SUCROSE

STRUCTURE DETERMINATION

Sucrose exists in two crystalline forms, a stable form, sucrose A, m.p. 184 -185 ⁰C, and the unstable form, sucrose B, m.p. 169 -170 ⁰C. Sucrose has been shown to be α-D-glucopyranosyl-β-D-fructofuranoside. Sucrose is hydrolysed by dilute acids or by the enzyme invertase to an equimolecular mixture of D (+)-glucose and D (-)-fructose. Methylation of sucrose gives octa-O-methylsucrose and this, on hydrolysis with dilute hydrochloric acid gives 2,3,4,6-tetra-O-methyl-D-glucose and 1,3,4,6-tetra-O-methyl-D-fructose. Thus glucose is present in the pyranose form, and fructose as the furanose.

Since sucrose is a non reducing sugar, both glucose and fructose must be linked via their respective reducing groups.

Sucrose is hydrolysed by the enzyme maltase, thus indicating the α-link is present. Again the glucose which is formed by hydrolysis of sucrose shows downward mutarotation. It therefore indicates that α-glucose is present in sucrose. There is an enzyme invertase which hydrolyzes methyl-β-fructofuranosides, and it has been found that it also hydrolyses sucrose. This suggest that fructose is present in sucrose in the β-form.

Oxidation of sucrose with periodic acid confirms its structure. Three molecules of periodic acids are consumed and one molecule of formic acid is produced. Consumption of three moles of HIO₄ indicates presence of six membered ring of a monosaccharide and five membered ring of another in sucrose. Formation of one mole of HCOOH confirms, it is the glucose which is six membered and fructose which is five membered.

Structure of sucrose

INVERSION OF SUCROSE

Hydrolysis of sucrose to D-glucose and fructose is very interesting in two ways. Firstly, the dextrorotatory sucrose gives the laevorotatory product on hydrolysis. The reason being the sucrose with  [α]_D^t = + 66.5⁰, on hydrolysis gives an equilibrium mixture of glucose, [α]_D^t = + 52.5⁰ and fructose [α]_D^t = - 92⁰. Now as the specific rotation value of fructose is high as compared to glucose as well as the parent compound sucrose, the mixture after hydrolysis will be, on the whole laevorotatory. Furthermore, since the direction of rotation is reversed (or inverted), the mixture of sugars formed on hydrolysis with a specific rotation of  - 20⁰ is known as invert sugar. Thus invert sugar is the equimolecular mixture of D-glucose and D-fructose obtained on hydrolysis of sucrose.

Hydrolysis of sucrose

The specific rotation of invert sugar is one half the sum of these of the individual monosaccharides:

1/2  [+ 52.5⁰ + (- 92⁰)] = - 20⁰ 

It is also interesting that the sucrose after hydrolysis is more sweeter than the sucrose itself. The reason being the presence of fructose in invert sugar which is the sweetest of all the sugars. This also explains why honey (containing a large proportion of invert sugar, that is formed by the hydrolysis of honey by the saliva of bees) is sweeter than sucrose. The relative sweetness of the common sugars as determined practically by taking the sucrose as an orbitary standard value of 100 are given below:

The relative sweetness of sugar

Secondly, hydrolysis of sucrose gives first of all α-D(+)-glucopyranose and β-D(+)- fructofuranose (this is believed to be dextrorotatory), but the letter is unstable and immediately changes into the stable form, D(-)-fructopyranose (the rotation of (-)-fructose is much greater than that of (+)-glucose).

 Ozone Layer

The ozone layer is mainly found in the lower region of the stratosphere in the atmosphere, the ozone layer lying between 15 km to 35 km from the earth's surface. Ozone is an allotrope of oxygen. Ozone is a pale blue colour gas with a distinctively pungent smell. The ozone layer is very useful for human beings, animals and plants living on the earth. The radiations emitted by the sun and reaching on the earth consists of ultra-violet radiations. These radiations are very harmful to the living organism. These radiations are of short wavelength and hence penetrate deep into our body and cause skin cancer. These radiations also damage crop plant. Ozone layer present in the stratosphere region of the atmosphere absorbs harmful ultra-violet radiations coming from the sun and thus prevent these radiations from reaching earth. Thus we are saved from the harmful effects cause by the ultra-violet radiations.

Formation of ozone layer

Ozone is an essential protective agent in the stratosphere. It is formed by photochemical dissociation of oxygen:

O₂ + hv -----> O• + •O*

(This requires light of  λ<242 nm, in the far UV.)

