UNIT 14 STRUCTURE ELUCIDATION BY …gpatpharmaplus.com/wp-content/uploads/2013/07/Structure...57 Structure Elucidation by Integrated Spectroscopic Methods • predict the spectral

56

Miscellaneous Methods

UNIT 14 STRUCTURE ELUCIDATION BY

INTEGRATED SPECTROSCOPIC

METHODS

Structure

14.1 Introduction Objectives

14.2 Molecular Formula and Index of Hydrogen Deficiency

14.3 Structural Information Available from Different Types of Spectra Mass Spectrum

UV-VIS Spectrum

IR Spectrum

1H-NMR Spectrum

14.4 Structure Elucidation of Organic Molecules

14.5 Summary

14.6 Terminal Questions

14.7 Answers

14.1 INTRODUCTION

This is the last Unit of this course. You have studied in details about the theoretical

aspects and the applications of different types of atomic and molecular spectroscopic

methods in the earlier Units. You have learnt that the molecular spectra contain

important information about the structure of the molecule. In this Unit you will learn

that how the information available from different types of molecular spectra can be

used for the structure elucidation of the Organic molecules. In this context we will use

the information available from UV-VIS, IR, 1H-NMR and mass spectrometry.

To facilitate this, we would first recapitulate the important structural information

available from different types of spectra and then, with the help of worked examples,

illustrate how to elucidate the structure of a molecule from its given spectral data. This

Unit, thus, aims to integrate the information available from different types of spectral

methods and apply this knowledge to arrive at the structure of a molecule. In other

words, you will be able to correlate the spectral signals of a molecule to different

structural units present in the molecule. Once you understand the correlation between

different spectral features and the structure of the molecule, you can also predict the

important spectral details for a given molecule from its structure.

Objectives

After studying this Unit, you should be able to:

• define and calculate the Index of Hydrogen Deficiency,

• discuss the importance of molecular formula and Index of Hydrogen Deficiency

in the structure elucidation of an organic compound,

• list important structural information available from UV-VIS, IR, 1H-NMR and

mass spectra of a molecule,

• correlate the signals and their intensities in different types of spectra of a

molecule to various structural units present in it,

• integrate the information available from different types of spectra to elucidate the

structure of a molecule, and

57

Structure Elucidation

by Integrated

Spectroscopic Methods

• predict the spectral data for a compound from its structural formula.

14.2 MOLECULAR FORMULA AND INDEX OF

HYDROGEN DEFICIENCY

We have mentioned that in this unit you will learn as to how the information available

from different types of molecular spectra can be used to elucidate the structure of a

molecule. It is interesting to know that in addition to spectra, the molecular formula can

also provide important structural leads. It is, therefore, worthwhile to know about the

kind of information available from the molecular formula.

You know that the molecular formula indicates the elements constituting the molecule

and the number of atoms of each element present in it. This can provide information

about the unsaturation and/or the ring structures in the molecule. The information can

be obtained by computing a parameter called Index of Hydrogen Deficiency (IHD). It

is defined as the number of pairs of hydrogen atoms that must be removed from the

corresponding saturated formula to give the molecular formula under consideration. Let

us learn about it with the help of some simple examples.

In case of ethane (C2H4), there are two i.e., one pair of hydrogen atoms less than that

required for the corresponding saturated compound having molecular formula C2H6.

Hence, its IHD will be 1. Similarly, for ethyne (C2H2), the number of hydrogen atoms

required for the saturated hydrocarbon would be C2H6 ‒ C2H2 = 4. Hence, the number

of pairs of H atoms needed would be 4/2= 2. So, IHD of ethyne is 2.

Let us now take the example of cyclopropane. What is its IHD? If we see the molecular

formula C3H6, we find that the numbers of hydrogen atoms are two less than that for the

corresponding saturated aliphatic hydrocarbon, i.e. propane. Thus, according to our

definition its IHD should be 1. You know, that this compound itself does not contain

any multiple bond or unsaturation per se. Then how do we account for the IHD of 1? In

this case, the IHD is due to the presence of a ring in the molecule. Thus, a ring

contributes to IHD in the same way as a double bond. Similarly, we can consider the

contributions of halogen atoms and nitrogen atoms on IHD. These considerations give

the following formula for the computation of the index of hydrogen deficiency for a

molecule from its molecular formula:

−=

2

atomshydrogen ofNumber atomscarbon ofNumber IHD

12

atomsnitrogen ofNumber

2

atomshalogen ofNumber +

+

−

You may note here that the divalent atoms such as oxygen and sulphur are not taken

into account while calculating the IHD. Thus, for a compound with the molecular

formula as C4H10O, the index of hydrogen deficiency would be zero as shown below:

IHD = 4 − 2

10+ 1 = 5 – 5 = 0

It implies that the molecule of C4H10O does not contain any unsaturation or a ring

structure. Why don’t you try to calculate the IHD for the compounds with molecular

formulas as given in the following SAQ.

