Intuition The big pictureA nitro group (− N O 2 -NO_2 − N O 2 most oxidised organic state. An amine (− N H 2 -NH_2 − N H 2 fully reduced . So the whole subtopic is one arrow:
R − N O 2 → [ H ] R − N H 2 R-NO_2 \xrightarrow{\;[H]\;} R-NH_2 R − N O 2 [ H ] R − N H 2 WHY care? Aromatic nitro compounds are cheap (just nitrate benzene), and reduction gives anilines — the gateway to dyes, drugs, and diazonium chemistry. Nitro = "amine in disguise."
Definition Nitro compoundA compound containing the ==nitro group − N O 2 -NO_2 − N O 2 nitrogen (not oxygen) to carbon.
Bonded through O instead → that's a nitrite ester (R − O − N = O R-O-N=O R − O − N = O
Intuition Why is the nitro group special?The N carries a formal positive charge and the two oxygens share the negative charge by resonance — both N–O bonds are equal length (≈1.22 Å). This resonance is WHY − N O 2 -NO_2 − N O 2 electron-withdrawing (both inductive and mesomeric).
N + O − = O ⟷ N + = O O − \overset{+}{N}\!\!\begin{smallmatrix}O^-\\\\=O\end{smallmatrix} \;\longleftrightarrow\; \overset{+}{N}\!\!\begin{smallmatrix}=O\\\\O^-\end{smallmatrix} N + O − = O ⟷ N + = O O −
Intuition Why does reduction need so many electrons?Going − N O 2 → − N H 2 -NO_2 \to -NH_2 − N O 2 → − N H 2 +3 to −3 : that's a 6-electron gain, plus we strip both oxygens and add hydrogens. So we always need a generous supply of [ H ] [H] [ H ]
In acidic conditions the reduction passes through:
A r − N O 2 → A r − N O ( nitroso ) → A r − N H O H ( hydroxylamine ) → A r − N H 2 Ar\!-\!NO_2 \to Ar\!-\!NO \;(\text{nitroso}) \to Ar\!-\!NHOH \;(\text{hydroxylamine}) \to Ar\!-\!NH_2 A r − N O 2 → A r − N O ( nitroso ) → A r − N H O H ( hydroxylamine ) → A r − N H 2
Intuition Why neutral/basic conditions are differentIn neutral or weakly acidic medium the intermediates can couple before full reduction, giving azoxy → azo → hydrazo compounds. So condition controls the product — acid drives all the way to the amine.
Reagent
Product
Key point / WHY
Sn/HCl A r - N H 2 Ar\text{-}NH_2 A r - N H 2 Classic, complete, lab-friendly
Fe/HCl A r - N H 2 Ar\text{-}NH_2 A r - N H 2 Cheaper ; HCl is catalytic (Fe regenerates it) → industrial favourite
H₂/Pt (Pd/Ni) A r - N H 2 Ar\text{-}NH_2 A r - N H 2 Clean, but reduces C = C C=C C = C C ≡ N C\equiv N C ≡ N C = O C=O C = O not chemoselective
S n / H C l Sn/HCl S n / H C l reduces both N O 2 NO_2 N O 2
strong, unselective
==( N H 4 ) 2 S (NH_4)_2S ( N H 4 ) 2 S N a 2 S Na_2S N a 2 S
reduces ONE of two N O 2 NO_2 N O 2
selective mild reductant
==L i A l H 4 LiAlH_4 L i A l H 4
R - N H 2 R\text{-}NH_2 R - N H 2 great for aliphatic nitro; with aromatic it gives azo compounds!
Worked example Predict: m-dinitrobenzene +( N H 4 ) 2 S (NH_4)_2S ( N H 4 ) 2 S Step: Sulfide is a mild, partial reductant → reduces only one nitro group.
Why this step? It hasn't enough driving power for both; statistically/electronically it stops at one.
Product: m-nitroaniline . (A 6[H] reagent like Sn/HCl would give m-phenylenediamine.)
Worked example Nitrobenzene → aniline → use it
C 6 H 5 N O 2 + 6 [ H ] → F e / H C l C 6 H 5 N H 2 C_6H_5NO_2 + 6[H] \xrightarrow{Fe/HCl} C_6H_5NH_2 C 6 H 5 N O 2 + 6 [ H ] F e / H C l C 6 H 5 N H 2 Why Fe/HCl? cheap full reduction.Aniline can then be diazotised — that's why we bothered making nitrobenzene in the first place.
C H 3 C H 2 N O 2 → CH_3CH_2NO_2 \to C H 3 C H 2 N O 2 → H 2 / N i H_2/Ni H 2 / N i or L i A l H 4 LiAlH_4 L i A l H 4 C H 3 C H 2 N H 2 CH_3CH_2NH_2 C H 3 C H 2 N H 2 Why L i A l H 4 LiAlH_4 L i A l H 4 It's aliphatic, so no azo coupling problem.
