[cherrybarry]

Ok, I attached a picture of the molecules in question (some of the lines disappeared but you know they are bonds). Bond B is predicted to be LONGER than Bond A, because there is some resonance structure that accounts for this. However, I am not sure if my reasoning is correct. My prediction is that the oxygen in Bond B can form a double bond with the carbon in the phenyl group, thus drawing the oxygen closer to the carbon and farther from the phosphorus. In Bond A, the oxygen cannot do this, since the carbon is sp3 hybridized. So what does everyone else think?

[Blueshawk]

When I first looked at it, my guess was that B is longer than A due to steric effects.  The phenyl group is a much larger group than Ch3, therefore the bond with phenyl would lenghten to decrease the steric hinderence.  

If a resonence structure were to occur to put a double bond on the Phenyl Oxygen with a C in the ring, then it would carry a positive charge due to three bonds, which would be in close proximity to - charges on the oxygens, of which two O have - charge and there would be one carbon in the ring with a - charge as well.

There is another resonence structure that would put the double bond b/w the O and P, then there would be - charges on the other 3 O's, and the ring would be as is.  

Each of these would cause the bond in B to be longer than A due to steric hinderence, and one resonence in B haveing a negative on the ring.

The resonence for A just puts a double bond b/t O and P, but CH3 is much smaller than Phenyl group, so A would be able to be shorter in bond length.

Does this help   ??  Sorry I have no structures, but you should be able to see them.  
 

[cherrybarry]

so would the double bond, or "1.5" bond (between O and C) draw the oxygen closer to the carbon? and would the negatively charged carbon in the ring attract the positively charged oxygen?

i've only taken 1 month of orgo so excuse my dumbness!!!

oh, how would the second resonance structure lengthen the P-O bond? It seems like it would shorten it, since a double bond is represented, and three O's in the ring are now negatively charged, drawing the O+ closer to the P, which would shorten the bond? But the same can be said for the first structure with the methyl group, so i'm not sure how the second resonance form you mentioned would apply?

[maxyoung]

My guess is that: in compound B, resonance effect between benzene ring and oxygen makes the oxygen atom less electron donating. Thus there would be less double bond character between O and P atoms.

[movies]

I agree with maxyoung on the electronic effects.  Sterics may play some role as well, but I think that you can make a good argument with electronics alone.

Basically, since you can draw resonance structures for compound B where the electrons on the O in question are delocalized into the aromatic system, there is less electron density available on that O to donate into the P-O bond, so there is less double bond character in that bond.

[Blueshawk]

That is what I tried to say in the first place, the phenyl group allows for more resonence structures that dont place a double bond between the O and P, while the only resonence for the methyl group is a double bond between the O and P.