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PT IA Chemistry > Reactions and Mechanisms in Organic Chemistry Pt. 2 > Flashcards

Reactions and Mechanisms in Organic Chemistry Pt. 2 Flashcards

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1
Q

what are the ways to reduce other carbonyl derivatives (not aldehydes/ketones)

A
  • Reduction of acid chlorides/anhydrides to alcohols using sodium borohydride (NaBH4)
  • Reduction of esters to alcohols using lithium aluminium hydride (LiAlH4)
  • Reduction of amides to amines using lithium aluminium hydride
  • Reduction of Carboxylic acids using Borane
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2
Q

How can we determine which species will/won’t react with sodium borohydride

A
  • using the reactivity series of carbonyls

Acyl Chlorides > acid anhydrides > aldehydes/Ketones > esters > amides

Aldehydes/Ketones are the lowest reactivity species that can be reduced by sodium borohydride

anything with a higher reactivity will be reduced

anything with a lower reactivity will not be

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3
Q

what is the mechanism for anhydrides/ acyl chlorides with sodium borohydride to form alcohols
(same mechanism for both)

A
  • the borohydride ion attacks the carbonyl carbon as before
  • But this time the oxygen attacks back into the molecule and the chlorine or alcohol like ion (for anhydrides)
  • this reforms a Ketone
  • this ketone is attacked again by a borohydride ion and is then protonated as in previous mechanisms to form an alcohol
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4
Q

what do we use to reduce esters to alcohols and why

A
  • lithium aluminium hydride (LiAlH4)
  • this is because esters are LESS reactive than ketones/aldehydes so a stronger reducing agent is needed
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5
Q

what is the mechanism for the reduction of esters to alcohols

A
  • very similar to previous mechanisms for reductions that used NaBH4 but just with LiAlH4
  • AlH4(-) ion attacks carbonyl
  • O(-) attacks back into the structure and the OR group is lost as a leaving group
  • this forms an aldehyde
  • this aldehyde is then attacked again by AlH4(-) and the mechanism continues as before to form an alcohol
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6
Q

Why, in the reduction of esters to alcohols, does the mechanism not stop at aldehydes

A

two reasons:
- LiAlH4 can be the source of more than one hydride (as in BH4(-) in previous booklet)

  • the aldehyde is more reactive than the starting material so it reacts in preference to the ester with LiAlH4
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7
Q

Why, in the reduction of esters to alcohols, is the reducing agent not added at the same time as a protic solvent (to protonate the resultant oxyanion in the final step)

A
  • LiAlH4 will react with a protic solvent
  • so the protonation is done during a work-up, after the reaction itself is complete
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8
Q

Give a brief overview (no mechanism required) of how carboxylic acids can be reduced to alcohols

A
  • they can be reduced using Borane (BH3)
  • 3 eq’s are required in the first step
  • it forms a triacylborate intermediate which reacts further with a BH3 to form an alcohol
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9
Q

what is used to reduce amides to amines and what is the mechanism

A
  • AlH4(-) attacks the carbon of the carbonyl and the C=O bond breaks by heterolytic fission
  • the O(-) attacks the spare p-orbital on the resultant AlH3 molecule (AlH3 acts as a Lewis acid)
  • the lone pair on the nitrogen attacks back into the structure and now OAlH3 is removed as it’s an ok leaving group
  • this forms an Iminium ion containing a C=N(+)R1R2 section
  • the AlH4(-) ion attacks the carbon of the C=N bond on the iminium ion forming an amine, this is analogous to the reduction of a ketone/aldehyde
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10
Q

Summarise which reducing agents can be used to reduce which species

A

(OOO) = (NaBH4, LiBH4, LiAlH4)

Acid Chlorides (YYY),
Acid Anhydrides (YYY),
Ketones/Aldehydes (YYY),
Esters (Slow YY),
Amides (NNY)
Carboxylic Acids (NN Slow so use BH3)

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11
Q

as a general rule, which reducing agent should be used

A
  • the mildest agent which will still reduce the substance
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12
Q

What are the main two reactions that are possible when an organometallic reagent (Grignard or Organolithium) react with carbonyls

A
  • Ketones can form from a single addition of the nucleophile
  • Alcohols can form from a double addition of the nucleophile
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13
Q

Which species can be reacted in the single addition of an organometallic reagent to carbonyls to form ketones, give the mechanism

A
  • Acid Halides and Acid Anhydrides react with 1eq of RMgBr or RLi to form ketones
  • the organometallic attacks the carbon of the carbonyl, the C=O bond breaks by heterolytic fission
  • the lone pair/ -ve charge on the oxygen attacks back into the structure and now the chlorine or OCOR (anhydride thing) is removed as chlorine/OCOR are good leaving groups
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14
Q

Which species can be reacted in the double addition of an organometallic reagent to carbonyls to form alcohols, give the mechanism

A
  • esters react with 2 eq’s of RMgBr or RLi followed by aqueous acid to form alcohols
  • the organometallic reagent attacks the carbon of the carbonyl and the C=O breaks by heterolytic fission
  • the lone pair/-ve charge on the oxygen attacks back into the molecule and the OR group of the ester is removed as it’s an ok leaving group
  • this forms a Ketone
  • this ketone is then further attacked by the organometallic reagent and an oxyanion forms as expected
  • this oxyanion is protonated by aqueous acid (HCl) which is only added AFTER the first part
  • this forms an alcohol
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15
Q

Why can we not make alcohols from acid anhydrides/acyl chlorides and organometallics but we can make ketones

Why can we not make ketones from esters and organometallics but we can make alcohols

A

For Acid Anhydrides/Acyl Chlorides:
- the starting material is more reactive than the ketone
- this means that the starting material will react quicker with the organometallic than the ketone so if 1eq of organometallic is added then the major product is the ketone

For Esters:
- the starting material is less reactive that the ketone
- so the ketone (formed after the first part of the reaction) will react quicker with the organometallics than the starting material and hence the reaction is forced through to completion
- if 1 eq is added then the alcohol is the major product and about 50% of the starting material remains

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16
Q

what occurs in the reaction of organometallic reagents with carbon dioxide

A

step 1) add RMgBr or RLi to CO2, nucleophilic attack
step 2) add HCl, protonation
carboxylic acid formed

