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A Level Chemistry Y1 > Haloalkanes > Flashcards

Haloalkanes Flashcards

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1
Q

what condition is necessary for halogens to react with alkanes

A
  • presence of ultraviolet light
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2
Q

what forms when a halogen reacts with an alkane

A
  • halogenoalkane and hydrogen halide
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3
Q

what is a radical

A
  • a reactive species due to the presence of an unpaired electron
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4
Q

what is the 1st stage of the free radical substitution mechanism

A

Initiation

  • ultraviolet light causes the covalent bond between the halogen atoms to break and create 2 halogen radicals
  • X₂→2X⚫
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5
Q

what is the 2nd stage of the free radical substitution mechanism

A

Propagation

1st propagation

  • the halogen radical reacts with the alkane and removes a hydrogen creating a hydrogen halide molecule
  • the alkane is left as an alkyl radical

2nd propagation

  • the alkyl radical reacts with a halogen molecule and creates a halogenoalkane and the remaining halogen atom is left as a halogen radical
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6
Q

what is the 3rd stage of the free radical substitution mechanism

A

Termination

  • two radicals react to make neutral compounds
  • alkyl + alkyl , alkyl + halogen or halogen + halogen
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7
Q

what happens to the halogen radical at the end of the second propagation

A
  • it goes on to react with another alkane molecule and 1st propagation is repeated
  • there will be a chain rection of propagations until you run out of halogen molecules or alkane
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8
Q

what are chlorofluorocarbons (CFCs)

A
  • halogenoalkanes containing both chlorine and fluorine atoms but no hydrogen
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9
Q

features and uses of short chain chlorofluorocarbons

A
  • gases at room temp
  • used in refrigerators
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10
Q

uses of longer chain chlorofluorocarbons

A
  • dry cleaning and as de-greasing solvents
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11
Q

why are CFCs harmful to the environment

A
  • they are very unreactive under normal conditions but in the atmosphere they are decomposed to give chlorine free radicals which destroy ozone
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12
Q

how do CFCs destroy ozone

A
  • the C-Cl bond breaks in the presence of UV light to produce a Cl radical
  • the Cl radical then decomposes ozone in the stratosphere
  • the resulting ClO radical decomposes more ozone and regenerates the Cl radical
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13
Q

what is the overall equation for ozone breakdown

A

2O₃→3O₂

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14
Q

what are alternatives to CFCs

A
  • hydrofluorocarbons which don’t contain chlorine and therefore can’t produce chlorine radicals
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15
Q

what is the general formula of halogenoalkanes

A
  • CnH2n+1X where X is the halogen
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16
Q

what is a primary halogenoalkane?

A
  • the halogen is bonded to a carbon that is bonded to only one other carbon
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17
Q

what is a secondary halogenoalkane?

A
  • the halogen is bonded to a carbon that is bonded to two other carbons
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18
Q

what is a tertiary halogenoalkane?

A
  • the halogen is bonded to a carbon that is bonded to three other carbons
19
Q

what is a mechanism

A
  • a representation of the movement of a pair of electrons during a chemical reaction
20
Q

how do you show a pair of electrons moving

A
  • a curly arrow starting from a bond or a lone pair of electrons and ending at the new location of the electron pair
21
Q

what is a nucleophile

A

electron pair donor

22
Q

how does nucleophilic substitution work

A
  • the halogen atom is more electronegative than the carbon atom its bonded to and so the electrons are more attracted to it
  • it becomes partially negatively charged while the carbon becomes partially positively charged
  • the nucleophile attacks the partially charged carbon and donates its lone pair of electrons forming a bond
  • the carbon halogen bond is broken forming a halide ion
23
Q

how do hydroxide ions work as a nucleophile (OH?)

A
  • when halogenoalkanes are put in warm aqueous sodium / potassium hydroxide, alcohols are formed
24
Q

how do cyanide ions work as a nucleophile (CN?)

