Problem+4

Ihara M, Taniguchi M,Tokumaga Y, Fukumoto K, JOC **1994**, //59//, 8092-8100

DIFFICULTY: 4/10 LEVEL: third semester undergraduate org chem. First year graduate student

If you find it too difficult, or just too easy to deserve your attention, simply go to HOME page and chose another problem

__SOLUTION & DISCUSSION:__ Some people may say that unearthing a 1994 paper is more archaeology than modern org chem, not worth wasting your eyes on. Dead wrong! Most of org chem recent developments are based on discoveries made several years ago (we will have some examples later on). Science at large works like this: reading, comparing, judging, improving, and sometimes discovering. Some among you who think of yourselves as smart guys will jump into the solution of this problem quickly by drawing arrows here and there and pushing electrons like crazy. Sure, a ring expansion at the expense of the four membered ring going the Wagner Meerwein way. But solving mechanism problems is much more than that. It is a deep learning experience of reasoning techniques that can help you with other more challenging problems. The discussion that follows shows why. Read on (I won't say 'please' because it's up to you!). This reaction seems easy enough to reason, but we will find some complications. For one, pay attention to the stereochemistry of 1 and 2, as shown in Fig 2:

This rendering demands a second look because arrows alone do not account for every thing you see here
 * Fig 2**

Going the systematic way we see that:

1) Aom accounting: Compound **1** (C 10 H 18 O 2 ) becomes cyclopropane product **2** (C 10 H 16 O). The difference in atomic composition is -H 2 O, yes, a water molecule must be lost in the process. This suggests that some sort of OH elimination must have occurred. Every time you suspect this, look for C=C bonds in products where the OH was.

2) There is one just C=C bond in **2** but far removed from the original carbinol C. This can be appreciated by using the C-C**OCH 3 ** group as tag, because alkoxides do not like to migrate but only rarely.

3) So your C=C must have either migrated from (**a**) an unstable intermediate, or (**b**) be formed after some sort of bond rearrangement.

4) //Bond rearrangement//? Indeed. One begins with a fused cyclobutane-cyclopentane combination and end up with a fused cyclopropane-cyclohexane set. One ring expands, the other shrinks. No carbons are lost so they must have undergone a deep-rooted rearrangement.

5) When you suspect a deep rooted rearrangement think about a carbonium ion taking part, because this is an energy demanding process and C + 's have only three destinies: substitution via SN1, elimination via E1, and rearrangement. Now, Where? At secondary or better yet tertiary C supporting a good leaving group. The carbinol C of **1** seems well suited for this. So take HO away and form C + . As shown on scheme 1, you should have now cation **3**.

6) Although HO- is potentially a leaving group, it is a lousy one by itself. It needs help from external agents such as abundant protons or Lewis acids in the medium. But our solvent is pyridine, a stubbornly basic compound. Ah, there is POCl 3 to help. P has a special liking for O atoms filling the role of a Lewis acid. It pulls away HO as HOPCl 3 (+) that eventually goes into POCl 2 and Pyridine.HCl. This HO take up drives the reaction forward. Besides the medium is polar enough to sustain Carbonium ions.

7) C+ is a powerful driving force, so much so that it is capable of perturbing proximal C-C bonds and cause their breakage. Cyclobutane has considerable ring tension and easily gives way to a cyclopropane as in **4a** by a symmetry allowed C-C [1,2] migration. Wasn’t it that cyclopropanes were also very tense? Yes, so this step requires some explanation (below). A new C+ is formed as **4a**, a tertiary one, thus favorable. Redrawing 4a as 4b clears the way for the final product by beta elimination via E1. By the way, 4b is the ‘migrated’ C+ we needed to create a new C=C bond as a result of water loss.

**Fig 3**:

 A comment on the cyclobutane → cyclopropane (**3** → **4**) transformation: 1.- Have a look at a 3D rendering of **1 (Fig 4)**. It is a congested V shaped molecule with all bulky substituents facing the outside, save for OH (the less bulky). This means that the molecular skeleton would be relieved if this OH departs, as in **3**. Decongestion is always a good driving force, a lot of bond electron repulsion is comforted, DH goes down.

2.- There are two things that this C+ 3 can do. Go back to **1** or promote rearrangement to a more comfortable molecule. This is **4**. Observe carefully the stereochemistry of **4b**. The angular methyl group on the cyclopropane is down, alpha, under the plane of the molecule so to speak. But in **3** it is on a trigonal (flat) C+, so how does it proceed to this particular configuration and not to a mixture of alpha an beta? Who is responsible for it? 3.- Have an answer? Yes? OK
 * Fig 4 **

But if not, the secret is here: the cyclobutyl C-C bond that breaks apart shown in yellow below [**Fig 5**] (**3a,** a simplified depiction of **3**). In particular, the way it opens up. As shown in the following **figure 5**, this C-C bond split takes place in a //conrotatory// manner (blue arrows in **3b**) as allowed by Woodward & Hoffmann symmetry rules: both atomic orbital lobes that constitute the sigma bond open like a two sleeve door. This movement forces down the CH3 group to the underside, the one you see in **3a** but is eclipsed in **3b** by one of the AO lobes.

At the same time the lobe moving upwards (B) ‘sees’ the gray lobe of the 2pz atomic orbital of C+ (dotted line) and close (green arrows). In sum, it is a kind of door opening here and closing there //on the same side of the molecule//.

This orbital dynamics is responsible for the observed stereochemistry.


 * Fig 5 **

Although some of you may argue that exactly the same thing might have occurred by opening this door towards the underside, the V shape would tend to close further on itself increasing repulsion. This would be a much higher energy path.

Did a fast electron pushing along quickly scribbled arrows tell you all this? I doubt it. Slow down and go to the bone marrow of things, if you want to really learn rather than show off. Did you enjoy this? I did, while explaining these tricks to you.

Prof. Miguel E Alonso-Amelot

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