1) C-Mg sigma protons attack delta+ve carbon on CO2, one of the CO2 bonds breaks by heterolytic fission, this forms RCOOMgBr

2) the lone pair on the single bonded oxygen attacks a proton on the H3O+ ion (formed from the acid)
this forms a carboxylic acid

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17
Q

Summarise what each carbonyl type (and carbon dioxide) forms on reaction with organometallic reagents then protonation

A

Acyl Chloride and ester:
ketone (with 1eq)
alcohol (with 2eq)

Ketone: Alcohol

Ester: Alcohol (with 2eq)

Carboxylic acid: Carboxylate anion metal salts

Carbon dioxide: Carboxylic acid

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18
Q

What is a generalised reaction of water with a carbonyl and what do the specifics depend on

A
  • oxygen on water attacks carbon on C=O bond
  • O(-) formed attacks back into molecule
  • proton from water group is removed, X group is removed to form carboxylic acid
  • the exact order in which this happens depends on the carbonyl
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19
Q

give the reaction conditions for each of the carbonyls (4) with water, what does this agree with

A
  • Acyl Chloride: Fast at 20 degrees
  • Acid Anhydride: Slow at 20 degrees
  • Ester: only on heating with an acid or base catalyst
  • Amide: only on prolonged heating with an acid or base catalyst
  • this confirms the three factors of strength of nucleophile, reactivity of carbonyl and leaving group ability for nucleophilic substitutions
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20
Q

explain how acid-mediated hydrolysis of esters and amides speeds up the reaction

A
  • it protonates the carbonyl oxygen making it more susceptible to attack from a nucleophile
  • it can protonate the leaving group which lowers the pKa and makes it a better leaving group
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21
Q

give the mechanism for the acid-mediated hydrolysis of an ester (and effectively an amide too)

A

1) the carbonyl oxygen is protonated
2) the H2O attacks the carbonyl carbon and the C=O pi bond breaks by heterolytic fission
3) +- H(+) to form an O(Me)(H)+ group and an OH group
4) let one of the lone pairs on one of the oxygens on one of the OH groups attack back into the molecule and the MeOH group is lost as it’s a good leaving group
5) Let an H2O deprotonate the product to form a carboxylic acid

this also reforms the catalyst so it is truly catalytic in acid

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22
Q

in what way is the acid-mediated hydrolysis of an amide different to the acid-mediated hydrolysis of an ester

A
  • in the ester case the acid catalyst is regenerated so it is truly catalytic in acid
  • in the amide case the amine (which was the leaving group) is protonated under meaning that one equivalent of acid is used up in the reaction so it is not catalytic in acid (but still acid-mediated)
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23
Q

explain how base-mediated hydrolysis of esters and amides speeds up the reaction

A
  • by creating a more reactive negatively charged nucleophile
  • by deprotonating the carboxylic acid product ‘pulling the equilibrium over irreversibly to the hydrolysis products
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24
Q

give the mechanism for the base-mediated hydrolysis of an amide (and effectively an ester too)

A
  • an OH(-) nucleophile is formed from the reaction of water with a base
  • this attacks the carbonyl carbon forming a tetrahedral product with an OH, O(-), OR/NH2 and R
  • this step is reversible with the majority returning to the starting product as OH(-) is a better leaving group than NH2(-)
  • in the small number of cases where the NH2 group is removed, the resulting carboxylic acid is immediately deprotonated in an irreversible reaction to form a carboxylate anion
  • this can then be protonated in a work up to form the carboxylic acid
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25
Q

in what way is the base-mediated hydrolysis of an ester different to the base-mediated hydrolysis of an amide

A
  • the only difference is that in the first step the equilibrium is equal and not such that the majority returns to the starting material
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26
Q

Give a summary of the hydrolysis of esters under acidic and basic conditions

A

Ester - Acid mediated - requires heat for a short period, H+ reformed so truly catalytic

Ester - Base mediated - requires heat for a short period, at least 1eq required to deprotonate carbox acid product, not catalytic

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27
Q

Give a summary of the hydrolysis of amides under acidic and basic conditions

A

Amide - acid mediated - usually requires heat for an extended period, AT LEAST one eq required to protonate amine generated at end of reaction, not catalytic in acid

Amide - Base Mediated - requires heat for an extended period of time, AT LEAST one eq of base required to deprotonate carbox acid at end of reaction, not catalytic in base

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28
Q

How do Carbonyls (excluding Ketones/aldehydes) react with alcohols

A
  • in almost exactly the same way as with water in hydrolysis so this time it forms esters not carbox acids
  • the mechanisms and principles are directly analogous
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29
Q

summarise the conditions and products for the reactions of carbonyls (not Aldehydes/ketones) with alcohols (R*OH)

A

Acyl Chlorides: Form RCOOR*, occurs quickly at 20 degrees

Acid Anhydrides: Form RCOOR*, occurs slowly at 20 degrees

Carboxylic acids: form RCOOR*, only occurs on heating with an acid catalyst

Esters: RCOOR1—> RCOOR*, a process called transesterification, requires heating with acid/base catalyst

Amides: Form RCOOR*, NOT USUALLY A GOOD WAY TO MAKE AN ESTER, difficult even with added acid/base

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30
Q

what mechanism does the reaction of acid chlorides/anhydrides with alcohols proceed by

A
  • the generalised mechanism for the hydrolysis of carbonyls (without catalyst)
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31
Q

what mechanism does the reaction of carboxylic acids with alcohols proceed by

A
  • the exact reverse of the acid mediated ester hydrolysis
  • Carbox. acid –> ester = Excess ROH and acid catalyst
  • Ester –> Carbox. acid = Excess water and acid catalyst

NOTE: Base mediated esterification of of acids CANNOT occur because the base will simply deprotonate the carbox acid

