Aldehydes and ketones Flashcards
(23 cards)
Key reactions of aldehydes and ketones?
- Oxidations
- Synthesis
- Hydration
- Adding alcohol
- Imine formation
- Enamine formation
- Grignard reagents
- Addition of sodium borohydride
Key types of oxidations?
With PCC, Tollen’s, dichromate, permanganate
Of 1° vs 2° alcohols.
Ways to synthesise a ketone?
- Oxidation of 2° alcohols
- Hydration of alkynes
- Friedel crafts acylation
Ways to synthesise an aldehyde?
- Oxidation of 1° alcohol using PCC
What’s a Friedel Crafts acylation?
acid chloride + aromatic compound in acidic conditions
- Forms phenyl - ester
What’s tautomerism?
Tautomerism describes constitutional isomers in eq with each other.
What the the properties of a Tollen’s reagent?
Tollen’s reagent is a very strong oxidant.
- Is selective: can be used to test for aldehydes, depositing silver as bi-product.
REACTS ONLY WITH ALDEHYDES
What makes aldehydes and ketones reactive?
Aldehydes and ketones are reactive due to their polar, electron rich C=O bond.
- Thus the O can react with an electrophile like H+ (the O is a nucleophile
- Thus the alpha carbon can react with a nucleophile like OH- (the C is faintly positive)
What makes a good nucleophile?
High e- density and low electronegativity.
- Anything that means it can give away e- more readily.
Acetal vs hemiacetal?
Hemiacetal has 1 ester, 2 Rs and a hydroxy group
Acetal has 2 esters and 2 Rs (swapped H for another R)
Under what circumstances can you oxidise a ketone?
A ketone can be oxidised to a diacid if it is in its enol form.
- Occurs at elevated temperatures.
What’s the more stable form in keto-enol tautomerism?
A keto form is more stable than an enol form.
Steps of an acid-catalysed hydration?
1) Protonation into activated electrophile (Add H3O+)
2) Add H2O (left after protonation)
3) Generates tetrahedral structure with +ve charge on -OH-H (H2O) group.
4) Deprotonation leaves gemdiol and reforms acid (H+ and H2O -> H3O+)
Steps of an alcohol addition?
1) Protonation into an activated electrophile
2) Add R-OH
3) Generates a tetrahedral structure with +ve charge on the RO-H group
4) Deprotonation leaves hemiacetal product and reforms acid.
PART TWO: HEMI-ACETAL to ACETAL:
1) Protonation of OH group into OH-H+ leaving group.
2) Elimination of water as a leaving group, reforming planar shape and double bond.
- Double bond exists from central carbon to ether linkage oxygen, this oxygen now carries the +ve charge
3) The fact that the O in the double bond is +ve means that the central carbon can act as an activated electrophile again: nuc attack by second alcohol unit occurs, reforming tetrahedral shape and leaving OR-H group with a +ve charge
4) This +ve charge is removed by deprotonation (addition of this lost H+ to water reforms the acid catalyst)
You now have your neutral di-ether.
What are the products of an alcohol addition to a carbonyl compound?
- Reacts with acid OR base and one unit of an alchohol (R-OH) group to make a hemi-acetal, which usually reacts with ANOTHER molecule of alcohol to react further into an acetal. This step needs an acid cat.
Steps to imine formation?
1) Direct attack of alpha carbon by R-NH2, leaves O with neg charge, -NRH2 with +ve and tetrahedral structure
2) Proton shift: H+ moves from -NRH2 to the -ve O to make an OH group and an -RNH group: now molecule is balanced of charge. This is a carbinolamine.
3) Protonation to generate water leaving group from OH
4) Elimination of water to form = bond to N, leaving the RN-H group with a +ve charge. This is iminium.
5) Deprotonation to leave R-N=alpha and regenerate catalyst. This is imine.
Different molecules within imine formation?
- Start with amine and carbonyl compound
- Carbinolamine formed after proton shift (-ol alcohol and amine group)
- Iminium ion formed after elimination of water: charged ion with = bond to N
- Imine: the neutral end point, nitrogen analogue of carbonyl.
Steps to enamine formation?
Step one: 2° amine to iminium
Step 2: iminium to enamine
Exact mechanism outside of course.
Involves taking a H off an R group and removing the +ve charge this way, then e- from the old double bond move to form the new one.
Enamine vs imine?
Imine: double bond between N and alpha carbon
Enamine: double bond between OG alpha carbon and its adjacent R group
Mechanism of grignard reagents?
With a carbonyl:
1) The +ve MgX fragment attracts to the carbonyl oxygen while the -ve R fragment attacks the alpha carbon. These happen simultanously and so breaking the C=O bond with MgX allows the carbanion to attack the central carbon and bond. Tetrahedral structure is formed.
2) Acidify to swap out Mg-X for H to leave an alcohol group.
What is a grignard reagent and how do you make one?
A grignard reagent behaves as a carbanion and its a R-Mg-X where X is a halide.
- R and X have -ve charge, Mg has +ve charge
This reversal of polarity (umpolung) occurs as a result of insertion of Mg across the C-X bond. Where originally there was a +ve carbon and a -ve halide, the Mg takes on the +ve charge of the carbon which in turn is now -ve.
What is the function of sodium borohydride?
Sodium borohydride acts as a hydride donor. This means that you can create a tetrahedral structure through nucleophilic (H-) attack and retain a -ve charge on the oxygen.
Carbonyl and sodium borohydride mechanism?
1) Nuc attack by hydride ion forms tetrahedral shape with -ve charge on the oxygen.
2) Remaining BH3 fragment attacks the -ve oxygen: this part of the molecule is still -ve.
3) Reaction with acid donates H+ which replaces BH3 and makes the molecules neutral. Leaves alcohol.
