[inash524]

Hi All,

I've attached two screenshots. The first has an IR, Mass spec, and NMR of a chemical compound. The second has its corresponding structure. I have a question about the position of the CH2 group in this structure. What gives away that it the CH2 group is next to the carbonyl versus the benzene ring?

Both structures seem equally valid to me. Any thoughts?

Thanks in advance!

[Babcock_Hall]

It is a forum rule that you must show an attempt to answer the question or share your thoughts before we can help you.  Maybe a good place to start is to talk about which technique or techniques is(are) most likely to provide an answer.

[inash524]

Hi Babcock_Hall, my apologies. Here is what I've been thinking so far:

I can see from the DEPT chart that there is only one methylene group in this structure. This is confirmed by integrating the peaks from the NMR. Now the question arises: where is this the methylene located? I see no peak splitting in the NMR, and therefore, we can assume that it is not situated close to a methine or methyl group. 

Using other evidence from the NMR (the slight shift of the CH3, no splitting), I've deduced that the CH3 is attached to the carbonyl (which is evidenced in the IR near 1700). But who's next to the carbonyl? It could either be the carbon with two methyl groups attached to it or it could be the methylene. We can see that the methylene group is noticeably shifted. However, it's not clear whether this shift is due to the benzene ring's deshielding effects or if it's from the carbonyl group. And honestly, I see no evidence to support either hypothesis.

[AWK]

Compare shift of CH2 with proton NMR spectrum of phenylacetone.

[inash524]

Hi Awk,

So this was a simple matter of knowing chemical shifts? There was no analytical component to it?

[Babcock_Hall]

I glanced at some shifts, and I am concerned that the shift from the aromatic ring versus the carbonyl would be similar.  What could you hope to learn from the mass spectrum?

[inash524]

Oh. Yes, that makes a lot of sense @Babcock_Hall. Thanks a lot!