Unit 5.3 Organic chemistry III (reaction mechanisms and aromatic compounds)Needs Unit 2.2/4.5 nomenclature, isomerism, bond polarity, bond enthalpy, reagents and reaction conditionsStructure of benzene Explain using the different types of covalent bonds, and bond enthalpy, the structure and stability of the benzene ringShow benzene as the electrophilic substitution in aromatic systems Recall, reagents and reaction conditions, reaction of:benzene with a nitrating mixture; bromine and chloroalkanes and acid chlorides in the presence of anhydrous aluminium chloridearomatic compounds with carbon-containing side chains with alkaline potassium manganate(VII) solution, resulting in the oxidation of the side-chainsphenol with sodium hydroxide, bromine and ethanoyl chloridearomatic nitro-compounds with tin and conc HCl acid reducing them to aminesphenylamine with nitrous acid; and the subsequent reaction of benzenediazonium ions with phenol

to represent movement of an electron pair or single electron, recall the following reaction mechanisms with reagents and conditions, and an overall equationhomolytic, free radical substitution (alkanes with chlorine) homolytic, free radical addition (polymerisation of ethene)iii heterolytic, electrophilic addition (symmetrical and unsymmetrical alkenes with halogens and hydrogen halides)

explanations of the orientation of addition should be in the context of the relative stability of the intermediate carbocation.

Markovnikov’s Rule will not be examinediv heterolytic, electrophilic substitution showing generation of the electrophile, eg NO2

+ (benzene with nitrating mixture, bromine, chloroalkanes, acid chlorides)

the orientation of substitution in benzene derivatives will not be examinedv heterolytic, nucleophilic substitution (halogenoalkanes with hydroxide ions and cyanide ions) SN1 and SN2.

students are not expected to carry out reactions involving cyanidevi heterolytic, nucleophilic addition (carbonyl compounds with hydrogen cyanide).

Benzene C6H6

3 C-C single bonds and 3 C-C double bonds.  All C-C bonds are the same1 electron left over for each C atom.  These 6 electrons are delocalised across all C atoms

Compound Reagent Conditions Product Reaction typearene  benzene  C6H6

Nitrating mixturenitric acid  HNO3

sulphuric acid H2SO4

heat under reflux  below 60oC 

nitrobenzeneC6H5NO2 + H2O

arene  benzene  C6H6

BromineBr2

Catalyst (dry) Anhydrous AlCl3

halogenoarenebromobenzeneC6H5Br(l)   + HBr(g)

arene  benzene  C6H6

ChloroalkaneChloroethane C2H5Cl

Catalyst (dry) Anhydrous AlCl3

ethylbenzeneC6H5C2H5(l) + HCl(g)

arene  benzene  C6H6

Acid chloride Ethanoyl chloride CH3COCl

Catalyst (dry) Anhydrous AlCl3

KetonephenylethanoneC6H5COCH3(l) + HCl(g)

arene methylbenzene  C6H5CH3

Potassium manganate VII KMnO4

alkaline conditions 

heat under reflux

Carboxylic acidbenzoic acid  C6H5COOH + H2O

Aromatic nitro compounds

C6H5NO2 + 6H+ + 6e- C6 H5NH2 + 2H2O nitrobenzene                         aminobenzene

(phenylamine)

heated under reflux with tin in conc. HCl as reducing agent

Amines

PhenolC6H5OH

Sodium hydroxideNaOH

Sodium phenoxideC6H5O-Na+(aq) + H2O(l)

PhenolC6H5OH

BromineBr2

C6H2Br3OH(aq) + 3HBr(aq)2,4,6-tribromophenol (TCP)

substitution

PhenolC6H5OH

ethanoyl chloride  CH3COC

Dry esterCH3COOC6H5 + HClphenylethanoate

Phenylamine nitrous acid  5oC C6H5NH2 + HNO2 + HCl

C6H5NH2 HNO2 NaNO2 and dil HCl situ

2H2O  +  C6H5N2+Cl-

Diazonium ionDiazonium ionC6H5N2

+Cl-phenol  C6H5OH

5oC Yellow azo dyeC6H5N2C6H5OH

Homolytic, free radical substitution (alkanes with chlorine)The reaction is exothermic but energy in the form of UV light must be supplied to initiate the reaction.Chlorine absorbs the UV light, energy supplied is greater than the bond strength of the Chlorine molecule which then splits into Cl atoms. 

