6.1.2 (a,b,c) Reactions of Carbonyl compounds

(a) oxidation of aldehydes using Cr₂O₇^2–/H⁺ (i.e. K₂Cr₂O₇/H₂SO₄) to form carboxylic acids

{In equations for organic redox reactions, [O] and [H] should be used.}

(b) nucleophilic addition reactions of Carbonyl compounds with:

(i) NaBH₄ to form alcohols

(ii) HCN [i.e. NaCN_(aq)/H⁺_(aq)], to form Hydroxynitriles

(c) the mechanism for nucleophilic addition reactions of Aldehydes and Ketones with NaBH₄ and HCN

{For NaBH₄, the nucleophile can be considered as being the hydride ion, H^–, with subsequent protonation of the organic intermediate from H₂O. For HCN, initial nucleophilic attack is from CN^– ions; subsequent protonation stage can be shown using H₂O or H⁺.}

What does this mean?

Carbonyl is a term which covers the Ketone and Aldehyde functional groups.

You should note that this only includes aldehydes and ketones even though Esters, Acyl Chlorides, Acid Anhydrides and Amides contain the C=O bond.

[O] represents any oxidising agent.

Examiners often ask for equations to include [O] as a reactant but they still expect you to balance the equation.

As it happens, oxidising an aldehyde to a carboxylic acid requires no balancing as the [O] is balanced by the additional O atom in the acid - this isn't true for oxidising alcohols though.

CH₃COH + [O] --> CH₃COOH

The Oxidising agent of choice is acidified Dichromate (Cr₂O₇^2–/H⁺)- the examiner will penalise you for not including the word acidified because the oxidation reaction won't work without the addition of conc. acid.

You should know that the dichromate ion is Cr₂O₇^2–.

And that it is reduced to Cr³⁺ when it reacts.

You can then work out the half-equation without having to memorise it

Cr₂O₇^2– --> 2 Cr³⁺

Balance O's with water: Cr₂O₇^2– --> 2 Cr³⁺ + 7 H₂O

Balance H's with H+: 14 H⁺ + Cr₂O₇^2– --> 2 Cr³⁺ + 7 H₂O

Balance charge with e^-: 6 e- + 14 H⁺ + Cr₂O₇^2– --> 2 Cr³⁺ + 7 H₂O

You won't be asked to use this half-equation to oxidise aldehydes but you need to know it for other parts of the syllabus.

The examiner expects you to know that aqueous NaBH₄ (sodium tetrahydridoborate) is the reducing agent of choice here (and in almost all other parts of the syllabus)

As with the oxidation, the equations should be written with [H] as a reactant.

CH₃COCH₃ + 2 [H] --> CH₃CHOHCH₃ --------- 2 [H] needed to balance

CH₃COH + 2 [H] --> CH₃CH₂OH --------- 2 [H] needed to balance

Unlike the Oxidation you are expected to know the mechanism.

Generally, Hydrogen atoms form H⁺ when they form ions.

But when reacted with metals, like Sodium in NaBH₄ , H^- ions (Hydride ions) can form.

The H^- ion acts as a nucleophile in Step 1.

Another H atom is needed by the RCH₂O^- in Step 2 but this can't be provided by more H^- because negative ions would obviously repel.

So the d+ H in a water molecule is used.

Overall, this reduction can also be thought of as Nucleohilic Addition.

The first step is the same as in the reduction/nucleophilic addition with NaBH₄

The examiner won't mind if the H required in Step 2 comes from the d+ Hydrogen in HCN (as above) or from H+ ions formed when HCN dissociates HCN --> H⁺ + CN^-.

The product is a Hydroxynitrile ( where Hydroxy signifies OH, and Nitrile signifies CN)

Overall, the reaction is CH₃COCH₃ + HCN --> CH₃COH(CN)CH₃

Although it's more usual to think of the reactant as H⁺/NaCN rather than HCN -- mostly since it's easier and safer to store NaCN.