infrared spectroscopy and mass spectrometry

Chapter 15Infrared Spectroscopy and Mass Spectrometry

Organic ChemistrySecond Edition

David Klein

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e

15.1 Introduction to Spectroscopy• Spectroscopy involves an interaction between matter

and light (electromagnetic radiation)• Light can be thought of as waves of energy or packets

(particles) of energy called photons• Properties of light waves include wavelength and

frequency• Is wavelength directly or inversely proportional to

energy? WHY?• Is frequency directly or inversely proportional to energy?

WHY?

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 15-2 Klein, Organic Chemistry 2e

15.1 Introduction to Spectroscopy• There are many wavelengths of light that can not be

observed with your eyes


15.1 Introduction to Spectroscopy• When light interacts with molecules, the effect depends

on the wavelength of light used

• This chapter focuses on IR spectroscopyCopyright © 2015 John Wiley & Sons, Inc. All rights reserved. 15-4 Klein, Organic Chemistry 2e

• Molecular bonds can vibrate by stretching or by bending in a number of ways

15.2 IR Spectroscopy

• This chapter will focus mostly on stretching frequencies• WHY do objects emit IR light?• WHY do some objects emit more IR radiation than

others?• WHERE does that light come from?


• Some night vision goggles can detect IR light that is emitted

• IR or thermal imaging is also used to detect breast cancer



• The energy necessary to cause vibration depends on the type of bond



• An IR spectrophotometer irradiates a sample with all frequencies of IR light

• The frequencies that are absorbed by the sample tell us the types of bonds (functional groups) that are present

• How do we measure the frequencies that are absorbed?

• Most commonly, samples are deposited neat on a salt (NaCl) plate. WHY is salt used?

• Alternatively, the compound may be dissolved in a solvent or embedded in a KBr pellet



• Analyze the units for the wavenumber,• ν = frequency and c = the speed of light



• A signal on the IR spectrum has three important characteristics: wavenumber, intensity, and shape



• The wavenumber for a stretching vibration depends on the bond strength and the mass of the atoms bonded together

• Should bonds between heavier atoms require higher or lower wavenumber IR light to stretch?

15.3 IR Signal Wavenumber


• Rationalize the trends below using the wavenumber formula

1.

2.



• The wavenumber formula and empirical observations allow us to designate regions as representing specific types of bonds

• Explain the regions above



• The region above 1500 cm-1 is called the diagnostic region. WHY?

• The region below 1500 cm-1 is called the fingerprint region. WHY?



DIAGNOSTIC REGION FINGERPRINT REGION

• Analyze the diagnostic and fingerprint regions below



• Compare the IR spectra



• Greater difference in masses of atoms attached, greaterthe greater the wavenumber

• C-H stretch ≈ 3000 cm-1

• O-H stretch ≈ 3400 cm-1

• Practice with conceptual checkpoint 15.1



• Compare the IR stretching wavenumbers below

• Are the differences due to mass or bond strength?• Which bond is strongest, and WHY?



• Note how the region ≈3000 cm-1 in the IR spectrum can give information about the functional groups present



• Is it possible that an alkene or alkyne could give an IR spectra without any signals above 3000 cm-1?

• Predict the wavenumbers that would result (if any) above 3000 cm-1 for the molecules below



HO


• Resonance can affect the wavenumber of a stretching signal

• Consider a carbonyl that has two resonance contributors

• If there were more contributors with C-O single bond character than C=O double bond character, how would that affect the wavenumber?



• Use the given examples to explain HOW and WHY the conjugation and the –OR group affect resonance and thus the IR signal?




• The strength of IR signals can vary

15.4 IR Signal Strength


• The more polar the bond, the greater the opportunity for interaction between the waves of the electrical field and the IR radiation

• Greater bond polarity = stronger IR signals



• Note the general strength of the C=O stretching signal vs. the C=C stretching signal

• Imagine a symmetrical molecule with a completely nonpolar C=C bond: 2,3-dimethyl-2-butene

• 2,3-dimethyl-2-butene does not give an IR signal in the 1500-2000 cm-1 region



• Some IR signals are broad, while others are very narrow

• O-H stretching signals are often quite broad

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?


• 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


• 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



• H-bonding is often more pronounced in carboxylic acids, because they can forms H-bonding dimers



• For the molecule below, predict all of the stretching signals in the diagnostic region



O

OH


• 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



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


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



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 spectra1. Focus on the diagnostic region (above 1500 cm-1)

a) 1600-1850 cm-1 – check for double bondsb) 2100-2300 cm-1 – check for triple bondsc) 2700-4000 cm-1 – check for X-H bondsd) Analyze wavenumber, intensity, and shape for each signal


15.6 Analyzing an IR Spectrum• Often, the molecular

structure can be identified from an IR spectra

2. Focus on the 2700-4000 cm-1 (X-H) region

• Practice with SkillBuilder 15.1


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 15.2Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 15-39 Klein, Organic Chemistry 2e

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?

1) H-Br

2) Et-OK

O3

(CH3)2SO

O



1) H-Br

2) Et-OK

O3

(CH3)2SO

O

15.6 Analyzing an IR Spectrum



1) H-Br

2) Et-OK

O3

(CH3)2SO

O

15.8 Into to Mass Spectrometry• Mass spectrometry is primarily used to determine the

molar mass and formula for a compound1. A compound is vaporized and then ionized2. The masses of the ions are detected and graphed


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.

• What is a radical cation?


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.


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. • Neutral fragments are not detected.


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?


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


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


15.9 Analyzing the M+• Peak

• Like most compounds, the M+• peak for pentane is NOT the base peak

• The M+• peak fragments easily


Base peak

M+.

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


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.


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.

• The natural abundance of deuterium is 0.015%. Will that affect the mass spec analysis?



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?


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 15.23 and 15.24


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


15.12 Analyzing the Fragments

• WHAT type of fragmenting is responsible for the “groupings” of peaks observed?


15.12 Analyzing the Fragments• In general, fragmentation will be more prevalent when

more stable fragments are produced• Correlate the relative

stability of the fragments here with their abundances on the previous slide


15.12 Analyzing the Fragments• Consider the fragmentation below

• All possible fragmentations are generally observed under the high energy conditions employed in EI-MS

• If you can predict the most abundant fragments and match them to the spectra, it can help you in your identification


15.12 Analyzing the Fragments• Alcohols generally undergo two main types of

fragmentation: alpha cleavage and dehydration


15.12 Analyzing the Fragments• Amines generally undergo alpha cleavage

• Carbonyls generally undergo McLafferty rearrangement

• Practice with conceptual checkpoints 15.25 – 15.28


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


CnH2n+2

15.16 Degrees of Unsaturation• Notice that the general formula for the compound,CnH2n+2, 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?


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)


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?


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?

• Forget this formula below (you can use if you want)


15.16 Degrees of Unsaturation

• NOTrogen for Nitrogen

•Halogen

•0xygen (the O is a zero)


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



infrared spectroscopy and mass spectrometry

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