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Reactions & Synthesis > Carbonyl Chemistry > Flashcards

Carbonyl Chemistry Flashcards

The different types of reactions involving the carbonyl group

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Organic Chemistry 101 flashcards
1
Q

Keto-enol tautomerization

A

Movement of the carbonyl double bond to the α carbon and the movement of the α hydrogen to the carbonyl oxygen.

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2
Q

Conjugate enolisation

A
  • Tautomerization that occurs through the conjugation of an alkyl chain.
  • Can be explained by the resonance forms of a ketone/enol.
  • Enolisation removes the γ-hydrogen.
  • The carbonyl is α-β unsaturated.
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3
Q

Electrophilicity/nucleophilicity of keto-enol forms

A

Keto form= electrophilic
Enol form= nucleophilic

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4
Q

Lithium diisopropyl amine (LDA)

A
  • A very strong (and bulky) base used to convert enols to enolates
  • Not nucleophilic due to bulk so is a terrible nucleophile but good base.
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5
Q

Malonate chemistry starting material

A

Diester (e.g. diethyl malonate)

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6
Q

Malonate chemistry product

A

Carboxylic acid

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7
Q

Malonate chemistry basic steps

A
  1. Enolization with a matching base (e.g. NaOEt for EtO-COCH2CO-OEt).
  2. Alkylation with haloalkane (R-X)
  3. Saponification with hydroxide then H+ (cleaves ester leaving -OH)
  4. Protonation (H3O+), then decarboxylation with heat of the diacid (cuts off one of the carboxylic acid groups)
  5. Enol tautomerises to form a ketone.
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8
Q

How to make carbocyclic carboxylic acids

A
  • Use malonate chemistry
  • Use an electrophile with two leaving groups (e.g. Br(CH2)3Br)
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9
Q

Retrosynthesis for malonate chemistry

A
  • α or α,α- disubstituted acetic acid starting material.
  • Cut at the α-carbon to:
    a. reform diester
    b. reform any alkyl side chains
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10
Q

How to deprotonate β-keto esters

A

Use a matching alkoxide.

Using the wrong alkoxide will:
- cause ester exchange creating a mixture of enolates
- causes saponification (OH- only)

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11
Q

Haloform reaction starting materials

A

Methyl ketone + halogen

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12
Q

Haloform reaction steps

A
  1. Convert a methyl ketone to an enolate (use a base like -OH)
  2. The enolate attacks the halogen and a halogen atom is added
  3. Tautomerisation restores the carbonyl group
  4. Repeat 2-3 more times to produce a CX3
  5. Addition of -OH forms a tetrahedral intermediate where the CX3 will leave in an elimination reaction
  6. CX3- removes a hydrogen from the newly formed carboxylic acid producing CXH (haloform) and a carboxylate.
  7. Protonate with H3O+

Remember: CX3 is a good leaving group as the C is slightly positive so can attract the electrons as it leaves

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13
Q

Deuteration

A
  • replace all α-hydrogens with deuterium
  • performed by tautomerisation then addition of deuterium to the double bond
  • LDA used as a base
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14
Q

Hell-Volhard-Zelinsky starting material

A

Carboxylic acid

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15
Q

Hell-Volhard-Zelinsky product

A

α-halogenated carboxylic acid (or ester if a alkoxide is used)

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16
Q

Hell-Volhard-Zelinsky Steps

A
  1. Halogen (from something like PBr3) replaces the hydroxyl group.
  2. Keto-enol tautomerism produces an enol.
  3. The enol can act as a nucleophile to add a halogen to the α-carbon.
  4. The oxygen containing species (e.g. H2O, OMe) attacks the carbonyl group forming a tetrahedral intermediate.
  5. The halogen on the carbonyl carbon leaves and the carbonyl group is reformed.
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17
Q

Addition of an alkyl group to ketone, ester or amide- alkyl group requirements.

A

The alkyl group must fulfil one of the following criteria:
- must have a good leaving group (e.g. halogen)
- primary alkyl group
- allylic alkyl group
- benzylic alkyl group

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18
Q

Addition of an alkyl group to ketone, ester or amide- steps

A
  1. React enol with LDA to form an enolate.
  2. The enolate acts as a nucleophile to attack the alkyl halide, replacing the hydrogen in the original compound with the alkyl group.

