Question

Below are the questions with the answer key. Please solve the problem and show work to support that answer.

For Problems 1-36, write a structure consistent with the molecular formula and 'H-NMR spectrum. Symmetry is important for many of them to reduce the number of NMR peaks expected. Show your tirals and tribulations in the space provided. 35. CeH11CI (A bit tricky.) 3 Н,с C-CH CH CH PPM 36. C HaBr2 (Diastereotopic protons.) Br PPM 37. In the area below, draw a reasonable picture of the proton NMR of ethyl acetate. Label the peaks with their assignment and relative areas. Make sure the relative peak heights and splittings are carefully drawn. Looking this up is easy, but teaches you nothing. Do it from basic principles and general NMR shifts. 10 PPM N 38. Below are the proton and carbon NMR spectra of the compound shown. Using the letters in each spectrum, label the structures to show which proton/carbon is which. In the carbon spectra, use letters that are close to multiple peaks for each peak in that group, i.e., c will be used twice, d will be used 4 times, and b will be used twice. C Peaks a, b, & c (both) are not present in DEPT 135 h e 200 160 140 120 20 180 80 60 100 PPM 40

Answer 1

Question 35.

Spin multiplicity = 2nI + 1

Where 'n' is the no. of coupling partners

I = nuclear spin quantum number of ¹H nucleus = 1/2

i) The 3H on the CH₃-group, which is on the left-top appears as singlet because it doesn't couple with any other proton, i.e. n = 0

Therefore, spin multiplicity for the corresponding 3H = 2*0*1/2 + 1 = 1 (singlet)

(ii) The 3H on the CH₃-group, which is on the right-bottom appears as doublet because it couples with one and only one adjacent proton, i.e. n = 1

Therefore, spin multiplicity for the corresponding 3H = 2*1*1/2 + 1 = 2 (doublet)

This signal appears down-field because the corresponding CH₃-group is attached to the sp²-carbon.

(iii) The 3H on the CH₃-group, which is on the left-bottom appears as doublet because it couples with one and only one adjacent proton, i.e. n = 1

Therefore, spin multiplicity for the corresponding 3H = 2*1*1/2 + 1 = 2 (doublet)

This signal appears up-field because the corresponding CH₃-group is attached to the sp³-carbon.

(iv) The one H (1H) of methine group which is connected to the sp² carbon appears as a quartet at down-field region because it is attached to sp²-carbon and it couples with 3 adjacent protons, i.e. n = 3

Spin multiplicity = 2*3*1/2 + 1 = 4 (quartet)

(v) The one H (1H) of methine group which is connected to the sp³ carbon appears as a quartet at comparatively up-field region because it is attached to sp³-carbon and it couples with 3 adjacent protons, i.e. n=3

Spin multiplicity = 2*3*1/2 + 1 = 4 (quartet)