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Chemistry A2 > Organic Chemistry > Flashcards

Organic Chemistry Flashcards

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1
Q

Optical Isomers/Enantiomers

A

Pairs of molecules that are non-superimposable mirror images of each other

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2
Q

Chiral

A

Asymmetric structure such that mirror images are non-superimposable

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3
Q

Condition for a molecule to be chiral

A

Must have four different substituent groups attached to chiral carbon

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4
Q

Racemate

A

50:50 mixture of two enantiomers

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5
Q

Measuring Optical Activity

A
  • polarimeter
  • polarised light passed through solution containing enantiomers
  • rotate plane of polarisation in OPPOSITE directions
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6
Q

Compare enantiomers of a molecule

A
  • same physical/chemical properties (except for with other chiral molecules)
  • rotate plane of polarisation in OPPOSITE directions
  • different reaction with other enantiomers
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7
Q

Physical Properties of Carbonyl Compounds

A
  • high boiling point due to dipole-dipole forces
  • soluble since hydrogen bonds form between lone pair on oxygen and water
  • very reactive due to C=O bond which is attacked by nucleophiles
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8
Q

Explain why nucleophilic addition of carbonyl molecules with KCN produces a racemic mixture

A
  • carbonyl molecule is planar
  • :CN has equal chance of attacking
  • C=O from above or below the plane
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9
Q

Fehling’s Test

A
  • distinguish between aldehyde and ketone
  • Fehling’s A = blue solution of copper (II) ions
  • Fehling’s B = alkali and complexing agent
  • brick red precipitate of copper (I) oxide
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10
Q

Tollen’s Reagent

A
  • distinguish between aldehyde and ketone
  • [Ag(NH3)2]+
  • silver mirror forms
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11
Q

Reduction of aldehydes and ketones

A
  • nucleophilic addition using hydride :H- ion written 2[H]
  • NaBH4 sodium tetrahydridoborate / methanol or water
  • LiAlH4 lithium tetrahydridoaluminate / ether or dry
    Aldehyde => primary alcohol
    Ketone => secondary alcohol
    Carboxylic Acids => primary alcohol (only using LiAlH4)
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12
Q

Suggest why NaBH4 does not react with carboxylic acids

A

Not strong enough to reduce carboxylic acids

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13
Q

Delocalisation of carboxylate ion

A
  • resonance
  • negative charge (e-) delocalised between two electronegative oxygen atoms
  • more stable
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14
Q

Methods of producing esters

A
  • carboxylic acid + alcohol (strong acid catalyst) by esterification
  • acyl chloride + alcohol by acylation
  • acid anhydride + alcohol by acylation
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15
Q

Naming Esters

A

Number carbons from -O- bond

Alcohol name then Acid/Acyl name

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16
Q

Methods of hydrolysis of esters

A

aqueous ester with

  • acid catalyst
  • base catalyst
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17
Q

Compare hydrolysis of esters using acid and base catalysts

A

acid catalyst
- reforms equilibrium mixture of acid and alcohol
base catalyst
- reacts with carboxylic acid forming salt of acid
- reaction goes to completion so equilibrium not established

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18
Q

IUPAC name of glycerol

A

Propane-1,2,3-triol

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19
Q

Common uses of esters

A
  • solvents
  • plasticisers
  • perfumes
  • food flavourings
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20
Q

Explain why sodium salts produced from hydrolysis of esters can be used in soaps

A
  • sodium salts are ionic and can dissociate to form Na+ and carboxylate ion
  • carbon chain is non polar so mixes with grease
  • carboxylate ion is polar so mixes with water
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21
Q

Biodiesel

A
  • mixture of methyl esters of long chain carboxylic acids

Triglyceride (Vegetable Oil) + 3 Methanol (NaOH catalyst) -> Fatty Methyl Esters + Glycerol

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22
Q

Acylation

A

Process by which acyl group ( R-C=O) is introduced into another molecule

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23
Q

Carboxylic Acid Derivatives

A

Compounds that can be converted to carboxylic acids via acidic/basic hydrolysis (acyl group + Z group)

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24
Q

Examples of Acid Derivatives

A
  • ester
  • acyl chloride
  • acid anhydride
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25
Q

Order of reactivity for addition-elimination reactions with any chlorides and acid anhydrides

