Organic Nitrogen Compounds — Practice Quiz

N-methyl propan-2-amine

Step 1: Identify the longest carbon chain attached to nitrogen: CH(CH3)2 is isopropyl = 3 carbons = propane. Step 2: The nitrogen is attached to C2 of the propane chain, so it is propan-2-amine. Step 3: There is a methyl group (CH3) directly on the nitrogen, so we add the prefix N-methyl. Step 4: The complete IUPAC name is N-methyl propan-2-amine. Step 5: 'N-isopropyl methanamine' would be incorrect because we should choose the longer chain (propyl, 3C) as the parent, not methyl (1C). The rule is to choose the longest chain as the parent amine.

The lone pair on nitrogen in aniline is delocalised over the benzene ring, making it less available for protonation

Step 1: In methylamine (CH3-NH2), the lone pair on nitrogen is fully available to accept a proton (H+), making it a good base. Step 2: In aniline (C6H5-NH2), the nitrogen's lone pair is in conjugation with the π electron system of the benzene ring. Step 3: This conjugation/delocalisation means the lone pair is spread over the entire ring through resonance, reducing its availability for accepting a proton. Step 4: Therefore, aniline is a much weaker base (pKb = 9.30) compared to methylamine (pKb = 3.38). Step 5: The -NH2 group in aniline is actually electron-donating (+M effect) to the ring, ma

Cuprous chloride (CuCl)

Step 1: In the Sandmeyer reaction, the diazonium group (-N2+) is replaced by a halide group using cuprous halide salts. Step 2: For chlorobenzene: C6H5-N2+Cl- + CuCl → C6H5-Cl + N2↑ + CuCl. Step 3: The key reagent is cuprous chloride (CuCl), which is the freshly prepared cuprous salt. Step 4: The Gattermann reaction also gives chlorobenzene but uses HCl/Cu powder — this is a different reaction with lower yield. Step 5: Cl2/FeCl3 would give electrophilic substitution (halogenation), not replacement of the diazonium group. NaCl is too weak a nucleophile for this reaction.

2,4,6-tribromoaniline (white precipitate)

Step 1: The -NH2 group in aniline is a very strong activating and ortho/para-directing group. Step 2: It strongly increases electron density at the ortho (2,6) and para (4) positions of the benzene ring. Step 3: Bromine water (Br2/H2O) without any Lewis acid catalyst is enough to brominate all three activated positions simultaneously due to the high reactivity. Step 4: The reaction gives 2,4,6-tribromoaniline as a white precipitate: C6H5-NH2 + 3Br2 → 2,4,6-Br3-C6H2-NH2 + 3HBr. Step 5: To obtain monobromo product, the -NH2 group must first be protected by acetylation to reduce its activating po