HOW TO PREPARE CHEMISTRY AS A SUBJECT FOR IIT-JEE
You know IIT JEE is the most prestigious & challenging examination for class XIIth students in India. It comprises of three subjects e.q. physics, chemistry & maths. Out of these physics & chemistry play a crucial role in the selection process of candidate. If you see the track record of previous years you will find that in chemistry & physics student score more marks than maths, because of the theoretical nature of these subjects Chemistry is one of the top most scoring subjects. Meticulous planning of preparation can give you a marvelous increase in your rank in IIT-JEE. Chemistry comprises of three branches physical, organic & inorganic. There are following tips which would be useful for you to prepare for chemistry as subject. Physical chemistry which comprises of about 40% of the entire paper can fetch you entire marks if you concentrate on understanding, the basic concept of each chapter. My personnel suggestion is that you should make your own notes for each concept. Each chapter has its own shortcuts, which can be understood with the help of a professional.
Important Topics of Physical Chemistry:
* Chemical Equilibrium
* Ionic Equilibria
* Solid State
* Volumetric Analysis
* Chemical Bonding
* Atomic Structure
Besides these there are other topics also which are less important with respect to examination. So you can prepare your studies accordingly.
Organic chemistry requires a clinical approach. You know IIT-JEE is an engineering entrance exam where more emphasis is given on synthetic organic chemistry rather than theoretical organic chemistry.
In order to have solid grip over the subject first of all mechanism portion should be understood.
After that chapterwise problems should be solved. The mechanism portion of organic chemistry can be classified as follows:
* Concept of acidity & basicity
* sp3-hybridised nucleophillic substitution
* sp2-hybridised nucleophillic substitution
* Elimination reactions
* Aromatic nucleophillic & Electrophillic substitution
* Free Radical mechanism
* Name Reactions
Use of Above Chapters:-
* When you understand sp3-hybridised carbon you can solve the question of alkyl halide, alcohol, ether, expoxide.
* When you understand sp2-hybridised carbon you can solve the questions of alkene, alkyne, aldehyde, ketone, acid & its derivatives.
* These chapters also help you to understand name reactions. These reactions are specific examples of a particular concept.
* In general organic chemistry one should understand two concepts very well.
1. Concept of acidity & basicity.
* This portion definetely fetches you 2 questions precisely questions on phenol, carboxylic acids & aniline should be practiced
2. Isomerism: In the field of isomerism more emphasis should be given to stereoisomerism more preciesely the optical isomerism. One should learn the formula's for calculating no. of isomers both for stereosiomerism as well as structural isomerism.
Inorganic Chemistry Course Topics:
Inorganic chemistry preparation should be more precise than exhaustive because it contribute to about 20% of your paper. Infact there are few topics which should be concentrated more than others.
* Qualitative Analysis :- This topic should be well prepared because it contributes to major portion of inorganic.
* Coordination Compound :- Now a days this topic is getting momentum. In every entrance examination you find at least 2 to 3 questions related to this topic.
Beside above I would like to suggest that in IIT-JEE chemistry, the more important fact is the applicative aspect of chemistry rather than the theoretical one. Hence an expert's advise will make you understand things in a clear manner. If you want to understand what I mean by expert advise read my next article. e.g.
* Do you understand Le-Chatlier's principle
* in volumetric analysis ?
* Can you solve all the problems on volumetric by single method ?
* Do you know the short cuts in Ionic Equilibrium ?
Do you understand Le-Chatlier's Principle ?
It is one of the most Important principles of our daily life. Le-chatlier's principle states that if a system at equilibrium is applied by an external stress (such as pressure, temperature, volume etc.) then the system adjusts itself in such a way so as to nullify the affect of applied external stress.
Most of the time to understand the affect of external stress gas eqn. PV=nRT is used but this equation can not handle more than two variable at a time. In order to understand the real application. of above law you can draw a projector equation comprising of all the external variables on one side & internal variables on the other side of the equation. Then apply the boundry condition on it to give the final result.
Example Question .;
#. For the following reaction what happens when
a) at const. Temperature pressure of the system is increased.
b) at const. Temperature volume of the system is increased.
c) at const. Temperature and pressure helium gas is added to the flask.
d) at const. Temperature and volume helium gas is added to the flask.
Let us assume that at equilibrium moles of SO2 be ;
SO3 be & O2 be & nT represents the total no of moles at equilibrium & total pressure be P atm.
SO2 + 1/2 O2 SO3
Above are two prejectors to project your answers; Let us see
Ans a : In equation (i) if temperature is constant KP is fixed pressure you are increasing & nT is fixed. Therefore lefthand side is increasing which means right hand side should also increase
nSO3 should increase or nSO2 & nO2 should decrease forward direction
Ans b : In equation (ii) if temperature fixed KP constant
Volume increased left hand side is decreases
nSO3 is decreased or nSO2 & nO2 is increased backward direction
Ans c : In equation (i) if T is fixed KP constant & P is fixed but 'He' is added means nT is
increased left hand side is decreased nSO3 is decreased backward direction
Ans d : In equation (ii) if T & V is fixed left hand side is fixed but there is no function of nT hence
addition of 'He' has no affect on it No change.
