13 Halogenoalkanes ●● TEST YOURSELF ON PRIOR KNOWLEDGE 1 1 a) i) A saturated carbon compound is one which contains only single bonds between the atoms in their molecules. ii) A hydrocarbon is a molecule which contains only carbon and hydrogen atoms. iii) A homologous series is a series of compounds which have the same general formula and the same functional group. They show a gradation in physical properties and display similar chemical properties and, in organic chemistry, each successive member differs from the next by a –CH2group. 2 a) C7H16. b) There are nine isomers with the molecular formula C7H16 (If you have studied some optical isomerism, there are eleven isomers counting all the enantiomers ) All of the isomers are chain isomers. 3 C6H14 + 9.5O2 → 6CO2 + 7H2O ●● TEST YOURSELF 2 1 a) Cl b) H c) H Cl H C C Cl H H Br Br H C C C C H H H H H Cl F C C C H H H H H H 1 2 H H H H C C Cδ+ C H H Br H H H 3 The carbon-bromine bond is polar as there is a difference in electronegativity between the two atoms. The carbon-chlorine bond is more polar as the difference in electronegativity between the carbon and chlorine atoms is larger than the difference in electronegativity between the carbon and bromine atoms. 4 a) 1,3-dibromopropane b) 3-chloro-2-iodopentane c) 1,1dichloro-1-fluorobutane d) 1,3-dibromo-3-methylbutane 5 a) C2H5Br b) C2H5Br c) CH3CH2CH2CH2Br ●● TEST YOURSELF 3 1 a) b) free radical substitution i) Cl2 → Cl● + Cl● ii) Cl● + C2H6 → ●CH2CH3 + HCl 13 HALOGENOALKANES ●CH CH 2 3 + Cl2 → Cl● +CH2ClCH3 iii) ●CH2CH3 + Cl● → CH2ClCH3 ●CH CH 2 3 c) + ●CH2CH3 → CH3CH2CH2CH3 There are several different tri-substituted halogenoalkanes you can draw. Ensure you have either two or four carbons in the molecule each with three chlorine atoms. You can place a maximum of three chlorine atoms on the end carbons and two on the carbons in the chain. ●● TEST YOURSELF 4 1 a) 2 A mechanism is a detailed step-by-step sequence illustrating how an overall reaction occurs. b) A curly arrow shows the movement of a pair of electrons. It must begin at a lone pair of electrons or in the centre of a covalent bond and end at an atom or in the centre of a bond. c) A nucleophile is an electron pair donor. d) A base is a proton acceptor. 2 a) H H C C Cδ+ C H H Brδ– H – H H H H H H H C C C C H H OH H H + Br– OH b) reagent: sodium hydroxide reaction conditions: aqueous REQUIRED PRACTICAL H H dissolve halogenoalkane in the minimum volume of ethanol reflux gently c) nucleophilic substitution d) CH3CH2CHBrCH3 + NaOH → CH3CH2CH(OH)CH3 + NaBr 3 a) The halogenoalkane is dissolved in a minimum volume of ethanol. The halogenoalkane-ethanol solution will mix with an aqueous solution. b) H H H H C C Cδ+ C H H Brδ– H – c) H C H H H H H H C C C C H H C H N H + Br– N 2-methylbutanenitrile d) CH3CHBrCH2CH3 + KCN → CH3CH2CHCNCH2CH3 + KBr 4 a) 2NH3 + CH3CHClCH3 → CH3CH(CH3)NH2 + NH4Cl b) 1-methylethylamine c) The second molecule of ammonia accepts a proton from the positively charged reactive intermediate. d) Excess ammonia in a sealed tube under pressure. e) The organic product is also a nucleophile and can react with the 2-chloropropane producing further substituted amines. ●● REQUIRED PRACTICAL Preparation of 1-bromobutane 1 C4H9OH + HBr → C4H9Br + H2O 2 a) H2SO4 + NaBr → NaHSO4 + HBr b) Sulfuric acid is reduced by the bromide ion 3 The reaction is exothermic, so this controls the reaction and prevents violent spitting. 4 Anti-bumping granules 5 Add a still head, add a thermometer, move condenser from upright to sideways position, put receiver on end of condenser. 3 6 a) Shake in a separating funnel, invert and release the pressure, remove the stopper and run off the layer. b) Sulfuric acid 7 water 8 Moles of butanol = 21.8/74.0 = 0.29 Moles of sodium bromide = 40/102.9 (excess) = 0.39 Moles 1-bromobutane = 0.29 = mass/136.9; mass = 40 g % yield = actual yield/theoretical yield = (3.55/40) × 100 = 8.9% 9 Practical – loss during transfer, loss during distillation. Theoretical – side reactions produce a different product, reaction does not go to completion ●● ACTIVITY Preparation of methylpropene gas 1 H 13 HALOGENOALKANES H 4 H H C H C C Cl H H C H H 2 It cannot form hydrogen bonds with water 3 a) C4H9Cl + KOH → C4H8 + KCl + H2O b) elimination 4 The volume of gas in the apparatus is less as it cools down, and water is sucked back due to pressure differences. Remove the delivery tube from the water after heating, or disconnect the delivery tube 5 Density = mass/volume; 0.8 = mass/2; mass = 1.6 g 1.6 = 0.0173 Moles = 92.5 Moles of methylpropene = 0.0173 (1 : 1 Ratio) Volume methylpropene = 24 × 0.0173 = 0.415 dm3 6 a) C4H9Cl + NaOH → C4H9OH + NaCl b) nucleophilic substitution.