3.3.4 ALKENES
Define hydrolysis.
A reaction where the molecule is split by the addition of water
Describe the mechanism for the hydration of alkenes.
Alcohols can be produced industrially by the acid-catalysed hydration (name of reaction) of alkenes (mechanism =electrophilic addition) Reagent: conc H3PO4 and excess water (aq) (provides H+ and water) - H+ regenerated (acts as a catalyst), none of the acid is used up, only the water. A- s with electrophilic additions, un unsymmetrical alkene will produce two different products (major + minor due to variable stability of the carbocation intermediate (1/2/3)) - Not to be confused with alkene + conc H2SO4. Alcohol can be produced by this but then add water. However this is a DIRECT reaction.
Q: State the meaning of the term stereoisomers. (2 MARKS)
- (Compounds / molecules with) the same structural formula - with atoms/bonds/groups arranged differently in space - OR atoms/bonds/groups that have different spatial arrangements / different orientation.
Q: Explain why bromine, a non-polar molecule, is able to react with propene. (2 MARKS)
- (High) e- dense or e- rich C=C or e- rich π bond or 4 e- between the C's (NOT just 'C=C') - causes induced dipole in Br2
How do you draw addition polymers?
- Break the double bond (pi), extend bonds out, add square brackets. Dont put repeating unit n unless given in the question. - Many monomers (single units) for polymers (repeating units). - Put a poly in front of the name of the alkene.
Q: In the manufacture of margarine, unsaturated vegetable oils such as sunflower oil are hardened. Soft and hard margarines are obtained from the same vegetable oil. How does the structure and the melting point of a soft margarine differ from that of a hard one? (2 MARKS)
- Difference in structure: soft margarine less hydrogenated or has more C=C bonds or is more unsaturated than hard margarine - Difference in melting point: soft has lower melting point
Describe and draw the mechanism for electrophilic addition of an alkene + X2.
- Dipole is induced by highly negative pi (𝝿) bond - Xδ+ is the electrophile - Pair of electrons attracted from the pi (𝝿) bond - Heterolytic fission of X-X bond - Carbocation intermediate formed with positive charge - X- now a nucleophile - Two halogen atoms added to the molecule RATIO = X2 : C=C (1:1)
Q: Propene reacts with bromine by a mechanism known as electrophilic addition. Explain what is meant by the term electrophile and by the term addition. (2 MARKS)
- Electrophile: e- pair / lone pair acceptor or e- deficient species or e- seeking species - Addition: reaction which increases number of substituents or convert double bond to single bond or where two molecules form one molecule
EQ: The alkene 3-methylpent-2-ene (CH3CH=C(CH3)CH2CH3) reacts with hydrogen bromide to form a mixture of 3-bromo-3-methylpentane and 2-bromo-3-methylpentane. Name and outline the mechanism for the formation of 3-bromo-3-methylpentane from this reaction of 3-methylpent-2-ene (CH3CH=C(CH3)CH2CH3) with hydrogen bromide. Explain why more 3-bromo-3-methylpentane is formed in this reaction than 2-bromo-3-methylpentane. (7 MARKS)
- Electrophilic addition - M2 must show an arrow from the double bond towards the H atom of the H-Br molecule - M3 must show the breaking of the H-Br bond - M4 is for the structure of the tertiary carbocation - M5 must show an arrow from the lone pair of electrons on the negatively charged bromide ion towards the positively charged atom (of either a secondary or) of a tertiary carbocation - 3-bromo-3-methylpentane is formed from 3y carbocation - OR 2-bromo-3-methylpentane is formed from 2y carbocation M7 - 3y carbocation more stable than 2y
Q: Alcohols can be prepared from alkenes in various ways. On a laboratory scale, a mixture of propan-1-ol and propan-2-ol can be prepared from propene in two steps. - In step 1, propene reacts with cold, concentrated sulfuric acid to form intermediate Compounds. - In step 2, the intermediate compounds react with water to form the mixture of alcohols. 1. Name and outline the mechanism for the reaction between propene and concentrated sulfuric acid to form the intermediate compound which gives propan-2-ol in step 2. 2. Explain why propan-2-ol is the major product of this preparation. (7 MARKS)
- Electrophilic addition - Major product/propan-2-ol formed via most stable carbocation/carbonium ion secondary carbocation/carbonium ion more stable (than primary) or reverse argument
EQ: Propene reacts with concentrated sulfuric acid to form two isomers, E and F. The structure of E is shown. Name and outline the mechanism for the formation of E in this reaction. (5 MARKS)
- Electrophilic addition - M2 must show an arrow from the double bond towards the H atom of the H2SO4 molecule - M3 must show the breaking of the H-O bond in H2SO4 - M4 is for the structure of the correct carbocation - M5 must show an arrow from the lone pair of electrons on the negatively charged oxygen of HSO4 - towards the positively charged atom of their carbocation drawn
Epoxyethane can react with ammonia to produce a range of compounds called..... They have a range of uses including....
