Question

Choose the molecule fit the 13C and H NMR from the list. Label the hydrogen
си, He 7 cu Т (н . Нst (13 сну сну И о н — С=С / Росан - Сиз 15 애 Н.С сн. NH₂ (130 'CH ЕС (СН2) 3 Сез ОН HBC Cr3 ABY І H2 (2 ) CH, HC - CH₂ L-3 애

Answer 1

NMR003:

In 13C-NMR 4 peaks seen only in aromatic region (120 to 140), that indicate compound is disubstituted phenyl ring with no aliphatic protons and 2 carbons are identicals.

In 1H-NMR , three peaks are seen with triplet (2 identical Hydrogen on adjacent carbon), doublet (one hydrogen on adjacent carbon)and singlet (no Hydrogen on adjacent carbon)

Hence this splitting fulfilled by structure 4 only.

1H singlet н Br Br Identical cabons -Identical carbons н 2H, doublet н 1H tiplet

NMR004:

In 13C-NMR 2 peaks seen only in aromatic region (120 to 140), that indicate compound is 1,4-disubstituted phenyl ring with no aliphatic protons and 4 carbons are identicals.

In 1H-NMR , only singlet is seen, hence only one type of proton present oo aromatic ring

Hence this splitting fulfilled by structure 15 only.

Al4 hydrogens are identicals anf gives singlet н Br Identical carbons н Br Н Identical carbons