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Organic Chemistry Mr Murphy
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Main Menu Menu Lessons 1-11: Lesson 1 – Homologous Series Lesson 1 – Homologous Series Lesson 2 – Isomers Lesson 2 – Isomers Lesson 3 – Meet the Families Lesson 3 – Meet the Families Lesson 4 – Alkanes Lesson 4 – Alkanes Lesson 5 – Alkenes Lesson 5 – Alkenes Lesson 6 – Alcohols Lesson 6 – Alcohols Lesson 7 – Halogenoalkanes Lesson 7 – Halogenoalkanes Lesson 8 – Reaction Pathways Lesson 8 – Reaction Pathways Lesson 9 – HL – Meet the Families (again) Lesson 9 – HL – Meet the Families (again) Lesson 10 – HL – S N 1 and S N 2 Revisited Lesson 10 – HL – S N 1 and S N 2 Revisited Lesson 11 – HL – More Nucleophiles Lesson 11 – HL – More Nucleophiles
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Main Menu Menu Lessons 12-18: Lesson 12 – HL – Elimination Reactions Lesson 12 – HL – Elimination Reactions Lesson 13 – HL – Condensation Reactions Lesson 13 – HL – Condensation Reactions Lesson 14 – HL – Condensation Polymerisation Lesson 14 – HL – Condensation Polymerisation Lesson 15 – HL – Geometric Isomerism Lesson 15 – HL – Geometric Isomerism Lesson 16 – HL – Optical Isomerism Lesson 16 – HL – Optical Isomerism Lesson 17 – HL – More Reaction Pathways Lesson 17 – HL – More Reaction Pathways Lessons 18-19 – Internal Assessment Lessons 18-19 – Internal Assessment Lesson 20 – Test Lesson 20 – Test Lesson 21 – Test Debrief Lesson 21 – Test Debrief
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Main Menu Lesson 1 Homologous Series
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Main Menu Overview Copy this onto a double-page spread. You should add to it as a regular review throughout the unit.
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Main Menu We Are Here
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Main Menu Homologous Series A homologous series is a family of compounds that differs only by the length of its hydrocarbon chain Members share: General formula Chemical properties Three such series are the: Alkanes Alkenes Alcohols
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Main Menu Homologous Series and Boiling Points What do you think will be the trend in melting/boiling points as you go down a homologous series? Why?
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Main Menu Formulas Draw the compound with the formula C 4 H 8 O
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Main Menu What did you get? Clearly a molecular formula is not enough!
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Main Menu Types of Formula Empirical Formula C 4 H 8 O C 4 H 8 O Molecular Formula C 4 H 8 O C 4 H 8 O Full Structural Formula Aka displayed formula Condensed Structural Formula Note the ‘=‘ used for the C=C double bond Skeletal formula Not required but v. useful Used in data booklet for complicated structures Do not use in exam answers! CH 2 =CHCH 2 CH 2 OH CH 2 =C(CH 3 )CH 2 OH
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Main Menu Thinking About Formulas Produce a table to summarise each of the formulas. Include columns for: What they show Pros Cons How you make them Draw full structural, condensed structural and skeletal formulas for at least 5 of the C 4 H 8 O compounds (not the cyclic ones)
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Main Menu Key Points Organic chemistry is the chemistry of carbon containing compounds A homologous series is a family of organic compounds differing only by the length of their carbon chains The melting and boiling point increases as you go down a homologous series Displayed formulas show the unambiguous arrangement of atoms in a compound
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Main Menu Lesson 2 Isomers
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Main Menu Refresh List two characteristics of a homologous series. Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 2: Isomers Objectives: Describe the term structural isomer Draw a name the non-cyclic alkanes Draw and name the straight-chain alkenes
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Main Menu Isomers Compounds with the same molecular formula but different structural formula The 20 different C 4 H 8 O compounds from last lesson are isomers of each other These are all structural isomers Same number of each atom, but bonded in a different order You would have even more if you included geometric and optical isomers
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Main Menu Structural Isomers of the Alkanes The (non-cyclic) alkanes have the general formula C n H 2n+2 Draw full and condensed structural formulas for every isomer of every one of the alkanes up to n = 6 If you finish early, draw each as a skeletal formula
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Main Menu Did you get them all?
