1. Assertion - Reason Type
Statement:In Williamson Synthesis, ter. alkyl halides cannot be used.
Reason:The reaction becomes reversible
2. A functional isomer of 2-methyl-2-propanol is
a. n-butyl alcohol
b. 2-butanol
c. 2-methoxy propane
d. ethyl methyl ether
3. The compound A (C2H7N) on treatment with chlorofom in alkaline medium gives a compound B. Compound B on reduction gives a compound C that responds to Liebermann's nitrosamine test. A and C are respectively:
a. ethylamine and ethyl methyl amine
b. secondary amine and primary amine
c. ethylamine and propylamine
d. primary amine and secondary amine
4. The volume of N2 liberated at S.T.P. by complete reaction of 4.5×10 kg of ethylamine with HNO2 is:
5. CH3(CH2)12COOH is the moelcular formula of ---- acid.
a. Lauric acid
b. Myristic acid
c. Stearic acid
d. Oleic acid
Tuesday, September 8, 2009
IIT-JEE-Chemistry - Final Revision Set 6
1.Colloids and crystalloids differ in -------
a. charge on particle
b. particle nature
c. particle size
d. concentration
2. which one of the following does not form lyophilic sol?
a. gum
b. starch
c. As2S3
d. gelatin
3. The half life of a radioactive isotope is 50 minutes. Its decay constant is ___.
4. The compound H4SiO4 represents
a. silicate
b. silicic acid
c. orthosilisic acid
d. orthosilicate
5. On treatment with cold hydroiodic acid, diethyl ether gives: ________ .
a. charge on particle
b. particle nature
c. particle size
d. concentration
2. which one of the following does not form lyophilic sol?
a. gum
b. starch
c. As2S3
d. gelatin
3. The half life of a radioactive isotope is 50 minutes. Its decay constant is ___.
4. The compound H4SiO4 represents
a. silicate
b. silicic acid
c. orthosilisic acid
d. orthosilicate
5. On treatment with cold hydroiodic acid, diethyl ether gives: ________ .
IIT JEE Chemistry Final Revision Set 5
1. Give the formula of Phenacite.
2. State true or false
Bromide is a better nucleophile than iodide.
3. The weakest oxyacid of chlorine is_________
4. Complete and balance
XeF6 + NH3 →
5. State true or false.
B6H10 is not a borane
2. State true or false
Bromide is a better nucleophile than iodide.
3. The weakest oxyacid of chlorine is_________
4. Complete and balance
XeF6 + NH3 →
5. State true or false.
B6H10 is not a borane
IIT JEE Chemistry Final Revision Set No. 4
1. Assertion-Reason
Statement: When nitrobenzene is treated Br2 in presence of FeBr3 the major product formed is m-bromonitrobenzene.
Reason: Easier loss of H+ to region aromaticity from the meta position than from ortho and para position.
2. Most stable carbonium ion is:
a. p-NO2-C6H4-+CH2
b. C6H5+CH2
c. p_Cl-C6H42-+CH2
d. p-CH3O-C6H4-+CH2
3. State true or false
Toluene reacts with bromine in the presence of light to give benzyl bromine while in presence of FeBr3 it gives p-bromotoluene.
4. Give the steps to carry out the tranformation
Ethylbenzene to benzene.
5. Assertion - Reason type.
Statement: Normally benzene give electrophilic substitution reaction rather than electrophilic addition.
Reason: It has no double bonds
Statement: When nitrobenzene is treated Br2 in presence of FeBr3 the major product formed is m-bromonitrobenzene.
Reason: Easier loss of H+ to region aromaticity from the meta position than from ortho and para position.
2. Most stable carbonium ion is:
a. p-NO2-C6H4-+CH2
b. C6H5+CH2
c. p_Cl-C6H42-+CH2
d. p-CH3O-C6H4-+CH2
3. State true or false
Toluene reacts with bromine in the presence of light to give benzyl bromine while in presence of FeBr3 it gives p-bromotoluene.
4. Give the steps to carry out the tranformation
Ethylbenzene to benzene.
5. Assertion - Reason type.
Statement: Normally benzene give electrophilic substitution reaction rather than electrophilic addition.
Reason: It has no double bonds
Monday, September 7, 2009
IIT JEE Chemistry - Revision Set 3
1. Potential difference of the electrical double layer formed in a colloidal sol is called
(a) Zeta potential (b) Nernst potential (c) EMF (d) Brownian potential
2. The nucleidic ratio, 13H to 11H in a sample of water is 8.0×10-8 : 1. Tritium undergoes decay with a half life period of 12.3 years. How many tritium atoms would 10.0 g of such a sample contain 40 years after the original sample is collected.
3. 1 gm of charcoal adsorbs 100 ml 0.5 M CH3COOH to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01×10² m²/gm
4.Hydrogen bonding is maximum in
(a) ethanol (b) Diethylether (c) Ethyl chloride (d)Triethylamine
5. The critical temperature of water is higher than that of O2 because the H2O molecule has
(a) few electrons than O2
(b) two covalent bonds
(c) dipole moment
(d)V-shape
(a) Zeta potential (b) Nernst potential (c) EMF (d) Brownian potential
2. The nucleidic ratio, 13H to 11H in a sample of water is 8.0×10-8 : 1. Tritium undergoes decay with a half life period of 12.3 years. How many tritium atoms would 10.0 g of such a sample contain 40 years after the original sample is collected.