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32
Q

what mechanism does the reaction of esters with alcohols proceed by

A
  • transesterification, this is a similar mechanism to acid/base mediated ester/amide hydrolysis
  • it only occurs with an acid or base catalyst
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33
Q

define/ give the term from

Retrosynthesis
TM
SM
FGI
Disconnection

A
  • Restrosynthesis - the process of dividing a molecule into simpler, precursors using known reactions
    TM = Target molecule
    SM = starting molecule
    FGI = functional group interconversion
    Disconnection = usually where C-C bonds are broken - shown as a ‘cut’ on the molecule
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34
Q

what can acid chlorides be made from and what is the type of synthesis

A
  • carboxylic acids, functional group interconversion
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35
Q

what can carboxylic acids be made from (5) and what are the types of synthesis

A
  • Acid Chlorides, Acid anhydrides, esters, amides, all FGI
  • or RMgBr + CO2 + HX, disconnection
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36
Q

what can esters be made from (4) and what are the types of synthesis

A
  • Acid Halides
  • Acid Anhydrides
  • Carboxylic acids
  • Esters (transesterification)
  • All FGI
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37
Q

what can Amides be made from (3) and what are the types of synthesis

A
  • Acid Halides
  • Acid Anhydrides
  • Esters
  • All FGI using ammonia
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38
Q

what can Ketones be made from (2) and what are the types of synthesis

A
  • Acetals, Hemiacetals and Hydrates, FGI
  • Acid Chloride or anhydride with RMgBr (1eq) or RLi (1eq), disconnection
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39
Q

what can primary alcohols be made from (5) and what is the type of synthesis

A
  • Aldehydes, Acid Halides, Acid Anhydrides, Esters, Carboxylic Acids + the right reducing agent
  • FGI
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40
Q

what can secondary alcohols be made from (2) and what are the types of synthesis

A
  • Ketone + NaBH4, FGI
  • Aldehyde + RMgBr or RLi, Disconnection
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41
Q

what can tertiary alcohols with 2 of the same group (3) and tertiary alcohols with 2 different groups (1) be made from and what are the types of synthesis

A
  • tertiary with 2 different groups: Ketone (with two different groups) and RMgBr (1eq) or RLi (1eq), disconnection
  • Tertiary with two of the same group: Acid Halides, Acid Anhydrides or Esters with RMgBr (2eq) or RLi (2eq), double disconnection
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42
Q

what can amines be made from and what is the type of synthesis

A
  • Amides + LiAlH4, FGI
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43
Q

what is a common mistake that is made when considering the reduction of amides with LiAlH4

A
  • in the first step the AlH4(-) attacks the carbonyl as expected
  • but in the second step the common mistake is for the oxygen to attack back into the structure and the NR1R2 group to be removed forming an aldehyde then eventually a carboxylic acid
  • what actually happens is that the negative oxygen will attack the spare p-orbital on the AlH3 which forms, then the lone pair on the nitrogen attacks back into the molecule and the OAlH3 is removed forming an Iminium ion
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44
Q

what is a common mistake when dealing with carboxylic acids and reducing agents

A
  • forgetting that the carboxylic acid is actually acidic
  • the AlH4(-) (or other reducing agent) will more likely act as a base than a reducing agent and simply pick up the acidic proton and a carboxylate salt will form
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45
Q

what is a common mistake that’s made when dealing with carboxylic acids and organometallic species

A
  • similarly to with reducing agents, the organometallic species will take the acidic proton NOT attack the carbonyl carbon
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46
Q

what’s a common mistake that’s made when dealing with carbox. acids under basic alcoholic conditions

A
  • the RO(-) ion which forms under basic alcoholic conditions will deprotonate the carbox. acid and not attack the carbonyl
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47
Q

what’s a common mistake that’s made when dealing with the reaction of a carboxylic acid with an amine

A
  • the easy mistake to make is to attack the carbonyl carbon with the lone pair on the nitrogen
  • what actually happens is that the amine simply deprotonates the carboxylic acid
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48
Q

what properties do large nucleophiles tend to have

A
  • React very quickly with saturated C but poorly at C=O groups
  • have high energy lone pairs, i.e. a high HOMO
  • are often uncharged or if they are charged then it is spread diffusely over large orbitals

e.g. RS(-), I(-), R3P

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49
Q

what properties do small nucleophiles tend to have

A
  • Attack C=O groups rapidly
  • have electrons concentrated close to the (usually electronegative) nucleus, i.e. a low HOMO
  • are usually charged and small so have a high charge density

e.g. RO(-), NH2(-), MeLi

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50
Q

what are large nucleophiles usually called and what effect does this have on the way that they react at carbons

A
  • large nucleophiles tend to be called soft nucleophiles
  • there is usually a small HOMO-LUMO gap (due to soft nucleophiles having a high HOMO)
  • Hence reactions tend to be FMO dominated and electrostatic effects are usually not important
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51
Q

what are small nucleophiles usually called and what effect does this have on the way that they react at carbons

A
  • small nucleophiles are usually called hard nucleophiles
  • there is usually a large HOMO-LUMO gap (due to the lower energy HOMO in hard nucleophiles)
  • Hence reactions tend to be more electrostatics driven but orbitals are still used for the reaction to occur
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52
Q

what are the key points that define an SN1 reaction

A
  • in an SN1 reaction the nucleophilic substitution is unimolecular
  • the formation of the reactive intermediate carbocation is the rate determining step
  • hence the rate determining step ONLY depends on the substrate conc. (not the nucleophile conc.)

Rate = k[RL]

  • over the reaction pathway we have an sp3 starting material to an sp2 intermediate carbocation back to an sp3 tetrahedral product
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53
Q

what are the key points that define an SN2 reaction

A
  • in an SN2 reaction the nucleophilic substitution is bimolecular
  • the formation of the product is the rate determining step
  • Hence the rate determining step depends on both the substrate AND the nucleophile

Rate = k[RL][Nu:]

  • in an SN2 reaction, the starting material is sp3 and is directly converted to an sp3 product through a transition state containing partial bonds to both the nucleophile and leaving group
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54
Q

how can we explain the presence of an intermediate and why the first step of the SN1 reaction is the slowest through an energy diagram

A
  • the energy diagram for the SN1 reaction shows the substrate then a large peak where there’s a transition state then a drop leading to a small dip (metastable equilibrium), this is where the intermediate is stable
  • from the intermediate there is a small peak leading to another transition state before returning to an even lower energy product
  • this metastable equilibrium point shows how an intermediate can form for short periods of time
  • the fact that the ‘peak’ between the substrate and the intermediate is the largest overall shows it needs the most energy and will hence be the rate determining step
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55
Q

how can we explain the lack of presence of an intermediate and rate determining step depends on the substrate and nucleophile in an SN2 reaction using an energy diagram

