Identifying Carbonyls

Question 1

Describe oxidation with Acidified di chromate

Answer 1

Aldehydes can be oxidised so a colour change of orange to green is observed. It is the Cr3+ ions that cause this colour change.

Question 2

What is the half equation for the reduction of sodium dichromate?

Answer 2

Cr2O7 2- + 14 H+ +6e- = 2Cr3+ + 7H2O

Question 3

Describe the oxidation with Fehlings/benedicts

Answer 3

Both contain Cu(II) complexes which are reduced to Cu(I) complexes. When reduced they go from blue to brick red.

Question 4

What is the half equation for fehlings/benedicts?

Answer 4

Cu2+ + e- = Cu1+

Question 5

Describe the oxidation with Tollens reagent

Answer 5

Aka the Silver mirror test. Silver nitrate and aqueous ammonia reduced to produced solid silver. This coats the boiling tube ruining it.

Question 6

Half equation for Tollens?

Answer 6

Ag(NH3)2 + e- = Ag (solid) + 2NH3

Question 7

Describe Brody’s reagent?

Answer 7

Reduction of 2,4 DNPH. If carbonyl present it becomes a bright orange precipitate. Each structure has dif. Melting points so can be identified.

Question 8

Properties of carbonyls

Answer 8

• relatively low boiling point

- small soluble, large not

Question 9

Why do carbonyls have a lower boiling point?

Answer 9

They cannot form H-bonds with themselves so only have London and dipoles.

Question 10

Why are only small carbonyls soluble in water?

Answer 10

Large carbonyls tails get in the way. H bonds with the water are made with the O.