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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: lalala123 on August 25, 2013, 08:59:22 AM

Title: Tetraphenylcyclopentadienone NMR
Post by: lalala123 on August 25, 2013, 08:59:22 AM

Please help....Important!!!

I came across a question while I was reading about the aldol reaction. I am not sure how to answer it. I have a test soon that can contain this question. The question is why H NMR is not a good enough way to characterize tetraphenylcyclopentadienone? Can anybody help?

Title: Re: Tetraphenylcyclopentadienone NMR
Post by: sjb on August 25, 2013, 09:46:55 AM

Quote from: lalala123 on August 25, 2013, 08:59:22 AM

Please help....Important!!!

I came across a question while I was reading about the aldol reaction. I am not sure how to answer it. I have a test soon that can contain this question. The question is why H NMR is not a good enough way to characterize tetraphenylcyclopentadienone? Can anybody help?

Without the full details, I'm not sure, but what would the proton NMR spectrum look like, do you think?

Title: Re: Tetraphenylcyclopentadienone NMR
Post by: darsinagoL on September 16, 2013, 11:20:13 AM

Only based on what I remember.  Tetraphenylcyclopentadienone is based around a 5-carbon ring with 4 phenyl groups.  Benzene rings usually show up as a multiplet on an H NMR, so with 4 it may be a bit hard to distinguish the signal(s) (may get different signals, or a strong signal if there is symmetry.)  Also since the ring in the middle contains no Hydrogens the signal for the inner ring may also be difficult to distinguish, if it's there at all, since H NMR is based on the bonds Hydrogens have to their substituents.