# Percentage composition

1. Jul 19, 2009

### leena19

1. The problem statement, all variables and given/known data
You are given a sample of cinnamon oil with some kerosene oil dissolved in it.Propose a method to determine atleast approximately the percentage of kerosene oil present in this sample,by volume.
N.B. 1) you are provided with facilities for various types of distillation
2) Clue :1. think of the chemical nature of the principal constituents of cinnamon oil.
2. Recollect that eugenol is a phenol and that camphor is a ketone

2. Relevant equations

3. The attempt at a solution
OK,so cinnamon oil has,
1.cinnamaldehyde (an aldehyde)
2.eugenol (a phenol)
3.camphor (a ketone)
and kerosene oil is just a long chain hydrocarbon we get during petroleum refining ?

First ,I thought of adding Brady's reagent to the sample.Then cinnamaldehyde and camphor would settle down as orange precipitates?
Then we're left with eugenol and kerosene.I think all phenols are solids unless they are dissolved in a base,but here we're using cinnamon oil so I don't really know how I'm supposed to seperate kerosene oil from eugenol.Would adding an acid like HCl work(to neutralise the base,maybe?)?
and how does distillation help?

Last edited: Jul 20, 2009
2. Jul 19, 2009

### symbolipoint

Does the example problems description allow for chromatography (the thought is for Gas Chromatography)? Bad question: Does kerosene contain alehyde or ketone or phenol functional groups?

3. Jul 19, 2009

### leena19

I don't think so.I think I've used chromatography paper or something like that in my lower grades when the teacher wanted to show us the different colours(I think?)in different solutions.other than that I haven't learnt anything about any type of chromatography.

I have a feeling it doesn't have any functional groups but I'm not very sure.
All I know is that kerosene obtained from petroleum refining has a large C skeleton ,about 12 to 16 C atoms and is used as a fuel,so I don't know if having any of those functional groups would help in burning ?

4. Jul 19, 2009

### Staff: Mentor

Boiling points?

5. Jul 20, 2009

### leena19

Ah!OK,so then I can use fractional distillation to seperate kerosene from eugenol.
Phenols usually have high boiling points(I can't find its exact value) due to the presence of Hydrogen bonds,but I don't think its as high as kerosene cause kerosene has a large C skeleton(whose boiling point my notes say is in the range of 175-325 degrees celsius ) ?

6. Jul 20, 2009

### leena19

I got down the answer for this from the markscheme,but I'm having trouble understanding it.I'm sure there are several methods of doing it ,but these were just 2 methods given.I really hope someone can explain it to me.
Method 1
1.Take a known volume of sample
2.Remove eugenol with aqueous NaOH(wouldn't this yield a solution of Na phenoxide ?and the other components in the mixture remain unreacted,right?so how can we seperate only eugenol here?)
3.Using a seperating funnel,seperate cinnamaldehyde and camphor by reacting the mixture with 2,4-DNPH.Then seperate the unreacted liquid
4.Measure the volume of unreacted liquid kerosene and calculate its % by volume

Method 2
1.Take a known volume of sample
2.Oxidise it using KMnO4 in either acid or alkaline medium(so here the cinnamaldehyde would get oxidised to a carboxylic acid solution and the ketone camphor,the phenol eugenol and kerosene would remain unreacted,right?if so then how can we)
3.Remove the unreacted liquid kerosene using a seperating funnel and calculate its percentage volume

Any help would be much appreciated!

Last edited: Jul 20, 2009
7. Jul 20, 2009

### chemisttree

So, your original problem was to isolate three different aromatic oils from kerosene! Eugenol forms Na+ salts upon treatment with aqueous NaOH and can be extracted into water. Cinnimaldehyde (aldehyde) and camphor (ketone) form 2,4-dinitrophenylhydrazone derivitaves, which can be removed by filtration. Distillation of the remainder yields pure kerosene. What don't you understand?

BTW, dilute basic KMnO4 reacts with alkenes to form glycols as well. BP of camphor is 204oC, eugenol is 256oC and cinnimaldehyde is 248oC and kerosene is between 150oC and 300oC. Can you separate these by distillation?

8. Jul 21, 2009

### leena19

I understand now.The problem was I didn't think the sodium salt would be extracted into water and the others would remain(which is kinda obvious,i don't know why i missed this,but)thanks for clearing it up for me!

I guess not.The boiling points of camphor,eugenol and cinnamaldehyde fall within the boiling point range of kerosene,so I don't think these can be seperated by distillation?

But i still find it difficult to understand this part of the solution given
I don't know much about glycols(just the chemical formula and that it's a diol) and wikipedia didn't help much(didn't understand a thing except that it's a syrupy liquid and is miscible with water like all alcohols,so the idea is if a glycol is formed we could extract it into water like we did with the Na+salt,right?)
http://en.wikipedia.org/wiki/Ethylene_glycol" [Broken]

Anyway when we add basic KMnO4 to cinnamaldehyde,the alkene part of it as you say would get oxidised to form a diol(forgive me if this is a stupid question but since the diol formed here is like a secondary alcohol,wouldn't it get further oxidised to a ketone?),
the alkene part in eugenol would also get oxidised to a diol,but what about camphor?
camphor has an IUPAC name of 1,7,7-trimethyl-bicyclo(2,2,1)heptan-2-one as given in this site http://www.3dchem.com/molecules.asp?ID=203#"
and wiki gave the structural formula in the line diagram-like (not sure what you call it)wayhttp://en.wikipedia.org/wiki/File:Camphor_structure.png which i wasn't very familiar with,but by looking at the IUPAC name,I don't think it has an alkene part in it,which would mean it wouldn't react with basic KMnO4?

I'm sorry I have so many annoying questions,but any help would be very very much greatly appreciated !
Already very grateful to all who helped!

Last edited by a moderator: May 4, 2017
9. Jul 21, 2009

### Staff: Mentor

My English failed me - I thought kerosene is much lighter fraction with lower boiling point. Now I have read that it is the same thing we call 'liquid parafine' here :grumpy:

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