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Chemistry IB HL (2016+) > Topic 10: Organic Chemistry > Flashcards

Topic 10: Organic Chemistry Flashcards

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1
Q

affixes for number of carbons

A 
1 = meth-
2 = eth-
3 = prop-
4 = but- 
5 = pent-
6 = hex-
7 = hept-
8 = oct-
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2
Q

affixes for bonding

A

single bond: -an-
double bond: -en-
triple bond: -yn-

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3
Q

affix of alkane

A

-ane

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4
Q

affix of alkene

A

-ene

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5
Q

affix of alkyne

A

-yne

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6
Q

affix of alcohol

A

-ol

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7
Q

affix of carboxylic acid

A

-oic acid

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8
Q

affix of ether

A

R1 = -oxy
R2 = -ane
R2 must be the longest chain (so it’ll be used as the parent name)

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9
Q

affix of halide

A

chloro-/bromo-/iodo-

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10
Q

affix of aldehyde

A

-al

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11
Q

affix of ketone

A

-one

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12
Q

affix of ester

A

R1: alkyl group name (e.g. methyl, ethyl)
R2: -oate

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13
Q

alkene + water -> ?

A

alcohol

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14
Q

trends in the alkene homologous series

A
  • increase in b.pt down the homologous series
  • increase in strength of Van der Waals/London/dispersion forces
  • increase in size of molecule/number of electrons
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15
Q

features of a homologous series

A
  • same general formula
  • successive members differ by a CH2 chain
  • same functional group
  • similar chemical properties
  • gradual change in physical properties (e.g. m.pt/b.pt)
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16
Q

catenation

A

carbon’s ability to link itself to form chains and rings

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17
Q

saturated compounds

A

contain only single bonds

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18
Q

unsaturated compounds

A

compounds containing double or triple bonds

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19
Q

aliphatics

A
  • compounds that don’t contain a benzene ring

- can be saturated or unsaturated

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20
Q

arenes

A
  • compounds that contain a benzene ring

- all are unsaturated

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21
Q

electrophile

A
  • electron-deficient species
  • attracted to electron-rich parts of molecules
  • positive ions or at least have partial positive charge
  • act as lewis acids
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22
Q

nucleophile

A
  • electron-rich species
  • attracted to parts of molecules that are electron-deficient
  • nucleophiles have a lone pair of e-s and may also have negative charge
  • act as lewis bases
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23
Q

effect of branching on b.pt

A
  • molecules become more spherical due to branching

- ↓ contact SA = ↓ no. of London dispersion forces = ↓ boiling point

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24
Q

organic compounds with London dispersion forces as their strongest intermolecular force

A
  • alkane
  • alkene
  • alkyne
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25
Q

organic compounds with dipole-dipole forces as their strongest intermolecular force

A
  • ester
  • aldehyde
  • ketone
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26
Q

organic compounds with hydrogen bonding as their strongest intermolecular force

A
  • amine
  • alcohol
  • carboxylic acid
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27
Q

factors affecting solubility in water of organic compounds

A
  • polarity of functional group:

- chain length:

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28
Q

factors affecting solubility in water of organic compounds: chain length

A
  • hydrocarbon chain is non-polar
  • and non-polar substance prefer to dissolve in non-polar solvents
  • lower members of alcohols, amines, aldehydes, ketones, and carboxylic acids are water soluble
  • but as the hydrocarbon chain increases in length, solubility in water decreases
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29
Q

ideal solvent for organic compounds

A

propan-1-ol

  • as it contains both polar and non-polar compounds
  • so it can dissolve both types (to some extent)
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30
Q

reactivity of alkanes

A
  • saturated hydrocarbons with strong C-C and C-H bonds
  • as C-H and C-C bonds are non-polar, they aren’t susceptible to attack by common reactants
  • so alkanes are generally stable under most conditions and can be stored/transported/compressed safely
  • only readily undergo combustion reactions with oxygen
  • only undergoes substitution rxns with halogens in UV light
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31
Q

combustion of alkanes

A
  • alkanes release a significant amount of energy when broken
  • so they are widely used as fuels
  • alkane combustion reactions are highly exothermic because of the large amounts of energy released when forming CO2 and H2O
  • the products are fully oxidized, so alkane undergoes complete combustion
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32
Q

combustion of hydrocarbons in limited oxygen conditions

A
  • incomplete combustion
  • CO and H2O produced instead
  • in extreme oxygen limitation, just C and H2O will be produced
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33
Q

combustion of hydrocarbons

A
  • under complete or incomplete combustion depending on oxygen availability
  • large amounts of energy are released generally
  • as C:H ratio increases with unsaturation, so does the smokiness of the flame due to unburned hydrocarbon
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34
Q

why is the combustion of hydrocarbons a problem?

