carbonyls - aldols Flashcards

(32 cards)

1
Q

what is the aldol reaction

A

a key C-C bond forming process between 2 carbonyl species , one acting as the electrophile with its C=O carbon and one as the enolate nucleophile with its α carbon, specifically

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2
Q

under what conditions can the aldol reaction take place?

A

acid or base catalysis

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3
Q

give 1 structural consequence of the aldol reaction

A

results in the formation of 2/more stereogenic centres

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4
Q

how does the aldol reaction work under base catalysis?

A
  • base protonates carbonyl group forming enolate little bit at a time (full deprotonated does not occur because equilibrium is so far on the RHS)
  • reactive carbonyl immediately reacts with any/all enolate generated, enolate forms C=O bond, nucleophilic C reacts with electrophilic carbonyl C and C=O is broken
  • -vely charged O reacts with water, protonating itself into hydroxyl + regenerating base
  • forms aldol product
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5
Q

give 1 advantage + 1 disadvantage of base catalysis of aldol reaction

A

as the base is a catalyst it regenerates itself so only a small amount is needed
however as this process is catalysed by base, it can go on to further react + side reactions

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6
Q

what happens when base re-reacts with aldol product?

A
  • attacks α hydrogen which causes C=O bond to break, via E1cB
  • double bond is formed as α proton leaves
    -vely charged O reforms C=O and double bond shifts adjacently and hydroxyl group leaves
  • this forms the dehydration prodcuct
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7
Q

how does the aldol reaction work under acid catalysis?

A
  • protonation via acid to form enol
  • protonated carbonyl immediately reacts with enolate generated, enolate forms C=O bond, nucleophilic C reacts with electrophilic carbonyl C and C=O is broken
  • deprotonation of carbonyl forms aldol product
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8
Q

give 1 disadvantage of performing the aldol reaction under acidic conditions

A

the elimination of water to form the dehydration product is easier under acid conditions, via E1 reaction
- OH group on aldol is protonated to H2O+, via E1 it leaves forming carbocation
- loss of proton forms second double bond = enone = dehydration product

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9
Q

is it possible for 2 of the same carbonyl compound undergo the aldol reaction + what is formed + 1 disadvantage?

A

yes, this is via self condensation = a dimerisation of a single carbonyl compound
- this product can be further deprotonated even if stoichiometry was perfect, leading to byproduct formation

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10
Q

give 2 ways byproduct formation be restricted

A
  • if enolisation can only occur at 1 carbon, onyl 1 product is possible
  • using specific enol equivalents which doesn’t allow the formation of certain byproducts, as specific enols are preprepared and then added to electrophilic carbonyl species, this means no base catalyst is needed for reaction therefore cannot go further
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11
Q

what is cross condensation + 1 disadvantage?

A

when 2 different carbonyl species are reacted together to form the aldol
this means that if both carbonyls can form enolates, there are 4 possible aldol products as there is 1 enolisable prpton per carbonyl and 1 electrophilic carbonyl per carbonyl, this is before considering second enolisations/condensation products
- serious lack of control

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12
Q

how do lithium enolates react to form aldols?

A

lithium enolates are made via reaction of LDA with ketone at low temperatures
- Li leaves O as C=O bond is formed, β carbon at the end of enolate C=C attacks δ+ C which breaks C=O which bonds with Li
- Li is removed via aq work up forming aldol

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13
Q

give 2 advantages of using lithium enolates for the aldol reaction

A
  • no base in the system so further reactions are prevented
  • electrophile is fully controlled as its added in small quantities to the lithium enolate mixture
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14
Q

why can the aldol reaction via lithium enolates be used with aldehydes?

A

aldehydes cannot be deprotonated under these conditions and if they are, they will be too reactive and self-condense/react

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15
Q

what are the 2 ways silyl enol ethers can be made?

A
  • quench lithium enolate with Me3SiCl
  • react ketone with weak base in the presence of Me3SiCl
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16
Q

what extra consideration needs to be made when using silyl enol ethers?

A

they are less reactive than lithium enolare, meaning electrophiles need to be made moer reactive, done via lewis acid which lowers the LUMO of the carbonyl sufficiently for the reaction to take place

17
Q

how are intramolecular aldol synthesised?

A

molecules containing 2 carbonyls can react via the aldol mechanism in the presence of acid or base to form an intramolecular aldol/cyclic aldol

18
Q

what are 2 similarities of the reactions forming intramolecular aldols and intermolecular aldols?

A
  • process can occur under both acidic + basic conditions
  • once enol/enolate forms it will quickly react with ketone
19
Q

are rings likely to form in aldol reactions?

A

yes - if it is possible for a ring to form it most likely will form + be the major product,, 5/6 membered rings especially

20
Q

why are other rings other than 5/6 membered rings less likely to form?

A

they do exist and can form, however they are much less stable and so more likely to reverse react back into starting products

21
Q

what is robinson annulation?

A

a 2 step process for forming 6 membered rings, micheal addition followed by intramolecular aldol reaction, then dehydration

22
Q

how are 6 membered rings formed via robinson annulation?

A

micheal addition: ketone + β unsaturated ketone
- ketone is deprotonated in the presence of base to form enolate
- -vely charged O reforms C=O bond causing double bond to attack β carbon of doble bond of β unsaturated ketone, causes C=O bond to break
- C=O bond reforms, double bond of β unsaturated ketone is protonated - generates enolate
aldol formation:
- base attacks α hydrogen, which attacks δ+ C causing C=O to break
- C=O reforms and enol double bond attacks carbonyl bond, causing it to break, this forms new ring
- -vely charged oxygen protonates into OH group
- base attacks α hydrogen which attacks δ+ C causing C=O to break + forms a double bond via E1cB
- C=O reforms, double bond attacks β carbon which causes OH to leave via dehydration
- final dehydrated product is formed

23
Q

what is the claisen reaction?

A

the acylation of α carbon via condensation reaction of 2 esters in base - similar to aldol reaction
occurs via ester enolate and forms 1,3 diketone

24
Q

give 1 limitation of the claisen reaction

A

similar issues to alcol cross condensation, the 2 esters should have very different reactivities to ensure success of the rection

25
how does functional group of esters affect claisen reaction?
in order of increasing reactivity: benzoate carbonate formate oxalate - this is because of increasing electrophilicity, as there are more EWGs attached which drop LUMO in energy as carbonyl c is made extra δ+ as so much electron density is pulled away from both sides
26
what is dieckmann condensation?
the intramolecular version of the claisen reaction, for forming rings
27
give 1 consideration that needs to be taken when deciding base + solvent for diekmann reaction
base + solvent alkyl groups must match ester otherwise transesterificaiton will occur
28
how are 1,3-diketones formed via claisen reaction?
- base attacks α hydrogen, which attacks δ+ C causing C=O to break, forms enolate - -vely charged O reforms C=O bond causing double bond to attack β carbon of doble bond of β unsaturated ketone, causes C=O bond to break - -vely charged O reforms C=O and alcohol ester group leaves
29
what is the knoevenagel reaction?
this is the aldol condensation reaction between 1,3-dicarbonyls an aldehydes/ketones - useful when a specific enol equivalent is used
30
what conditions are needed for the knoevenagel reaction?
works under many temperatures/mild conditions, uses amine bases buffered with acetic acid
31
how does the base used affect the knoevenagel reaction?
the type of amine base changes the mechanism - primary/secondary amines for iminium ions with aldehydes/ketones, tertiary amine do not
32
how do 1,3-dicarbonyls and aldehydes/ketones react via the knoevenagel reaction?
- amine base attacks α hydrogen on 1,3-dicarbonyl, which attacks δ+ C causing C=O to break, forms enolate - carbonyl reforms and enolate double bond attacks δ+ C of ketone/aldehyde, forming new C-C bond - -vely charged O on added aldehyde protonates - base attacks α hydrogen on 1,3-dicarbonyl again, which attacks δ+ C causing C=O to break, forms enolate again - C=O reforms, C=C bond moves and OH leaves, forming product