In a HWE reaction, a 1, 2-disubstituted alkene was formed in which the olefinic protons exhibited...

Question:

In a HWE reaction, a 1, 2-disubstituted alkene was formed in which the olefinic protons exhibited a vicinal coupling of 9.0 Hz in the 1H NMR. What is the stereochemistry of the double bond?

Proton NMR couplings

The interaction or coupling between two non-equivalent hydrogens in a proton NMR spectrum can be expressed in terms of a coupling constant (J). The value of J can be calculated by subtracting the values of the chemical shifts of two adjacent peaks in a multiplet (bigger with the smaller). The value is always expressed in terms of Hz, so often the ppm values of the chemicals shifts needs to be converted.

Answer and Explanation:

The product in this Horner Wadsworth Emmons (HWE) reaction is a 1,2-disubstituted alkene.

Either the cis or trans stereochemistry can be formed as shown in the image.

Trans coupling is normally bigger than cis coupling on a proton NMR.

For trans configuration, the value of J (coupling constant) is between 11 - 19 Hz, while it is between 5 and 14 Hz for cis.

Based on this information we can assume a cis configuration since a J value of 9 Hz is mentioned in the question.


Learn more about this topic:

How to Read NMR Spectra of Organic Compounds
How to Read NMR Spectra of Organic Compounds

from Organic Chemistry: Help & Review

Chapter 2 / Lesson 2
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