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प्रश्न
Arrange the following.
In increasing order of basic strength C6H5NH2, C6H5NHCH3, C6H5NH2, p-Cl-C6H4-NH2
उत्तर
Chlorine atom has both – I effect and + R effect since – I effect out weights the + R effect, therefore p-chloro aniline is weak base than aniline. Alkyl groups are electron-donating groups.
As a result, the electron density on the nitrogen atom increases in the ethylamine, and thus they can donate lone pair of electrons more easily. Therefore Ethylamine is more base than aromatic amines.
Due to delocalization of the lone pair of electrons of the N-atom over the benzene ring, C6H5NH, and C6H5NHCH3 are far less basic than C2H5NH2. Further due to +I effect of the CH3 group, C6H5NHCH3 is a little more basic than C6H5NH2. Therefore increasing order basic strength is p-Cl-C6H4-NH2 < C6H5NH2 < C6H5NHCH3 < C6H5NH2.
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संबंधित प्रश्न
Which of the following reagent can be used to convert nitrobenzene to aniline?
\[\ce{CH3CH2Br ->[aq NaOH][\Delta] A ->[KMnO4/H^+][\Delta] B ->[NH3][\Delta] C ->[Br2/NaOH] D}\] ‘D’ is:
Secondary nitro alkanes react with nitrous acid to form ____________.
How will you convert nitrobenzene into N-phenylhydroxylamine?
Identify compounds A, B and C in the following sequence of reaction.
\[\ce{C6H5NO2 ->[Fe/HCl] A ->[HNO2][273 K] B ->[C6H5OH] C}\]
Identify compounds A, B and C in the following sequence of reaction.
\[\ce{C6H5NH2 ->[(CH3CO2)O][pyridine] A ->[HNO3][H2SO4, 288 K] B ->[H2O/H^+] C}\]
Account for the following:
Aniline does not undergo Friedel-Crafts reaction.
Arrange the following.
In increasing order of basic strength aniline, p-toluidine and p-nitro aniline
Identify A, B and C.
Account for the following.
Aniline does not undergo Friedel – Crafts reaction.