Determine % Volume of Kerosene Oil in Cinnamon Oil Sample

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SUMMARY

The discussion focuses on determining the percentage volume of kerosene oil in a cinnamon oil sample using various chemical methods. Key components of cinnamon oil include cinnamaldehyde (an aldehyde), eugenol (a phenol), and camphor (a ketone). Two proposed methods involve using aqueous NaOH to extract eugenol and employing KMnO4 for oxidation, allowing for the separation of kerosene from the other components. The boiling points of the substances indicate that fractional distillation is a viable method for isolating kerosene from the mixture.

PREREQUISITES
  • Understanding of fractional distillation techniques
  • Knowledge of chemical functional groups: aldehydes, phenols, and ketones
  • Familiarity with aqueous NaOH and its reactions with phenols
  • Basic principles of oxidation reactions, particularly with KMnO4
NEXT STEPS
  • Research the process of fractional distillation and its applications in separating liquid mixtures
  • Study the chemical properties and reactions of phenols, specifically eugenol
  • Learn about the oxidation reactions involving KMnO4 and their implications in organic chemistry
  • Explore the use of chromatography, particularly Gas Chromatography, for separating complex mixtures
USEFUL FOR

Chemistry students, organic chemists, and laboratory technicians involved in the analysis and separation of chemical mixtures, particularly in the context of essential oils and petroleum products.

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Homework Statement


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

Homework Equations





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 separate kerosene oil from eugenol.Would adding an acid like HCl work(to neutralise the base,maybe?)?
and how does distillation help?
 
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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?
 
symbolipoint said:
Does the example problems description allow for chromatography (the thought is for Gas Chromatography)?

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 learned anything about any type of chromatography.

Does kerosene contain alehyde or ketone or phenol functional groups?

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 ?
 
Boiling points?
 
Ah!OK,so then I can use fractional distillation to separate 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 ) ?
 
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 separate only eugenol here?)
3.Using a seperating funnel,separate cinnamaldehyde and camphor by reacting the mixture with 2,4-DNPH.Then separate 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!
 
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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?:wink:
 
chemisttree said:
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?

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!

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?
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 separated by distillation?

But i still find it difficult to understand this part of the solution given
leena19 said:
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?)

chemisttree said:
dilute basic KMnO4 reacts with alkenes to form glycols as well.
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"

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!
 
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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
 

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