A
  • the energy diagram shows one large peak between the starting substrate and the product, i.e. the nucleophile has started to bond before the leaving group has fully left
  • this means that the highest energy ‘peak’ (activation energy) is the part of the reaction involving both the substrate and the nucleophile so the rate of reaction depends on both
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56
Q

what are the four major categories of variable which need to be discussed when considering nucleophilic substitutions at a saturated carbon

A
  • substrate structure
  • nucleophile
  • leaving group
  • solvent
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57
Q

what are the 5 main points to consider when thinking about substrate structure in SN1/SN2 reactions

A
  • steric hindrance = ease of nucleophile approach and transition state formation
  • factors which influence carbocation intermediate stability in SN1 reactions = hyperconjugation, pi-bond and lone pair donation
  • factors which influence transition state stability in SN2 reactions = adjacent double bonds and carbonyl groups
  • Stereochemical consequences of the SN1 and SN2 reaction processes = formation of racemic mixtures vs single enantiomers
  • Hybridisation state = no SN1 or SN2 at sp2 centres
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58
Q

what are the HOMO and LUMO involved in an SN2 reaction and how does this link to the angle at which the nucleophile attacks

A
  • HOMO = Nu l.p.
  • LUMO = C-Br sigma*
  • this explains why the nucleophile attacks the ‘back-side’ of the carbon, because this is where the biggest component of the C-Br sigma* is
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59
Q

explain how the number of substituents on a saturated carbon affects the likelihood of an SN2 reaction (2)

A

“SN2 reactions become less likely the more substituents there are on the C centre being attacked by the nucleophile”

  • this is because the increased steric hindrance makes approach of the nucleophile more difficult
  • equally in the sp3 starting carbon, all angles are 109.5 but in the transition state, some increase to 120, some decrease to 90, if there are large substituent groups, these will have steric clashes on reduction of angle
  • the more numerous the substituents, the harder to form the transition state
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60
Q

in substrate structure, what are the two factors that affect the carbocation stability in an SN1 reaction

A
  • hyperconjugation (sigma-bond conjugation)
  • pi-bond and lone pair donation
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61
Q

what is the hybridisation of the positive carbon in the intermediate carbocation of an SN1 reaction and the two reasons for it

A
  • sp2 NOT sp3
  • this is because the bond angle of 120 keeps the C-R bonds as far as possible from each other meaning the repulsion from pairs of electrons is minimised
  • the electrons are also placed into lower energy orbitals with more s-character
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62
Q

How does the rate of an SN1 reaction change with the addition of methyl groups to a carbocation intermediate

A
  • the rate of the SN1 reaction increases with number of substituent alkyl groups
  • this is partly because they increase steric hindrance so SN2 is less likely to occur
  • but also because they can provide additional stability from a relatively weak donation of adjacent sigma-bond electrons into the empty p-orbital of the carbocation
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63
Q

how do adjacent alkyl groups help to stabilise an intermediate carbocation in SN1 (hyperconjugation)

A
  • electrons held in C-H or C-C sigma bonds on substituent alkyl groups are at an angle to the vertical empty p-orbital on the positive carbon of the intermediate carbocation
  • however partial donation still occurs i.e. the C-H sigma acts as a partial HOMO and the C+ 2p acts as a partial LUMO
  • this overall lowers the energy of the carbocation and increases its stability, allowing the SN1 reaction to take place at a faster rate as the activation energies are lower
  • this cannot occur on a CH3+ cation as the hydrogens do not bond to anything so no hyperconjugation can occur
64
Q

what is an oversimplification which should be avoided when drawing/explaining hyperconjugation in an SN1 carbocation

A
  • drawing a diagram where arrows are placed on the sigma bonds
  • this makes it appear as though the stabilisation is going through the sigma bonds- it is not
65
Q

how can adjacent pi-bonds stabilise a carbocation in the intermediate of an SN1 reaction, how does the stabilising effect of this compare to other stabilising effects

A
  • pi-bonds can donate electrons through conjugation
  • the positive charge is stabilised by delocalisation
  • the more double bonds into which the positive charge can delocalise into (i.e. the more resonance forms that can form) the more stable the carbocation
  • this effect is more stabilising than hyperconjugation but less stabilising than lone pair donation
66
Q

how can atoms containing lone pairs adjacent to the positive carbon on a carbocation help stabilise in in an SN1 reaction, how does this stabilising effect compare to other stabilising effects, which reaction can we link this to

A
  • lone pair donation is the strongest of all stabilising effects
  • the lone pair on an atom next to the substituting carbon can donate its lone pair and assist in the removal of the leaving group
  • this also forms a delocalised ion
  • an example of this is the formation of an oxonium ion
  • this links to the formation of an acetal from a hemiacetal under acidic conditions
  • where when the ester group oxygen attacks back into the structure to remove the H2O+ group
  • this is the strongest of all the stabilising effects
67
Q

summarise the stabilising effects of a carbocation intermediate in and SN1 reaction and how they compare

A

1) adjacent sigma bonds (hyperconjugation) - the more adjacent sigma bonds there are (e.g. alkyl groups), the more stabilised the carbocation - weakest effect

2) Adjacent pi-bonds - positive charge can be delocalised over more atoms, increasing stability - adjacent benzene rings are better than isolated double bonds - middle strength effect

3) adjacent lone pair - where there are two atoms containing lone pairs attached to the same carbon atom then the mechanism can be done by the lone pair on one atom assisting in the departure on another, and by stabilising the carbocation through resonance forms

overall

nN > nO > Aryl/C=C > sigma(C-Si) > sigma (C-C or C-H) > sigma (C-X) (where X = N>O>S>Hal)

68
Q

What is the main factor in the stability of an SN2 transition state, explain why this is the case

A

“Both electron-donating and electron-withdrawing substituents are able to stabilise the SN2 transition state”

  • the partial bonds in the transition state are formed from mixing the filled and vacant HOMO and LUMO orbitals together
  • thus there is some character of filled and vacant orbitals in each partial bond
  • therefore, stabilising delocalisation is possible from either filled or vacant orbitals
69
Q

what are the two main types of group which can stabilise the transition state in an SN2 reaction

A
  • Adjacent C=C double bonds, e.g. vinyl and phenyl
  • adjacent carbonyl C=O
70
Q

why do adjacent C=C bonds help to stabilise the transition state in an SN2 reaction

