Q3. Ethoxyethane boils far below butan-1-ol, yet dissolves in water almost as well. Explain both facts.
Ether molecules have no O–H group, so they cannot hydrogen bond to each other, and their boiling point stays close to pentane’s. But the ether’s oxygen can accept a hydrogen bond from water, so it dissolves about as well as an alcohol of the same molecular mass: 7.5 g against 9 g per 100 mL (NCERT page 217, which asks you to explain the solubility).
Q4. Give the products when 1-methoxy-2-methylpropane, (CH3)2CHCH2–O–CH3, is heated with concentrated HI.
CH3I and 2-methylpropan-1-ol, (CH3)2CHCH2OH. Neither group is tertiary, so iodide attacks the less substituted carbon, the methyl group, by SN2 (reasoned from NCERT page 218).
Q5. To make ethoxybenzene, C6H5–O–C2H5, would you heat sodium phenoxide with bromoethane, or bromobenzene with sodium ethoxide? Why?
Sodium phenoxide with bromoethane: the phenoxide ion attacks the primary carbon by SN2 (NCERT page 216). Bromobenzene does not undergo nucleophilic substitution under these conditions: its C–Br bond has partial double bond character and its carbon is sp2 (reasoned from NCERT Unit 6, pages 181–182). Compare NCERT Intext Question 7.11 (page 220) and Exercise 7.24 (ii), page 224.
Q6. What are the moles of ethoxyethane and of butan-1-ol that dissolve in 100 mL of water, by NCERT’s figures? (C = 12, H = 1, O = 16.)
Both are C4H10O, 74 g mol−1. Ethoxyethane: 7.5/74 = 0.10 mol; butan-1-ol: 9/74 = 0.12 mol (reasoned from NCERT page 217).
Q7. Why does anisole react with bromine in ethanoic acid without an iron(III) bromide catalyst, and what is the main product?
The –OCH3 group activates the ring by resonance, as –OH does in phenol, so no Lewis acid catalyst is needed. The main product is p-bromoanisole, 1-bromo-4-methoxybenzene, formed in 90% yield (NCERT page 219).