CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

Subject Chemistry

Paper No and Title Paper 12: Organic Spectroscopy

Module No and Title 31: Combined problem on UV, IR, 1H NMR, 13C NMR and

Mass - Part III

Module Tag CHE_P12_M31

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

TABLE OF CONTENTS

1. Learning Outcomes

2. Introduction

3. Problems and their solutions

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

1. Learning Outcomes

After studying this module, you shall be able to

Solve problem related with electronic transitions

Learn how to differentiate molecule on the basis of IR spectroscopy

Correlate spectra with structure of compound

Interpret the spectroscopic data

2. Introduction

The knowledge and concepts of UV-visible, IR, 1H NMR, 13C NMR and Mass help us in

solving problems based on the experimental data. It will help us in analysing the

experimental data to elucidate the structure of any organic compound.

While analysing the data the following point must be kept in mind:

In UV-visible spectroscopy; the types of bonds and electrons plays important role in

understanding the electronic transitions.

UV-visible spectroscopy gives information regarding the presence of conjugation,

carbonyl group etc.

The IR values gives information regarding the functional group present in the

molecule

The 1H NMR tells us the number and environment of neighbouring hydrogens

present.

The 13C NMR helps in getting the information about the type of carbon atom(s)

present in the molecule.

Mass spectral data gives information about the total mass and fragmentation pattern of

the molecule.

By combining all the information one can find the structure of the molecule.

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

3. Problems and their solutions

Q. 1: How is PMR used to establish on which carbon, monochlorination of methyl ethyl ether

occurs?

A. 1: The three possible products with their spectra are described as follows:

a.) CH3-CH2-O-CH2Cl (a triplet, a more downfield quartet and a most downfield singlet);

(t) (q) (s)

b.) CH3-CHCl-O-CH3 (a doublet, a quartet downfield and a singlet between them);

(d) (q) (s)

c.) ClCH2-CH2-O-CH3 (a singlet and two triplets, the downfield order is: t1 < s < t2)

(t2) (t1) (s)

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

Q. 2: Problem: Find the structure of an unknown compound of molecular formula C9H12,

using its 1H NMR spectum and its IR spectrum.

1H NMR Spectrum:

IR Spectrum:

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

A. 2:

Step 1: Determine the degrees of unsaturation.

Just plug the formula C9H12 into the equation for degrees of unsaturation: Degrees of

Unsaturation = [((# of Carbons x 2)+2) - # of Hydrogens]/2.

[((9 x 2)+2) - 12]/2 = [20-12]/2 = 4.

(For non hydrocarbon elements: we treat nitrogen atoms as ½ of a carbon in the above

equation, halides as 1 hydrogen, and ignore oxygens.)

There are 4 degrees of unsaturation. With four or more degrees of unsaturation, you can often

expect to find an aromatic ring in your unknown compound.

Step 2: Look at the IR to find what functional groups are in the molecule.

You have a jumble of peaks between 1800 and 2000 cm-1 in the IR, indicative of an aromatic

ring. You can confirm the presence of the ring by looking at the NMR, at the peaks around 7

ppm.

Step 3: Find out how many hydrogens each set of peaks represents.

To get the relative ratios, it's best to use a ruler. Since that's difficult on a computer screen, I

can tell you that the relative integration for the peaks are 5:1:6 for the three sets of signals.

Since this adds up to the number of hydrogens in the molecular formula (12), the peaks at

around 7 ppm represents 5 hydrogens, the peak at 3 ppm represents 1 hydrogen, and the peak

at around 1 ppm represents 6 hydrogens.

Step 3: Find all of the compound fragments.

To find the compound fragments, look at the NMR integration. First, you notice the benzene

peaks at around 7.0 ppm, which is the piece you have already identified from the IR. It

integrates for 5H, so from the integration you know that the benzene ring is substituted

exactly once (6H for unsubstituted benzene - 1H), since one of the ring hydrogens is must be

replaced by a substituent. Write down this fragment (if you haven't already).

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

fragment worth C6H5

Now take the molecular formula and subtract out the aromatic piece you just figured out to

see what you have left. The benzene substituent is a C6H5 fragment, so if you subtract that

from C9H12 you get C3H7 (note that the benzene ring contains all 4 degrees of unsaturation:

one for the ring and three for the three double bonds. This means you can have no more rings

and no more double or triple bonds).

C3H7 is what you have left. Now take a look at the other integrations. You have a set of peaks

that integrates for 1H, so write that fragment down:

The bonds coming off marked as lines are simply bonds not attached to hydrogens. The other

peak is a doublet that integrates for 6H. Trouble you say? A carbon can't have six hydrogens

coming off of it? This is really no problem. It just means that there are two symmetric 3-

hydrogen groups (methyls) that are chemically equivalent in the NMR, so draw those .

, symmetric

That's all of the fragments of the molecule. You can check to make sure all atoms are

accounted for by subtracting out all of the fragments from the molecular formula. C9H12 -

(Benzene fragment C6H5) - (CH fragment) - (2 CH3 fragments) = C0H0. Good, all atoms are

accounted for.