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Main Menu And skeletally
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Main Menu Naming Straight-chain alkanes Suffix: Tells us the functional group of the molecule For alkanes it is ‘-ane’ Prefix: Tells us the length of the longest carbon chain: 1 carbon: meth- 2 carbons: eth- 3 carbons: prop- 4 carbons: but- 5 carbons: pent- 6 carbons: hex- Example 1: ethane Example 2: butane: Task: write in the names of the 4 straight chain alkanes next to your diagrams from last slide
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Main Menu Naming branched-chain alkanes Start by naming the longest chain Add extras to say the size of a branch, its position and how many of that branch Branch Size: 1 carbon: methyl- 2 carbons: ethyl- 3 carbons: propyl- Position: Number the carbons in the longest chain Choose numbers to minimise the total numbers used Number of same branches One branch – nothing Two branches – di- Three branches – tri- Four branches – tetra- Example 1: 2-methylpropane Example 2: 2,3-dimethylbutane Task: name the remaining alkanes
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Main Menu The straight-chain alkenes Alkenes are the same as alkanes but have one C=C double bond. The suffix for the alkene homologous series is ‘-ene’ Task: draw full structural and skeletal formulas for each of the straight-chain alkenes up to C6 and name them. Do the branched ones as well if you have time Hint: you need to state the position of the double bond, but only if there is the possibility of multiple isomers: i.e. ‘but-2-ene’ or ‘hex-1-ene’ but only ‘ethene’ not ‘eth-1-ene’
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Main Menu Did you get them?
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Main Menu Key Points Structural isomers have the same number of each atom but they are connected differently When naming compounds The longest carbon chain forms the prefix The functional group tells you the suffix Sometimes numbers need to be used to tell you where this functional group is Side chains and other groups are named according to what they are, how many there are and their position
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Main Menu Lesson 3 Meet the Families
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Main Menu We Are Here
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Main Menu Lesson 3: Meet the Families Objectives: Meet and learn to recognise the 7 functional groups required for the SL course Produce a mind-map summarising each of the homologous series
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Main Menu Functional Groups Table (landscape) You need to research and produce a mind- map summarising the following functional groups: Alkane Alkene Alcohol Aldehyde Ketone Carboxylic acid Halide/Halogenoalkane Your table should have four columns including: Name of functional group General structural formula (use ‘R’ to signify a carbon chain) Rules for naming them (including the position where relevant) A named example Relative volatility Relative solubility in water For alcohols and halides you should include a branch to explain the difference between 1 o, 2 o and 3 o You should also have a branch called ‘Other Functional Groups’ that just allows you to recognise the groups: Amine Ester Benzene If HL you should leave space for four more functional groups
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Main Menu Building Organic Compounds Use molecular models to make any of the compounds mentioned in your mind-map: Draw it (structural and skeletal) Name it Give it to a friend and challenge them to do the same Only go up to 6 carbons Only include branched-chains for the alkanes
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Main Menu Review You should know and be able to recognize the functional groups and their naming conventions
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Main Menu Key Points There are 7 functional groups we need to know in detail and 3 extra we need to be able to recognise We will look at each in detail over the rest of the unit
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Main Menu Lesson 4 Alkanes
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Main Menu Refresh The following is a computer-generated representation of the molecule, methyl 2- hydroxy benzoate, better known as oil of wintergreen. a) Deduce the empirical formula of methyl 2-hydroxy benzoate and draw the full structural formula, including any multiple bonds that may be present…The computer-generated representation shown does not distinguish between single and multiple bonds. b) Name all the functional groups present in the molecule. H H H H H H H H C C C C C C C C O O O Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 4: Alkanes Objectives: Explain the stability of the alkanes Observe the combustion of alkanes Describe the free-radical substitution reactions of alkanes and its mechanism Observe the free-radical substitution of hexane
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Main Menu Combustion of Alkanes (5 mins) The alkanes really don’t do much Combustion is of one of two notable reactions (this is why we use them for fuels) Complete combustion: alkane + oxygen carbon dioxide + water Incomplete combustion: Alkane + oxygen carbon + carbon monoxide + carbon dioxide + water The amounts of C, CO and CO 2 will vary depending on conditions Task: Observe the combustion of the gas from the gas taps (propane/butane mix) and of a small amount hexane (in spirit burners). Hold the end of a clean boiling tube just over the flame for 15 seconds, this will collect soot from the flame. Record all observations clearly and try to account for them Include balanced equations to describe the (complete) combustion
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Main Menu Why Good Fuels? To do: Use Table 12 in the data booklet to help you determine the trend in energy released per gram by combustion of the alkanes. Use bond enthalpies to help you explain the trend noted above. What do you think should be the characteristics of a good fuel? Use the above to decide and explain which out of methane and octane is a better fuel.
