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Question 106 of 108

Q.(a) Write reasons for the following :

(i) Ethylamine is soluble in water whereas aniline is insoluble.
(ii) Amino group is o- and p-directing in aromatic electrophilic substitution reactions, but aniline on nitration gives a substantial amount of m-nitroaniline.
(iii) Amines behave as nucleophiles.
(OR)
(b) How will you carry out the following conversions :
(i) Nitrobenzene to Aniline
(ii) Ethanamide to Methanamine
(iii) Ethanenitrile to Ethanamine
Punjab PsebCBSE Class XII Board 2022Subjective· 3mImportance★★★★★
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Part (a): ethylamine is water-soluble (effective H-bonding, small chain) while aniline is not (bulky hydrophobic ring); in acidic nitration aniline is protonated to the meta-directing −NHX3X+-\ce{NH3+}, giving substantial m-nitroaniline; amines are nucleophilic because of the N lone pair. Part (b): nitrobenzene → aniline (Sn/HCl then NaOH); ethanamide → methanamine (Hofmann bromamide, one C less); ethanenitrile → ethanamine (LiAlH₄).

Part (a)

  1. Ethylamine soluble, aniline insoluble. Ethylamine's small –NH₂ group hydrogen-bonds strongly with water, and its short ethyl chain barely disrupts the water structure — so it is very soluble. In aniline the same –NH₂ can H-bond, but the large hydrophobic benzene ring dominates and its lone pair is partly delocalised into the ring (less available for H-bonding), so aniline is only sparingly soluble.
  2. o/p-directing but gives m-nitroaniline. Free –NH₂ donates its lone pair by resonance and is strongly activating, o/p-directing. Nitration, however, is done in a strongly acidic HNOX3/HX2SOX4\ce{HNO3/H2SO4} mixture that protonates the amine:

    CX6HX5NHX2+HX+→CX6HX5NHX3X+\ce{C6H5NH2 + H+ -> C6H5NH3+}

    The −NHX3X+-\ce{NH3+} group is electron-withdrawing, deactivating and meta-directing, so a substantial fraction of the product is m-nitroaniline (the o/p isomers come from the small amount of free aniline present). …

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