- Ethanolamines - Cosmetics, soaps, and detergents.
How do you identify a geometrical isomer?
- Focus on C=C - Summarise groups - Focus on the 'heavier' groups (priority) - Notice where they are in relation to each other - Fixed in position due to non-rotational C=C
What are geometrical isomers?
- Form of stereoisomerism - Geometric isomers have groups that occupy different relative positions in space Specific to alkenes as the C=C is non-rotational
Describe the C=C bond.
- Functional group - Area of high electron density (ie very negative (NOT ELECTRONEGATIVE)) - Open to electrophilic attack - Have one double bond and 2 regular bonds → trigonal planar
Describe the properties of alkenes.
- General formula: CnH2n (assuming one C=C bond_ - Unsaturated: contain C=C so tend to undergo addition reactions - Non-polar: no distinct dipole moment present - Water insoluble: as they are non-polar - Reactive: C=C is an area of high electron density. Open to electrophilic attack. - Varied melting/boiling point: increases with increased Mr as stronger I.D / VDW's IMF's → more energy needed to separate - Can show geometrical isomerism: C=C is non-rotational - Can get cyclical alkenes
Describe Maokovnikov's rule (asymmetrical alkenes)
- Given 'the choice' the hydrogen bonds to C with most hydrogens - Forming a carbocation - Carbocation more stable in major product due to electron inducing effect of carbons (alkyl group) on either side of it. - Electrons are drawn to it making it more stable. it is bonded to only 1 carbon the electron inducing effect will be less, less electrons drawn to it, so is less stable. - The greater the no. of alkyl chains attache to the C+, the more stable it becomes Rules: most to least stable 3° > 2° > 1°
EQ: But-1-ene reacts with a reagent of the form HY to form a saturated compound. 1. Suggest a reagent of the form HY which reacts with but-1-ene. (1 MARK) 2. Name and draw a mechanism for the reaction in part (1) (5 MARKS) 3. Explain how three isomeric products are formed when HY reacts with but-1-ene. (3 MARKS)
- HBr OR HCl OR H2SO4 - Electrophilic addition - The major product exists as a pair of enantiomers - The third isomer is 1-bromobutane (minor product) - Because it is obtained via primary carbocation
EQ: Propene reacts with concentrated sulfuric acid to form two isomers, E and F. The structure of E is shown. 1. Draw the structure F. (1 MARK) 2. Explain why more of isomer E than isomer F is formed in this reaction. (2 MARKS)
- Look at image (E top, F bottom) - idea that E is formed from/via more stable carbocation - idea that 2y carbocation is more stable than 1y carbocation
How do you find the monomer from a given polymer?
- Look for the simplest repeating unit - Ignore side groups above and below (bonds are single so are rotational and can be up or down) - Reverse the process, ie: remove the square brackets, unextend bonds, rebond the double bond (pi) - Find name of monomer and therefore deduce name of polymer (by adding poly infront)
Naming geometrical isomers: Cahn-Ingold Prelog system:
- Look up the atomic no.s of the atoms directly bonded to each carbon in the C=C - Highest atomic no. takes priority - If two atomic no.'s are the same on one carbon in the C=C, then we move to the next atom in the chain and compare - Where are the priority groups relative to each other? - Priority groups on ze zame zide of the C=C → Z isomerism - Priority groups on opposite sides of the C=C → E isomerism
Q: The following table gives the names and structures of some structural isomers with the molecular formula C5H10. The minor product in this reaction mixture is 2-bromo-3-methylbutane. Explain why this bromoalkane is formed as a minor product. (2 marks)
- M1 Reaction goes via intermediate carbocations / carbonium ions - Tertiary carbocation / carbonium ion is more stable (than the secondary carbocation / carbonium ion) OR Secondary carbocation / carbonium ion is less stable (than the tertiary
What are the reactions of epoxyethane?
