Nitro compounds — preparation, reduction to amines (Sn - HCl, Fe - HCl, H₂ - Pt): Foundations

Before you can follow that journey you must own every symbol it uses. Below, each idea is built from nothing, drawn as a picture, and justified by why the topic needs it. Read top to bottom — each rung of the ladder rests on the one below.

1. An atom, a bond, a formula

Why the topic needs this: every reagent ( $H N O_{3}$ , $H_{2} S O_{4}$ , $H_{2}$ ) and every product ( $- N O_{2}$ , $- N H_{2}$ ) is written as a formula. If you can't read the counting, you can't read the topic.

2. What " $R$ " and " $A r$ " mean

Before we draw any group we need the placeholder that stands in for "whatever the group is attached to," so nothing in the pictures uses an undefined symbol.

3. The two star groups, drawn

Now that $R$ is defined, here are the two groups as pictures — refer to them for the rest of the page.

Figure — Nitro compounds — preparation, reduction to amines (Sn - HCl, Fe - HCl, H₂ - Pt)

Figure s01 (alt-text): on the left, the nitro group in red — a central $N$ joined by two bonds to two $O$ atoms, with the leftover bond reaching left to a carbon group $R$ ; on the right, the amine group — an $N$ joined to two $H$ atoms and one bond to $R$ . Both attach to carbon through the nitrogen.

4. Resonance — why the two N–O bonds are twins

We meet resonance before charges, because the way $- N O_{2}$ shares its charge only makes sense once you can blend two drawings.

Figure s03 (alt-text): two structures joined by a double-headed arrow — drawing A has the double bond on the top oxygen and $O^{-}$ on the bottom, drawing B is the mirror; below, in red, the resonance hybrid shows two identical dashed N–O bonds and a single half-minus charge marked on each oxygen.

$N + O^{-} = O ⟷ N + = O O^{-}$

Intuition What the picture shows — and where the negative charge really sits

In the figure, drawing A puts the double bond on the top oxygen (so the bottom O keeps the full $O^{-}$ ); drawing B is the mirror image. Neither is the truth alone. The real state is the hybrid (red, bottom of the figure): the double-headed arrow tells us to merge the two drawings, so both N–O bonds become identical, halfway between single and double (about $1.22 \overset{˚}{A}$ each).

How to draw the charge in the hybrid: since drawing A put the $-$ on the bottom O and drawing B put it on the top O, in the hybrid each oxygen carries half of one negative charge. We show this as a small $-$ (or $\frac{1}{2} -$ ) marked on each oxygen instead of a full $-$ on just one — it is delocalised equally over both, a direct consequence of the two resonance drawings.

Why the topic needs this: this smearing is why $- N O_{2}$ is such a powerful electron-withdrawing group, and it's the origin of the group's stability. See Ambident nucleophiles for how the related $N O_{2}^{-}$ ion uses the same two-ended personality.

5. Charge, and why nitrogen wears a " $+$ "

Figure s02 (alt-text): the nitro group with a red $+$ marked beside the nitrogen and, thanks to the resonance of the previous section, a half-minus charge marked on each of the two oxygens, illustrating that N is electron-short and the negative charge is shared over both O atoms.

Why the topic needs this: the $+$ on nitrogen is exactly why $- N O_{2}$ pulls electrons out of a benzene ring (electron-withdrawing), and why reduction has to pour electrons in to reverse it.

6. Oxidation state — the number the whole topic counts

Worked example Counting N in

$- N O_{2}$ — one clean rule, no fudge factor The rule: for each bond, decide which atom is more electronegative (hungrier). The hungrier atom is charged $- 1$ for that bond; its partner is charged $+ 1$ . Add up all of nitrogen's bonds.

