Chapter 14: IR and Mass Spectroscopy Flashcards
Diagnostic region
IR Spectroscopy
Above 1500 cm−1
Analyzing an IR spectrum
IR Spectroscopy
Focus on the diagnostic region; analyze wavenumber, intensity, and shape for each signal
- 1600-1850 cm−1− check for double bonds
- 2100-2300 cm−1− check for triple bonds
- 2700-4000 cm−1− check for X−H bonds
Focus on the region above 3000 cm-1
Double bond region
IR Spectroscopy
1600−1850 cm−1
Triple bond region
IR Spectroscopy
2100−2300 cm−1
Bonds to hydrogen
IR Spectroscopy
Greater than 2700 cm−1
sp3 C−H bonds
IR Spectroscopy
~2900 cm−1
sp2 C−H bonds
IR Spectroscopy
~3100 cm−1
sp C−H bonds
IR Spectroscopy
~3300 cm−1
Effect of resonance on wavenumber
IR Spectroscopy
Lowers wavenumber by 20−40 cm−1
More resonance structures stabilize a bond and thus decrease its energy
Effect of symmetry on wavenumber
IR Spectroscopy
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
Hydroxyl signal
IR Spectroscopy
−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
Carboxylic acid signal
IR Spectroscopy
−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
Primary amine signal
IR Spectroscopy
−NH2 groups produce two signals at 3350 and 3450 cm−1
Secondary amine signal
IR Spectroscopy
−NH groups produce one signal ~3450 cm−1
Carbonyl signals
IR Spectroscopy
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
Predicting molecular formula from (M+1)+● peak
Mass Spectrometry
- 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
- 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
Chlorine (M+2)+● peak
Mass Spectrometry
(M+2)+● peak of chlorine will be approximately one-third the height of the molecular ion (M+)+● peak
Bromine (M+2)+● peak
Mass Spectrometry
(M+2)+● peak of bromine will be approximately the same height as the molecular ion (M+)+● peak
Analyzing IR fragments
Mass Spectrometry
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
Alcohol fragments
Mass Spectrometry
A signal at M−18 is characteristic of an alcohol
The hydroxyl group itself will not usually exhibit a molecular ion peak
Calculating degree of unsaturation
A fully saturated alkane will have 2n+2 hydrogen atoms where n is the number of carbon atoms
Calculation of HDI
- Count the total number of hydrogens in molecule; look at the molecular formula
- Add one H for each halogen
- Ignore oxygen
- Subtract one H for each nitrogen
- Subtract this number from 2n+2 and divide by 2
HDI patterns
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
