1. IR/MS Key

2. 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

3. 2
C=O
stretch
m/z = 128 (M)
m/z = 113
m/z = 58
m/z = 43

4. 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.

5. 4 From MS: contains Br From IR: carbonyl Two possibilities: m/z = 43
m/z = 150 (M)
m/z = 107
m/z = 43

6. 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

7. 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

8. Propose a structure consistent with the following IR spectrum and molecular formula7
UN = (16-6)/2 = 5
C7H6O
sp2C-H
C=O
stretch
aldehyde
C-H
aromatic
ring

9. 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?

10. 9 m/z = 69  contains N Formula = C4H7N UN = 2 From IR:
From IR:
triple bond (2246)
No N-H

11. 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