AQA AS Chemistry C14 Alkenes Kerboodle Answers

This page contains the AQA AS Chemistry C14 Alkenes Questions and kerboodle answers for revision and understanding Alkenes.This page also contains the link to the notes and video for the revision of this topic.

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
  1. 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

DISCLAIMER

Disclaimer: I have tried by level best to provide the answers and video explanations to the best of my knowledge. All the answers and notes are written by me and if there is any similarity in the content then it is purely coincidental. But this is not an alternative to the textbook. You should cover the specification or the textbook thoroughly. This is the quick revision to help you cover the gist of everything. In case you spot any errors then do let us know and we will rectify it. References: BBC Bitesize AQA GCSE Science Kerboodle textbook Wikipedia Wikimedia Commons Join Our Free Facebook Group : Get A* in GCSE and A LEVEL Science and Maths by Mahima Laroyia: https://www.facebook.com/groups/expertguidance.co.uk/ For Free Tips, advice and Maths and Science Help