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Main Menu Why so boring stable? There are at least two reasons why alkanes are so unreactive Task: Think back to your knowledge of molecular structure, and look at the tables of bond-enthalpies in the data booklet to see if you can work out why.
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Main Menu Halogenation Alkanes will undergo halogenation if reacted with a halide in the presence of u.v. light. For example: C 2 H 6 (g) + Cl 2 (g) CH 3 CH 2 Cl(g) + HCl(g) ethane chloroethane This reaction is an example of free radical substitution u.v.
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Main Menu Radicals Radicals are species with unpaired electrons They are crazy reactive Halogens form radicals when hit by uv light of the right frequency: Cl 2 2 Cl The dot after the Cl represents the unpaired electron and tells us we have a radical This process is called homolytic fission – the bond breaks equally with one electron going to each chlorine Task: draw Lewis structures for the Cl 2 molecule and each of the Cl radicals u.v.
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Main Menu Reaction Mechanism: Free Radical Substitution Cl 2 2 Cl Cl + C 2 H 6 C 2 H 5 + HCl C 2 H 5 + Cl 2 C 2 H 5 Cl + Cl Cl + Cl Cl 2 Cl + C 2 H 5 C 2 H 5 Cl C 2 H 5 + C 2 H 5 C 4 H 10 Initiation Radicals formed by homolytic fission Propagation These steps feed each other the radicals needed to continue Termination Any two radicals can combine to terminate the reaction Concentration of radicals is low so this is a rare event A single radical can cause thousands of cycles of the propagation stage before it reaches termination This same mechanism applies to all of the halogens The alkane can be substituted multiple times, until every H has been replaced u.v.
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Main Menu Try it Yourself Place approx 1 cm 3 of hexane into two test tubes Add roughly 1 cm 3 of bromine water to each and stopper them, then give them a good shake. Leave one test-tube in the classroom but take the other outside and shake it in direct sunlight. Record and explain all observations Write equations showing the mechanism of the reaction Draw full structural and skeletal formulas of at least 6 possible products, and name each one.
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Main Menu Extension: Research the role of free radical reactions in the depletion of the ozone layer
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Main Menu Key Points Alkanes are pretty unreactive They release a lot of energy on combustion, and are easy to handle which makes them good fuels Undergo free radical substitution to form halogenoalkanes and a hydrogen halide in the presence of UV light
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Main Menu Lesson 5 Alkenes
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Main Menu Refresh Write an equation for the reaction between methane and chlorine to form chloromethane. Explain this reaction in terms of a free- radical mechanism. Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 5: Alkenes Objectives: Describe the main addition reactions of the alkenes Extract an alkene from a citrus fruit
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Main Menu Reactivity of Alkenes Alkenes are considerably more reactive than alkanes and are a major industrial feedstock The reactivity is due to the double bond: The double bond contains 4 electrons This is a significant amount of charge which: Makes it attractive to electrophiles Enables it to polarise approaching molecules Most reactions of alkenes are addition reactions where two molecules come together to make one new one
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Main Menu Alkenes and hydrogen Alkene + hydrogen alkane Reaction conditions: Hot Ni catalyst This is an addition reaction, in which the hydrogen adds across the double bond
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Main Menu Alkenes and hydrogen halides Alkene + hydrogen halide halogenoalkane Reaction conditions: This reaction occurs very readily and needs no special conditions This is an addition reaction, in which the hydrogen halide adds across the double bond
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Main Menu Alkenes and halogens Alkene + halogen dihalogenoalkane Reaction conditions: This reaction occurs very readily and needs no special conditions If the halogen used is an aqueous solution of bromine (bromine water), the orange-brown colour of bromine solution is decolourised. This is the standard test for alkenes.
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Main Menu Alkenes and water Alkene + water alcohol Reaction conditions: Water must be steam Phosphoric or sulphuric acid catalyst This is the process used to make industrial ethanol Fermentation from sugar would be far too expensive!
