National Science Day - 28th February

Today is National Science Day of India.

Learn your science subjects with more commitment. Try to understand things in depth. As an engineer, you need to find new solutions based on existing scientific as well as new theories.

http://knol.google.com/k/narayana-rao-k-v-s-s/national-science-day-india-28th-february/2utb2lsm2k7a/2346#

You need to keep interest in science even after you become an engineer.

Polymers and Their Monomers

Polymer - Monomers


Homopolymers (Only one monomer)

Cellulose – Glucose
Natural rubber – Isoprene (2-metyl-1, 3-butadiene)

Neoprene - Chloroprene
Nylon-6 - caprolactum

Polyacrylonytrile (PAN) or Orlon – Vinyl cyanide (acrylonitrile)
Polyethelene - Ethylene (Ethene)
(Popularly polythene)
Polyethyl acrylate (PEA) - ethyl acrylate
Polymethyl methacrylate (PMMA) – Methyl methacrylate
Polymonochlorotrifluoroethylene (PCTFE) - chlorotri-fluoroehtylene
Polypropylene – Propylene
Polystyrene or Styron – styrene
Polytetrafluoroethylene (Teflon or PTFE) - tetrafluoroethylene
Polyvinyl chloride – Vinyl chloride
Proteins - Alpha-Amino acids

Starch - Glucose


Copolymers (Two different monomers)

Bakelyte (Phenol formaldehyde resin) - Phenol and Formaldehyde
Buna-S - 1,3-butadiene and styrene
Dacron (Polyster) - Terephthalic acid and ethylene glycol
Glyptal - ethylene glycol and phthalic acid
Melamine polymer (Melamine formaldehyde) - Melamine and fromaldyhyde
Nylon-66 - Hexamethylenediamine and adipic acid
Saran - vinyl choride and vinylidene chloride
Styrene-butadiene rubber - styrene and 1,3-butadiene
Thiokol - 1,2 dichlorlethene (ethylene dichloride) and sodium polysulphide
Terylene - ethylene glycol and terephthalic acid

Difference Between a Metal and Polymer

In the chapter in polymers, it is explained that polymers are compounds of very high molecular masses formed by the combination of a large number of simple molecules. Simple example of a polymer is polyethelene. This polymer is formed by the compound formation between molecules of ethelene. The molecules of ethylene in the case of formation of polymer are referred to as monomers.

But what is the difference between a metal and polymer?

Metal

At a molecular level, the basic units that make up a metallic structure are relatively small and uniform in size. Consequently, they readily arrange into a structure that is very regular and predictable. Material scientists call these structures as crystalline solids.


Polymer

The smallest unit in a polymeric material is a very large molecule with an extended chain shape. (Remember linear chains, branched chains and cross linked chains). Even the smallest polymer molecule will be about 50 times more massive than the heaviest naturally occurring unit in a metallic structure. These polymer molecules can twist, turn, fold, and entangle in almost unlimited variations (Remember the point regarding straightening of chains due to stretching). In addition, not all of these molecules are of the same size. The smallest molecule in any random sample of plastic is often a thousand times smaller than the largest, which increases the possibilities for local variation in the structure of a material.


In a metal there ionic forces between free flowing electrons and the positive charged nucleuses that hold the metal together. In a polymer there are covalent bonds between molecules of various sizes, with each molecule being polymer of number of monomers.

Top Down and Bottom Up Learning

Top Down learning is studying a lesson first and then solving problems. Most of the time you are doing it in all complex subjects and topics.

Bottom Up learning is trying to solve a problem and finding new issues which are to be used in solving the problem. Everybody needs to engage in bottom up learning also. Top down is not sufficient.

Solve all problems of past JEE examinations at least five years 2009, 2008,2007,2006 and 2005 and engage in bottom up learning.

Now it is the time.

Chemistry Specific Topics Tested in IIT JEE 2009 Paper II

1. Migration of H and CH3 groups in intermediate reaction compound when a positive carbon ion is present.


Migrating nature of H is more than CH3.

2. The spin only magnetic moment value


The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is -
(A) 0 (B) 2.84
(C) 4.90 (D) 5.92

Answer A


CO is a strong field ligand.

3. Stability order of resonance structures


The species with incomplete octet are less stable.

4. k = Ae^(–Ea/RT)


For a first order reaction A→P, the temperature (T) dependent rate constant(k) was found to follow the equation logk = – (2000)(1/T) + 6.0
The pre-exponential factor A and the activation energy E_aa, respectively, are -

Answer:
1.0 × 10⁶6 s^-1 and 38.3 kJ mol^-1



5. Identifying bonding patterns and single and double bonds


The nitrogen oxide(s) that contain(s) N-N bond(s) is (are) -


(A) N2O (B) N2O3
(C) N2O4 (D) N2O5

Answer: A,B, and C

6. Sugars : Identifying reducing sugar and nonreducing sugar and glucosidic linkages in the given compounds

7. Identifying the amine in a given reaction


In the reaction
2X + B2H6 → [BH2(X)2]⁺ [BH4]^-

amine (X) is (are)

8. Identifying state functions


Among the following, the state function(s) is (are)-
(A) Internal energy (B) Irreversible expansion work
(C) Reversible expansion work (D) Molar enthalpy

Answer A and D

9 Identifying the scope for spontaneous reaction

10. Reactions of metals with acids


Column I
(A ) Cu + dil HNO3
(B) Cu + conc HNO3
(C) Zn + dil HNO3
(D) Zn + conc HNO3

