BIOMOLECULES CHEMISTRY CLASS 12

                           BIOMOLECULES ( CHEMISTRY )     

• Carbohydrates: Polyhydroxy aldehydes or polyhydroxy ketones or compounds on hydrolysis give carbohydrates.
• Classification of carbohydrates:
  Monosaccharides
  (a) Simplest carbohydrates
  (b) It cannot be hydrolysed into simpler compounds
  (c) Examples – Glucose, mannose
  Oligosaccharides
  (a) Carbohydrates which gives 2 to 10 monosaccharide units on hydrolysis
  (b) Examples – Sucrose, Lactose, Maltose
  Polysaccharides
  (a) Carbohydrates which on hydrolysis give large number of monosaccharide units.
  (b) Examples – Cellulose, starch
• Anomers: Pair of optical isomers which differ in configuration only around C1 atom are called anomers. Examples – ALPHA-D-glucopyranose and BETA-D-glucopyranose.
• Epimers: Pair of optical isomers which differ in configuration around any other C atom other than C1 atom are called epimers. E.g. D-glucose and D- mannose are C2epimers.
  

Preparation of glucose (also called dextrose, grape sugar):

• From starch

• Structure of glucose

• Structure elucidation of glucose:

a) D – glucose with HI

b) D – glucose with HCN

c) D – glucose with NH2OH

d) D- glucose with Fehling’s reagent

e) D – glucose with Tollen’s reagent

f) D – glucose with nitric acid

 

g) D – glucose with (CH3CO)2O and ZnCl2

h) D – glucose with bromine water

i) Glucose with phenylhydrazine to form osazone

Glucose and fructose gives the same osazone because the reaction takes place at C1 and C2 only.

• Other Reactions of Glucose (Presence of ring structure)

Glucose does not give Schiff’s test and does not react with sodium bisulphite and NH3. Pentaacetyl glucose does not react with hydroxyl amine. This shows the absence of –CHO group and hence the presence of ring structure.

• Cyclic structure of glucose:

• Haworth representation of glucose:

• Cyclic structure of fructose:

• Haworth representation of fructose

• Glycosidic linkage: The oxide linkage formed by the loss of a water molecule when two monosaccharides are joined together through oxygen atom is called glycosidic linkage.

• Sucrose (invert sugar):

a) Sucrose is a non-reducing sugar because the two monosaccharide units are held together by a glycosidic linkage between C1 of ALPHA-glucose and C2 of BETA– fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar.

b) Sucrose is dextrorotatory but on hydrolysis it gives dextrorotatory & laevorotatory and the mixture is laevorotatory.

• Haworth Projection of Sucrose:

• Maltose:

1. Maltose is composed of two α-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II).
2. The free aldehyde group can be produced at C1 of second glucose in solution and it shows reducing properties so it is a reducing sugar.
   

• Haworth projection of maltose:
  
• Lactose (Milk sugar):It is composed of β-D-galactose and β-D-glucose. The linkage is between C1 of galactose and C4 of glucose. Hence it is also a reducing sugar.
  
• Haworth projection of lactose:

• Starch: It is a polymer of -glucose and consists of two components — Amylose and Amylopectin.

• Amylose:

1. It is a water soluble component
2. It is a long unbranched chain polymer
3. It contains 200 – 1000 ALPHA-D-(+)- glucose units held by 1,5– glycosidic linkages involving C1 – C4glycosidic linkage
4. It constitutes about 15-20% of starch

• Amylopectin

1. It is a water insoluble component
2. It is branched chain polymer
3. It forms chain by C1 – C4glycosidic linkage whereas branching occurs by C1 – C6glycosidic linkage
4. It constitutes about 80-85% of starch

• Cellulose:

1. It occurs exclusively in plants.
2. It is a straight chain polysaccharide composed only of ALPHA-D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

• Glycogen:

1. The carbohydrates are stored in animal body as glycogen.
2. It is also known as animal starch because its structure is similar to Amylopectin.
3. It is present in liver, muscles and brain.
4. When the body needs glucose, enzymes break the glycogen down to glucose.

• Amino acids:

Amino acids contain amino (–NH2) and carboxyl (–COOH) functional groups.

Where R – Any side chain
Most naturally occurring amino acids have L – Config.

• Types of amino acids:

a). Essential amino acids: The amino acids which cannot be synthesised in the body and must be obtained through diet, are known as essential amino acids. Examples: Valine, Leucine

b). Non-essential amino acids: The amino acids, which can be synthesised in the body, are known as non-essential amino acids. Examples: Glycine, Alanine

• Zwitterion form of amino acids:

1. Amino acids behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxyl group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and negative charges.
2. In zwitterionic form, amino acids show amphoteric behaviour as they react both with acids and bases.

