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Chemistry IB HL > Topic 10 - Organic Chemistry > Flashcards

Topic 10 - Organic Chemistry Flashcards

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1
Q

alkenes

A
  • have reactive double bonds

- form addition polymers

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2
Q

alkene + water -> ?

A

alcohol

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3
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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4
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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5
Q

catenation

A

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

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6
Q

saturated compounds

A

contain only single bonds

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7
Q

unsaturated compounds

A

compounds containing double or triple bonds

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8
Q

aliphatics

A
  • compounds that don’t contain a benzene ring

- can be saturated or unsaturated

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9
Q

arenes

A
  • compounds that contain a benzene ring

- all are unsaturated

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10
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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11
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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12
Q

addition reaction

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

- characteristic of unsaturated compounds

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13
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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14
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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15
Q

fission

A

bond-breaking reactions

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16
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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17
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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18
Q

homologous series

A

families of compounds used to classify organic compounds

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19
Q

predicting the next member in a homologous series

A

each successive member differs by a -CH2- group

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20
Q

alkanes general formula

A

CnH(2n+2)

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21
Q

alcohols general formula

A

CnH(2n+1)OH

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22
Q

trends in alkane

A
  • increasing boiling point down the group
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23
Q

main features of homologous series

A
  • each successive member differs by a -CH2- group
  • members of a homologous series are represented by the same general formula
  • members show a gradation in physical properties
  • members of a series have similar chemical properties
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24
Q

full structural formula

A
  • shows every bond and atom

- usually 90/180 degree (and 120 degree) angles are used as this is the clearest possible representation in 2D

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25
Q

condensed structural formula

A
  • often omits bonds where they can be assumed
  • groups atoms together
  • contains the minimal info required to describe the molecule unambiguously
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26
Q

stereochemical formula

A
  • shows relative positions of atoms and groups around carbon in 3D
  • bonds sticking forwards are shown as a wedge
  • bonds sticking backwards are shown as hashed lines
  • bonds along the plane of the paper are shown as solid lines
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27
Q

IUPAC nomenclature guidelines

A
  1. Identify the longest straight chain of carbon atoms
  2. Identify the functional group
  3. Identify subchains and substituent groups
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28
Q

substituent groups in organic molecules

A
  • side chains or functional groups in addition to the one used as the suffix
  • they are given as the prefix
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29
Q

structural isomers

A
  • molecules with the same molecular formula but with differing arrangements of atoms
  • each isomer is a distinct compound with distinct physical and chemical characteristics
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30
Q

primary carbon atom

A

attached to:

  • the functional group
  • 2+ H atoms
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31
Q

secondary carbon atom

A

attached to:

  • functional group
  • 1 H atom
  • 2 alkyl groups
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32
Q

tertiary carbon atom

A

attached to:

  • functional group
  • 3 alkyl groups
  • 0 H atoms
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33
Q

arene

A
  • class of compounds derived from benzene (C6H6)

- they form a special branch of organic compounds known as aromatics

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34
Q

how does benzene behave differently from other unsaturated molecules?

A
  • 1:1 ratio of C:H indicates high degree of unsaturation (greater than alkenes/alkynes)
  • but unlike other unsaturated molecules, benzene has no structural isomers
  • benzene is also unwilling to undergo addition reactions
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35
Q

structure of benzene

A
  • cyclic structure
  • a framework of single bonds attaches each C to the one on either side and to a H atom
  • each of the 6 Cs are sp2 hybridized
  • forms 3 sigma bonds with angles of 120 degrees
  • planar shaped
  • stable arrangement
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36
Q

why is benzene’s structure stable?

A
  • there’s 1 unhybridized p e- on each carbon atom
  • their dumbbell shapes are perpendicular to the ring
  • instead of forming discrete alternating pi bonds, they effectively overlap in both directions
  • thus spreading themselves out evenly and forming a delocalized pi e- cloud
  • electron density is concentrated above & below the plane of the ring
  • this lowers the internal energy of the molecule
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37
Q

benzene bond lengths

A
  • all C-C bond lengths in benzene are equal and intermediate in length
  • because each bond contains a share of 3 e-s between the bonded atoms
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38
Q

in what way does the experimental enthalpy of hydrogenation for C6H6 + 3H2 –> C6H12 differ from the theoretical value?

