Question

NMR Day 1 4) C,H,0 HOP-04-01 12 11 10 9 ppm (signal at -2.2ppm is considered a triplet; -1.5ppm is considered a multiplet)

Answer 1

Solution :

4) The given NMR spectrum belongs to 1-butanal (CH3CH2CH2CHO).

The four signals are due to four types of protons (a to d).

CH3 ^a CH2 ^b CH2 ^c CH ^d O

>> A triplet at 1.0 ppm is due to proton a of CH3 group.

>> A multiplet at 1 5 ppm is due to proton b of CH2.

>> A triplet at 2.2 ppm is due to proton c of CH2 group.

>> The chemical shift of proton d at 9.8 ppm due to presence of alhehydic proton.

5) The C-13 spectrum suggest the presence of 3-pentanone.

CH3CH2COCH2CH3

Both CH3 are equivalent, hence gives quartet at 10 ppm and both CH2 gives triplet at 25 ppm.