SAQ 1

Calculate the IHD for the compounds having the following molecular formulae:

The determination of IHD

alongwith other tests/data

helps us to

choose between alternative

structures proposed for an

unknown compound.

The index of hydrogen

deficiency helps us to know

about the number of

multiple bonds (i.e. double

or triple) or rings present in

a molecule.

58


(a) C4H4 (b) C5H10O (c) C6H12O6

…………………………………………………………………………………………..

…………………………………………………………………………………………..

Having understood the importance of molecular formula in structure elucidation and the

method of computing the index of hydrogen deficiency, let us now take up a

comprehensive account of the information about the structural features/units of a

molecule available from the different types of spectra exhibited by it.

14.3 STRUCTURAL INFORMATION AVAILABLE FROM

DIFFERENT TYPES OF SPECTRA

In this section, you will learn about important structured information about a molecule

which can be obtained from its different types of spectra. Let us begin with the

information available from the mass spectrum of a molecule.

14.3.1 Mass Spectrum

i) The m/z value of the molecular ion, +•M gives the molecular mass and can also

be used for generating the molecular formula (subsec. 13.4.1).

ii) The relative intensities of +•+ 1][M and

+•+ 2][M peaks can be related to the

number and nature of hetero atoms present in a molecule. You would recall from

Unit 13 that a typical pattern of M+1 and M+2 peaks is observed if a chlorine or

bromine atom is present in the molecule.

iii) The odd molecular mass is indicative of the presence of a nitrogen atom in the

molecule. However, this has to be further confirmed by other means or by

analysing the fragmentation patterns for the typical nitrogen containing functional

groups.

iv) The characteristic peaks arising from typical fragmentation patterns of various

classes of functional groups such as α ‒ cleavage, loss of small molecules such

as H2O, C2H4, etc. are quite useful.

v) Certain peaks which may be attributed to the rearrangement of the molecular ion

or its fragments ions also give significant structural leads.

You are suggested to have a relook at the subsection 13.4.1 of Unit 13. Table 13.2

containing commonly lost fragments and stable fragment ions observed in the mass

spectrum is being reproduced here so as to facilitate you in the interpretation of the

mass spectra of the examples being taken up in the next section.

Table 13.2: Some commonly lost fragments and the stable fragments in the mass

spectrum

Commonly lost fragments

Fragment lost Peak obtained Fragment lost Peak obtained

CH3

.

+.M 15

OCH3

.

M - 31+.

OH.

+.M 17

Cl.

M - 35+.

CN.

+.M 26

CH3C O

.

+.M - 43

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by Integrated


H2C CH

2 +.M 28

OCH2CH

3

.

+.M - 45

CH2CH

3.

+.M 29

CH2

.

+.M - 91

Common stable ions

m/z values Ion

m/z = 43 CH

3 C O+

m/z = 91 CH

2

.+

.+

+.m/z = M - 1

CH

O .+

R C OR

14.3.2 UV-VIS Spectrum

i) An absorption in the UV-VIS region of the spectrum, i.e., the UV-VIS spectrum

of a molecule indicates the presence of certain functional groups that have

characteristic *π π *, π n →→ transitions. You may refer to Table 2.1 of Unit 2.

These transitions are so characteristic that the absence of a UV-VIS spectrum for

a molecule eliminates the presence of a number of functional groups in the

molecule.

ii) The λ max and the intensity of the absorption bands are indicative of the extent of

conjugation in the molecule; larger the wavelength, greater the conjugation.

You can refer back to Table 2.2 of Unit 2 for details of the characteristic UV

absorptions of various structural units in the molecules.

14.3.3 IR Spectrum

You would recall from Unit 3 of Block 1 that the region of IR spectrum ranges from

600-4000 cm‒1 and the infrared spectrum can be broadly divided into two regions as

explain below.

i) The region spanning from 3600 to 1200 cm‒1 is called the functional group region

and includes the stretching vibrations or group frequencies of common functional

groups. The absorptions in this region provide information about the presence of

characteristic functional groups.

ii) The region that includes all frequencies below 1200 cm‒1

is called the fingerprint

region and includes molecular vibrations, characteristic of the entire molecule or

large fragments of the molecule. This region is useful for confirming the identity

of a particular molecule by comparing with the spectral absorptions of its

authentic sample.

iii) In addition to the presence (or even the absence) of a particular functional group,

the analysis of IR spectrum gives the following information:

a) Presence of hydrogen bonding in the molecule (3200-3600 cm‒1

)

b) Geometry (cis or trans) in case of alkenes

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c) Substitution pattern of aromatic compounds (690-850 cm‒1)

Table 3.1 containing characteristic group frequencies of some important functional

groups is being reproduced here for your reference.