The activated oxygen atoms, O*, react with molecular oxygen to form ozone:

•O* + O₂ + M ------> O₃ + M 

The ozone formed in this way absorbs ultraviolet radiation with λ < 340 nm, regenerating molecular oxygen:

O₃ + hv -----> O₂ + O•

followed by

O• + O₃ ------> 2O₂  

This mechanism filters out much of the sun's ultraviolet radiation, protecting plant and animal life on the surface of the earth from other damaging photochemical reactions.

Causes of the ozone hole formation

The ozone hole formation occurs due to the compounds added to the atmosphere by humans. Chlorofluorocarbons, especially CF₂Cl₂  and CCl₃F, known as CFC 12 and 11 respectively are the most important type of these compounds. These compounds widely used as refrigerants, blowing agents for the manufacture of plastic foams.

The destruction of ozone by these compounds is caused, paradoxically, by their extreme stability and lack of reaction under ordinary conditions. Due to their stability, they remain in the atmosphere indefinitely and finally diffuse to the stratosphere. The intense high energy ultraviolet radiation in the stratosphere causes dissociation and forms chlorine atoms, which then undergo a series of reactions that destroy ozone:

CCl₂F₂ + hv -------> Cl• + •CClF₂ 

(This requires ultraviolet radiation with  λ = 200 nm)

Cl• + O₃ ------> ClO• + O₂

ClO• + O• -------> Cl• + O₂ 

(These two reactions remove  O₃ and oxygen atoms without reducing the number of chlorine atoms, Cl•.)

Other compounds, such as NO and NO₂, also contribute to the chain of events:

NO• + ClO• -------> •Cl + NO₂•

NO₂• + O₃ -------> NO₃• + O₂

NO• + O₃ ------> NO₂• + O₂

NO₂• + NO₃• + M -------> N₂O₅ + M

The chains are terminated by reactions such as

•Cl + CH₄ -------> HCl + •CH₃

•Cl + H₂ ------> HCl + H•

That are followed by combinations of the new radicals to form stable molecules such as CH₄, H₂, C₂H₆, and by a reaction that ties up the chlorine:

ClO• + NO₂• + M --------> ClONO₂ + M 

During the winter, a combination of air flow pattern and low temperature create stratospheric clouds of ice particles. The surface of these particles is an ideal location for reaction of NO₂, OCl, and O₃. These clouds contain nitric acid hydrate, formed by

N₂O₅ + H₂O -------> 2 HNO₃

ClONO₂ + H₂O ------> HOCl + HNO₃

These reactions, plus

HCl + HOCl ------> Cl₂ + H₂O

HCl + ClONO₂ ------> Cl₂ + HNO₃

remove chlorine from the air and generate Cl₂ on the surface of the ice crystals. In the spring, increased sunlight splits these molecules into chlorine atoms, and the decomposition of ozone proceeds at a much higher rate. The reaction with NO₂ that would remove ClO from the air is prevented because the NO₂ is mostly tied up as HNO₃ in the ice. The polar vortex prevents mixing with air containing a higher concentration of ozone, and the result is a reduced concentration of ozone over the Antartic. As the air warms in the summer, the circulation changes, the clouds dissipate, and the level of ozone returns to a more nearly normal level.

Volcanic activity that injects SO₂ into the atmosphere also has an effect that depends on temperature and on the height of the SO₂ injection. The SO₂ reacts with air to form SO₃, which then reacts with water to form sulfuric acid aerosols. These volcanic aerosols particularly at cold polar temperatures, reduced the nitrogen oxide concentration of the air and active chlorine species that destroy ozone. Because these aerosols are stable at warmer temperatures (200 K) than the natural stratospheric clouds, and because they can exist at lower altitudes, they can have significant effects. Until the level of chlorine is reduced to preindustrial levels, low temperatures and volcanic activity are likely to create Arctic ozone holes each spring as a result of reactions during the winter.

Within two years of the discovery of the ozone hole, the international community had accepted this as evidence of a global problem, and the Montreal Protocol on Substances that Deplete the ozone layer was signed. it set a schedule for decreasing use and production of CFCs and eventually for their complete ban.

