Identify the compound with molecular formula C5H11Cl
with the following 1H and 13C NMR spectra. Identify
protons that correspond to every peak in the 1H NMR. Clearly
indicate what information you get from the 13C NMR spectrum.
The molecular formula tells us that there are NO degrees of
unsaturation so no rings or double or triple bonds.
So the 13C NMR has four peaks, indicating that
there are four unique carbons in the molecule. Therefore, two of the five
carbons must be the same!
In the 1H NMR, there are three signals, a triplet
at about 1 ppm, a singlet at about 1.55 ppm, and a quartet at 1.8. The triplet
at 1ppm must be a methyl group with an adjacent CH2 group (or CH2CH3).
The quartet at 1.8 is for a CH2 with an adjacent methyl (or CH2CH3).
Piecing this together we get 2-chloro-2-methylbutane:
where the (a) hydrogens are found
as the singlet at 1.55ppm, the (b) hydrogens
are the quartet at 1.8ppm and the (c) hydrogens are
the triplet at 1.0ppm.