The above problem works out nicely because the energy differences of the alpha a

Question

The above problem works out nicely because the energy differences of the alpha and beta states are similar and so the gaps between peaks (coupling constants, J values) are about the same (7Hz). In some situations, particularly those with little rotational averaging and low symmetry, the coupling constants of hydrogens (U values) are different. Consider This nitrosubstituted alkene. It has 3 Hydrogens. Each hydrogen is in a different environment, and each hydrogen couples the other 2. The difference here is that the coupling constants between each of the types of H are different as indicated in the picture below. Jac = 20 Hz Jab 10 Hz &=6.5 Your task is to show the signal made by Ha in the NMR spectrum as it couples first with He and then with Hb. To aid you n your work. We will define 1Hz 1 mm so you can use a ruler to make signal A. Repeat the process for signal C. lif you are lost... you can see a vinyl pattern in one of the upcoming problems) 1·Match the isomers of CH10 to their respective it spectra by placing the correct letter in the space to the right. OH

Explanation / Answer

21. Matching NMR spectrum with structure

First compound from left : spectrum B

it has 4 types of non-equivalent carbons

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Second compound from left : spectrum A

it has 3 types of non-equivalent carbons

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third compound from left : spectrum D

it has 2 types of non-equivalent carbons

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Last compound on right : spectrum C

it has 4 types of non-equivalent carbons

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