# For the following reaction at equilibrium
A(g) + 2B(g) C(g) + 3D(g)
What happens when at constant Temperature, Pressure of the system is increased but simultaneously some amount of C was added so as to maintain the partial pressure of A constant.
c. no change
d. data insufficient
A + 2B C + 3D
Let the moles be nA, nB , nC and nD respectively & total pressure be P atm.
In above function on left hand side both numerator & denominator is increased
Hence unless the rate is known the answer cannot be predicted
Hence option (4) is correct.
Do you know that there are only 4P4 = 24 no. of questions in volumetric analysis :
Volumetric Analysis means titration of compounds which means mixing of compound A with compound B but that is possible ony when compound A reacts chemically with compound B. In volumetric Analysis questions can be classified under following four categories.
1. Redox titration
2. Iodometric titration
3. Acid Base Titration
4. Precipitation method
All these are manual titrations a question can be formed only on the concept of manual titration. Since there are only four methods therefore the total no. of questions in volumetric Analysis will be only
There is a very interesting fact that you can solve all the questions by a single method called n factor method.
Moreover this method doesnot require to remember any reaction. You can solve any question without writing the chemical reaction because molar ratio can be calculated with the help of n factor. If you want to understand this, read my next article on n-factor.
Diels Alder Reaction:
The reversible and thermal (4, 2) cyclo addition reaction between a diene and a dienolphile (dieneoving species usually contains an active multiple bond) in which the dienophile adds 1, 4 to the diene to produce a six membered cyclic product (adduct) is known as Diels Alder Reaction. Here the two units the diene and the dienophile, get joined by two bonds created from the two bonds in two units, a new bond is also formed. The reaction may also be caused by the catalytic amount of light or lewis acids.
The equilibrium lies over to right; if the temp. is raised, the equilibrium swifts to the left and the reverse diels Alder product are obtained.
* However the dienes may be acylic, cyclic and aromatic conjugated systems. Even non conjugated Diens of appropriate geometry may undergo the Diels Alder. Diens noted below:-
Several types of Dienophiles may be used in the Diels Alder reaction:-
(d) Diens : -C = C - C = C-, -C = C = C- (e) other then C-C :- CO2Me-N = N - CO2Me
Tetra cyano ethene is known to be the best dienophile till to date.
The procedure for Diels Alder Reaction is very simple only the mixture of the Diene and the dienophile needs heating with or without solvent, the common solvent is benzene.
It has been found that the electron donating groups in the dienes accelerate the reaction while the electron withdrawing group retard it, on the other hand reverse effects have been found in the cases of dienophiles - the electron withdrawing group in them accelerate the reaction whereas the electro donating group retards the same. However the opposite electronic effects on the diene and dienophile accelerate the rate but similar deccelerate the same. Cyclic dienes in the cisoid form react more readily then the corresponding acylic dienes. In general with the increasing sub. in the diene the rate of the reaction decreases, tetra sub. dienes in 1, 4- position do not reacts. In fact stearic effects have pro found influence on the reaction. Large groups on 2,3 positions make it unreactive. The cisoid conformation of such a diene gets frozen to the transoid form because of the steric strain developed by the large groups. Bulky cis, cis 1,4 - disub. dienes also do not react. Among the four 1,4 diphenyl butadienes only the trans isomer gives the Diels Alder Reaction.
Since the diene and the dienophile during addition can orient in two ways, a mixture of products are obtained, of course one of them is the major product.
The discussion on the stereochemistry involves several steps :-
(a) Only conjugated dienes in the cisoid conformation undergo the reaction. Any factor that makes the diene assume the transoid conformation will hinder the reaction.
(b) The reaction is stereospecific with respect to the dienophiles and the addition is always cis, the cis group. Remains cis and trans group remains trans in the product.
Q1. Explain why :-
(a) Cyclo pentadiene remains as dimer but the former can be obtained from the latter on distilling at room temperature at 170°C - 200°C.
(b) Tetraphenyl cyclopentadiene is monomeric at room termperature.
(a) Cyclo pentadiene is highly reactive as a diene and it is moderately so as a dienophile when cyclo pentadiene is kept at room temperature the self Diels Alder Reaction occurs and a dimer is formed. However Diels Alder Reaction is a reversible reaction. For this reason on heating the monomer, cyclo pentadiene forms.
(b) The moecule has four bulky substituents, its dimeric form experiences high steric repulsion so dimer can't exist at room temperature.