Table 3.1: Characteristic group frequencies of some important functional groups

1H-NMR Spectrum

Functional

group

Structure (IR

absorbing bond

in bold)

IR frequency

range

(cm‒1

)

Functional

group

Structure (IR

absorbing bond

in bold)

IR frequency

range

(cm‒1

)

Alkane (C-H) R3C-H

(R= H or C) 3000-2800

Alkyne

(C≡C)

RC CR (R=H or C)

2260-2100

Alcohol (O-

H) RO-H

3600-3200

(broad band)

Nitrile

(C≡N)

RC N (R=C)

2260-2200

Amines,

Amide

(N-H)

R

N H

R

------

3500-3300

(One band)

3500-3300

(two bands),

1650-1560

(bending)

Alkene

(C≡C)

R2C CR

2 (R=H or C)

1600-1680

Carboxylic

acid (O-H)

C

O

HR O

3200-2500

(broad band)

Benzene ring

(C=C)

(C=C ring

“breathing”)

1450-1600

2 to 3 bands

Aldehydes or

Ketones

(C=O)

O

C

RR

1750-1705

Alcohol or

Ether

(C-O)

---

(R=H or C)

1300-1000

Carboxylic

acid (C=O)

R

O

C

OH (R=H or C)

1790-1680

Alkyne

(C-H)

RC C H

(R=C)

3300

Amide

(C=O) R

C

O

NR2

(R-H or C)

1850-1800 and

1790-1730

Acid chloride

(C=O)

O

C

R Cl (R=H or C)

1815-1790

Acid

anhydride

(C=O)

C

O

O O

C

R R

(R=H or C)

1850-1800

and

1790-1740

(two bands)

Nitro

Compound

(NO2)

R

O

O

N

+

(R = C)

1660-1500

(asymmetric

stretch)

1390-1260

(symmetric

stretch)

61


by Integrated


You would recall from Unit 12, the following spectral features of the 1H-NMR and the

structural information available from them. Spectral data tells us about the following

important aspects of a molecule.

i) The number of different signals in the 1H-NMR spectrum indicates about the

different types of protons present in the molecule.

ii) The position of the signals i.e. their chemical shift values, tells about the

electronic environment of a particular proton. The chemical shifts of different

types of protons are given in Fig. 14.1.

Fig. 14.1: Range of Chemical shift values for different types of protons

iii) The area under the peaks obtained from the integrals for the signals of various

types of protons provides information about the ratio of the numbers of different

types of protons present in a molecule.

iv) The spin-spin splitting pattern of a particular signal gives information about the

number of neighbouring protons present around the given type of protons.

You can also refer back to Unit 12, particularly Sections 12.6, 12.7 and 12.10 for the

details of the above aspects. Having learnt about the information available from

different types of spectral data, we can now take up how to use it to arrive at the

structure of a given compound. You can study about this in detail in the next section

after achieving the following SAQ.

SAQ 2

Which spectrum you will be referring to in order to look for the presence of the

following in a molecule?

i) Conjugation

ii) Hydrogen bonding

iii) Hetero atom

…………………………………………………………………………………………..

…………………………………………………………………………………………..

…………………………………………………………………………………………..

…………………………………………………………………………………………..

14.4 STRUCTURE ELUCIDATION OF ORGANIC

MOLECULES

62


In this section, we will discuss the structural elucidation by taking some organic

compounds as examples. We will analyse the data available from various types of

spectra for a given compound and integrate the information about various structural

units present in the compound to arrive at its structure.

Example 1

A compound A, having molecular formula C3H6O, shows the following spectra. Deduce

its structure.

Fig. 14.2: Mass spectrum of compound A

Fig. 14.3: IR spectrum of compound A

63


by Integrated


Fig. 14.4: 1H-NMR spectrum of compound A

Solution

There is no unique way of solving a problem of structure elucidation. What we need to

do is to pick up all the important information available from the spectra and then

integrate them to arrive at the structure. Sometime one or two spectra give enough clues

to the structure. In such cases we suggest a tentative structure and check that whether it

supports the other data. Let us make a beginning.

IHD: We start with calculation of IHD.

−=

2


12


2


+

−

Therefore, IHD=3‒6/2 = 0

The IHD of zero clearly shows that the molecule does not have any unsaturation or a

ring structure.