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Main Menu Polymerisation Under the right conditions, alkene molecules will add to each other creating a polymer In this case, 1-bromo-2-fluoroethene polymerises to form poly-1- bromo-2-fluroethene Conditions: Vary from alkene to alkene but often include high pressure, temperature and a catalyst The carbons in the C=C double bonds form the carbon chain, everything else hangs off this chain
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Main Menu Drawing polymers Draw three-monomer lengths of the polymers formed by: Propene Styrene Pent-2-ene
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Main Menu Homework Research the economic importance of alkenes including: Manufacture of margarine Manufacture of ethanol Polymerisation
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Main Menu Key Points Alkenes undergo addition reactions with: Hydrogen Hydrogen halides Halogens Water (steam) Alkenes undergo addition polymerisation Alkenes are very economically important due to the range of products they can make
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Main Menu Lesson 6 Alcohols
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Main Menu Refresh Which products can be potentially obtained from crude oil and are economically important? I. Plastics II. Margarine III. Motor fuel A. I and II only B. I and III only C. II and III only D. I, II and III Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 6: Alcohols Objectives: Explain the relative ease of combustion of the alcohols Describe the oxidation reactions of the alcohols Investigate the oxidation reactions of the alcohols
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Main Menu Alcohols as Fuels Alcohols combust more readily than equivalent alkanes but release less energy since they are already partially oxidised Alcohol + oxygen carbon dioxide + water Alcohols are used as fuels: As a fuel for cars – either pure or blended with petrol Methanol as fuel for competitive motorsports including dragsters and monster trucks Much fuel ethanol is fermented from crops…crops that could otherwise be eaten, forcing up food prices. Is this ok?
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Main Menu Oxidation of alcohols The most important reactions of the alcohols are their oxidations A range of compounds will oxidise them so the oxidiser is often represented as [O] One oxidising agent you need to know is potassium dichromate, K 2 Cr 2 O 7. When using this, orange Cr (VI) is reduced to green Cr (III) More on what this means in the oxidation and reduction unit See next slide for details
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Main Menu Oxidation reaction scheme
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Main Menu Key Points Alcohols are highly combustible Primary alcohols oxidise to form aldehydes, which oxidise to form carboxylic acids Secondary alcohols oxidise to form ketones Tertiary do not oxidise due to the 3 strong C-C bonds surrounding the –OH carbon
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Main Menu Lesson 7 Halogenoalkanes
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Main Menu Refresh a) Draw four structural isomers of molecular formula C 4 H 10 O which contain the –OH group. b) On reaction with acidified potassium dichromate(VII), two of the isomers are oxidized in two steps to produce different products. Draw the structural formula of the two products formed from one of the isomers. Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 7: Halogenoalkanes Objectives: Describe the substitution reactions of halogenoalkanes with a strong base Understand the S N 1 and S N 2 mechanisms for nucleophilic substitution Produce an animation showing the two different mechanisms
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Main Menu Nucleophilic Substitution One of the most important reactions undergone by halogenoalkanes is nucleophilic substitution A nucleophile is a ‘nucleus-loving’ species that is attracted to positive charges. Nucleophiles have either full negative charges or delta-negative charges Water and hydroxide are both nucleophiles In this case we can also call the reaction ‘hydrolysis’ The carbon in the carbon-halogen bond has a + charge due to the greater electronegativity of the halogen This makes it susceptible to attack by nucleophiles
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Main Menu Halogenoalkanes and strong bases A substitution reaction takes place, where the halogen atom is displaced by the hydroxide ion halogenoalkane + sodium hydroxide alcohol + sodium chloride Conditions: Aqueous base Gently warmed (can at room temperature, but may be quite slow) This is a nuclephilic substitution. The C attached to the halogen is + due to the high electronegativity of the halogen The OH - ion (our nucleophile) is attracted to the + carbon A nucleophile is a species with a negative charge or a lone pair that is attracted to positive/delta-positive atoms
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Main Menu S N 1 – Unimolecular nucleophilic substitution – animation hereanimation here Unimolecular because only one molecule is involved in the rate determining step The rate determining step involves the spontaneous breaking of the carbon-halogen bond and is a heterolytic fission, forming a halide ion and a carbocation intermediate The stability of the carbocation intermediate is a key factor in S N 1 The attack by the nucleophile (OH - ) is very fast, but does need the carbocation to be formed first The rate is only dependent on the concentration of the halogenoalkane: Rate = k[halogenoalkane] Note: the curly arrows show the movement of pairs of electrons
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Main Menu S N 2 – Bimolecular nucleophilic substitution – animation hereanimation here Bimolecular because two molecules are involved in the rate determining step In the rate determining step, the nucleophile (OH - ) attacks at the same time as the carbon-halogen bond breaks. The reaction passes through a negative transition state where the carbon has a ‘half-bond’ to both the –OH and the –Br with an overall negative charge The rate is dependent on both the concentration of the halogenoalkane and the nucleophile Rate = k[halogenoalkane][nucleophile]
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Main Menu S N 1 or S N 2? 1 o halogenoalkanes predominantly undergo S N 2 2 o halogenoalkanes undergo a mix of S N 1 and S N 2 3 o halogenoalkanes predominantly undergo S N 1 You do not need to know why at SL, but will find out more at HL
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Main Menu Understanding Check Write balanced equations, using condensed structural formulas, and name the products of the reactions of the following with dilute sodium hydroxide: Bromoethane 1-chloro-2-iodopropane 3-chloro-3-ethylhexane Name and draw the mechanism for the first and third one.