Rank the hunger (electronegativity): $O > N > C$ . So oxygen wins every N–O pair; nitrogen wins the N–C pair.
Count nitrogen's bonds in $- N O_{2}$ : nitrogen has three bonds to oxygen in total (one double bond = 2 bonds, one single bond = 1 bond) and one bond to carbon.
Score each bond for N:
- each of the 3 bonds to O: nitrogen loses → $+ 1$ each → $+ 1 + 1 + 1 = + 3$ .
- the 1 bond to C: nitrogen wins → $- 1$ .
Total: $+ 3 + (- 1) = + 2$ . $oxidation state of N in - N O_{2} = + 2.$ Keep the takeaway: each bond to O drives N up (+1), the bond to C drives it down (−1); oxygen wins three times against one loss to carbon, so N lands at $+ 2$ . (No formal charges enter this tally at any point.)

7. Reduction, oxidation, and the symbol $[H]$

8. The reaction arrow and its labels

Why the topic needs this: every equation in the parent — nitration, reduction, the intermediates — is written in exactly this grammar. Once you can read the arrow, the whole page becomes sentences.

9. How the pieces feed the topic

The figure below is a prerequisite map — a rendered flowchart. Read it like a ladder: each box is one idea from this page, and an arrow means "you need the box it comes from before the box it points to makes sense." Start at the top with atoms, follow the arrows downward; every path funnels into the bottom box — the actual topic (4.4.3). If you can trace a route from any box down to "Nitro → amine," you're seeing exactly which foundations that step rests on.

Figure s04 (alt-text): a top-down flowchart. "Atoms & symbols" feeds "Bonds," which feeds "Formula reading"; "Formula reading" and "R / Ar placeholder" feed "Groups NO2 & NH2"; "Groups" feeds "Resonance," which feeds "Formal charge (+ on N)"; "Groups" also feeds "Oxidation state (+2 → −3)," which feeds "Reduction gains electrons," which feeds "6[H] = 3H2." "Reaction arrow grammar," "6[H] = 3H2," and "Groups" all feed the bottom box "Nitro → amine (topic 4.4.3)," drawn in red.

Equipment checklist

Cover the right side and answer each before starting the parent note.

What does the subscript in

C_{6} H_{6}

tell you?

How many of that atom are present — six carbons, six hydrogens.

How do you tell a bond-dash from a minus-charge in a formula?

A dash between two atoms is a bond; a small stroke written as a superscript on one atom is a charge.

What do

R

and

A r

stand for?

R

= any attached group built on carbon (may carry O, N, halogens);

A r

= an aromatic (benzene-type) ring.

What is the difference between

- N O_{2}

and

- O - N = O

- N O_{2}

nitrogen attaches to carbon; in the nitrite ester an oxygen attaches. Different bridging atom, different chemistry.

In the resonance hybrid of

- N O_{2}

, where is the negative charge and how is it drawn?

Split equally — half a minus on each oxygen, because the two drawings put the full

-

on opposite O's.

How does formal charge differ from oxidation state?

Formal charge splits each bond's electrons equally between the two atoms; oxidation state gives both electrons to the hungrier atom. Different accounts — never mix them.

Tally the oxidation state of N in

- N O_{2}

by the electronegativity rule.

Three bonds to O each give

+ 1

(

= + 3

); one bond to C gives

- 1

; total

+ 2

. (No formal charge used.)

What is the oxidation state of N in

- N H_{2}

and why?

- 3

; N wins all its bonds (to 2 H and 1 C), each counting

- 1

By how much does nitrogen's oxidation number change

- N O_{2} \to - N H_{2}

, and why is the reagent count still

6 [H]

Nitrogen falls by 5 (

+ 2 \to - 3

); the count is

6 [H]

because balancing the whole group also carries off both oxygens as

2 H_{2} O

What does

[H]

mean and what is

6 [H]

as molecular hydrogen?

One hydrogen atom of reducing power (one electron + one proton);

6 [H] = 3 H_{2}

- N O_{2} \to - N H_{2}

oxidation or reduction?

Reduction — nitrogen gains electrons (goes from

+ 2

- 3