[nj_bartel]

 http://img527.imageshack.us/my.php?image=aaakzp.jpg

Why is the answer not 6?

[CopperSmurf]

I think it should be only 4 peaks (all singlets). Remember that proton NMR peaks looks at a certain hydrogen and then looks for any hydrogens on the carbon directly next to it. So it looks for coupling.

An easier way to look at it is how many different environments does each hydrogen experience? there are 4 different environments. For example, the two H on CH₂ is in the same environment, not different ones.

[James Newby]

If the 5 membered rings are flat then the protons will be in identical environments, but id have thought the esters line up with the aromatic ring.  More than likely the chair is puckered slightly and there are pseudo-axial and pseudo-equatorial positions.

So i count 4 environments,

i).two doublets from the lactone protons (small geminal coupling)
ii) two phenyl protons (singlets) at slightly different signals

MODIFY: three environments, the protons on the aromatic ring are in identical environments and there will be a singlet

[nj_bartel]

Ok, I guess I'm having a little trouble seeing when opposite faces of a ring will be in the same/different environments.

For example, the lactone ring's protons, one looks like it would be seeing the ester on it's left, and the other looks like it would be seeing the ester on it's right.  Why does the fact that the ring is flat make these two equivalent?

[CopperSmurf]

if it's flat then you get equivalent hydrogens (by symmetry), one pointing away and one pointing towards you. If it's a bent ring, then you got a problem of axial vs. equitorial (similar idea to cis vs trans).

[James Newby]

All saturated rings with sp3 carbons want to have a bond angle of 109.5.  Combine this with the fact that the ester will want to overlap its pi electrons with the aromatic ring.  The CH2 unit will pucker either up or down to get as close to the 109.5 as possible.  One proton will be closer to the ester than the other, hence the pseudo positions.

I may be wrong in my previous post, if the flipping of the CH2 is faster than the NMR timescale then only one environment will be seen.

[nj_bartel]

But as the rings are flat, both protons are equidistant from everything else, making them equivalent, correct?  And are you sure the arene protons are equivalent?  The substituents on the arene aren't the same.

Thanks to both of you.

[CopperSmurf]

The arene protons, it "could" potentially be equivalent if it's exactly symmetrical but it's not. If they were exactly symmetrical, you'd have a different problem, cis/trans lactone rings.

They're not equivalent in this case because on one side, one of the oxygens on the lactone ring is connected to the aromatic ring directly, but that's not the case in the other lactone ring, which you pointed out.

I think what was meant earlier (I might be wrong) but the protons on the aromatic ring aren't exactly equidistant nor equivalent to each other. But the two protons on the CH₂ on the lactone rings are generally equivalent, but it is not equivalent to the CH₂ on the other end of the molecule.

[nj_bartel]

Ok, that makes sense, thanks 

[James Newby]

sorry, coppersmurf is right the aromatic positions arent equivilent