Química Orgánica 2 · Sustitución electrofílica aromática
3-Nitrobenzoic acid from benzoic acid
Condiciones: HNO3, H2SO4, Δ
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Arrastra desde los electrones que se mueven hasta donde van, o toca uno y luego el otro. Cada flecha se revisa; Pista te ayuda cuando te atascas. No necesitas cuenta.
El mecanismo, paso a paso
Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
Leer los 8 pasos escritos
Protonate nitric acid
Sulfuric acid, the stronger acid, protonates the OH oxygen of nitric acid so that it can leave as water.
Las flechas curvas
- The OH oxygen of nitric acid uses a lone pair to take a proton from sulfuric acid.
- The O–H electrons stay on the sulfuric acid oxygen, giving hydrogen sulfate.
Lose water to form nitronium
Water leaves as the negatively charged oxygen forms a second N=O bond, giving the linear nitronium ion, NO₂⁺ — the electrophile.
Las flechas curvas
- The N–O bond to the protonated oxygen breaks, releasing water.
- A lone pair on the negative oxygen forms the new N=O π bond so nitrogen keeps its octet.
The ring attacks nitronium
The carbon meta to the carboxyl group attacks NO₂⁺. The carboxyl group withdraws electron density, so heating is needed.
Las flechas curvas
- The ring π pair at the carbon bearing the drawn hydrogen forms the new C–N bond; the neighboring carbon becomes cationic.
- One N=O π pair moves onto oxygen as the C–N bond forms, keeping eight electrons on nitrogen.
Delocalize the charge to the para carbon
The charge moves para to the new bond. None of the contributors puts positive charge on the carbon bearing the electron-poor carboxyl carbon.
Las flechas curvas
- The neighboring ring π pair shifts toward the cationic carbon, moving the positive charge to the carbon para to the new bond.
Return the charge beside the sp³ carbon
The arenium ion is one hybrid; redraw the contributor with the positive charge back on the carbon next to the sp³ C–H. That contributor shows which π bond the C–H electrons will re-form in the deprotonation.
Las flechas curvas
- The π pair shifts back toward the para carbon, leaving the positive charge on the ortho carbon next to the sp³ C–H.
Why meta?
A carbonyl directly on the ring makes the attached ring carbon electron poor.
Hydrogen sulfate removes the ring proton
Hydrogen sulfate removes the proton from the sp³ carbon that gained the nitro group; the C–H pair restores aromaticity and sulfuric acid is regenerated.
Las flechas curvas
- The negative oxygen of hydrogen sulfate takes the ring proton, regenerating sulfuric acid.
- The C–H bond electrons form the ring π bond to the neighboring cationic carbon, restoring aromaticity.
3-Nitrobenzoic acid, water and sulfuric acid
The nitro group enters meta to the carboxyl group, giving 3-nitrobenzoic acid; sulfuric acid is regenerated.
Qué vas a practicar
Draw nitration of benzoic acid at the meta position and explain why the carboxyl group directs meta.
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