RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
Amines are the organic cousins of ammonia, and swapping hydrogens for alkyl or aryl groups changes their basicity in ways that surprise most students at first. This chapter works through how to rank 1°, 2° and 3° amines by basicity, how a diazonium salt becomes a gateway to a dozen other functional groups, and how the Hofmann bromamide reaction quietly removes a carbon atom on its way from an amide to an amine.
Which of the following is the weakest base in aqueous solution?
Answer
Aniline. The nitrogen lone pair in aniline is delocalised into the benzene ring by resonance, so it is much less available to accept a proton, and the resulting anilinium cation loses this resonance stabilisation entirely. Methylamine and dimethylamine are both more basic than ammonia in water because their alkyl groups donate electron density inductively without any competing resonance withdrawal.
Aniline is treated with excess bromine water at room temperature. The product is:
Answer
2,4,6-Tribromoaniline as a white precipitate. The –NH₂ group is a powerful activator that donates its lone pair into the ring by resonance, so much so that no Lewis-acid catalyst is needed and substitution occurs at all three positions activated by resonance (both ortho positions and the para position), immediately at room temperature.
Benzenediazonium chloride is treated with CuCN in the presence of KCN. This is an example of the:
Answer
Sandmeyer reaction. Copper(I) salts such as CuCl, CuBr and CuCN catalyse the replacement of the diazonium (–N₂⁺) group by Cl, Br or CN respectively, providing a reliable route to substituted benzenes that could not easily be made by direct electrophilic substitution. Here the diazonium group is replaced by –CN to give benzonitrile, C₆H₅CN.
In the Hinsberg test, a colourless liquid amine gives a precipitate with benzenesulfonyl chloride that is insoluble in excess NaOH. The amine is:
Answer
A secondary amine. A secondary amine reacts with benzenesulfonyl chloride to form N,N-disubstituted sulfonamide with no N–H hydrogen remaining, so it has no acidic proton left to be removed by NaOH and the precipitate stays insoluble. A primary amine instead forms a sulfonamide that still has one acidic N–H, which NaOH deprotonates to give a soluble salt, while a tertiary amine has no N–H to begin with and does not react to form a solid at all.
Assertion (A): The Gabriel phthalimide synthesis cannot be used to prepare aniline. Reason (R): Aryl halides do not undergo the nucleophilic substitution step required by the mechanism, because the C–X bond in an aryl halide has partial double-bond character and the carbon is not accessible to backside attack.
Answer
Both A and R are true and R is the correct explanation of A. The Gabriel synthesis requires the phthalimide anion to displace a halide from an alkyl halide by an SN2 mechanism. Aryl halides resist nucleophilic substitution by this route because the halogen is conjugated with the ring (partial double-bond character strengthens the C–X bond) and the sp² carbon is shielded from backside attack, so aniline cannot be made this way; aliphatic primary amines are the only amines this method can cleanly produce.
Write the equation for the Hofmann bromamide degradation of propanamide, and state how many carbon atoms the product amine contains compared with the starting amide.
Answer
CH₃CH₂CONH₂ + Br₂ + 4NaOH → CH₃CH₂NH₂ + Na₂CO₃ + 2NaBr + 2H₂O. Propanamide has three carbons; the product, ethylamine, has only two. The carbonyl carbon is lost as sodium carbonate during the rearrangement (via an isocyanate intermediate that is hydrolysed), so the amine formed always has one fewer carbon than the parent amide.
Explain why ethylamine is a stronger base than aniline, and why aniline is a stronger base than diphenylamine.
Answer
Ethylamine's nitrogen lone pair is fully available for bonding to a proton, and the ethyl group actually pushes extra electron density onto nitrogen by the +I effect, making the lone pair even more available; this gives ethylamine a strongly basic character comparable to or exceeding ammonia. In aniline, the lone pair is partly delocalised into the single benzene ring by resonance, reducing its availability and making aniline a much weaker base than ethylamine. In diphenylamine, the same nitrogen lone pair is now delocalised into two benzene rings instead of one, spreading it even more thinly and reducing its availability for protonation even further, so diphenylamine is a weaker base than aniline.
How would you convert aniline to benzene in two steps? Give reagents and equations.
Answer
Step 1 (diazotisation): C₆H₅NH₂ + NaNO₂ + 2HCl --273–278 K--> C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O, giving benzenediazonium chloride. Step 2 (deamination): C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ + N₂↑ + H₃PO₃ + HCl. Hypophosphorous acid reduces the diazonium salt, replacing the –N₂⁺ group with hydrogen and releasing nitrogen gas, so the net effect of the two steps is removal of the –NH₂ group from the ring.
(i) Explain, with resonance structures described in words, why aniline undergoes electrophilic substitution faster than benzene and predominantly at the ortho and para positions. (ii) Describe the diazo coupling reaction of benzenediazonium chloride with aniline, giving the product and explaining the colour. (iii) Why must diazotisation be carried out below 5 °C?
