Benzene

Lessons

Senior 6 · Chemistry

Benzene

4 lessons

Lesson 3 of 4

Electrophilic Substitution: Nitration and Halogenation

25 min

Why benzene substitutes instead of adding

Shake cyclohexene with bromine water and the orange colour vanishes in seconds: the alkene adds bromine across its double bond. Shake benzene with bromine water and nothing happens. Benzene reacts with bromine only when a catalyst is present, and then the product is bromobenzene, $\mathrm{C_6H_5Br}$ , not an addition product. A hydrogen atom has been replaced. The reason is the delocalisation energy of Lesson 1: addition would destroy the delocalised ring and lose about 152 kJ mol⁻¹ of stability, but substitution keeps the ring intact.

Definition

Electrophile: a species that accepts a pair of electrons to form a new covalent bond; it is usually positive or has a positive dipole. Examples: $\mathrm{NO_2^+}$ , $\mathrm{Br^+}$ , $\mathrm{CH_3CO^+}$ .

Electrophilic substitution: a reaction in which an electrophile replaces an atom (usually hydrogen) on an aromatic ring. Example: $\mathrm{C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2O}$ .

Arenium ion (σ-complex): the positively charged intermediate formed when an electrophile bonds to one ring carbon. That carbon is $s p^{3}$ , so only five carbons share the remaining four π electrons.

The two-step pattern

Attack. The electron-rich π system of the ring donates a pair of electrons to the electrophile $\mathrm{E^+}$ , forming a C–E bond. The ring becomes the arenium ion. This step breaks the delocalisation, so it has a high activation energy and is the slow step.
Loss of $\mathrm{H^+}$ . A base removes the hydrogen from the carbon that now carries E. The C–H bonding pair moves back into the ring, restoring the delocalised system.

Every electrophilic substitution in this unit follows this pattern; only the electrophile, and the way it is made, change. In curly-arrow mechanisms, an arrow always starts at an electron pair (a lone pair, a bond or the π system) and ends where the pair goes (an atom or a new bond).

Nitration

Reagents and conditions: a mixture of concentrated nitric acid and concentrated sulfuric acid (the "nitrating mixture"), warmed to no more than about 55 °C.

Making the electrophile. Sulfuric acid is the stronger acid. It protonates nitric acid, which then loses water to form the nitronium ion, $\mathrm{NO_2^+}$ :

$\mathrm{HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-}$

Overall reaction:

$\mathrm{C_6H_6(l) + HNO_3(l) \rightarrow C_6H_5NO_2(l) + H_2O(l)}$

Nitrobenzene is a pale yellow oil. Above about 55 °C a second nitro group goes on, giving 1,3-dinitrobenzene, so the temperature must be controlled. The sulfuric acid is a catalyst: the $\mathrm{H^+}$ removed in step 2 combines with $\mathrm{HSO_4^-}$ to re-form $\mathrm{H_2SO_4}$ .

Figure 3.1: Mechanism of nitration. The horseshoe shows that the intermediate has lost part of its delocalised ring.

Worked example

Problem. 15.6 g of benzene is nitrated and 18.5 g of pure nitrobenzene is obtained. Calculate the percentage yield. ( $M_{r}$ : benzene 78.11, nitrobenzene 123.11)

Solution. One mole of benzene gives one mole of nitrobenzene.

$n(\mathrm{C_6H_6}) = \frac{15.6}{78.11} = 0.1997\,\mathrm{mol} \qquad \text{theoretical mass} = 0.1997 \times 123.11 = 24.59\,\mathrm{g}$

$\text{percentage yield} = \frac{18.5}{24.59} \times 100 = 75.2\,\%$

Answer: 75.2%. Some benzene is lost by evaporation and some product by dinitration and during separation and washing.

Halogenation

Reagents and conditions: dry bromine (or chlorine) with a halogen carrier catalyst, iron(III) bromide (or aluminium chloride for chlorine), at room temperature, away from sunlight. Iron filings can be used, because they react with bromine to form the catalyst: $\mathrm{2Fe + 3Br_2 \rightarrow 2FeBr_3}$ .

Making the electrophile. The bromine molecule is not polar enough to attack the stable ring. Iron(III) bromide, a Lewis acid, accepts a lone pair from bromine and polarises it until it gives up a bromine cation:

$\mathrm{Br_2 + FeBr_3 \rightarrow Br^+ + FeBr_4^-}$

$\text{overall: } \mathrm{C_6H_6(l) + Br_2(l) \rightarrow C_6H_5Br(l) + HBr(g)}$

Steamy fumes of hydrogen bromide are given off. In step 2, $\mathrm{FeBr_4^-}$ takes the proton: $\mathrm{H^+ + FeBr_4^- \rightarrow FeBr_3 + HBr}$ , regenerating the catalyst.

Figure 3.2: Mechanism of bromination. The iron(III) bromide catalyst is regenerated in step 2.