- Reacts with water to form ethane-1,2-diol; used as antifreeze in cars and other vehicles. The mechanism of the reaction involves protonation of the oxygen followed by a nucleophilic attack of an OH- ion.
Q: The equation for the first reaction in the conversion of hex-3-ene into hexan-3-ol is shown below. CH3CH2CH=CHCH2CH3 + H2SO4 → CH3CH2CH2CH(OSO2OH)CH2CH3 Outline a mechanism for this reaction. (4 MARKS)
- M1 must show an arrow from the double bond towards the H atom of the H - O bond OR HỌ on a compound with molecular formula for H2SO4 M1 could be to an H+ ion and M2 an independent O- H bond break on a compound with molecular formula for H2SO4 - M2 must show the breaking of the O-H bond. - M3 is for the structure of the carbocation. - M4 must show an arrow from the lone pair of electrons on the correct oxygen of the negatively charged ion towards a correct (positively charged) carbon atom. - NB The arrows here are double-headed
Q: Consider the following reaction scheme. Explain why 1-bromopropane is only a minor product in Reaction 1. (3 MARKS)
- M1 the reaction to form 1-bromopropane goes via the primary carbocation OR 1o carbocation - M2 primary carbocations are less stable than secondary carbocations
EQ: Concentrated sulfuric acid reacts with alkenes, alcohols and sodium halides. 1. Outline the mechanism for the reaction of concentrated sulfuric acid with propene to show the formation of the major product. (4 MARKS) 2. Draw the structure of the minor product of the reaction between concentrated sulfuric acid and propene. (1 MARK) 3. Explain why the product shown in your answer to Question 2.2 is the major product. (2 MARKS)
- M1: must show an arrow from = of C=C towards the H atom of the HO bond or HO that is part of HOS... on a compound with molecular formula H2SO4 M1 could have arrow to H+ in which case M2 would be for an independent HO bond break on a compound with formula H2SO4 - M2: must use an arrow to show the breaking of the HO bond - M3: is for the correct carbocation structure - M4: must show an arrow from a lone pair of electrons on the correct oxygen of the negatively charged ion towards the positively charged carbon atom - NB: The arrows are double-headed - minor product = CH3CH2CH2OSO3H - (major) product formed via more stable carbocation OR secondary carbocation more stable (than primary) - Due to electron-releasing character / (positive) inductive effect of two alkyl / methyl groups (as opposed to one)
How are high density polyethene's formed?
- Made at temperatures and pressures little greater than room conditions, uses a Ziegler-Natta catalyst, - Polymer with less chain branching - Chains can pack together well. - Density of the plastic greater + melting temperature higher. - Uses: milk crates, buckets
How are low density polyethene's formed?
- Made by polymerising ethene at high pressure and high temperature via a free-radical mechanism. - Produces a polymer with a certain amount of chain branching. - Consequence of random nature of free-radical reactions. - Branched chains do not pack together well - Product is flexible, stretches well, low density. - Properties make it suitable for packaging (plastic bags), sheeting and insulation for electrical cables.
EQ: A student carried out an experiment to determine the number of C=C double bonds in a molecule of a cooking oil by measuring the volume of bromine water decolourised. The student followed these instructions: - Use a dropping pipette to add 5 drops of oil to 5.0 cm3 of inert organic solvent in a conical flask. - Use a funnel to fill a burette with bromine water. - Add bromine water from a burette to the solution in the conical flask and swirl the flask after each addition to measure the volume of bromine water that is decolourised. The student's results are shown in Table 2. In a trial experiment, the student failed to fill the burette correctly so that the gap between the tap and the tip of the burette still contained air. Suggest what effect this would have on the measured volume of bromine water in this trial. Explain your answer. (2 MARKS)
- Measured volume would be greater - Level in burette falls as tap is filled before any liquid is delivered
EQ: Pent-1-ene reacts with hydrogen bromide to produce 2-bromopentane as the major product. Explain why 2-bromopentane is the major product of this reaction. (2 MARKS)
- Ml 2-bromopentane is formed via the secondary (or 2°) carbocation - OR 1-bromopentane is formed via the primary (or 1°) carbocation - M2 a secondary carbocation is more stable than a primary carbocation award this mark only if the quality of language justifies the award.
