Halogenoalkanes

Question 1

describe the steps of naming a halogenoalkane

Answer 1

1) find the longest carbon chain - this will form the last part of the name
2) the names and positions of the halogens on the molecule forms the prefix
3) halogens must be in alphabetical order if more than one present
4) if more than one of the same type of halogen is present add di/tri/tetra in front of the prefix

Question 2

describe the trend in boiling point for halogneoalkanes

Answer 2

• increase down group
• strength of imfs determines bp
• number of electrons in the halogen atoms inc going down group
• more electrons = stronger vdw

Question 3

what makes halogenoalkanes polar?

Answer 3

• halogens more electronegative than carbon so they pull electrons in covalent bond towards themselves
• halogen becomes partially negative and carbon becomes partially positive

Question 4

describe how chloromethane can be made by free radical substitution

Answer 4

1) initiation: Cl-Cl bond broke by UV light, producing 2 highly reactive radicals
2) propagation:
- Cl radical reacts with a methane molecule to make a methyl radical
- methyl radical reacts with a Cl₂ molecule forming a chloromethane and Cl radical
- this Cl radical can react with more methane
3) termination: 2 radicals react to make a stable non-radical molecule

Question 5

what is a nucleophile?

Answer 5

an atom/group of atoms which is attracted to partially positive/electron deficient atoms where it donates a pair of electrons

Question 6

list some examples of nucleophiles. what makes them nucleophiles?

Answer 6

• H₂O
• OH^-
• OR^-
• X^-
• NH₃
• ^- CN
• they have a lone pair which they can donate forming a new bond

Question 7

name the mechanism by which halogenoalkanes react with hydroxide ions

Answer 7

nucleophilic substitution

Question 8

what happens when halogenoalkanes react with hydroxide ions by nucleophilic substitution? what conditions are required?

Answer 8

• a nucleophile will attack the partially positive carbon and will replace the halogen on the halogenoalkane
• the C-X (x=halogen) bond breaks both electrons move from the bond to the halogen
• new bond formed between OH^- ion and carbon, forming an alcohol
• conditions:
• warm aqueous sodium hydroxide (source of OH^-)
• carried out under reflux

Question 9

what happens when halogenoalkanes react with cyanide ions? what conditions are required?

Answer 9

• a nucleophile will attack the partially positive carbon and will replace the halogen on the halogenoalkane
• the c-x breaks, both electrons move from the bond to the halogen
• a new bond is formed between the CN^- ion and carbon, forming a nitrite
  - conditions:
  
  • warm ethanolic potassium cyanide
  • carried out under reflux

Question 10

what happens when halogenoalkanes react with ammonia? what conditions are required?

Answer 10

• ammonia will attack the partially positive carbon and will replace the halogen on the halogenoalkane forming an intermediate
• another molecule of NH₃ acts as a base by reacting with hydrogen (base = proton acceptor)
• an amine and an ammonium ion is produced
• conditions:
• heat with ethanoic ammonia
• must have excess ammonia (for 2nd phase to take place)

Question 11

describe the reactivity of halogenoalkanes. what determines the reactivity?

Answer 11

• more reactive going down group
• bond strength/bond enthalpy determines reactivity

Question 12

what happens when halogenoalkanes react with hydroxide via elimination? what conditions are required?

Answer 12

• OH^- will attack the hydrogen on a carbon adjacent to the carbon with the halogen on
• OH^- acts as a base forming water
• electrons in bond move to form a double bond between two carbons
• the C-X breaks and both electrons move from the bond to halogen, forming an alkene
• conditions:
• warm ethanoic NaOH
• carried out under reflux
• ethanol solvent used

Question 13

what determines whether a reaction is elimination or substitution? support answer using an example

Answer 13

• solvent used
• when reacting NaOH with a halogenoalkane can make either an alkene (by using ethanol as a solvent), an alcohol (by using water as a solvent), or both an alkene and an alcohol (by using a mixture of ethanol and water as a solvent)

Question 14

what are CFCs?

Answer 14

• molecules that have had all their hydrogens replaced by chlorine and fluorine
• stable molecules but are broken down by UV

Question 15

why are C-Cl bonds more likely to be broken down by UV than C-F?

Answer 15

they have the lowest bond enthalpy

Question 16

describe how CFCs destroy ozone via free radical substitution

Answer 16

• initiation:
  sunlight breaks the C-Cl bond in a CFC molecule and produces 2 radicals which will react with ozone molecules
  CCl₃F_(g) + hv –> °CCl₂F_(g) + Cl°_(g)
• propagation:
  1) Cl° reacts with O₃ to form the ClO° intermediate and O₂
  2) the ClO° reacts with more O₃ to make O₂ and Cl°
  3) as Cl° is reformed it acts as a catalyst
  Cl°_(g)+ O_3(g) –> O_2(g) + ClO°_(g)
  ClO°_(g) + O_3(g) –> 2O_2(g) + Cl°_(g)
• termination:
• 2 radicals react
• overall equation:
  2O₃ –> 3O₂