A
  • CO2 and H2O are greenhouse gases
  • thus they contribute to global warming and climate change
  • CO is a toxin as it combines irreversibly with blood hemoglobin, preventing it from carrying oxygen
  • unburned carbon is released into the air as particulates
  • they have a direct effect on human health
  • they also catalyze the formation of smog in polluted air
  • they are also the source of global dimming
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35
Q

substitution reaction

A
  • main reaction undergone by alkanes
  • occurs when another reactant (halogen) takes the place of a hydrogen atom in the alkane
  • UV rays needed to provide energy to break the covalent bonds in the halogen molecule
  • energy splits the halogen molecule into free radicals
  • the radicals start a chain reaction in which a mixture of products (including the halogenoalkane) is formed
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36
Q

homolytic fission

A
  • when a covalent bond breaks by splitting the shared pair of e-s between the 2 products
  • produces 2 free radicals, each with 1 unpaired e-
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37
Q

heterolytic fission

A
  • when a covalent bond breaks and the shared pair of e-s go to one product
  • produces 2 oppositely-charged ions
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38
Q

free radical substitution

A
  1. Initiation:
    Cl2 → 2Cl•
    - in UV light
  2. Propagation: (formation of new radicals)
    CH4 + Cl• → CH3• + HCl
    CH3• + Cl2 → CH3Cl + Cl•
  3. Termination (when 2 radicals react together)
    CH3• + Cl• → CH3Cl
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39
Q

addition reaction

A
  • occurs when 2 reactants combine to form a single product

- characteristic of unsaturated compounds

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40
Q

substitution reaction

A
  • occurs when 1 atom or group of atoms in a compound is replaced by a different atom or group
  • characteristic of saturated and aromatic compounds
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41
Q

addition-elimination reaction

A
  • AKA condensation reaction
  • occurs when 2 molecules join together (addition) and in the process small molecules are lost (elimination)
  • reaction occurs between a functional group in each reactant
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42
Q

addition reaction (alkene + H2)

A

alkane

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43
Q

addition reaction (alkene + hydrogen halide)

A

halogenoalkane

44
Q

addition reaction (alkene + halogen)

A

dihalogenoalkane

45
Q

addition reaction (alkene + water)

A

alcohol

catalyst: H2SO4

46
Q

uses of addition reactions

A
  • bromination
  • hydrogenation
  • hydration
47
Q

uses of addition reactions: bromination

A
  • reaction used to distinguish between alkanes and alkenes
  • bromine water changes from brown to colorless
  • for alkanes: occurs in UV light only
  • for alkenes: will occur rapidly at room temp
48
Q

uses of addition reactions: hydrogenation

A
  • used in margarine industry
  • converts unsaturated hydrocarbon chains into saturated compounds with higher melting points
  • so that margarine is solid at room temp
49
Q

uses of addition reactions: hydration

A
  • can be used to form an important solvent (ethanol)

- addition of steam to ethene

50
Q

addition polymerization reactions

A
  • under certain conditions, alkenes can undergo addition reactions with themselves
  • to form a long chain polymer made up of thousands of C atoms
  • can also be extended to other substituted alkenes to give a wide variety of different addition polymers
51
Q

reactivity of alcohols

A

as the -OH group is polar, it increases alcohols’ solubility in water relative to alkanes

52
Q

combustion of alcohols

A
  • like hydrocarbons, alcohol combustion produces significant amounts of energy along with CO2 and H2O
  • the amount of energy released per mole increases down the homologous series
53
Q

incomplete combustion of alcohols

A

like alkane, the incomplete combustion of alcohol produces CO and H2O instead of CO2 and H2O

54
Q

oxidation of alcohols

A
  • combustion completely oxidizes alcohol molecules
  • but alcohols can also react with oxidizing agents that selectively oxidize the O atom in the -OH group
  • this keeps the carbon skeleton intact
  • allows alcohols to be oxidized into other organic compounds
  • the most common oxidant used as acidified potassium dichromate (VI) [K2Cr2O7], and the bright orange solution is reduced to just green Cr (III) over the course of the rxn
  • oxidants are often represented as + [O] in diagrams
55
Q

oxidation of primary alcohols

A
  • primary alcohols are oxidized to form aldehydes, then further oxidized to form carboxylic acids
  • typically oxidized by acidified K2Cr2O7 (orange –> green)
  • the second oxidation (aldehydes to carboxylic acids) needs to be heated under reflux
56
Q

why can’t wine be left exposed to air?