What is the criteria for a geometrical isomer?
- Molecule must have C=C - Must have two different groups bonded to each carbon in the C=C
How can poly(chloroethene), commonly known as PVC, have it's properties be modified?
- PVC (plasticised) is flexible. - Esters make good plasticisers. They get in between the polymer strands, disruption the IMFs that hold them together and push them apart, molecular forces are weakened, allowing them to "slide" over each other, making it flexible.
Describe and draw the mechanism for electrophilic addition of an alkene + conc. H2SO4, and name and describe reaction that comes after it.
- Permanent dipole on Oδ--Hδ+ - Hδ+ is the electrophile - Pair of electrons attracted from the pi (𝝿) bond - Heterolytic fission of X-X bond - Carbocation intermediate formed with positive charge - OSO3H- now a nucleophile - Hydrogensulfate now added to molecule - Markovnikov's rule also applies here Product isn't useful so goes through secondary hydration/hydrolysis reaction after (+H2O) - Alcohol produced - H2SO4 regenerated (so acts as a catalyst)
What is a 1° carbocation?
- Primary - Where 2H's are bonded to the C (involved in the double bond)
Describe the test or alkenes.
- Qualitative (yes/no, not how much) Bromine water: Br2 (aq) → orange solution Test: Add Br2 (aq) to sample at room temperature (dropwise). This is an electrophilic addition reaction Observation: - Positive = orange to colourless (decolourises) - Negative = orange remains Ratio: 1 C=C : 1Br2
What is the affect of reaction conditions on forming addition polymers?
- Reaction conditions used in the production of polymer chains can be altered to give the plastics produced different properties. - High pressures and temperatures produce branched chain polymers with weak intermolecular forces. - Lower pressures and temperatures produce straight chain polymers with strong intermolecular forces.
Q: In the electrophilic addition of Isomer 4: (CH3)2C=CHCH3 with hydrogen bromide, 2-bromo-2-methylbutane if formed as the major product. The minor product in this reaction mixture is 2-bromo-3-methylbutane. Explain why this bromoalkane is formed as a minor product. (2 MARKS)
- Reaction goes via intermediate carbocations / carbonium ions - Tertiary carbocation / carbonium ion is more stable (than the secondary carbocation / carbonium ion) - OR Secondary carbocation / carbonium ion is less stable (than the tertiary carbocation / carbonium ion)
Q: The reaction scheme below shows the conversion of compound A, 2-methylbut-1-ene, into compound B and then into compound C. State the reagent and condition used in Step 2. Name compound C. (3 MARKS)
- Reagent: H2O or water OR steam, Or dilute sulphuric acid - Condition: heat, or warm, or boil or reflux [50-100°C] - Name of compound C: 2-methylbutan-2-ol
Q: In the manufacture of margarine, unsaturated vegetable oils such as sunflower oil are hardened. State the reagent and conditions used in this process. (3 MARKS)
- Reagent: Hydrogen of H2 - Conditions: Ni (catalyst) (Ignore Pt) - 100 - 200 °C or heat (1)
What is a 2° carbocation?
- Secondary - Where 1H is bonded to the C (involved in the double bond)
What is a 3° carbocation?
- Tertiary - Where 0 H's are bonded to the C (involved in the double bond)
Describe the structure of epoxyethane.
- Three-membered ring - All the bond angles in the ring are approximately 60°, while normal angle for C-C-O and C-O-C bonds is 109.5º. - Ring tends to spring open ('ring strain') and makes epoxyethane highly reactive and potentially hazardous. - So not normally stored - converted to other useful products
What are the uses of polymers and the disadvantages of using them?
