Hydrocarbons: Question 5

Worked solution

Part (a): Generating the electrophile

Concentrated sulfuric acid protonates concentrated nitric acid, and a second mole of sulfuric acid then removes the resulting water molecule as $\text{H}_3\text{O}^+$, releasing the electrophile:

$\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$

Check (atom and charge balance): N: 1 = 1. O: $3 + 2(4) = 11$ on the left; $2 + 1 + 2(4) = 11$ on the right. H: $1 + 2(2) = 5$ on the left; $3 + 2(1) = 5$ on the right. S: $2 = 2$. Charge: $0$ on the left; $(+1) + (+1) + 2(-1) = 0$ on the right. All balance.

The electrophile generated is the nitronium ion, $\text{NO}_2^+$.

Part (b): The substitution mechanism

The delocalised $\pi$ electrons of the aromatic ring in ethylbenzene attack the electrophilic nitrogen atom of $\text{NO}_2^+$, forming a new $\text{C}-\text{N}$ bond. The carbon that bonds to the nitro group becomes $\text{sp}^3$-hybridised, so the ring’s aromatic delocalisation is broken at that point: this produces the arenium ion, a non-aromatic, positively charged intermediate in which the remaining positive charge is spread (delocalised) over the other ring carbons.

Because this intermediate has lost the extra stability of full aromatic delocalisation, it is short-lived. A base present in the mixture, such as the hydrogensulfate ion $\text{HSO}_4^-$, removes the proton attached to the same carbon that now bears the $\text{NO}_2$ group. This restores the ring’s full, delocalised aromatic system, giving the neutral nitro-substituted product and regenerating $\text{H}_2\text{SO}_4$:

$\text{arenium ion} + \text{HSO}_4^- \rightarrow \text{nitro product} + \text{H}_2\text{SO}_4$

Part (c): Explaining the directing effect and the rate difference

Directing effect. The ethyl group is an alkyl substituent, and alkyl groups release electron density towards the ring carbon they are attached to (an inductive, electron-donating effect, reinforced by hyperconjugation). When the incoming electrophile attacks the 2- or 4-position, the resulting arenium ion has one of its major contributing (resonance) structures carrying the positive charge on the ring carbon directly bonded to the ethyl group. Exactly where the extra electron density from the ethyl group can stabilise it most effectively. When the electrophile instead attacks the 3-position, none of the arenium ion’s contributing structures places the positive charge next to the ethyl-substituted carbon, so this intermediate receives no such extra stabilisation. Since attack at the 2- and 4-positions gives a more stable (lower-energy) arenium ion, these positions are strongly favoured, and only a small amount of the 3-nitro isomer forms via the less-stabilised pathway.

Rate effect. Because the ethyl group donates electron density into the ring as a whole, ethylbenzene’s aromatic ring is more electron-rich than benzene’s. This means the $\pi$ electrons are more readily available to attack the electrophile $\text{NO}_2^+$, so ethylbenzene undergoes electrophilic substitution faster than benzene does under the same conditions. The ethyl group is described as ring-activating as well as 2,4-directing.

Final answers

• (a) $\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$; electrophile $= \text{NO}_2^+$ (nitronium ion).
• (b) Ring attacks $\text{NO}_2^+$ forming a non-aromatic arenium ion; loss of $\text{H}^+$ (to e.g. $\text{HSO}_4^-$) restores aromaticity, giving the nitro product.
• (c) The ethyl group’s electron-donating (inductive) effect stabilises the arenium ion most when the charge sits next to the alkyl-substituted carbon (true only for 2-/4-attack, not 3-attack) so substitution is 2,4-directed; the same electron donation makes the whole ring more reactive, so ethylbenzene reacts faster than benzene.