A
  • bacteria will slowly oxidize the ethanol to ethanoic acid

- i.e. alcohol to carboxylic acid

57
Q

obtaining aldehydes from primary alcohols

A
  • the oxidation of primary alcohols can be stopped on the first step of oxidation (alcohols to aldehydes)
  • distillation can be utilized to remove the aldehydes from the reaction mixture
  • this is possible because aldehydes have lower boiling points than alcohols and carboxylic acids
58
Q

obtaining carboxylic acids from primary alcohols

A
  • simply leave the aldehyde alone with the oxidant for an extended period of time
  • preferably heated under reflux
59
Q

oxidation of secondary alcohols

A
  • secondary alcohols are oxidized to ketone
  • preferably heated under reflux
  • typically oxidized by acidified K2Cr2O7 (orange –> green)
60
Q

oxidation of tertiary alcohols

A
  • tertiary alcohols don’t readily oxidize
  • their carbon structure needs to be broken and this requires a lot more energy than the oxidation of other alcohols
  • so unlike the other 2 oxidations, oxidizing tertiary alcohols with K2Cr2O7 won’t result in a color change as there is no reaction
61
Q

esterification reaction

A

carboxylic acid + alcohol → ester + water

  • reversible reaction
  • type of condensation reaction
  • catalysed by concentrated H2SO4
  • the acid becomes the -oate while the alcohol becomes the alkyl group
62
Q

separating and distinguishing esters in an esterification reaction

A
  • as the ester has the lowest boiling point, they can be removed via distillation
  • the presence of esters can be identified by their distinct fruity, sweet smell
  • as they don’t have -OH groups (unlike the reactants), they can’t form hydrogen bonds and will remain as an insoluble layer on the surface of water
63
Q

structure of halogenoalkanes

A
  • basically like alkanes except 1 halogen atom is substituted for 1 of the hydrogen atoms
  • as halogenoalkanes are saturated molecules, they undergo substitution reactions
64
Q

reactivity of halogenoalkanes

A
  • halogenoalkanes contain a polar bond
  • this makes them more reactive than alkanes
  • the halogen atom is more electronegative than the C atom
  • so they exert a stronger pull on the shared e-s in the C-H bond
  • thus halogen exerts a partial -tive charge while the C atom exerts a partial +tive charge
  • the C atom is then said to be e- deficient
  • this e- deficient C atom defines much of a halogenoalkane’s reactivity
65
Q

organic importance of nucleophilic substitution

A

allows organic compounds to be converted into many other classes of organic compounds

66
Q

why do carboxylic acids have higher boiling points than alcohols?

A
  • both carboxylic acids and alcohols form H bonds as well as dipole-dipole interactions and London dispersion forces
  • however, H bonding can occur between 2 carboxylic acids twice (as their functional group have H and OH)
67
Q

why is benzene unusual among unsaturated compounds?

A
  • it’s an alkene that’s less saturated than alkynes (its C:H ratio is 1:1)
  • it has no isomers
  • it doesn’t like addition reactions
68
Q

importance of benzene

A

used in the synthesis of drugs, dyes, and plastics

69
Q

what makes benzene such a stable compound?

A
  • all C atoms are sp2 hybridized
  • at “ring level” they form sigma bonds of 120°
  • each C contains one unhybridized p orbital that spread out evenly and overlap to form pi bonds above and below “ring level”
  • this forms delocalized pi e- clouds with e- density concentrated in donut-shaped rings parallel to the plane of the ring (above and below)
  • this lowers the molecule’s internal energy, making it more stable
70
Q

bond lengths of benzene

A
  • all C to C bond lengths are equal (both - and = bonds)

- each bond contains a share of 3 e-s between the bonded atoms

71
Q

why is benzene’s ΔH(hyd) lower than standard enthalpy values would suggest?

A
  • benzene is very stable…
  • delocalization of e-s minimizes repulsion
  • this lowers internal energy by the amount of difference calculated between the actual ΔH(hyd) value and the theoretical value
  • this effect is called resonance energy/stabilization energy
72
Q

why does benzene prefer substitution reactions despite being unsaturated?