3. 1 gm of charcoal adsorbs 100 ml 0.5 M CH3COOH to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01×10² m²/gm
4.Hydrogen bonding is maximum in
(a) ethanol (b) Diethylether (c) Ethyl chloride (d)Triethylamine
5. The critical temperature of water is higher than that of O2 because the H2O molecule has
(a) few electrons than O2
(b) two covalent bonds
(c) dipole moment
(d)V-shape
Monday, August 24, 2009
IIT JEE Chemistry - Revision Set 2 - Past JEE Questions
1. The dipole moment of KCl is 3.336×10-29 Coulomb metre which indicates that it is a highly polar molecule. The interatomic distance between K+ and Cl- in this molecule is 2.6×10-10 m. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.
2. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n =4 to n=2 of He+ spectrum?
3. The standard reduction potential for the half-cell,
NO3-(aq)+ 2H+(aq)+ e- → NO2(g) + H2O
is 0.78 V. Calcualte the reduction potential in 8 M H+.
4. Specify the products in this reaction
C6H5COOH + CH3MgI → ? + ?
5. Give the structure of A if A(C3H9N) reacts with benzenesulphonyl chloride to give a solid insoluble in alkali.
Source (1993 JEE paper)
2. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n =4 to n=2 of He+ spectrum?
3. The standard reduction potential for the half-cell,
NO3-(aq)+ 2H+(aq)+ e- → NO2(g) + H2O
is 0.78 V. Calcualte the reduction potential in 8 M H+.
4. Specify the products in this reaction
C6H5COOH + CH3MgI → ? + ?
5. Give the structure of A if A(C3H9N) reacts with benzenesulphonyl chloride to give a solid insoluble in alkali.
Source (1993 JEE paper)
Friday, August 21, 2009
Selected Past II JEE questions - Set 1 (for JEE 2010)
1. Give reasons - two or three sentences
a. Phenyl group is known to exert negative inductive effect. But each phenyl ring in biphenyl (C6H5-C6H5) is more reactive than benzene towards electrophilic substitution.
b. The species [CuCl4]2- exists while [CuI4]2- does not.
c. Bond dissociation energy of F2 is less than that of Cl2.
d. Hydrogen peroxide acts an an oxidising as well as a reducing agent.
e. Sulphur dioxide is a more powerful reducing agent in an alkaline medium than in acidic medium.
2. A light bluish green crystalline compound responds to the following tests:
i. Its acqueous solution gives a brown precipitate or coloration with alkaline K2[HgI4] solution.
ii. Its acqueous solution gives a blue colour with K3[Fe(CN)6] solution.
iii. Its solution in hydrochloric acid gives a white precipitate with BaCl2 solution.
Identify the ions present and suggest the formula of the compound.
3. Write tautomeric forms of phenol
4. At 27°C, hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure as that of H2 is leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litres, what is the molar mass of the unknown gas.
5. An organic compound 'A' on treatment with ethyl alcohol gives a carboxylic acid 'B', and compound 'C'. Hydrolysis of 'C' under acidic conditions gives 'B' and 'D'. Oxidation of 'D' with KMnO4 also gives 'B'. 'B' on heating with Ca(OH)2 gives 'E' (molecular formula, C3H6O). 'E' does not give Tolleng test and does not reduce Fehling's solution but forms a 2, 4-dinitrophenylhydrazone. Identify 'A', 'B', 'C', 'D', and 'E'.
a. Phenyl group is known to exert negative inductive effect. But each phenyl ring in biphenyl (C6H5-C6H5) is more reactive than benzene towards electrophilic substitution.
b. The species [CuCl4]2- exists while [CuI4]2- does not.
c. Bond dissociation energy of F2 is less than that of Cl2.
d. Hydrogen peroxide acts an an oxidising as well as a reducing agent.
e. Sulphur dioxide is a more powerful reducing agent in an alkaline medium than in acidic medium.
2. A light bluish green crystalline compound responds to the following tests:
i. Its acqueous solution gives a brown precipitate or coloration with alkaline K2[HgI4] solution.
ii. Its acqueous solution gives a blue colour with K3[Fe(CN)6] solution.
iii. Its solution in hydrochloric acid gives a white precipitate with BaCl2 solution.
Identify the ions present and suggest the formula of the compound.
3. Write tautomeric forms of phenol
4. At 27°C, hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure as that of H2 is leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litres, what is the molar mass of the unknown gas.
5. An organic compound 'A' on treatment with ethyl alcohol gives a carboxylic acid 'B', and compound 'C'. Hydrolysis of 'C' under acidic conditions gives 'B' and 'D'. Oxidation of 'D' with KMnO4 also gives 'B'. 'B' on heating with Ca(OH)2 gives 'E' (molecular formula, C3H6O). 'E' does not give Tolleng test and does not reduce Fehling's solution but forms a 2, 4-dinitrophenylhydrazone. Identify 'A', 'B', 'C', 'D', and 'E'.
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