Homolytic fission – 2 electrons in the bond are given one by one to the two species, making 2 free radicalsFree radical - An atom or group of atoms which posses an unpaired electron

Free radical substitution has 3 steps

step 1: Initiation Cl---Cl ----> 2Cl. Each atom retains one electron from the

covalent bond between the atoms

step 2: Propagation H3C---H +    Cl.

-----> CH3. + HCl Each chlorine atom reacts with a molecule of

methane giving a methyl radical.

H3C. +     Cl---Cl ----> CH3Cl + Cl

. The methyl radical reacts with a

molecule of Chlorine to form chloromethane and a Cl atom

step 3: Termination H3C.

+        Cl.

---> CH3Cl Radicals combine

Homolytic, free radical addition (polymerisation of ethene)At high pressure and temperature low density polythene is made by free radical addition polymerisation.  This reaction can proceed in the gas phase, liquid phase or in solution.

Initiation An agent such as benzoyl peroxide splits to form radicals (C6H5COO)2 ----> 2C6H5COO. -----> 2C6H5

.  + 2CO2

Propagation C6H5

.  + H2C=CH2 ---> C6H5CH2CH2

.   C6H5CH2CH2

.  + H2C=CH2 --->  C6H5CH2CH2CH2CH2

. 

Termination Termination results in chains hundreds of monomer units long

C6H5(CH2CH2)nCH2CH2

. +  C6H5(CH2CH2)mCH2CH2

. ---> C6H5CH2CH2(CH2CH2)n+mCH2CH2H5C6

Heterolytic, electrophilic addition (alkenes)Heterolytic fission – 2 shared electrons in the bond are split unequally between the two atoms. One of the atoms keeps both electrons, giving ions. Electrophile – ‘electron seeking’ A species which attacks a carbon atom by accepting an electron pair. It is thus a Lewis acid Carbonium ions – Where the carbon atom bears the positive charge eg C2H5 +

Addition reaction Alkene provides 2 electrons for the new bond Br/Cl is electrophilic movement of a pair of electrons

CH2                                   CH2+ CH2

+                     CH2Br ||    + Brd+-Brd- --------->      |             +  Br - |          + Br ----------> |

CH2                                         CH2Br CH2Br                        CH2BrEthene Bromine 1,2-dibromoethane

Ethene reacts with hydrogen chloride to give chloroethane: This is also an addition reaction. The first step is the formation of two ions, the ethyl cation and the chloride anion which combine to form the product.

Heterolytic, electrophilic substitution (benzene)Electrophile takes 2 of the delocalised [pi] electrons on the benzene ring An unstable [pi] complex containing both an electrophile and a leaving group is formed as an intermediate. Nitration carried out under reflux at 55-60oC using a nitrating mixture (equal amounts of conc. nitric & sulphuric acid which react to generate the nitryl cation NO2+)

HNO3 + H2SO4 -------> NO2+ + H2O + HSO4

-

           step 1                ----->       +     H+     step 2

Aluminium chloride is used as a catalyst which helps form the electrophile                             

Br-Br  + AlCl3 ---> Br+---Br---AlCl3

- CH3-Br  + AlCl3 ---> +CH3---Br---AlCl3-

CH3COCl  +  AlCl3 ---> CH3C+=O + Cl-AlCl3-

Heterolytic, nucleophilic substitution (SN1 and SN2)Nucleophile - Attacks a carbon atom with a partial positive charge by donating an electron pair.               

