Spectra Problem #2 Solution

Chapter 13: Spectroscopy

SPECTRA PROBLEM #2 SOLUTION

In the MS, the molecular ion occurs at m/z = 60 indicating the MW = 60 g / mol.
The IR shows an -OH (3350 cm^-1) and C-O (1075 cm^-1) but no C=O (around 1700 cm^-1), or C=C (around 1600 cm^-1)
13 C-nmr shows 3 types of C, including a deshielded C (69 ppm), and 2 other hydrocarbon C (27 and 10 ppm)
H-nmr

d/ppm	multiplicity	integration	assignment
3.6	triplet	2	CH₂ coupled to 2H, deshielded
2.3	singlet	1	isolated H (-OH ?)
1.6	multiplet	2	CH₂ coupled to many H
0.9	triplet	3	CH₃ coupled to 2H

The H-NMR gives us the following pieces -OH (helped by the IR) -CH₃, and 2 different -CH₂-.
We can start to fit them together based on the coupling patterns as: CH₃CH₂CH₂-

Summary....

The MS indicated MW = 60 g/mol.
The IR showed the presence of an O-H and C-O bonds.
H nmr gives a CH₃CH₂CH₂- group
From the data, we can get a molecular formula of C₃H₈O = 3 x 12 + 8 x 1 + 1 x 16 = 60 g/mol
This has an IHD = 0, so no p bonds or rings.

Altogether....

With the pieces we have : CH₃CH₂CH₂- and -OH
IR an consistent with an alcohol.
Given these pieces, there is only one way they can fit together.

Checking the H-nmr, the middle -CH₂- is actually six lines since it has a total of 5 similar neighbours. The -OH deshields the other -CH₂- but does not couple to it.

1-propanol