[fobbz]

I am currently stuck on two questions. In my orgo class we have started to learn about reaction mechanisms in Sn2 and Sn1 reactions. I am a little unclear as to how to start number 1, but I beleive 3 is right I just want to make sure I am understanding.

Question 3

I drew the enantiomer of C5H11Br, assigning Br as priority 1 , CH2CH2CH3 as priority 2, CH3 as priority 3 and H as priority 4 (outward from page). This would be R confromation, but because H is not going into the page, it is S.

The product has (CH₂)₃CH₃ as priority 1, C≡H as priority 2, CH₃ as priority 3 and finally H as 4. Here H is going into page and the conformation is S.


Question 1

I am really unsure how to begin this question. I am confused as to what the compound is reacting with, H₃O⁺ or CH₃OH. Or is it reacting with H₂O?

Not I am sorry for the bad lighting.

[Altered State]

I think Q3 is right if it undergoes Sn2 as you say, but I'm not totally sure as long as we do not know disolvent and other conditions.

Q1 definetly does not occur by Sn2 reaction. You've got a tertiaty alcohol.
Hint, -OH in acid media will be protonated and can leave the molecule as water.
The two different products formed may help you aswell: how many sustitution products would we expect in a Sn2 reaction? Why is that?

[fobbz]

In Sn2 we would expect only one right?... The inversion at the substitution... So it has to be SN1. So would the CH3OH act as the nucleophile? Then CH3 will attack and OH will leave and protonate to water? In this case H30+ would bond with OH- and get water?

What is the significance of the top and bottom compounds at the reaction arrow? I cannot find in my notes or in my text what they mean! I am sure I have seen it before though.

[Altered State]

About your last question, I don't really think that a compound role in the reaction changes for being in the bottom or top of the arrow, it's just a notation of compounds or conditions used in the reaction. But I'm also a student, please correct me.

[Altered State]

Correct, it will be Sn1.
Check out Sn1 mechanism for tertiary alcohols in acid media in any orgchem book.

Hints:
Sn1 occurs in 2 steps, first the leaving group leave the molecule and you get a carbocation.
Then a nucleophile can attack that possitively charged carbon. Also, for how many ways can the nuc
 attack? That should explain the 2 products.

Remember also acid media, something in the molecule may be protonated by its action.

[fobbz]

Okay so I couldn't find the tertiary alchohol reaction in acid media in my text, but I found this page: http://www.mhhe.com/physsci/chemistry/carey5e/Ch08/ch8-2.html

I tried to apply the same concepts to the SN1 mechanism for reaction of alcohols with HBr, for question 1.

Sorry again for the bad lighting. Can you perhaps point me to the text you are using?

[Altered State]

That's right, thats the Sn1 mechansim. A group leaves the molecule (slow sten, as you've written), so you get a carbocation that can be attacked by a nucleophile, left-way or right-way, giving a racemic mixture.

Sn1 general mechanism must be in any given general organic chemistry book.
Is there anything that you do not understeand?