Question

1. mix 2.5 Ml cyclohexanone, 7.5g of KMnO4, 65 ml H2o in a round bottom flask. Use steam bath and heat to 30 degree C. 2. Add 1 ml of 3M NaOH solution, stirring with glass rod. When the temperature reach 45 degree C. slow the oxidation process by ice-cooling, and keep the temperature at 45 degree C for 30 mins. 3. Then heat the picture on heating mantle, adjust to 70V, wi reflux to complete the oxidation (55 degree C) 4. Make a spot test by withdrawing reaction mixture on the top pf the stirring rod and touching it to a filter paper. KMnO4 if present, will show purple ring around the spot. 5. Keep heating and make the spot test, unit there is no KMnO4 present. 6. Separate the solid and liquid mixture by suction filtration, rinse the brown precipitate well with water. 7. Add a boiling chip and evaporate the filtration to a volume of 17 ml. (while evaporating my cup tipped over and all of my product spilled, I only had about 3ml left.) 8. Acidify the hot solution with concentrated HCl to pH 1-2, add 5 ml acid in excess, and let the solution stand to crystallize. 9. Collect the crystals with suction filtration, rinse the solid with water. Turn on vacuum for few minutes to dry the solid completely. 10. measure the madd of the solid. and determine the melting point.

Calculate the theoretical yield of adipic acid.

What is the limiting reagent in this reaction? Why do you think this reagent rather than the other made limiting?

Why does acidification of the concentrated reaction mixture cause precipitation of adipic acid?

Why does the initially purple solution of KMnO4 change color as the oxidation progresses?

Answer 1

Cyclohexanone is a 6-membered cyclic ketone with the structure (CH2)5CO

Alkaline permanganate oxidation will open the 6-membered ring to give a straight chain dicarboxylic acid,
hexanedioic acid (adipic acid)

Reaction:
(CH2)5CO + [O] → HOCO(CH2)4COOH

Since this is an alkaline reaction medium, the hexanedioic acid will react with sodium hydroxide (NaOH) to
form a water SOLUBLE disodium salt:

HOCO(CH2)4COOH + 2NaOH → [-OCO(CH2)4COO-] 2Na + 2H2O

Some of the hexanedioic acid will also form a water INSOLUBLE pink manganous salt that will precipitate:

HOCO(CH2)4COOH + Mn^2+ → [-OCO(CH2)4COO-] Mn

Acidification, after oxidation, of the alkaline solution with hydrochloric acid will precipitate the slightly
soluble hexanedioic acid:

[-OCO(CH2)4COO-] 2Na + HCl → HOCO(CH2)4COOH + NaCl

Since the manganous hexanedioate is insoluble, it will not be neutrallised and will co-precipitate with your
product contaminating it.

The molar mass of cyclohexanone is 98.15 g/mol
The molar mass of hexanedioic acid is 146.14 g mol−1
Density of cyclohexanone is 0.95 g/mL

Starting from 2.5 mL cyclohexanone, the theory yield of hexanedioic acid is:

(2.5 x 0.95) ÷ 98.15 x 146.14 g
= 0.02419 x 146.14 g = 3.53 g

Melting point: 152 °C

Here, the potassium permanganate is reduced, meaning it gains electrons, and the sugar is oxidized, meaning
it loses some.

This happens in two steps. In the first step, the permanganate ion (the part of the potassium permanganate
that changes) is reduced to the manganate ion:

MnO4- + e- → MnO4-2

The compound on the left is purple, and the one on the right is green. As this reaction is going, there is
some purple and some green in the solution and these combine to make it look blue at the
beginning.

Next, the green manganate is reduced again into manganese dioxide:

MnO4-2 + 2H2O + 2e- → MnO2 + 4OH-

The manganese dioxide is a brown solid, but it’s in such tiny particles that it appears to make the liquid
turn yellow.