A
  • addition reactions would disrupt the delocalized pi e- clouds
  • this would involve supplying the resonance energy needed to disrupt the cloud
  • plus without the cloud, the product would be less stable
  • substitution reactions would preserve the delocalized e- clouds and therefore retain stability
73
Q

why does benzene have no isomers?

A
  • benzene is symmetrical with alternating single and double bonds
  • so all adjacent positions in the ring are equal
74
Q

SN1

A
  • nucleophilic substitution unimolecular
    rate = k [halogenoalkane]
  • undergone by tertiary halogenoalkanes
  • forms carbocation intermediate before forming final product
  • tertiary halogenoalkanes have 3 alkyl groups attached to the central C atom, which causes steric hindrance (this makes it difficult for nucleophiles to attack)
  • so instead, the halogenoalkane typically detaches the halide heterolytically, leaving an opening for attack
  • in the carbocation intermediate state before the nucleophile attacks, the 3 alkyl groups have a stabilizing (positive inductive) effect to help the carbocation remain in the unstable state before being attacked
  • in total forms 2 transition states (between reactant –> carbocation intermediate, then between carbocation intermediate –> product)
75
Q

SN2

A
  • nucleophilic substitution bimolecular
    rate = k [halogenoalkane][nucleophile]
  • undergone by primary halogenoalkanes
  • forms unstable transition state before forming final product
  • stereospecific: the geometric arrangement of the product is the inverse of the reactant
  • favored by polar, aprotic solvents i.e. polar solutions that can’t form H bonds (e.g. propanone)
76
Q

NOTES: how to tell if an organic compound is primary/secondary/tertiary

A
  • look at the C atom attached to the functional group
  • the number of C atoms attached to it tells you whether it’s primary/secondary/tertiary (e.g. the C atom attached to a primary alcohol’s functional group has a bond with one C atom)
77
Q

NOTES: how to tell if an organic compound is chiral

A
  • if a chiral carbon is present, the compound is chiral
  • a chiral carbon is a C atom with 4 different groups attached
  • (CH3)3 does NOT count as “different groups”
78
Q

factors affecting SN rates

A
  • effect of mechanism (SN1 vs SN2): SN1 is faster, so rate of reaction: tertiary > secondary > primary
  • secondary alcohols undergo a mix of SN1 and SN2
  • polarity of C-X bond: more polar = the C is more e- deficient = more vulnerable to attack (F > … > I)
  • strength of C-X bond: weaker bond = easier to overcome = more vulnerable to attack (I > … > F)
  • effect of strength of bond > effect of polarity of bond, so iodoalkanes are generally favored for faster SN reaction rates
79
Q

electrophilic addition reaction

A
  • undergone by alkenes

-

80
Q

characteristics of alkenes that allow them to undergo many synthetic pathways

A
  • the atoms of the C=C bond are sp2 hybridized
  • planar triangular bond geometry with angle of 120
  • pi bond creates 2 areas of electron density (one above and one below the plane of the bond axis)
  • pi bond e-s are attractive to electrophiles
  • the pi bond e-s are much weaker than the sigma bond e-s so they’re much more easily broken during addition
81
Q

mechanism of ethene + Br2

A
  • bromine becomes polarized by the e- rich region of the alkane (due to e- repulsion)
  • Br2 splits heterolytically to form Br+ and Br-
  • Br+ attacks the pi bond and breaks it, bonding with a C atom
  • the other C atom is left with an incomplete octet, forming an unstable carbocation intermediate
  • it reacts with the Br- to form 1,2-dibromoethane
82
Q

mechanism of ethene + HBr

A
  • HBr is already polar so it doesn’t get polarized by the e- rich region of the alkane
  • HBr splits heterolytically to form H+ and Br-
  • H+ attacks the pi bond and breaks it, bonding with a C atom
  • the other C atom is left with an incomplete octet, forming an unstable carbocation intermediate
  • it reacts with the Br- to form bromoethane
83
Q

considering unsymmetric electrophilic additions (mechanism of propane + HBr)

A
  • 2 possible pathways that produce isomers of bromopropane
  • remember that the preferred pathway is the one that gives a more stable carbocation intermediate
  • also remember that alkyl groups around a C atom give a stabilizing effect
  • tip: H+ will bond to the C atom that is already bonded to the greater no of Hs
84
Q

why does benzene undergo electrophilic substitution?

