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Intermediate — requires understanding of electronic effects, reaction conditions, and exceptions in aromatic vs aliphatic systems.
Trap: Assuming tertiary amines are always strongest bases in water. Avoid: In aqueous phase, secondary amines are strongest due to balance of inductive effect and solvation; tertiary amines have steric hindrance.
Trap: Thinking aniline undergoes electrophilic substitution at ortho/para without protection. Avoid: Aniline must be acetylated first because –NH? gets protonated in acid, making it meta-directing.
Trap: Believing aliphatic diazonium salts are stable like aromatic ones. Avoid: Aliphatic diazonium salts are highly unstable and immediately decompose to alcohols and N? gas.
Q1. Which of the following is the strongest base in aqueous solution? A) CH?NH? B) (CH?)?NH C) (CH?)?N D) NH?
Answer: B Explanation: (CH?)?NH has the lowest pKb (3.27) among amines due to combined inductive and solvation effects. Why others fail: (CH?)?N has strong +I effect but poor solvation and steric hindrance, reducing basicity.
Q2. Which amine gives carbylamine test? A) Dimethylamine B) Aniline C) N-Methylaniline D) Trimethylamine
Answer: B Explanation: Only primary amines (aliphatic or aromatic like aniline) give carbylamine test. Why others fail: Secondary and tertiary amines do not react with CHCl?/KOH to form isocyanide.
Q3. What is the product when benzenediazonium chloride reacts with CuBr? A) Bromobenzene B) Benzene C) Phenol D) Nitrobenzene
Answer: A Explanation: This is Sandmeyer reaction: ArN + CuBr-Ar–Br + N?. Why others fail: Phenol forms only when diazonium salt is warmed with water, not with CuBr.
Q4. Which of the following has the highest pKb value? A) Methylamine B) Ammonia C) Aniline D) Dimethylamine
Answer: C Explanation: Aniline (pKb-9.38) is weakest base due to resonance; higher pKb means weaker base. Why others fail: Dimethylamine has lowest pKb (strongest base), so it's not correct.
Q5. What is the major product when aniline is treated with acetyl chloride followed by nitration and then hydrolysis? A) o-Nitroaniline B) m-Nitroaniline C) p-Nitroaniline D) 2,4,6-Trinitroaniline
Answer: C Explanation: Acetylation protects –NH? group, making it ortho-para directing; para product dominates due to steric hindrance. Why others fail: Without protection, nitration gives meta product; but here protection changes orientation.
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