135A organic - substitution, elimination and addition Flashcards

1
Q

Alcohols are bad leaving groups. What can they be activated with (protonation of OH)

A

Acids/ HCl

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2
Q

how many substances are involved in the RDS

A

2 (varying the conc of both changes the rate)

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3
Q

how many substances are involved in the RDS

A

1(one one conc changes rate)

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4
Q

What areas on the molecule could nucleophilic sub (OH-) occur

A
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5
Q

What areas on the molecule could nucleophilic sub (OH-) occur

A
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6
Q

What areas on the molecule could nucleophilic sub (OH-) occur

A
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7
Q

What areas on the molecule could nucleophilic sub (OH-) occur

A

nowhere on the molecule

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8
Q

what does SN1 stand for

A
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9
Q

does SN1 have a carbonation intermediate

A

yes

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10
Q

primary/ secondary/ tertiary undergos SN1

A

tertiary (+ stable secondary cations with conjugation)

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11
Q

What is a carbocation structure

A

planar and sp2 hybridised carbon

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12
Q

What makes a carbocation stable? / conjugation and hyper conjugation

A
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13
Q

Could these molecules undergo SN1

A

1- (slow but) has conjugation due to double bond so could form a stable primary carbocation
2- very fast; conjugation and methyl groups

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14
Q

What factors effect wether something can undergo SN1 or SN2

A
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15
Q

Why do SN1 reactions give racemic mixtures

A

because the nucleophile has equal chance of attacking from the above or below/ planar carbocation

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16
Q

Which carbocation is the nucleophile more likely to attack/ which product will be the major

A

nucleophile will attack least sterically hindered/ primary will be major product

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17
Q

What does SN2 stand for

A
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18
Q

Does an SN2 have a carbocation intermediate

A

no- has transition state

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19
Q

What energy determines how fast a reaction will go

A

activation energy (higher Ea the slower the reaction)

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20
Q

In substitution how does the LUMO and HOMO of the reactants interact (nucleophile and electrophile)

A

HOMO of nucleophile and LUMO of electrophile. For a Y- the HOMO is the lone pair

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21
Q

The HOMO of the nucleophile goes to what bonding orbital of the electrophile

A

the σ* anti bonding orbital (when electrons fill it the bond breaks)

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22
Q

In SN2, what angle must the nucleophile and electrophile orbitals align in order to react (transition state where they leave and join simultaneously)

A

180°

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23
Q

primary/ secondary/ tertiary undergos SN2

A

primary and some secondary (minimal steric hinderance so nucleophile can get to σ* c-x bond)

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24
Q

Of primary carbocations what will increase the rate of SN2 substitutions

A

Reaction is faster if the carbocation is more stable/ conjugation

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25
Q

What is SN2 inversion

A

The leaving group and going group will not be of same orientation if the reaction takes place at the chiral centre

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26
Q

What is SN2 inversion

A

The leaving group and going group will not be of same orientation if the reaction takes place at the chiral centre

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27
Q

What would be the transition state for this reaction

A
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28
Q

Why are primary carbocations with less CH3 groups on it faster at SN2 than ones with

A

Less steric hinderance

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29
Q

What is torsional strain

A

Strain when hydrogens are on the same side of a cyclic ring

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30
Q

Why are ring compounds predicted ROR always different than recorded

A

Predictions assume rings always flat/ planar, but rings are actually arranged to have minimum energy conformation (chair)

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31
Q

How is torsional strain minimised

A

puckering

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32
Q

What are the bond angles/ structure shapes of the carbons

A

tetrahedral 109°

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33
Q

Which C-H bonds are axial and which ones are equatorial

A
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34
Q

What does a “ring flip” do

A

Equatorial and axial swap positions

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35
Q

Is the biggest group more favourable in the axial or equatorial position

A

equatorial

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36
Q

Are the hydrogens more favourable in the axial or equatorial position

A

equitorial because they are less sterically hindered

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37
Q

Why does the first reaction have a faster SN2

A

The leaving group is OTS. It is more favourable to leave/ easier accessible in the equatorial position

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38
Q

Why does t-Bu stay equatorial and its the other group that flips

A

t-bu can only be equatorial (conformational lock)

