IR/NMR Spectroscopy

Question 1

What should you do first with an IR spectrum?

Answer 1

Draw a line at 3000 and look for peaks immediately to the left (sp2 C-H’s) and the right (sp3 C-H’s). Then look for O_H peaks at around 3600-2800, and C=O double bonds around 1800.

Question 2

Where does a carboxylic acid show up in NMR? An aldehyde proton?

Answer 2

Around 10-12ppm and 8-10ppm respectively.

Question 3

What solvent peaks are present in NMR?

Answer 3

CDCl3 produces a singlet at 7.3 ppm in HNMR and a triplet at 7.7 in CNMR, and water produces a broad singlet at about 1.6ppm.

Question 4

What are the units of IR or vibrational spectroscopy?

Answer 4

The wavenumber, or cm-1

Question 5

What are the two fundamental IR vibrations?

Answer 5

Stretching and bending.

Question 6

What are different qualitative aspects of peaks?

Answer 6

How large (intense) they are, aka strong or weak, and how much distance they cover (are they sharp or broad).

Question 7

Where can you find a table of IR vibrational frequencies for functional groups?

Answer 7

page 313 on the pdf.

Question 8

What is characteristic of carboxylic acids in IR spectra?

Answer 8

They have an unusually large O-H peak, from around 3200-2500 wavenumbers.

Question 9

Where do sp3, sp2, and sp C-H’s occur in an IR spectrum?

Answer 9

Around 2800-3000, 3000-3200, and 3300 respectively.

Question 10

Where are the triple bonds for CN and CC located in IR spectroscopy?

Answer 10

In the 2280-2100 region, strong to medium for the CN bond, and weak to medium for the alkyne. They are the only possible peaks in this region.

Question 11

Where are C=O bonds found in IR spectroscopy?

Answer 11

Around 1850-1630 wavenumbers, with a strong and sharp peak.

Question 12

Where are amide absorptions located on an IR spectrum?

Answer 12

Primary and secondary amides exhibit N—H stretching absorption bands in the 3500–3150 cm1 region, two bands for primary amides and one band for secondary amides.

Question 13

Where is the C=C region in an IR spectrum?

Answer 13

Around 1440-1680 wavenumbers, with intensities varying from weak to medium.

Question 14

Where is the C-O stretching bond located in an IR spectrum?

Answer 14

In the fingerprint region, so it will be very difficult to find.

Question 15

Where can we find good information on diagnostic bonds in the textbook?

Answer 15

Page 323 in the pdf.

Question 16

How are NMR spectrometers referred to by?

Answer 16

By their operating frequency, such as 60MHz.

Question 17

Where does an additional noise peak for deuterated chloroform occur?

Answer 17

At 7.26ppm, because some of the deuterium is actually protium.

Question 18

What happens in NMR spectroscopy as a carbon becomes more substituted?

Answer 18

The hydrogen attached to it becomes more deshielded, and is moved downfield (away from TMS).

Question 19

What affects the intensity of a peak in IR spectroscopy, and how is this relevant to peaks that should exist but do not?

Answer 19

The dipole moment, and since only stretching vibrations that affect the dipole moment are seen on the IR spectrum, we do not see stretching vibrations that do not affect the dipole moment, such as molecules that have perfect symmetry.

Question 20

Where can we find tables to predict NMR chemical shift of protons?

Answer 20

On page 351 in the pdf.

Question 21

What is a vinyl proton?

Answer 21

A proton attached to a C=C double bond.

Question 22

What is a geminal group?

Answer 22

A group attached to a =C double bonded carbon, which must also have a proton attached to the doubly bonded carbon.

Question 23

How can H-bonding be prevented?

Answer 23

By excessive steric hindrance in a molecule. This will lead to a sharp O-H peak rather than a broad one.

Question 24

Why does carboxylic acid have different IR spectra than other O-H bonds?

Answer 24

Because there are multiple sites for H-bonding, so the signal is much broader as opposed to other O-H bonds, and will often obscure the C-H bond stretch.

Question 25

What happens in IR to the location of bonds when you weaken the bond?

Answer 25

The wavenumber decreases, and moves to the right.

Question 26

Where would you get a signal for a secondary amine in an IR spectrum? A primary amine?

Answer 26

A single peak at around 3300 wavenumbers for a secondary amine, and two signals at around 3300 and 3400 for a primary amine. The reason for this is due to symmetric stretching in the secondary amine vs asymmetric (3400, takes more energy) and symmetric stretching (3300, takes less energy) in the primary amine.

Question 27

What does resonance do to a carbonyl bond in IR spectrum?

Answer 27

It reduces the strength of the carbonyl bond, which reduces the energy of the bond and the spring constant K, which slightly reduces the wavenumber at which the bond will be found.

Question 28

How do inductive effects affect a carbonyl bond in an IR spectrum?

Answer 28

Inductive effects pull electrons towards the bond and increase the strength of a double bond, thus increasing the spring constant, thus increasing the wavenumber. Inductive effects often compete with resonance effects.