C2H5NH2 , C6H5NHCH3, (C2H5)2NH and C6H5NH2
For the above compounds, decreasing order of the pKb values (in gaseous phase) is -
1. C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH
2. C2H5NH2 > C6H5NHCH3 > C6H5NH2 > (C2H5)2NH
3. C6H5NHCH3 > C6H5NH2 > C2H5NH2 > (C2H5)2NH
4. (C2H5)2NH > C6H5NHCH3 > C2H5NH2 > C6H5NH2
The increasing order of boiling point of :
C2H5OH, (CH3)2NH, C2H5NH2 is-
1. (CH3)2NH < C2H5NH2 < C2H5OH
2. (CH3)2NH > C2H5NH2 > C2H5OH
3. (CH3)2NH < C2H5NH2 > C2H5OH
4. (CH3)2NH < C2H5OH< C2H5NH2
The increasing sequence of solubility of the following molecules ;C6H5NH2, (C2H5)2NH, C2H5NH2
in water is -
1. C6H5NH2 < (C2H5)2NH < C2H5NH2
2. C6H5NH2 > (C2H5)2NH < C2H5NH2
3. C6H5NH2 > (C2H5)2NH > C2H5NH2
4. C6H5NH2 < (C2H5)2NH > C2H5NH2
The A and B in the above reactions are -
1. | Alcohol, Diazonium salt | 2. | Aldehyde, Diazonium salt |
3. | Diazonium salt, Alcohol | 4. | Diazonium salt, Aldehyde |
Amines are less acidic than alcohols of comparable molecular masses because :
1. N is more electronegative than O.
2. O is more electronegative than N.
3. N can accomodate negative charge easily than O
4. Amides are more stable than alcohols
An aromatic compound ‘A’ on treatment with ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. The Substrate A is -
1. Aniline
2. Benzoic acid
3. Benzamide
4. Benzamine
A method for the identification of primary, secondary and tertiary amines is:
1. Azo-dye test
2. Carbylamine test
3. Hinsberg’s test
4. None of these
Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because :
1. | Alkyl halides do not undergo electrophilic substitution with the anion formed by the phthalimide |
2. | Alkyl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide |
3. | Aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide |
4. | Aryl halides do not undergo electrophilic substitution with the anion formed by the phthalimide |
:
The product from the above reaction would be -