of the oxygen. This can occur by several species. This reaction likely occurs in aqueous phase, so the proton could be removed by hydroxyl ions from the autoionization of water, the conjugate base of the acid used to protonate ethanol, or another ethanol (I used another ethanol in my example).
4. Final product

Edit: A large majority of the time, you will see this reaction using Sulfuric acid (H2SO4) and Ethanol. I can draw this out if you want just let me know.

This post was edited by Zpot on Nov 10 2017 07:48pm

acid is a catalyst.

Tre first step is protonation of the alkene, which is easier using a strong acid than ethanol, a very weak acid.