The production of HCl gas on adding phosphorus pentachloride is a useful test for the alcohol( presence of hydroxyl group) since the blue litmus experiences a colour change from red to blue due to the formation of an acidic gas( hydrogen chloride).
Nb: Phosphorus pentachloride contains the ions PCl4+ and PCl6-. The PCl4+ ion performs a similar role to the H+ ion in the process of nucleophilic substitution By HBr. It weakens the carbon-oxygen bond and assists in the nucleophilic substitution reaction. The PCl4+ ions and the H+ ion both act as Lewis acids, accepting a lone pair from the oxygen atom in the hydroxyl group on the alcohol molecule.
3i)Pour a little ethanol in to a watch glass and set it on fire using a blue Bunsen burner flame.
Observations
- the colourless ethanol lights/ignites immediately.
- it burns with a clean blue(non-luminous) flame and produces no soot.
Inference:
Alcohols burn in air to form carbon dioxide and water. The molecular structure is completely destroyed and the constituent atoms oxidize carbon dioxide and water. The reaction for the combustion of ethanol is as follows:
CH3CH2OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
The reaction is strongly exothermic, and hence ethanol is used as a fuel in areas where it can be used cheaply.
The blue flame suggested that there was a good supply of oxygen and hence complete combustion of ethanol took place. There was no carbon monoxide or unburnt carbon(soot) formed due to incomplete combustion of Ethanol.
3ii)To 5 cm3 of ethanol in a clean dry test tube add 2-3 drops of concentrated H2SO4(aq) carefully. Add a drop at a time. Test any gas evolved by setting it on fire while in an inverted test tube.
Observations:
- Few bubbles seen in the mixture.
- In the presence of a lighted splint, the splint went off producing a ‘pop’ sound.
Inference:
This involves elimination of water. Alcohols are manufactured by adding water to an alkene, using phosphoric acid as a catalyst. This reaction may be reversed to eliminate water from alcohol and produce an alkene. The elimination process involves dehydration i.e. removal of a molecule of water. Alcohols are dehydrated by warming with concentrated sulphuric acid at about 180 degrees:
CH3CH2OH(l) ) → CH2 CH2 + H2O(l)
In this case the ethanol is protonated by the acid. Water is then lost to form a carbocation( in which a positive charge is very briefly located on a carbon atom). Any nucleophile could now attack this carbon atom.
However, if no nucleophile is present, two electrons from the neighboring C—H bond form another C—C bond( making a double bond) and the hydrogen is lost. An alkene is formed.
4i) to 2cm3 of ethanol, add a few drops of acidified potassium dichromate followed by warming.
Observations:
The orange solution turns dirty green and light green precipitate is formed that sediments at the bottom of the test tube.
Inference( partial oxidation of primary alcohols):
Ethanol is a primary alcohol. Primary alcohols can be oxidized to aldehydes or carboxylic acids depending on the reaction conditions.
Through partial oxidation an aldehyde is obtained if you use an excess of the alcohol, and distil off the aldehyde as soon as it forms. The excess of the alcohol means that there isn't enough oxidising agent present to carry out the second stage. Removing the aldehyde as soon as it is formed ensures that it can not be further oxidised.
If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, CH3CHO.
The electron half equation for the reaction is shown below:
A simplified version of the equation can be used to concentrate on what is exactly happening. To do that, oxygen from an oxidising agent is represented as [O]. That would produce the much simpler equation:
The simple structures below illustrate the relationship between the ethanol and ethanal. In ethanol, the O—H group is attached to a carbon atom bearing two hydrogen atoms. Oxidation removes one of these hydrogen atoms together with the hydrogen atom in the O-H group.
Nb: If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. Hence the solution turned from orange to green colour.
The electron-half-equation for this reaction is
4II)Repeat with H+/ KMnO4-
Observations:
The solution changes colour from colourless to purple to red brown and on heating to dark brown solution. There is also formation of a brown precipitate.
Inference:
The acidified potassium manganate( VII) solution being a powerful oxidizing agent oxidizes the primary alcohols to acids hence the ethanol forms ethanoic acid. It is too powerful to form aldehydes and hence it forms carboxylic acids.
The aldehyde( in this case ethanal) functional group –CHO consists of a hydrogen atom attached to a carbonyl C=O carbon atom. This structure may be oxidized further. This results in a carboxylic acid( ethanoic acid) with the structure RCOOH.
Overall in this case, the primary alcohol ethanol oxidizes in two stages: first to the aldehyde(ethanal) and then to the carboxylic acid( ethanoic acid).
You could write separate equations for the two stages of the reaction - the formation of ethanal and then its subsequent oxidation.
This is what is happening in the second stage:
Nb: there was an overall colour change from colourless to brown. This was because of the formation of manganese(IV) oxide:
MnO4-(aq) + 4 H+(aq) + 3e-==>MnO2(s) + 2 H2O(l)
5) to 2cm3 of propan-2-ol, add a few drops of acidified potassium dichromate followed by warming. Repeat with H+/ KMnO4-
Observations:
It becomes a honey coloured solution and turns green on heating.
