Transcript
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 conditions Structure of benzene Explain using the different types of covalent bonds, and bond enthalpy, the structure and stability of the benzene ring Show 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 chloride aromatic compounds with carbon-containing side chains with alkaline po tassium manganate(VII) solution, resulting in the oxidation of the side-chains phenol with sodium hydroxide, bromine and ethanoyl chloride aromatic nitro-compounds with tin and conc HCl acid reducing them to amines phenylamine 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 equation homolytic, 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. not be examined Markovnikov’s Rule will will not + iv heterolytic, electrophilic substitution showing generation of the electrophile, eg NO 2 (benzene with nitrating mixture, bromine, chloroalkanes, acid chlorides) the orientation of substitution in benzene b enzene derivatives will not be examined v heterolytic, nucleophilic substitution (halogenoalkanes with hydroxide ions and cyanide ions) S N1 and SN2. students are not expected to carry out reactions involving cyanide vi 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 same same 1 electron left over for each C atom. atom. These 6 electrons are delocalised delocalised across all C atoms Compound arene benzene C6H6
Conditions heat under reflux o below 60 C
Product nitrobenzene C6H5NO2 + H2O
arene benzene C6H6
Reagent Nitrating mixture nitric acid HNO3 sulphuric acid H2SO4 Bromine Br2
Catalyst (dry) Anhydrous AlCl 3
arene benzene C6H6 arene benzene C6H6
Chloroalkane Chloroethane C2H5Cl Acid chloride Ethanoyl chloride CH3COCl
Catalyst (dry) Anhydrous AlCl 3 Catalyst (dry) Anhydrous AlCl 3
arene methylbenzene C6H5CH3 Aromatic nitro compounds
Potassium manganate VII KMnO4
alkaline conditions heat under reflux heated under reflux with tin in conc. HCl as reducing agent
halogenoarene bromobenzene C6H5Br(l) + HBr(g) ethylbenzene C6H5C2H5(l) + HCl(g) Ketone phenylethanone C6H5COCH3(l) + HCl(g) Carboxylic acid benzoic acid C6H5COOH + H2O Amines
+
-
Phenol C6H5OH Phenol C6H5OH
C6H5NO2 + 6H + 6e C6 H5NH2 + 2H2O nitrobenzene aminobenzene (phenylamine) Sodium hydroxide NaOH Bromine Br2
Phenol C6H5OH
ethanoyl chloride CH3COC
Dry
Phenylamine
nitrous acid
5C
o
Sodium phenoxide + C6H5O Na (aq) + H2O(l) C6H2Br3OH(aq) + 3HBr(aq) 2,4,6-tribromophenol (TCP) ester CH3COOC6H5 + HCl phenylethanoate C6H5NH2 + HNO2 + HCl
Reaction type
substitution
C6H5NH2
HNO2
Diazonium ion + C6H5N2 Cl
phenol C6H5OH
NaNO2 and dil HCl situ o 5C
+
-
2H2O + C6H5N2 Cl Diazonium ion Yellow azo dye C6H5N2C6H5OH
Homolytic, free radical substitution (alkanes with chlorine) The reaction is exothermic but energy in the form of UV li ght 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 – fission – 2 2 electrons in the bond are given one by one to the two species, making 2 free radicals Free 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 agent such as benzoyl benzoyl peroxide splits to form radicals
.
.
(C 6H5COO) 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 – fission – 2 2 shared electrons in the bond are split unequally between the two atoms. One of the atoms keeps both electrons, giving ions. Electrophile – Electrophile – ‘electron ‘electron seeking’ A species which attacks a carbon atom by accepting an electron pair. It is thus a Lewis acid + Carbonium ions – ions – Where Where the carbon atom bears the positive charge eg C 2H5
Addition reaction Alkene provides 2 electrons for the new bond movement of a pair of electrons CH2 ||
CH2 d+
d-
+ Br -Br --------->
CH2 Ethene Bromine
+
| CH2Br
Br/Cl is electrophilic
+
CH2 + Br
-
| CH2Br
CH2Br -
+ Br ---------> | CH2Br 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. o Nitration carried out under reflux at 55-60 C using a nitrating mixture (equal amounts of conc. nitric & sulphuric acid which react 2+ to generate the nitryl cation NO )
+
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 + CH3COCl + AlCl3 ---> CH3C =O + Cl-AlCl3
CH3-Br + AlCl3 ---> CH3---Br---AlCl3
Heterolytic, nucleophilic substitution (S N1 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 the rate determining step, the mechanism mechanism must involve at least two steps.