A
  • it has been experimentally proven that benzene is more stable than the Kekule structure predicts
  • this is bc delocalization minimizes the repulsion between electrons
  • this gives benzene a more stable structure by reducing its internal energy by (experimental value - theoretical value) and thus its resonance energy by extension
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39
Q

resonance energy

A
  • AKA stabilization energy

- energy required to overcome the stability of the delocalized ring

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40
Q

benzene reactivity

A
  • benzene is reluctant to undergo addition reactions
  • more likely to undergo substitution reactions
  • bc addition reactions are energetically not favored
  • they would disrupt the entire cloud of delocalized electrons
  • resonance energy would be required
  • without the delocalized ring of e-s, the product would be less stable as well
  • substitution reactions are preferred as they preserve the stable ring structure
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41
Q

why does benzene only have 1 isomer?

A
  • only one isomer exists of each compound
  • because benzene is a symmetrical molecule with no alternating single/double bonds
  • so all adjacent positions in the ring are equal
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42
Q

trends in the physical properties of organic compounds

A
  • a framework consisting of carbon and hydrogen only (this is known as the hydrocarbon skeleton)
  • functional group (differs between homologous series)
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43
Q

how does branching affect the volatility of compounds?

A

more branching = lower boiling point

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44
Q

reactivity of alkanes

A
  • saturated hydrocarbons with strong C-C and C-H bonds
  • thus require high activation energy to break those bonds
  • so alkanes are generally stable under most conditions and can be stored/transported/compressed safely
  • as C-H and C-C bonds are non-polar, they aren’t susceptible to attack by common reactants
  • so alkanes are generally very unreactive
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45
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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46
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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47
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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48
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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49
Q

substitution reaction

A
  • main reaction undergone by alkanes
  • occurs when another reactant (halogen) takes the place of a hydrogen atom in the alkane
  • light-dependent as energy from UV rays is needed to break 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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50
Q

reaction mechanism

A
  • the chain reaction of the substitution reaction

- it occurs as a sequence of steps

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51
Q

substitution reaction mechanism steps

A
  • initiation
  • propagation
  • termination
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52
Q

substitution reaction mechanism: initiation

A
  • AKA photochemical homolytic fission
  • the bond between the 2 halogen atoms in the diatomic halogen molecule is broken
  • photochemical refers to the light dependency of the rxn
  • homolytic refers to the the fact that the 2 products have an equal assignment of bond e-s upon splitting
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53
Q

substitution reaction mechanism: propagation

A
  • AKA chain reaction

- series of reactions that use and produce free radicals

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54
Q

substitution reaction mechanism: termination

A
  • remove free radicals from the rxn mixture

- by causing them to react together and pair up their e-s

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55
Q

reaction used to distinguish between alkanes and alkenes

A
  • bromine water changes from brown to colorless
  • for alkanes: occurs in UV light only
  • for alkenes: will occur rapidly at room temp
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56
Q

general formula of alkanes

A

C(n)H(2n+2)

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57
Q

general formula of alkenes

A

C(n)H(2n)

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58
Q

structure of alkenes

A
  • unsaturated hydrocarbons
  • double bond is made up of 1 sigma and 1 pi bond
  • the C atoms are sp2 hybridized
  • trigonal planar shape
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59
Q

reactivity of alkenes

A
  • more reactive than alkanes
  • as its double bond is the site of reactivity of the molecule
  • the pi bond is broken relatively easily
  • so the double bond can be readily broken for a reaction
  • 2 new bonding positions are created on the C atoms
  • this enables alkenes to undergo addition reactions
  • they can form a range of differing saturated products
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60
Q

addition reaction (alkenes + hydrogen)