A
  • transition state is stabilised by pi(C=C) or pi*(C=C)
  • a benzyl group has the same effect but has better stabilisation
  • it’s possible to invoke both electron withdrawal and donation as a stabilising factor
71
Q

why do adjacent C=O bonds/carbonyl groups help to stabilise the transition state in an SN2 reaction

A
  • the transition state is stabilised by adjacent pi*(C=O)
  • only possible to invoke electron withdrawal as a stabilising factor
72
Q

why is it best not to put a positive sign on the central carbon of the transition state in an SN2 reaction

A
  • whilst stabilisation of a +ve carbon is feasible by filled pi-bonds, it does not make sense for adjacent C=O groups which can only withdraw electrons
73
Q

what is a further reason why adjacent carbonyl groups can help SN2 reactions to proceed with speed (apart from transition state stabilisation)

A
  • the pi* of the carbonyl can combine with the sigma* of the carbon at the centre of the nucleophilic substitution to form a lower LUMO than either of the two individually, this is energetically more stable
74
Q

What are the key points to note about the stereochemical consequences of SN1 and SN2 reaction

A
  • SN1 reactions form a racemic mixture (mixture of enantiomers)
  • SN2 reactions form a single enantiomer
75
Q

What about SN1 reactions means you obtain a mixture of enantiomers

A
  • the reaction proceeds via an intermediate carbocation
  • this means in the second step of an SN1 reaction
  • attacking from the top of the planar intermediate forms one enantiomer
  • attacking from the bottom of the planar intermediate forms another enantiomer
  • this forms a 50-50 mix
76
Q

explain why (if starting with a single enantiomer) an SN2 reaction forms only a single enantiomer product

A
  • the SN2 process gives inversion stereochemistry at the carbon which has been attacked (turns inside out like umbrella)
  • this means if you start with a single enantiomer, you MUST end with a single enantiomer
77
Q

what is the key thing to remember about hybridisation state and how nucleophilic substitution reactions can occur at them

A
  • There can be no SN1 or SN2 nucleophilic substitution at sp2 centres
78
Q

why can an SN1 reaction not occur at an sp2 centre

A
  • a stabilised carbocation would have to form
  • consider a benzene ring, it appears the double bonds around can help to stabilise the positive charge
  • however, because of the sp2 HAO directions, the positive charge sits perpendicular to the p-orbitals forming the double bonds so cannot be stabilised
  • Hence a stable carbocation cannot form so SN1 cannot occur
79
Q

why can and SN2 reaction not occur at an SP2 centre (3 reasons)

A

1) the approach of the nucleophile is repelled by electron density in the double bonds

2) the approach of the nucleophile into the sigma* LUMO is blocked by the rest of the aromatic ring

3) the inversion of the centre attacked is impossible

80
Q

are nucleophiles important in SN1 and SN2 reactions

A
  • Not important in SN1
  • VERY important in SN2
81
Q

explain why the nucleophile is not important in an SN1 reaction

A
  • the rate determining step (RDS) is the loss of the leaving group so good and bas nucleophiles all give products
  • the nucleophile doesn’t even have to be deprotonated to be reactive as it is reacting with a highly reactive carbocation
82
Q

Explain why the nucleophile is very important in an SN2 reaction

A
  • the nucleophile is involved in the RDS and so it is very important as it directly affects the rate of reaction
83
Q

in SN2 reactions, can we simply base nucleophile strength off of pKa value like we did for nucleophilic substitution

A
  • No
  • in nucleophilic substitution at a saturated carbon it is a bit more complicated
84
Q

what are the two situations that we should consider when thinking about the strength of a nucleophile at a saturated carbon - SN2

A

1) where the atom forming the bond is the same of a series of nucleophiles

2) where the atom forming the bond is not the same over a series of nucleophiles

85
Q

in SN2 reactions, what is the trend for the strength of a nucleophile where the atom forming the bond is the same over the series of nucleophiles

A
  • the same as for a carbonyl carbon substitution
  • the higher the pKa of the nucleophiles, the better the nucleophile because it will represent a weaker acid so it will be more likely to form bonds with a carbon/hydrogen
86
Q

in SN2 reactions, why is the trend for nucleophiles where the atom forming the bond is not the same over a series NOT as simple as pKa values

A
  • in this situation we cannot only use pKa as a guide to determine nucleophile strength
  • this occurs because in SN2 nucleophilic substitution at a saturated carbon, the difference in electronegativity between the carbon and the leaving group is low
  • this means that the bond is not very polar so the reaction is dominated more by HOMO-LUMO interactions than by electrostatic interactions
  • Hence we need to consider soft/hard nucleophiles more so than pKa
87
Q

what is the trend in SN2 reactions for the strength of a nucleophile where the atom forming the bond is not the same over the series (of nucleophiles)

A
  • the fact that SN2 reactions are dominated by HOMO-LUMO interactions means that soft nucleophiles will often react best
  • if the electrophile is the same in both cases then generally the lower down the periodic table the atom is found: the larger the atom, the higher in energy its lone pair (HOMO), the closer the energy of the HOMO and LUMO (sigma*), the better the HOMO-LUMO interaction so the more favourable the SN2 reaction

e.g. sulphur is a better nucleophile than oxygen in SN2 as its HOMO is higher in energy so closer in energy to the sigma* LUMO

88
Q

is the leaving group important in SN1 and/or SN2 and why

A

Both, it is in the rate determining step of both SN1 and SN2 reactions

89
Q

what are the two factors which should we should think about when considering the leaving group ability of the halides, summarise the overall outcomes

A

1) C-X bond strength
2) pKa of HX

  • the C-X bond strength decreases down the group
  • the pKa of H-X increases down the group (meaning the anion becomes more stable)
  • Hence the halides become better leaving groups as you move down the group
90
Q

what is the reason why alcohols will not usually react with nucleophiles, explain why (3 reasons)

A
  • alcohols don’t usually react with nucleophiles because -OH is not a good leaving group

OH is not a good leaving group for 3 main reasons:
1) the pKa of water is quite high at 14 (i.e. OH(-) is not a stable anion)
2) the nucleophile will usually be basic enough to just remove the proton instead
3) even if the nucleophile did displace the hydroxide anion, hydroxide is basic enough to remove the proton from other alcohols, preventing further reaction

