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प्रश्न
Arrange the following:
In increasing order of basic strength:
C6H5NH2, C6H5NHCH3, C6H5CH2NH2
उत्तर
C6H5NHCH3 is more basic than C6H5NH2 due to the presence of an electron-donating −CH3 group in C6H5NHCH3.
Again, in C6H5NHCH3, −C6H5 group is directly attached to the N-atom. However, it is not so in C6H5CH2NH2. Thus, in C6H5NHCH3, the −R effect of −C6H5 group decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2
संबंधित प्रश्न
Arrange the following:
Aniline, p-nitroaniline, p-methylaniline - in the increasing order of their basic strength
Arrange the following:
In increasing order of basic strength:
Aniline, p-nitroaniline and p-toluidine
Write the structures of the main products of the following reactions:
The following reaction takes place in the presence of:
Arrange the following in decreasing order of their basic strength:
C6H5NH2, C2H5NH2, (C2H5)2NH2, NH3
The correct increasing order of basic strength for the following compounds is ______.
(I)
(II)
(III)
Explain why \[\ce{MeNH2}\] is stronger base than \[\ce{MeOH}\]?
When methyl iodide is heated with ammonia, what is the product obtained?
When ethanol is mixed with ammonia and passed over alumina the compound formed is which compound?
Which of the following statement is true about methyl amine?
Which of the following is most basic?
By the presence of a halogen atom in the ring, what is the effect of this on basic property of aniline?
Give reasons for the following observation:
pKb of aniline is lower than the m-nitroaniline.
Which of the following compound cannot be produced if 1-propane amine is treated with NaNO2 and HCl?
Among the following, which has the highest value of pKb?
Arrange the decreasing order of pKb values.
\[\ce{C6H5NH2, C6H5NHCH3, C6H5CH2NH2, CH3NH2, NH3}\]
The correct order of the increasing basic nature of Ammonia, Methylamine and Aniline is: