Joseph K. Myers


Reactions of Alcohols: Experiment 9

(Preparation of cyclododecanone from the oxidation of cyclododecanol)


The determination of the oxidation state of a carbon atom can be used to find a change in the oxidation state. The rules used assign carbon a formal oxidation state based on +1 for any atom less electronegative than carbon, -1 for atoms more electronegative, and 0 for C-C bonds. The carbon must bear an oxidation number that makes this sum equal zero.

For example, a carbon bonded to one oxygen (-1) , one hydrogen (+1), and two carbons (0), must bear a formal oxidation number of -1 + 1 + 0 = 0. If this oxgen bond is changed to a double bond, then the carbon is bonded to the equivalent of two oxygens (-2) and two carbons (0). Therefore, the oxidation number of the carbon atom must be +2.

Increasing the oxidation state or oxidation number of the carbon atom is called an oxidation reaction. This is done in two ways:

1) Increasing the number of carbon-O bonds (or any more electronegative atom than C).

2) Decrease number of C-H bonds.

Aldehydes and ketones can be synthesized by the oxidation of primary or secondary alcohols.

Hypohalites are one option for oxidizing alcohols. The mechanism by which these oxidize alcohols (creating a two-electron change in the oxidation number of the functionalized carbon atom) probably involves initial formation of an alkyl hypohalite.

This arises from reaction of the alcohol with the hypohalous acid that is in equilibrium with hypohalite ion in the aqueous medium.

E2 elimination of the elements of H-X from the alkyl hypochlorite leads now to either an aldehyde or ketone.

Reactions and mechanisms

(Attached on separate paper.)

There is more than one step in the reaction.

Table of reagents and reactants

(Table 01)

compoundmolecular formulamw (g)fpmp (deg-C)bpdensity (g/cm3)
glacial acetic acidC2H4O260.05244016.6117.91.0492
sodium hypochloriteClNaO74.44217 18401.209
anhydrous sodium sulfateNa2O4S142.03714 884 2.68
sodium bisulfiteHNaO3S104.05587 150 1.48
sodium bicarbonateCHNaO384.00687 2708512.159
sodium chlorideClNa58.44277 80114132.165
diethyl etherC4H10O74.1224-40-116.334.60.7134

Table of Reactants

(Table 02)


(*The density of cyclododecanol is not given in literature, and so the volume could not be computed.)

Flowchart or procedure

1) Put 0.5 g cyclododecanol + 1.2 mL acetone + 0.4 mL CH3COOH into RB flask. Reflux. Maintain T 45-deg-C.

2) Add 4.5 mL NaOCl using a pipet from the top of the condenser, drop by drop throughout reflux, for 15 min. Wait for 15 min. Let cool.

3) See two layers and remove water layer using a pipet.

4) Check organic layer using starch paper. Color goes to black if you have added enough NaOCl. Otherwise, add drop by drop 0.4 mL NaOCl and heat 2-3 min. Repeat this until it the color test changes, but repeat at most three times.

5) Cool. Transfer into separatory funnel (decant or use pipet). Add 5 mL of ether to RB and rinse it, and add this to the separatory funnel.

6) Separate organic layer.

7) Add 5 mL ether to aqueous layer (which was just separated from the organic layer). Then the aqueous layer will separate and form a new organic layer within itself.

8) Combine these two organic layers.

9) Wash organic layer: a) 5 mL Na2SO3, and then b) 5 mL NaCl.

10) Dry over Na2SO4 for 10 min.

11) Decant ether to RB flask. Then evaporate ether to obtain final product. Weigh the product and calculate percent yield and find the mp.


Cyclododecanol is the limiting member of the reaction. Therefore, the number of moles of theoretical product should be equal to the number of moles of cyclododecanol used.

This means that the theoretical yield of cyclododecanone is 2.9839 mmols. Based on the molecular weight of the product, the theoretical yield should have a mass of 0.54398 g.

Using the equation %y = ME/MT * 100, we get

%y = 0.53/0.54398 * 100 = 97.4%.

The results are summarized in this table:

(Table 03)

compoundtheoretical yield (g)experimental yield%ymelting point exp (deg-C)melting point lit.


Solid of the product dried onto surfaces of containers which contained product-bearing solution.

The constitution of the product made it very difficult to obtain a large enough sample for determining its melting point. The product would not easily tamp into the capillary melting point tubes.

Discussion and conclusion

A large ratio of product was lost in solution transfers, because the wet solution would dry and form a residue of product on any sides of funnels, beakers, or other containers used to contain the product.

It is unlikely that such a good yield could have been obtained under the laboratory conditions, without some explanation. There are three reasons for the large yield.

First, there may have been a tiny upwards variation in the amount of cyclododecanol used in the reaction, which would raise the yield because this is the limiting compound in the reaction. Secondly, conditions in the lab may have made cyclododecanol more dense than standard conditions under which its literature density is taken. Since the cyclododecanol was measured by volume, this would have increased the number of moles available for the reaction. Thirdly, there was probably a slight residue of unevaporated solvent under the surface of the product when it was spread out on the watch glass and weighed.

In general, these observations and results show that the experiment was carried out quite well, and that information was gained about the reactions of alcohols.


1. Gilbert JC, Martin SF. Experimental Organic Chemistry. 2002. 3rd Edition. p500-504.

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