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39
Q

What effects the strength of a leaving group

A

C-X bond strength and stability of leaving group anion X-

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40
Q

What makes a good leaving group

A
  1. good leaving group is the conjugate base of a strong acid pKa is low
  2. weak C-X bond so X can easily leave
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41
Q

What makes a good leaving group

A

good leaving group is the conjugate base of a strong acid pKa is low

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42
Q

Which halide will be the best leaving group

A

Iodine- has weakest C-X bond

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43
Q

Is OH a good leaving group

A

no- H2O (con acid) has a pKa of 14, so it is a very weak acid

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44
Q

How does this alcohol get activated by the strong acid

A

H2O is a good leaving group, so Br can be installed on the alkyl instead

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45
Q

is water a good leaving group

A
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46
Q

Why does this react via Sn1

A

tertiary substrates cannot do Sn2

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47
Q
A
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48
Q

Draw a curly arrow mechanism for the following activation of an alcohol

A

no stereochemistry to worry about because its primary

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49
Q

What is the shorthand for a sultanate ester

A

TsOH

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50
Q

Are sultanate esters good leaving groups

A

yes

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51
Q

An alcohol being activated by being turned into a sultanate ester. What are the steps in this reaction

A
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52
Q

An alcohol being activated by being turned into a sultanate ester. What are the steps in this reaction

A
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53
Q

Why is this reaction favourable

A
  1. Strong S-O bond formed
  2. SO2 gas is entropically favourable
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54
Q

What is the structure of SOCl2

A
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55
Q

What is the mechanism of this reaction

A
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56
Q

What does this mean

A

determine the stereochemistry

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57
Q
A

Sn2 inversion where there is stereochemistry

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58
Q

What conditions are needed for an Sn2 inversion

A

reaction needs to happen at a stereocentre (e.g.)

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59
Q

What is an ether

A

oxygen with two carbons either side

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60
Q

Are ethers good leaving groups

A

no- but can be protonated with a strong acid to make a better leaving group (oxygen gets protonated)

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61
Q

Why would there be no reaction here

A

ester is a bad leaving group

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62
Q

What is the mechanism for an ether being protonated by a strong acid to make it a better leaving group

A

the carbon on the other side of the oxygen is sp2

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63
Q

What are epoxides

A

cyclic ether

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64
Q

Are epoxides good nucleophiles

A

yes- do not need activation/ they want to be nucleophiles to remove ring strain

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65
Q

Why do epoxides want to remove ring strain and undergo SN2 reactions

A

all bond angles are 60°, but all atoms are sp3 hybridised so want to be 109°

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66
Q

What do epoxides turn into when undergoing SN2 reactions

A

alcohols

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67
Q

ring opening of an epoxide

A
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68
Q

Why does a nucleophile effect the rate of an SN2 reaction but not Sn1

A

because nucleophile is involved in the RDS of an Sn2 reaction

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69
Q

What makes a good nucleophile

A
  1. bases of a weak conjugate acid
  2. lower down the periodic table (this can disrupt weak con acid argument)
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70
Q

Which would be the best nucleophile

A

OH because it has the weakest con acid

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71
Q

If Y is a good nucleophile, where does the equilibrium lie

A

to the right

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72
Q

If Y is a good nucleophile what does that say about its conjugate acid Me-Y

A

it is a weak acid

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73
Q

Why are elements further down the period table better nucleophiles

A

less charge density/ electrons more available to do reaction

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74
Q

Why would a sulphur based nucleophile be a better nucleophile than an oxygen based one

A
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75
Q

note- if the reaction was Sn1, the nucleophile would have no effect on rate so the reactions would go at the same speed

A

bottom anion is more stable; electron withdrawing =O, and you can draw resonance structures where electrons are shared so less to use for reaction

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76
Q

Which reaction would go quicker

A

I is further down the periodic table, and less electronegative/ less charge density/ more e- available for reaction

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77
Q

what is a polar protic solvent

A

solvent that can undergo hydrogen bonding(H-N,O,F)

