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A level Chemistry AQA > Organic pt2 (Halogenoalkanes and alkenes) > Flashcards

Organic pt2 (Halogenoalkanes and alkenes) Flashcards

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1
Q

What are the three steps of the free radical substitution mechanism (chronological)

A

initiation

propagation

termination

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2
Q

What are the properties of halogenoalkanes?

A
  • insoluble = only contain dipole-dipole attractions and van der waal forces ( not polar enough to attract water)
  • The boiling point increases with increased chain length and has you go down group 7 = increased van der Waal forces because the larger the molecules, the greater the number of electrons (and therefore the larger the van der Waals forces)
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3
Q

What is the most reactive halogenoalkane and what is the least reactive halogenoalkane?

A

iodo-compounds are the most reactive and fluro-compounds are the least reactive. This is because as you go down group 7, the molecular mass of the halogen atom increases, the attraction of the electrons to the nucleus gets weaker, so the bond enthalpies fall.

this means C-I would be a weak bond = more reactive

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4
Q

Define free radical substitution

A

when radicals swap H for halogen, so a haloalkane is made

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5
Q

Describe what happens in the free radical substitution mechanism in chlorine and methane, step by step

A

Initiation = The covalent bond in Chlorine (Cl2 because its a diatomic atom) is broken when introduced to UV light to produce 2 Cl radicals

Propagation = This starts a chain reaction: The Cl radical collides with methane, taking one hydrogen from methane = CH3 radical and HCL

CH3 radical collides with Cl2 in test tube = CH3Cl+ Cl radical

Termination= 2 radicals eventually collide to give a stable molecule and stops the chain reaction

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6
Q

Why is UV light an essential condition required for this reaction to occur?

A

because it provides energy to break Cl-Cl, which starts the chain reaction

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7
Q

What is the overall equation for the decomposition of ozone into oxygen?

A

2O3 → 3O2

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8
Q

How is the ozone layer formed?

A

U.V light from the sun provides the enrgy to break the O=O bond in oxygen molecules:

O2 → 2 oxygen free radicals

These radicals are very reactive and react with other oxygen molecules to form ozone

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9
Q

Why is the ozone layer beneficial to us

A

it absorbs harmful ultraviolet light

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10
Q

What happens to the ozone layer when it absorbs UV light

A

it dissociates to form an O2 and an oxygen atom

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11
Q

How do CFCs damage the ozone layer

A

When CFCs rise to the stratosphere, the UV light from the sun provides energy and breaks the bond between carbon-chlorine to produce chlorine radical

The chlorine radical then reacts with the ozone layer to produce ClO radical. The ClO radical then reacts with the ozone layer also.

The Cl radicals act as a catalyst so they are not destroyed and keep on decomposing the ozone

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12
Q

What is a nucleophile and what do they always have?

A

an electron pair donor

a negatively charged ion or an atom with a S- charge

Nucleophiles always have a lone pair

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13
Q

What three nucleophiles can be involved in nucleophilic substituton?

A

Hydroxide ion (-OH), ammonia, cynaide ion

:OH, :CN & :NH3

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14
Q

Name the condition & product for the nucleophile :OH- (nucleophilic substitution)

A

Warm, aqueous NaOH or KOH → Alcohol

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15
Q

Outline nucelophillic substittion with bromoethane and sodium hydroxide

A
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16
Q

Name the condition & product for the nucleophile :CN- (nucleophilic substitution)

A

Warm halogenoalkane with ethanolic KCN(aq) → Nitrile

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17
Q

Outline nucelophillic substittion with bromoethane and potassium cyanide

A
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18
Q

Name the condition & product for the nucleophile :NH3 (nucleophilic substitution)

A

Warm halogenoalkane with excess ethanolic ammonia → Amine

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19
Q

Outline nucelophillic substittion with bromoethane and ammonia

A
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20
Q

Why is the reaction between hydroxide ions & halogenoalkanes sometimes called hydrolysis?

A

∵ in the mechanism, the exact same reaction will happen with water

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21
Q

In nucleophillic substituition, why is there a dative bond between the carbon and the nucleophile

A

When C-halogen bond breaks, the halogen takes the electrons (net 1 gains an electron). The C atom, therefore, has to form a dative bond with the lone pair of the nucleophile

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22
Q

When halogenoalkanes react with ammonia, what can the amine group in the product act as and why & what can this lead to?

