AQA AS Chemistry C13 Halogen Alkanes Kerboodle Answers

This page contains the AQA AS Chemistry C4 Energetics Questions and kerboodle answers for revision and understanding Energetics.This page also contains the link to the notes and video for the revision of this topic.
 

C13.1 Introdction AQA AS Chemistry C13 Halogen Alkanes Kerboodle Answers: Page No. 207

1.Ans A 1-iodobutane C 1-chloroburane B 2-bromopropane D 2-bromobutane c .A, because ii has the highest Mr, and 1hercfore mos1 elccirons and highest van der Waals forces. 2 Ans.Because 1hc C-X bond becomes stronger.

13.2 Nucleophilic substitution in halogenoalkanes AQA AS Chemistry C13 Halogen Alkanes Kerboodle Answers: Page No. 210

1.Ans- R-X+OH–>ROH+X a.Because haloalkanes do not dissolve in aqueous solutions b. OH– c Because the OH group replaces the halogen atom d X– e R-1 2 CN– b c Propanenitrite.  

13.3 Elimination reactions in halogenoalkanes AQA AS Chemistry C13 Halogen Alkanes Kerboodle Answers: Page No. 213

1 A A base B An acid C Anudeophile D An electrophile A base 2 CCl2F2 3. Propan-2-ol, propene. b. Show that it decolourises a solution of bromine. c.

Practice questions: Page No. 214-215

1. M2 must show an arrow from the lone pair on the oxygen of a negatively charged hydroxide ion to a correct H atom M3 must show an arrow from a correct C–H bond adjacent to the C–Br bond to a correct C–C bond. Only award if an arrow is shown attacking the H atom of a correct adjacent C–H bond in M2 M4 is independent provided it is from their original molecule, BUT CE=0 for the mechanism (penalise M2, M3 and M4 only) if nucleophilic substitution mechanism is shown Award full marks for an E1 mechanism in which M4 is on the correct carbocation NB These are double-headed arrows (ii) (iii) Position(al) (isomerism or isomer) (b) Displayed formula for 3-methylbutan-2-ol (ii)
  • Lower / decreased temperature OR cold
  • Less concentrated (comparative) OR dilute KOH
  • Water (as a solvent) / (aqueous conditions)
(iii) Nucleophilic substitution 2 a) (i) Nucleophilic substitution M1 must show an arrow from the lone pair of electrons on the oxygen atom of the negatively charged hydroxide ion to the central C atom. M2 must show the movement of a pair of electrons from the C-Br bond to the Br atom. Mark M2 independently. Award full marks for an SN1 mechanism in which M1 is the attack of the hydroxide ion on the intermediate carbocation. (ii) 2-bromopropane only (iii) Polar C–Br OR polar carbon–bromine bond OR dipole on C–Br (iv) Moles of halogenoalkane = = 0.0814 Theoretical mass of organic product = 0.0814 × 60.0 = 4.88 g Percentage yield = 4.6 / 4.88 = 94.3% Student was correct (b) Elimination 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 C-C bond and should only be awarded if an attempt has been made at M1 M3 is independent. Award full marks for an E1 mechanism in which M2 is on the correct carbocation.
  1. c)
Any one condition from this list to favour elimination;
  • Alcohol(ic) / ethanol(ic) (solvent)
  • High concentration of KOH / alkali / hydroxide OR concentrated KOH / hydroxide
  • High temperature or hot or heat under reflux or T = 78 to 100#°C
3 b1-bromopropane; propan-1-ol (c) Br- d) Substitution
  1. e)
It has a lone pair of electrons and a negative charge {f)
  1. g)
Faster, the C-I bond is weaker and breaks more easily than the C-Br bond
  1. h)
They are poorer nucleophiles as they are neutral rather than negatively charged
  1. i) A proton (H+ ion) has to be lost.
Baneer 6

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