SN1 mechanism: (CH3)3C-Br        ------> (CH3)3C+ + Br-                   (CH3)3C+  +   OH -

------> (CH3)3COH                 2-methyl-2-bromopropane nucleophile 2-methylpropan-2-ol

Rate = k[R-Br]     R=the alkyl group (CH3)3CBr + NaOH ------> (CH3)3COH + NaBr Unimolecular reaction. Since the nucleophile is not involved in the rate determining step, the mechanism must involve at least two steps.

     SN2 mechanism: HO - + CH3------Br --------> [ HO - - - CH3 - - - Br] - [ HO - - - CH3 - - - Br] -----------

> HO-----CH3 + Br -

Rate = k[CH3Br][OH-] CH3Br + OH - -----> CH3OH + NaBr Bimolecular reaction.  It is thought to proceed in a single step involving a transition state. The cyanide ion CN- can also take part in nucleophilic substitution 

Heterolytic, nucleophilic additionThis is a nucleophilic addition reaction with a two step mechanism CH3CHO + HCN ------> CH3CH2CH(OH)CN

ethanal                        2-hydroxypropanonitrile (a cyanohydrin)

               CH3                              CH3              CH3                                              CH3          |                          |                |                                 |     N=C-   C=O   ------> N=C-C-O-      N=C-C-O-    H+       ------>  N=C-C-O-H                  |                              |                |                                            |                 H                            H               H                                          H

2. (a) Equations for the hydrogenation of three compounds are given below, together with the corresponding enthalpy changes.

Explain, in terms of the bonding in benzene, why the enthalpy change of hydrogenation of benzene is not –360 kJ mol-1

Delocalisation / π-system due to overlap of six p-orbitals OR due to overlap of p-orbitals around the ring Confers stability/ benzene at a lower energy level / more energy needed to break bonds compared with having three separate π / double bonds / cyclohexatriene, Kekule structure

(b) Benzene can be converted into phenylamine, C6H5NH2 in two stages. Give the reagents needed for each step and identify the intermediate compound formed

1st step: sulphuric and nitric acid

concentrated Intermediate: Nitrobenzene /

C6H5NO2 2nd Step: Tin / iron and conc

HCl (followed by addition of alkali)

(c) Benzene, C6H6, reacts with bromoethane, CH3CH2Br, in the presence of a catalyst, to form ethylbenzene, C6H5CH2CH3, and HBr(i) Give the formula of a catalyst for this reaction.

AlBr3/FeBr3 / AlCl3 / Al2Cl6 / FeCl3 / Fe2Cl6

(ii) Give the mechanism for the reaction between benzene and bromoethane, including the formation of the species that reacts with the benzene molecule.AlBr3 + CH3CH2Br CH3CH2+ + AlBr4−

(iii) Name the type of mechanism involved in this reactionElectrophilic substitution

(d) A mixture of ethylbenzene (boiling point 136°C) and benzene (boiling point 80°C) can be separated by fractional distillation.A labelled boiling point/composition diagram for this mixture is shown below.

Use the diagram to explain what happens when a mixture containing 60% ethylbenzene and 40 % benzene is fractionally distilled

At least two horizontal and two vertical tie-lines drawn from 60% ethylbenzene Vapour condensed and then reboiled Vapour (from 60% ethylbenzene liquid) gets richer in the more volatile component (benzene) / residue gets richer in ethylbenzene Pure benzene distilled off / ethylbenzene left as residue

1. (a) Pent-1-ene, CH3CH2CH2CH=CH2, polymerises in a similar manner to ethene.(i) Draw enough of the chain of poly(pent-1-ene) to make the structure of the polymer clear.

(ii) Give the mechanism for the polymerisation of pent-1-ene, using a peroxide initiator RO–OR that produces RO. radicals. Show only the initiation and two propagation steps. Include the use of an appropriate type of arrow to show the movement of an electron.