A

hydrogen + alkene –> alkane

  • AKA hydrogenation
  • catalyzed by nickel
  • ideal temp: 150°C
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61
Q

applications of 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
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62
Q

concerns about application of hydrogenation

A

trans fats are produced by partial hydrogenation

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63
Q

addition reaction (alkenes + halogens)

A

halogen + alkene –> dihalogenoalkane

  • occur quickly at room temp
  • color change observed (color to colorless)
  • alkene’s double bond is broken and halogen atoms attach to each of the 2 C atoms that were in a double bond
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64
Q

addition reaction (alkenes + hydrogen halides)

A

hydrogen halide + alkene –> halogenoalkane

  • occur rapidly at room temp
  • reactivity in order: HI > HBr > HCl
  • this is because the strength of the halide bond decreases down the group
  • so as HI has the weakest bond it reacts most readily
65
Q

addition reaction (alkenes + water)

A

water + alkene –> alcohol

  • AKA hydration
  • ideal condition when heated with steam
  • conc. H2SO4 as catalyst
  • involves an intermediate/transition state
  • in which the double bonds are broken, then H+ and HSO4- ions bond to the C atoms that were in a double bond
  • final state occurs with hydrolysis and replacement of HSO4- with OH-, and H2SO4 then reforms
66
Q

industrial significance of the hydration of alkenes

A
  • used to manufacture ethanol on a large scale

- as ethanol is a very important solvent

67
Q

addition polymers

A
  • long chains of alkenes
  • forms when alkenes are joined together in addition reactions
  • this occurs bc alkenes can readily break their double bonds to react
68
Q

repeating unit

A
  • used to symbolize polymer structures
  • it’s a single unit with repeating bonds at each end
  • put in a bracket with n in subscript
    e.g. ethene in polymer form is called polythene (polyethene)
    propene in polymer form is polypropylene (polypropene)
69
Q

general formula of alcohols

A

C(n)H(2n+1)

70
Q

alcohol functional group

A

-OH

71
Q

reactivity of alcohols

A

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

72
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
73
Q

incomplete combustion of alcohols

A

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

74
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
75
Q

oxidation of primary alcohols

A
  • primary alcohols are oxidized to form aldehydes, then further oxidized to form carboxylic acids
  • the second oxidation (aldehydes to carboxylic acids) needs to be heated under reflux
76
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

77
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
78
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
79
Q

oxidation of secondary alcohols

A
  • secondary alcohols are oxidized to ketone

- preferably heated under reflux

80
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
81
Q

esterification reaction

A

carboxylic acid + alcohol ester + water

  • reversible reaction
  • type of condensation reaction
  • catalysed by concentrated H2SO4
82
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
83
Q

general formula of halogenoalkanes

A

C(n)H(2n+1)X

84
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
85
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
86
Q

nucleophilic substitution reactions

A
  • reactions in which substitution of the halogen (From halogenoalkane) occurs
  • due to a nucleophile being attracted to the e- deficient C atom
87
Q

reactivity of benzene

A
  • stable because of the delocalized ring of e-s
  • that delocalized ring of e-s represents an area of e- density and is the site of reactivity
  • prefers substitution reactions
  • where a H atom is replaced by an incoming group
  • preferred because the arene ring is conserved
  • doesn’t favor addition reactions
  • as addition reactions would lead to the loss of the stable arene ring
  • this is because the products would have higher energy than the reactants
88
Q

electrophilic substitution reactions

A
  • electrophiles are attracted to the e- rich benzene ring

- results in the substitution of one of the H atoms with an ion from the electrophilic molecule

89
Q

general mechanics of a nucleophilic substitution reaction (halogenoalkane)

A
  • polar C-Halogen bond results in e- deficient C
  • C is therefore attacked by a nucleophilic ion (e.g. OH-)
  • the C-Halogen bond breaks and the halogen is the leaving group (leaves in the form of a halide)
  • as the halide takes both e-s from the bond, this is heterolytic fission
90
Q

heterolytic fission

A

when a covalent bond breaks and one product takes both e-s in the bond

91
Q

leaving group

A

the group that gets displaced in a reaction

92
Q

can water act as a nucleophile in a nucleophilic substitution reaction?

