Chapter 13: Spectroscopy

SPECTRA PROBLEM #5 SOLUTION

• In the MS, the pair of peaks of about 1:1 ratio at 122 and 124 suggest Br due to the isotopes ⁷⁹Br and ⁸¹Br
• The molecular ion occurs at m/z = 122 indicating the MW = 122 g / mol (based on ⁷⁹Br)
• The IR shows has no absorbtions for the main functional groups, looks like C-H and C-C only.
• 13 C-nmr shows 3 types of C, including a slightly deshielded C (35 ppm) plus 2 other hydrocarbon C (26 and 13 ppm)
• H-nmr

d/ppm	multiplicity	integration	assignment
3.4	triplet	2	CH2 deshielded, coupled to 2H
1.8	sextet	2	CH2 coupled to 5H
1.0	triplet	3	CH3 coupled to 2H

Summary....

• The MS indicated MW =  122 g/mol.
• The H nmr gives a total of 7 H, C nmr 3 C
• Use this to check the molecular formula : C₃H₇Br = 3 x 12 + 7 x 1 + 1 x 79 = 122 g/mol
• This confirms that we have the molecular formula

Altogether....

The coupling patterns in the H nmr are the critical information  The patterns give a CH3 next to a CH2 and a CH2 next to another CH2 Since we only have 3C, this means we have a CH3CH2CH2-  With the pieces we have :  CH3CH2CH2-  and -Br  This can only be put together to give CH3CH2CH2-Br

1-bromopropane

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