A

because its highly stable aromatic ring prefers substitutions, but its e- dense regions are attractive to electrophiles

85
Q

electrophilic substitution mechanism

A

slow rxn: benzene + E+ –> non-aromatic carbocation with the +tive charge distributed over the molecule
fast rxn: non-aromatic carbocation –> benzene with E + H+
- the loss of the H+ ion restores the product’s electrically neutral state
- because 2 e-s from the C-H bond move to regenerate the aromatic ring
- the product of an electrophilic substitution is always more stable than the reactant

86
Q

nitrating mixture

A
  • mixture of conc H2SO4 and conc HNO3
  • as the stronger acid, H2SO4 protonates HNO3 to H2NO3+
  • H2NO3+ then breaks down to produce H2O + NO2 +
  • this is used to supply NO2 + for the nitration of benzene
87
Q

nitration of benzene

A
  • electrophilic substitution that adds -NO2 in place of -H
  • the NO2 is generated using a nitrating mixture
    CONDITIONS: heat at 50°C with conc H2SO4 (part of the nitrating mixture)
  • the released H+ reforms HSO4 - back to H2SO4
88
Q

reduction of carbonyls

A
  • carbonyl compounds: those with the C=O group (i.e. alcohols, ketones, aldehydes, and carboxylic acids)
  • their oxidations can be reversed using NaBH4 (aqueous/alcoholic solutions) or LiAlH4 (anhydrous conditions)
89
Q

when should NaBH4 or LiAlH4 be used?

A
  • NaBH4 is generally preferred as it’s safer

- but LiAlH4 is used where NaBH4 is not strong enough (with carboxylic acids)

90
Q

reduction with NaBH4

A

heat aldehydes/ketones in the presence of NaBH4

91
Q

reduction with LiAlH4

A
  • heat carboxylic acids in the presence of LiAlH4 and dry ether
  • the rxn can’t be stopped at the aldehyde stage and will proceed until the primary alcohols are synthesized
  • this is bc aldehydes react too readily with LiAlH4
92
Q

reduction of nitrobenzene

A
  1. React C6H5NO2 with Sn and HCl to form C6H5NH3 +

2. React C6H5NH3 + with NaOH. The OH- group will remove the additional H to form H2O and C6H5NH2 (phenylamine)

93
Q

cis isomers

A

isomers with the same groups on the same sides of the bond plane

94
Q

trans isomers

A

isomers with the same groups on opposite sides of the bond plane

95
Q

E isomers

A

isomers with the two highest priority groups on opposite sides of the bond plane

96
Q

Z isomers

A

isomers with the two highest priority groups on the same sides of the bond plane

97
Q

Cahn-Ingold-Prelog rules of priority

A
  • for atoms: higher atomic no = higher priority
    Br > Cl > F
  • for alkyl groups: longer chain = higher priority
    C3H7 > C2H5 > CH3
98
Q

characteristics of chiral compounds

A
  • asymmetric/stereocentre

- can be arranged in 2 different 3D configurations that are mirror images of each other (optical isomerism)

99
Q

enantiomers

A
  • optical isomers
  • i.e. mirror images of each other
  • they have opposite configurations at all their chiral centres
100
Q

racemate

A

AKA racemic mixture

- mixtures containing equal amounts of 2 enantiomers

101
Q

diastereomers

A
  • many chiral compounds have more than one chiral centre
  • enantiomers are when optical isomers have opposite configs at every chiral centre
  • diastereomers are when isomers have opposite configs at more than one (BUT NOT ALL) chiral centre
  • therefore they are not mirror images of each other
102
Q

properties of diastereomers

A

unlike enantiomers, they differ in physical and chemical properties

103
Q

properties of enantiomers

A
  • optical activity: when a beam of plane-polarized light is passed through a solution containing enantiomers, the plane of the polarized light is rotated by a certain angle
  • a pair of enantiomers will react differently with a common reactant (e.g. a single enantiomer of another compound)
104
Q

polarimeter

A
  • measures the amount and direction of rotation when a beam of plane-polarized light is passed through a solution of enantiomers
  • ordinary light is passed through a polarizer, then through the solution, then through an analyzer
  • the results are used to compare different solutions
  • separate solutions of enantiomers with similar concentrations will rotate plane-polarized light in equal amounts but in opposite directions
  • thus, a racemic mixture is optically inactive
105
Q

why is the differing reactivity of a pair of enantiomers to other chiral compounds significant?

A
  • biological systems are chiral compounds
  • e.g. one enantiomer in a drug can be therapeutic while the other can cause malformations in a foetus
  • this resulted in the invention of asymmetric synthesis, a process for the manufacture of a single enantiomer using a chiral catalyst

Chemistry IB HL (2016+) (13 decks)

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