A
  • it can
  • but it lacks a negative charge
  • so it’s a relatively weak nucleophile
  • the rxn occurs more slowly
93
Q

SN2 mechanism

A
  • stands for substitution nucleophilic bimolecular
  • favored by polar, aprotic solvents
  • 1-step concerted reaction with a transition state
  • stereospecific
  • has inversion effect (like umbrella being blown inside out) on the final product

reactants -> transition state -> products

94
Q

bimolecular reaction

A

when the rxn mechanism of a rxn is dependent on both reactants

e.g. for halogenoalkanes
rate = k [halogenoalkane] [nucleophile]

95
Q

stereospecific reaction

A
  • when the 3D arrangement of the reactants determines the 3D arrangement of the products
  • occurs because in transition state, the bond formation occurs before fission
  • so the stereochemistry of the C attacked isn’t lost
96
Q

aprotic solvents

A
  • solvents unable to form hydrogen bonds
  • due to not containing -OH or -NH groups
  • they still may have strong dipoles though
  • in a reaction, they will solvate a metal ion over a nucleophile
  • thus increasing the rate of rxn
97
Q

SN of primary halogenoalkanes

A
  • SN2
  • as H atoms are small, the C atom is open to attack by nucleophiles
  • an unstable transition state forms in which the C atom is weakly bonded to both the halogen and the nucleophile
  • the C-Halogen bond then breaks heterolytically
  • this releases the halogen and forms an alcohol product
98
Q

why does an inversion effect occur on an SN reaction involving a primary halogenoalkane?

A
  • the nucleophile attacks on the opposite side from the leaving group
  • thus the arrangement of atoms become inverted when the halogenoalkane undergoes fission
  • like an umbrella blowing inside out)
99
Q

SN1 mechanism

A

rate = k [halogenoalkane]

  • substitution nucleophilic unimolecular
  • 2 steps
  • the rate-determining step is determined only by the halogenoalkane concentration
  • favored by polar, protic solvents
  • non-stereospecific

reactants –> carbocation intermediate –> products

100
Q

steric hindrance

A
  • when large groups bonded to a specific atom makes it difficult for nucleophiles to attack that atom
  • due to their size
101
Q

carbocation intermediate

A
  • occurs when the C atom breaks its C-Halogen bond (heterolytic fission)
  • and the C atom is left with a temporary +tive charge
  • this phenomenon is called the carbocation intermediate
  • carbocation intermediate molecules are unstable as they don’t fulfil the octet rule
102
Q

positive inductive effect

A
  • when a group has an electron-donating effect

- this helps stabilize an unstable structure

103
Q

SN of tertiary halogenoalkanes

A
  • SN1
  • the tertiary C atom is bonded to 3 alkyl groups and 1 halogen
  • the 3 alkyl groups provide steric hindrance

1st step (slow):

  • the C atom breaks its C-Halogen bond heterolytically
  • carbocation intermediate forms
  • the 3 alkyl groups have a positive inductive effect
  • stabilizes the carbocation to persist long enough for the 2nd step to occur
  • as the carbocation intermediate has a planar shape, the nucleophile can attack from any position
2nd step (fast):
- the C atom bonds with the nucleophile
104
Q

protic solvent

A

e. g. water
- contains an -OH or -NH group
- can form hydrogen bonds
- thus are effective in stabilizing the carbocation intermediate

105
Q

SN of secondary halogenoalkanes

A

can be SN1, can be SN2

106
Q

factors affecting rate of SN

A
  • type of SN
  • polarity of C-Halogen bond
  • strength of C-Halogen bond
  • choice of solvent
107
Q

factors affecting rate of SN: type of SN

A
  • SN1 faster than SN2

- overall speeds: tertiary > secondary > primary

108
Q

factors affecting rate of SN: polarity of C-Halogen bond

A
  • less polar bonds would result in C being less e- deficient
  • thus the more polar the bonds, the more vulnerable C is to a nucleophilic attack
  • overall speeds: fluoroalkane > … > iodoalkane
109
Q

factors affecting rate of SN: strength of C-Halogen bond

A
  • less reactive halogen = weaker bond

- overall speeds: iodoalkane > … > fluoroalkane

110
Q

factors affecting rate of SN: clash between polarity and strength of C-Halogen bond

