[Aznhmonglor]

A sample of E-4,5-Dimethyl-4-octene reacts with CH3CO3H to produce "X" as the highest MW Organic product.

This question is so confusing, stereochemistry matters but I dont even know how to start it. Help would be appreciated.

[azmanam]

Have you seen things with the formula RCO₃H before?  What kind of reagent is that?  What types of reactions does it do?

Or to start from the other side: what types of functional groups are present in the starting material?  What sorts of reactions do you know for those types of functional groups?

[Aznhmonglor]

So confusing lol...I know its a nucleophilic reaction...Has to be an SN2 reaction also....The alkene in the starting material must be converted into an alkane and the CH3CO3H must attack the starting material but I dont know where it should go....

[azmanam]

I find if I'm confused, then focus on a different part of the problem. 

Let's refocus on the CH₃CO₃H.  Hit the Googles and interwebs (or your textbook and lecture notes) to find out some information about this reagent.  It's key to unlocking this question.

(not all "nucleophilic reactions" are S_N2)

[Aznhmonglor]

Okay so I find that CH3CO3H looks like this



Can the alcohol group lose the H and attach to the starting material that way?

[azmanam]

Good.  That's the right structure.  What's its name?  What kind of molecule is it?  What 'class' does it belong to?  What reactions does that class of molecules like to do?

Have you seen molecules like this in your book or lecture notes before?

(it's not an alcohol group in this case)

[Aznhmonglor]

Oh wow you made me realize its a peracetic acid...lol I remember learning about it....but that was with another set of reactions and not the ones we're learning now which is either SN1 or SN2.

[azmanam]

Right.  This is not an S_N1 or S_N2 reaction.  Gotta keep those old reactions handy in the toolbox.  At this point in the semester professors like bringing back old reactions and stringing them together with new reactions to make more complex compounds over a series of steps.  I suspect this might be the first step in a multistep synthesis?

So do you know what's going on here?  What's your highest MW product?

[Aznhmonglor]

OMG are you my professor?!!!!! you speak as if you are, HOLY!...LOL I find this so amazing...you ask the right questions at the right times!....I swear you're like my guardian to this class...LOL!....yes it wants the highest MW product...now I looked through my notes and what peracetic acid does is that the oxygen with the hydrogen attacks the double bond of the starting material. What this does is then transfer one pair of the lone pairs to the other oxygen conneted to it. This leaves the starting material with an OH group connected to the starting material. Correct?

[azmanam]

Not sure I'm following what you're saying, but I don't think it gets to the right product.  What class of molecule are we making in the end?  It's not an alcohol.

Peracids are 'code reagents' for this level of organic.  Whenever you see a peracid (mCPBA, peracetic acid, or anything than ends -CO₃H, you should immediately think this class of product.  For this level of organic chemistry, peracids do one and only one transformation.  Alkenes to... ?

[Aznhmonglor]

Epoxide?! Please tell me thats right!

[azmanam]

yes.  good work.

now, as you said stereochemistry is important here.  This is a stereospecific reaction and you will get a specific orientation in your product as a result.  Make sure you account for that when you draw the product.

for reference, here's what we're talking about.  Click on the "change equation" button.  i don't like the first mechanism they propose.

http://www.cem.msu.edu/~reusch/VirtTxtJml/addene2.htm#add4c

[Aznhmonglor]

Okay so this is what I have come to



I didn't quite understand if I should take out the methyl groups, where they a part of the epoxide?, did I get my stereochemistry correct?

[azmanam]

you have drawn one enantiomer of the product.  The product will be formed in a racemic mixture.  You can actually keep the methyl groups in the plane of the board.  But they definitely have to stay in.

[Aznhmonglor]

Okay this is good...now you said that I drew one of the enantiomer....what could the other one be? cis and trans of the epoxide oxygen?



Dont think this is possible?