Arenes Flashcards
Explain the concept of resonance stability. Note: include explanation of (1) hybridisation for the c atoms in benzene, (2) p orbital overlap and (3) delocalisation
Each carbon atom in benzene has 3 sigma bonds and 1 pi bond. All carbon atoms in benzene are sp2 hybridised. Each carbon atom in benzene has one unhybridised p orbital which lies perpendicular to the plane of the benzene ring.
The p orbitals of benzene are able to overlap sideways with eo, creating a continuous overlap of p orbitals throughout the ring. This enables pi electrons to delocalise across the ring, creating delocalised pi electron cloud which gives benzene resonance stability
(FOR FULL COMPREHENSIVE EXPLN LOOK AT CRUCIBLE NOTES)
Explain why benzene undergoes Esub instead of Eadd reactions (hint: resonance stability)
Esub reactions enable benzene to retain its aromaticity and preserve its resonance stability. E add reactions would disrupt the stable aromatic ring and result in a loss of resonance stability —> hence this is energetically unfavourable
Explain why concentrated H2SO4 catalyst is needed for Esub of benzene, but not for Eadd of alkenes?
Alkenes contain an electron rich C=C bond which ATTACKS Ephiles with partial positive charge (such as HX or X2).
However, due to resonance stability of benzene, Ephiles with partial positive charge are too weak to react with benzene.
Hence, concentrated H2SO4 catalyst is needed to protonate HNO3 to NO2+, which is a much stronger electrophile for reaction w benzene
What is the geometry of the carbocation intermediate formed in the first step of the Esub reaction?
Tetrahedral, non planar
What is the difference between the RNC for Electrophilic substitution with halogens and Friedel-crafts alkylation?
Esub:
X2(g/l), anhydrous AlX3 or FeX3 or Fe, rtp
Friedel Craft alkylation:
RX, anhydrous AlX3 or FeX3 or Fe, rtp
Describe what is inductive and resonance effect in your own words (need to hit some key points though)
Inductive effect is the withdrawal/donation of electrons via a SIGMA bond due to differences in electronegativity of the carbon atom in benzene and the substituent bonded directly to it.
Resonance effect is the withdrawal/donation of electrons via the overlap of p orbitals of the substituent and a carbon atom of benzene.
(Resonance ED effect): substituent bonded directly to benzene has a lone pair of electrons in its p orbital. Can delocalise into ring
(Resonance EW effect): ATOM bonded directly to benzene has no lone pairs of electrons in the p orbitals. However, it does contain multiple polar bonds with an atom/component that is more electronegative than itself, which can withdraw electron density from the ring via resonance
Why is resonance effect stronger than inductive effect for —OH and —NH3 substituents?
(Think abt what is the resonance and inductive effect for these substituents first)
resonance: electron donating
Inductive: electron withdrawing
Generally, resonance effect is stronger than inductive effect as electrons are delocalised through the PI electron cloud, while for inductive effect, electrons only move within one sigma bond
For —Cl/Br/I, why is inductive effect stronger than resonance effect?
Resonance effect: Donation
Inductive effect: withdrawal
Lone pair of electrons in Cl/Br/I is in the 3p orbital. 3p and 2p orbital overlap is less effective, thus resonance effect weaker. Hence, inductive effect (electron withdrawing) is stronger than the electron donating resonance effect
Explain why the electrophilic substitution of methylbenzene requires a
lower temperature than that of benzene.
The presence of the electron-donating alkyl group in methylbenzene increases electron
density of the benzene ring.
Hence, methylbenzene is more susceptible to Esub than benzene and is more
reactive.
Milder conditions are needed for the electrophilic substitution of methylbenzene. (e.g.
room temperature, 30 °C)
What is the RNC for nitration of methylbenzene and benzene? (Check this)
Benzene: conc HNO3, conc HCl, heat under reflux @ 55 degrees
Methylbenzene: conc HNO3, conc HCl, heat under reflux @ 30 deg
Why must nitration of benzene be carried out at a specific temp?
Higher temps would result in multisubstitution. Subsequent substitutions will require more vigorous conditions as NO2 is deactivating.
What are the observations for side chain oxidation of methyl/alkylbenzene? acidic vs alkaline?
Purple KMnO4 decolorises.
White ppt of benzoic acid formed
Alkaline:
purple KMnO4 decolorises
sodium benzoate is formed, which is soluble in water
brown ppt of MnO2 is formed
Can you use K2Cr2O7 for side chain oxidation?
NO
What happens if side chain oxidation of alkylbenzene is carried out in alkaline conditions instead? Ie, KMnO4, dilute NaOH, HUR
Sodium benzoate formed instead of benzoic acid. No white ppt observation, but KMnO4 will still decolourise. Additionally, brown ppt of MnO2 will be formed