• Isoelectronic point: The pH at which the dipolar ion exists as neutral ion and does not migrate to either electrode cathode or anode is called isoelectronic point.

• Proteins: Proteins are the polymers of ALPHA-amino acids and they are connected to each other by peptide bond or peptide linkage. A polypeptide with more than hundred amino acid residues, having molecular mass higher than 10,000u is called a protein.

• Peptide linkage: Peptide linkage is an amide linkage formed by condensation reaction between –COOH group of one amino acid and –NH2 group of another amino acid.

Peptide link age

• Primary structure of proteins: The sequence of amino acids is said to be the primary structure of a protein.

• Secondary structure of proteins: It refers to the shape in which long polypeptide chain can exist. Two different types of structures:

ALPHA-Helix:

1. It was given by Linus Pauling in 1951
2. It exists when R- group is large.
3. Right handed screw with the NH group of each amino acid residue H – bonded to – C = O of adjacent turn of the helix.
4. Also known as 3.613 helix since each turn of the helix hasapproximately 3.6 amino acids and a 13 – membered ring is formed by H – bonding.
5. C = O and N – H group of the peptide bonds are trans to each other.

BETA-pleated sheet:

1. It exists when R group is small.
2. In this conformation, all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by hydrogen bonds.

• Tertiary structure of proteins: It represents the overall folding of the polypeptide chain i.e., further folding of the 2° structure.

• Types of bonding which stabilize the 3° structure:

1. Disulphide bridge (-S – S-)
2. H – bonding – (C = O … H – N)
3. Salt bridge (COO– … + Na+)
4. Hydrophobic interactions
5. van der Waals forces

• Two shapes of proteins:

Fibrous proteins
a) When the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, then fibre– like structure is formed.
b) These proteins are generally insoluble in water
c) Examples: keratin (present in hair, wool, silk) and myosin (present in muscles), etc

Globular proteins
a) This structure results when the chains of polypeptides coil around to give a spherical shape.
b) These are usually soluble in water.
c) Examples: Insulin and albumins

• Quaternary structure of proteins:

1. Some of the proteins are composedof two or more polypeptide chains referred to as sub-units.
2. The spatial arrangement of these subunits with respect to each other is known as quaternary structure of proteins.

• Denaturation of proteins:

1. The loss of biological activity of proteins when a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH. This is called denaturation of protein.
2. Example: coagulation of egg white on boiling, curdling of milk.

• Nucleoside:

1. Base + sugar

• Nucleotide:

1. Base + sugar + phosphate group

• Nucleic acids (or polynucletides):

1. Long chain polymers ofnucleotides.
2. Nucleotides are joined by phosphodiester linkage between 5’ and 3’ C atoms of a pentose sugar.

• Two types of nucleic acids:DNA

1. It has a double stranded ALPHA-helix structure in which two strands are coiled spirally in opposite directions.
2. Sugar present is –D–2-deoxyribose
3. Bases:
   i) Purine bases: Adenine (A) and Guanine (G)
   ii) Pyrimidine bases: Thymine (T) and cytosine (C)
4. It occurs mainly in the nucleus of the cell.
5. It is responsible for transmission for heredity character.RNA

1. It has a single stranded ALPHA-helix structure.
2. Sugar present is ALPHA–D–ribose
3. Bases:
   i) Purine bases: Adenine (A) and Guanine (G)
   ii) Pyrimidine bases: Uracil (U) and cytosine (C)
4. It occurs mainly in the cytoplasm of the cell.
5. It helps in protein synthesis.

Double helix structure of DNA:

1. It is composed of two right handed helical polynucleotide chains coiled spirally in opposite directions around the same central axis.
2. Two strands are anti-parallel i.e., their phosphodiester linkage runs in opposite directions.
3. Bases are stacked inside the helix in planes PERPENDICULAR to the helical axis.
4. Two strands are held together by H – bonds (A = T, G=- C).
5. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.
6. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.
7. Diameter of double helix is 2 nm.
8. Double helix repeats at intervals of 3.4 nm. (One complete turn)
9. Total amount of purine (A + G) = Total amount of pyramidine (C + T)

• Vitamins: Vitamins are organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism.

• Classification of vitamins: Vitamins are classified into two groups depending upon their solubility in water or fat.

