Organic Spectroscopy, Chemistry tutorial

Introduction:

Modem spectroscopy processes or methods have revolutionized the study of organic chemistry. Almost all compounds which one encounters in an introductory organic chemistry course can be quickly recognized by using readily available spectroscopic methods. This contrasts with the traditional chemical characterization methods utilized before modem spectroscopy was available. It usually needs several hours, in several cases days, of laboratory work to completely characterize a compound by using chemical methods, while the same degree of characterization can frequently be accomplished in minutes by using spectroscopy.

Particularly helpful to practicing organic chemist are Infrared (IR) and Nuclear Magnetic Resonance (NMR) Spectroscopy. These two methods alone are often enough to find out the structure and purity of the compound once the molecular formula is found out by other MOMS.

Interaction of Light and Matter: Physical Basis of Spectroscopy

As simple as it sounds, spectroscopy comprises of shining light on matter and monitoring which wavelengths are absorbed and that wavelengths are reflected or transmitted. However the perception is very simple, it wasn't until the twentieth century that scientists start to understand the astonishing analytical power of probing the structure of matter by employing a broad range of wavelengths of light within and outside the visible range. The main breakthrough that led to this understood the quantum nature of light and matter and the quantum mechanical nature of their interaction. Once such relationships were understood, it was comprehended that photons of light can serve as the quantum probes, giving back accurate information regarding energy levels in molecules.

Properties of Light: Wavelength, Frequency and Energy:

However the word 'light' is routinely employed to refer to visible light, the part of the electromagnetic radiation spectrum that the human eye is sensitive to, the word light in its wider definition comprises the whole electromagnetic radiation spectrum, as illustrated in the figure shown below. The spectrum is continuous and covers a very broad range of wavelengths from radio waves having greater than 10 meter wavelength to gamma rays having wavelengths shorter than 10-12 meters.

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Fig: Electromagnetic spectrum-Wavelength, Frequency and Energy

As the speed of light is similar for all wavelengths, then for any particular wavelength, the relationship between frequency (v = number of waves per second) and wavelength is represented by:

v = c/λ

Here, c is the speed of light. According to this equation, wavelength and frequency are inversely proportional. The shorter the wavelength, the higher the frequency.

Moreover, to its wave properties, all light is known to exist as the small indivisible units, or particles termed as photons. For any particular wavelength of light, the energy of each photon is precisely similar. The photon energy can be readily computed by employing the relationship:

E = hv = hc/λ

Here, h = Plank's constant. Thus, the photon energy is inversely proportional to the wavelength and directly proportional to the frequency.

Energy States of Matter:

If matter is observed at the atomic or molecular level, all the energy changes are discontinuous; they occur in quantized steps or jumps. This is true, for all kinds of energy changes, whether they involve, for illustration, changes in the rotation of molecules, the vibration of atoms joined though chemical bonds, or the energy states of electrons in molecular orbitals. This quantum nature of matter taken altogether by the quantum nature of light gives interesting consequences if light interacts with matter. Whenever the energy of a photon of light precisely matches the difference between two energy levels in a molecule, the photon can be absorbed through the molecule and the photon's energy fully transferred to the molecule resultant in the disappearance of the photon. This absorption of light photons by a change in energy of molecules is the common base for all forms of spectroscopy.

Infrared Absorption Spectroscopy:

Infrared Light Absorption and molecular structure:

The photon energy of light in infrared range corresponds to differences in the vibrational energy levels of molecules. Absorption of infrared light yields in an increase in the amplitude of different molecular vibrations. To interpret infrared spectra, bonds between the atoms in molecules can be thought of as springs, and the atoms as point masses at the ends of the springs. According to this model, portion of the molecule can stretch and bend independently of the rest of the molecule. The frequency of stretching and bending can be associated to the atomic masses and the force constants of the springs as represented below:

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Fig: Frequency of stretching and bending

Here, v is the frequency of stretching or bending, 'k' is the force constant (that is, a measure of the stiffness of the spring) and 't' is the reduced mass (that is, incorporating the masses at the two ends of the spring). The reduced mass is stated as:

Here, M1 and M2 are the total masses at the two ends of spring.

Uses of Infrared Spectra:

Infrared spectra are employed for the identification and for structural determination. The simplest use of infrared spectra is to find out whether two samples are similar. The infrared spectrum of an organic compound gives such a detailed and unique pattern of vibrational absorption bands that it serves as a 'fingerprint' of the compound. If two samples are similar, their IR spectra, obtained under the similar conditions, must be the similar. If the samples are different, their spectra will be dissimilar.

Whenever synthesizing a known compound in the laboratory, comparison of the IR spectrum of the newly synthesized sample by a sample of known structure and purity is a convenient and sensitive manner to establish the identity and purity of the laboratory sample. The comparison of IR spectra of different fractions obtained in the fractional distillation or of material before and after Recrystallisation is one manner to follow the growth of purification.

