Q3. Outline how to make iodobenzene from benzene.
Nitrate benzene to nitrobenzene, reduce it to aniline (for example with a metal and acid, NCERT page 262), diazotise at 273–278 K, then treat the benzenediazonium chloride with KI (NCERT page 275): C6H5N2+Cl− + KI → C6H5I + KCl + N2.
Q4. In NCERT Example 9.5, what job does the nitro group do, and how is it got rid of?
It occupies the para position, so bromine is strongly directed ortho to –CH3; it is then reduced to –NH2, diazotised and replaced by H with H3PO2 (NCERT page 277; reasoned).
Q5. Why does brominating toluene not selectively give 3-bromotoluene, and how can a diazonium salt give it?
–CH3 sends bromine ortho and para, giving 2- and 4-bromotoluene (NCERT Class 11 Chemistry, Unit 9, pages 324–325). Start instead from 3-methylaniline: diazotise at 273–278 K, then treat it with Cu2Br2/HBr (Sandmeyer reaction, NCERT page 275) to get 3-bromotoluene (reasoned).
Q6. At most, what mass of 1,3,5-tribromobenzene can 2.79 g of aniline give by the route in Q1? (C = 12, H = 1, N = 14, Br = 80.)
M(C6H5NH2) = 72 + 7 + 14 = 93 g mol−1, so 2.79/93 = 0.0300 mol. Each step is one mole to one mole, and M(C6H3Br3) = 72 + 3 + 240 = 315 g mol−1: 0.0300 × 315 = 9.45 g (reasoned).