A
  • when halogenoalkanes are put in warm aqueous ethanol solution / potassium cyanide, nitriles are formed
  • this adds an extra carbon to the chain
25
Q

how can you show the rate of hydrolysis of the haloalkanes

A
  • gently heat seperate samples of 1-chloropropane, 1-bromopropane and 1-iodopropane
  • add an equal amount of aqueous sodium hydroxide and aqueous silver nitrate
  • time how long it takes for a precipitate (AgCl, AgBr, AgI) to form
26
Q

what is the trend of reactivity as you go down the group of haloalkanes and why

A
  • reactivity increases as C-F is the strongest bond
  • Fluorine is the smallest atom of halogens therefore the shared electrons are much closer to the positive nucleus which results in a stronger bond
27
Q

how does excess ammonia (NH₃) work as a nucleophile

A

halogenoalkanes warmed with excess ammonia in sealed container, primary amines are formed (NH₂)

  • overall equation:
    CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
28
Q

how many further substitutions are there

A

3

29
Q

what is a secondary amine

A

-A secondary amine has the general formula R2NH.
- an ammonia molecule (NH3) in which two of the hydrogens have been replaced by alkyl groups.

30
Q

how are secondary amines formed

A
  • The lone pair on the nitrogen in the primary amine attacks the + carbon exactly the same as the ammonia did. Bromine is lost as a bromide ion, and the immediate product is a salt called diethylammonium bromide
31
Q

what is a tertiary amine

A
  • A tertiary amine has the general formula R3N.
    -an ammonia molecule in which all three of the hydrogens have been replaced by alkyl groups.
32
Q

how are tertiary amines formed

A
  • the secondary amine still has an active lone pair of electrons on the nitrogen atom. That, in turn, can attack bromoethane if it happens to collide with it.
  • the organic product of this reaction is the tertiary amine, triethylamine
33
Q

what is a quaternary ammonium salt

A
  • a quaternary ammonium salt is an ammonium salt (for example, NH4+ Br-) in which all the hydrogens have been replaced by an alkyl group
34
Q

how is a quaternary ammonium salt formed

A
  • the tertiary amine still has an active lone pair on the nitrogen that can attack the + carbon in the bromoethane
  • this time there is nowhere else for the reaction to go. There is no longer a hydrogen atom on the nitrogen that an ammonia molecule could remove, and so the reaction finally comes to an end.
35
Q

what leads to a better yield of the primary amine

A
  • a large excess of ammonia
36
Q

what leads to a high yield of quaternary ammonium salt

A
  • a large excess of halogenoalkane
37
Q

what is usually obtained during these further substitution reactions

A
  • a mixture of primary secondary tertiary and quaternary ammonium salts
38
Q

how can hydroxide ions act as a base during alternative elimination

A
  • the hydroxide ion removes a hydrogen atom from an adjacent carbon to the halogen atom
  • next, the electrons in the carbon hydrogen bond move onto the carbon-carbon bond to create a double bond
  • finally, the electrons in the carbon-halogen bond move onto the halogen atoms breaking the bond and creating a halide ion
39
Q

what can happen during the elimination of asymmetrical halogenoalkanes

A
  • two different alkene products being formed
  • (positional isomers)
40
Q

what factors can affect if a reaction undergoes elimination or substitution

A
  • the structure of the halogenoalkane
  • the base strength of the nucleophile
  • the reaction conditions
41
Q

how does the structure of the halogenoalkane affect if a reaction undergoes elimination or substitution

A
  • primary halogenoalkanes favour substitution
  • secondary halogenoalkanes favour both
  • tertiary halogenoalkanes favour elimination
42
Q

how does the base strength of the nucleophile affect if a reaction undergoes elimination or substitution

A

as the strength of the base increases the chances of elimination increases
- using aqueous solution substitution is favoured
- using ethanolic solution elimination is favoured

43
Q

how does the reaction conditions affect if a reaction undergoes elimination or substitution

A

the higher the temp, the greater the chance for elimination
- elimination is favoured by hot ethanolic conditions
- substitution is favoured by warm aqueous conditions

A Level Chemistry Y1 (8 decks)

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