How do you convert:
i) Chlorobenzene to biphenyl
ii) propene to 1-iodopropane
iii) 2-bromobutane to but-2ene
Or
Write the major products(s) in the following:
Give reasons for the following:
i) Aniline does not undergo Friedal-Crafts reaction.
ii) (CH3)3N is more basic than (CH3)3N in an aqueous solution.
iii) Primary amines have higher boiling point than tertiary amines.
Due to hectic and busy schedule,Mr.Singh started taking junk food in the lunch break and slowly became habitual of eating food irregularly to excel in his field. One day during the meeting he felt severe chest pain and fell down. Mr Khanna, a close friend of Mr. Singh took him to doctor immediately. The doctor diagnosed that Mr Singh was suffering from acidity and prescribed some medicines. Mr Khanna advised him to eat home made food and change his lifestyle by doing yoga,meditation and some physical exercise. Mr Singh followed his friend's advice and after few days he started feeling better.
after reading the above passage, answer the following:
i) what are the values (at least two) displayed by Mr Khanna?
ii) What are antacid? Give one example
iii) would it be advisable to take an antacid for a long period of time ? Give
i) Mn Shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
ii) Cr2+ is a strong reducing agent.
iii) Cu2+ salts are coloured while Zn2+ salts are white.
b) Complete the following equations:
i) Mn Shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4 because of the ability of oxygen to form multiple bonds with Mn metal.
ii) Cr2+ is strongly reducing in nature. It has a d4 configuration. Cr2+ is a stronger reducing agent because it can lose one of its electrons to become Cr3+ in which the t2g level of d-orbital is half filled and the eg level is empty which is a more stable configuration.
iii) The electronic configuration of Zn = 3d10 4s2
Zn2+ = 3d10
where as the electronic configuration of Cu = 3d10 4s1
Cu2+ =3d9
In the case of Zn fully filled d orbital is present therefore no d-d transition can be possible in this case and it is colourless.
In the case of copper 3d9 because of d-d transition electrons emits light in the visible range and hence they are coloured compounds.
Distinguish between:
i) C6H5-COCH3 and C6H5-CHO
ii) CH3COOH and HCOOH
c) Arrange the following in the increasing order of their boiling points:
CH3CHO,CH3COOH,CH3CH2OH
Or
a) Write the chemical reaction involved in Wolff Kishner reduction.
b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
C6H5COCH3, CH3-CHO, CH3COCH3
c) why carboxylic acid does not give reactions of carbonyl group ?
d) Write the product in the following reaction
e) A and B are two functional isomers of compound C3H6O. On heating with NaOH and I2, isomers B forms a yellow precipitate of iodoform whereas isomer A does not form any precipitate. Write the formulae of A and B.