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 Multiple Choice QuestionsShort Answer Type

201.

Give one chemical test to distinguish between the following pairs of compounds.
Aniline and N — methlyaniline.

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202.

Account for the following:
pKb of aniline is more than that of methylamine.


It is because, in aniline, the NH2 group is attached directly to the benzene ring. It results in the unshared electron pair of the nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation.

on another hand, in the case of methylamine (due to the +I effect of a methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pkb of aniline is more than that of methylamine 
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203.

Account for the following:
Ethylamine is soluble in water, whereas aniline is not.

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204.

Account for the following:
Methylamine in water reacts with ferric chloride to precipitate ferric hydroxide.

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205. Account for the following:
Although amino group is o- and p-directing for aromatic electrophilic substitution reaction, aniline on nitration gives a substantial amount of m-nitro aniline.
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206. Account for the following:
Aniline does not undergo Friedel Craft's reaction.
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207. Account for the following:
Diazonium ions of aromatic amines are more stable than those of aliphatic amines.
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208. Account for the following:
Gabriel phthalimide synthesis is preferred for the synthesizing amines.
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209.

Arrange the following:
In decreasing order of the pKb values C2H5NH2, C6H5 NHCH3, (C2H5)2NH and C6H5NH2.

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210.

Arrange the following:
In decreasing order of basic strength:
C
6H5NH2, C6HsN(CH3)2, (C2H5)2NH and CH3NH2

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