A
  • the strength dominates the polarity

- so overall speeds: iodoalkane > … > fluoroalkane

111
Q

factors affecting rate of SN: choice of solvent

A
  • SN1 favored by polar protic solvents

- SN2 favored by polar aprotic solvents

112
Q

how to identify the end of an SN reaction

A
  • the halide may appear
  • if AgNO3 is added, it will react with the halide
  • forms a silver halide precipitate with a distinct color
113
Q

nature of C=C bonds in alkenes

A
  • Cs are sp2 hybridized
  • they form a planar triangular shape
  • bond angle 120
  • open structure, makes it easy for incoming groups to attack
  • 1 central sigma bond and 2 pi bonds above & below
  • as pi bonds are areas of e- density, they attract electrophiles
  • pi bonds are much weaker than sigma bonds so they break a lot more easily during addition reactions
114
Q

electrophilic addition reactions

A
  • reactions in which an electrophile molecule attacks 2 double-bonded C atoms
  • breaks one of the bonds so the = bond becomes single bond
  • the molecule will split to attach to each C atom
115
Q

electrophilic addition reactions: alkene + halogen (symmetric)

A

alkene + halogen –> dihalogenoalkane

e. g. ethene + bromine –> dibromoethene
- ethene gas is bubbled through bromine

step 1 (slow):

  • although bromine is non-polar, a temporary dipole can be induced by electron repulsion as bromine approaches the pi bond (which is e- dense)
  • the bromine closer to the pi bond gains a partial +tive charge and splits heterolytically
  • the +tive bromine acts as an electrophile
  • results in unstable carbocation intermediate that doesn’t follow octet rule
step 2 (fast): 
- the carbocation reacts with the -tive bromide ion
116
Q

electrophilic addition reactions: evidence for the bromine-splitting reaction mechanism

A
  • rxn between ethene + bromine in the presence of Cl-
  • no dichloro compounds formed
  • thus confirming that the initial attack was by the electrophile Br+
117
Q

electrophilic addition reactions: alkene + hydrogen halide (symmetric)

A

alkene + hydrogen halide –> halogenoalkane

  • ethene gas is bubbled through hydrogen bromide
  • same mechanism as alkene + halogen
118
Q

electrophilic addition reactions: alkene + hydrogen halide (asymmetric)

A

alkene + hydrogen halide –> halogenoalkane

  • there are multiple potential stereoisomers of each halogenoalkane depending on the stereoisomer of alkene and on which C atom is attacked by the electrophile
  • as all alkenes after ethene have stereoisomers
  • each stereoisomer reactant has 2 potential stereoisomer products (although only 1 will be produced)
  • if the electrophile attacks a primary C, in carbocation state there’s a positive inductive effect from the 1 alkyl group
  • positive inductive effect doubles for secondary C and triples for tertiary C, etc
  • the more stable the carbocation, the more likely it is that the reaction will complete with a halogenoalkane
119
Q

Marknovnikov’s Rule for electrophilic addition reactions (asymmetric alkene + hydrogen halide)

A

the more electropositive part of the reacting species will always bond to the C atom that has the least bonds with other C atoms

120
Q

why does benzene undergo electrophilic substitution reactions?

A
  • despite high unsaturation, benzene doesn’t behave like alkenes
  • as its stable aromatic ring will be altered in addition reactions, substitution is preferred
  • but it’s still attracted to electrophiles
  • so benzene undergoes electrophilic substitution
121
Q

electrophilic substitution reactions: benzene

A
  • mostly specific to arenes
  • this reaction requires high activation energy
  • benzene’s delocalized cloud of pi electrons seeks electrophiles and forms a new bond as a H atom is lost

1st step (slow):

  • e- pair from benzene is attracted to electrophile
  • symmetry of delocalized pi e-s is disrupted
  • unstable carbocation intermediate forms
  • both the entering group and leaving group are temporarily bonded

