Aldehydes and Ketones Rxns Flashcards
Nucleophilic acyl addition (2 mechanisms)
- Deprotonated nucleophile plus ketone
- Nu adds to C, double bond to O breaks to single
- Acid adds
- H from acid attacks -Ve charge on O; changes it to -OH
- Protonated Nucleophile
- Adds to C, double bond to O breaks into a single bond
- Proton transfer; one H from the nucelophile breaks off and attaches to the O atom.
- -H20
- Final product- Nu is attached to the central C atom with a double bond (remaining proton and OH group are removed as water)
Formation of acetals
- Ketone + H+ -> H adds to the O group and gives it a positive charge, making the central carbon atom even more delta +
- Methanol attacks the +Ve carbon centre (the O atom is the one that bonds and also gains a +Ve charge). The double bond breaks and the OH group looses its positive charge.
- A base steals one of the H atoms attached to the methanol group, getting rid of it’s positive charge (OHCH3+ -> OCH3). The product formed is a hemiacetal + BH
- H+ is added to the OH group (OH-> OH2+)
- OH2+ is eliminated as water. The first methanol group becomes attached with a double bond (which increases delta +Ve charge on the central atom, as in step one)
- Methanol is added. As in step 2, the O atom attaches to the central carbon (attacking the + charge). The O of the new group carries a positive charge.
- The extra H atom on the new methanol group is picked up by a base, getting rid of the positive charge on the O. The resulting product is an acetal.
Formation of imines
Nucleophilic addition followed by the elimination of water
- Amine + ketone -> double bond to O breaks and amine attaches to central carbon through the N atom (which has a positive charge)
- Negatively charged O atom picks up one of the protons from the amine, eliminating both the negative charge and the positive charge on the nitrogen
- Addition of H+. Attaches to the OH group.
- Elimination of water + a double bond forms to the nitrogen, which now has a positive charge again (C=N+RH)
- -H+ eliminates the positive charge. The final species is 2R–C=N-R
Addition of amines
Aka reductive amination
- Ketone + amine -> imine intermediate
- Reduction of imine with H2 and Ni catalyst yields amine
Enol-Keto tautomerism
- Proton transfer between the alpha carbon (opposite end to double bond O) and the carbonyl oxygen
- C-C=O to C=C-OH
- The keto form (with double bond O) is generally favoured
- Product of reaction is a racemic mixture
Alpha-halogenation
Ketone + X2 = C=OCH2X
Oxidation of aldehydes
- Aldehyde + H20 -> Hydrate (2 OH groups, one H and one methyl attached to the central carbon)
- Hydrate + oxidant (eg. H2CrO4) -> intermediate species (reduced oxidant species connected to one of the O atoms, base comes in and removes extra H attached to CC)
- Final product; carboxylate acid (-COH-> -COOH)
Oxidation of ketones
Ketone Oxidation is more resistant; only works for cyclic species and usually a strong oxidant is needed eg. HNO3 (chromate doesn’t usually work)
1. Cyclic Ketone + strong oxidant at high temperature = diacid (ring breaks)
Reduction
Both ketone and aldehydes can be reduced in the presence of a metal catalyst (Ni/Pt) at 2atm with H2. The species formed is an alcohol
Metal hydride reduction (selective)
Aka nucleophilic acyl addition
Why use it? Because it’s selective for carbonyl groups (won’t reduce double bonds or other groups)
1. Metal hydride (eg. NaBH4) plus ketone-> double bond breaks and H+ is added to the CC
2. Addition of H+attacks the O-, forming an -OH group
3. Final product: =O -> -OH
Addition of Grignard reagents
- organometallic compounds that are strongly basic and nucleophilic, eg. H3C (delta minus)-Mg (delta plus)Br
- addition with protic solvent doesn’t work because the GR reacts preferentially with the acid
1. Formation of grignard reagent (H3C-Br + Mg with ether = H3C-MgBr
2. Reagent plus ketone -> break O double bond, methyl group added to CC
3. Plus H+ -> Three methyl groups and a hydroxyl bonded to CC