- Unreactive hydrocarbon chains with multiple strong, non-polar covalent bonds: useful for manufacturing shopping bags (poly(ethene)). - The unreactive nature of the bonds means they are not biodegradable and cannot be broken down by species
EQ: A student carried out an experiment to determine the number of C=C double bonds in a molecule of a cooking oil by measuring the volume of bromine water decolourised. The student followed these instructions: - Use a dropping pipette to add 5 drops of oil to 5.0 cm3 of inert organic solvent in a conical flask. - Use a funnel to fill a burette with bromine water. - Add bromine water from a burette to the solution in the conical flask and swirl the flask after each addition to measure the volume of bromine water that is decolourised. The student's results are shown in Table 2. Outline how the student could improve this practical procedure to determine the number of C=C double bonds in a molecule of the oil so that more consistent results are obtained. (4 MARKS)
- Use a larger single volume of oil - Dissolve this oil in the organic solvent - Transfer to a conical flask and make up to 250 cm3 with more solvent - Titrate (25 cm3 ) samples from the flask
Describe the sigma bond within a C=C bond.
- Very strong sigma (σ) bond - Form between overlap of two S orbitals - The overlap between the two S orbital is directly between the two nuclei so the electrostatic attraction between the positive nuclei and the overlap of electrons is very strong - Bond doesn't get broken
Describe the pi (𝝿) bond within a C=C bond.
- Weaker pi (𝝿) bond - Form between overlap of two p orbitals - Weaker as they are above and below the two nuclei / the plane, so weaker attraction. - It is this pi (𝝿) bond that gets involved in reactions, not sigma. Presence of pi orbital means bond cannot rotate as forms orbital above and below the single bond. - Highly negative overlaps are open to electrophilic attack as it is an area of high electron density and is exposed. It is these pi (𝝿) bonds that break during addition reactions.
EQ: Chloroethene can be polymerised to form poly(chloroethene), commonly known as PVC. This polymer can be used to make pipes, window frames and electrical insulation. Plasticisers can be added to change the properties of PVC A section of poly(chloroethene) is shown. Use your understanding of the properties of PVC to explain whether you would expect to find a plasticiser in the PVC used to insulate electrical cables. (1 MARK)
- YES it is present - because PVC needs to be flexible / bendy
EQ: Oleic acid (C18H34O2) is a straight-chain fatty acid obtained from plant oils. Isooctane can be made from oleic acid. The skeletal formula of oleic acid is shown in Figure 3. Identify a reagent that could be used in a chemical test to show that oleic acid is unsaturated. State what would be observed in this test. (2 MARKS)
- bromine (water or in organic solvent or CCl4) / Br2 (aq) / Br2 - (orange/yellow to) colourless / decolourised / loses its colour
EQ: Chloroethene can be polymerised to form poly(chloroethene), commonly known as PVC. This polymer can be used to make pipes, window frames and electrical insulation. Plasticisers can be added to change the properties of PVC A section of poly(chloroethene) is shown. Chloroethene has a melting point of −154 ºC All types of PVC melt at temperatures over 100 ºC Explain why PVC melts at a higher temperature than chloroethene. (2 MARKS)
- it / PVC is bigger/longer molecule / has more electrons / has bigger surface area / greater Mr - it / PVC has stronger (van der Waals' / dipole-dipole) forces between molecules / intermolecular forces
What are the typical properties of poly(chloroethene), commonly known as PVC?
- uPVC (Unplasticised): very hard (but brittle) and sturdy / rigid; due to strong intermolecular bonding between polymer chains prevents them moving over each other - Covalent bonds between Cl & C atoms are polar, Cl more electronegative. - δ- charges on the chlorine atoms and δ+ charges on the carbon atoms - permanent dipole-dipole forces between the polymer chains - Waterproof, doesn't react with acids
Q: The following table gives the names and structures of some structural isomers with the molecular formula C5H10. Use Table A on the Data Sheet when answering this question. Isomer 3 and Isomer 4 have similar structures. 1. State the infrared absorption range that shows that Isomer 3 and Isomer 4 contain the same functional group. (1 MARK) 2. State one way that the infrared spectrum of Isomer 3 is different from the infrared spectrum of Isomer 4. (1 MARK)
