Question

Describe the proton-coupled ¹³C NMR spectra for compounds 1, 3, and 5 indicating the relative positions of the signals.

Short answer

1. Triplet-triplet-quartet.
2. Doublet-quartet.
3. Doublet-doublet-quartet.

Step-by-step solution

Step-by-Step SolutionStep 1: 1-bromopropene relative positions of signal in 13C NMR spectra

There are three carbon centers in 1-bromopropane with a bromine group. CH₃ group is attached far away from electron withdrawing group, thus, Br gives lowest frequency signal. The next increasing frequency signal is given by the CH₂groups, and the highest frequency signal is given by the carbon attached near the Br group. In the ¹³C NMR spectra, frequency gets decreased from right to left. Hence, the highest frequency signal ‘a’ comes first, followed by ‘b’, then ‘c’.

Signal splitting is triplet-triplet-quartet in this compound.

2-bromopropene relative positions of signal in 13C NMR spectra

The presence of equivalent carbons ‘a’ and non-equivalent carbon produce 2 signals from the compound. Two methyl groups not near to electron withdrawing group produces lowest frequency signal; and carbon atom, which is near to the electron withdrawing group, gives highest frequency signal. Thus, relative position of signals will interpret from increasing to decreasing order in spectra.

Signal splitting is doublet quartet in this compound.

Isobutraldehyde relative positions of signal in 13C NMR spectra

This compound gives 3 signals form carbon b, c, and a. The two equivalent protons not near the electron withdrawing group produces lowest frequency signal. The next position is for CH group, and the last carbon which is attached to the withdrawing group gives the highest frequency signal. Thus, the position of signals in ¹³C NMR spectra will be higher to lower frequency.

Signal splitting is doublet-doublet-quartet in this compound.