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प्रश्न
Arrange the following.
Increasing order of basic strength C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2.
उत्तर
Due to the delocalization of the lone pair of electrons of the N-atom over the benzene ring, all aromatic amines are less basic than alkylamines i.e., CH3NH2. The presence of electron-donating groups (– CH3) on the N-atom increases the basicity of substituted aniline with respect to C6H5NH2.
In (C6H5)NH2, the lone pair of electrons on the N-atom is delocalized over two benzene rings instead of one in C6H5NH2, therefore (C6H5)NH2 is much less basic than C6H5NH2. Combining all the three trends together, the basic strength of the four amines increasing in order.
(C6H5)NH2 < (C2H5)2NH < C6H5N(CH3)2 < CH3NH2
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संबंधित प्रश्न
Choose the most correct option.
Identify ‘B’ in the following reactions
\[\ce{CH3 - C ≡ N ->[Na/C2H5OH] A ->[NaNO2/dilHCl]B}\]
\[\ce{Aniline + benzoylchloride ->[NaOH] C6H5 - NH - COC6H5}\] this reaction is known as ____________.
\[\ce{C6H5NO2 ->[Fe/HCl] A ->[NaNO2/HCl][273 K] B ->[H2O][283 K] C}\] ‘C’ is:
IUPAC name for the amine is:
\[\begin{array}{cc}
\phantom{.}\ce{CH3}\\
|\phantom{..}\\
\ce{CH3 - N - C - CH2 - CH3}\\
\phantom{.}|\phantom{.....}|\phantom{........}\\
\phantom{}\ce{CH3}\phantom{..}\ce{C2H5}\phantom{....}
\end{array}\]
How will you convert nitrobenzene into m-nitro aniline?
Write a short note on the following.
Carbylamine reaction
The reaction of NHO2 with 'A' gives quartering ammonium salt. A is which of the following?
Assertion A: Aniline on nitration yields ortho, meta and para nitro derivatives of aniline.
Reason R: Nitrating mixture is a storng acidic mixture.
In the light of the above statements, choose the correct answer from the options given below:
Define Amines.
Write a short note on the following.
Ammonolysis