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78
Q

what is a polar aprotic solvent

A

have a dipole but has no H available for hydrogen bonding

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79
Q

What does a higher dielectric constant mean (ε)

A

a solvent is more polar

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80
Q

Sn1/Sn2 reactions favour which solvents polar protic/ aprotic

A

Sn1 faster in polar protic
Sn2 faster in polar aprotic

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81
Q

Why are SN1 reactions faster in polar protic solvents

A

polar protic solvents stabilise cations and anions. Intermediate has dropped E more than the starting material because it is charged and starting material is neutral
Their transition states will have lower energy; therefore lower activation energy

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82
Q

For an SN2 reaction why is the energy higher/ more favourable in a polar aprotic solvent

A

smaller Ea than polar protic

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83
Q

Why is the transition state less polar than the staring material

A

in transition state the charge is distributed over 2 atoms instead of localised on one in the starting

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84
Q

what are the conditions for an elimination reaction

A
  1. base to attack beta hydrogen
  2. form a relatively stable carbocation
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85
Q

What is a beta hydrogen

A

carbon next to group of interest/ leaving group is alpha, must be a hydrogen on the carbon next to it

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86
Q

What factors effect elimination reaction rates

A
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87
Q

What type of leaving group is needed for an elimination reaction

A

good leaving group/ con base of a strong acid

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88
Q

Are quaternary ammonium ions good leaving groups in eliminations

A

yes- not so good for substitutions because it is quite spherically demanding

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89
Q

Why do elimination reactions result in an increase in entropy

A

Elimination reactions with 2 reactants, give 3 molecules of products

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90
Q

Why are elimination reactions more favourable at higher temperatures

A

△S (entropy) is always positivie △G will always be more negative

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91
Q

What is the difference between a E1 and E2 reaction

A

E1= 1 species in RDS
E2= 2 species in RDS

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92
Q

What are the secondary carbocations that are stable enough to do an E1 elimination reaction

A

benzylic and allylic substances. Have additional π systems for conjugation

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93
Q

What type of bases can be used for E1 elimination reactions

A

weak bases; bases do not effect ROR of E1. Base must not act as a nucleophile to avoid competing substitution

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94
Q

How does H2SO4 react as a base in this reaction

A

H2SO4 protonates the OH leaving, HSO4- which is a poor base, but it doesn’t have to be a good base as its E1 and still can accept a proton

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95
Q

What is the difference between a base and a nucleophile, and why for elimination reactions do you only want a base

A

base= specifically attacks protons
nucleophile= donates e- to all other atoms
You only want a base for an elimination, so it will attack the β hydrogen to form an alkene. If you have a base that is also a nucleophile, the reaction will have competing substitution

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96
Q

What is the mechanism for this reaction

A
97
Q

In an E1 reaction, for an alkene to form (π bond to be created) what needs to be true about the orbitals of the reactant where the cation charge is, and where the C-H (β bond) that is breaking is

A

empty p-orbital of cation and breaking C-H bond orbitals must be parallel. Assumption that the e- from the breaking C-H bond must move into the p orbital of the + charge from the diagram

98
Q

The starting material doesn’t have specified stereochemistry. This is one possibility for what this molecule looks like with given sterics. What is the second way it could be arranged via bond rotation

A
99
Q

This is for an elimination reaction where the beta hydrogen leaves. Which steric has the lowest energy conformation and will therefore lead to the major product

A

second as it is **least sterically hindered
** First one- Ph and Me are too close to each other

100
Q

In E1 elimination reactions, will E or Z isomers for the product be favoured

A

E isomers have lower energy conformations in the intermediate carbocations

101
Q

more/less substituted alkenes are more stable

A

more substituted alkenes are more stable/ transition state leading to more substituted alkene is lower in energy

102
Q

Why are more substituted alkenes more stable

A

stabilised through hyperconjugation of their empty π* anti-bonding orbital with adjacent C-H or C-C orbitals

103
Q

In an E1 elimination, which product had the most stable transition state

A

the one most on the right/ more substituted products will have had more stable transition states because of hyperconjugation stabalisation (adjacent alkyl groups give adjacent C-C and C-H bonds)