1. (Both infrared spectra show an absorption in range) 1620 to 1680 (cm−1) 2. The fingerprint (region) / below 1500 cm−1 will be different or its fingerprinting will be different OR different absorptions / peaks are seen (in the reg
Q: The conversion of compound A into compound B can be achieved in two steps as shown below. The intermediate compound, X, has an absorption at 1650 cm-1 in its infra-red spectrum. 1. The intermediate compound, X, has an absorption at 1650 cm-1 in its infra-red spectrum. (2 MARKS) 2. For each step in this conversion, give the reagents and essential conditions required and outline a mechanism. (11 MARKS)
1. - Identity of X; 2-methylpropene - Absorption at 1650 cm-1 indicates an alkene present 2. Reagents - Step 1 KOH (allow NaOH) - alcoholic - warm - Step 2 HBr - MECHANISMS IN IMAGE ATTACHED
Q: Consider the following reaction scheme. Reaction 2 proceeds in two stages. Stage 1: CH3CH=CH2 + H2SO4 → CH3CH(OSO2OH)CH3 Stage 2: CH3CH(OSO2OH)CH3 + H2O → CH3CH(OH)CH3 + H2SO4 1. Outline a mechanism for Stage 1 of Reaction 2, using concentrated sulphuric acid. 2. State the overall role of the sulphuric acid in Reaction 2. (5 MARKS)
1. - M1 arrow from double bond to H of H - O bond - M2 arrow from bond to oxygen atom to show H - O bond breakage - M4 arrow from lone pair of electrons to carbon atom of carbocation 2. Catalyst ONLY
Q: Concentrated sulfuric acid is used in a two-stage process to convert 2-methylpropene into 2-methylpropan-2-ol. Stage 1 (CH3)2C=CH2 + H2SO4 → (CH3)2C(OSO2OH)CH3 Stage 2 (CH3)2C(OSO2OH)CH3 + H2O → (CH3)2C(OH)CH3 + H2SO4 1. Deduce the type of reaction in Stage 2 of this conversion. 2. State the overall role of sulfuric acid in this conversion. (2 MARKS)
1. Hydrolysis 2. Catalyst
What is the atom economy in addition polymers?
100%. All alkenes can be made into longer, saturated hydrocarbon chains.
Q: 0.50kg of non-2,5-diene was completely saturated in a hydrogenation reaction. What mass of H, was needed to achieve this?
16.13g
Q: How many of the C=C bonds in this molecule able to show geometrical isomerism?
2 The first two double bonds have 2 priority groups on each of the double bonds. No priority group on the last carbon from the third C=C
EQ: This structure shows a molecule that has been used as a plasticiser in PVC. Deduce the number of hydrogen atoms in this molecule. (1 MARK)
38
Q: Hydration of ethene is used in the alcoholic drinks industry. 300kg of ethene was completely hydrated. What mass of ethanol would be produced in this reaction?
493 kg
Describe and draw the mechanism for electrophilic addition of an alkene + HX.
Alkene → halogenoalkane - HX has permanent dipole as Br is more electronegative than H. - Hδ+ is the electrophile - Pair of electrons attracted from pi (𝝿) bond - Heterolytic fission of H-X bond - Carbocation formed - X- now a nucleophile - Single halogen atom added to the molecule Ratio = HX: C=C (1:1)
EQ: Dodecane (C12H26) is a hydrocarbon found in the naphtha fraction of crude oil. Dodecane can be used as a starting material to produce a wide variety of useful products. The scheme in Figure 2 shows how one such product, polymer Y, can be produced from dodecane. Name the homologous series that both C2H4 and C4H8 belong to. Draw a functional group isomer of C4H8 that does not belong to this homologous series. (2 MARKS)
Alkenes
What are addition polymers formed from?
Alkenes and substituted alkenes.
EQ: Thermal cracking of hydrocarbons produces molecules that are attacked by electrophiles because they have a region of high electron density. Draw the structure of one of these molecules that contains four carbon atoms. (1 MARK)
Any structural representation of alkene with 4C atoms, either: - But-1-ene - But-2-ene - Methylpropene
Q: Which statement about E-1,2-dichloroethene is correct? A It has the same boiling point as Z-1,2-dichloroethene. B It forms a polymer with the same repeating unit as Z-1,2-dichloroethene. C It has the same IR spectrum as Z-1,2-dichloroethene in the range 400-1500 cm−1 . D It has a molecular ion peak different from that of Z-1,2-dichloroethene in its mass spectrum. (1 MARK)
B;
Q: The alcohol, butan-2,3-diol was the major product formed by hydration of an alkene. Which alkene was used as the feedstock for this reaction?
But-1,3-diene
How is expoyethane made?
