Infrared Spectroscopy – used to analyze the presence of functional groups (bond types) in organic molecules It is the study of the interaction of infrared energy with organic molecules; the process analyzes molecules for molecular bond vibrations, measuring how fast bonds vibrate. How IR spectroscopy works:

1. A sample is placed in the infrared instrument (“IR”) and allowed to interact with infrared energy. When the molecule has a bond vibration that matches that of the infrared energy, the molecule absorbs that energy, preventing all of the energy from just passing through the molecule.

2. When the energy is absorbed, the instrument records that energy as not being entirely transmitted through a molecule. The energy that is absorbed has a specific frequency, thus identifying the (same) frequency of vibration of the bond in the molecule. This process works only because every type of bond we commonly look at in organic molecules has a different vibrational frequency. So the FREQUENCY VALUE tells you what TYPE OF BOND is vibrating.

3. When the analysis is complete, every frequency (and therefore bond type) has been noted on the spectra.

4. With every frequency identified, every bond type is identified, and functional groups can then also be identified.

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When the sample is placed in the instrument and the instrument scans a range of frequencies, looking to match a frequency with a frequency of vibration within the molecule, two things could happen: • When nothing matches, 100% of the energy just

passes through the molecule (100% photon energy is transmitted)

• When the frequency of the energy from the instrument matches that of the frequency of the vibration of the bond, the molecule absorbs the energy (less than 100% is transmitted).

The IR shows the range of frequencies of energy (X-axis) and how much energy is passing through (Y-axis). The peaks are those shown at frequencies when less than 100% energy is being transmitted (upside down peaks!).

Infrared Spectroscopy:

How to use the 5 zone approach to identify functional groups

Definition: Infrared Spectroscopy is the study of the Infrared Spectrum. An Infrared

Spectrum is the plot of photon energy (x axis) versus the amount of photons (y axis)

• X axis: the stretching frequency (wavelength)

• Y axis: the amount of photons absorbed (% Transmittance)

An example of an Infrared Spectrum:

The Infrared Spectrum is divided into 5 Zones

• Zone 1

o 3700-3200 cm-1

o Alcohol O-H

o Amide/Amine N-H

o Terminal Alkyne ! C-H

• Zone 2

o 3200 – 2700 cm-1

o Alkyl C-H (peak < 3000 cm-1

)

o Aryl or vinyl C-H (peak > 3000 cm-1

)

o Aldehyde C-H

o Carboxylic Acid O-H

• Zone 3

o 2300-2000 cm-1

o Alkyne C ! C

o Nitrile C ! N

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Y Axis: Percent Transmittance: Amount of energy that passes through the molecule • If no bonds are vibrating at frequencies of energy,

100% energy photon transmittance occurs • If bonds are vibrating at same frequency, energy is

absorbed, less than 100% is transmitted. X Axis: Wavenumbers: the frequency of infrared energy, expressed as reciprocal wavelength in centimeters (1/cm or cm-1) There are TWO Factors that affect the FREQUENCY of vibration (X axis): 1. Size of the atoms attached to the bond. Larger atoms

vibrate slower, at a lower frequency. Smaller atoms vibrate faster, higher frequency.

2. Bond length (the strength of the bond). Shorter stronger bonds vibrate faster, at a higher frequency. Longer, weaker bonds vibrate slower, lower frequency.

C-H versus C-Br

versusC C C C

sp C-H versus sp2 C-H sp3 C-Hversus

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The INTENSITY (Size) of the Peak (Y axis): 1. As far as molecule structure is concerned, the intensity of the peak (strong, medium, weak) is affected by the extent that the vibrational process changes the DIPOLE of the molecule. When the molecule absorbs infrared energy that matches the vibrational frequency, the bond that is affected undergoes significant bond stretching and compressing to dissipate the excess energy. The greater the stretching effect has on the dipole moment while vibration occurs, the larger the peak will appear on the IR spectrum. Non-polar:

Polar:

Polar covalent bonds will often be the largest (strongest) peaks on the IR spectrum. 2. The intensity of a peak is also affected by the number of any specific type of bond in a molecule. The more bonds of the same type will result in a larger number of data points acquired at a specific frequency, which will result in

C C When stretched: C C

non-polar no change to dipole

C O When stretched: C O

polar Large change to dipole moment

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the appearance of a larger peak than one would normally expect. This explains why sp3C-H bonds of alkyl groups can produce large peaks, even though they are non-polar. 3. Also remember that using too much or too little sample changes the amount of material available during acquisition of data on your molecule. Too much sample results in an overly large amount of data for EVERY bond, resulting in a spectrum where every peak looks strong (big). Too little sample will produce a spectrum where every peak appears weak (small) and nothing looks significant. Sample Infrared Spectrum:

Analysis of an IR Spectrum: The Infrared Spectrum is divided into 6 Zones: • Zone 1: 3700-3200 cm-1

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o Alcohol O-H o Amine or Amide N-H o Terminal Alkyne sp C-H (CΞC-H)

• Zone 2: 3200-2700 cm-1

o Aryl or vinyl sp2C-H (>3000 cm-1) o Alkyl sp3C-H (<3000 cm-1) o Aldehyde sp2C-H o ALSO: Carboxylic Acid O-H

• Zone 3: 2300-2100 cm-1

o Alkyne CΞC o Nitrile CΞN

• Zone 4: 1850-1650 cm-1

o Carbonyl (C=O) functional groups • Zone 5: 1680-1450 cm-1

o Alkene C=C o Aromatic ring pseudo double bonds

• Zone 6: “the Fingerprint Region” – 1500-600 cm-1

o Alcohol, Ester, Carboxylic Acid, Ether C-O (1300-1050 cm-1)

o Amine, Amide C-N o Aromatic substitution patterns (680-860 cm-1)

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1. Alkanes do not have a functional group but they do contain sp3C-H (3000-2800 cm-1) as do alkyl groups so these are often very prominent peaks on an IR spectrum.

Each functional group in the infrared region has distinct characteristics with bonds that you need to recognize and identify Alcohols, Phenols, Amines – all of these functional groups are capable of “hydrogen bonding”. Due to the intermolecular forces, the bond lengths for O-H/N-H vary in length resulting in “broad peaks”. Zone 1 (3500-3200 cm-1)

3-methylpentane

sp3C-H

OH OHNH

NH2

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2. Alcohols: “U” shaped peak, Zone 1 (O-H, 3500-3200 cm-1)

Look what happens to the “U” shaped peak when the alcohol is too sterically hindered to hydrogen bond to another molecule:

4-methyl-2-pentanol

O-H

sp3C-H

C-O

OH

O-H, NOThydrogen bonding

OH

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3. Amines: 1º “W” shaped peak, Zone 1 (2 N-H, 3500-3200 cm-1)

4. Amines: 2º “V” shaped peak, Zone 1 (N-H, 3500-3200

cm-1)

butylamine

NH

H

NH2

sp3C-H

diethylamine

N

H

N-H

sp3C-H

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5. Amines: 3º no visible peak in Zone 1 (no N-H, 3500-3200 cm-1)

Note: Amides will also have “W” and “V” peaks for NH2 and NH, respectively, as necessary. 6. Ethers: C-O peak, Zone 6 (1250-1050 cm-1)

N,N-dimethylethylamine

N

sp3C-H

O

C-Osp3C-H

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7. Alkyl Halides - C-X in Zone 6 o C-Cl 550-850 cm-1 o C-Br 500-650 cm-1

8. Alkenes

o C=C in Zone 5 (1680-1620 cm-1) o Sp2C-H in Zone 2, unless tetrasubstituted (3100-

3000 cm-1) Each of the following contains an alkene, except for the aromatic ring, which has a different pi bond system.

