Questions on alkenes: structure, electrophilic addition and Markovnikov's rule

Question 1

Draw the mechanism for the reaction of 2-methylbut-1-ene, CH₂=C(CH₃)CH₂CH₃, with hydrogen bromide to form the major product, and explain why this product, rather than its isomer, is the major one.

Accepted Answer

Indicative content

H–Br drawn with δ+ on H and δ− on Br, and a curly arrow from the C=C double bond to the H atom; allow the dipole to be omitted if the arrow is correct
a curly arrow from the middle of the H–Br bond to the Br atom, showing heterolytic fission and the formation of the bromide ion
the tertiary carbocation (CH₃)₂C⁺CH₂CH₃, drawn with the positive charge on the carbon carrying the methyl and ethyl groups
a curly arrow from a lone pair on Br⁻ to the positive carbon, giving 2-bromo-2-methylbutane, (CH₃)₂CBrCH₂CH₃, as the major product
if the hydrogen bonds to the other carbon of the double bond, a primary carbocation forms instead, leading to 1-bromo-2-methylbutane as the minor product
the tertiary carbocation is more stable because three alkyl groups release electron density towards the positive carbon, so it forms faster by the route of lower activation energy

Levels

Level 3 (5–6 marks) every curly arrow, charge and intermediate of the mechanism is correct, the major product is drawn and named, the alternative primary carbocation is identified, and the preference is explained through carbocation stability from the inductive effect of alkyl groups
Level 2 (3–4 marks) the dipole, the first two arrows and the tertiary carbocation are right and the major product is given; the bromide arrow is absent or starts from the wrong place, or the two carbocations are not compared
Level 1 (1–2 marks) one correct arrow or the correct product, typically with the primary carbocation drawn as the intermediate or the positive charge on the CH₂ carbon, and no reason for the major product

Question 2

Outline the mechanism for the reaction of propene with hydrogen bromide to form 2-bromopropane.

Accepted Answer

B1 a curly arrow from the C=C double bond to the H atom of H–Br; allow the arrow to the δ+ H if dipoles are shown B1 a curly arrow from the H–Br bond to the Br atom, the bond breaking heterolytically to give Br⁻ B1 the structure of the secondary carbocation, CH₃C⁺HCH₃, with the positive charge on the middle carbon B1 a curly arrow from a lone pair on the bromide ion to the positively charged carbon; reject an arrow starting at the minus sign rather than a lone pair

Question 3

Outline the mechanism for the reaction of ethene with bromine to form 1,2-dibromoethane.

Accepted Answer

B1 the induced dipole on Br₂, δ+ on the bromine nearer the double bond, with a curly arrow from the C=C double bond to that bromine atom B1 a curly arrow from the Br–Br bond to the further bromine atom B1 the structure of the carbocation, CH₂BrC⁺H₂, with a bromine on one carbon and the positive charge on the other B1 a curly arrow from a lone pair on Br⁻ to the positively charged carbon

Question 4

Suggest the major organic product when hydrogen bromide adds to each of pent-1-ene, CH₂=CHCH₂CH₂CH₃, and 2-methylpent-2-ene, (CH₃)₂C=CHCH₂CH₃, naming the type of carbocation each product forms through.

Accepted Answer

B1 pent-1-ene gives 2-bromopentane, CH₃CHBrCH₂CH₂CH₃, as the major product B1 it forms through the secondary carbocation CH₃C⁺HCH₂CH₂CH₃, the hydrogen having bonded to the CH₂ end B1 2-methylpent-2-ene gives 2-bromo-2-methylpentane, (CH₃)₂CBrCH₂CH₂CH₃, as the major product B1 it forms through the tertiary carbocation (CH₃)₂C⁺CH₂CH₂CH₃, more stable than the secondary alternative; ignore the name of the minor product

Question 5

Describe a test to show that a liquid hydrocarbon is unsaturated, and explain why the main product of cyclohexene with bromine water is not 1,2-dibromocyclohexane.