IR: A strong absorption around 1700 cm‒1

is indicative of a carbonyl group (>C = O);

this may account for the oxygen in the molecule.

NMR: The NMR spectrum provides two important bits of information as given below.

i) A single signal indicates that only one type of protons is present in the molecule

i.e., all the protons are equivalent.

ii) The chemical shift of δ = 2.2 indicates that the protons are slightly deshielded;

may be due to being in the neighbourhood of an anisotropic group (>C = O)

Mass: The mass spectrum shows prominent peaks at m/z 15, 43, 58. The M+ ion

appears at m/z 58 while the base peak is at m/z 43. The M+ peak is in accordance with

the molecular formula and the base peak (M‒15) is indicative of a stable 3OCHC+

ion

64


which is obtained by the loss of 3HC•

radical from the molecular ion. Similarly the peak

at m/z =15 can be attributed to the loss of 3OCHC•

(M‒ 43) from the molecular ion.

Thus, the fragmentation pattern in mass spectrum indicates presence of CH3 and

COCH3 groups.

On the basis of the above information i.e. the presence of >C = O group (from IR),

− COCH3 and – CH3 (from MS) we may propose the structure of the compound A to be

the following.

CH3CH

3C

O

Compound A

This structure also explains the equivalence of all the protons observed in 1H-NMR

spectrum. Thus, we can say that the compound A is acetone having the structural

formula as shown above.

Example 2

A compound B, having molecular formula C7H8, shows the following spectra. From the

given spectra, we will now see what could be its possible structure.

Fig. 14.5: Mass Spectrum of Compound B

Fig. 14.6: IR Spectrum of Compound B

65


by Integrated


Fig. 14.7: 1H-NMR Spectrum of Compound B

Solution

IHD: Let us again start with the molecular formula and calculate its IHD, we get

−=

2


12


2


+

−

= 7 − 2

8+1

= 7− 4 +1

= 4

The IHD of four suggests that the compound does contain a ring and/or double bonds.

NMR: The NMR spectrum is quite useful in this case and provides the following

information.

i) There are two types of protons

ii) The integration of the peaks shows that the number of the two types of

protons is in the ratio of 5:3.

iii) The signal at δ =7.2 indicates the presence of benzene ring.

iv) The other signal for three protons at (δ =2.2) can be attributed to a –CH3

group attached to a benzene ring.

=> the presence of benzene ring accounts for the IHD of 4 (a ring and three

double bonds)

IR: There are very important pieces of information available from the IR spectrum.

i) The bands in the region of 1400‒1600 cm‒1

point to the presence of a

benzene ring; the signal at 1605 cm‒1 is quite characteristic of the same.

66


ii) Two strong peaks near 690 cm‒1 and 750 cm‒1 indicate monosubstituted

benzene ring. These are due to out of plane bending of aromatic = C − H.

The overtones of bands are observed between 2000 ‒1667 cm‒1

.

iii) The signal at about 3050 cm‒1

can be ascribed to sp2 C − H stretching,

while at about 2900 cm‒1

is due to sp3 C − H stretching.

Thus, the IR spectrum clearly points to a monosubstituted benzene.

Mass: The mass spectrum given in Fig. 14.5 shows the M+ peak at m/z 92 and the base

peak at m/z 91. Looking at the table we can say that this peak could be due to benzyl

carbocation, as shown in the following structure.

CH2

+

Benzyl carbocation

The spectral data evidences indicate that compound B to have the following structure

CH3

Toluene

And, hence the compound B is toluene.

Example 3

A compound C, having molecular formula C7H6O, exhibits the following spectra:

Fig. 14.8: Mass spectrum of compound C

67


by Integrated


Fig. 14.9: IR spectrum of compound C

Fig. 14.10: 1H‒ NMR spectrum of compound C

Solution

To arrive at its structure, we start with the determination of IHD from its molecular

formula. We get,

IHD = 7 − 2

6+ 1

= 7−3 + 1

= 5

Again, there may be a possibility that the ring and/or the double or triple bonds may be

present in the structure of the molecule.

IR: Similar to the previous compound the IR spectrum, suggests for a monosubstituted

benzene.

i) The aromatic C = C stretch is observed between 1600 −1450 cm‒1

.

68


ii) Two strong peaks near 690 cm‒1 and 750 cm‒1 are indicative of out of plane

= C − H bending for monosubstituted benzene.

iii) In addition, the strong signal at 1700 cm‒1

suggests for the carbonyl group.

iv) The doublet, between 2850 cm‒1

and 2750 cm‒1

is characteristic of C− H stretch

of aldehydes. This further qualifies the nature of carbonyl group.

Thus, an IHD of 4 out of 5 is accounted for by the presence of benzene unit (which

contains one ring and three double bonds) and the fifth by the carbonyl group.

