1. 15.5 IR Signal Shape
Some IR signals are broad, while others are very narrow
O-H stretching signals are often quite broad
Copyright 2012 John Wiley & Sons, Inc.

2. 15.5 IR Signal Shape
When possible, O-H bonds form H-bonds that weaken the O-H bond strength
The H-bonds are transient, so the sample will contain molecules with varying O-H bond strengths
Why does that cause the O-H stretch signal to be broad?
The O-H stretch signal will be narrow if a dilute solution of an alcohol is prepared in a solvent incapable of H-bonding
WHY does H-bonding affect the O-H bond strength?
Copyright 2012 John Wiley & Sons, Inc.

3. 15.5 IR Signal Shape
In a sample with an intermediate concentration, both narrow and broad signals are observed. WHY?
Explain the cm-1 readings for the two O-H stretching peaks
Copyright 2012 John Wiley & Sons, Inc.

4. 15.5 IR Signal Shape
Consider how broad the O-H stretch is for a carboxylic acid and how its wavenumber is around 3000 cm-1 rather than 3400 cm-1 for a typical O-H stretch
Copyright 2012 John Wiley & Sons, Inc.

5. 15.5 IR Signal Shape
H-bonding is often more pronounced in carboxylic acids, because they can forms H-bonding dimers
Copyright 2012 John Wiley & Sons, Inc.

6. 15.5 IR Signal Shape
For the molecule below, predict all of the stretching signals in the diagnostic region
Practice with conceptual checkpoint 15.9
Copyright 2012 John Wiley & Sons, Inc.

7. 15.5 IR Signal Shape
Primary and secondary amines exhibit N-H stretching signals. WHY not tertiary amines?
Because N-H bonds are capable of H-bonding, their stretching signals are often broadened
Which is generally more polar, an O-H or an N-H bond?
Do you expect N-H stretches to be strong or weak signals?
See example spectra on next slide
Copyright 2012 John Wiley & Sons, Inc.

8. 15.5 IR Signal Shape
Copyright 2012 John Wiley & Sons, Inc.

9. 15.5 IR Signal Shape
The appearance of two N-H signals for the primary amine is NOT simply the result of each N-H bond giving a different signal
Instead, the two N-H bonds vibrate together in two different ways
Copyright 2012 John Wiley & Sons, Inc.

10. 15.5 IR Signal Shape
A single molecule can only vibrate symmetrically or asymmetrically at any given moment, so why do we see both signals at the same time?
Similarly, CH2 and CH3 groups can also vibrate as a group giving rise to multiple signals
Practice with conceptual checkpoint 15.10
Copyright 2012 John Wiley & Sons, Inc.

11. 15.6 Analyzing an IR Spectrum
Table 15.2 summarizes some of the key signals that help us to identify functional groups present in molecules
Often, the molecular structure can be identified from an IR spectra
Focus on the diagnostic region (above 1500 cm-1)
cm-1 – check for double bonds
cm-1 – check for triple bonds
cm-1 – check for X-H bonds
Analyze wavenumber, intensity, and shape for each signal
Copyright 2012 John Wiley & Sons, Inc.

12. 15.6 Analyzing an IR Spectrum
Often, the molecular structure can be identified from an IR spectra
Focus on the cm-1 (X-H) region
Practice with SkillBuilder 15.1
Copyright 2012 John Wiley & Sons, Inc.

13. 15.7 Using IR to Distinguish Between Molecules
As we have learned in previous chapters, organic chemists often carry out reactions to convert one functional group into another
IR spectroscopy can often be used to determine the success of such reactions
For the reaction below, how might IR spectroscopy be used to analyze the reaction?
Practice with SkillBuilder 15.2
Copyright 2012 John Wiley & Sons, Inc.

14. 15.7 Using IR to Distinguish Between Molecules
For the reactions below, identify the key functional groups, and describe how IR data could be used to verify the formation of product
Is IR analysis qualitative or quantitative?
Copyright 2012 John Wiley & Sons, Inc.

15. 15.8 Into to Mass Spectrometry
Mass spectrometry is primarily used to determine the molar mass and formula for a compound
A compound is vaporized and then ionized
The masses of the ions are detected and graphed
Can you think of ways to get an organic molecule to ionize?
Will the molecule need to absorb energy or release energy?
Copyright 2012 John Wiley & Sons, Inc.

16. 15.8 Into to Mass Spectrometry
The most common method of ionizing molecules is by electron impact (EI)
The sample is bombarded with a beam of high energy electrons (1600 kcal or 70 eV)
EI usually causes an electron to be ejected from the molecule. HOW? WHY?
What is a radical cation?
Copyright 2012 John Wiley & Sons, Inc.