Determination of Molecular Structure:

A more complicated level of interpretation of an ER spectrum is to utilize it to establish the structure of an unknown material. The comparison of IR spectra of substances of known structure has led to the establishment of numerous correlations between the wavelength (and frequency) of IR absorption and characteristics of molecular structure. The presence or absence of absorption at certain wavelengths points out the presence or absence of some functional groups or structural features. Establishing the presence or absence of carbonyl groups and alcohol groups is among the simplest and most definitive of functional group assignments which can be made by using IR. The C=O stretching absorption of any carbonyl group generally exhibits a distinctive strong band in the region of 1730-1670 cm-1. The figure below describes the IR spectrum of methyl benzoate (an ester).

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Fig: Determination of Molecular Structure

The carbonyl stretching band is the strong peak somewhat above 1700 cm-1. The O-H stretching absorption of an alcohol group illustrates up as a medium to strong band in the region 3700-3300 cm-1. The very strong, broad absorption band in the range of 3300 to 3500 cm-1 is the O-H stretching band. The broad appearance of this band is because of the extensive hydrogen bonding.

Sample Preparation:

For infrared absorption spectrometry the sample should be positioned in a beam of infrared radiation. The container or support for the sample should be transparent to infrared radiation and should thus be made up of one of a small number of materials that can't include glass. The most generally used material is sodium chloride, that is transparent between 5000 and 600 cm-1

Liquid Samples:

The spectrum of a pure liquid is most simply found out as a liquid film between a pair of sodium chloride plates. A drop of liquid is positioned in the center of one of the salt plates, and the second plate is put on top. The pair of plates having thin liquid film in between is then carefully positioned in a sample holder and the top of the holder is gently pressed in place to hold the plates altogether. The sample holder is then positioned in the sample beam of the instrument, after a background scan has been obtained. Whenever the strongest peak in the spectrum absorbs more than around 98% of the light, then the amount of sample is reduced via separating the plates and wiping one of the plates clean by a Chem-wipe tissue. After eliminating the sample from one side of the plates, the plates might be positioned altogether again in the holder and the other scan taken.

Solid Samples: Thin Film Method

By modem, high resolution FTIR instruments, good IR spectra can be obtained on the solid samples as thin films on sodium chloride plates. This is the easiest and simplest means to get the IR spectrum of a solid sample. A small quantity of sample (< 20 mg) is dissolved in a minimum amount (< 5 ml) of a dry, volatile solvent (example: dichloromethane). Some drops of the solution are put on the face of a sodium chloride plate and allowed to evaporate leaving a thin film of solid on the face of the plate. The second clean salt plate is placed over the film, the plates are put in a sample holder and the spectrum is obtained as with liquid thin films. Whenever the IR absorption bands in the spectrum are too weak, the salt plates are separated and some more drops of solution are evaporated to the plate having the film. If a NMR spectrum has been obtained on a solid sample dissolved in the typical NMR solvent (example: CDCl3) some drops of this solution evaporated onto a salt plate is usually sufficient to get the IR spectrum.

Solid Samples: The Solution Method

The IR spectra of solids are very often found out in solution. The concentration must be adjusted, if essential in such a way that the strongest peak will absorb between 90 and 98% of the light. The absorbance of a sample is proportional to both its concentration and thickness.

The perfect solvent would be transparent over the whole infrared range of wavelengths, however no such liquids are acknowledged. Due to this reason, the spectrum of the solvent should be subtracted from that of the sample dissolved in this solvent. This is completed in a double beam instrument by putting a second cell which includes pure solvent in the reference beam or in a single beam instrument via recording the spectrum of the pure solvent as background and then subtracting this background spectrum. Carbon tetrachloride and chloroform are the solvents most frequently employed for the infrared spectroscopy.

Solid Samples: KBr Method

The other method generally used to get the infrared spectrum of a solid is to prepare a potassium bromide salt pellet having the solid sample evenly dispersed all through the pellet. This process incorporates the solid sample right to the salt window. KBr pellets can be made by grinding 1-2 mg of sample with around 100-400 mg of anhydrous potassium bromide in a clean mortar and to press the resultant mixture into a translucent wafer, by using a die press. The wafer is then mounted in the sample beam.

Absorption in the -OH region in the spectrum obtained through the KBr pellet method should be interpreted by great caution, as it is hard to ensure that no water gets into the sample throughout the preparation.

Nuclear Magnetic resonance Spectroscopy:

Nuclear magnetic resonance spectrometry (NMR) assesses the absorption of light energy in the radio-frequency portion of the electromagnetic spectrum. The method of energy absorption comprises reorientation of magnetic nuclei with respect to the applied magnetic field. NMR spectrometry is mainly based on the fact that the energy of the photon of absorbed radio-frequency (rf) radiation goes into changing the orientation of the magnetic nucleus from being 'with' the magnetic field to being 'against' the field.