2nd step (fast):

  • leaving group leaves
  • 2 e-s from the C-H bond move to regenerate the symmetry of the aromatic ring
122
Q

nitrating mixture

A

mixture of concentrated nitric acid and sulfuric acid

123
Q

electrophilic substitution reactions: nitration of benzene

A
  • substitution of H by -NO group
  • catalyst: conc H2SO4
  • ideal temp: 50°C
  • nitronium is used as electrophile
  • generated by using a nitrating mixture at 50°C
  • sulfuric acid is stronger so it protonates nitric acid
  • the nitrating mixture then loses H2O to form nitronium ions
  • NO2+ is a strong electrophile and reacts with pi e-s to form carbocation intermediate
  • the loss of a proton (H+) leads to reformation of the arene ring
  • final product: C6H5NO2 + H2O
124
Q

carbonyl compounds

A

compounds containing C=O groups

125
Q

redox reactions: reduction of carbonyl compounds

A

oxidation of:
primary alcohol –> aldehyde –> carboxylic acid
secondary alcohol –> ketone

these oxidation reactions can be reversed using:

  • sodium borohydride (NaBH4) in aqueous/alcoholic soln
  • lithium aluminium hydride (LiAlH4) in anhydrous conditions
  • both reagents produce a hydride ion (H-) to act as a nucleophile
  • NaBH4 is safer but not strong enough to reduce carboxylic acids
  • ## LiAlH4 can reduce carboxylic acids
126
Q

redox reactions: reduction of nitrobenzene

A
  • nitrobenzene can be reduced to form phenylamine
  • 2-stage reduction process
    • C6H5NO2 reacts w a mixture of tin (Sn) and conc HCl
      - heated under reflux (boiling water bath)
      - produces C6H5NH3+ (phenylammonium ions) which is protonated because of the acidic conditions
    • reacted w NaOH to remove H+
      - forms C6H5NH (phenylamine)
127
Q

synthetic route

A
  • series of discrete steps used to convert compounds from one form to another
  • by organizing reactions in a sequence so that the product of one reaction is the reactant of the next reaction
  • its basis lies in the interconversions of functional groups
128
Q

retro-synthesis

A
  • technique for working out synthetic routes

- works backwards from the target molecule through precursors all the way to the starting material

129
Q

isomer

A

compounds with the same molecular formula but different arrangement of atoms

130
Q

types of isomerism

A
  • structural isomerism

- stereoisomerism

131
Q

structural isomerism

A

atoms and functional groups are attached in different ways

132
Q

stereoisomerism

A
  • similar attachments

- different spatial (3D) arrangements

133
Q

types of stereoisomerism

A
  • configurational isomerism

- conformational isomerism

134
Q

configurational isomerism

A

can only be interconverted by breaking covalent bonds

135
Q

conformational isomerism

A
  • can be interconverted by free rotation about sigma bonds
  • usually spontaneously interconvert via rotation so cannot be isolated separately
  • but some conformers are more stable than others (these are favored)
136
Q

types of configurational isomerism

A
  • cis-trans or E-Z isomerism

- optical isomerism

137
Q

cis-trans or E-Z isomerism

A
  • exists where rotation is restricted around atoms

- so that they become fixed in space relative to each other

138
Q

cis isomers

A

isomers with the same groups on the same side of the double bonds/ring (i.e. reference plane)

139
Q

trans isomers

A

isomers with the same groups on opposite sides of the double bonds/ring (i.e. reference plane)

140
Q

reference plane

A

the plane of the ring in a cyclic compound

141
Q

situations where cis-trans or E/Z isomerism may arise

A

DOUBLE BONDS

  • if the double bond consists of 1 sigma and 1 pi bond (the pi bond must be formed by sideways overlap of p orbitals)
  • thus free rotation is not possible as the p orbitals would be pushed out of position, breaking the pi bond

