Question

What protons in alcohol A give rise to each signal in its ¹HNMR spectrum? Explain all splitting patterns observed for absorptions between 0–7 ppm.

Short answer

H_a : 1 + 1 = 2 ,doublet and exhibits a peak at $~1.4\mathrm{ppm}$.

H_b: singlet (as it is attached to an oxygen atom). It exhibits a peak at $~2.7\mathrm{ppm}$.

Hc: 3 + 1 =4, quartet and exhibits a peak at ~ 4.7 ppm .

H_d : The aromatic protons exhibit peaks in the region 7.4 ppm.

Step-by-step solution

Splitting pattern of  1HNMR signals

The splitting of ¹HNMR signals is determined/predicted using the n+1 rule.

Here n represents the number of hydrogen atoms connected to the carbon atom adjacent to the carbon attached to the hydrogen atom under consideration.

If the number of splitting is one, then it's called singlet; if two, then doublets, three (triplet), four (quartet), five (pentet), and more than five (multiplet).

Chemical shift of 1HNMR signals

Alkyl hydrogens have the least chemical shift ~ 0.8 to 1.7 and the hydrogen atoms on alkyl groups adjacent to a double or triple bonded carbon atom exhibit a slightly higher chemical shift.

The chemical shift of aromatic protons is around 6 to 8 ppm.

The hydroxide group exhibits a signal around 1.5 to 2.5 ppm.

The splitting pattern of 1HNMR signals of the given compound

The splitting pattern of the ¹HNMR signals of the given compound is as follows:

Non-equivalent protons in the given alcohol

H_a : 1 + 1 = 2, doublet and exhibits a peak at 1.4 ppm .

Hb : singlet (as it is attached to an oxygen atom). It exhibits a peak at 2.7 ppm

Hc: 3 + 1 = 4, quartet and exhibits a peak at 4.7 ppm.

Hd : The aromatic protons exhibit peaks in the region 7.4 ppm.