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Main Menu Refresh Halogenoalkanes undergo substitution with strong bases to form alcohols The reaction has two possible mechanisms: S N 1: the C-X bond breaks and then the nucleophile attacks S N 2: the nucleophile attacks at the same time as the C-X bond breaks The mechanism depends on the halogenoalkane: 1 o - S N 2 2 o - S N 1 and S N 2 3 o - S N 1
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Main Menu Refresh Predict the mechanism (S N 1 or S N 2) expected for the reaction of the following halogenoalkanes with aqueous KOH. 1-chlorobutane to form butan-1-ol 2-chloro-2-methylpropane to form 2 methylpropan-2-ol. Describe the mechanism for one of the above reactions using diagrams with curly arrows to show electron movements.
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Main Menu Lesson 8 Reaction Pathways
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Main Menu We Are Here
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Main Menu Key Points Inter-converting between organic compounds is the bread and butter of an organic chemist Reaction pathways are the map that help you navigate from your point of origin to your destination
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Main Menu Reaction Pathways Review all the reactions covered to date Produce a diagram summarising all of the reactions in one place including: Reactants and products for reactions (use general names) Reaction conditions It will look something like the diagram right Leave space for more if you are HL
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Main Menu Some Challenges Use your reaction pathways diagrams to help you solve the following problems Give the reactants and conditions to produce butan-2-one from a halogenoalkane in two steps.
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Main Menu Lesson 9 Meet the Families Part 2 – The Revenge of Father-in-Law
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Main Menu Refresh Propene is converted to propanone in a two stage process. Propene → X → Propanone What is the formula of compound X? A. CH 3 CHBrCH 3 B. CH 3 CH 2 CH 2 Br C. CH 3 CHOHCH 3 D. CH 3 CH 2 CH 2 OH Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 9: Meet the Families….again Objectives: Learn to recognise and name a further 4 functional groups
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Main Menu Functional Groups Mind map You need to be able to recognize these functional groups : Amide Amine Nitrile (R-C triple bonded to N) Ester – with the full detail now For each one you should include: General structural formula Rules for naming them (including the position where relevant) With an example Relative volatility and solubility in water For amines you should include a branch to explain the difference between 1 o, 2 o and 3 o
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Main Menu Lesson 10 S N 1 and S N 2 Revisited
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Main Menu Refresh What is the IUPAC name of the compound CH 3 CH 2 COOCH 2 CH 3 ? A. Ethyl ethanoate B. Propyl ethanoate C. Ethyl propanoate D. Pentyl propanoate Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 10: S N 1 and S N 2 Revisited Objectives: Understand why the nature of the halogenoalkane affects the mechanism of nucleophilic substitution reactions Understand why OH - is a better nucleophile than H 2 O Understand the effect of the halogen on the rate of nuclephilic substitution reactions
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Main Menu S N 1 and S N 2 Recap
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Main Menu S N 1 and the carbocation The carbocation is an unstable species, and will often immediately attract the halide ion straight back Alkyl groups surrounding the carbocation donate electron charge to it and stabilise it In this diagram, the arrows on the bonds represent the charge donated by the surrounding alkyl groups 3 o carbocations have most surrounding alkyl groups and therefore are most stable, thus S N 1 is preferred for 3 o halogenoalkanes
