IR/MS Key

C3H4O UN = (6+2-4)/2 = 2 Practice Problem 1 C=C triple bond stretch O-H stretch C3H4O UN = (6+2-4)/2 = 2 sp3 C-H stretch

2 C=O stretch m/z = 128 (M) m/z = 113 m/z = 58 m/z = 43

3 UN = 0 Odd mass – contains N No N-H stretches m/z = 73 (M) m/z = 58 UN = 0 Odd mass – contains N No N-H stretches Must be tertiary amine.

4 From MS: contains Br From IR: carbonyl Two possibilities: m/z = 43 m/z = 150 (M) m/z = 107 m/z = 43

Propose structures for compounds that meet the following descriptions: An optically active compound C5H10O with an IR absorption at 1730 cm-1 A non-optically active compound C5H9N with an IR absorption at 2215 cm-1 5 UN = (12-10)/2 = 1 IR – must be carbonyl Only one optically active structure can be drawn: UN = (12-9+1)/2 = 2 IR – must be triple bond IR – no mention of N-H stretch

You are carrying out the dehydration of 1-methylcyclohexanol to yield 1-methylcyclohexene. How could you use IR to determine when the reaction is complete? 6 Look for: Loss of O-H stretch ~3300 (broad) Gain of C=C stretch ~1650 Gain of sp2C-H stretch >3000

Propose a structure consistent with the following IR spectrum and molecular formula 7 UN = (16-6)/2 = 5 C7H6O sp2C-H C=O stretch aldehyde C-H aromatic ring

8 Nitriles undergo a hydrolysis reaction when heated with aqueous acid. What is the structure of the product of hydrolysis of propanenitrile (CH3CH2CN) if it has IR absorptions at 2500 to 3100 cm-1 (br) and 1710 cm-1 (s) and has a parent peak in MS at 74?

9 m/z = 69  contains N Formula = C4H7N UN = 2 From IR:   From IR: triple bond (2246) No N-H

10 From IR: C=O (1716) No aldehyde CHO MW 98  C6H10O UN = 2   MW 98  C6H10O UN = 2 Back to IR – no alkene (1650) must be ring