104
Q

What is an E2 elimination

A

elimination with 2 species in RDS. bonds break and form simultaneously

104
Q

What is an E2 elimination

A

elimination with 2 species in RDS. bonds break and form simultaneously

105
Q

What does the word antii-periplanar mean in terms of an E2 reaction

A

hydrogen atom and the leaving group must be anti-periplanar to each other for maximum orbital overlap (180° to each other). This is given by a staggered conformation

106
Q

Why can primary alkyl substances only undergo E2 reactions

A

because they form a very unstable carbocation (nothing to conjugate or hyper conjugate with)

107
Q

What is needed for an E2 reaction

A
  1. good leaving group
  2. beta hydrogen
  3. strong base usually sterically demanding
  4. can be primary tertiary or secondary
108
Q

tertiary alkyl substances can undergo SN2/E2 but cannot undergo SN2/ E2

A

Cannot undergo SN2, but can do E2

109
Q

Why can only certain substituted cyclohexanes undergo E2 elimination

A

leaving group must be axial for E2 elimination. Beta hydrogens must be anti-periplanar to leaving group

110
Q

Elimination is nucleophilic attack on carbon/hydrogen while substitution is attacking carbon/hydrogen

A

elimination attacks hydrogen
substitution attacks carbon

111
Q

more basic nucleophiles favours E2/SN2. Basic=weak con. acid/ more affinity for attacking a proton

A

elimination

112
Q

What reaction will a Br- nucleophile typically undergo

A

SN2

113
Q

Will a higher concentration of a base impact the rate of an E2 elimination reaction (base attacks the H and removes it to form a C=C)

A

higher conc of base- accelerated reaction

114
Q

Charged bases/nucleophiles will tend to perform E1/SN1/SN2/E2 reactions.
Wheras neutral bases/nucleophiles will be E1/SN1/E2/SN2

A

Charged bases/nucleophiles will tend to perform SN2/E2 reactions.
Reactions where neutral bases/nucleophiles are involved tend to go through carbocations (i.e. they tend to be SN1/E1).

115
Q

What does the measure of basicity tell you about a nucleophile

A

the ability of a species to accept a proton

116
Q

What makes a strong base

A

weak conjugate acid (high pKa)

117
Q

Bases are used in elimination and substitution reactions. More sterically demanding bases favour elimination/substitution

A

sterically demanding bases favour elimination, easier for them to not get as close and remove the H to form C=C than to substitute/attach onto a substance. It is difficult for a large base to get in by the carbon to do SN2

118
Q

Why does this reaction only do E2 not SN2

A

KOt-Bu is a large base/nucleophile and so favours more accessible β hydrogen than trying to attach itself

119
Q

Why does this reaction favour substitution not elimination. However there is a small amount of elimination at would it be E1/E2

A
  1. most sub- base/nucleophile is quite small so can substitute instead of just take β hydrogen
  2. Would be E2 as its primary (E1 is tertiary and some secondary)
120
Q

How do each of these effect wether a reaction is E1/E2

A

E1- can have any base/ usually weak because strong bases go E2. Electrophile must be very stable carbocation (tertiary and some secondary). leaving group must be good.
E2- strong bases. Primary, secondary or tertiary. β hydrogen atom. Good leaving group.

121
Q

What do both of these look like as a newman projection and which ones the least sterically hindered so more likely to go on and form a major product in an E2 reaction

A

2 is more sterically hindered so will make a minor product. Elimination can happen at both because both have a Br and H anti-periplanar, but there will be more product from projection 1

122
Q

What would this newman projection look like drawn out with stereospecifics (forget its going to a transition state)

A
123
Q

Explain why each transition state produces and E or a Z isomer (either Me are on same side or on opposite)

A

First one gives Me on opposite sides because one Me is hashed and one is blocked
Second one gives Me on same side because both Me’s are hashed

124
Q

Which transition state has the higher energy

A

bottom one- most sterically hindered because the Me’s are on the same side

125
Q

This is an E2 reaction. Why does the bromine leave but not the flourine

A

bromine is a better leaving group (further down table)

126
Q

Why are bulky groups better leaving groups (ie. things further down the periodic table)

A

Bulky groups are better leaving groups as when they leave, the steric factor gets stabilised