1. Water soluble vitamins i) These vitamins are soluble in water.
   ii) Water soluble vitamins must be supplied regularly in diet because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body.
   iii) Example: Vitamin C, B group vitamins.
2. Fat soluble vitamins
   i) These vitamins are soluble in fat and oils but insoluble in water.
   ii) They are stored in liver and adipose (fat storing) tissues.
   iii) Example: Vitamin A, D, E and K

Important vitamins, their sources and their deficiency diseases:

Name of vitamins	Sources	Deficiency diseases
Vitamin A	Fish liver oil, carrots, butter and milk	xerophthalmia (hardening of cornea of eye) Night blindness
Vitamin B1 (Thiamine)	Yeast, milk, green vegetables and cereals	Beriberi (loss of appetite, retarded growth)
Vitamin B2 (Riboflavin)	Milk, egg white, liver, kidney	Cheilosis (fissuring at corners of mouth and lips), digestive disorders and burning sensation of the skin.
Vitamin B6 (Pyridoxine)	Yeast, milk, egg yolk, cereals and grams	Convulsions
Vitamin B12	Meat, fish, egg and curd	Pernicious anaemia (RBC deficient in haemoglobin)
Vitamin C (Ascorbic acid)	Citrus fruits, amla and green leafy vegetables	Scurvy (bleeding gums)
Vitamin D	Exposure to sunlight, fish and egg yolk	Rickets (bone deformities in children) and osteomalacia (soft bones and joint pain in adults)
Vitamin E	Vegetable oils like wheat germ oil, sunflower oil, etc.	Increased fragility of RBCs and muscular weakness
Vitamin K	Green leafy vegetables	Increased blood clotting time

CAPACITORS CLASS 12

 

                                 CAPACITORS              

A capacitor is a device that stores electrical energy. It is an arrangement of two conductors carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.

Note the following points about capacitors:
1) The net charge on the capacitor as a whole is zero. When we say that a capacitor has a charge q, we mean the positively charged conductor has a charge +q and negatively charged conductor has a charge -q.
2) The positively charged conductor is at a higher potential than the negatively charged conductor.
3) The potential difference V between the conductors is proportional to the charge q. The ratio q/V is known as capacitance C of the capacitor. Thus,

C=qV

4) Capacitance depends on the size and shape of the plates and the material between them. It does not depend on q or V individually.
5) The SI units of capacitance are farad(F), which is equivalent to coulomb/volt. Practical values of capacitances are usually measured in microfarad (μF).

1μF=10−6F

6) It is a scalar, having dimensions

[C]=[QV]=[Q2W]       [asV=WQ]
or,   [C]=[A2T2ML2T−2]=[M−1L−2T4A2]

Isolated Conducting Sphere as a Capacitor

A conducting sphere of radius R carrying a charge q can be treated as a capacitor. The high-potential conductor is the sphere itself and the low potential conductor is a sphere of infinite radius. The potential difference between these two sphere is

V=q4πε0R−0=q4πε0R

Hence, its capacity is (the capacitance of an isolated conductor is normally called capacity)

C=qV=4πε0R

The capacity of a spherical conductor is directly proportional to its radius. As the potential of earth is assumed to be zero, the capacity of the earth or of any conductor connected to earth (irrespective of its shape or charge on it) will be

C=qV=q0=∞

However, if we assume the earth to be a conducting sphere of radius 6400 km, its capacitance will be,

C=4πε0R=6400×1039×109=711μF

Energy Required to Charge a Conductor

When a conductor is charged its potential changes from 0 to V. In this process, work is done against repulsion between charge stored on the conductor and charge coming from the charging body. This work is stored as electrostatic potential energy U. So, if dq charge is given to a conductor at potential V,

dU=dqV=dqqC
⸫   U=1C∫q00qdq=q202C  (where q0 is the total charge given to the conductor)

Since q=CV, we can say that the energy stored in a charged (conductor) is

U=12q2oC=12qoV=12CV2

Note that if the charging source (say, battery) supplies charge at constant potential (say V), the work done by the charging source W=qV, whereas the energy stored in the charged conductor is U=12qV. Thus, in charging a body 50 % of the energy is wasted as heat.

Application 1
Small identical droplets of distilled water (radius 0.1 mm) are found to have a charge 2 pc each. If 64 of these coalesce to form a single drop, calculate (a) the charge on it, and (b) its potential.

