Created by yannycollins
almost 10 years ago
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Question | Answer |
Describe the delocalised model of benzene | - Cyclic hydrocarbon (C6H6) - The p-orbitals on the carbon atoms overlap sideways to produce a pi-cloud of electron density above and below the plane of the carbon atoms containing 6 electrons. - Delocalised pi-cloud spread over whole molecule - Therefore stable molecule |
Nitration of benzene: Type of reaction and essential conditions and reagents | - Electrophilic substitution - Nitrating mixture of conc. H2SO4 and HNO3 - 50 degrees C - Conc. sulphuric acid = catalyst |
Halogenation of benzene: Type of reaction and conditions and reagents | - Electrophilic substitution - Room temp. and pressure - Halogen carrier (e.g. FeCl3) |
Equation for formation of NO2+ ion (nitryl cation) for nitration of benzene | HNO3 + H2SO4 --> NO2+ + HSO4- + H2O |
Equation for formation of Br+ electrophile in halogenation of benzene - Equation for reformation of FeBr3 (catalyst) | Br2 + FeBr3 --> Br+ + FeBr4- H+ + FeBr4- --> FeBr3 + HBr |
Why does bromine react more readily with cyclohexene than with benzene? | - The electrons in the pi-bond of cyclohexene are localised between two carbon atoms and therefore have a region of high electron density. This can polarise the halogen. -Benzene has a delocalised pi-cloud spread over whole ring, which has a low electron density. Cannot polarise non-polar molecule so a halogen carrier is required to generate powerful electrophile. |
-Equation for phenol + sodium hydroxide -Name the salt produced salt | 6H5OH + NaOH --> C6H5O-Na+ + H2O Salt produced: Sodium Phenoxide |
- Equation for phenol + sodium - Name organic product | 2C6H5OH + 2Na --> 2C6H5O-Na+ + H2 -Sodium phenoxide |
Phenol + 3Br2 = | 2,4,6-tribromophenol + 3HBr |
Explain why bromine reacts more readily with phenol that with benzene | - In phenol: There is a lone pair of electrons (in p-orbital) on the oxygen atom in the phenol group, which is drawn into the benzene ring - This creates a higher electron density and activates ring -Increased electron density can polarise a bromine molecule |
Write the equation for the oxidation of ethanol and conditions | CH3CH2OH + [O] --> CH3CHO + H2O (forms an aldehyde- distilled immediately) CH3CH2OH + 2[O] --> CH3COOH + H2O (oxidation further to a carboxylic acid) - oxidising agent: H+/Cr2O7(2-) from potassium dichromate Orange to Green |
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