Step 5: Piecing it all together.

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

At this point there is only one way that the pieces can go together, and that is as shown. The 4

fragments come together to make isopropyl benzene:

the four fragments come together to make

Step 6: Checking the answer.

You know this structure is correct because there was only one way to put the pieces together,

but let's just check the peak splitting to make sure. For the tertiary carbon with one hydrogen,

you would expect a septet (multiplet) that integrates for 1H. For the CH3s you would expect a

doublet that integrates for 6. For the aromatic ring, you would expect several peaks at around

7 ppm that integrates for 5H.

Q. 3: A compound (molecular formula: C4H10O) gave PMR spectrum consisting of two group

of lines (multiplets) with relative intensities in the ratio of 3:2. The PMR spectrum of another

compound having the same formula exhibited two lines with a relative area of 9:1. Identify

these substances.

A. 3: Diethyl ether and t-butylether

Q. 4: The PMR spectra of dimethyl formamide shows two signals at 2.84 and 3.0 δ for the

methyl protons at room temperature but a single sharp line appears at higher temperatures

(165 °C).

A. 4: At the room temperature, the lone pair of electrons on the nitrogen atom increases

sufficiently the double bond character of the (C-N) to restrict rotation at room temperature

and the spectrometer senses two different methyl groups (cis and trans) w.r.t lone hydrogen.

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

At elevated temperatures, however, the rotation around the C-N bond is so rapid that eacg=h

methyl group experiences the same time averaged environment.

Q. 5: Identify the unknown compound using the following spectral information.

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

A.

A. 5:

Mass Spectrum Analysis: The integral molecular weight of the unknown is 88. Because this

is an even number, the compound has either no nitrogen atoms or an even number of N

atoms. The highest m/z peaks do not show evidence of the presence of atoms such as Cl or Br

(which would have two peaks in 75:25 and 50:50 ratios, respectively).

Elemental Analysis: None was given

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

IR Analysis: The absorption at 3000 cm–1 (and not to the left) shows only aliphatic

(saturated) C–H groups. The carbonyl, C=O, absorbs strongly at 1740 cm–1. The two intense

peaks between 1000-1300 cm–1 indicate the carbonyl belongs to the ester group.

13C-NMR analysis: The NMR shows 4 different kinds of carbon atoms. The absorption at

172 ppm is typical of the carbonyl carbon. The upfield absorptions at 14 and 21 ppm are

likely –CH3 and/or –CH2– groups (although no C– H splitting patterns are given with this

spectrum). The absorption at 61 ppm could be a C next to an electronegative atom.

1H-NMR analysis: The NMR shows three different kinds of CH groups. The integration of

2:3:3 suggests a methylene (– CH2–) and two methyls (–CH3). The CH group absorbing as a

singlet cannot be next to a neighboring CH group (n+1=1 and so n=0). The upfield triplet

must be next to a C bonded to 2H’s (n+1=3) and the downfield quartet must be next to a C

bonded to 3 H’s (n+1=4). The triplet integrates for 3H’s and so must be a –CH3 group next to

a methylene; the quartet integrates for 2 H’s and so must be a –CH2– next to a methyl. Thus,

there is an ethyl group present, –CH2CH3.

Combined Analysis: The IR shows the only functional group present is an ester. There is no

aromaticity or other unsaturation in the molecule (except for the C=O). This means that the

absorption in the 13C-NMR at 61 ppm cannot be due to a C=C group and the carbon causing

the absorption is next to the oxygen atom of the ester. The presence of a C=O is confirmed in

the 13C-NMR. The ester group accounts for 44 amu of the 88 integral mass. The ethyl group

is on one side of the ester functional group and accounts for 29 mass units. The remaining

mass (88-44-29 = 15 and belongs to the singlet methyl group. The methyl is on the other side

of the ester functional group. Because the quartet is so far downfield in the range for

methylene 1H-NMR absorption, the methylene must be bonded to the O of the ester.

Unknown Identity: Thus, the unknown is identified as ethyl acetate.

Q. 6: Biphenyl exhibits a very intense absorption band (Ɛmax = 19000) at 250 nm but its 2,2’

derivative shows absorption almost similar to o-xylene (λmax 262, Ɛmax = 270).

A. 6: Since the angle of twist in biphenyl is small, therefore conjugation between the rings is

not affected. Biphenyl thus shows a very intense absorption band at 250 nm (K’ band). 2,2’-

dimethylbiphenyl with methyl substituents in ortho positions, however, is more stable in

twisted conformations as compared to biphenyl and suffers serious non-bonded compressions

from the juxtaposed substituents. Thus, 2,2’- dimethylbiphenyl may be regarded as two moles

CHEMISTRY

PAPER No. 12: Organic Spectroscopy

MODULE No. 31: Combined problem on UV, IR, 1H NMR, 13C NMR

and Mass- Part III

of o-xylene and shows absorption similar to o-xylene, due to loss of conjugation in the

twisted conformation.