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Main Menu Mechanism and rate For S N 1:rate = k[halogenoalkane] For S N 2:rate = k[halogenoalkane][nucleophile] Since S N 1 only depends on one reactant, it tends to be faster (all else being equal) than S N 2 Therefore, if we consider the rates of hydrolysis / substitution, as a rule of thumb: Tertiary > Secondary > Primary
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Main Menu S N 2 and steric hindrance Alkyl groups are physically bulky, and make it difficult for a nucleophile to attack the carbon: this is called steric hindrance 1 o halogenoalkanes only have one surrounding alkyl group so steric hindrance is low and SN2 is favourable 3 o halogenoalkanes have three surrounding alkyl groups so steric hindrance is high and SN2 is unfavourable The black arrows on the diagram are supposed to show possible avenues of approach by the nucleophile, red crosses show where they are blocked
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Main Menu Changing the Nucleophile Water can act as our nucleophile: Halogenoalkane + water alcohol + hydrogen halide However, hydroxide is much better. Why? Explain why using ideas from the bonding unit
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Main Menu Changing the Halogen The rate of substitution / hydrolysis varies greatly depending on the halogen atom As a rule, with all else being equal, the rate changes as follows: Iodine > Bromine > Chlorine Explain why using ideas from the bonding unit
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Main Menu Key Points The substitution mechanism followed depends on: Stabilisation of the carbocation by surrounding alkyl groups Steric hindrance of the carbocation by surrounding alkyl groups Hydroxide ions are better nucleophiles than water due to their strong negative charge Iodoalkanes react faster than chloroalkanes due to the C-I bond being weaker than the C-Cl bond
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Main Menu Lesson 11 More Nucleophiles
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Main Menu Refresh Which statements about substitution reactions are correct? I. The reaction between sodium hydroxide and 1-chloropentane predominantly follows an S N 2 mechanism. II. The reaction between sodium hydroxide and 2-chloro-2-methylbutane predominantly follows an S N 2 mechanism. III. The reaction of sodium hydroxide with 1- chloropentane occurs at a slower rate than with 1-bromopentane. A. I and II only B. I and III only C. II and III only D. I, II and III Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 11: More Nucleophiles Objectives: Understand the reaction of halogenoalkanes with ammonia Understand the reaction of halogenoalkanes with potassium cyanide Describe and explain the reduction of nitriles
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Main Menu More nucleophiles So far we have met two nucleophiles: Hydroxide Water This lesson we meet two more: Ammonia, NH3 Cyanide, - CN Draw the Lewis structure of cyanide These undergo participate in nucleophilic substitution in exactly the same way as hydroxide and water.
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Main Menu Key Points Halogenoalkanes react with: Ammonia to form primary amines Further reactions can for secondary and tertiary amines and quaternary ammonium salts Potassium cyanide to form a nitrile This is useful as it creates a new C-C bond…no mean feat! The nitrile can be reduced to an amine by reaction with H 2 and Ni catalyst
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Main Menu Lesson 12 Elimination Reactions
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Main Menu Refresh What is the product of the following reaction? CH 3 CH 2 CH 2 CN + H 2 A. CH 3 CH 2 CH 2 NH 2 B. CH 3 CH 2 CH 2 CH 3 C. CH 3 CH 2 CH 2 CH 2 CH 3 D. CH 3 CH 2 CH 2 CH 2 NH 2 Ni. cat Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 12: Elimination Reactions Objectives: Describe with equations the elimination reactions of halogenoalkanes Describe the mechanism of elimination reactions Complete an experiment investigating elimination reactions
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Main Menu Complete the equation 1-bromobutane + sodium hydroxide
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Main Menu The Correct Answer What are the conditions?
