Physics-
General
Easy
Question
Fig., here shows P and Q as two equally intense coherent sources emitting radiations of wavelength 20m The separation PQ is 5m, and phase of P is ahead of the phase of Q by 90o A, B and C are three distant points of observation equidistant from the mid-point of PQ The intensity of radiations of A, B, C will bear the ratio
- 0 : 1 : 4
- 4 : 1 : 0
- 0 : 1 : 2
- 2 : 1 : 0
The correct answer is: 2 : 1 : 0
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chemistry-General
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chemistry-General
chemistry-
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There are two paths (I and II) for the preparation of phenyl-2,4-dinitro phenyl ether
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Path I is feasible, whereas path II is not
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There are two paths (I and II) for the preparation of phenyl-2,4-dinitro phenyl ether
Which of the following statements is true?
Path I is feasible, whereas path II is not
Path II is feasible, whereas path I is not
iii. The Cl of (a) undergoes SN reaction because it is activated by the two EWG groups
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chemistry-General
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physics-General
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Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
Arrange the following free radicals in the decreasing order of their stability.
Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
Arrange the following free radicals in the decreasing order of their stability.
chemistry-General
chemistry-
Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
3, 5–dimethylcyclopentene reacts with N–bromosuccinimide (NBS) in CCl4 in presence of light or peroxide. Identify the product.
Homolytic bond fission of a covalent single bond gives rise to free radicals. Owing to the presence of an odd electron, free radicals are highly reactive. They have planar to pyramidal geometry depending upon the groups attached to the C-atom having odd electron. Alkyl free radicals are stabilised by hyperconjugation whereas allyl and benzyl free radicals are stabilised by resonance. They are formed as intermediates in the reaction mixture either in the gaseous phase or in non-polar solvents. Addition of HBr to alkenes in presence of peroxide, the substitution of allylic or benzylic hydrogen by ‘Cl’ at high temperature or by ‘Br’ in presence of NBS are examples of reactions involving free radical intermediates.
3, 5–dimethylcyclopentene reacts with N–bromosuccinimide (NBS) in CCl4 in presence of light or peroxide. Identify the product.
chemistry-General