A

Primary amine ( = N substituted amine)
Ammonia ( = Amide)
Alcohol ( = Ester)
Water ( = Carboxylic acid)

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26
Q

Advantages of ethanoic anhydride over ethanoyl chloride in the manufacture of aspirin

A
  • cheaper
  • forms ethanoic acid rather than hydrogen chloride
  • safer / less corrosive
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27
Q

Order of boiling points for functional groups with similar Mr

A

Carboxylic Acids - hydrogen bonding dimerise
Alcohols - hydrogen bonding (+ primary have larger surface area so more contact points therefore more van der Waals)
Ketones - stronger dipole-dipole due to two electron donating alkyl groups around C=O group
Aldehydes - dipole-dipole
Alkanes - only van der Waals

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28
Q

Flaws in Kekulé model of benzene

A
  • should undergo electrophilic addition but rarely does
  • reaction with bromine should give rise to three isomers but only one produced
  • hexagon should not be symmetrical but x-ray diffraction shows it is a perfect hexagon
  • thermochemical data for hydrogenation implies lower in energy so more stable than thought (shows reaction is less exothermic than predicated based on cyclohexene)
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29
Q

Suggest how to differentiate benzene from cyclohexene

A

Bromine water - benzene does not decolourise since it rarely undergoes electrophilic addition

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30
Q

Explain the stability of benzene

A
  • resonance (lower energy state)

- delocalised electrons in p orbitals above and below the plane of the aromatic ring

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31
Q

Suggest why benzene has a higher boiling point than hexane

A
  • flat structure rather than puckered

- molecules can pack closer together so stronger vdw

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32
Q

Suggest why reactions with benzene are most likely to be electrophilic substitutions

A
  • delocalised e- prone to attack by electrophiles

- aromatic ring is very stable so often remains in tact

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33
Q

Suggest why a smoky flame is an indication of an aromatic compound

A

Benzene ring has a high carbon:hydrogen ratio

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34
Q

Conditions for nitration of benzene

A
  • concentrated sulphuric acid
  • concentrated nitric acid
  • 20-60 degrees C
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35
Q

Suggest how reaction between sulphuric acid and nitric acid creates nitronium electrophile

A

H2SO4 + HNO3 -> NO2+ + H2O + HSO4-

  • sulphuric acid is stronger acid so donates H+ to nitric acid
  • sulphuric acid is catalyst
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36
Q

Describe what would happen if nitration of benzene occurred above 55 degrees C

A

Multiple points of substitution (nitration)

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37
Q

Suggest uses of nitrated arenes

A
  • explosives (TNT)
  • industrial dyes
  • formation of amines
38
Q

Describe structure of benzene

A
  • planar
  • stable due to delocalised system in aromatic ring
  • bond length intermediate between C-C and C=C
39
Q

Conditions for Friedel-Crafts Acylation

A
  • AlCl3 / FeCl3 catalyst
  • aycl chloride
    CH3COCl + AlCl3 -> AlCl4- + CH3CO+
40
Q

Suggest why ammonia is a better nucleophile than water

A

Nitrogen is less electronegative than oxygen so donates lone pair of electrons more readily

41
Q

Suggest why a primary amine is a better nucleophile than ammonia

A

Primary amine has electron-donating alkyl groups attached to nitrogen so donates lone electron pair more readily due to positive inductive effect

42
Q

Suggest why geometrical isomers are formed from dehydration of alcohols

A

Loss of H+ from end of C next to C-OH

43
Q

Conditions for esterfication

A
  • strong acid, e.g. sulfuric acid
  • warm (<100 degrees C)
  • reflux
44
Q

Conditions for acylation with acyl chloride

A
  • (alcohol reagent)
  • no acid
  • room temperature (since acyl chloride highly reactive)
45
Q

