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AQA A2 Benzene

Why was Kekules Model replaced with the delocalised model for Benzene structure? Kekules model suggested that there was alternating c=c and c–c, hence you'd expected 3 bonds of each of the two lengths. However, x–rays showed that each of the six bonds were the same length with a length inbetween that of a c=c and a c–c.
Describe the delocalised model Each carbon donates one e-from its p orbital; all the p orbitals overlap leading to two rings of delocalised electrons one above and one below the plane of the carbon. This is a region of high electron density. Each of the bonds are the same length .
What effect does the ring of delocalised electrons have on the stability of benzene? Very stable, will not undergo addition reactions since this would destroy benzene ring and decrease stability. However it will undergo substitution and is prone to attack by electrophiles due the delocalised ring being a region of high electron density.
Give the reagents and conditions for nitration, the equation for the formation of the electrophile h2s04 + hn03 –––> +h2no3 + hs04–\n+h2no3––––> +no2 +h20 (no2=electrophile)\nreagents: conc h2s04, conc hno3 conditions: t< 50 (for adding on nitro)
Why would you nitrate Benzene? (2 reasons/uses) 1)Nitro compounds can be reduced to aromatic amines – used as dyes 2) Nitro compounds decompose violently when heated so can be used in explosives
Why would use friedel crafts acylation? Adds acyl group which can be further modified to produce useful products. Better yield than FC alkylation
Give reagents and conditions, as well as the formation of the electrophile for friedel–crafts acylation? ch3cocl + alcl3 –––> +ch3co + alcl4– alcl4– + H+ –––> hcl + alcl; reagents: acyl chloride conditions: heated under reflux, anhydrous, alcl3 catalyst
What is the name of the reaction between benzene and Br2 Electrophilic substitution
What is the salt formed when sodium and phenol react? Sodium phenoxide
Name two used of phenol Antiseptic, adhesives, Bakelite plastics, polymer production
A student obtained 6.80g phenylamine starting from 10.0g nitrobenzene. Calculate the percentage yield of phenylamine. Give your answer to 3sf. (4) Moles of nitrobenzene= 10/123=0.0813;Mole ratio is 1:1 so theoretical yield of phenylamine is 0.0813x93=7.5609g;% yield= 6.8/7.5609x100=90.6%
Explain why bromine acts more readily with phenol than with benzene. a lone pair of electrons on the oxygen atom in the phenol group is drawn into ring. This creates high e- density in the ring and ring is activated. increased e- density polarises Br2, when are then attracted more strongly towards Phenol than in benzene.
Name the two steps in the industrial preparation of dyes. diazotisation and coupling reaction.
State the conditions and reagents for the nitration of benzene. Conc. HNO3, conc. H2SO4, 50°C.
State 3 reasons why Kekule’s model of benzene did not work Carbon-carbon bond lengths; Hydrogenation of Benzene; Benzene’s low reactivity (reaction with bromine)
State features of the delocalised model of benzene It is a cyclic hydrocarbon with 6 carbon atoms and 6 hydrogen atoms; Hexagonal ring; 120 degrees, trigonal planar; Ring of electron density above and below the plane of carbon atoms
Give the equation of nitration of benzene and its conditions C6H6 + HNO3 -> C6H5NO2 + H2O ; H2SO4 and 50 degrees
State the 4 halogen carriers used in the halogenation of benzene FeCl3, FeBr3, AlCl3, AlBr3
Define electrophilic substitution It is a type of substitution reaction in which an electrophile is attracted to an electron-rich centre or atom where it accepts a pair of electrons to form a new covalent bond
Explain why benzene is unable to react with bromine without the help of a halogen carrier catalyst Benzene has delocalised Pi electrons spread all over the 6 carbons, alkenes only has above and below 2 carbons in the double bond. Benzene has a lower Pi electron density than alkenes.A halogen carrier is needed to generate a more powerful electrophile
What are the conditions/ catalyst for the reduction of nitrobenzene to make phenylamine? Mixture of tin and conc HCl
What is the first step in the industrial preparation of dyestuffs and the conditions required? Diazotisation; HNO2, HCl, below 10 degrees
What is the second step in the industrial preparation of dyestuffs and the conditions required? Coupling. NaOH, under alkaline conditions
How do you get from a diazonium salt to an azo dye? The dizonium salt reacts with aqueous NaOH at room temperature to couple with a coupling reagent (usually phenol) to form the azo dye and NaCl
State two use's of an azo dye. A dye and An indicator (methyl orange)
Explain the bonding in benzene. Benzene has a pi electron cloud above and below the ring. This pi electron cloud has 6 electrons. There is sigma bonding between the carbon atoms.
Write an equation to show how the nitronium is formed for the nitration of benzene H2SO4 + HNO3 -> HSO4- + H2NO3+ and H2NO3+ -> H20 + NO2+
What are the conditions required for the halogenation of benzene and why? Halogen carrier such as FeCl3 or AlBr3 required in order to generate the electrophile due to the low reactivity of benzene.
Explain the resistance to the bromination of benzene. The delocalized pi electron cloud is spread across the whole structure of benzene, making it more stable and giving it a lower electron density. The low electron density makes it less able to induce a dipole and attract the Br+ to the ring.
Show how nitrous acid (HNO2) is made "in situ" with an equation. NaNO2 + HCl --> HNO2 + NaCl
Describe what an electrophile is An electrophile is an atom or group of atoms that is attracted to an electron- rich centre, where it accepts a pair of electrons to form a new covalent bond.
State the conditions for the nitration of benzene reaction 50 degrees with concentrated nitric acid and concentrated sulphuric acid (catalyst).
Outline the mechanism of electrophilic substitution in arenes The electrophile accepts a pair of pi electrons from the delocalised ring to form a new covalent bond. One carbon is ring becomes C+; adjacent C-H bond breaks and the two electrons from C-H bond are returned to the delocalised ring.
Give the steps for making an azo dye starting with phenylamine 1)Phenylamine -> diazonium salt, using HNO2 and HCl below 10 C degrees to form a benzenediazonium chloride salt and water.2)diazonium salt is coupled with phenol and NaOH under alkaline conditions and below 10 degrees.
State the reagents and conditions for converting nitrobenzene into phenylamine. Sn/conc. HCl and reflux
State the reagents and conditions for converting phenylamine to the benzenediazonium salt. NaNO2/HCl <10 degrees.
Created by: UrsulineChem