Module 6: Carbonyl Compounds (6.1.2) Flashcards
What is an aldehyde?
- Carbonyl (C=O) functional group at the end of the chain.
- Structural formula written as CHO.
- Suffix is -anal
- Carbon atom of carbonyl group always designated C1.
What is a ketone?
- Carbonyl functional group joined to 2 C atoms in the chain.
- Structural formula written as CO.
- Suffix is -one.
Oxidation of Aldehydes
- Oxidised to a carboxylic acid by reflux, using K2Cr2O7 and dilute H2SO4
- Orange to green colour change.
- EXAMPLE: C4H8O + [O] ——-> C3H7COOH
- K2Cr2O7, H2SO4 and reflux over the arrow
Oxidation of Ketones
- Do NOT undergo oxidation reactions due to lack of reactivity.
The C=O Bond
- A σ bond between the Carbon and Oxygen
- A π bond above and below the plane of Carbon and Oxygen.
- C=C bond in alkenes non polar.
- C=O bond in carbonyls are polar.
- Oxygen more electronegative than Carbon, electron density of double bond lies closer to O.
- Carbon end slightly more +ve and Oxygen end slightly -ve. (Dipoles)
Nucleophilic Addition
- Due to polarity of C=O bond, nucleophile is attracted to and attacks the slightly +ve Carbon atom.
- Resulting in addition across the bond.
Reduction of Aldehydes with NaBH4
- Warmed with NaBH4 in an aqueous solution.
- Aldehydes reduced to primary alcohols.
- EXAMPLE: C4H8O + 2[H] ——-> C4H9OH
- NaBH4 and H2O over the arrow.
Reduction of Ketones with NaBH4
- Warmed with NaBH4 in an aqueous solution.
- Ketones reduced to secondary alcohols.
- EXAMPLE: C3H6O+ 2[H] ——–> C3H8O
- NaBH4 and H2O over the arrow.
Mechanism for NaBH4
- Lone pair of e- from Hydride ion, :H- is attracted to the δ+ C atom.
- Dative covalent bond formed between Hydride and C of C=O bond.
- π bond in C=O breaks by heterolytic fission.
- Oxygen donates lone pair to H in H2O.
- Intermediate has been protonated to form an alcohol, with OH- as a byproduct.
Reduction of Aldehydes with NaCN
- NaCN and H2SO4 used to provide HCN.
- HCN is a colourless poisonous liquid and cannot be used in an open lab.
- Provides a mean of increasing C chain length.
- Product has 2 functional groups, OH and C TRIPLE BOND N
- EXAMPLE; C3H6O + HCN ——-> C3H9OCN
- NaCN and H2SO4 above the arrow
Mechanism for NaCN
- Lone pair of e- from Cyanide ion, :CN- is attracted to the δ+ C atom.
- Dative covalent bond formed between Cyanide and C of C=O bond.
- π bond in C=O breaks by heterolytic fission.
- Intermediate protonated by Hydrogen ion to form a hydroxynitrile.
Testing for C=O Bond with 2,4-DNP
Method:
1. Add 5cm of 2,4-dinitrophenylhydrazine/Brady’s reagent to test tube.
2. Add 3 drops of unknown solution using a dropping pipette and leave to stand.
3. If no crystals form, add a few drops of HS2O4.
4. Yellow-orange precipitate indicates presence of an aldehyde or ketone.
Reaction:
1. Nucleophilic-addition elimination.
2. Condensation reaction, removal of water.
3. NH2 adds across C=O bond.
Melting Point of 2,4-DNP
Method:
- Impure yellow/orange ppt filtered to separate it from the solution.
- Recrystallised to from pure sample of crystals.
- Melting point measured and compared against known to identify original carbonyl compound.
Distinguishing Between Aldehydes and Ketones
Method:
- Add 3cm of AgNO3 (aq).
- Add NaOH (aq) until brown precipitate of silver oxide, Ag2O formed.
- Add dilute ammonia solution until brown ppt dissolves to form colourless solution
- Add 2cm of unknown solution and equal volume of freshly prepared Tollen’s reagent.
- Leave tube to stand in 50°C for 10 minutes.
- Observe whether silver mirror is formed.
Why Does The Silver Mirror Form?
- Ag+ acts as an oxidising agent presence of ammonia.
- Ag+ reduced to Ag.
- Aldehyde oxidised to carboxylic acid.
