Chapter 14: IR and Mass Spectroscopy

Question 1

Diagnostic region

IR Spectroscopy

Answer 1

Above 1500 cm⁻¹

Question 2

Analyzing an IR spectrum

IR Spectroscopy

Answer 2

Focus on the diagnostic region; analyze wavenumber, intensity, and shape for each signal

1. 1600-1850 cm⁻¹− check for double bonds
2. 2100-2300 cm⁻¹− check for triple bonds
3. 2700-4000 cm⁻¹− check for X−H bonds

Focus on the region above 3000 cm-1

Question 3

Double bond region

IR Spectroscopy

Answer 3

1600−1850 cm⁻¹

Question 4

Triple bond region

IR Spectroscopy

Answer 4

2100−2300 cm⁻¹

Question 5

Bonds to hydrogen

IR Spectroscopy

Answer 5

Greater than 2700 cm⁻¹

Question 6

sp³ C−H bonds

IR Spectroscopy

Answer 6

~2900 cm⁻¹

Question 7

sp² C−H bonds

IR Spectroscopy

Answer 7

~3100 cm⁻¹

Question 8

sp C−H bonds

IR Spectroscopy

Answer 8

~3300 cm⁻¹

Question 9

Effect of resonance on wavenumber

IR Spectroscopy

Answer 9

Lowers wavenumber by 20−40 cm⁻¹

More resonance structures stabilize a bond and thus decrease its energy

Question 10

Effect of symmetry on wavenumber

IR Spectroscopy

Answer 10

Symmetry dampens a signal making it invisible

Tetrasubstituted double bonds and internal triple bonds do not produce a signal when R groups are identical due to no net dipole moment

Question 11

Hydroxyl signal

IR Spectroscopy

Answer 11

−OH groups produce broad signals from 3200−3600 cm⁻¹ when in concetrated solutions due to hydrogen bonding

−OH groups produce narrow signals around 3600 cm⁻¹ when in when diluted in a solvent that cannot form hydrogen bonds with the alcohol

Question 12

Carboxylic acid signal

IR Spectroscopy

Answer 12

−COOH groups produce a very broad signal from 2200−3600 cm⁻¹ corresponding to the carboxylic −OH

A strong narrow signal at ~1700 cm⁻¹ will also be present corresponding to the carbonyl group

Question 13

Primary amine signal

IR Spectroscopy

Answer 13

−NH₂ groups produce two signals at 3350 and 3450 cm⁻¹

Question 14

Secondary amine signal

IR Spectroscopy

Answer 14

−NH groups produce one signal ~3450 cm⁻¹

Question 15

Carbonyl signals

IR Spectroscopy

Answer 15

C=O groups produce a strong narrow signal around or above 1700 cm⁻¹

Typically the most prominent signal

Conjugation will lower carbonyl signals by about 20−40 cm⁻¹

Question 16

Predicting molecular formula from (M+1)^+● peak

Mass Spectrometry

Answer 16

1. Determine number of carbon atoms in the compound by analyzing the relative abundance of the M+1 peak− (M+1)^+● relative height divided by the (M+)^+● relative height times 100; divide this value by 1.1 to find the number of carbons
2. Analyze the remaining mass of the molecular ion to determine what other atoms may be present; likely oxygen or nitrogen (will give an odd molecular weight)

Simplified below

Question 17

Chlorine (M+2)^+● peak

Mass Spectrometry

Answer 17

(M+2)^+● peak of chlorine will be approximately one-third the height of the molecular ion (M+)^+● peak

Question 18

Bromine (M+2)^+● peak

Mass Spectrometry

Answer 18

(M+2)^+● peak of bromine will be approximately the same height as the molecular ion (M+)^+● peak

Question 19

Analyzing IR fragments

Mass Spectrometry

Answer 19

MS can only detect charged species; including fragments

Thorough analysis can give structural information

Charged species will still prefer most stable cation or anion; the more abundant ion will be the more stable one

Question 20

Alcohol fragments

Mass Spectrometry

Answer 20

A signal at M−18 is characteristic of an alcohol

The hydroxyl group itself will not usually exhibit a molecular ion peak

Question 21

Calculating degree of unsaturation

Answer 21

A fully saturated alkane will have 2n+2 hydrogen atoms where n is the number of carbon atoms

Calculation of HDI

1. Count the total number of hydrogens in molecule; look at the molecular formula
2. Add one H for each halogen
3. Ignore oxygen
4. Subtract one H for each nitrogen
5. Subtract this number from 2n+2 and divide by 2

Question 22

HDI patterns

Answer 22

HDI 0

Compound is fully saturated and has no π bonds or rings

HDI 1

• One π bond
• One non-aromatic ring

HDI 2

• Two π bonds
• One triple bond
• Two non-aromatic rings
• One non-aromatic ring and one π bond

HDI 4

Potentially indicates an aromatic ring