Question

All work must be very neat and organized. you need to organize your thoughts, please use o separate sheet of paper. Below are six IH NMR spectra from six different compounds. The molar mass for each compound is given. Determine the structure of each compound. For each 'H NMR spectra below, draw (on the spectra) the determined structure and annotate (correlate the equivalent protons with the corresponding NMR signal(s) in the spectrum). CHEM 144 Lab 16 S2 NMR Compound C, M = 102.13 g/mol group/#H10 Split molecule: 2 PPM Compound D, M = 136.15 g/mol group #HI Isont oso molecule: 108 PPM

"Determine the structure of each compound. For each NMR below, draw the determined structure and annotate( correlate the equivalent protons with the corresponding NMR signals in the spectrum)."

Answer 1

In HNMR , equivalent hydrogens give a single signal. The splitting of the signals is determined by the number of neighbouring hydrogens. In generat the splitting is equal to n + 1 (n is the number of neighbouring hydrogens).

Compound 1

There are 2 geminal methyl groups (2 x CH3) at 1.3 delta. These methyl groups have 1 neighbouring proton (they are split into a doublet)

There is 1 methyl group (CH3) that has no neighbouring hydrogen (it gives a singlet) at delta 2. This means that there is one carbon atom that does not have any hydrogen bonds attached to this CH3

There is 1 CH group with more than 3 neighbouring hydrogens (it is spit into a multiplet) at delta 5. As this CH has a quiet downfield shift, it is bonded to a carbon that is directly attached to a hetero atom.

So till now we have 5 carbons and 10 hydrogens. This brings the mass to 70. The remaining mass is 32. In all probability, the mass of 32 corresponds to two Oxygen atoms in the molecule. This can not be carboxyli acid as there is no acid proton. So it will be an ester.

CH3 1 > Singlet at SQ 46-6-O-CHO 2 Multiplet at ss CH₃ 3a3b Doublet at sl.2 3 M = 102.13 g/mol

Compound 2

There is 1 methyl group (CH3) that has no neighbouring hydrogen at delta 2.3 (singlet)

There are 4 aromatic hydrogens that are seen as doublets at 7.2 and 8 delta. These 4 Hydogens are present in 2 set of equivalent hydrogens. This means that the benzene ring has 2 substituents at para position (if there were only 1 substituent, then 5 hydrogens would have been seen. Since there are 2 sets of signals having 2 Hydrogens each, the substituents on the ring are symmetrical. This will happen only for para substitution).

There is one hydrogen at around 12.8 delta. This corresponds to a carboxylic acid hydrogen (COOH).

So now there are 8 carbons ( 6 from benzene ring, 1 from methyl group, 1 from carboxylic acid group), 8 hydrogens and 2 oxygens. So the mass is 136.15 g/mol as required.