Aromatics Flashcards
What are arenes?
Hydrocarbons based on benzene
Describe the carbon to carbon bond lengths in benzene
Intermediate between those expected for a C-C single bond and a C=C double bond.
Describe the electron arrangement in benzene
Each carbon has three covalent bonds. The fourth electron of each carbon is in a p orbital and there are six of these electrons in benzene. The p orbitals overlap and the electrons in them are delocalised. They form a region of electron density above and below the ring of carbons. The delocalised system (pi system) makes benzene unusually stable.
Why is enthalpy change for hydrogénation evidence for stability of benzene?
Enthalpy change for cyclohexene is -120kj/mol. So expected enthalpy for three double bonds is -360kj/mol. Actual value is -208kj/mol so is 152kj/mol more stable than expected.
Problems with Kekule’s model of benzene
Hydrogenation enthalpy. Doesn’t react with bromine. His model would not be planar or symmetrical as double bonds are shorter than single bonds.
Why is benzene’s melting point much higher than hexane?
Benzene’s flat, hexagonal molecules pack together very well in the solid state so are harder separate which must happen for the solid to melt.
Solubility of benzene
Not with water but do mix with other non-polar solvents
What prefix is given to compounds where the benzene ring is not the priority?
Phenyl
Why do aromatic molecules favour electrophilic substitution mechanisms?
The delocalised system means there is a high electron density so is attacked by electrophiles. The aromatic ring is very stable so needs energy to be put in before the system can be destroyed which is called the delocalisation energy. So ring almost always stays intact in reactions. Substitution leaves the aromatic system unchanged.
Why do arenes burn with a smoky flame?
High C:H ratio so usually unburnt carbon remaining when they burn in air which produces soot.
Mechanism for electrophilic substitution (general)
Arrow from ring to electrophile. Broken circle with + ends before two adjacent carbons to on bonded to electrophile and H. Arrow from C-H bond to +. Forms normal aromatic ring with one C bonded to electrophile and separate H+.
Formation of electrophile for nitration
Concentrated sulfuric acid (1) reacts with concentrated nitric acid (1) to form H2NO3+ + HSO4-. Because sulfuric acid is stronger acid so donates a proton. H2NO3+ loses molecule of water (1) to give NO2+ (1) which is a nitronium ion. Overall equation:
H2SO4 + HNO3 -> NO2+ + HSO4- +H2O
Uses of nitrated arenes
First step in making aromatic amines to make industrial dyes. Important step in production of TNT.
Formation of electrophile in Friedel-Crafts acylation
RCOCl + AlCl3 -> RCO+ + AlCl4-
The aluminium chloride is a catalyst and is reformed by the ion reacting with a H+ from the benzene ring and HCl is formed.
RCO+ has double bond C=O
Conditions for Friedel-Crafts acylation
AlCl3 catalyst, heat gently