Solution:

(a)  From conservation of charge, we have
Q=nq=64×2×10−12C=128×10−12C
(b) From conservation of mass, we have
n×(4/3πr3)ρ=1×(4/3πR3)ρ
or,  R=(n)1/3r=(64)1/3×0.1×10−3=0.4×10−3m
⸫   V=Q4πε0R=128×10−12×9×1090.4×10−3=2880  V

Sharing of Charge

Let us have two isolated spherical conductors of radii R1 and R2, charged at potentials V1 and V2.
q1=C1V1     and   q2=C2V2
where  C1=4πε0R1   and  C2=4πε0R2
The combined charge is q1 + q2 and combined capacitance is C1 + C2.
Now if they are connected through a wire, charge will flow from conductor at higher potential to that at lower potential till both acquire the same potential,

V=(q1+q2)(C1+C2)=C1V1+C2V2C1+C2=R1V1+R2V2R1+R2

And hence, if q'1 and q'2 are the charges on the two conductors after sharing,
q′1=C1V   and    q′2=C2V   with  (q′1+q′2)=(q1+q2)=q
So,  q′1q′2=C1C2=R1R2
Thus, the charge is shared in proportion to capacity.
Some energy is lost in sharing charges. This energy is lost mainly as heat when charge flows from one body to the other through the connecting wire and also as light and sound if sparking takes place. The loss in energy is
W=UI−UF=(12C1V21+12C2V22)−12(C1+C2)V2
=C1C22(C1+C2)(V1∼V2)2

Application 2
Two isolated metallic solid spheres of radius R and 2R are charged such that both of these have same charge density σ. The spheres are located far away from each other and connected by a thin conducting wire, find the new charge density on the bigger sphere.

Solution:

As charge density on both spheres is same, the total charge,

q=q1+q2=4π(R)2σ+4π(2R)2σ=20πR2σ   ....(i)

Now in sharing, the charge is shared in proportion to capacity (i.e., radius), so the charge on the bigger sphere,

q′2=R2(R1+R2)q=2RR+2Rq=23q
⸫   σ′2=q′24π(2R)2=(2/3)q16πR2=q24πR2=56σ   [using Eqn. (i)]

Capacitance of some Capacitors

(1)  Parallel Palate Capacitor
σ=qA   and      E=σε0=qε0A
⸫   V=Ed=qdε0A
⸫   C=qV=ε0Ad
Note that the capacitance is independent of charge given, potential raised, nature of metal or thickness of plates.

(2)  Spherical Capacitors
The induced charge q' on shell B is equal and opposite to charge q on inner sphere A.
As the charges on the two conductors are equal and opposite, the system is a capacitor.
The electric field at a point P between the shells,

E=EA+EB=14πε0qr2      [as EB=Ein=0]
or    −dVdr=14πε0qr2         [asE=−dVdr]
⸫   V=−∫0VdV=q4πε0∫badrr2=q4πε0[1a−1b]
⸫   C=qV=4πε0ab[b−a]

Note that
(1)  As b→∞, the capacitance reduces 4πε0a. This shows that a spherical conductor is a spherical capacitor with its other plate of infinite radius.
(2)  As a and b both become very large, maintaining the difference a−b=d (finite), the expression for C reduces to C=ε0Ad. This shows that a spherical capacitor behaves as a parallel plate capacitor if its spherical surfaces have large radii and are close to each other.

(3)  Cylindrical Capacitor :

The field at a point P is
E=14πε02λr
But   E=−dVdr
⸫   −∫v0dV=2λ4πε0∫badrr    ⇒V=2λ4πε0ln(ba)
⸫   C=qV=λL(λ/4πε0)ln(b/a)=2πε0Lln(b/a)

Energy Stored in a Capacitor

If dq charge is given to a capacitor at potential V, the work done is

dW=dq(V)          [as q=CV]
or,   W=∫q0(q/C)dq=12(q2/C)=12CV2=12qV          [as q=CV]

This work is stored as electrical potential energy,

U=W=CV2=12q2C=12qV

This energy is not localized on the charges or the plates but is distributed in the field.
In case of a parallel plate capacitor, the field is limited between the plates, in a volume A x d. We can determine the energy density uE in this volume,

uE=UVolume=12CV2Ad=12[ε0Ad]V2Ad[asC=ε0Ad]
uE=12ε0(V/d)2=12ε0E2       [as  Vd=E]

Force Between the Plates

The plates carry equal and opposite charges. There is a force of attraction between them. To calculate this force, we use the fact that electric field is conservative, for which F = -(dU/dx).
In case of a parallel plate capacitor

U=12q2C=12q2xε0A                 [asC=ε0Ax]
⸫    F=−ddx[12q2ε0Ax]=−12q2ε0A

The negative sign implies that the force is attractive. The force per unit area is

∣∣FA∣∣=12q2ε0A2=σ22ε0=12ε0E2                 [asqA=σandE=σεo]