2-chloro-2-methylpropane

Cl

sp3C-H C-Cl

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Note how this disubstituted alkene shows a medium sized C=C peak, even though it is a non-polar bond. With two alkyl groups on the same carbon of the alkene, the electron cloud of this double bond is electronically lopsided, being electron-rich on one side (as drawn). Take a look at an alkene with a more electronically symmetrical double bond. The C=C peak is smaller between 1680-1620 cm-1 due to more even distribution of electron density, caused by having two alkyl groups attached, one on each end of the alkene:

2-methyl-1-pentene

C=C

sp2C-H

sp3C-H

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An even more electronically symmetrical alkene – can’t even see the peak for C=C between 1680-1620 cm-1:

9. Alkynes

o CΞC in Zone 3 (visible in asymmetric alkynes) (2300-2100 cm-1)

Terminal Internal

H H

H

H

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o Sp C-H in Zone 2 (terminal alkyne only; 3300 cm-1)

Like the alkenes, the more asymmetrical the electron distribution of an alkyne, the more obvious the CΞC peak. This makes the terminal alkyne typically the easiest CΞC peak to view on a spectrum. With just one methyl group on one end of an alkyne, see the size difference in the alkyne peak for 6-methyl-2-heptyne (Tiny CΞC peak between 2300-2100 cm-1):

1-hexyne

spC-H

spC-H

C C

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Now see what having an ethyl group does: The peak for CΞC peak is pretty much non-existent between 2300-2100 cm-1:

Spectral Database forOrganic Compounds SDBS

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10. Nitriles o CΞN in Zone 3 (more intense than CΞC)

Carbonyl-Containing Functional Groups – Ketones, aldehydes, carboxylic acids, esters, acid anhydrides, acid halides

o The carbonyl group, C=O, is typically the most intense (largest, strongest) peak on an IR spectrum

o Value shifts 30-50 cm-1 LOWER when next to aromatic ring due to resonance causing a lengthening of the C=O bond

NC

N

hexanenitrile

sp3C-H

C N

N

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11. Ketones § C=O only (1725-1705 cm-1)

12. Aldehydes

o Two major types of bonds:

§ C=O in Zone 3 (1695-1725 cm-1) § Sp2C-H “fangs” in Zone 2 (2830-2810 and

2740-2720 cm-1)

O O

4-methyl-2-pentanone

O

C=Osp3C-H

H

O

H

O

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13. Carboxylic Acids

o Three major types of bonds: § Extremely broad OH (3500-2500 cm-1) § C=O in Zone 3 (1680-1710 cm-1) § C-O in Zone 6 (1050-1250 cm-1)

Note that the spectrum below has OVERLAPPING PEAKS – the peak for the OH and the peaks for the sp3C-H stretches are overlapped to form a very odd shaped peak. [The pink box contains the portion from the OH bond.]

C=O

sp2C-H

sp3C-H

H

O

OH

O

HO

O

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It is not unusual to have overlapping peaks on an IR. Another example: The aldehyde peak between 2830-2810 cm-1 is blending in with the sp3C-H, making it harder to distinguish this as containing an aldehyde. Always easiest if you focus on finding the peak between 2740-2720 cm-1.

HO

O

C-O

C=O

sp3C-H

O-H

sp2C-H (visible)sp2C-H (buried bypeaks from sp3C-H)

H

O

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In this example the alcohol peak is oddly “pointed” due to the overlap of the OH peak with the sp C-H of the terminal alkyne:

14. Esters

o C=O in Zone 3 (1730-1750 cm-1) o C-O in Zone 6 (1150-1300 cm-1)

O-H

spC-H

O

O

O

O

OH

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15. Acid Anhydrides

o Two C=O in Zone 3 (1710-1770 ; 1780-1850 cm-1) o C-O in Zone 6 (1150-1250 cm-1)

O

O

sp3C-H

C=O C-O

O

O O

O

O O

propionic anhydride

O

O O

C-O2 C=O

sp3C-H

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16. Amides o C=O in Zone 3 (1640-1680 cm-1) o NH peak(s) in Zone 1 (“V” or “W” shaped) (3200-

3500 cm-1)