91
Q

what can be done to make OH a better leaving group (2)

A

1) protonate it
2) do a sulfonate ester formation

92
Q

how can an OH group be protonated to make it a better leaving group

A
  • add an acid (often conc. HCl)
  • this will protonate the OH group to H2O
  • pKa(H2O) = 14, pKa(H3O(+)) = -1.7, this is a big difference
  • this allows the H2O(+) group to leave forming a positive carbocation intermediate in an SN1 reaction
93
Q

what is used in sulfonate ester formation, how does it make OH a better leaving group

A

para-toluenesulfonyl chloride ‘tosyl chloride’ and pyridine,
it converts OH to OTs, pKa(TsOH) = -1.3 so it is a much better leaving group

94
Q

Explain the mechanism for sulfonate ester formation

A

1) the lone pair on the OH group attacks the sulfur atom on the TsCl, the pi bond to an oxygen breaks forming TsClOH(-)

2) the O(-) group attacks back into the structure reforming its double bond and the Cl is removed

3) the lone pair on the nitrogen of pyridine removes the proton left on the oxygen to form the product RCH2OTs (or similar), PyHCl also forms

4) the resultant OTs group is a much better leaving group than OH

95
Q

what is a common mistake that’s made in the mechanism for the formation of a sulfonate ester

A
  • the steps are reversed
  • it’s easy to try to deprotonate the alcohol first using pyridine
  • THIS DOESN’T HAPPEN
  • because pKa(ROH) =~16, pKa(Py) =~5
96
Q

what are the 3 main reasons why ethers don’t tend to react in nucleophilic substitution reactions, what has to happen for them to react

A

1) there is very little bond angle strain
2) the C-O bond is very strong
3) the alkoxide anion is similar in its leaving group ability to the hydroxide ion, i.e. it is NOT a good leaving group

  • in order for ethers to undergo nucleophilic substitution, they must be protonated AND reacted under vigorous conditions (strong acids and high temperatures)
97
Q

What is an epoxide and give the main reasons why they will undergo nucleophilic substitution despite ethers not

A
  • a three membered ring which is an ether, i.e. a carbon at two corners and an oxygen at the other
  • although the leaving group is still the alkoxide ion the C-O bond is weaker due to worse overlap and sever bond angle strain (60 degrees not 109.5)
  • this means the release of ring strain helps to make the epoxides very good electrophiles in SN2 reactions
98
Q

what is something that must be considered when doing SN2 reactions with epoxides fused to other rings

A
  • stereochemistry, the nucleophile must attack from bottom to ensure inversion holds
99
Q

summarise which solvents should be used for SN1 / SN2 reactions

A

SN1 = polar, protic solvents
SN2 = polar, aprotic solvents

100
Q

explain the reasons for why SN1 reactions require polar, protic solvents

A
  • ions are formed during the SN1 reaction process, the intermediate carbocation and the leaving group anion

the polar, protic solvent can solvate both of these:
- the (-) part of the solvent molecule can stabilise the carbocation
- the (+) part of the solvent molecule (protic part) can stabilise the anion/LG or and can surround it and prevent it from acting as a nucleophile and reforming the SM

101
Q

explain why polar, aprotic solvents are best for SN2 processes

A
  • no ions are formed during an SN2 process
  • polar, aprotic solvents are good at solvating cations (as they have a negative part) but not anions (no protons to pair with anions)
  • this means the delta(-) part of the solvent can solvate the cation of the nucleophile making the anion more reactive to be a nucleophile in the process
102
Q

give some examples of protic, polar solvents in decreasing polarity

A

water, methanol, ethanol, acetic acid

103
Q

give some examples of polar, aprotic solvents, in decreasing polarity

A

DMSO, DMF, MeCN, Acetone, CH2Cl2, THF, Ethyl Acetate

104
Q

give some examples of non-polar solvents

A

Chloroform, Diethyl Ether, Toluene, Benzene, Cyclohexane, Hexane etc.

105
Q

what are the two types of elimination reaction - why are they significant

A

E1, first step is identical to SN1, forms alkene, RDS only contains SM

E2, first step is similar to SN2, also forms alkene, RDS contains SM and Nu

106
Q

give the general mechanism for E1

A
  • the leaving group (something like a halide) leaves the carbon and forms an intermediate carbocation
  • the ‘nucleophile’ Y acts as a base and not a nucleophile, it deprotonates the proton on the carbon adjacent to the positive charge carbon
  • the electrons from the C-H sigma bond are donated into the empty p-orbital on the +ve carbon to form an alkene and HY
107
Q

give the mechanism for E2

A
  • the ‘nucleophile’ Y still acts as a base, it deprotonates the carbon adjacent to the carbon with the leaving group
  • the C-H sigma bond electrons are donated into the C-X sigma* (LUMO)
  • this forms a transition state and eventually the alkene and HY
108
Q

compare the rates of reaction for E1 and E2, give their rate equations

A
  • it’s analogous to the SN1 and SN2 reaction rates
  • E1 reaction rates are dependent only on the formation of the carbocation as this is the RDS so the rate equation only includes the starting material

Rate = k [RL]

  • E2 reaction rates are dependent on both the nucleophile and the starting material as these are both involved in the RDS

Rate = k [RL][Nu]

109
Q

give an overview of the orbitals involved in an E1 reaction and hence how the orbitals must be arranged

A
  • for the alkene forming step to proceed, the C-H sigma bond attached to the proton being removed must be parallel to the empty P-orbital for best overlap
  • the two main HOMO-LUMO interactions occurring are
    1) Lone pair on B: (HOMO) with C-H sigma* (LUMO)
    2) C-H sigma (HOMO) with empty p-orbital (LUMO)
110
Q

give an overview of the orbitals involved in an E2 reaction and hence how the orbitals must be arranged

A
  • in an E2 elimination, for the most efficient overlap to take place the C-H sigma and the C-LG sigma* must lie antiperiplanar to each other

antiperiplanar = at 180 degrees to each other and in the same plane

  • this arrangement gives the most efficient elimination process, there are 2 main HOMO-LUMO interactions occurring
    1) Lone pair on B: (HOMO) with C-H sigma* (LUMO)
    2) C-H sigma (HOMO) with C-LG sigma* (LUMO)
111
Q

what is the main thing to note about the stereochemical consequences of E1 and E2