By partially oxidising ethene; mixture of ethene and oxygen passed over a catalyst of finely-divided silver: CH2=CH2 +½ O2→ CH2OCH2 A second reaction is possible which doesn't give epoxyethane CH2=CH2 + 3O2→ 2CO2 + 2H2O But with a suitable catalyst and control of conditions, 90% of the first reaction can be obtained.
What are plasticised PVC's used for?
Commonly used for electrical insulation (cables), waterproof clothing
What are the typical uses of poly(chloroethene), commonly known as PVC?
Commonly used for windows and doors and housing for electricals (TVs, computers etc).
Q: State what is meant by the term unsaturated as applied to an alkene. (1 MARK)
Contains a C=C bond
EQ: Concentrated sulfuric acid reacts with alkenes, alcohols and sodium halides. Name the mechanism for the reaction of concentrated sulfuric acid with an alkene. (1 MARK)
Electrophilic addition
Define asymmetrical alkenes.
Have different no. of hydrogen on each carbon atom in C=C
Describe what hydration is, and describe the reaction along with the conditions required.
Hydration (+H2O (g)) → H2O, conc H3PO4 catalyst, >100 C (steam), high pressure Alkene → alcohol Markovnikov's rule applies here no mechanism needed but know what the major and minor product is.
Describe what hydrogenation is, and describe the reaction along with the conditions required.
Hydrogenation (adding H2) → H2, Ni/Pt catalyst, 150C Alkene → alkane The more double bonds the more hydrogen RATIO: 1C=C : 1H2 Very important reaction in the food industry. Unsaturated vegetable oils are converted by this process to saturated ones, eg, margarine (makes it taste better)
Whaat is the CIS / Trans system for naming geometrical isomers?
If the two prioirity groups in a molecule are teh same eg CH3 then teh CIS / TRANS notation can be used. - Ie, zame zide = z = CIS - Opposite side = e = trans
What is epoxyethane?
Intermediate chemical from which useful products can be made.
Describe the overall affect of one sigma bond and one pi (𝝿) bond in C=C.
Look at image.
Draw the displayed formula of H2SO4.
Look at image.
EQ: A section of the polymer poly(chloroprene), a synthetic rubber, is shown. Draw the displayed formula for the repeating unit of poly(chloroprene). (1 MARK)
Look at image.
EQ: Compound J, known as leaf alcohol, has the structural formula CH3CH2CH=CHCH2CH2OH and is produced in small quantities by many green plants. The E isomer of J is responsible for the smell of freshly cut grass. Give the structure of the E isomer of J. (1 MARK)
Look at image.
EQ: Draw the displayed formula of C5H11Br that is the major product of the reaction of 2-methylbut-2-ene with hydrogen bromide. (1 MARK)
Look at image.
EQ: Draw the skeletal formula of 3-methylbutanal. (1 MARK)
Look at image.
What is the displayed formula of epoxyethane?
Look at image.
Q: Bromine reacts with alkenes, even though bromine is a non-polar molecule. Explain why bromine molecules react with the double bonds in alkenes. (2 MARKS)
M1 Double bonds are electron-rich OR electron pair donors OR centres of electron density. - M2 Bromine becomes polarised/becomes polar OR forms an induced dipole OR becomes δ+/δ-
What are strereoisomers?
Molecules that have the same molecular + structural formula, but the atoms occupy different relative positions in space / different spatial arrangements
What products are made during the electrophilic addition of asymmetrical alkenes?
One major and one minor product
Epoxyethane can react with alcohols, RO-H to form compounds such as...
