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Chem > AS organic chemistry > Flashcards

AS organic chemistry Flashcards

1
Q

What is a homologous series?

A

Family of molecules that share similar properties, share the same general formula and share the same functional group.

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2
Q

Heterolytic fission

A

A covalent bond breaks and both electrons in the bonding pair go to one atom, creating two oppositely charged ions. Endothermic.

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3
Q

Homolytic fission

A

A covalent bond breaks and one electron each in the bonding pair goes to each atom, creating two free radicals. Exothermic.

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4
Q

Which out of homo/heterolytic fission is more likely?

A

Homolytic fission as it requires less energy.

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5
Q

Nucleophile

A

Electron pair donors. Negatively charged.

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6
Q

Electrophile

A

Electron pair acceptors. Positively charged.

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7
Q

Free radical

A

Unpaired electron. Will react with anything.

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8
Q

Chain isomers

A

Carbon skeleton arranged in a different way, e.g. straight chain or branched. Similar chemical properties but different physical.

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9
Q

Positional isomers

A

Skeleton and functional group same, functional group attached to different carbon atom. Different chemically and physically.

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10
Q

Functional group isomers

A

Same atoms arranged into different functional groups. Different chemically and physically.

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11
Q

General formula of cycloalkane

A

CnH2n.

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12
Q

Initiation step of chlorination of methane

A

UV light provides energy to break Cl-Cl bond to form two free radicals (homolytic fission).

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13
Q

Propogation step of chlorination of methane

A

Cl free radical attacks H-C bond forming a methyl radical (CH3) and HCl.
Methyl radical attacks another Cl2, forming CH3Cl and Cl free radical.
This Cl free radical can repeat propogation step.

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14
Q

Termination step

A

Two free radicals react together, forming stable molecule. This is rare because free radicals are sparse.

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15
Q

Further reaction of chloromethane in free radical subsitution

A

CH3Cl can react with another Cl free radical to form dichloromethane and again to form trichloromethane and then to tetrachloromethane. This means end product of free radical substitution is a mixture of molecules. Also forms structural isomers.

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16
Q

How to maximise chloromethane in free radical sub

A

Excess of methane. More chance of chlorine radical colliding only with methane.

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17
Q

Naming molecules, priority of functional groups

A

OH > double bond > alkyl group, halogen

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18
Q

Electrophilic addition of HBr on symmetrical alkenes

A

Positive dipole on H is attracted to the electron rich double bond.
Bond made between C and H, so bond broken between H and Br.
Carbocation intermediate formed which reacts with :Br- to form bromoalkane.

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19
Q

Electrophilic addition of Br-Br on symmetrical alkene

A

The double bond on the alkene polarises the Br-Br bond which forms an induced dipole.
The Br with delta+ is attracted to the double bond and forms a bond with the carbon.
Carbocation intermediate formed reacts with :Br- to form dibromoalkane.

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20
Q

How does the test for double bond work?

A

Add the orange bromine water to organic molecule sample. The Br2 will react via electrophilic addition with the double bond and form dibromoalkane which is colourless.

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21
Q

Hydrogenation conditions

A

Ni catalyst and 170C.

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22
Q

Hydrogenation

A

H2 added to alkene to make alkane. Ni catalyst and 150C.

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23
Q

What is E and Z in a geometric isomer?

A

Apart and together

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24
Q

How to assign priority in geometric isomer?

A

According to atomic number.

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25
Q

When is geometric isomer?

A

When there is a double bond because you cannot rotate a double bond.

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26
Q

Markovnikov’s rule

A

Addition of HX to an asymmetrical alkene will result in major product having the H attached to the C with the most H.

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27
Q

Markovnikov’s rule explanation

A

Major product is formed more readily because it forms a secondary carbocation which is more stable than a primary (+ve charge will be on the other C with more alkyl groups).

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28
Q

Conditions to form poly(ethene)

A

High temp and pressure or moderate temp, moderate pressure and catalyst.

29
Q

Two types of poly(ethene)

A 
Low density (formed w/o catalyst).
High density (formed w/ catalyst).
30
Q

Properties of low density poly(ethene) and why

A

Branches mean polymer cannot pack closely together, therefore weaker London forces and lower melting point.

31
Q

Properties of high density poly(ethene) and why

A

No branches, can pack, greater London forces, high melting point.

32
Q

How to oxidise alcohol to aldehyde

A

With acidified oxidising agent K2Cr2O7 with a condenser while warmed as aldehyde boils off easiest.

33
Q

How to oxidised alcohol to carboxylic acid

A

With acidified oxidising agent K2Cr2O7 heated under reflux to prevent escape of organic vapours.

34
Q

Oxidation of secondary alcohol

A

Can only be oxidised to ketone as no more H attached to C to remove.

35
Q

Oxidation of tertiary alcohol

A

Cannot be oxidised as no H attached to C to remove.

36
Q

Making RCl from ROH with PCl5(s) and test for -OH group

A

ROH + PCl5(s) -> RCl + POCl3 + HCl

HCl steamy fume given off. Turns damp litmus paper red. Test tube has to be dry as water would also have formed HCl.

