Carbonyl compounds (3) More on Aldehydes and Ketones

 Carbonyl compounds (3) More on Aldehydes and Ketones

Let’s look at how aldehydes and ketones behave with oxidising and reducing agents and how the nucleophilic addition reaction compares with the electrophilic addition reaction in alkenes.

First a comparison between electrophilic addition in alkenes and nucleophilic addition in carbonyls.

	Electrophilic addition of H–Br to propene CH2=CHCH3	Nucleophilic addition of HCN to ethanol CH3CHO
Nature of the bond that is attacked	Planar Non-polar Carbon-carbon double bond A σ and a π bond Equal distribution of electrons	Planar Polar Carbon-oxygen double bond A σ and a π bond with an unequal distribution of electrons because oxygen is more electronegative than carbon.
Attacking species	Electrophile Polar H—Br            δ+   δ–	Nucleophile  :CN–   the cyanide ion. A nucleophile because it has a lone pair of electrons on the carbon atom.
Site of attack	The δ+ H atom of the H–Br attacks the π electron cloud above or below the plane of the propene molecule.	The δ+ charge on the carbon atom of the carbonyl C=O group. The :CN– ion attacks from above or from below the plane of the ethanal molecule.
Intermediate form	A carbocation forms with + charge on the middle carbon atom e.g: This is the more stable carbocation.	An anion forms with a negative charge on the oxygen atom e.g:
Resultant product	This forms when the Br– ion attacks the carbocation. The product is 2–bromopropane: This product is not optically active.	This forms when the O– abstracts a proton from an HCN molecule. The product is ethanal cyanohydrin: Note each individual product is optically active but the two isomers that form are equally present so that the product mixture is racemic and not optically active.

How aldehydes and ketones behave with oxidising and reducing agents:

Oxidising agents:

Ketones cannot be oxidised further since they do not possess an α hydrogen on the central carbonyl carbon.

Aldehydes can be oxidized up to carboxylic acids since they do possess an α hydrogen atom on the carbonyl carbon. 

So heated under reflux with acidified sodium dichromate solution propanal becomes propanoic acid.

Oxidation half equation:

CH₃CH₂CHO   +   H₂O     ➝   CH₃CH₂COOH +  2H+   +  2e–

Reduction half-equation:

Cr₂O₇^2-   +  14H+   +   6e–   ➝   2Cr³⁺   +   7H₂O   

Can you add up the two half equations and balance them to build the full ionic equation for the oxidation of propanal?

Reducing agents:

Typical reducing agents such as lithium tetra hydridoaluminate(III) LiAlH₄ in dry ether will act on both aldehydes and ketones and reduce them down to their corresponding alcohol e.g:

Propanal