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Magazine Chapter 13: Q 58P (page 419)
Question: Propose structures for the three compounds whose NMR spectra are shown.
Short Answer
Step by step solution
Step-by-Step SolutionStep 1:
Nuclear magnetic resonance is the technique which is used to identify the hydrogen in compound and produce the corresponding peaks for its detection.
Non-Equivalent and Equivalent proton
The proton which is having a different environment and all protons give different peaks is the non-equivalent proton whereas the proton sharing the same environment and all protons giving the same singles or peaks are the equivalent protons.
Signal production
All equivalent proton produces a single signal and all the non-equivalent proton produces a separate signal for each proton.
Structure of compound A
As per the given graph, it is clearly visible that the long peak is absorbed at 2.17 that must be a carbonyl and three peaks are observed due to the methylene at 2.46 and then one the right side of carbonyl the methyl group must be present at 0.95 and the peak absorbed at 1.64 due to the adjacent methylene group so the structure must be as follow:
Compound A
Structure of compound b
As per the graph, it is clearly visible that the peak absorbed at the 7.65 is due to the ring proton so there must be a ring present and the long peak shown at 2.31 is the peak due to alkane means the methyl group attached to the ring and the peak at 7.0 is due to bromine so the structure is:
Compound B
Structure of compound c
As per the graph shown it鈥檚 clearly visible that the peak absorbs at the 7.29 indicates the presence of the ring and the absorbance at the 3.16 with a side peak that the methylene group is present adjacent to each other and long peak absorbance is the indication of bromine attach to them.
Compound C
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