Conditions for acylation with acid anhydride

A
  • (alcohol reagent)
  • no acid
  • heat/reflux
46
Q

Suggest why a tertiary alcohol cannot be oxidised

A

No H attached to same carbon as -OH group/ x3 alkyl groups attached to C-OH

47
Q

State mechanism for esterification

A

Addition-Elimination

48
Q

Test to differentiate between ethanoyl chloride and chloroethane

A

Silver Nitrate
Ethanoyl Chloride - white precipitate
Chloroethane - no reaction

49
Q

Explain why amines have lower boiling points than alcohols

A
  • hydrogen bonding is not as strong as in alcohols

- nitrogen is less electronegative than oxygen

50
Q

Explain why phenylamine is not very soluble

A
  • benzene ring cannot form hydrogen bonds
51
Q

Suggest why amines are often supplied as hydrochlorides in medication

A
  • insoluble amine converted to soluble hydrochloride salt

- more soluble in bloodstream

52
Q

Suggest how amines can be identified

A
  • strong fishy smell
53
Q

Suggest why secondary/tertiary amines are relatively insoluble in water and describe how to covert them into soluble amine salts then back to amines

A
  • long hydrophobic carbon tails cannot form H bonds with water
  • add HCl to form hydrochloride which contains ions that dissolve in water
  • add strong base like NaOH to regenerate insoluble amine
54
Q

Suggest why tertiary amines are weaker bases than primary and secondary amines

A

lone pair on N is inaccessible to H+

55
Q

Suggest why phenylamines are the weakest bases

A
  • lone pair on N
  • delocalised into aromatic ring
  • least able to donate electron pair
56
Q

Suggest difference in major product formed when ammonia is in excess compared to halogenoalkane, and vice versa

A

ammonia in excess => primary amine

halogenoalkane in excess => quaternary ammonium salt

57
Q

Suggest two methods of producing a primary amine from a halogenoalkane

A
  • KCN/aq+ethanol => nitrile
  • 2H2 or 4[H] / Ni catalyst => primary amine
    OR
  • EXCESS NH3 => primary amine
58
Q

Suggest why reduction of nitriles is preferred to nucleophilic substitution with halogenoalkanes in manufacture of primary amines

A
  • purer product since only primary amine can be formed

- several nucleophilic substitutions can occur

59
Q

Suggest how phenylamine can be formed from nitrobenzene

A
  • reduction
  • Sn/HCl
  • room temperature
  • tin and hydrochloric acid react to form H2 which reduces NO2 to 2H20
  • 6[H] required
60
Q

Suggest how N substituted phenyl amine can be formed from phenylamine

A
  • nucleophilic substitution

- halogenoalkane

61
Q

Common uses of amines

A
  • manufacture of nylon, polyurethane, dyes and drugs

- synthesis of quaternary ammonium compounds to act as cationic surfactants in hair and fabric conditioners

62
Q

Describe how cationic surfactants work

A
  • ions cluster with charged ends in water
  • hydrocarbon tail on surface
  • form a coating preventing build up of static charge
63
Q

Suggest why secondary amines have a lower boiling point than primary amines

A
  • form hydrogen bonds

- nitrogen atom in the middle of the chain rather than at the end so have a weaker dipole

64
Q

Suggest why tertiary amines have a lower boiling point compared to primary and secondary

A
  • no H attached to N so no hydrogen bonding

- only dipole-dipole

65
Q

Suggest why LiAlH4 is used rather than NaBH4 to reduce nitriles to primary amines

A
  • NaBH4 is not a strong enough reducing agent
66
Q

Suggest combinations of molecules which can join to make condensation polymers

A
  • dicarboxylic acids and diols (=polyesters)
  • dicarboxylic acids and diamines (=polyamides)
  • amino acids
67
Q

Suggest what is meant by a condensation reaction

A

reaction in which two molecules combine to form a larger molecule while eliminating a small molecule, e.g. water

68
Q

Describe the structure of cisplatin

A
  • Pt(NH3)2Cl2

- square planar

69
Q

Explain how cisplatin works as an anti-cancer drug

A
  • bonds to nitrogen atom (lone N on smaller ring) on two adjacent guanine bases on DNA strand
  • by ligand substitution reaction
  • lone pairs on nitrogen form dative covalent bond with Pt
  • Cl- displaced
  • prevents transcription of DNA
  • DNA replication cannot occur
  • cells cannot replicate so die
70
Q

Suggest why transplatin would be ineffective as an anti-cancer drug

A
  • Cl- are not on the same side of Pt
  • only one Cl- displaced by lone pair of nitrogen on guanine
  • one bond is weak / could easily break
  • not effective is stopping transcription of DNA
71
Q

Suggest possible side effects of using cisplatin and how these could be minimised

A
  • death of healthy cells that replicate rapidly, e.g. hair cells
  • small doses
  • targeted treatment
72
Q