17. 15.8 Into to Mass Spectrometry
How does the mass of the radical cation compare to the original molecule?
If the radical cation remains intact, it is known as the molecular ion (M+•) or parent ion
Often, the molecular ion undergoes some type of fragmentation. WHY?
Copyright 2012 John Wiley & Sons, Inc.

18. 15.8 Into to Mass Spectrometry
The resulting fragments may undergo even further fragmentation
The ions are deflected by a magnetic field
Smaller mass and higher charge fragments are affected more by the magnetic field. WHY?
Neutral fragments are not detected. WHY?
Copyright 2012 John Wiley & Sons, Inc.

19. 15.8 Into to Mass Spectrometry
Explain the units on the x and y axes for the mass spectrum for methane
The base peak is the tallest peak in the spectrum
For methane the base peak represents the M+•
Sometimes, the M+• peak is not even observed in the spectrum, WHY?
Copyright 2012 John Wiley & Sons, Inc.

20. 15.8 Into to Mass Spectrometry
Peaks with a mass of less than M+• represent fragments
Subsequent H radicals can be fragmented to give the ions with a mass/charge = 12, 13 and 14
The presence of a peak representing (M+1) +• will be explained in section 15.10
Copyright 2012 John Wiley & Sons, Inc.

21. 15.8 Into to Mass Spectrometry
Mass spec is a relatively sensitive analytical method
Many organic compounds can be identified
Pharmaceutical: drug discovery and drug metabolism, reaction monitoring
Biotech: amino acid sequencing, analysis of macromolecules
Clinical: neonatal screening, hemoglobin analysis
Environmental: drug testing, water quality, food contamination testing
Geological: evaluating oil composition
Forensic: Explosive detection
Many More
Copyright 2012 John Wiley & Sons, Inc.

22. 15.9 Analyzing the M+• Peak
In the mass spec for benzene, the M+• peak is the base peak
The M+• peak does not easily fragment
Copyright 2012 John Wiley & Sons, Inc.

23. 15.9 Analyzing the M+• Peak
Like most compounds, the M+• peak for pentane is NOT the base peak
The M+• peak fragments easily
Copyright 2012 John Wiley & Sons, Inc.

24. 15.9 Analyzing the M+• Peak
The first step in analyzing a mass spec is to identify the M+• peak
It will tell you the molar mass of the compound
An odd massed M+• peak MAY indicate an odd number of N atoms in the molecule
An even massed M+• peak MAY indicate an even number of N atoms or zero N atoms in the molecule
Give an alternative explanation for a M+• peak with an odd mass
Practice with conceptual checkpoint 15.19
Copyright 2012 John Wiley & Sons, Inc.

25. 15.10 Analyzing the (M+1)+• Peak
Recall that the (M+1)+• peak in methane was about 1% as abundant as the M+• peak
The (M+1)+• peak results from the presence of 13C in the sample. HOW?
Copyright 2012 John Wiley & Sons, Inc.

26. 15.10 Analyzing the (M+1)+• Peak
For every 100 molecules of decane, what percentage of them are made of exclusively 12C atoms?
Comparing the heights of the (M+1)+• peak and the M+• peak can allow you to estimate how many carbons are in the molecule. HOW?
The natural abundance of deuterium is 0.015%. Will that affect the mass spec analysis?
Practice with SkillBuilder 15.3
Copyright 2012 John Wiley & Sons, Inc.

27. 15.11 Analyzing the (M+2)+• Peak
Chlorine has two abundant isotopes
35Cl=76% and 37Cl=24%
Molecules with chlorine often have strong (M+2)+• peaks
WHY is it sometimes difficult to be absolutely sure which peak is the (M)+• peak?
Copyright 2012 John Wiley & Sons, Inc.

28. 15.11 Analyzing the (M+2)+• Peak
79Br=51% and 81Br=49%, so molecules with bromine often have equally strong (M)+• and (M+2)+• peaks
Practice with conceptual checkpoints and 15.24
Copyright 2012 John Wiley & Sons, Inc.

29. 15.12 Analyzing the Fragments
A thorough analysis of the molecular fragments can often yield structural information
Consider pentane
Remember, MS only detects charged fragments
Copyright 2012 John Wiley & Sons, Inc.

30. 15.12 Analyzing the Fragments
WHAT type of fragmenting is responsible for the “groupings” of peaks observed?
Copyright 2012 John Wiley & Sons, Inc.

31. 15.14 High Resolution Mass Spec
MS is suited for the identification of pure substances
However, MS instruments are often connected to a gas chromatograph so mixtures can be analyzed
Copyright 2012 John Wiley & Sons, Inc.