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Main Menu Elimination With warm, aqueous sodium hydroxide, halogenoalkanes undergo substitution With hot, ethanolic sodium hydroxide an elimination reaction will take place: halogenoalkane + sodium hydroxide alkene + water + sodium halide
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Main Menu The mechanism Similar to substitution, there are unimolecular (E1) and bimolecular (E2) mechanisms: They work in a similar way to S N 1/2 and are affected by similar factors: E1: E2: You only need remember one (go with E2…shorter), and do not need to know the same level of detail as for substitution
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Main Menu Key Points Halogenoalkanes will undergo elimination on reaction with sodium hydroxide if: It is dissolved in ethanol Hot The elimination proceeds by both bimolecular and unimolecular mechanisms
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Main Menu Lesson 13 Condensation Reactions
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Main Menu Refresh But-1-ene can be prepared from 1- bromobutane. a) State the equation (using structural formulas) and the necessary reagents and conditions for the preparation of but-1-ene from 1-bromobutane. b) Identify the type of reaction and explain the mechanism for the preparation of but-1-ene from 1-bromobutane using curly arrows to represent the movement of electron pairs. Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 13: Condensation Reactions Objectives: Describe with equations the reaction of alcohols with carboxylic acids Describe with equations the reaction of amines with carboxylic acids Complete a lovely-smelling esters experiment
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Main Menu Condensation Reactions A reaction in which two molecules join together, and produce a molecule of water as a by-product. Nuff said
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Main Menu Esterification – the reaction of alcohols and carboxylic acids Carboxylic acids react with alcohols to make esters: The ester linkage is outline in red Carboxylic acid + alcohol ester + water You may need to review your mind-map for details on naming esters…basically the alcohol gives the ‘-yl’ part and the acid the ‘- oate’ part H + cat
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Main Menu Amides – the reaction of amines and carboxylic acids Carboxylic acids react with amines to make amides: The amide linkage is outlined in red Carboxylic acid + amine amide + water You may need to review your mind-map for details on naming esters The amine gives the ‘-yl’ part and the acid the ‘-anamide’ part, the N- signifies that the alkyl group is attached to the nitrogen Note the similarity to esterification
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Main Menu Draw and name the products of the following reactions: Ethanoic acid with ethanol Methanoic acid with butylamine Pentanoic acid with phenol (C 6 H 5 OH) Propanoic acid with methylamine
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Main Menu Key Points Carboxylic acid + alcohol ester + water Carboxylic acid + amine amide + water
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Main Menu Lesson 14 Condensation Polymerisation
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Main Menu Refresh What is the IUPAC name of CH 3 CH 2 CONH 2 ? A. Aminopropanal B. Ethanamide C. Propylamine D. Propanamide Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 14: Condensation Reactions Objectives: Describe with equations the condensation reactions of diacids with diols Describe with equations the condensation reactions of diacids with diamides Perform the nylon rope-trick
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Main Menu Polyesters Polyesters form from monomers containing two functional groups For example dicarboxylic acids and dialcohols Each end of each molecule forms an ester linkage, allowing for long chains to build up
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Main Menu Polyamides Similar to polyesters, polyamides form from monomers containing two functional groups For example dicarboxylic acids and diamines Each end of each molecule forms an amide linkage, allowing for long chains to build up
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Main Menu A quick note You can also get condensation polymerisation from monomers containing one of each of the functional groups. The best example is amino acids and proteins:
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Main Menu Key Points Polyesters form from monomers containing two acid and two alcohol groups Polyamides form from monomers containing two acid and two amine groups
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Main Menu Lesson 15 Geometrical Isomerism
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Main Menu Refresh Which formula represents a polyamide? A. ( CH 2 –CHCl ) n B. ( NH–(CH 2 ) 6 –NH–CO–(CH 3 ) 4 – CO ) n C. ( CF 2 –CF 2 ) n D. ( O–(CH 2 ) 2 –O–CO–--–CO ) n Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 15: Geometrical Isomerism Objectives: Understand the term stereoisomerism Understand and identify geometrical isomerism Understand the chemical significance of cis-trans isomerism
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Main Menu Molecular Modelling Make a molecule of but-2-ene Compare your model with your neighbours
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Main Menu Stereoisomerism Stereoisomers are compounds with the same structural formula but different 3D arrangement of atoms There are two types of stereoisomerism: Geometrical isomerism Optical isomerism
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Main Menu Geometrical (cis-trans isomerism) It happens because double bonds are not free to rotate Each C=C carbon must have two different groups attached to it. In the trans isomer, the substituents (the –CH 3 groups) are on opposite sides of the double bond In the cis isomer, the substituents are on the same side of the double bond Geometric isomerism involves the arrangement of groups around a double bond Or a single bond that can’t rotate freely such as in a cyclic compound
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Main Menu Properties Chemical and physical properties will often be similar but there can be important differences as you are about to see….