Thus, you may see here that the IR spectrum has provided enough leads to suggest a

plausible structure. In the molecule, an aldehyde (−CHO) group is linked to the

monosubstituted benzene ring. Hence, the compound has the structure C6H5CHO. Let

us look into other spectra and see do they support the assignment.

NMR: The NMR spectrum also supports the proposed structure. Let us see the

available information.

i) The multiplet between δ 7.4 – 7.9 is suggestive of aromatic protons.

ii) The one proton signal at δ 9.8 is quite useful as it is suggestive of an aldehydic

proton.

The complex multiplet between δ 7.4 – 7.9 when resolved under high resolution shows

that there are three different types of aromatic protons and these are in the ratio of

2:1:2. The multiplet when observed at higher resolution (at 300 MHz) shows a doublet

(2H) and two triplets. The doublet at δ =7.5 is due to two ortho protons. The triplet due

to (1H) at ∼ δ 7.6 is due to para proton and the other triplet is due to 2 meta protons

appears at about δ 7.9. Thus, the 1H-NMR spectral data also supports that the

compound is benzaldehyde. Let us see what does the mass spectrum say.

Mass: The mass spectrum also supports the suggested structure. Let us analyse the

evidence.

i) The molecular ion gives an intense M+ peak at m/z = 106.

ii) An equally intense M – 1 peak at m/z 105 is suggestive of the loss of one

hydrogen via the α -cleavage process – a very favourable fragmentation

mechanism.

iii) The characteristic peak at m/z 77 can be due to the loss of – CHO to

give +56HC ion.

iv) A further loss of HC = CH from +56HC ion yields +

34HC ion; the peak for which

is observed at m/z 51 in the mass spectrum.

Thus, all the above spectral data confirm that the compound C is benzaldehyde and it

has the following structure. CHO

Benzaldehyde

Example 4

A compound D having molecular formula C3H6O2 shows the following spectral data.

Find out the structure.

69


by Integrated


Fig. 14.11: Mass spectrum of compound D

Fig. 14.12: IR spectrum of compound D

Fig. 14.13: 1H ‒NMR spectrum of compound D

Solution

You can compute the IHD for the molecule to be 1 suggesting a single double bond or a

ring.

IR: The IR spectrum of this molecule shows a feature that we have not observed so far.

Let us see what information do we get from IR.

i) The intense peak close to 1700 cm‒1

is characteristic of the carbonyl group.

ii) The broad band around 3000 cm‒1

is characteristic of – OH stretching; probably

hydrogen bonded – OH group.

70


These two pieces of information indicate towards the presence of a – COOH functional

group. Let us see other spectra for more information.

NMR: We can arrive at the following observations from its NMR spectrum.

i) The NMR spectrum gives three signals indicative of three different types of

protons.

ii) The small peak at a highly downfield δ = 10.5 which is likely to be due to the

acidic proton of the − COOH.

iii) The typical pattern of a two proton quartet and three proton triplet at about δ 1.2

and δ 2.3 respectively are suggestive of – CH2CH3 group.

On the basis of the two spectra, the tentative structure of the molecule can be proposed

to be CH3CH2COOH. Let us check whether the mass spectrum supports it or not.

Mass: The mass spectrum contains the following information.

i) The molecular ion peak M+ is at m/z 74.

ii) A M −17 peak at m/z 57 is indicating the loss of – OH.

iii) The intense peak at m/z 29 could be due to the loss of COOH.

iv) The peak at m/z 45 probably arises because of COOH+ ion.

As the mass spectrum clearly accounts for the –COOH group we can say that the

proposed structure is correct. The given compound D is propanoic acid with the

following structure.

CH3CH2COOH (Propanoic acid)

Example 5

A compound E having molecular formula C5H9O2Cl exhibits the following spectra.

Suggest its structure.

Fig. 14.14: Mass spectrum of compound E

71


by Integrated


Fig. 14.15: IR spectrum of compound E

Fig. 14.16: 1H-NMR spectrum of compound E

IHD: As usual we first calculate the IHD from the molecular formula.

IHD = 5 − 2

9−

2

1 + 1

= 5−5 + 1

= 1

So the molecule has a double bond or a ring. Let us see what do the spectra suggest?

IR: The following observations from IR spectrum are given below.

i) The strong band in the region 1750 – 1735 cm‒1

indicates C = O stretch.

ii) This along with two strong stretching absorptions in the range 1300 – 1000 cm‒1

due to C – O indicate the presence of ester function (-O-C=O)

72


Let us look into other spectra to identify the groups attached to the ester function.