Q.2. Explain why the following compounds do not undergo Diels Alder reaction as dienes
(a) Dienes in the cisoid form undergoes the Diels Alder Reaction. Transoid forms do not undergo the reaction because this forms leads to a highly strained six member. T.S. whose energy content is very high. Compound (i) and (iv) experience high strain in the cisoid form and they remain in the transoid form. So compounds (i) and (iv) do not undergo. Compound (iii) is locked in the transoid form, so it also not undergo the reaction. Compound (ii) 1,3- butadyne is a linear molecule Hence all the C's are in sp hybridised state. The p.A. O's on and carbons of the compound when overlap the p.A. O's of the dienophile, a highly strained 1, 2, 3 cyclo hexatriene will be formed the triene moleity will be linear and corresponding T.S. will also have trienyl like linear moeity with high strain and energy. It is difficult to over come the energy Barrier of the T.S. so the reaction does not occur.
For (v) the two double bonds of the molecule are nearly at right angle to one another so in this case also the T.S. will be strained but less strained than that of compounds (i) to (iv). Hence reaction takes slowly.
Q.3. Why does anthracene undergo the Diels Alder Reaction but naphthelene does not ?
The energy gained by the cyclo addition reaction being less then the energy lost by the naphthalene molecule owing to the loss of aromaticity by one of the two benzene rings, naphthelene does not take part in the Diels Alder. In the case of anthracene the energy gained by the reaction is more than the energy lost by the loss of aromaticity of the middle ring and so the reaction occurs.
Problems for Practice
1. Reactivity order of the following Dienophiles :
higher the electron withdrawing effects higher will be reactivity.
2. Give the structure of the product of the Diels-Alder reaction between.
a. maleic anhydride and isoprene b. maleic anhydride and 1,1'-bicyclohexenyl(I)
c. maleic anhydride and 1-vinyl-1-cyclohexene
d. 1,3-butadiene and methyl vinyl ketone
e. 1,3-butadiene and crotonaldehyde f. 2 mol 1, 3-butadiene and dibenzalacetone
g. 1,3-butadiene and -nitrostyrene (C6H5CH=CHNO2)
h. 1,3-butaddiene and 1,4-napththaoquinone (II)
i. p-benzoquinone and 1,3-cyclohexadiene
j. p-benzoquinone and 1,1'-bicyclohexenyl (I)
k. p-benzoquinone and 2 mol 1, 3-cyclohexadiene
l. p-benzoquinone and 2 mol 1, 1'-bicyclohexenyl (I)
m. 1,3-cyclopentadiene and acrylonitrile n. 1,3-cyclohexadiene and acrolein
3. From what reactants could the following be synthesized by the Diels-Alder reaction ?
4. The following observation illustrate one aspect of the stereochemistry of the Diels-Alder Reaction: maleic anhydride + 1, 3-butadiene A(C8H8O3)
A + H2O, heat B(C8H10O4)
B + H2, Ni C(C8H12O4), m.p. 192°C
fumaryl chloride (trans-ClOCCH = CHCOCl) + 1, 3-butadiene D(C8H8O2Cl2)
D + H2O, heat E(C8H10O4)
E + H2, Ni F(C8H10O4), m.p. 215°C
F can be resolved; C cannot be resolved.
Does the Diels-Alder reaction involve a syn-addition or an anti-addition ?
5. Give the stereochemical formulas of the products expected from each of the following reactions. Label meso compounds and racemic modifications.
a. crotonaldehyde (trans-2-butenal) + 1, 3-butadiene
b. p-benzoquinone + 1,3-butadiene
c. maleic anhydride + 1, 3-butadiene, followed by cold alkaline KMnO4
d. maleic anhydride + 1, 3-butadiene, followed by hot KMnO4 C8H10O8.
(b) benzil (C6H5COCOC6H5)+dibenzyl ketone (C6H5CH2COCH2C5H5)+base A(C29 H20O), ''tetracyclone''
A + maleic anhydride B(C33H22O4)
B + heat CO + H2 + C(C32H20O3)
(c) A + C6H5C CH D(C37H26O)
D + heat CO + E(C36H26)
(d) hexachloro-1, 3-cyclopentadiene + CH3OH + KOH F(C7H6Cl4O2)
F + CH2 = CH2, heat, pressure (C9H10Cl4O2)G
G + Na + t-BuOH (C9H14O2)H
H + dilute and I(C7H8O), 7-ketonobronene
7. Give the likely structures for A to E, 1,3-butadiene + propiolic acid (HC = CCOOH)A(C7H8O2)
A + 1mol LiAlH4 B(C7H10O)
B + methyl chlorocarbonate (CH3OCOCl) C(C9H12O3)
C + heat (short time) toluene + D(C7H8)
D + tetracyanoethylene E(C13H8N4)
Compound D is not toluene or 1,3,5-cycloheptatriene; on standing at room temperature it is converted fairly rapidly into toluene.