A

Can acts as nucleophile∵ it has lone pair of electrons ∴ may react with halogenoalkane molecules = giving mixture of products

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23
Q

In the second step of nucleophilic substitution with ammonia, why isn’t the ammonia molecule a nucleophile anymore? What does it act as instead?

A

When the ammonia in the first step binds to the halogenalkane, it forms a dative bond with the carbocation (which forms when the halogen is displaced and takes some of the electrons). This means that the N in NH3 becomes positive because it has given its electrons to the carbon.

This means the ammonia that reacts with the compound in the second stage is a base because it removes hydrogen which is H+ due to the hydrogens electrons moving to the nitrogen and therefore is a proton acceptor

it instead acts as a base ( accepts a proton from substituted ammonium ion)

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24
Q

What do curly arrows show?

A

The movement of electrons

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25
Q

What is elimination and when does it happen?

A

When a small molecule is eliminated from a haloalkane , in the presence of a strong base = alkene is produced

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26
Q

Name the conditions of elimination reaction?

A

hot, ethanolic solution

27
Q

Outline the elimination reaction of 2-bromopropane and NaOH/KOH

A
28
Q

Describe what happens to the electrons in elimination reactions

A

The OH- attacks the hydrogen attached to a carbon adjacent to the carbon with halogen

When the OH- attacks hydrogen, the shared pair of electrons shift, so the hydrogen net loses 1 electron (H+). The OH- removes H+ ion to make H2O

The pair of electrons shift to the carbons, to make a double carbon bond

This results in a pair of electrons from one of the carbon in the newly formed C=C to shift to the halogen atom, so the halogen atom net gains 1e- and the bond breaks

29
Q

Which temperatures are best for elimination and which are best for substitution?

A

High = elimination

Low = subtitution

30
Q

What type of hydrocarbon is an alkene

A

unsaturated

31
Q

Why are alkenes more reactive than alkanes

A

because they have a double bond = high concentration of electrons between the two carbon atoms

32
Q

What are the bond angles in alkenes

A

120

33
Q

Why can’t double bonds rotate?

A

because the pi bond can’t rotate = if the bond were to be rotated, the overlap between the two p-orbitals will be broken and the pi bond will no longer exist

34
Q

What is a sigma bond?

A

Its a form of chemical covalent bond between atoms. It consists of the overlap between two s atomic orbitals of the atoms. This is the type of covalent bond found in single bonds

35
Q

What is a pi bond?

A

Its a form of covalent bond that occurs between atoms. It involves the overlapping of two p orbitals of the bonding atoms, which produces a cloud of electrons above and below the molecule. This is perpendicular to the plane on the molecule

Denoted by a double line (=) between atoms in their dispalyed formulae.

36
Q

What is stereoisomerism?

A

The atoms making up the isomers are joined up in the same order, but still manage to have a different spatial arrangement.

37
Q

What type of stereoisomerism is E/Z isomerism?

A

Geometrical isomers

38
Q

What does E isomer mean and what does Z isomer mean?

A

e = opposite

z= zusammen

39
Q

What is the criteria for E/Z isomerism?

A
  • C=C cannot rotate
  • Each C in C=C has 2 different groups on it
40
Q

What are the rules we use to work out E/Z isomerism

A

Cahn-Ingold-Prelog rules

41
Q

What are the physical properties of alkenes?

A
  • Boiling and melting points depend on carbon chain length NOT DOUBLE BONDS, due to van der waal forces
  • not soluble in water
42
Q

Describe and explain the reactivity of alkenes?

A

very reactive: The C=C forms an electron- rich area in the molecule, which can easily be attacked by electrophiles

43
Q

What is an electrophile?

