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Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: CJ2347 on February 17, 2014, 09:34:13 PM

Title: 1,4-diphenyl-1,3-butadiene NMR
Post by: CJ2347 on February 17, 2014, 09:34:13 PM

Recently in lab, we synthesized the above named product though the reaction of cinnamaldehye with benzyl triphenylphosphonium chloride. And then analyzed the product with an NMR in order to determine which stereoisomer was formed. The coupling constants indicate that the E,E isomer was formed. The integration given on the readout for the aromatic peaks is 7.15. Isn't this value to high? Shouldn't it be around 5? Why is it higher?

Title: Re: 1,4-diphenyl-1,3-butadiene NMR
Post by: discodermolide on February 17, 2014, 10:27:33 PM

Obviously 2 of the C=C protons appear under the aromatic signals, increasing the integral by 2.

Title: Re: 1,4-diphenyl-1,3-butadiene NMR
Post by: CJ2347 on February 18, 2014, 10:19:33 PM

I don't understand. I thought the the two doublets of doublets represented the vinylic protons.

Title: Re: 1,4-diphenyl-1,3-butadiene NMR
Post by: discodermolide on February 19, 2014, 12:43:09 AM

Looks like all the aromatics are more or less together.

Title: Re: 1,4-diphenyl-1,3-butadiene NMR
Post by: clarkstill on February 19, 2014, 02:26:18 AM

If it's a weak NMR sample in CDCl3 there may also be a large CHCl3 peak underneath the aromatic region, which you could be including in your integration by accident.