Question

Problem #3: C4H100 (Assign hydrogens in your structure with the same labels used in the NMR spectra) Transmittance Warenumber(cm) doublet, 6H broad singlet, IH (d) doublet, 2H nonet, IH (b) 1H-NMR WORKSHEET PEM

Answer 1

#3

Molecular formula of the compound = C₄H₁₀O

Degree of unsaturation = (2c + 2 - h)/2    (where c, h are the number of carbons and hydrogens in the compound)

                                      = (2 x 4 + 2 - 10)/2

                                      = 0

Hence, the compound is saturated. This means that the compound is not ketone or aldehyde.

The broad U-shaped peak in IR spectrum at 3200-3600 cm^-1 range suggests the presence of H-bonded O-H group in the compound.

¹H NMR peak assignments:

doublet, 6H broad singlet, 1H (d) H2 1 CH2 H_CH2 HO doublet, 2H nonet, IH (b) PPM

The broad singlet over 3.5 ppm is due to the exchangeable hydrogen of alcoholic O-H group. The tall doublet corresponding to 6H below 1 ppm and a nonet corresponding to 1H at ~1.5 ppm and short doublet corresponding to 2H at ~3.5 ppm suggest that the compound has -CH₂CH(CH₃)₂ moiety.

The peaks split following n + 1 rule. Here n is the number of inequivalent hydrogens in the adjacent carbon. The 6H of the two methyl groups of peak a has only one hydrogen atom in the adjacent carbon. Hence, it splits in 1 + 1 = 2 (doublet).

The hydrogen of peak b has 8 hydrogens in adjacent carbon. Hence, it splits in 8 + 1 = 9 (nonet).

The hydrogen of peak c has 1 hydrogen in adjacent carbon. Hence, it splits in 1 + 1 = 2 (doublet).

The O-H hydrogens of peak d are exchanged rapidly (faster than NMR data acquisition timescale) via H-bonding. Hence, it is appearing as a broad peak.