A

an electron pair acceptor

44
Q

Draw the mechanism for the electrophillic addition of ethene with bromine . Explain what is happening

A

The electron rich area of the double bond in the alkene induces a dipole in the bromine molecule. The S+ bromine atom in the bromine molecule behaves as an electrophile

Electrons from the C=C double bond are attracted to the S+ bromine atom, and a new bond is formed between the carbon atom and S+ bromine atom

At the same time, the bond between the two bromine atoms breaks releasing a bromide ion

The intermediate carbocation attracts the bromide ion and a bond is formed by the lone pair electrons on the bromide ion and the carbon atom (the bromide ion acts as a nucleophile here)

45
Q

Draw the mechanism for the electrophillic addition of ethene with hydrogen bromide

A
46
Q

Why do halogenoalkanes have higher boiling points than alkanes with similar chain lengths

A

because they have higher relative molecular masses and they are more polar

47
Q

Draw the mechanism for the electrophillic addition of ethene with Sulfuric acid and name the product

A

ethyl hydrogensulfate

48
Q

What are the conditions for electrophillic addition of an alkene with sulfuric acid and what are the products?

A

cold, concentrated sulfuric acid = alkyl hydrogensulfates

49
Q

What happens when alkyl hydrogensulfate is warmed with water? Draw what happens with ethyl hydrogen sulfate and water

A

hydrolysis occurs to produce an alcohol and sulphuric acid is regenerated

50
Q

Explain Markownikoff’s Rule (positive inductive effect)

A

Electrons in Alkyl groups shift towards the positive carbon in a carbocation, to stabilize the positive charge. This means the more alkyl group a carbocation has, the more stable it is. Therefore tertiary carbocation > primary carbocation

51
Q

Explain why 2-bromopropane is a more major product than 1-bromopropane

A

2- bromopropane is the major product because it is made via a secondary carbocation intermediate

the other product made via a primary carbocation intermediate

a secondary carbocation is more stable than primary carbocation

because the secondary carbocation has more R groups releasing electron density towards the c+ = more stable

this means the product lasts for longer, hence this is the product that is more likely to form.

52
Q

Identify the characteristic of a haloalkane molecule that enables it to undergo elimination and nucleophilic addition

A

the polar C–halogen bond

53
Q

Define hydrolysis

A

The splitting of molecules using water

54
Q

The rate of reaction of 2-chloropropane with ammonia is lower than the rate of reaction of 2-bromopropane with ammonia. Explain why. (2 marks - exam question)

A
  • C-cl is shorter due to less shielding
  • the C-Cl bond in 2-chloropropane will therefore have a higher bond energy
55
Q

Explain why the sulfur compounds found in crude oil should be removed from the fractions before they are used for central heating fuel?

A
  • On combustion SO2 is produced
  • which causes acid rain
56
Q

Why can alkenes make up addition polymers?

A

Can act as monomers and form polymers ∵ double bond can open up & join together to make long chains

57
Q

Draw poly(phenylethene) from phenylethene

A
58
Q

What is rigid PVC [poly(chloroethene] used for?

A

window frames and drainpipes

59
Q

What is flexible PVC [poly(chloroethene] used for?

A

electrical cable insulation, flooring tiles and clothing

60
Q

Why are addition polymers very unreactive?

A
  • Polyalkene chains = saturated molecules
  • Main carbon chain of polyalkene = non-polar
61
Q

What are the properties of long polyalkenes and short polyalkenes? Explain why the properties are like this.

A

long= strong and rigid = van der waals are closer together, so the forces are stronger between chains

short = weaker and more flexible= less van der waals forces

62
Q

What happens when you add plasticiser to a polymer and why?

A

the polymer becomes more flexible as platiciser molecules get between te polymer chains and push them apart = reduces strength of the intermolecular forces between the chains , so they can slid around more = polymer easier to bend

63
Q

Explain how the Cahn-Ingold- Prelog priority rules can be used to deduce the full IUPAC name of this compound (6 marks)

A
  • First figure out which functional group has a higher priority in each carbon in the C=C bond, by seeing which one has a higher atomic number.
  • One the right hand side (the first carbon) CH2OH is the highest priority group as the C has a higher atomic number (6) than H (1)
  • On the LHS, both groups have a carbon atom, so we need to consider atoms one bond further away. the CH3CH” is the highest priority group because the next bond away from its carbon is another carbon so the total atomic number is 12 compared to the methyl group with a combined atomic number of 7
  • Both priority groups are on the same side (above the plane), so the isomer is Z
  • IUPAC name : z-3-methylpent-2-en-1-ol.

A level Chemistry AQA (30 decks)

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