RO-CH2-CH2-O-H
Q: Cyclohexene can be converted into amine W by the two-step synthesis shown below. Suggest an identity for compound V. For Reaction 3, give the reagent used and name the mechanism. For Reaction 4, give the reagent and condition used and name the mechanism. Equations and mechanisms with curly arrows are not required. (6 MARKS)
Reaction 3 - M2 HBr 1 - M3 Electrophilic addition - Reaction 4 - M4 Ammonia if wrong do not gain M5 - M5 Excess ammonia or sealed in a tube or under pressure - M6 Nucleophilic substitution
EQ: A student carried out an experiment to determine the number of C=C double bonds in a molecule of a cooking oil by measuring the volume of bromine water decolourised. The student followed these instructions: - Use a dropping pipette to add 5 drops of oil to 5.0 cm3 of inert organic solvent in a conical flask. - Use a funnel to fill a burette with bromine water. - Add bromine water from a burette to the solution in the conical flask and swirl the flask after each addition to measure the volume of bromine water that is decolourised. The student's results are shown in Table 2. The oil has a density of 0.92 g cm-3 and each of the 5 drops of oil has a volume of 5.0 × 10-2 cm 3 . The approximate Mr of the oil is 885 The concentration of bromine water used was 2.0 × 10-2 mol dm-3 . Use these data and the results from experiment 1 to deduce the number of C=C double bonds in a molecule of the oil. Show your working. (5 MARKS)
Stage 1 - Mass of oil = 0.92 × (5.0 × 10^-2 × 5) = 0.23 (g) - Mol of oil = 0.23 / 885 = 2.6 × 10^-4 Stage 2 - Mol bromine = 2.0 × 10^-2 × (39.4 / 1000) = 7.9 × 10^-4 Stage 3 - Ratio oil : bromine - 2.6 × 10-4 : 7.9 × 10-4 - Simplest ratio = (2.6 × 10-4 / 2.6 × 10-4) : (7.9 × 10-4 / 2.6 × 10-4) - = 1 : 3 Hence, 3 C=C bonds
EQ: Compounds A, B and C all have the molecular formula C5H10 A and B decolourise bromine water but C does not. B exists as two stereoisomers but A does not show stereoisomerism. Use this information to deduce a possible structure for each of compounds A, B and C and explain your deductions. State the meaning of the term stereoisomers and explain how they arise in compound B. (6 MARKS)
Stage 1 - deduces which compounds are saturated/unsaturated - states that A & B are unsaturated / do contain C=C / alkenes (this can be obtained from the structures) - as they decolourise bromine water - states that C is saturated / does not contain C=C / is (cyclo)alkane (this can be obtained from the structures) - as it does not decolourise bromine water Stage 2 - deduces the structures - suggests suitable name/structure for A x pent-1-ene, x 2-methylbut-1-ene, x 3-methylbut-1-ene, x 2-methylbut-2-ene - B = pent-2-ene (name/structure) - suggests a suitable name/structure of C (cyclopentane, methylcyclobutane, any dimethylcyclopropane) Stage 3 - can explain the stereoisomerism - explains what stereoisomerism is in terms of molecules with the same structural formula but a different arrangement of atoms/bonds/groups in space - explains how it arises by discussing that C=C cannot rotate, - explains how it arises by discussing that each C in C=C has two different groups (ignore reference to Mr in this context) or by drawing the E and Z isomers of B
EQ: Another structural isomer of J is shown below. Explain how the Cahn-Ingold-Prelog (CIP) priority rules can be used to deduce the full IUPAC name of this compound. (6 MARKS)
Stage 1: consider the groups joined to right hand carbon of the C=C bond - Consider the atomic number of the atoms attached - C has a higher atomic number than H, so CH2OH takes priority Stage 2: consider the groups joined to LH carbon of the C=C bond - Both groups contain C atoms, so consider atoms one bond further away - C, (H and H) from ethyl group has higher atomic number than H, (H and H) from methyl group, so ethyl takes priority Stage 3: conclusion - The highest priority groups, ethyl and CH2OH are on same side of the C=C bond so the isomer is Z - The rest of the IUPAC name is 3-methylpent-2-en-1-ol
Q: If on one of the carbons you had OH bonded and C2HCH2CH3 bonded. Would OH have priority because O has a bigger mr then C even though the overall mr do CH2CH2CH3 is bigger?
The O would take priority.
What are alkenes?
Unsaturated hydrocarbons
A Further reaction of epoxyethane after it has reacted with water to form an alcohol can produce a range of polymers of different chain lengths which have a range of applications. Name these.
n =1: production of polyesters such as PET, the plastic used for soft drinks bottles n = 2: used to make polyurethane plastics n = 3: used as a plasticiser to make plastics such as PVC more flexible n = 4+: (polyethylene glycols (PEGS) used in detergents, cosmetics, and paints
Q: The following diol would be the major product when hydrating which alkene?
pent-1,4-diene
What feature of the double bond prevents a z geometrical isomer from changing into a e geometrical isomer?
restricted rotation OR no rotation OR cannot rotate
Why are polymers unreactive?
they are saturated
