Q3. How can you tell ethylamine from diethylamine with one of these tests? (Compare Lab Manual Discussion Question (xiv), page 99.)
The carbylamine test. Ethylamine, a primary amine, gives the foul-smelling ethyl isocyanide, C2H5NC; diethylamine, a secondary amine, does not (NCERT page 271).
Q4. Aniline has limited solubility in water but becomes much more soluble in sufficient dilute HCl. Why? (Compare Lab Manual Unit 10, Discussion Question (xi), page 113.)
Aniline is a base: with HCl it forms the salt anilinium chloride, C6H5NH3+Cl−, and amine salts are soluble in water (Lab Manual, pages 97–98; NCERT page 267).
Q5. Why must the product of the carbylamine test be destroyed at once, and how?
The isocyanide is highly toxic. The Lab Manual’s answer: cool the tube and carefully add an excess of concentrated hydrochloric acid (page 97). Our safety advice: the instructor does this in a fume hood and follows the laboratory’s waste procedure; students do not improvise it.
Q6. In the azo dye test, 0.2 g of aniline is treated with 2 mL of 2.5% sodium nitrite solution. Taking 2.5% to mean 2.5 g in 100 mL, which is used up first, and at most what mass of dye, C16H12N2O, can form? (C = 12, H = 1, N = 14, O = 16, Na = 23.)
NaNO2: 2 × 2.5/100 = 0.050 g, and M = 23 + 14 + 32 = 69 g mol−1, so 0.050/69 = 7.2 × 10−4 mol. Aniline: 0.2/93 = 2.2 × 10−3 mol. Each mole of dye needs one of each, so the nitrite is used up first; the 0.2 g of β-naphthol, C10H8O (M = 144 g mol−1), is 0.2/144 = 1.4 × 10−3 mol, also more than the nitrite. M(dye) = 192 + 12 + 28 + 16 = 248 g mol−1: 7.2 × 10−4 × 248 = 0.18 g at most (reasoned; reading “2.5%” as mass per volume is an assumption).