127
Q

Draw this molecule as a newman projection

A
128
Q

Draw this out as a molecule

A
129
Q

Why could this newman projection transition state not be used in an E2 reaction and must be rotated

A

H and Br need to be on opposite sides/ anti-periplanar/ at 180°; projection must be rotated to get the correct transition state that will give the major product

130
Q

What is a stereoselective reaction vs a stereospecific reaction

A

stereoselective- no specific stereochemistry given so can get multiple different product stereochemistry
stereospecific- starting material determines stereochemistry of product

131
Q

Fill in with either less/more

A

hindered bases attack most accessible hydrogen atoms

132
Q

A hindered base attacks a hydrogen with lots of substituted groups around it and one without as many. Which transition state will be higher

A

hindered bases attack most accessible hydrogen atoms; transition state on way to more substituted product will be higher in energy due to more steric demand

133
Q

What is an E1cB reaction

A
  1. deprotonation to make an anion
  2. By its own accord the anion moves electrons to kick out the leaving group to leave an alkene
134
Q

This is an E1cB reaction. After the anion is formed, how would this form an alkene

A
135
Q

substrates that undergo E1cB reactions must be able to form what

A
136
Q

How do you decide which bromine leaves

A

must be the right hand Br because it is on an sp3 hybrided carbon; cannot have substation on sp2 hybridised (if cation did form there, there would be not additional stabilisation)

137
Q

How to decide whether its SN1 or SN2

A

acetone is polar aprotic so must be SN2

138
Q

What can alkenes act as nucleophile/electrophile

A

nucleophile π-bond (HOMO) of alkene with LUMO of electrophile forms addition products

139
Q

What is the mechanism for this simple alkene addition

A

bromine acts as a nucleophile once the carbocation is made

140
Q

Why does this only have the one product and not the other carbocation

A

the most stable carbocation route is take

141
Q

What is the Markovnikov’s Rule (alkene HX addition rule)

A

on addition of HX to an alkene, the hydrogen adds to the least substituted carbon atom

142
Q
A
143
Q

Draw “curly arrow” mechanisms to determine the major products from the following reactions

A
144
Q

What needs to be present for an anti-Markovnikov’s product to be formed (hydrogen doesn’t go on the least substituted carbon)

A

peroxide -the least-substituted alkyl halide is formed as the major product

145
Q

How can a radical be formed

A

homolyses high heat or light

146
Q

What are alkyl radicals stabilised by

A

hyper-conjugation and conjugation

147
Q

What are the three steps of a radical reaction

A

Initiation- making the radical
Propagation- radical reacting with non-radical
termination- radical and radical

148
Q

Why is it the O-O bond that breaks

A

O-O bonds are very weak

149
Q

How would this radical react with HBr in a propagation step

A
150
Q

Is a primary or tertiary methyl radical more stable; which radical is more likely to be formed in a propagation step

A

tertiary- therefore in a propagation step, if the methyl radical is being made again the tertiary one will be made/ primary would have higher energy transition state

151
Q

How would this propagation step react and form

A

forms tertiary radical rather than primary as it is more stable/ transition state lower in energy

152
Q

propagation step product

A
153
Q

What would each of these termination steps look like

A
154
Q

What does an alcohol and H2SO4 make

A

alkene (elimination reaction)

155
Q

halogenoalkane and CN

A

nitrile

156
Q

What is a meso compound

A

non-optically active member of a set of stereoisomers, at least two of which are optically active

156
Q

What is a meso compound

A

non-optically active member of a set of stereoisomers, at least two of which are optically active

157
Q

What would the the transition state for this

A
158
Q

Why does this reaction make a racemic mixture

A

bromine could attack from either face

159
Q

Draw the ORBITALS on of Br- adding onto the cation (Br attacks 180° to Br+)

A
159
Q

What is the curly arrow mechanism for this (Br2 ring)

A
160
Q

Does Br2 have fumes

A

yes

161
Q

What does NBS look like

A
162
Q

Why is NBS used in a reaction where you need bromine instead of bromine

A

bromine fumes, and NBS provides a constant, low concentration of Br2 through reaction with trace HBr