Initiation Propagation step 1

Propagation step 2

(b) Pent-1-ene reacts with hydrogen bromide to give 2-bromopentane as the major product. (i) Give the mechanism for this reaction.

(ii) By considering the nature of the intermediates in this reaction, explain why the major product is 2-bromopentane rather than 1-bromopentane

Secondary cation is more stable than primary CONDITIONAL on reference to cations Structures of the 2 intermediate carbocations / intermediate cation giving 2-bromopentane is

secondary and primary for 1-bromopentane (c) Molecules of 2-bromopentane are chiral. If a single isomer of 2-bromopentane is reacted with hydroxide ions, the SN1 reaction that results gives pentan-2-ol, but the product mixture shows no optical activity. (i) How would you test for optical activity?

Sample in polarimeter / use of crossed polaroids / pass polarised light through sample

Rotates the plane of (polarisation of plane)-polarised (monochromatic) light(ii) Explain, in terms of the reaction mechanism, why the product mixture does not show optical activity.

intermediate (carbocation) planar

equal (probability of) attack from either side

(leads to) racemic/ 50:50 / equimolar mixture5. (a) Give the structural formula of the organic product when phenol is reacted with:(i) sodium hydroxide solution

C6H5O – Na+ / C6H5ONa/ C6H5O –

(ii) aqueous bromine

OR (iii) ethanoyl chloride.

(b) An azo dye can be made from benzenediazonium chloride.(i) State the reagents and conditions needed to make benzenediazonium chloride from phenylamine.

NaNO2 / sodium nitrite / nitrate(III)conc aq / dil HCl / hydrochloric acid, NOT HCI Any temperature between 0 – 10 oC OR range between 0-10 oC

(ii) Write an equation, using structural formulae, to show the reaction between benzenediazonium ions and phenol to give the azo dye.

(iii) What condition is required for the reaction in (ii) above?

Alkaline/alkali/sodium hydroxide/ NaOH /KOH/potassium hydroxide/ sodium carbonate/sodium hydrogencarbonate3. This question concerns the following reaction scheme starting from benzene, C6H6

(a) Explain why the low resolution n.m.r. spectrum of benzene contains only a single peak All hydrogen nuclei/atoms/protons in same (chemical) environment(b) (i) Identify the reagent and the catalyst needed for reaction 1. Reagent = ethanoyl chloride

Catalyst (anhydrous)

aluminium chloride / AlCl3/Al2Cl6(ii) What type of reaction is reaction 1? Electrophilic substitution(iii) Draw the mechanism for reaction 1.

OR

(c) (i) Identify the reagents needed for reaction 2. HCN + KCN

OR KCN + Acid

OR HCN + Base/alkali

OR HCN/KCN pH 5 - 9(ii) What type of reaction is reaction 2?

Nucleophilic addition(iii) Draw the mechanism for reaction 2.

OR

(d) The product of reaction 2 is a mixture of two isomers. (i) Draw the structures of these two isomers, making their three-dimensional shape clear.

(ii) What would be the effect of this mixture on monochromatic plane-polarised light? Give reasons for answer with reference to your mechanism in (c)(iii).

(No effect) as ketone planar Attack possible from top or bottom Producing racemic/50:50 mixture (of enantiomers) / rotations cancel out(e) How could infra-red spectroscopy be used to show that the product of reaction 2 did not contain traces of the reactant phenylethanone?

No absorption corresponding to C=O / carbonyl OR No absorption around 1700 cm-1

propenal

(a)Give the structural formula of the compound formed in the reaction when propenal reacts with 2,4-dinitrophenylhydrazine

(c) Propenal reacts with hydrogen cyanide as shown by the following equation CH2—CHCHO + HCN CH2—CHCH(OH)CN(i) Write the mechanism for the reaction.

