AQA AS Chemistry C14 Alkenes Kerboodle Answers

C14.1 Alkenes AQA AS Chemistry C14 Alkenes Kerboodle Answers: Page No. 219

1.Ans- Hex-2-ene 2.Ans- CH₃CH₂CH₂CH₂CH==CH₂ 3.Ans- 4.Ans- Electrophiles 5 –Ans Electron-rich

14.2 Reactions of alkenes AQA AS Chemistry C14 Alkenes Kerboodle Answers : Page No. 223

1.Ans- 2 Ans- Electrophilic additions 3.Ans- 1-bromopropane and 2-bromopropane b 2-bromopropane c.It is formed from the more stable of the two intermediate carbocations. 4.Ans Chloroethane 5.Ans- Decolourises bromine solution

14.3 Addition polymers AQA AS Chemistry C14 Alkenes Kerboodle Answers : Page No. 227

1.Ans-

1. B. and D

2.a b Vinyl chloride C Poly(chloroethene) 3.Ans-. CF₂=CF₂ 4.Ans- CH₂=CHCl

Practice questions: Page No. 228-229

1.Ans-  (a) Molecular formula: C₄H₈ Empirical formula: CH₂ (b) (i) Name of mechanism: electrophilic addition (ii) Structure: Explanation: major product formed via tertiary carbocation Or minor product formed via primary carbocation Primary carbocation less stable than tertiary carbocation (c) 2 (a) Copy and complete the mechanism below by drawing appropriate curly arrows. Curly arrow from lone pair on oxygen of hydroxide ion to H atom on C–H adjacent to C–Br Curly arrow from single bond of adjacent C–H to adjacent single bond C–C Curly arrow from C–Br bond to side of Br atom 3.Ans-  (a) (i) (ii) CH₃CH₂CH₂ 4.Ans- (a) (b) 5.Ans- (a) Position(al) (isomerism) (b) M1 must show an arrow from the double bond towards the H atom of the H–Br molecule M2 must show the breaking of the H–Br bond. M3 is for the structure of the secondary carbocation. M4 must show an arrow from the lone pair of electrons on the negatively charged bromide ion towards the positively charged carbon atom of either a primary or secondary carbocation. NB The arrows here are double-headed (c) M1 must show an arrow from the lone pair on oxygen of a negatively charged hydroxide ion to a correct H atom M2 must show an arrow from a C-H bond adjacent to the C-Br bond towards the appropriate C-C bond. Only award if an arrow is shown attacking the H atom of an adjacent C-H (in M1) M3 is independent provided it is from their original molecule. Award full marks for an E1 mechanism in which M2 is on the correct carbocation. NB The arrows here are double-headed 6.Ans- (a) (i)  (3 marks) M1 must show an arrow from the lone pair on oxygen of a negatively charged hydroxide ion to the correct H atom M2 must show an arrow from the correct C-H bond to the correct C-C bond. Only award if an arrow is shown attacking the H atom of the correct C-H bond in M1 M3 is independent but CE=0 if nucleophilic substitution N.B these are double- headed arrows (ii) (iii) M1 (Compounds / molecules with) the same structural formula M2 with atoms/bonds/groups arranged differently in space OR atoms/bonds/groups that have different spatialarrangements / different orientation.

1. b)

M1 must show an arrow from the double bond towards the H atom of the H – O bond OR HO 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 7.Ans-

1. a)

Electrophile: electron pair / lone pair acceptor or electron-deficient species Addition: reaction which increases number of substituents or converts a double bond to single bond (b) Mechanism: 8 (a) (i) But-1-ene (ii) Two H on one carbon of double bond (iii) CH₃CH=CHCH₃ (iv)

1. b) (i) Electrophilic addition

(ii) (iii) via more stable carbocation which is secondary 9.Ans-

1. a) (i) 3-bromo-3-methylpentane only

(ii) Electrophilic addition (reaction) (iii) M1 Displayed formula of 2-bromo-3-methylpentane (iv) Structure of (E)-3-methylpent-2-ene

10. Ans-

Stage 1 Consider the groups to the right hand carbon ofthe C=C bond. Consider the atomic number of the atoms attached. Carbon has a higher atomic number than hydrogen, so CH2CH2OH takes priority. Stage 2 Consider the groups joined to the left hand carbon of the C+C bond. Both groups contain carbon atoms, so continue to consider atoms further away. The propyl group has a higher priority than the ethyl group. Stage 3 The highest priority groups, propyl and CH2CH2OH, are on the same side of the C=C bond so the isomer is z. The rest of the name is 4-ethylhept-3-en-1-ol. Baneer 6

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