Table 3.1: Bromine with an alkene and with benzene

	Cyclohexene	Benzene
Conditions	bromine water, room temperature, no catalyst	dry bromine, $\mathrm{FeBr_3}$ catalyst, room temperature
Type of reaction	electrophilic addition	electrophilic substitution
Product	1,2-dibromocyclohexane (or a bromo-alcohol in water)	bromobenzene + HBr
Why	a localised C=C bond is electron-rich enough to polarise $\mathrm{Br_2}$	the delocalised ring is less reactive; a catalyst must make $\mathrm{Br^+}$ ; substitution keeps the ring's stability

Worked example

Problem. Benzene reacts with chlorine in the presence of aluminium chloride. Write (a) the equation for the formation of the electrophile, (b) the overall equation, and (c) the equation that regenerates the catalyst.

Solution.

(a) $\mathrm{Cl_2 + AlCl_3 \rightarrow Cl^+ + AlCl_4^-}$

(b) $\mathrm{C_6H_6 + Cl_2 \rightarrow C_6H_5Cl + HCl}$

The product is chlorobenzene. The mechanism is the same as Figure 3.2 with Cl in place of Br.

Warning

Safety. The nitrating mixture is extremely corrosive and a strong oxidising agent; nitrobenzene is toxic by skin contact and inhalation. Bromine causes severe burns and its vapour is toxic. These reactions are carried out only in a fume cupboard, with gloves and goggles, and the acid mixture is always kept cool while it is made.

Tip

For every mechanism, write three things: (1) the equation that makes the electrophile; (2) the two-step mechanism with correct curly arrows and the horseshoe intermediate with its + charge; (3) the equation that regenerates the catalyst. Start each curly arrow on electrons, never on a positive charge.

Chemistry Student: Nitration of Benzene (6 min). Watch how the nitronium ion is formed, and follow each curly arrow in the two-step mechanism.

✦

Practice

Define an electrophile and explain why benzene reacts with electrophiles.
Explain, in terms of energy, why benzene undergoes substitution rather than addition.
State the reagents and conditions for the nitration of benzene. Why must the temperature be kept below about 55 °C?
Write the equation for the formation of the nitronium ion. Which substance acts as an acid and which as a base?
Draw the mechanism for the bromination of benzene with $\mathrm{FeBr_3}$ , showing all curly arrows and the intermediate.
Iron filings can be used instead of iron(III) bromide. Explain why.
7.80 g of benzene gives 11.0 g of bromobenzene ( $M_{r}$ = 157.0). Calculate the percentage yield.
A student draws the first curly arrow of nitration starting at the + charge on $\mathrm{NO_2^+}$ and ending on the ring. Explain the error.

Show answersHide answers

An electrophile accepts an electron pair to form a new covalent bond. Benzene has a region of high electron density (the delocalised π electrons above and below the ring), which attracts electrophiles.
Addition would destroy the delocalised π system and lose about 152 kJ mol⁻¹ of stability. In substitution the delocalised ring is restored in the second step, so the product keeps that stability.
Concentrated nitric acid and concentrated sulfuric acid, warmed to no more than about 55 °C. Above this, further nitration gives 1,3-dinitrobenzene.
$\mathrm{HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-}$ . Sulfuric acid is the acid (proton donor); nitric acid acts as a base (proton acceptor).
Step 1: curly arrow from the π electrons of the ring to $\mathrm{Br^+}$ , forming C–Br and the arenium ion (horseshoe over five carbons, + charge). Step 2: $\mathrm{FeBr_4^-}$ takes the $\mathrm{H^+}$ ; curly arrow from the C–H bond into the ring, restoring delocalisation. Products: $\mathrm{C_6H_5Br}$ , $\mathrm{FeBr_3}$ and HBr (as in Figure 3.2).
Iron reacts with bromine to form iron(III) bromide, $\mathrm{2Fe + 3Br_2 \rightarrow 2FeBr_3}$ , which is the real catalyst (halogen carrier).
$n(\mathrm{C_6H_6}) = \frac{7.80}{78.11} = 0.09986$ mol; theoretical mass = $0.09986 \times 157.0 = 15.68$ g; yield = $\frac{11.0}{15.68} \times 100 = 70.2\,\%$ .
Curly arrows show electron pairs moving. The electrons come from the ring's π system (the source) and go to the nitrogen of $\mathrm{NO_2^+}$ (the sink). A + charge has no electron pair to move, so the arrow must start at the ring and point to the N.

Sources and credits

Image: Sample of Nitrobenzene.jpg by LHcheM, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Nitration of Benzene (A-Level Chemistry), Chemistry Student, embedded from YouTube.
The text, mechanisms, table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Why benzene substitutes instead of adding

2The two-step pattern

3Nitration

4Halogenation

Why benzene substitutes instead of adding

The two-step pattern

Nitration

Halogenation