Question

1. Signal Assignment. Below are a set of ranges where a specific H' NMR signal might be found. 0.5-2.0ppm 2.0-3.0ppm 3.0-5.Oppm 0.5-6.Oppm 5.0-7.0ppm 7.0-8.Oppm 9.0-10.0ppm 10.0-12.0ppm Assign to the slots above where the following highlighted protons might be found. a) CH3-CH2- b) CH-CH2-CH2- c) HC-0 d) H-C(O)-R e) H-O-R ) H-O-C(O)-R g) H;C-Ar h ) H-C-C- j) CH3-CEC ...some groups appear in the same ppm range. [2pts each group correctly assigned) 2. Signal Splitting. Decide what splitting pattern (singlet, doublet, triplet, quadruplet, or multiplet) you might see for the following highlighted protons under a 100MHz H NMR. [5pts each) a) CH-CH2- b) CH3-CH-CH2- c) HC-O CH2 d) H-CH=C(CH3)2 e) the protons on a mono-substituted benzene ring 3. Proton Environments. Draw 1-buten-1-ylbenzene with all the hydrogens shown. Label each hydrogen, starting from the methyl group, using "a", "b", and so on. If two or more hydrogens are expected to be in the same electromagnetic environment, label them with the same letter. [15pts] Proton Count, Integral. Draw 1,1'-(1,2-ethanedyl)bis(4-methylbenzene) with all the hydrogens shown. Label each hydrogen type using "a", "b", etc. starting with the hydrogens on the methyl group on one of the benzenes. If we set the integral of these methyl protons at the value 3.00, what are the integral values for all the other hydrogen types. (Hint: there are only 4 different hydrogen environments.) [20pts]

Answer 1

Solution:

1H NMR signal range:

i) 0.5 to 2.0 ppm = option a and b

ii) 2.0 to 3.0 ppm = option j and g

iii) 3.0 to 5.0 ppm = option c

iv) 0.5 to 6.0 ppm = option e

v) 5.0 to 7.0 ppm = option h

vi) 7.0 to 8.0 ppm = option i

vii) 9.0 to 10.0 ppm = option d

viii) 10.0 to 12.0 ppm = option f

2) Multiplicity of signals:

i) CH3-CH2- = triplet

ii) CH3CH2CH2- = multiplet

iii) CH3-O-CH2- = singlet

iv) H-CH=C(CH3)2 = singlet