Multi-Step Conversions Using Amines and Diazonium Salts
Chemistry · Amines · NEET
A multi-step conversion is a chain of reactions where the product of one step becomes the starting material for the next. In amines, the key idea is that aniline is a "hub": reduce a nitro group to get aniline, turn aniline into a diazonium salt with HNO2 at 273-278 K, and then swap the diazonium (-N2+) group for almost anything (-Cl, -Br, -I, -CN, -OH, -H, -NO2, or an azo dye). Memory hook: "NO2 to NH2 to N2+ to whatever you want" - the diazonium salt is the traffic circle that connects all the exit roads.
The aromatic hub: reduce a nitro group to aniline, diazotise it cold (273-278 K) to the diazonium salt, then swap the -N2+ group for almost any group. Warming instead gives phenol plus N2 gas.
Your doubts, answered
How do I decide the order of reagents in a multi-step conversion question?
Work backwards from the target. Ask: what is the last group I need, and what reaction gives it? Then ask what that step needs as its starting material, and keep going back to the given compound. For aromatic questions the usual chain is nitro to amine (Sn/HCl or Fe/HCl or H2/Ni), amine to diazonium (NaNO2 + HCl at 273-278 K), then diazonium to the final group. A classic NEET trap: if you must place a substituent at a certain ring position, do the ring reaction FIRST (while the -NH2 or -NO2 is still directing traffic), and only remove the nitrogen group at the end by deamination (H3PO2 or H2O).
In these conversions, does the number of carbon atoms stay the same, go up, or go down?
It depends on the tool. Diazonium swaps (-N2+ becoming -Cl, -Br, -OH, -H, -NO2) keep the carbon count the same. Adding -CN (nitrile) and then reducing it ASCENDS the series by one carbon (R-X to R-CN to R-CH2-NH2). Hofmann bromamide degradation (amide + Br2/NaOH) DESCENDS by one carbon (R-CONH2 to R-NH2, losing one C as carbonate). NEET loves to test this: an amide with 4 carbons gives an amine with only 3 carbons.
Why make the diazonium salt cold at 273-278 K and use it right away?
Aromatic diazonium salts (Ar-N2+) are only mildly stable because the positive charge is spread into the benzene ring by resonance. Even so, if you warm them they decompose, releasing N2 gas and giving phenol. So you prepare them cold (0-5 degrees C = 273-278 K) and use the fresh solution immediately in the next step. Aliphatic diazonium salts (R-N2+) are so unstable they break down at once, which is why this whole diazonium toolbox only works for aromatic amines.
How can I turn aniline into bromobenzene when direct bromination gives tribromoaniline instead?
You cannot brominate aniline directly to bromobenzene because -NH2 is strongly activating and gives 2,4,6-tribromoaniline. The diazonium route solves this: aniline to benzenediazonium chloride (HNO2, cold), then Cu2Br2/HBr (Sandmeyer or Gattermann) replaces -N2+ with a single -Br. This controlled, single substitution is exactly why diazonium salts are so important in aromatic synthesis.
Aniline to phenol - what are the exact steps?
Two steps. First diazotise: aniline + NaNO2/HCl at 273-278 K gives benzenediazonium chloride. Second, warm the diazonium salt with water (hydrolysis, warm dilute acid); -N2+ leaves as N2 gas and -OH takes its place, giving phenol. Starting from nitrobenzene, just add a reduction step first (Sn/HCl to aniline).
⚠️ The NEET trap ✗ Reduce nitro to amine first, then brominate to put -Br on the ring, then diazotise and deaminate. Also assuming Hofmann degradation keeps the same number of carbons. ✓ Do the ring substitution while the directing group is still present, and remove the nitrogen function LAST. Remember Hofmann bromamide (Br2/NaOH) removes ONE carbon: butanamide (4C) gives propylamine (3C), not butylamine. 🧠 Reagent order and carbon-count changes are where students lose the mark.