NOTE: this occurs via the SN2 pathway

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19
Q

Aldol reaction starting materials

A
  1. Aldehyde= electrophile
  2. Enolate= nucleophile
20
Q

Aldol reaction product

A

β-hydroxy-aldehyde (aldol)

21
Q

Aldol reaction steps

A
  1. Enolate acts as a nucleophile and attacks the carbonyl of the aldehyde producing a tetrahedral intermediate.
  2. Water reacts with the intermediate to form the hydroxy group on the β-carbon.
22
Q

Aldol condensation starting material

A

Aldol

23
Q

Aldol condensation product

A

α-β unsaturated carbonyl

24
Q

Aldol condensation steps

A
  1. Aldol reacts with H3O+ and heat to protonate the hydroxyl group.
  2. Under basic conditions the base removes an α-hydrogen causing:
    - a double bond to form between the α and β carbons
    - the H2O+ to leave
25
Q

Retrosynthesis for the α-β unsaturated carbonyl and β-hydroxy-carbonyls

A
  1. Disconnect the α, β C-C or α, β C=C bond.
  2. Add hydrogens to the α-carbon
  3. Turn β carbon into a C=O
26
Q

Claisen condensation starting materials

A
  1. ester enolate
  2. carbonyl from a second ester
27
Q

Claisen condensation product

A

1,3-dicarbonyl compound

28
Q

Claisen condensation steps

A
  1. Formation of an ester enolate using a matching alkoxide.
  2. Addition of a second molecule of the ester to form a tetrahedral intermediate.
  3. Elimination of the alkoxide group of the second ester.
  4. Alkoxide group takes an α-hydrogen causing enolisation and producing an alcohol.
  5. Protonation forms the 1,3-dicarbonyl.
29
Q

Diekmann reaction starting material

A

Diester

30
Q

Diekmann reaction product

A

β-keto ester (with the ketone belonging to a cycle)

31
Q

Diekmann reaction steps

A
  1. Use a matching alkoxide to produce an enolate at one end.
  2. Double bond of the enolate end attacks the other carbonyl producing a tetrahedral intermediate.
  3. The alkoxide group attached to the same carbon as the O- leaves and the carbonyl is reformed.
32
Q

Retrosynthesis for the Claisen/Diekmann

A
  1. Disconnect the α-β C-C bond.
  2. Add one hydrogen to the α-carbon.
  3. Add -OR to the β-carbon to give an ester
33
Q

Acetoacetate chemistry starting material

A

methyl-keto-ester (?)

34
Q

Acetoacetate chemistry product

A

Methyl ketone

35
Q

Acetoacetate chemistry steps

A
  1. Enolisation of the ester carbonyl using a matching alkoxide.
  2. The double bond acts as a nucleophile to attack the an alkyl group- adding it to the α-carbon.
  3. Ester hydrolysis (saponification) of the ester through a tetrahedral intermediate that eliminates the alkoxide and restores the carbonyl.
  4. Treatment with alkoxide (matching original ester) to deprotonate the hydroxyl group.
  5. Protonation and heating to “eliminate” the sideways carboxyl group.
  6. Keto-enol tautomerisation to restore the carbonyl.
36
Q

Retrosynthesis for acetoacetate chemistry

A
  1. Cut the alkyl chains attached to the α-carbon.
  2. Add an ester group to the α-carbon.
  3. Add halogens to the alkyl chains.
37
Q

Michael Reaction starting materials

A
  1. Enolate
  2. α-β-unsaturated carbonyl
38
Q

Michael reaction product

A

1,5-dicarbonyl

39
Q

How does the Michael Reaction proceed?

A

Through conjugate addition

40
Q

Michael Reaction steps

A
  1. Enolate acts as a nucleophile, attacking the β- carbon of the α-β-unsaturated carbonyl.
  2. Compounds combine in acidic conditions, reforming the carbonyl.
41
Q

1,2 vs 1,4 addition

A

1,2-addition:
- less stable as negative charge is localised on oxygen
- limited to very reactive nucleophiles
- fastest and dominates reactions with strong nucleophiles.
1,4-addition:
- can occur with weak nucleophiles (enolates).
- reversible and under thermodynamic control.

42
Q

Retrosynthesis for 1,5-dicarbonyls

A
  1. Number from one carbonyl to the other
  2. Disconnect C2-C3 bond
  3. At add a H to C2 and make C3-C4 a double bond

(It is possible to number in the reverse direction to give different starting materials)

43
Q

Mannich Reaction starting materials

A
  1. Formaldehyde
  2. Enolizable ketone
  3. Secondary amine
44
Q

Mannich Reaction product

A

Mannich Base

45
Q

Mannich Reaction Steps

A
  1. Protonation of formaldehyde
  2. Nucleophilic attack at the carbonyl by the secondary amine.
  3. A second protonation of the hydroxy group
  4. Elimination of the H2O+ and formation of a double bond (C=N).
  5. Nucleophilic attack on imminium by enolizable ketone.
46
Q

Betti Reaction

A

A Mannich reaction using a phenol rather than an enolisable ketone; the product is a Betti base.

Reactions & Synthesis (1 decks)

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