NMR

i) The 1H-NMR spectrum shows the presence of four different types of protons.

ii) The two quartets at δ = 4.23 and at δ = 4.39 are for one proton and two protons

respectively.

iii) The other two signals, a doublet at δ = 1.68 and a triplet at δ = 1.31 are for two

protons each.

iv) One triplet and one quartet could be attributed to ethyl group (−CH2CH3).

v) The other quartet and a doublet may be due to CH3CH(Cl)‒ unit.

We are still not very sure but the tentative structure at this stage is given below.

CH3 CH C O CH

2CH

3

Cl O

let us see how does the mass spectrum help us.

Mass: The interpretation of the mass spectra is given below.

i) The M+ peak at m/z 136 is very-very small.

ii) The base peak at m/z = 29 suggests for the presence of – CH2CH3.

iii) The peak at m/z = 91 (M ‒ 45) is suggestive of the loss of – OCH2CH3 fragment;

the peak at m/z =45 corroborates this.

iv) The reasonably intense peak at m/z = 63 does go into support the proposed

structure as this may correspond to CH3CH(Cl)+ moiety which may be obtained

from the loss of CO from the m/z = 91 peak of CH3CH(Cl)CO+

v) The peak at m/z = 26 supports this.

Thus, we can say that the structure proposed above is correct and the given compound

has the following structure

CH3 CH C O CH

2CH

3

Cl O

Ethylchloropropionate

Example 6

A compound F has a molecular formula C4H11N and it exhibits the following spectra.

Determine its structure on the basis of these spectra.

73


by Integrated


Fig. 14.17: Mass spectrum of compound F

Fig. 14.18: IR spectrum of compound F

Fig. 14.19: 1H-NMR spectrum of compound F

Solution

If we calculate the IHD from the molecular formula of compound F by the formula

given below,

IHD = 1atomsnitrogenof.No

2

atoms hydrogenof.Noatomscarbonof.No ++−

2

We get,

IHD = 12

1

2

114 ++−

= 2

105 −

= 5 − 5 = 0

74


Thus, the compound has no unsaturation or double/triple bond or rings.

Mass: The compound contains nitrogen as is evident from its molecular formula. The

molecular ion M+ peak is at m/z at 73 which is in accordance with the nitrogen rule.

The base peak at m/z 30 indicates that the compounds could be a primary amine as

primary amines gives this characteristic peak due to α -cleavage as shown below.

[R CH2 − ] 22.

2 HNCHRNH+

• =+→

+

m/z = 30

IR: The interpretation of the IR signal is as follows.

i) The two bands in the range of 3500 – 3300 cm‒1

are supportive of the presence of

a primary amine as these bands could be arising due to the N – H stretching.

ii) The other amines are ruled out as the secondary amines show only one band in

this region and tertairy amines do not show any band here.

iii) The presence of amino group is supported by the broad band near 800 cm‒1

due

to the out of plane N − H bending and the C − N stretching band around 1090

cm‒1

.

NMR: The 1H-NMR spectrum shows 4 sets of signals at δ= 0.92 (triplet, 3H), δ= 1.16

singlet (2H, broad), δ= 1.3 – 1.5 (multiplet, 4H) and δ= 2.7 (triplet, 2H).Let us now try

to identify the structural units present in the compound on the basis of these signals and

their splitting patterns.

i) The three proton triplet at δ 0.92 could be due to the protons of a methyl group,

next to a − CH2 group, i.e. CH3 – CH2 – .

ii) The broad singlet due to two H at δ 1.16 could be due to –NH2 protons which do

not show coupling with neighbouring protons.

iii) The two proton triplet at δ= 2.7 is quite downfield and is likely to be for a – CH2

group attached to – NH2 group. The splitting into triplet, could be due to the

presence of a neighbouring –CH2 group. This yields the following structural unit.

− CH2 − CH2 − NH2

iv) The remaining 4 protons appear as a multiplet between δ 1.3-1.5.

If we put together this unit and the CH3 − CH2 − unit, we get the structure indicated

below:

CH3CH

2CH

2CH

2NH

2

Butanamine

Thus, on the basis of the spectra the compound is found to be butanamine with the

structure given above.

14.5 SUMMARY

The determination of molecular formula is an important aspect in the structure

elucidation of an organic compound. From the molecular formula, the Index of

Hydrogen Deficiency can be calculated. The knowledge of IHD helps to know about the

possibility of presence of unsaturation and/or rings in a molecule.