163
Q

fill in gaps with more/less

A

more substituted/ Z isomers/ 1,1

164
Q

Which alkenes will have a faster rate of reaction

A

more substituted/ Z isomers/ 1,1 are more reactive

165
Q

What are bromohydrins

A

Br and OH

166
Q
A

ring opening Br+, but instead of Br- attacking, water attacks

167
Q

A bromohydrin alcohol can be treated with a base to get an epoxide. What Is the mechanism for this (deprotonation)

A

this reaction is fast

168
Q

brohydrin + base =

A

epoxide

169
Q

Alkene+ Br2 +H2O =

A

bromohydrin

170
Q

alkene + H —> alkane but under what conditions because it has a high activation energy

A

catalyst

171
Q

With an alkene —-> alkane reaction, what is special about where the hydrogens add across the double bond

A

on the same face

172
Q

alkene + H2O (in acidic conditions so H20 makes H30+) =

A

alcohols (OH adds to one end, last H adds to other) (H adds to least substituted end

173
Q

How would this alkene get H off the surface of a catalyst

A

-absorbed onto metal surface
-H transferred onto one face of the alkene
▪ Products are single diastereoisomers (but either racemic or meso)

174
Q

What is the mechanism for this

A

must form a stabilised tertiary carbocation intermediate

175
Q

What reactant could be used for an oxidation reaction

A

hydrogen peroxide H2O2

176
Q

Alkenes can undergo hyboronation (reacting with BH3) and then oxidation. What does this form

A

anti-maokovnikov alcohols ( substituent is bonded to a less substituted carbon, rather than the more substitued carbon)

177
Q

BH3 is electrophilic/ nucleophilic

A

electrophilic/ can act as Lewis acid (electron acceptor) because of empty p orbital

178
Q

hyboronation of alkenes can be due to reacting with BH3, or other derivatives such as

A
179
Q

hyboronation has a ** four-membered transition state**; what would the this transition state

A
180
Q

if H2 adds to the same face of an alkene, how could the products produced be meso

A
181
Q

How do you form an epoxides from alkenes

A

peroxycarboxylic acid (peracid)

182
Q

Are epoxides good nucleophiles

A

yes

183
Q

What is the mechanism for an alkene + mCPBA to form an epoxide

A

HOMO of alkene (πC-C) interacts with LUMO of mCPBA (σ*O-O). The product will be racemic because the syn addition could add to either face

184
Q

Which compound is meso

A

meso due to line of symmetry

185
Q

Alkene + mCPBA = epoxide (include transition state)

A

ring could have attacked from bottom/ arbitrary

186
Q

Alkene + mCPBA = epoxide (include transition state)

A

ring could have attacked from bottom/ arbitrary

187
Q

epoxides + base =

A

SN2 (due to steric hindrance) ring opening/ good nucleophiles at the least hindered position

188
Q

In acidic conditions, where each substituent goes isn’t as rigid as it is in basic conditions. What product will be the major product and which will be the minor

A
189
Q

What reactant can ring open an epoxide to form 1,2 diols

A

NaOH (SN2 reaction)

190
Q

What blocks would correspond with this drawing

A
191
Q

What is mCPBA (alkene reactions)

A

electrons from O-H bond to go alkene and double bond goes to that O

192
Q

What is dihydroxylation of alkenes

A

-syn addition
-(OsO4)

193
Q

What is the structure of OsO4

A
194
Q

OsO4 makes 1,2 diol

A
195
Q

OsO4 makes 1,2 diol

A
196
Q

OsO4 is used to make 1,2 diols, but is toxic and expensive. A catalytic variant was developed using N-methylmorpholine N-oxide (NMO) to oxidise the Os(VI) back into Os(VIII) which means the OsO4 is reusable. What is the structure of NMO

A
197
Q

OsO4 + alkene =

A

1,2 diol

198
Q

1,2 diols can be oxidised (into their ketones and break the C-C bond) with what reactant

A

▪ Can also use lead acetate, Pb(OAc)4, which is known as a Criegee oxidation

199
Q

mechanism for making 1,2 diols (cyclic electrons)

