Draw the correct structure from the MS, 1H NMR, 13C NMR, and IR data given below. Both of the peaks on the 1H NMR spectrum are doublets (J = 2.5 Hz) and the integration ratio is 1:1.

Please select the correct answer choice(s) from the following The signal at 1.4 ppm is a [s/ d/ t/ q ] and is the [ TWO METHYL GROUPS OF AN ISOPROPYL/ METHYL GROUP OF AN ETHYL/ METHYL GROUP ATTACHED TO AN AROMATIC RING ].

The HNMR spectra shown below contains six singlets that correspond to the six different proton groups identified in structure 1. Use an HNMR table to help match each signal to a proton group.

Assign all of the signals in the ¹H NMR spectrum for propyl acetate. Fill in the table below with the 'H NMR data. Number each proton (or set of protons) to match the corresponding peak in the NMR. Integration Splitting Propyl acetate structure with protons labeled Signal - Chemical shift (ppm) 1 2 3 4 11 propyl acetate 10 9 8 7 6 5 ppm 2H, triplet 4 3 3H, singlet 3H, triplet 2H, sextet 2 1 Figure 3.7 ¹H NMR spectrum for propyl acetate. 0

Can you explain this problem? I got it wrong on my exam and don't understand why.

13C NMR is less sensitive than 'H NMR because of the greater mass of carbon. True False