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Organic Chemistry and HNMR Spectrum

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1. A proton that absorbs radio frequency radiation at higher field strength will appear _____ in a 1H NMR spectrum.
a. at 0 ppm
b. downfield (away from TMS)
c. upfield (closer to TMS)
d. at 7.27 ppm just like CDCl3

2. The carbonyl peak of benzaldehyde appears as a very strong and intense peak at 1703 cm^-1. Normally the carbonyl peak of an aldehyde appears at 1725 cm^-1. Why does the carbonyl peak of benzaldehyde appear at a lower frequency?
a. resonance
b. electronegativity
c. conjugation
d. induce dipole

3. Predict the splitting pattern for each kind of hydrogen in the following molecule. (Make splitting assignments starting with the methyl group on the left end with the methyl group on the right). CH3CH2COOCH3
a. triplet, quartet, singlet
b. multiplet, multiplet, triplet
c. singlet, triplet, quartet
d. doublet, doublet, singlet

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Solution Summary

This solution identifies the correct answer for each multiple choice question but it also provides justification on what radio frequency the proton will adsorb in higher field strength, why the carbonyl peak of benzaldehyde appears at a lower frequency, and the splitting pattern of each kind of hydrogen.

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A proton that absorbs radio frequency radiation at higher field strength will appear upfield (closer to TMS) in a H nmr spectrum. This is important to remember. Upfield is to the right. Decreasing ppm from TMS is upfield. Upfield goes with increased field strength. The easier it is to flip protons, the more downfield (left) ...

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