Uses of condensation polymers

A
  • clothing, carpets and rugs from nylon 6,6
  • fire-resistant clothing from terylene
  • bullet proof vests from kevlar
73
Q

Compare properties of addition polymers with condensation polymers

A
  • AP have more branching due to random free radical addition mechanism
  • CP have long, straight chains so have a higher tensile strength
  • CP have hydrogen bonds between chains so can form helical structures
  • CP are biodegradable due to C=O POLAR bonds which are readily attacked by nucleophiles (hydrolysed)
  • AP are non-biodegradable due to long non-polar chains AND strong C-C bonds
74
Q

Suggests methods of disposing of polymers

A
  • landfill sites
  • incinerate (release green house gases and toxic fumes)
  • recycling
75
Q

Evaluate recycling as a method of disposal for polymers

A

Advantages
- saves raw materials
- reduces landfill
- lower cost for recycling than making from raw materials
- reduces CO2 emission by not being incinerated
Disadvantages
- difficult and costly to sort and collect
- easily contaminated/not always suitable for purpose
- (limited number of times to recycle before polymer chains degrade)

76
Q

Suggest what determines if an amino acid is alpha or beta

A

number of carbon atoms between amino and carboxylic acid group

77
Q

Explain what is meant by a zwitterion and why this means all amino acids are solid at room temperature

A
  • double ion which has separate positive and negatively charged groups
  • formed when H+ from carboxylic acid group on amino acid is transferred to amine group
  • amino acids exits as zwitterions so have ionic bonding between them so form solids at room temperature
78
Q

Define isoelectric point

A

pH at which zwitterion exists

79
Q

Describe what happens to amino acids at high and low pHs

A
  • deprotonated at high pH
  • protonated at low pH
  • zwitterion only forms at isoelectric point so amino acid is either acidic or basic
80
Q

Geometrical Isomerism

A
  • or E/Z isomerism
  • type of stereoisomerism
  • refers to positions of substituents at either side of carbon-carbon double bond
81
Q

Suggest advantages and disadvantages of producing esters with acyl chlorides rather than carboxylic acids

A

Advantages
- greater yield/reaction goes to completion (irreversible)
- faster
Disadvantages
- dangerous/ vigorous (exothermic)
- HCl produced rather than water which is toxic
- acyl chlorides are more expensive

82
Q

State by products formed when producing esters from

a) carboxylic acids
b) acyl chlorides
c) acid anhydrides

A

a) water
b) hydrogen chloride
c) carboxylic acid based in acid anhydride

83
Q

Give the structural formula of the first three acid anhydrides

A

methanoic anhydride - (HCO)2O
ethanoic anhydride - (CH3CO)2O
propanoic anhydride - (CH3CH2CO)2O

84
Q

Explain why melting point of aminoethanoic acid is higher than hydroxyethanoic acid

A
  • ionic bonding in aminoethanoic acid due to presence of zwitterions
  • bonding in aminoethanoic acid is stronger than in hydroxyethanoic acid
85
Q

Give the structure of the amide functional group and describe how to name a N-substituted amide

A 
N-C=O
1) N-(alkyl group)
2) name carbon chain and remove "e" from end
3) -amide suffix 
e.g. CH3-CONH-CH2CH2CH3
N-propylethanamide
86
Q

Suggest disadvantages of producing primary amines by nucleophilic addition with KCN followed by hydrogenation

A
  • KCN is toxic/poisonous

- lower yield since two step process

87
Q

Give reagents and conditions to produce hydroxynitriles from carbonyl molecules

A
  • KCN

- aqueous AND ethanolic

88
Q

Explain why bromobenzene does not react with ammonia to produce phenylamine

A
  • benzene ring only attacked by electrophiles
  • ammonia is a nucleophile
  • benzene repels nucleophiles (so nucleophilic substitution cannot occur)
89
Q

Compare different secondary structures of proteins

A
  • alpha helix : polypeptide forms hydrogen bonds with itself
  • beta pleated sheet : hydrogen bonds form between adjacent polypeptides
90
Q

Explain how ionic bonds and disulphide bridges form in tertiary structure of protein

A
  • ionic bonds between -NH3+ and -COO-

- disulphide bridges between S on different R groups in presence of oxygen (eliminate water)

Chemistry A2 (12 decks)

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