Problem+1

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 * HERE IS A SOLUTION TO PROBLEM 1**:



Before tackling this problem, let us give it some level of difficulty DIFFICULTY: medium, 4/10 LEVEL: third semester of undergraduate org chem. First year graduate student.

__ SOLUTION & DISCUSSION __

Let's take this step by step to obtain the most from this reaction. 1) Watch for differences between starting material (SM) & product (P). These are: a.- The oxa ring disappears but the rest of the Carbon backbone remains in place b.- There are two conjugated C=C bonds in P, which may have been involved in the C-O bond breakage, suggesting an elimination reaction. (Look for an acidic proton somewhere). This suggests that a base substracts this proton. This base is likely MeLi, indeed a powerful base. This particular acidic proton must be on the chloromethylene section. c.- The C=O group becomes a tert alcohol, suggesting the addition of a methyl group. This ought to come from MeLi, indeed also a powerful alkylating agent. 2) After this superficial analysis, it is clear that MeLi is acting as a base and as a methylating agent. However, MeLi does not attack the CH2Cl group in an SN reaction, but limits itself to abstract the methylene proton. 3) After the gamma-elmination step where the C-O-C bridge gives way, and the C=O alkylation (formally an addition reaction of Me-Li), we end up with a doucle Li alkoxide in a highly basic medium. So we need to add protons to render everything neutral, and this is the role of the hydronium ion.

The reaction sequence would look like this:

Unsolved questions in this problem: 1) We have two reactions taking place, which conceivably may not take place simultaneously. Can you reason why not? I'll tell you: Both reactions require two reagents, SM and MeLi. This means that there must be an effective clash (enough energy, propper orientation of reactants, right stereochemistry) for each one of these reactions to take place. Statistically it would be very unlikely that a single molecule of SM gets a double successful hit from two MeLi molecules.

2) If there are two reactions here, you may wish to know which one occurs first? To answer this we would have to have an idea, however rough, of the relative energy barrier of each process. This depends on several variables including ahrd/soft relationships between reaction centers, acidity of protons on the -CH2Cl group and how difficult it is for a small rat like MeLi to approach the C=O group,

Also, once the first reaction takes place the resulting intermediate will be a Li alkoxide, a strong negatively charged sector that will discourage the approach of a secon Methyl anion (to either alkylate or substract protons).

So the question above is not as simple as it looks, is it? A 3D view of SM and a first unalkylated intermediate may show that the carbonyl carbon is not as exposed as one may surmise after a first look. Compare these two 3D views: SM (starting material) at left shows that the lower (alpha) side of the ring is congested by 3 protons, making the alkylation step on C=O difficul from that side. The upper side of the 'plane' fins also steric impediment. meanwhile the CH2Cl group appears well exposed.

However, if proton abstraction and the gamma-elimination step occurs first to yield intermediate **B** the ring 'relaxes' and the C=O carbon is more exposed from below while the beta side continues to be somewhat hindered. As a consequence, the absolute configuration of the tertiary carbinol will be modified depending on which step (proton abstraction or alkylation) is more favorable. Interesting...Yes. Intersting to realize the amount of information one relatively simple problem may conceal. You are invited to read Prof Morris paper, of which this is just one of the various reactions and enticing account he gives in this paper: JOC 2003; 68:9648. By the way, we will soon solve the question of loading up Jmol structures in this wiki so you can play around with them on screen, turn around and see molecules from every possible angle. Mike

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