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Fig: Nuclear Magnetic resonance Spectroscopy

Spin:

Only atoms having spin can be viewed through NMR. Spin is stated by the mass number (that is, the number of protons and neutrons) and the atomic number (that is, the number of protons).

Spin 1/2 atom: Mass number is odd. Illustrations: 1H and 13C.

Spin 1 atom: Mass number is even. Illustrations: 2H and 14N.

Spin 0 atom: Mass number is even. Illustrations: 12C and 16O.

For the aim of this course, proton NMR (in which hydrogen nuclei are noticed) will be illustrated. Though, any nuclei with spin can be noticed (carbon-13 and fluorine-19 are as well common).

Shielding:

Electrons around the nuclei can 'shield' the nucleus from the consequences of the external magnetic field, lowering ΔE. Therefore, the magnetic field felt via any particular nucleus can be deduced an in equation:

Hnuc = Hext - Hshielding

Here, Hnuc is the field felt through the nucleus, Hext is the applied external magnetic field and Hshielding is the shielding or opposing magnetic field because of electrons in the vicinity of the nucleus.

Chemical Shift:

ΔE is proportional to the chemical shift, δ (delta) that is deduced in units of parts per million (ppm). The less shielding around nuclei, the larger the delta E and bigger the chemical shift. Tetramethylsilane (TMS) is employed as a reference compound (δ = 0 ppm) since all of the hydrogen nuclei are extremely strongly shielded by the large and uniform cloud of electrons in this molecule.

The chemical shift of a proton signal can provide important information regarding the shielding that in turn can give an indication of the kinds of nearby atoms or functional groups. Electronegative atoms like O, N, X, or double builds can reduce shielding ('dcshield') resultant in an increase in chemical shift.

Equivalent Nuclei:

The symmetry of molecules plays an extremely significant role in finding out the number of signals noticed in the NMR spectrum. Whenever two or more hydrogen nuclei are equivalent via symmetry they will encompass the similar chemical shift and they will give mount to one signal (however not essentially a single peak) in the NMR spectrum. For illustration, the hydrogen nuclei in 2-Propanone (acetone) are all equivalent via symmetry and give mount to one NMR signal while the hydrogen nuclei in 3-Methyl-2-butanone (that is, isopropyl methyl ketone) can be divided into three different groups, where the nuclei in each group are equivalent through symmetry, however the groups are not equivalent to one other via symmetry. The NMR spectrum includes 3 signals, one for each group. A NMR signal might be a single peak (that is, singlet) or a closely spaced group of peaks (that is, doublet, triplet, quartet, or other multiplet) that consists of a distinct chemical shift and represents one or more equivalent nuclei.

NMR Tube:

The spectrum is found out by the sample in an NMR tube, a thin-walled glass tube 5mm in diameter and 180mm long, sealed at the bottom. A tight-fitting plastic cap closes the top of the tube.

To prepare the NMR sample:

A) Put a small quantity of compound in a small disposable vial.

B) Transfer solvent through glass pipette into the vial to dissolve the compound.

C) Then transfer the solvent via the similar glass pipette into the NMR tube.

If there is certain insoluble material in the vial, then filter the solution via a pipette having glass wool or cotton.

Solving structures using NMR/IR:

1) If the molecular formula is given, compute the number of double bond equivalents (as well termed units of Unsaturation).

Remember: The number of double bond equivalents it the number of double bonds and/or rings present in the compound. If the # of DBE is 4 or greater, consider a benzene ring substructure.

For hydrocarbons or compounds having only C, H, and O:

# of DBE = [(2n + 2) - (actual # of H's)]/2

For each and every halogen in formula, add 1 to actual # of H's.

For each and every nitrogen in formula, subtract 1 from the actual # of H's.

2) If the infrared spectrum is given, look for the presence (or absence) of common functional groups. In specific, look for the C=O stretch of ketones, aldehydes, esters and so on (1650 - 1850 cm-1). The O-H stretch of alcohols and carboxylic acids, the N-H stretch of amines and the C=C stretch of alkenes as well encompass characteristic absorption peaks.

3) Observe the integrations in the NMR spectrum. Find out how many protons are symbolized by each signal (that is, each and every peak or multiplet).

4) Observe the chemical shifts of each set of peaks and compare them to those of common structure groups as illustrated in an NMR correlation table.

5) Look at the splitting patterns. For simple multiplets (that is, doublets, triplets, quarlets and so on) use the n + 1 rule to find out the number of nearest neighbor H's.

6) Write down the partial structures that can he deduced from the chemical shifts, integrations and splitting patterns.

7) Connect the pieces (that is, partial structures) altogether.

8) If there is more than one possible structure, draw each and every possible structure and find out which one is more completely consistent with all of the NMR and IR data.

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