-
SN2 mechanism: HO + CH3------Br --------> [ HO - - - CH3 - - - Br] -> HO-----CH3 + Br -
-
-
[ HO - - - CH 3 - - - Br] ---------
-
Rate = k[CH3Br][OH ] CH3Br + OH -----> CH3OH + NaBr Bimolecular reaction. It is thought to proceed in a single step step involving a transition state. The cyanide cyanide ion CN can also take part in nucleophilic substitution Heterolytic, nucleophilic addition This is a nucleophilic addition reaction with a two step mechanism
CH3CHO + HCN ------> CH3CH2CH(OH)CN ethanal 2-hydroxypropanonitrile (a
cyanohydrin) CH3
CH3
| | N=C C=O ------> N=C-C-O | | H H
CH3
CH3 | -
N=C-C-O | H
+
H
| ------> N=C-C-O-H | H
2. (a) Equations for the hydrogenation of three compounds are given below, together with the corresponding enthalpy changes.
-1
Explain, in terms of the bonding in benzene, why the enthalpy change of hydrogenation of benzene is not – 360 360 kJ mol -system Delocalisation / π -system due to overlap of six p-orbitals due to overlap of p-orbitals around the ring OR 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 p henylamine, C 6H5NH2 in two stages. Give the reagents needed for each step and identify the intermediate compound formed st
1 step: sulphuric and nitric
concentrated Intermediate: Nitrobenzene /
2 Step: Tin / iron and conc
acid
C6H5NO2
nd
HCl (followed by addition of alkali) (c) Benzene, C6H6, reacts with bromoethane, CH 3CH2Br, in the presence of a catalyst, to form ethylbenzene, C 6H5CH2CH3, and HBr (i) Give the formula of a catalyst for this reaction. AlBr /FeBr / AlCl / Al Cl / FeCl / Fe Cl 3
3
3
2
6
3
2
6
(ii) Give the mechanism for the reaction between benzene and bromoethane, including the formation of the species that reacts with the benzene molecule. +
AlBr + CH CH Br CH CH + AlBr 3
3
2
3
2
− 4
(iii) Name the type of mechanism involved in this reaction Electrophilic 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.
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 r icher in ethylbenzene Pure benzene distilled off / ethylbenzene left as residue
Use the diagram to explain what happens when a mixture containing 60% ethylbenzene and 40 % benzene is fractionally distilled
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 – RO – OR 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 1bromopentane Secondary cation is more stable than pr imary 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 S N1 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 mixture 5. (a) Give the structural formula of the organic product when phenol is reacted with: (i) sodium hydroxide solution –
+
C H O Na / C H ONa/ C H O 6
5
6
5
6
–
5
(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. NaNO / sodium nitrite / nitrate(III) 2
conc aq / dil HCl / hydrochloric acid, NOT HCI o
o
Any temperature between 0 – 0 – 10 10 C OR range between 0-10 C (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 hydrogencarbonate
3. This question concerns the following reaction scheme starting from benzene, C 6H6
(a) Explain why the low resolution n.m.r. spectrum of benzene contains only a single peak nuclei/atoms/protons in same (chemical) environment (b) (i) Identify the reagent and the catalyst needed for reaction 1 . ethanoyl chloride
All hydrogen Reagent = Catalyst (anhydrous)
aluminium chloride / AlCl3/Al2Cl6 (ii) What type of reaction is reaction 1? 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 alkali
HCN/KCN pH 5 - 9 OR (ii) What type of reaction is reaction 2? Nucleophilic addition (iii) Draw the mechanism for reaction 2 .
OR
Electrophilic
(d) The product of reaction 2 is a mixture of two isomers. dimensional shape clear.
(i) Draw the structures of these two isomers, making their three-
(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 r eaction when propenal reacts with 2,4-dinitrophenylhydrazine (c) Propenal reacts with with hydrogen cyanide as shown by the following equation (i) Write the mechanism for the reaction.
CH2 — CHCHO CHCHO + HCN
CHCH(OH)CN CH2 — CHCH(OH)CN
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 CH 2 — CHCHO CHCHO + HBr (i) Write the mechanism for the reaction.
CH3CHBrCHO
(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 electrophile 4. 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/CH 3Cl
Catalyst: (anhydrous) aluminium chloride/AlCl 3/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•
OR 2Cl• Cl2 2PhCH2• PhCH2CH2Ph OR PhCH 2• + Cl• PhCH2Cl (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 – condenser – need need not be shown as separate ite ms [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 H 2SO4(aq) OR - NaOH(aq) and boil
-
Boil/heat (under reflux)/reflux 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 PCl5 /PCl3 /SOCl2 OR Add ethanol PCl5 and ethanol (1) PCl5 in ethanol (0) (d) (i) An isomer, X, of phenylethanoic acid has the molecular formula C 8H8O2. 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) pentachloride/phosphorus(V) chloride OR PCl3 / Phosphorus trichloride/phosphorus(III) trichloride/phosphorus(III) chloride OR 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