A
200
Q

What reactants can be used to turn an alkene into 2 aldehydes

A

NaIO4 and O3

201
Q

O3 is a good oxidant and can turn an alkene into its ketones. What are the products of this reaction

A
202
Q

What is the structure of the O3 reactant

A
203
Q

What is the mechanism for an alkene reacting with O3 (cyclic electrons) and then Me2S to make its subsequent aldehydes

A

SMe2 basically takes one of the oxygens away

204
Q

Alkene reacting with O3 and Me2S can make its aldehydes, but Ozonide can also be reacted with other oxidants to create other products. What do these make

A

54

205
Q

How to go from an alkene —> acid

A

O3, H202

206
Q

How to go from an alkene —> aldehyde

A

O3, Me2S

207
Q

How to go from an alkene —> alcohol

A

O3, NaBH4

208
Q

What reactant forms chloropropanes from alkenes

A

addition of carbenes to alkenes

209
Q

what are alkynes

A

triple bond

210
Q

What are the table trends (s-character means the anion [made when deprotonation] is closer to the nucleus and therefore has greater stability)

A

Alkynes are deprotonated to give anion in an sp-hybridised orbital
table and answer in photos

211
Q

what is a terminal alkyne

A

carbon-carbon triple bond is at the end of the carbon chain

212
Q

Why do alkynes have a lower pKa than alkanes and alkenes

A

alkynes are easier to deprotonate because their anions are more stable/ have higher s character

213
Q

What can be used to deprotonate a terminal alkyne

A

strong base (must be pKa >25 for complete deprotonation) /bases attack hydrogen

214
Q
A
215
Q

1,2-Dibromoalkanes can be used to make alkynes, also with a strong base with two E2 reactions. What is the mechanism for this

A
216
Q

What is NaOt-Bu

A
217
Q

which will be the stronger base/ could be instead of NaOt-Bu

A
218
Q

Alkynes + water and strong acid —->

A

ketones

219
Q

What is the mechanism

A

makes most stable carbocation

220
Q

like other alkenes, alkynes can undergo hydroboration. What dictates where the Boron adds to and what must the boron have in order for this reaction to go ahead instead of side reactions

A

least hindered end
Hydroboration of terminal alkynes to give aldehydes requires a sterically demanding borane to avoid side reaction

221
Q

What are the conditions for hydrogenation/ adding to H to any alkene/ alkyne to make an alkane

A

hydrogen and a palladium/ carbon catalyst

222
Q

Hydrogenation of alkynes will go to completion at an alkane. How would you stop it an alkene

A

H with poisoned palladium catalyst/ Poison reduces the activity of the palladium and prevents further reduction

223
Q

What is the most popular palladium catalyst poisoned with lead

A

Lindlar’s catalyst

224
Q

hydrogenation of an alkyne is a syn addition and will therefore lead to a (Z) alkene. What conditions do you do the reaction in to leave to the (E) isomer instead

A

dissolving metal conditions/ Birch reduction (radical chemistry and makes single electron)

225
Q

What is the mechanism for this

A
226
Q

A more substituted molecule on the C with the leaving group will undergo faster/slower E1 elimination

A

faster

227
Q

What types of molecules can undergo E1

A

tertiary and some secondary

228
Q

What type of molecule will always be the fastest in E1 AND E2 eliminations

A

tertiary, secondary then last primary

229
Q

Examples of strong bases

A

OH-, n-BuLi, NaNH2,

230
Q

Examples of weak bases

A

NH3, CH3NH2

231
Q

weak base = strong conj acid
Which will be the strongest base/ what factors effect how strong a base is

A

base accepts a proton; CH3- because it is the least stable.
1) Across the table (C,N,O,F) EN increases, so F will be the most stable as it wants the electrons close to it and less available for the reaction
2)Down periodic table polorizability increases so stronger bases

232
Q

Will F, Cl, Br or I be the best base

A

Flourine is the best base as it is least stable- Iodine is the most stable as charge is more stabilised as it is concentrated in the largest area

233
Q

Which anion is the most stable

A
234
Q
A
235
Q

Is acetone polar protic/polar aprotic

A

polar aprotic