Different kind of information is available from various spectra of a molecule. The mass

spectral data helps us to know the molecular mass of a compound and the fragment ions

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of the molecule. An analysis of the fragment ion peaks indicates the type of functional

groups or structural units that may be present in the molecule. The UV spectral data

indicates about the extent of conjugation present in a compound. The IR spectral data

tells about different types of functional groups present in a compound. It also gives

information about the presence/absence of hydrogen bonding, substituent pattern of

aromatic rings etc. The 1H-NMR reveals the number of different types of protons

present in a molecule, their number and the electronic environment, etc.

Together, the variety of information available from different types of spectra leads us to

the elucidation of structure of an unknown compound. The structure-spectra correlation

also helps us to predict the spectral data for a given compound. The spectral studies,

thus, help a lot in structure determination and identification of compounds.

14.6 TERMINAL QUESTIONS

1. Calculate IHD for the following molecules:

(a) C6H6 (b) C4H8O2

(c) (d)

O

2. A compound G having molecular formula C6H6O exhibits the following spectra

as given in Fig. 14.20 to 14.22.

Fig. 14.20: Mass Spectrum of compound G

Fig. 14.21: IR Spectrum of compound G

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Fig.14.22: 1H-NMR Spectrum of compound G

Arrive at its structure and correlate different signals with structural units present in the

molecule.

3. Predict the important spectral signals which you would expect in the IR,

1H-NMR and mass spectra of the following compounds:

i) Ethyl ethanoate

ii) Benzoic acid

4. Which type of spectral data will you use to differentiate between isomeric

compounds having molecular formula C3H6O?

14.7 ANSWERS

Self Assessment Questions

1. (a) 3 (b) 1 (c) 1

2. (i) UV (ii) IR (iii) Mass

Terminal Questions

1. The IHD can be calculated with the help of following formula.

−=

2


+

−

2


2

atomshalogen ofNumber +1

i) IHD= (6-6/2) + 1 = 3+1 = 4

ii) IHD= (4-8/2) + 1 = 1

iii) The molecular formula for the given compound is C6H8. The corresponding

IHD is

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IHD= (6-8/2) + 1= 2+1=3

iv) The molecular formula for the given compound is C4H6O. The corresponding

IHD is

IHD= (4-6/2) + 1= 1+1=2

2. The structure of the compound G on the basis of the given spectra is

benzyl acetate

3. The most significant spectral signals expected in the IR, 1H-NMR and mass

spectra are as follows:

i) Ethylethanoate, CH3COOC2H5

Mass :

The molecular ion peak would appear at m/z = 88.

The fragment CH3CO+ is very stable and is likely to give an intense signal at m/z=43.

accordingly, we would expect a signal at M-43 i.e., at m/z =45. Other important

fragments are expected at m/z =29 and 15 (for ethyl group and methyl group).

IR :

The most important signal in the IR spectrum of ethyl eethanoate is for the carbonyl

absorption. As this is a part of the ester linkage it is expected to appear at around 1740

to 1750 cm-1.

NMR:

The NMR spectrum is expected to be quite simple. The ethyl part of the molecule

would give a combination of a triplet ( for CH3) and a quartet (for CH2). The methyl

group triplet is expected to be around δ = 1-1.5 and the CH2 quartet is likely to go

downfield to about 4 ppm.

As the other methyl group protons cannot have any coupling interactions, it is expected

to appear as a singlet around 2 ppm (due to the deshielding effect of the carbonyl

group).

ii) Benzoic acid, C6H5COOH

Mass

i) The molecular ion M+ peak is expected at m/z = 122.

ii) The loss of –OH is quite favorable and is expected to give a peak at

m/z=105

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iii) Similarly, the loss of –COOH is also favourable and is expected to give a

peak at m/z=77; the expected ion being +56HC

iv) A further loss of HC = CH from +56HC ion would yield +

34HC ion; the

peak for which may be observed at m/z 51.

IR

In benzoic acid we can expect 3-4 important signals/ features in the IR spectrum. These

are as follows:

i) Characteristic bands in the range of 1600 cm-1 for the benzene ring

ii) The carbonyl group at around 1700-1750 cm-1

ii) A broad signal about 3000 cm-1

for hydrogen bonding

iii) Characteristic pattern in the range of 700-750 cm-1 for the mono substituted

benzene.

NMR

The aromatic protons are expected as a multiplet between δ 7.4 – 7.9 due to the

difference in chemical shift positions of the ortho, meta and para protons and the

coupling between them.

The acidic proton of the –COOH group is expected to be quite downfield around 11-

12 ppm.

4. Let us first write down the structural formulae for the possible isomers of the

compound with the molecular formula, C3H6O.