N a N O 2 + R − X NaNO_2 + R-X N a N O 2 + R − X Why it feels right: N a N O 2 NaNO_2 N a N O 2 The fix: N O 2 − NO_2^- N O 2 − sodium salt gives mostly the O-attached nitrite ester (R - O - N = O R\text{-}O\text{-}N\text{=}O R - O - N = O A g N O 2 AgNO_2 A g N O 2
L i A l H 4 LiAlH_4 L i A l H 4 Why it feels right: L i A l H 4 LiAlH_4 L i A l H 4 The fix: With aromatic nitro it stops at coupled azobenzene (P h - N = N - P h Ph\text{-}N\text{=}N\text{-}Ph P h - N = N - P h H 2 / P t H_2/Pt H 2 / P t
H 2 / P t H_2/Pt H 2 / P t Why it feels right: clean, no acid, atom-economical.
The fix: It is not chemoselective — it also hydrogenates C = C C=C C = C C = O C=O C = O − C N -CN − C N
( N H 4 ) 2 S (NH_4)_2S ( N H 4 ) 2 S Why it feels right: it is a reducing agent.
The fix: It's weak — selectively reduces just one N O 2 NO_2 N O 2
Recall Self-test (hide & answer)
How many [ H ] [H] [ H ] A r - N O 2 → A r - N H 2 Ar\text{-}NO_2 \to Ar\text{-}NH_2 A r - N O 2 → A r - N H 2
Why A g N O 2 AgNO_2 A g N O 2 N a N O 2 NaNO_2 N a N O 2
Name the 3 acidic-reduction intermediates in order.
Which reagent reduces only one of two nitro groups?
Why is L i A l H 4 LiAlH_4 L i A l H 4
Recall Feynman: explain to a 12-year-oldA nitro group is like a nitrogen wearing a heavy backpack of two oxygens — it's "loaded up." Reducing it means handing the nitrogen six hydrogen helpers so it can drop the oxygens (which leave as water) and grow two friendly H arms (− N H 2 -NH_2 − N H 2 make those hydrogen helpers; or you can blow pure hydrogen gas over a metal sponge (Pt) that splits the H 2 H_2 H 2
Mnemonic Remember the reduction toolkit"Sn, Fe, Pt — Complete; Sulfide — Single; LiAlH₄ — Aliphatic (Azo if Aromatic)."
And the electron count: "Nitro to amine, count to SIX" (6[H] = 3H₂).
How many [H] are needed to reduce ArNO₂ to ArNH₂ and why? 6[H]; N goes +3→−3 (6 electrons), 2 O leave as 2H₂O (4H) and 2H add to N.
Equivalent of 6[H] as molecular hydrogen? 3H₂ (with Pt/Pd/Ni catalyst).
Why use AgNO₂ instead of NaNO₂ to make a nitroalkane? NO₂⁻ is ambident; Ag salt gives N-bonded nitroalkane, Na salt gives O-bonded nitrite ester.
Three intermediates in acidic reduction of nitrobenzene, in order? Nitroso (Ar-NO) → hydroxylamine (Ar-NHOH) → amine (Ar-NH₂).
Which reagent selectively reduces only ONE nitro group in a dinitro compound? Ammonium sulfide / Na₂S ((NH₄)₂S) — mild partial reductant.
What does LiAlH₄ give with an aromatic nitro compound? Azobenzene (Ph-N=N-Ph), NOT aniline.
Why is H₂/Pt not always the best nitro reducer? It is not chemoselective; it also reduces C=C, C=O, and –CN.
Why does nitration need conc. H₂SO₄? To generate the electrophilic nitronium ion NO₂⁺ from HNO₃.
Role of HCl in Fe/HCl reduction (industrial advantage)? HCl acts catalytically (regenerated), making Fe/HCl cheaper than Sn/HCl.
Product of m-dinitrobenzene + (NH₄)₂S? m-Nitroaniline (only one NO₂ reduced).
Resonance equal N-O bonds
Strongly electron-withdrawing
Intuition Hinglish mein samjhoDekho, nitro group (− N O 2 -NO_2 − N O 2 − N H 2 -NH_2 − N H 2 H N O 3 HNO_3 H N O 3 H 2 S O 4 H_2SO_4 H 2 S O 4 N O 2 + NO_2^+ N O 2 +
Reduction me ek hi cheez yaad rakho: + 3 +3 + 3 − 3 -3 − 3 6 electron ka kaam, isliye humein 6[H] chahiye, ya 3 molecule H 2 H_2 H 2 H 2 / P t H_2/Pt H 2 / P t C = C C=C C = C C = O C=O C = O
Do trick wale points exam me aate hain. Pehla: nitroalkane banane ke liye A g N O 2 AgNO_2 A g N O 2 N a N O 2 NaNO_2 N a N O 2 N O 2 − NO_2^- N O 2 − L i A l H 4 LiAlH_4 L i A l H 4 azobenzene banega! Aur agar do nitro group hain aur sirf ek reduce karna ho, toh mild ( N H 4 ) 2 S (NH_4)_2S ( N H 4 ) 2 S
