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Draw the structure of the compound that is consistent with the 1H NMR data below. (Assume...

Draw the structure of the compound that is consistent with the 1H NMR data below. (Assume that long-range coupling is not observed.) *Suggestion: Find \"end groups\" (like methyls) and work as far as possible towards the middle *When \"neighboring\" hydrogens are unequal, the splitting gets quite complex and multiplets result. This is especially true if a signal set is coupled by both alkyl and alkenyl hydrogens, which are very different from each other.Draw the structure of the compound that is consist

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The structure of the compound is as shown below. For simplicity to explain I have labbelled the protons as a, b, c and d.

First let us calculate the degree of unsaturation in the compound.

Given molecular formula : C5H9Br. Replace Br with an H and recalculate the molecular formula. It will be C5H10, which is short of 2 H's from the saturated molecule with 5-carbon's in it. Thus, the molecule has 1 degree of unsaturation. Must be an alkene.

The assignment of peaks is as follows,

a : This will appear as a triplet for 3H at 1.06 ppm

b : will appear as a multiplet for 2H at 2.00 ppm

c : will appear as a triplet for 1H at 5.87 ppm. The downfield shift for this proton is the result of Br attached to the carbon.

d : will appear as a singlet for 3H at 2.26 ppm. The downfield shift of CH3 signal is due to the fact that it is attached to an sp2 carbon.

The structure of the compound is shown below in the figure.

H2C-CH2-CHEC CH Br

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