CH2 CH CH

2 OH

CH3

C

OH

CH2

CH3 C

O

CH3

CH3

CH2 C

H

O

i)

ii)

iii)

iv)

These compounds can be differentiated by using UV-VIS, IR, NMR or Mass

spectrometry; NMR being the most convenient. The NMR of the compound

i) would give just one signal in NMR whereas the compounds ii) and iv)

would give three different signals each. On the other hand the compound iii)

would give as many as four signals. The compounds ii) and iv) can be further

differentiated by the location of the signals and the splitting pattern.

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SOME USEFUL BOOKS AND WEBSITES

1. Robert M. Silverstein, Francis X. Webster and David Kiemle (2005).

Spectrometric Identification of Organic Compounds, 7th edition. Wiley.

2. William Kemp (1991). Organic Spectroscopy, 3rd edition. W.H. Freeman &

Company.

3. Douglas A. Skoog, F. James Holler and Timothy A. Nieman. (1998) Principles

of Instrumental Analysis, 5th edition. Brooks/Cole.

4. S.M. Khopkar (2008) Basic Concepts in Analytical Chemistry, 3rd edition. New

Age International publishers.

http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr2.htm#nmr11

http://www.chemistry.nmsu.edu/studntres/chem435/Lab8/intro.html

http://www.chem.uic.edu/web1/ocol/spec/MS.htm

http://base-peak.wiley.com/

http://www.chem.ucla.edu/~webspectra/Solvingspectralproblems.html

80

Miscellaneous Methods INDEX

Absorption signals 26

Acetone 26, 63

Analysers 43

Anisotropic 19, 21, 31

Applications of mass spectrometry Qualitative applications 46

Quantitative applications 51

Applications of NMR spectroscopy

Qualitative applications 28 Ethanol 28

Methanol 30

Benzyl alcohol 30

Phenol 31

Base peak 39

Benzaldehyde 67, 68

Boltzmann constant 9

Boltzmann distribution law 9, 12

Broad or wide line 27

Butanamine 73

Characteristics of mass spectrum 38

Chemical exchange 30, 32

Chemical ionisation 42

Continuous wave spectrometry 12

Coupling constant 14, 23

Deshielding 15, 16, 17

Desorption sources 41

Diamagnetic effect 16, 17

Direct insertion probe 41

Double focussing analysers 44

Electromagnets 25, 43

Electron ionisation 42

Ethychloropropionate 71

Factors affecting chemical shift Anisotropy of chemical bonds19, 20, 33

Electronegativity 19

Hydrogen bonds 21

Field ionisation 42

Field sweep method 12

Fourier transform spectroscopy 14

Fourier transformation 14

Fragment ions 39

Fragmentation by cleavage at a single bond 48

by α-cleavage in molecules with heteroatoms 48

due to intramolecular rearrangement 48, 49

Free induction decay 14, 32

Frequency sweep method 12

Gas phase sources 41

Index of hydrogen deficiency 57, 58

Induced magnetic field 15, 16, 19

Instrumentation for mass spectrometry Inlet device 40, 41

Ionisation chamber 40

Analyser 40, 43

Detector 40, 44, 45

Processing and output devices 40, 45

Instrumentation for NMR spectroscopy Magnet 25

Sample probe 25

Detector system 25

Sample handling 26

Integral modes 26

Integration trace 26

Isotopic peaks 39

Larmor frequency 11

Larmor precession 10, 12

Longitudinal relaxation 13

Magnetic moment 7, 8, 9

Magnetic sector analyzer 43

Magnitude of coupling constants 23

Mass spectrum 38, 40

McLafferty rearrangement 48

Mechanism of resonance 11

Molecular ion 38

Multiplicity of spin-spin interactions 23

Natural abundance 39

Nitrogen rule 46

Nuclear g-factor 7

Paramagnetic effect 17

Pascal’s triangle 22, 23

Permanent magnets 25

Precessional motion 10, 11, 12, 34

Propanoic acid 69

Quantisation 8

Quantitative applications Analysis of multicomponent mixtures 27

Elemental analysis 27

Organic functional group analysis 28

Radical cation 38

Radical ion 38

Relaxation mechanisms 12, 13

Representation of NMR 26

Resolution or sharp line spectrum 27

Scan 14, 15

Shielding 15

Shielding constant 16

Shielding mechanism 16

Spin angular momentum quantum number 6, 7, 13, 32

Spin down 9

Spin flip 12

Spin up 9

Spin-lattice relaxation 13, 32

Spin-spin coupling 21, 23

Spin-spin relaxation 13

Standard for chemical shift 17

Structure elucidation of organic compounds 61

Superconducting magnets 25

Tetramethylsilane 17, 21, 26

Theory of mass spectrometry 38

Theory of NMR spectroscopy 6

Toluene 65

Transducer 41

Transverse relaxation 13

Types of nuclei 6

Unit of chemical shift 17

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