[sven222]

Firstly, here's the question,

When the cyclohexanone shown is dissolved in d6-benzene, 3JAB = 3.0 Hz.  When the same molecule is dissolved in d4-methanol, 3JAB = 11.0 Hz.  What are the conformations in these two solvents and why do the conformations differ. *Structure attached*

This is what i'm thinking so far,

So when methanol is the solvent, the coupling constant is 11Hz, thus the bond angle between Ha and Hb is 180, thus they are each in axial positions. The CHMe2 group is in the equatorial possie. We'll call this structure A.

When benzene is the solvent, the coupling constant is 3Hz, thus the bond angle between Ha and Hb is ~60, thus they are each in equatorial positions. The CHMe2 group is in the axial possie. We'll call this structure B.

So ring flip causes the coupling constant to change from 3 to 11, or vice versa. Obviously this has something to do with the solvent. One of the major differences between benzene and methanol is methanols ability to form hydrogen bonds. Does forming hydrogen bonds somehow allow the conversion from B to A? This would explain the change in coupling constants when methanol is used over benzene.

However as you may know, the CHMe2 group is much more stable in the equatorial possie, due to a number of reasons. So wouldn't structure B just flip into structure A automatically, given the added stability associated with structure A? Thus not needing the methanol?

I think i'm either a/ missing something basic here, or b/ looking too far into the question.

Any assistance is appreciated.

[sven222]

So after a long night of deep thought *cough*, I have come up with the following,

In methanol, we know that both of the hydrogens are in axial positions, thus putting the CHMe2 group in the equatorial position, which it quite enjoys. The structure is further stabilised by hydrogen bonding between solvent molecules and particular regions of the ring (ie the carbonyl oxygen).

When methanol os removed as the solvent, and replaced by benzene, hydrogen bonding between solvent molecules and the ring no longer takes place. In order to compensate for this, the ring adopts a twisted boat conformation, allowing hydrogen bonding between the carbonyl oxygen and the adjacent equatorial (now) hydroxyl group and the carbonyl oxygen and the adjacent deshielded proton (CH2). This keeps the CHMe2 group in the favoured equatorial position, while still keeping the extra stability provided by the original methanol hydrogen bonds.

By conforming to the twisted boat formation, Ha becomes equatorial, thus explaining the 3Hz coupling constant. (as Hb is still axial).

Sound good?


(I know it's hard to visualise without pictures, however does the theory seem sound? ie changing to the boat conformaiton to provide extra hydrogen bonding that was taken away by the removal of methanol as a solvent)