Matching is to be done with output of the reaction (various compounds)

11. Identifying compounds that are produced from given reactions


(p) Nucleopilic substitution
(q) Elimination
(r) Nucleophilic addition
(s) Esterification with acetic anhydride
(t) Dehydrogenation

Compounds are presented as line diagram format

12. coordination number


The coordination number of Al in the crystalline state of AlCl3 is –

13. Oxidation number
The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is –

14. Emission of alpha and Beta particles in a nuclear reaction


The total number of α and β particles emitted in the nuclear reaction U[238,92]
→ Pb[214,82] is –

15. pH of a solution


The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10^-4. The pH of a 0.01M solution of its sodium salt is –

16. organometallic compounds structure


The number of water molecule(s) directly bonded to the metal centre in CuSO4. 5H2O is –

17. Molecular weight and kinetic energy


At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is -

18. Number of Isomers


The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular formula C5H10 is-

19. Enthalpy of Combustion of Gas


In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K^-1, the numerical value for the enthalpy of combustion of the gas in kJ mol^-1 is-

Answer 9

Chemistry Topics Tested in IIT JEE 2009 Question Paper I

1. Van Der Waal's Equation for Real Gases

2. Writing compound name of 2-Bromo-5-hydroxybenzonitrile given the pictorial view of the molecule.

(Cyanide group has higher priority than phenol. Hence correct IUPAC name is

2-Bromo-5-hydroxybenzonitrile)

3. Atomic mass of an element having isotopes

Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5 %, respectively, the atomic mass of Fe is -

(A) 55.85 (B) 55.95

(C) 55.75 (D) 56.05

[Ans. B]

4. Polymers

Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular

force of attraction is weakest is -

(A) Nylon (B) Poly(vinyl chloride)

(C) Cellulose (D) Natural Rubber

[Ans. D]

Sol. Natural rubber is an example of elastomers, which has weakest intermolecular forces (dispersion forces).

5. Henry's Law

The solubility of a gas is related to its pressure by Henry's law.

The mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure of the gas in equilibrium with the solution.

m = Kp

m = mass of the gas dissolved in a unit volume of the solvent.

p = is the pressure of the gas in equilibrium with the solution

K = proportionality constant

Alternative statement

The pressure of a gas over a solution in which the gas is dissolved is proportional to the mole fraction of the gas dissolved in the solution.

p = Kx

p = partial vapour pressure

x = mole fraction of the gas in the solvent

K = Henry's law constant

Problem

The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 10⁵

atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm

pressure is -

Answer: 4.0 × 10^-4

6. Acidity order of aromatic carboxylic acids and phenol

Benzoic acid (–COOH) is more acidic than phenol. Electron withdrawing group increase acidic nature &

electron donating group decrease acidity.

7. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for

Sb2S3 sol is -

Sb2S3 is negatively charged sols. Because of the formation higher positive charge ions coagulating power

of Al2(SO4)3 is the highest

8. Reactions of Phosphorous

The reaction of P4 with X leads selectively to P4O6. The X is -

(A) Dry O2 (B) A mixture of O2 and N2

(C) Moist O2 (D) O2 in the presence of aqueous NaOH

Answer B

P4 is converted to P4O6 in the presence of limited supply of oxygen.

9. Reactions of Sodium

The compound(s) formed upon combustion of sodium metal in excess air is(are) -

10. Defects of Solids

The correct statement(s) regarding defects in solids is(are) -

(A) Frenkel defect is usually favoured by a very small difference in the size of cation and anion

(B) Frenkel defect is a dislocation defect

(C) Trapping of an electron in the lattice leads to the formation of F-center

(D) Schottky defects have no effect on the physical properties of solids

Answers: B,C

11. Geometrical isomerism of organometallic compounds

12. Stereoisomers

The correct statement(s) about the compound H3C(HO)HC–CH=CH–CH(OH)CH3 (X) is (are)

(A) The total number of stereoisomers possible for X is 6

(B) The total number of diastereomers possible for X is 3

(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4

(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2

Answer: A,D

13. Anilines

p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated

with a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue.

Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate (II) leads to the

formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent.

Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a

brown coloration due to the formation of Z.

Questions: Find compound X,Y and Z

14. Carbonyl Compounds

A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed

by dehydration to give an olefin Q. Ozonolysis fo Q leads to a dicarbonyl compound R, Which undergoes

intramolecular aldol reaction to give predominantly S.

Questions asked on Structure of P,Q,R and S

15. Attaching these properties to elements in matching

(p) Paramagnetic

(q) Undergoes oxidation

(r) Undergoes reduction

(s) Bond order ≥ 2

(t) Mixing of 's' and 'p' orbitals

16. Matching of characteristic reactions with compounds

(p) Reduction with H2/Pd-C

(q) Reduction with SnCl2/HCl

(r) Development of foul smell on treatment with chloroform and

alcoholic KOH

(s) Reduction with disobutylaluminium hydride (DIBAL-H)

(t) Alkaline hydrolysis

Compounds

(A) CH3CH2CH2CN, (B) CH3CH2OCOCH3, (C) CH3 – CH = CH – CH2OH, (D) CH3CH2CH2CH2NH2

For question paper

http://ankurmittal.com/solutions/paper1_solutioin.pdf