Page 1

ORGANIC SPECTROSCOPY Chem.345

BY

Dr. Ahmed El Taher

e-mail: taher2211@yahoo.com


Chemist needs to be able to determine the structures of compounds


CHAPTER 1 WHAT CAN BE LEARNED FROM MOLECULAR FORMULA? The purpose of this chapter is to describe how the molecular formula of a compound is determined and how structural information may be obtained from that formula

CHAPTER 2 INFRARED SPECTROSCOPY Infrared (IR) spectroscopy is used to determine the kinds of functional groups in an organic compound


CHAPTER 3 MASS SPECTROMETRY Mass spectrometry is used to determine the molecular mass and the molecular formula of an organic compound and to identify certain structural features of the compound

CHAPTER 4 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY Nuclear magnetic resonance (NMR) spectroscopy helps to identify the carbon-hydrogen framework of an organic compound


CHAPTER 1

WHAT CAN BE LEARNED FROM MOLECULAR FORMULA?


WHAT CAN BE LEARNED FROM MOLECULAR FORMULA?

You can determine the number of rings and / or double bonds


Determination of the molecular formula of a substance involves: Molecular weight determination

Quantitative elemental analysis

Qualitative elemental analysis

Types of atoms

Relative numbers of atoms


The first two steps establish an empirical formula for the compound The empirical formula the simplest whole-number ratios of the elements and may need to be multiplied by an integer to obtain the true molecular formula

Exercise: An amino acid has the percentage composition C 32.00%, H 6.71%, and N 18.66%. Calculate the empirical formula of this substance.


Index of Hydrogen Deficiency Unsaturation index It is the number of π bonds and / or rings a molecule contains.

Isopentyl acetate C7H14O2 Isopentyl acetate: C7H14O2

Nicotine: C10H14N2


 Each ring has one IHD  Each double bond has one IHD  Each triple bond has two IHD

IHD = 1

IHD = 6

IHD = 1

IHD = 3

IHD = 3

IHD = 5

IHD = 5


A hydrocarbon has a molecular formula of C6H8. It will react with hydrogen and a palladium catalyst to give a compound of formula C6H12. Give three possible structures.

Hydrogenation shows only two double bonds. Therefore, there must also be a ring. Pd C6H8 + 2 H2 C6H12


CHAPTER 2

INFRARED SPECTROSCOPY


What is Spectroscopy? Spectroscopy is the study of interaction of electromagnetic radiation with atoms and molecules

With light, you aren’t looking directly at the molecule but its “ghost”

Each type of spectroscopy, different light frequency, gives a different picture  the spectrum.


C = λν C ν λ E h ν

is is is is is is

,

E = hν

and

1 ν= λ

the speed of light (3 x 108 m/s) the frequency (Hz or s-1) the wavelength (m) the energy absorbed or emitted (J) Plank's constant (6.626 x l0-34 J.s) the wave number (cm-1)

High frequencies, large wave numbers, and short wavelengths are associated with high energy.


UV

NMR

Ms

IR


IR Spectroscopy When a molecule absorbs IR radiation, it will be excited to higher vibration energy state due to increase in the amplitude of the vibration motion of the bond in the molecule

Modes of vibrations There are two types of vibrations

which are IR active giving rise to absorption

Stretching vibration

Symmetric stretch

this is due to the change in bond length

Asymmetric stretch


Bending vibration

this is due to the change in bond angle

in-plane Scissoring

Rocking

out–of–plane Wagging :Note

ν asymmetric stretching

Twisting

> ν symmetric stretching

> ν bending


Functional Groups and Fingerprint Regions IR spectrum can be divided into two areas:

ď ľ In the range 4000-1400 cm-1 is where most of the functional groups show absorption bands, this is called the functional group region Each bond in the molecule has its own vibration frequency. Thus it will absorb a quantized IR radiation which is recorded on IR chart as specific bands

ď ś In the range 1400-600 cm-1 which is corresponding for the compound as a whole, this is called fingerprint region


The Position of Absorption Bands The amount of energy required to stretch a bond depends on the strength of the bond and the masses of the bonded atoms. Hooke’s Law:

ν = 4.12  Reduced mass = µ =

K µ

cm −1

M1 M2 , where M1 and M2 are atomic weights M1 + M2

 K is the force constant and its value depend on the type of the bond


 Stronger bonds show absorption bands at a larger wave numbers

 Lighter atoms show absorption bands at a larger wave numbers

 Hybridization affects the force constant also. As (s) character

increases, the bond becomes more strong so it will absorbs at higher wave number


The Intensity of Absorption Bands

 The intensity of an absorption band depends on the size of the

change in dipole moment associated with the vibration: The greater the change in dipole moment, the more intense the absorption.

 The intensity of an absorption band depends on the number of

bonds responsible for the absorption: C–H stretch will be more intense for a compound such as octyl iodide (17 C–H) than for methyl iodide (3 C–H).


The intensity of an absorption band depends on the concentration of the sample: Concentrated samples have greater numbers of absorbing molecules and, therefore, more intense absorption bands.

ď ˇ


Intensities referred to as: strong (s), medium (m), weak (w), broad, and sharp.


Not all bonds in a molecule are capable of absorbing IR energy, even if the frequency of the radiation exactly matches that of the bond motion Only those bonds which have dipole moment (polar bonds) are capable of absorbing IR radiation. Thus symmetric bonds such as H2 , Cl2 , ‌ symmetrically substituted alkenes or alkynes have no IR absorption bands and i.e. IR inactive

Lecture 1  

spectroscopy lecture , introdution , molecular formula, IHD, IR