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Main Menu To do (individually, not pairs): Build, draw and label the cis and trans isomers of: Pent-2-ene 2,3-dichloro-but-2-ene 1,2-dichlorocyclopropane 1,2-dichlorocyclobutane Research the melting/boiling points of cis/trans but-2-ene, hex- 3-ene and oct-4-ene. What pattern do you notice between the cis and trans isomers? Why do you think this happens? Draw the cis and trans isomers of 1,2-dichloroethene. Predict, with a reason, which you think will have the higher boiling points. Research online to determine whether you were right and think of a reason if you were wrong.
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Main Menu Chemical properties of geometric isomers Can be different For example: Why do you think this happens? Hint: think about the previous couple of lessons
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Main Menu Research Research online the importance of geometrical isomerism Find three examples of commercial compounds (medicines, plastics etc) where geometrical isomerism plays a key role in their properties
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Main Menu Key Points Cis isomers: the substituents are on the same side Trans isomers: the substituents are on opposite sides Cis alkenes have lower mp/bp than trans due to ‘rounder’ shape Cis halogenoalkenes have higher mp/bp due to polarity Chemical properties can be different where cis brings groups close enough to react
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Main Menu Lesson 16 Optical Isomerism
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Main Menu Refresh The molecular formula, C 3 H 4 Cl 2 represents several isomeric compounds. Some isomers are cyclic and some are unsaturated. a) Draw the structures of two cyclic compounds that are structural isomers and state the names of both isomers. b) Two of the non-cyclic compounds have geometrical isomers. Draw the structures of these compounds and their geometrical isomers (4 structures needed). Reviewing Your Notes You should spend 60 seconds reviewing your notes from last lesson before attempting this. Your notes and mind-map must be ready for me to inspect.
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Main Menu We Are Here
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Main Menu Lesson 16: Optical Isomerism Objectives: Understand and identify optical isomerism Understand the chemical significance of optical isomerism
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Main Menu Modelling Build a model of 2-methylbutan-2-ol Compare your model with your neighbours Have you all made the same thing?
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Main Menu Optical Isomerism Optical isomerism occurs when you have at four different groups all bonded to a central atom. For example: amino acids Build these for yourself to see that they are different
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Main Menu Some terminology The central carbon is referred to as ‘chiral’ You will see books refer to the ‘chiral carbon’ or ‘chiral centre’ You may see books talking about the ‘chirality’ of a molecule A molecule might be described as ‘chiral’ if it has a ‘chiral centre’ A carbon must have 4 different groups to be chiral The two optical isomers are referred to as enantiomers A racemic mixture is a one with a 50:50 mix of the two enantiomers Enantiomers can be referred to as right or left handed If you see ‘R-’ or ‘L-’ in a name, this is what they are referring to There is a standard way to work this out but we don’t need it!
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Main Menu Properties of optical isomers Enantiomers share virtually identical chemical and physical properties However: Two enantiomers will rotate plane- polarised light in opposite directions…hence the term ‘optical isomerism’ More on the next slide Chemical properties differ significantly in chemical systems where 3D-shape is important, particularly biochemistry The image above right shows an enzyme, these are sensitive to the exact shape of a molecule
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Main Menu Rotation of plane-polarised light Light normally vibrates in many different directions Plane-polarised light only vibrates in one direction Pure solutions of enantiomers rotate the plane-polarised light in opposite directions A racemic mixture will not rotate light as the rotations from each enantiomer cancel each other out This rotation can be detected using a polarimeter
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Main Menu To do: Assess whether or not the following compounds display optical isomerism. If they do, draw each of the two enantiomers in 3D and place a * next to the chiral carbon. Build them from molecular models if it helps: 2-chloropropane 2-chlorobutane 2-chloropropanal chlorocyclopropane 1,2-dichlorocyclobutane Research the importance of optical isomerism in biochemistry. Give details of two important examples. Draw a Venn diagram summarising geometrical and optical isomerism
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Main Menu Key Points Optical isomers (enantiomers) rotate plane-polarised light in opposite directions To display optical activity, a compound must have at least four different groups attached to the same atom The central atom is referred to as chiral
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