37
Q

Making RBr with HBr

A

HBr made in situ with KBr and 50% H2SO4 (to avoid oxidising Br- to Br2): KBr + H2SO4 -> KHSO4 + HBr
ROH + HBr -> RBr + H2O

38
Q

Making RI

A

2P (red phos) + 3I2 -> 2PI3 (making PI3 in situ)

PI3 + 3ROH -> H3PO3 + 3RI

39
Q

Dehydration of alcohol with H3PO4

A

Elimination reaction. The alcohol group is protonated in order to make water available as a leaving group. Then a beta proton is removed from beta carbon with the most alkyl groups attached, replacing lost H+ on acid.
H3PO4 is used as it is a weak acid so most likely not to cause reverse reaction.

40
Q

Why is RI more reactive than RF

A

C-F bond is stronger because F is smaller atom therefore less shielding between electron pair and nucleus, stronger electrostatic forces of attraction.

41
Q

Nucleophilic substitution in halogens

A

OH-, CN-, NH3 (nucleophiles) replace the halogen.`

42
Q

Nucleophilic substitution CH3CH2Br with OH- product and conditions

A

Ethanol

Warm aq. NaOH or aq. KOH

43
Q

Nucleophilic substitution CH3CH2Br with CN- product and conditions

A

Propionitrile

Hot ethanolic KCN under reflux

44
Q

Nucleophilic substitution CH3CH2Br with ammonia overall equation

A

CH3CH2Br + NH3 -> CH3CH2NH2 + HBr
HBr reacts with NH3 (HBr + NH3 -> NH4Br) therefore two moles of NH3 needed.
CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br

45
Q

Nucleophilic substitution CH3CH2Br with ammonia product and conditions for most yield in primary

A

Primary amine product

Warm with excess NH3 to prevent getting an ammonium salt in a sealed container under pressure.

46
Q

How does an amine react further after nucleophilic sub

A

Lone pair on the amine means that it acts as a nucleophile, meaning it can react with CH3CH2Br again and jettison a proton again, giving another lone pair to react again. It does this, forming secondary, tertiary amines until there are no more protons to jettison and you are left with a quaternary ammonium salt R4NBr.

47
Q

Sn1 (and draw example)

A 
Tertiary/secondary haloalkanes react with a nucleophile.
Step 1 (Rate Determining Step, requires energy) and step 2 (faster, electrostatic forces of attraction).
48
Q

Sn2 (and draw example)

A 
Primary/secondary haloalkanes react with a nucleophile.
One step (Rate determining step, bond half formed and broken, transition state).
49
Q

Why can a primary haloalkane not undergo Sn1

A

A primary carbocation would have to be formed but they are too unstable to exist, hence a T.S.

50
Q

Why can a tertiary haloalkane not undergo Sn2

A

Bulky alkyl groups block the nucleophile from attacking the carbon atom at 180 degrees.

51
Q

Order of reactivity of haloalkanes (primary, secondary, tertiary)

A

3, 2, 1 (not necessary to know but because Sn1 is faster than Sn2 due to the increased stability of the tertiary carbocation).

52
Q

Conditions for nucleophilic sub and elimination in haloalkanes

A

Warm aqueous KOH and hot ethanolic KOH

53
Q

Test for halides with haloalkanes

A

Nucleophilic substitution with NaOH to get halide.
Addition of HNO3 to remove excess OH-.
Addition of AgNO3 to form AgX.
AgCl is white precipitate.
AgBr is cream “.
AgI is yellow “.
Can also check further with ammonia solution.

54
Q

Formula for chain alkanes and cyclic alkanes

A

Chain - CnH2n+2

Cyclic - CnH2n

55
Q

How is alkane fuel obtained

A

Fractional distillation, cracking and reforming of crude oil.

56
Q

Why are chain alkanes reformed?

A

Too explosive, therefore combustion properties are changed to improve combustion in engines (making cyclic alkanes etc.)

57
Q

Which pollutants are formed by combustion of alkane fuels.

A

CO, C, NOx, unburned hydrocarbons, SO2/SO3.

58
Q

Risks of CO

A

Binds to haemoglobin, asphyxiation and death.

59
Q

Risks of C particulate

A

Small particulates cause respiratory diseases, cancer.

60
Q

Risks of NOx

A

Forms smog, greenhouse gas, acidic, aggravation of asthma and irritation of respiratory system.

61
Q

Risks of unburnt hydrocarbons

A

Smog formation and harmful effects on nervous system and headaches.

62
Q

Risks of SO2/SO3

A

Acid rain, changes in lung function, decreased fertility, bronchitis.

63
Q

How does a catalytic converter solve some pollutant problems.

A

Provides huge surface area for chemical reactions to take place to convert polluting chemicals to safer ones. Platinum catalyst.

64
Q

Which reactions happen in a catalytic converter

A

2CO + 2O2 -> 2CO2
Carbon is combusted to CO2.
2NO -> N2 + O2
Hydrocarbons are burnt.

65
Q

Biofuels

A

Produced from organic materials, fewer toxic emissions, energy used for processing must be carbon neutral.

66
Q

Why is biofuel better?

A

The release of the CO2 is similar to fossil but with fossil, the carbon was captured millions of years ago and the release of this carbon recently imbalances the carbon cycle.

67
Q

Test for OH group

A

Add PCl5, HCl steamy fumy given off that turns damp litmus paper red.

68
Q

Importance of CN- nucleophilic sub

A

Adds another C to chain

69
Q

Requirements for ez isomerism

A

C=C and each c attached to 2 different groups

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