Ozone Layer Depletion

Raman Spectroscopy 

If a beam of light is passed through a transparent substance, a small part of the radiation energy is scattered, this scattering persisting even if all dust particles are rigorously excluded from the substance. If the very narrow frequency band radiation or monochromatic radiation is used, the scattered energy will consist almost entirely of radiation of the incident frequency (Rayleigh scattering) but, in addition, certain distance frequencies above and below that of the incident beam will be scattered, this is referred to as Raman scattering.

Quantum Theory of Raman Scattering

Raman spectroscopy deals with the scattering of light and not with its absorption, like other conventional branches of spectroscopy. Consider a photon of frequency v falling on a molecule, if there is an elastic collision, then the scattered photon will have the same energy as the incident photon if there is an inelastic collision, the scattered photon will have either a higher or a lower energy than the incident photon. If we assume that the kinetic energy of both that is the photon and the molecule remains unchanged before and after the collision, then we write from the law of conservation of energy:

hv + E = hv' + E' ----- (1) 

Where hv is the energy of incident photon before the collision and hv' is the energy of the scattered photon after the collision, E is the molecular energy before the collision, and E' is the energy of the molecule after the collision. Rearranging equation 1, we get ----

(E'-E)/h = v-v' ----- (2)

Now consider the following cases :

Case I. v=v'         so that E=E'

 Case  II(a). v<v'   so that E>E'

   Case  II(b).  v>v'  so that E<E' 

Here case I is referred to as Rayleigh scattering, while the cases II(a) and II(b) referred to Raman scattering. In the case of Rayleigh scattering, the frequency of the scattered photon and the frequency of the incident photon is the same, while in the case of Raman scattering, when the collision occurred between the photon and the molecule there are two possibilities, one is energy from photon transferred to molecule and other is energy taken away from the molecule. The Rayleigh scattering and the Raman scattering are schematically represented as:

The Rayleigh scattering and the Raman scattering

If the molecule, excited to the higher unstable state when the molecule returns to the original vibrational state, we get Rayleigh scattering. If the molecule returns to a different vibrational state, we get Raman scattering (stokes lines). If the molecule, initially in the first excited vibrational state, excited to the higher unstable vibrational state, and finally returns to the ground state, this again gives rise to Raman scattering (anti-Stokes lines). Thus, the Raman spectrum of a molecule consists of Stokes lines and anti-Stokes lines, situated symmetrically about the Rayleigh line.

The Rayleigh line is far more intense than the Stokes line, and the Stokes line has a greater intensity than the anti-Stokes lines. The anti-Stokes lines are very difficult to observe in conventional Raman spectroscopy because these molecules are returned from an unstable excited vibrational state to the ground state and there are initially very few molecules in the excited vibrational state.

Classical Theory of Raman Scattering

When a molecule is put into a static electric field it suffers some distortion, the negative pole attracted the positively charged nuclei, and the positive pole attracted the electrons. This separation of charge centres causes an induced electric dipole moment to be set up in the molecule and the molecule is said to be polarized. The size of the induced dipole µ, depends both on the magnitude of the applied field, E, and on the case with which the molecule can be distorted. We can write

µ = αE ----- (3)

Where α is the polarizability of the molecule. The electric field vector E itself is given by:

E = E₀ sin 2πνt ----- (4)

Where E₀ is the amplitude of the vibrating electric field vector and v is the frequency of of the incident light radiation. Thus from equation (3) and (4), we get 

µ = αE₀ sin 2πνt ----- (5)

Such an oscillating dipole emits radiation of its own oscillation with a frequency v, giving the Rayleigh scattered beam, if , however, the polarizability varies slightly with molecular vibration, we can write

α = α₀ + (Әα/Әq) q ----- (6)

Where the coordinate q describe the molecular vibration. We can also write q as 

q = q₀ sin 2πν_mt ----- (7) 

Where q₀ is the amplitude of the molecular vibration and v_m is its frequency. From equation (6) and (7), we have

α = α₀ + (Әα/Әq) q₀ sin 2πν_mt ----- (8)

Substituting for α in equation (5), we have

µ = α₀E₀ sin 2πνt + (Әα/Әq) q₀E₀ sin 2πνt sin 2πν_mt ----- (9)

Making use of trigonometric relation :

sin x sin y = 1/2 [cos (x-y) - cos (x+y)]

This equation reduces to

 µ = α₀E₀ sin 2πνt + 1/2 (Әα/Әq) q₀E₀ cos 2π (ν-ν_m) t – 1/2  (Әα/Әq) q₀E₀ cos 2π (ν +ν_m) t ----- (10)

= α₀E₀ sin 2πνt + 1/2 (Әα/Әq) q₀E₀ x [cos 2π (ν-ν_m) t – cos 2π (ν+ν_m) t] ----- (11) 

Thus the oscillating dipole has three distinct frequency components : (i) the exciting frequency v with amplitude α₀E₀ (ii) v-v_m and (iii) v+v_m with very small amplitudes = (1/2) (Әα/Әq) q₀E₀. Hence, the Raman spectrum of a vibrating molecule consists of a relatively intense band at the incident frequency and two very weak bands at frequencies slightly above and below that of the intense band.