17. Acid Halides

o C=O in Zone 3 (1790-1810 cm-1) o C-X in Zone 6

NH

N

O

NH2

O O

pentanamide

NH2

O

C=O2N-H sp3C-H

Cl

O

Br

O

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18. Aromatic Rings

o “Pseudo double bond” peaks – one must appear ~1600 cm-1 and 1-2 more ~1500-1450 cm-1 (Zone 5)

o Sp2C-H peak (Zone 2, 3100-3000 cm-1) o Substitution patterns for monosubstituted and

disubstituted aromatic rings (Zone 6) • Mono Substituted (Zone 6: 680-710, 730-770 cm-1)

acetyl chloride

Cl

O

C=O

sp3C-H

C-Cl

Monosubstituted 1,2-disubstituted 1,3-disubstituted 1,4-disubstitutedortho meta para

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• Ortho Substituted (Zone 6: 735-770 cm-1)

sp3C-H

sp2C-H pseudoC=C

monosubstituted

pseudo C=C

Ortho

sp3C-H

sp2C-H

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• Meta Substituted (Zone 6: 810-750, 725-680 cm-1)

• Para Substituted (Zone 6: 860-800 cm-1)

19. Nitro (-NO2) – Spans Zones 5 and 6

o N=O/N-O (1300-1390 and 1500-1600 cm-1)

sp3C-Hpseudo C=C

Meta

sp2C-H

Para

pseudo C=C

sp3C-Hsp2C-H

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So – How do you SOLVE an IR? You need to break the IR up into the basic Zones, moving from left to right and only delving into the Fingerprint Zone if need be (i.e. you think you have a functional group that applies).

NO

O

N

O

O

NO2

sp3C-H

N=O/N-ON=O/N-O

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Example 1:

Zone 1: (3700-3200 cm-1) Alcohol OH present (“U”) (note the C-O in Zone 6) Amine or Amide absent – no “V” or “W” No sharp stretch at 3300 cm-1

Zone 2: (3200-2700 cm-1) No Aryl/vinyl sp2C-H (>3000 cm-1)

Sp3C-H present (<3000 cm-1) No aldehyde “fangs” between 2800-2700 No broad OH of carboxylic acid (no C=O in Zone 4)

Zone 3: (2300-2100 cm-1) No CΞC or CΞN

Zone 4: (1850-1650 cm-1) No C=O peak (must be longest, strongest peak)

Zone 5: (1680-1450 cm-1) No peak at 1600, etc – no aromatic ring No peak 1680-1620 – no alkene

Functional group? Alcohol

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Example 2:

Zone 1: (3700-3200 cm-1) Alcohol OH absent (“U”) Amine or Amide absent – no “V” or “W” No spC-H - sharp stretch at 3300 cm-1

Zone 2: (3200-2700 cm-1) No Aryl/vinyl sp2C-H (>3000 cm-1)

Sp3C-H present (<3000 cm-1) No aldehyde “fangs” between 2800-2700 No broad OH of carboxylic acid (no C=O in Zone 4)

Zone 3: (2300-2100 cm-1) No CΞC or CΞN

Zone 4: (1850-1650 cm-1) C=O present (note the C-O in Zone 6) Zone 5: (1680-1450 cm-1) No peak at 1600, etc – no aromatic ring No peak 1680-1620 – no alkene

Functional group? Ester

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Example 3:

Zone 1: (3700-3200 cm-1) Alcohol OH absent (“U”) Amine or Amide absent – no “V” or “W” No spC-H - sharp stretch at 3300 cm-1

Zone 2: (3200-2700 cm-1) Aryl/vinyl sp2C-H present (>3000 cm-1) Sp3C-H present (<3000 cm-1) No aldehyde “fangs” between 2800-2700 No broad OH of carboxylic acid (no C=O in Zone 4)

Zone 3: (2300-2100 cm-1) No CΞC or CΞN

Zone 4: (1850-1650 cm-1) No C=O present

Zone 5: (1680-1450 cm-1) Peak at 1600, etc – Aromatic Ring No peak 1680-1620 – no alkene

Zone 6: (850-680 cm-1) Peaks ~690, 750 – monosubstituted aromatic ring Functional group? Monosubstituted aromatic ring