OR

(ii) Name the type of mechanism involved in this reaction. Nucleophilic addition

(d) Propenal reacts with hydrogen bromide as shown by the following equation CH2—CHCHO + HBr CH3CHBrCHO(i) Write the mechanism for the reaction.

(ii) Name the type of mechanism involved in this reaction.Electrophilic addition

(e) C=O and C=C bonds have the same electronic structure but their reactions occur by different mechanisms. Explain why this is so.

• C = O is a polar bond OR O more electronegative than C

• C = C has high electron density OR C = C is electron rich

• Cδ+ can be attacked by a nucleophile OR (C in) C = O can be attacked by nucleophile OR C = C attacked by electrophile4. Phenylethanoic acid occurs naturally in honey as its ethyl ester: it is the main cause of the honey’s smell. The acid has the structure

Phenylethanoic acid can be synthesised from benzene as follows:

(a) State the reagent and catalyst needed for step 1.Reagent: chloromethane/CH3Cl

Catalyst: (anhydrous) aluminium chloride/AlCl3/Al2Cl6(b) (i) What type of reaction is step 2?

Free radical substitution

(ii) Suggest a mechanism for step 2. The initiation step, the two propagation steps and a termination step. You may use Ph to represent the phenyl group, C6H5.

- Cl2 2Cl• - PhCH3 + Cl• PhCH2• + HCl - PhCH2• + Cl2 PhCH2Cl + Cl• - 2PhCH2• PhCH2CH2Ph OR PhCH2• + Cl• PhCH2Cl OR 2Cl• Cl2

(iii) Draw an apparatus which would enable you to carry out step 2, in which chlorine is bubbled through boiling methylbenzene, safely. Do not show the uv light source.

- flask and vertical condenser – need not be shown as separate items [Ignore direction of water flow; penalise sealed condenser]

- gas entry into liquid in flask [allow tube to go through the side of the flask, but tube must not be blocked by flask wall]- heating from a electric heater/heating mantle/sandbath/water bath/oil bath

(c) (i) Give the structural formula of compound A.

(ii) Give the reagent and the conditions needed to convert compound A into phenylethanoic acid in step 4.HCl (aq) OR dilute H2SO4(aq) - Boil/heat (under reflux)/reflux OR - NaOH(aq) and boil - Acidify

(iii) Suggest how you would convert phenylethanoic acid into its ethyl ester.- ethanol and (conc) sulphuric acid - heat/warm/boil/reflux conditional on presence of ethanol

OR - PCl5 /PCl3/SOCl2 - Add ethanol PCl5 and ethanol (1) PCl5 in ethanol (0)

(d) (i) An isomer, X, of phenylethanoic acid has the molecular formula C8H8O2. This isomer has a mass spectrum with a large peak at m/e 105 and a molecular ion peak at m/e 136. The ring in X is monosubstituted. Suggest the formula of the ion at m/e 105 and hence the formula of X.

X is OR

(ii) Another isomer, Y, of phenylethanoic acid is boiled with alkaline potassium manganate(VII) solution and the mixture is then acidified. The substance produced is benzene-1,4-dicarboxylic acid:

Suggest with a reason the structure of Y.

Side-chain(s) oxidised to COOH

(e) Benzene-1,4-dicarboxylic acid can be converted into its acid chloride, the structural formula of which is

This will react with ethane-1,2-diol to give the polyester known as PET.(i) What reagent could be used to convert benzene-1,4-dicarboxylic acid into its acid chloride?

PCl5 /Phosphorus pentachloride/phosphorus(V) chlorideOR PCl3/ Phosphorus trichloride/phosphorus(III) chlorideOR SOCl2/Thionyl chloride/sulphur oxide dichloride

(ii) Give the structure of the repeating unit of PET.

(iii) Suggest, with a reason, a type of chemical substance which should not be stored in a bottle made of PET. (concentrated) acid/alkali (ester link) would be hydrolysed OR polymer would react to form the monomers/alcohol and acid