If, however, the molecular vibration does not change the polarizability of the molecule, then (Әα/Әq) = 0 so that the dipole oscillates only at the frequency of the incident (exciting) radiation. The same is true of the molecular rotation. We conclude that for a molecular vibration or rotation to be active in the Raman spectrum, it must cause a change in the molecular polarizability, i.e., Әα/Әq ≠ 0. --- (12)

Homonuclear diatomic molecules such as H₂, N₂, O₂, which do not show IR spectra because they do not possess a permanent dipole moment, do show Raman spectra since their vibration is accompanied by a change in polarizability of the molecule. As a consequence of the change in polarizability, there occurs a change in the induced dipole moment at the vibrational frequency.

Rotational Raman Spectra

For the vibrational Raman spectrum of a diatomic molecule, the selection rule is

∆ν = ±1 ----- (13)

For the rotational Raman spectrum of a diatomic molecule, the selection rule is

∆J = 0, ±2 ----- (14) 

For the rotational selection rule, the operative part is

∆J = +2 ----- (15)

Since ∆J = 0 corresponds to Rayleigh scattering and ∆J = -2 transition can be ignored as the rotational quantum number of the upper state must be greater than that of the lower state. Using the selection rule with the energy level expression for the rigid diatomic rotor, viz,

F(J) = BJ (J+1) cm^-1 (J = 0, 1, 2, 3 ---)

We find that ∆F(J) = F(J+2) - F(J) = B (4J+1) cm^-1 

Since ∆J = +2, we may level these lines S branch lines and thus,

∆F(J) = B(4J+6) cm^-1 (J = 0, 1, 2, 3 ---) --- (16)

Where J is the rotational quantum number in the lower state.

Thus if the rotational energy gains by the molecule from the photon during collision we have a series of S branch lines to the low wavenumber side of the exciting line (Stokes line), while if the molecule loses energy to the photon the S branch lines appear on the high wavenumber side (anti-Stokes lines). The wavenumber of the corresponding spectral lines are given by :

ν̅_s = ν̅_exc ± ∆F (J) = ν̅_exc ± B (4J +6) cm^-1 ----- (17)

Where the plus(+) sign refers to anti-Stokes lines, the minus (-) sign to Stokes lines, and ν̅_exc is the wavenumber of the exciting radiation.

The energy level diagram and the rotational Raman spectrum for a diatomic molecule are shown in the figure. We see that the first Stokes line (or the first anti-stokes line) appears at a distance of 6B from the exciting Rayleigh line. This is evident from equation 16, where putting J=0 we obtain ∆F (J) = 6B. The separation of the successive Raman lines is, however, 4B, as can be easily verified by putting J = 1, 2, 3, etc. in equation 17.

Rotational Raman spectrum of a rigid diatomic molecule

Rotation-Vibration Raman Spectrum

For the rotation-vibration Raman spectrum, the selection rules involve the changes in both the vibrational and the rotational quantum numbers. For the rotation vibration Raman spectrum of a diatomic molecule the selection rules are:

∆ν = +1;  ∆J = 0, ±2 ----- (18)

Since at room temperature most of the molecules are in the ground vibrational state (v = 0), so we see, only the vibrational transition from v = 0 to v = 1. The transition with ∆J = 0 form a Q-branch, the transition with ∆J = +2 form an S-branch, and the transition with ∆J = -2 form an O-branch. The rotational Raman transitions accompanying a 0→1 vibrational transition are shown in the figure. Here, ∆ν̅  measures the displacement from the exciting mercury line. The Q-branch exhibits an intense narrow line at ν̅_exc ± B that is usually unresolved while the S and O branches form weak wings which extend to lower and higher wave numbers, respectively, from the intense narrow line.

The rotation-vibration transitions in the Raman spectrum of a diatomic molecule