Answer
(i) The nitrogen lone pair on aniline's –NH₂ group conjugates with the ring π system: resonance structures can be drawn placing extra electron density (and formal negative charge) specifically on the ortho carbons and the para carbon, while the meta carbons are never enriched in this way. This makes the ortho and para positions unusually electron-rich compared with benzene, so an electrophile attacks there preferentially, and the arenium ion intermediate formed at ortho or para is further stabilised by direct resonance donation from nitrogen, lowering the activation energy and making aniline react far faster than benzene overall. (ii) Benzenediazonium chloride couples with aniline in a mildly acidic medium at the para position of the aniline ring (or on the ring nitrogen if para is blocked) to give 4-aminoazobenzene, C₆H₅–N=N–C₆H₄–NH₂. This azo compound is intensely coloured (yellow-orange) because the extended conjugation through the –N=N– azo linkage between the two aromatic rings creates a chromophore that absorbs visible light, an effect exploited in azo dyes. (iii) The diazonium ion is thermodynamically unstable at room temperature; above about 5 °C it rapidly decomposes, losing N₂ gas and reacting with the water solvent to form phenol (C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂ + H⁺). Keeping the reaction mixture in an ice bath below 5 °C slows this hydrolysis enough that the diazonium salt can be isolated in solution and used promptly for further reactions such as Sandmeyer or coupling reactions.
(i) Compare the carbylamine reaction and the Hinsberg test as methods for distinguishing primary, secondary and tertiary amines, stating what each test reveals. (ii) An unknown amide C₃H₇CONH₂ is subjected to Hofmann bromamide degradation. Identify the product, name it, and give the overall equation. (iii) Explain why methylamine is soluble in water but aniline is only sparingly soluble.
Answer
(i) The carbylamine (isocyanide) test is specific to primary amines only: a primary amine warmed with chloroform and alcoholic KOH gives the foul-smelling isocyanide RNC, while secondary and tertiary amines give no reaction at all, so this test can identify a primary amine but cannot distinguish secondary from tertiary. The Hinsberg test, using benzenesulfonyl chloride and then NaOH, distinguishes all three: a primary amine gives an NaOH-soluble sulfonamide (N–H retained, acidic), a secondary amine gives an NaOH-insoluble sulfonamide (no N–H left), and a tertiary amine gives no solid product at all since it has no N–H to react with the sulfonyl chloride. The Hinsberg test is therefore the more complete diagnostic of the two. (ii) C₃H₇CONH₂ is butanamide. Hofmann degradation removes the carbonyl carbon as carbonate, so the product is propan-1-amine, CH₃CH₂CH₂NH₂, an amine with one fewer carbon than the starting amide. CH₃CH₂CH₂CONH₂ + Br₂ + 4NaOH → CH₃CH₂CH₂NH₂ + Na₂CO₃ + 2NaBr + 2H₂O. (iii) Methylamine, like ammonia, has a small alkyl group and an N–H bond that can both donate and accept hydrogen bonds with water molecules, so it mixes freely and is highly soluble at ordinary concentrations. Aniline has a large, hydrophobic benzene ring attached to nitrogen; although the N–H can still hydrogen-bond with water, the bulky non-polar ring dominates the molecule's interaction with the solvent and severely limits how much of it water can accommodate, so aniline dissolves only sparingly.
A perfumer wants to convert p-toluidine (4-methylaniline) into 4-methylphenol for a fragrance base, and separately wants a pure sample of an amine with no risk of over-alkylation contamination for a second product. (a) Write the two-step sequence (reagents and equations) to convert p-toluidine to 4-methylphenol. (b) Explain briefly why direct treatment of 4-methylbromobenzene with NaOH would be a poor alternative route to the same phenol. (c) Name the synthesis method that would give the perfumer a pure primary amine with no secondary or tertiary amine contamination, and briefly state why it avoids that problem. (d) If the amine needed were aniline itself, explain why the method in (c) would fail and suggest what should be used instead.
Answer
(a) Step 1 (diazotisation): 4-CH₃C₆H₄NH₂ + NaNO₂ + 2HCl --273–278 K--> 4-CH₃C₆H₄N₂⁺Cl⁻ + NaCl + 2H₂O. Step 2 (hydrolysis on warming): 4-CH₃C₆H₄N₂⁺Cl⁻ + H₂O --warm--> 4-CH₃C₆H₄OH + N₂↑ + HCl, giving 4-methylphenol directly, since warming a diazonium salt in water is exactly the reaction that must be avoided during diazotisation but is deliberately exploited here. (b) Aryl halides such as 4-methylbromobenzene are extremely unreactive towards nucleophilic substitution under ordinary conditions because the C–Br bond has partial double-bond character from conjugation with the ring and the sp² carbon resists backside attack; NaOH would require very harsh, high-pressure/high-temperature conditions to force any substitution, making it a poor practical route compared with diazonium hydrolysis. (c) The Gabriel phthalimide synthesis. The alkyl halide reacts with the phthalimide anion inside a ring system that permits only one substitution per nitrogen, and the amine is released only in a later, separate hydrolysis step, so there is no possibility of the amine reacting further with unreacted alkyl halide the way ammonia would in a direct alkylation, which eliminates over-alkylation entirely. (d) The Gabriel synthesis would fail for aniline because it needs an aryl halide to undergo the initial SN2 displacement by phthalimide anion, and aryl halides are unreactive to SN2 for the same reasons given in (b). Aniline should instead be made by reducing nitrobenzene (Sn/HCl or Fe/HCl followed by NaOH work-up, or catalytic H₂/Ni), a route that does not depend on nucleophilic substitution at an aromatic carbon at all.
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