Real NEET questions
2023
Identify the final product in the following reaction sequence: C6H5N2+Cl- --(i) Cu2Br2/HBr (ii) Mg/dry ether (iii) H2O--> Product
A · Phenol, C6H5OH
B · Benzene, C6H6 ✓
C · Phenylmagnesium bromide, C6H5MgBr
D · 4-Bromophenol, HO-C6H4-Br
Solution: Step (i) is a Gattermann/Sandmeyer-type replacement: Cu2Br2/HBr swaps the diazonium group for -Br, giving bromobenzene (N2 lost). Step (ii): bromobenzene + Mg in dry ether gives the Grignard reagent phenylmagnesium bromide. Step (iii): the Grignard reacts with water (acidic H) and is protonated to the parent hydrocarbon, benzene. So the final product is benzene. C6H5MgBr is only the intermediate, so (C) is a distractor.
2023
Choose the correct sequence of reagents in the conversion of 4-nitrotoluene to 2-bromotoluene.
A · Br2; Sn/HCl; NaNO2/HCl; H2O/H3PO2 ✓
B · Sn/HCl; Br2; NaNO2/HCl; H2O/H3PO2
C · NaNO2/HCl; Sn/HCl; Br2; H2O/H3PO2
D · Sn/HCl; NaNO2/HCl; Br2; H2O/H3PO2
Solution: Brominate FIRST: on 4-nitrotoluene, Br2 goes ortho to -CH3 (the carbon that will finally carry -Br), because the -NO2 and -CH3 direct it there. Then reduce -NO2 to -NH2 (Sn/HCl), diazotise (NaNO2/HCl, cold), and finally deaminate with H3PO2 to replace -N2+ by -H, leaving 2-bromotoluene. Doing the ring bromination before reducing/diazotising is the key: the directing group must still be present when you brominate.
2016
A nitrogen-containing aromatic compound A reacts with Sn/HCl, then with HNO2, to give an unstable compound B. B with phenol forms a coloured compound C, C12H10N2O. The structure of A is:
A · Aniline, C6H5NH2
B · Nitrobenzene, C6H5NO2 ✓
C · Benzonitrile, C6H5CN
D · Benzamide, C6H5CONH2
Solution: The chain is reduction to diazotisation to azo coupling. Sn/HCl reduces a nitro group to an amine, so A must be nitrobenzene, giving aniline. HNO2 (NaNO2 + HCl, 273-278 K) diazotises aniline to the unstable benzenediazonium salt B. B couples at the para position of phenol to give the orange azo dye p-hydroxyazobenzene, C12H10N2O = C. Aniline (A) is wrong because it is the intermediate, not the starting material that needs Sn/HCl reduction.
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Why are diazonium salts so central to aromatic multi-step synthesis?
Because a single -N2+ group can be replaced by -F, -Cl, -Br, -I, -CN, -OH, -NO2, -H, or coupled into an azo dye. It is one intermediate with many exit routes, letting you place groups on the ring that direct electrophilic substitution cannot reach cleanly.
Which reagents ascend the amine series and which descend it?
Ascend by one carbon: alkyl halide to nitrile (KCN/NaCN) then reduce (LiAlH4, Na(Hg), or H2/Ni). Descend by one carbon: Hofmann bromamide degradation of an amide (Br2/NaOH). Same carbon count: Gabriel synthesis, ammonolysis, and diazonium swaps.
Can I use this diazonium toolbox on aliphatic (open-chain) amines?
No. Aliphatic diazonium salts are extremely unstable and decompose immediately to release N2, giving a mixture of alcohols and alkenes. The stable, useful diazonium chemistry works only with aromatic primary amines like aniline.
What temperature is needed to make a benzenediazonium salt?
273-278 K, that is 0-5 degrees C, in the presence of HCl and NaNO2 (which generate HNO2 in situ). Above this temperature the salt hydrolyses to phenol.
How do I convert a nitro compound all the way to a hydrocarbon that has one less nitrogen group?
Reduce -NO2 to -NH2, diazotise to -N2+, then deaminate: warm the diazonium salt with hypophosphorous acid (H3PO2) or ethanol, which replaces -N2+ by -H. This effectively removes the nitrogen function from the ring.