CYCLIC MOLECULES

  • most cycloalkanes exhibit aromaticity and the ring of C atoms restricts rotation
  • bond angles are strained from the tetrahedral angles (e.g. cyclobutane has bond angles of 90°)
  • when the molecule contains 2+ different groups attached to the double bonds of the ring, they can be arranged to give 2 different isomers (cis or trans)
  • substituted groups don’t have to be on adjacent carbon atoms (it’s their position relative to the reference plane that matters)
142
Q

E/Z isomers

A

used where cis-trans can’t explain the structure

e.g. if all groups attached to the double bond are different, or if the C atoms are bonded to 2+ different substituents

143
Q

Cahn-Ingold-Prelog rules of priority

A
  • out of all the atoms bonded to C-atoms of the = bond, higher atomic number = higher priority
  • longer carbon chains have higher priority
    e. g. C3H7 > C2H5 > CH3 > H & Br > Cl > F
  • E-isomers: higher priority groups on opposite sides of the = bond
  • Z-isomers: higher priority groups on the same side of the = bond
144
Q

chiral molecules

A

AKA stereocentre OR asymmetric compounds

  • when a C atom is attached to 4 different atoms/groups
  • the 4 attached atoms/groups have bond angles of 109.5°
  • the compound has no plane of symmetry so is non-superimposable
145
Q

optical isomerism

A
  • a chiral molecule has 2 alternative 3D configurations (both mirror images of each other)
  • they are said to be optical isomers
  • most molecules have multiple chiral atoms so they’ll have more than just 2 optical isomers
146
Q

non-superimposable mirror images (not a flashcard)

A
  • think of it like hands
  • hands are mirror images but non-superimposable
  • as the fingers and thumbs don’t line up
  • similarly, isomers of chiral molecules are mirror images yet non-superimposable
147
Q

types of optical isomers

A
  • enantiomers

- diastereomers

148
Q

enantiomers

A
  • optical isomers
  • mirror images
  • they have opposite configurations at each chiral centre
149
Q

racemic mixture

A

AKA racemate

  • mixture containing equal amounts of 2 enantiomers (must be mirror images of each other)
  • optically inactive
150
Q

diastereomers

A
  • molecules that have opposite configurations at 1+ (but not all) chiral centres
  • so not mirror images of each other
  • common sugars are diastereomers of each other (e.g. glucose, galactose)
151
Q

properties of enantiomers

A
  • optically active

- reacts with other chiral molecules

152
Q

plane-polarized light

A
  • light that has been passed through a polarizer
  • so that only light oscillating along a specific plane passes through
  • while all other planes are blocked out
153
Q

properties of enantiomers: optical activity

A
  • optical isomers will show a difference in a specific interaction with light
  • when plane-polarized light is passed through optical isomers, the plane of polarization is rotated
  • different enantiomers of the same compound (& at same concs) will rotate plane-polarized light in equal amounts but in opposite directions
154
Q

what can be used to detect optical activity?

A
  • polarimeter
  • measures direction of rotation & how much the light has been rotated

PROCEDURE

  • light is plane-polarized
  • then passed through a solution
  • then passed through a second polarizer (the analyser)
  • the analyser is rotated until the light can pass through it (thus identifying the extent and direction of rotation)
155
Q

symbols used in classifying optical activity

A
  • rotation to the right is denoted by +

- vice versa for left (-)

156
Q

properties of enantiomers: reactivity with other chiral molecules

A
  • if a racemic mixture reacts with an enantiomer of another chiral compound, the 2 enantiomers of the racemic mixture will react to produce diff products
  • the products have distinct chemical and physical properties and thus can be easily separated
  • this is one way to separate a racemic mixture into its enantiomer constituents
157
Q

importance of the differing reactivity between a pair of enantiomers

A
  • biological systems are chiral environments
  • when ingesting drugs, one enantiomer may produce a helpful product while the other may produce a life-endangering product
158
Q

separating a racemic mixture into its constituents

A
  • if a racemic mixture reacts with an enantiomer of another chiral compound, the 2 enantiomers of the racemic mixture will react to produce diff products
  • the products have distinct chemical and physical properties and thus can be easily separated

Chemistry IB HL (11 decks)

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