A
  • they can both form cis or trans isomers from restricted rotation about a double bond
112
Q

explain the stereochemical consequences of an E1 reaction where there’s only 1 R group on the carbon with the leaving group and 1 on the adjacent carbon

A
  • the trans (or E-isomer) is the major product
  • the cis (or Z-isomer) is the minor product
  • this is because the arrangement of the R groups in the carbocation intermediate can be such that the two R groups are on the same side of the molecule or on opposite sides
  • where the two R groups are on the same side this can form the Z isomer but this gives a greater steric clash as the two bigger groups are closer to each other, it also gives a higher energy carbocation intermediate
  • where the two R groups are on opposite sides of the molecule, this is a lower energy carbocation intermediate, it also gives less steric clash in the E-isomer that forms so overall this forms the major product
113
Q

explain the stereochemical consequences of an E1 reaction where there’s more than 1 R group on the carbon with the leaving group and 1 on the adjacent carbon

A
  • the same principle occurs as with just 1 R group on each carbon but this time instead of considering the steric clash between the two R groups, consider the steric clash between the biggest R groups and minimise this
114
Q

explain the stereochemical consequences of an E2 reaction

A
  • the situation with E2 is analogous to that with E1 but now given it all occurs in one step and that the C-LG sigma* must be antiperiplanar to eliminate, the outcome only depends on the arrangement on groups in the starting material
  • e.g. if R1 and R2 are on the same side of the molecule in the SM then they the cis isomer WILL form regardless of whether this gives the biggest steric clash
115
Q

are halides good leaving in groups in E1 and E2

A
  • good for both
116
Q

is OH(-) a good leaving group in E1 or E2

A
  • Never good in either
117
Q

is OH2(+) a good leaving group in E1 or E2

A
  • Good for E1 (OH under acidic conditions)
  • Never for E2 (E2 is always carried out under basic conditions)
118
Q

is TsO(-) a good leaving group in E1 or E2

A
  • Good in E1
  • Good in E2 (can be formed from OH)
119
Q

is NR3(+) a good leaving group in E1 or E2, explain how it can be made

A
  • Good in both
  • it can be formed by reacting a tertiary amine with an alkyl halide, this forms a quaternary amine salt
  • it can then leave in exactly the same way as normal leaving groups in E1 and E2
120
Q

are there links between SN1/E1 and SN2/E2 in terms of substrate structure

A
  • sometimes but its not as simple as good substrate for SN1 = good substrate for E1 etc.
121
Q

why is it that many elimination processes can proceed by either E1 or E2

A
  • steric factors which disfavour SN2 do not disfavour E2 in the same way

e.g. a tertiary halide would only do SN1 and NOT SN2, but it can do either E1 or E2

122
Q

how well do primary substrates react via E1 or E2

A
  • They can’t react via E1 (too unstable intermediate)
  • they react well via E2
123
Q

how well do secondary substrates react via E1 or E2

A
  • they are ok at reacting in both
124
Q

how well do tertiary substrates react via E1 or E2

A
  • they react well via E1 (stable tertiary carbocation)
  • they can also react via E2 as there are many protons to be deprotonated, they may need favourable conditions though
125
Q

how well do alpha-alkoxy substrates react via E1 or E2

A
  • they react well in E1 reactions
  • they can also react via E2
126
Q

how well do allylic substrates react via E1 or E2

A
  • they react well via E1
  • it can also react via E2
127
Q

how well do benzyllic substrates react via E1 or E2

A
  • E1 reacts well
  • E2 is also possible
128
Q

what are some key substrates that cannot react by either E1 or E2

A

Me-X
Ar-CH2-X
C(CH3)3-CH2-X

129
Q

how can the basicity of the nucleophile affect the outcome of a reaction in relation to elimination or substitution

A

Strong bases mainly give elimination whereas weak ones give substitution

130
Q

why is potassium tert-butoxide tBuOK a common choice of base/nucleophile when requiring elimination reactions

A
  • it is a very strong base so will favour elimination over substitution
  • it is also bulky, this makes it less effective at ‘squeezing’ through the structure in substitution reactions
131
Q

how can the size of a nucleophile impact whether SN2 or E2 occurs

A
  • SN2 reactions are affected lots by steric hindrance as the the nucleophile must attack the sigma* of the carbon so must get very close
  • E2 reactions are not affected much by steric hindrance as the nucleophile/base must only access the alpha-protons on a molecule
  • Hence the more ‘bulky’ a nucleophile is, the more likely it is to react via E2 over SN2
132
Q

How can temperature affect whether E1/E2 or SN1/SN2 occurs, explain why

A
  • in E1/E2, two molecules become 3
  • in SN1/SN2, two molecules become two
  • hence the entropy increase of E1/E2 is greater
  • Thus, due to Gibbs free energy, E1/E2 are favoured at higher temperatures
133
Q

summarise how methyl substrates will react for poor, weakly basic, stronger basic (unhindered) and strongly basic (hindered) nucleophiles

A

Poor = no reaction

Weakly basic = SN2

Strongly basic (unhindered) = SN2

Strongly basic (hindered) = SN2

134
Q

summarise how primary (unhindered) substrates will react for poor, weakly basic, stronger basic (unhindered) and strongly basic (hindered) nucleophiles

A

Poor = no reaction

Weakly basic = SN2

Strongly Basic (unhindered) = SN2

Strongly Basic (hindered) = E2

135
Q

summarise how primary (hindered) substrates will react for poor, weakly basic, stronger basic (unhindered) and strongly basic (hindered) nucleophiles

A

Poor = No reaction

Weakly Basic = SN2

Strongly Basic (unhindered) = E2

Strongly Basic (hindered) = E2

136
Q

summarise how secondary substrates will react for poor, weakly basic, stronger basic (unhindered) and strongly basic (hindered) nucleophiles

A

Poor = SN1, E1(slow)

Weakly Basic = SN2

Strongly Basic (unhindered) = E2

Strongly Basic (hindered) = E2

137
Q

summarise how tertiary substrates will react for poor, weakly basic, stronger basic (unhindered) and strongly basic (hindered) nucleophiles

A

Poor = SN1, E1

Weakly Basic = SN1, E1

Strongly Basic (unhindered) = E2

Strongly Basic (hindered) = E2

138
Q

What types of double bond can H-X react with, what is the mechanism

A
  • Symmetrical Alkene
  • Asymmetrical Alkene
  • Electrophilic Addition
139
Q