32. 15.14 High Resolution Mass Spec
GC-MS gives two main forms of information
GC-MS is a great technique for detecting compounds such as drugs in solutions such as blood or urine
The chromatogram gives the retention time
The Mass Spectrogram (low-res or high-res)
Copyright 2012 John Wiley & Sons, Inc.

33. 15.15 MS of Large Biomolecules
To be analyzed by EI mass spec, substances generally must be vaporized prior to ionization
Until recently (last 30 years), compounds that decompose before they vaporize could not be analyzed
In Electrospray ionization (ESI), a high-voltage needle sprays a liquid solution of an analyte into a vacuum causing ionization
HOW is ESI relevant for analyzing large biomolecules?
ESI is a “softer” ionizing technique. WHAT does that mean?
Copyright 2012 John Wiley & Sons, Inc.

34. 15.16 Degrees of Unsaturation
Mass spec can often be used to determine the formula for an organic compound
IR can often determine the functional groups present
Careful analysis of a molecule’s formula can yield a list of possible structures
Alkanes follow the formula below, because they are saturated
Verify the formula by drawing some isomers of pentane
Copyright 2012 John Wiley & Sons, Inc.

35. 15.16 Degrees of Unsaturation
Notice that the general formula for the compound,
, changes when a double or triple bond is present
Adding a degree of unsaturation decreases the number of H atoms by two
How many degrees of unsaturation are there in cyclopentane?
Copyright 2012 John Wiley & Sons, Inc.

36. 15.16 Degrees of Unsaturation
Consider the isomers of C4H6
How many degrees of unsaturation are there?
1 degree of unsaturation = 1 unit on the hydrogen deficiency index (HDI)
Copyright 2012 John Wiley & Sons, Inc.

37. 15.16 Degrees of Unsaturation
For the HDI scale, a halogen is treated as if it were a hydrogen atom
How many degrees of unsaturation are there in C5H9Br?
An oxygen does not affect the HDI. WHY?
Copyright 2012 John Wiley & Sons, Inc.

38. 15.16 Degrees of Unsaturation
For the HDI scale, a nitrogen increases the number of expected hydrogen atoms by ONE
How many degrees of unsaturation are there in C5H8BrN?
You can also use the formula below
Copyright 2012 John Wiley & Sons, Inc.

39. 15.16 Degrees of Unsaturation
Calculating the HDI can be very useful. For example, if HDI=0, the molecule can NOT have any rings, double bonds, or triple bonds
Propose a structure for a molecule with the formula C7H12O. The molecule has the following IR peaks
A strong peak at 1687 cm-1
NO IR peaks above 3000 cm-1
Practice with SkillBuilder 15.4
Copyright 2012 John Wiley & Sons, Inc.

40. Study Guide for sections 15.5-15.16
DAY 20, Terms to know:
Sections diagnostic region, fingerprint region, primary amine, secondary amine, tertiary amine, mass spectrometry (MS), ionization, electron impact (EI), radical cation, molecular ion, fragmentation, base peak, electrospray, soft vs. hard ionization, degrees of unsaturation or hydrogen deficiency index (HDI)
DAY 20, Specific outcomes and skills that may be tested on exam 3:
Sections
Be able to explain how shape in IR spectra relates to the type of bond absorbing the radiation.
Be able to explain how H-bonding affects signals involving stretching of H atoms bonded to heteroatoms.
Be able to explain how N-H stretching works for both primary and secondary amines.
Given a molecule, be able to predict the key IR stretching peaks that would be observed.
Be able to explain how two molecules can be distinguished by differences in their IR spectra.
Be able to explain how ionization and detection of mass occurs in MS.
Be able to explain how the molecular ion forms in MS and how and why fragmentation often occurs.
Be able to explain why only charged molecules and fragments are observed in MS.
Be able to explain why the (M+1)+ peak is seen in MS and how its height relative to the M+ peak relates to the number of carbons in the molecule.
Be able to recognize the presence of Cl or Br in a compound based on MS data and the isotopic distribution.
Be able to use given data about the MS and IR to conclude information about a compound’s formula, degrees of unsaturation, and ultimately its structure.
Klein, Organic Chemistry 2e

41. Practice Problems for sections 15.5-15.16
Complete these problems outside of class until you are confident you have learned the SKILLS in this section outlined on the study guide and we will review some of them next class period
Klein, Organic Chemistry 2e

42. Day 21: EXAM 3 Prep for Day 22 Must Watch videos:
(how NMR works)
(number of NMR signals)
(shielding)
(chemical shifts)
(diamagnetic anisotropy)
Other helpful videos:
(1H NMR chemical shift)
(lecture 17-18)
Read sections
Klein, Organic Chemistry 2e