Q3. Arrange NH3, CH3NH2, (CH3)2NH and (CH3)3N in decreasing order of basic strength in water.
(CH3)2NH > CH3NH2 > (CH3)3N > NH3; the pKb values are 3.27, 3.38, 4.22 and 4.75 (NCERT pages 267 and 269).
Q4. What are the products when propan-1-amine reacts with nitrous acid made from NaNO2 and HCl?
NCERT’s scheme (page 271) is R–NH + HNO → [R–NCl] → ROH + N + HCl, with the nitrous acid made from NaNO and HCl. For propan-1-amine the balanced overall equation is CHCHCHNH + HNO → CHCHCHOH + N + HO (reasoned from NCERT page 271); write it in an exam. In practice this does not guarantee pure propan-1-ol: the unstable intermediate can rearrange or eliminate, giving propan-2-ol and propene as well (reasoned). Nitrogen evolution, rather than a unique alcohol, is the useful distinction here.
Q5. Write the equation for the carbylamine reaction of aniline and name the organic product.
C6H5NH2 + CHCl3 + 3KOH → C6H5NC + 3KCl + 3H2O, on heating with ethanolic KOH. The product is phenyl isocyanide (isocyanobenzene), the foul-smelling carbylamine (reasoned from NCERT page 271).
Q6. How is sulphanilic acid made from aniline, and in what form does NCERT draw it?
Aniline and concentrated sulphuric acid give anilinium hydrogensulphate, C6H5NH3+HSO4−; heated with sulphuric acid at 453–473 K it gives p-aminobenzenesulphonic acid, sulphanilic acid, as the major product. NCERT draws it in equilibrium with its zwitter ion, H3N+–C6H4–SO3− (NCERT page 273).
Q7. Which is the stronger base, aniline or 4-nitroaniline? Why? (Compare NCERT Exercise 9.4 (iii)(a), page 278.)
Aniline: the electron-withdrawing –NO2 group decreases the basic strength of a substituted aniline (NCERT page 269).
Q8. What does N-methylaniline give with ethanoyl chloride in the presence of pyridine?
N-methyl-N-phenylethanamide, C6H5N(CH3)COCH3, and HCl, which pyridine removes: a secondary amine is acylated at its one N–H (reasoned from NCERT page 270).