Give the mechanism of H-X with C=C double bonds

A

1) double bond attacks delta+ve H
- H-X bond breaks by heterolytic fission
- SLOW step

2) X(-) ion attacks the positive carbon of the carbocation
- FAST step

140
Q

what do we need to consider when reacting an asymmetrical alkene with X-H that we don’t when reacting a symmetrical alkene with X-H, how is this done

A
  • two different products will be able to form when reacting with an asymmetrical alkene, consider which the major and minor products are
  • this will be given by which is the more stable carbocation
  • use concepts from previous topics to determine this
141
Q

why does the more stable intermediate carbocation give the major product

A
  • the more stable carbocation intermediate requires a lower activation energy so overall it will proceed with a quicker rate of reaction
  • Hence it forms the major product
142
Q

What are the bromination and iodination reacts of alkenes

A
  • alkenes will react with bromine or iodine (in an inert solvent) to form a 1,2 di-substituted product
143
Q

describe the mechanism of iodination and bromination of alkenes/describe the HOMO/LUMO interactions

A
  • this mechanism works by initially creating a bromonium or iodonium ion
  • the initial HOMO/LUMO interaction is of the pi-bond of the alkene and the X-X sigma*

1) the pi bond attacks the closer X and the X-X bond breaks by heterolytic fission
2) the lone pair on the X that is attacked, attacks one of the carbons of the double bond
3) this forms a bromonium or iodonium ion
4) the X(-) ion that forms from the breaking of the X2 molecule attacks the delta(+ve) carbon adjacent to the X(+) on the bromonium ion and the C-X bond breaks
5) this forms the di-substituted product

144
Q

what is the stereochemical outcome of the bromination/iodination reaction on a fused ring

A
  • the attack of the halide on the bromonium/iodonium ion occurs on the back face, this gives rise to the trans product
  • no cis isomer is formed
145
Q

under what conditions is a halohydrin formed

A
  • a halohydrin is formed when bromination/iodination occur with a large amount of water present
146
Q

what is the mechanism for halohydrin formation

A
  • similar to bromination/iodination
  • but instead of the halide ion attacking the bromonium/iodonium ion, it is the water molecule which attacks
  • it is then deprotonated to form a halohydrin (halide group on one carbon, alcohol group on another)
147
Q

what is the stereochemical outcome of a halohydrin formation in a fused ring system

A
  • the same as in bromination or iodination, the attack is on the back face so the trans product is formed
148
Q

give the mechanism/describe how epoxides can be made from trans-halohydrins

A
  • if a halohydrin is treated with a base then the alcohol is deprotonated and a rapid SN2 reaction occurs
  • the -ve charge on the deprotonated alcohol group attacks the carbon with the X attached, the X group leaves and an epoxide is formed
149
Q

give the overall reaction for the one step epoxidation of alkenes

A

alkene + m-CPBA —> epoxide + meta-chloro benzoic acid

m-CPBA = meta-chloro peroxybenzoic acid

150
Q

give the mechanism for the epoxidation of alkenes

A
  • it is a one step mechanism
  • there are 4 arrows
    1) arrow from double bond to oxygen of OH on m-CPBA (C=C pi is HOMO, O-O sigma* is LUMO)
    2) arrow from O-O to C-O
    3) arrow from C=O to the H of the OH group, i.e. C=O pi breaks by heterolytic fission and bonds to H
    4) arrow from OH bond to a carbon of the double bond
  • this, in one step, forms an epoxide and meta-chloro benzoic acid
151
Q

what is the stereochemical outcome of the epoxidation of alkenes

A
  • because both of the new C-O bonds are formed on the same face of the alkenes pi-bond (i.e. it’s a one step reaction), the geometry of the alkene is reflected in the stereochemistry of the epoxide, cis gives cis, trans gives trans
152
Q

What is a reaction that gives the opposite regioselectivity to a standard electrophilic addition

A
  • hydroboration, this is a reaction which puts an alcohol group on the less likely position on an alkene
153
Q

give the overall reaction (i.e. reactants/products) of hydroboration

A

alkene —–> alcohol (with OH in less likely position)

1) BH3
2) H2O2, NaOH, H2O

154
Q

Give the mechanism for hydroboration

A

1) addition of borane over double bond, this occurs all in one step, draw full arrow from double bond to the empty p-orbital on Boron of BH3, draw partial (dashed) arrow from B-H bond to the carbon on double bond with the greater charge stability

2) this forms a transition state see booklet for details)

3) this forms a saturated alkane with a BH2 group on the less charge stable and less sterically hindered carbon

4) this same reaction occurs twice more to form a BR3 molecule

5) the next step is the breakdown of trialkylborane with hydroperoxide ion, the OH(-) from the NaOH deprotonates the H2O2

6) the lone pair/negative charge on the O(-)-OH ion attacks the empty p-orbital on the BR3 forming BR3OOH(-)

7) the O-O bond is weak so breaks at the same time as a strong C-O bond is formed, one of the alkyl groups migrates from B to O (the electrons of the B-C bond hop across to O-O sigma*)

8) this forms an R-O-BR2 molecule, the OH(-) ion attacks the empty p-orbital on boron

9) the B-O bond breaks and the resultant molecule is protonated to form the product

155
Q

comment on the regioselectivity of hydroboration/ why is it that it doesn’t do what usually happens

A
  • two main reasons
    1) electronics, the pi-bond electrons are being donated to the empty p-orbital of the boron, the partial positive charge this forms needs to go onto the carbon that’s more able to stabilise it

2) sterics, the BH2 group is large so goes onto the less sterically hindered group

156
Q

what are the stereochemical outcomes of hydroboration

A
  • because the boron and hydrogen add to the double bond in an essentially concerted process, the stereochemical outcome from the reaction process overall is the syn addition of H and OH to the double bond i.e. they add to the same face
157
Q

what is something to note about steric hindrance adjacent to a ring double bond in electrophilic addition (bromination m-CPBA or borane)

A
  • they can cause steric hindrance for the incoming electrophile
  • so overall the less hindered face of the double bond is favoured
  • that is to say, if the large adjacent group points up, the addition groups will point down and vice versa

PT IA Chemistry (9 decks)

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