Accepted Answer

B1 shake the liquid with bromine water at room temperature; accept bromine dissolved in an organic solvent B1 an unsaturated hydrocarbon turns the bromine water from orange to colourless; accept 'decolourised', reject 'clear' B1 a saturated hydrocarbon, with no C=C bond, leaves the bromine water orange B1 water molecules, in large excess, compete with bromide ions to bond to the carbocation, so the main product is 2-bromocyclohexan-1-ol

Question 6

Outline the mechanism for the reaction of ethene with concentrated sulfuric acid to form ethyl hydrogensulfate.

Accepted Answer

B1 a curly arrow from the C=C double bond to the H atom of an O–H group in H₂SO₄ B1 a curly arrow from that O–H bond to the O atom B1 the structure of the carbocation CH₃C⁺H₂, and the hydrogensulfate ion with its negative charge on an oxygen B1 a curly arrow from a lone pair on the negatively charged oxygen of HSO₄⁻ to the positively charged carbon

Question 7

Describe the bonding in the C=C double bond of ethene, and state the bond angle about each carbon atom.

Accepted Answer

B1 one σ bond, formed by the direct overlap of orbitals, with its electron density between the two carbon nuclei B1 one π bond, formed by the sideways overlap of a p orbital on each carbon, with its electron density above and below the plane of the molecule B1 each carbon has three regions of electron density, so the arrangement is trigonal planar with bond angles of about 120°

Question 8

Explain why alkenes react with electrophiles but alkanes do not.

Accepted Answer

B1 the π bond of an alkene is a region of high electron density, above and below the plane of the molecule B1 an electrophile is an electron pair acceptor, so it is attracted to that region, and the π bond is weaker than a σ bond so it breaks readily B1 alkanes have only σ bonds, with their electron density held between the nuclei, so there is no exposed electron-rich region to attract an electrophile

Question 9

Explain how a bromine molecule, which has no permanent dipole, is able to act as an electrophile towards ethene.

Accepted Answer

B1 as a Br₂ molecule approaches the double bond, the high electron density of the π bond repels the electron pair in the Br–Br bond B1 the electrons move towards the far bromine atom, so a dipole is induced, with the nearer bromine δ+ and the far bromine δ− B1 the δ+ bromine atom can then accept a pair of electrons from the π bond, so it acts as an electrophile

Question 10

Explain why 2-bromopropane, not 1-bromopropane, is the major product when propene reacts with hydrogen bromide.

Accepted Answer

B1 2-bromopropane forms via a secondary carbocation, CH₃C⁺HCH₃, and 1-bromopropane via a primary carbocation, CH₃CH₂C⁺H₂ B1 alkyl groups release electron density towards the positive carbon, and the secondary carbocation has two alkyl groups against one, so it is more stable; ignore any reference to the stability of the products B1 the more stable carbocation forms faster, by a route with a lower activation energy, so more product comes by that route

Question 11

Suggest why three different organic products form when ethene is bubbled into bromine water that also contains dissolved sodium chloride.

Accepted Answer

B1 the first step gives the same carbocation, CH₂BrC⁺H₂, whatever then attacks it B1 in the second step any species with a lone pair can bond to the positive carbon: a bromide ion, a chloride ion or a water molecule B1 the products are 1,2-dibromoethane, 1-bromo-2-chloroethane and 2-bromoethanol, the last after the added water molecule loses H⁺

Question 12

Calculate the mass of bromine that reacts with 2.10 g of propene. Mr of propene is 42.0 and Mr of Br₂ is 159.8.

Accepted Answer

M1 amount of propene = 2.10 ÷ 42.0 = 0.0500 mol M1 propene and bromine react in a 1 : 1 ratio, so 0.0500 mol of Br₂ reacts A1 mass of bromine = 0.0500 × 159.8 = 7.99 g, to three significant figures

Question 13

Calculate the number of C=C double bonds in one molecule of limonene, C₁₀H₁₆, from the titration result given in the table.

Accepted Answer

M1 amount of limonene = 0.272 ÷ 136.0 = 2.00 × 10⁻³ mol M1 mole ratio of bromine to limonene = 4.00 × 10⁻³ ÷ 2.00 × 10⁻³ = 2.00 A1 each C=C adds one Br₂, so each limonene molecule has 2 double bonds, a whole number

Question 14

Compare the organic products formed when hydrogen bromide reacts with but-1-ene and with but-2-ene.

Accepted Answer

B1 but-2-ene is symmetrical, so either carbon of the double bond gives the same secondary carbocation and the only product is 2-bromobutane B1 but-1-ene is unsymmetrical, so a secondary and a primary carbocation can both form, and there are two products B1 from but-1-ene the major product is 2-bromobutane, via the more stable secondary carbocation, and 1-bromobutane is the minor product

Question 15

State the catalyst, the temperature and the pressure used to make ethanol industrially from ethene and steam.

Accepted Answer

B1 a phosphoric acid catalyst, H₃PO₄, on a silica support B1 a temperature of about 300 °C B1 a pressure of about 60 to 70 atm

Question 16

Explain how hot, concentrated, acidified potassium manganate(VII) could be used to distinguish pent-1-ene from pent-2-ene.

Accepted Answer

B1 the hot concentrated reagent breaks the C=C bond, and each carbon of the double bond becomes part of a C=O group B1 pent-2-ene, CH₃CH=CHCH₂CH₃, gives two carboxylic acids, ethanoic acid and propanoic acid B1 pent-1-ene, CH₂=CHCH₂CH₂CH₃, gives butanoic acid, and its terminal CH₂ is oxidised to carbon dioxide, which turns limewater milky

Question 17

Describe the bonding in the C=C bond of ethene in terms of hybridised orbitals.

Accepted Answer

B1 each carbon mixes its 2s orbital with two 2p orbitals to form three sp² hybrid orbitals B1 the sp² orbitals lie in one plane at 120° and overlap end on to form the σ bonds, to the other carbon and to two hydrogens B1 the unhybridised 2p orbital on each carbon, at right angles to that plane, overlaps sideways with the other to form the π bond

Question 18

State the general formula of the alkenes, and state what is meant by the term unsaturated.

Accepted Answer

B1 the general formula is CₙH₂ₙ, for alkenes with one double bond and no ring B1 unsaturated means the molecule contains at least one carbon–carbon double bond, so more atoms can be added to it

Question 19

Explain why a tertiary carbocation is more stable than a primary carbocation.

Accepted Answer

B1 alkyl groups release electron density along their σ bonds towards the positive carbon, the inductive effect B1 a tertiary carbocation has three alkyl groups and a primary one has one, so the positive charge is spread more in the tertiary ion and it is more stable

Question 20

State the colour change, and name the organic product, when ethene is shaken with cold, dilute, acidified potassium manganate(VII).

Accepted Answer

B1 the solution changes from purple to colourless; accept 'decolourised' B1 the product is ethane-1,2-diol, HOCH₂CH₂OH, with an OH group on each carbon of the old double bond

Question 21

Suggest the structure of an alkene, C₅H₁₀, that gives propanone and ethanoic acid when it is heated with hot, concentrated, acidified potassium manganate(VII).

Accepted Answer

B1 propanone, CH₃COCH₃, comes from a carbon of the double bond carrying two CH₃ groups, and ethanoic acid from a carbon carrying one CH₃ group and one hydrogen B1 joining those two carbons by the double bond gives (CH₃)₂C=CHCH₃, 2-methylbut-2-ene

Quantity	Value
Mass of limonene	0.272 g
Mr of limonene	136.0
Bromine that reacts completely with the sample	4.00 × 10⁻³ mol