Inference:
The structure of a secondary alcohol is RR’CHOH. The –OH group is attached to a carbon atom bearing one hydrogen atom. As in the case of primary alcohols, oxidation removes this hydrogen atom together with the hydrogen atom in the O—H group. This results in a ketone RR’C=O.
if you heat the secondary alcohol propan-2-ol with potassium dichromate(VI) solution acidified with dilute sulphuric acid, you get propanone formed.
Altering reaction conditions makes no difference whatsoever to the product.
Using the simple version of the equation and showing the relationship between the structures:
CH3CHOH(aq)→(CH3)2CO(aq)
In the second stage of the primary alcohol reaction, an oxygen is "slotted in" between the carbon and the hydrogen in the aldehyde group to produce the carboxylic acid. In this case, there is no such hydrogen - and the reaction has nowhere further to go. As a result the final product is propanone.
6) to 2cm3 of 2-methylpropan-2-ol, add a few drops of acidified potassium dichromate followed by warming. Repeat with H+/ KMnO4-
Observations:
The solution turns dark orange on heating and the flame goes off producing a ‘pop’ sound.
Inference:
Tertiary alcohols aren't oxidised by acidified potassium dichromate(VI) solution. There is no reaction whatsoever.
If you look at what is happening with primary and secondary alcohols, you will see that the oxidising agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom attached to the -OH. The structure of a tertiary alcohol is RR’R’’COH. In this case, there is no hydrogen atom attached to the carbon atom bearing the hydroxyl group.
You need to be able to remove those two particular hydrogen atoms in order to set up the carbon-oxygen double bond.
Due to this tertiary alcohols like 2-methylpropan-2-ol cannot be oxidized readily( except under extreme conditions when carbon-carbon bonds break).
Nb: a powerful acidic oxidizing agent converts tertiary alcohols in to a mixture of carboxylic acids.
7i) to pea shaped flask of quick fit apparatus pour in 20 cm3 of ethanol. Drop in a few pieces of anti-bumping chips. Add 10 cm3 of H+/ Cr2O7(aq) and immediately distill off the product as it forms.
Observations:
- The solution changes colour from orange to dark green.
- Clear colourless distillate is collected in the receiver( the beaker)
- reducing property.
Inference:
Ethanol which is a primary alcohol is oxidized by the chromate ions in two stages first to the aldehyde ethanal. The colour change is caused by the reduction of potassium dichromate to green chromium(iii) ions. The electron half equation is shown below:
7ii)Carry out a mirror test on the sample of the distillate.
Observations:
- silver particles are formed on the walls of the test tube
Inference:
Tollens’ reagent is sometimes called “ammoniacal silver nitrate”. It is made as follows. A small quantity of aqueous sodium hydroxide is added to aqueous silver nitrate, forming a precipitate of hydrated silver oxide Ag2O. Addition of aqueous ammonia causes the precipitate to form a solution containing the silver ammine complex [Ag(NH3)2]+(aq). On warming, aldehydes(like ethanal in this case) reduce Tollens’ reagent, which is colourless, to a grey precipitate of metallic silver: In a thoroughly cleaned glass container, the precipitate forms a silver mirror. Ketones do not react with Tollens’ reagent.
8)To 5 cm3 of ethanol in to a boiling tube, add an equal quantity of glacial ethanoic acid, followed by 3 drops of sulphuric acid( H2SO4(aq)). Warm a bit. Transfer and pour the resultant mixture in to a 100mL half-filled beaker with cold water.. Repeat with 2 methyl propan-2-ol. Explain the difference in smell if any.
Observations:
-Effervescence occurs when it is transferred to the cold water.
-The gas turns blue litmus red and there is formation of a sweet smelling product.
Inference:
Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid.
The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R'OH (where R and R' can be the same or different) is:
Ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid as a catalyst to produce the ester, ethyl ethanoate. The reaction is slow and reversible. To reduce the chances of the reverse reaction happening, the ester is distilled off as soon as it is formed.
In this reaction, it is the C(O)—H bond in the acid that breaks and not the C—OH bond in the alcohol. The reaction mechanism is confirmed by using the alcohol that is labeled with the isotope 18O. On separation of the reaction products it is found that it is the ester that contains the labeled oxygen and not the water.
8ii)Repeat with propan-2-ol
Observations:
- Effervescence is observed.
- Gas produced turns blue litmus red
- Sweet smelling product
8iii)Repeat with 2 methyl propan-2-ol.
Observations:
- Effervescence is observed.
- Gas produced turns blue litmus red
- Very sweet smelling product
Combined inference(Explain the difference in smell if any):
The explanations for the reactions of propan-2-ol and 2 methyl propan-2-ol are the same as those of the primary alcohol ethanol. The only difference is that the ester that is formed gives a sweeter smell as we move from ethanol to propan-2-ol and to 2 methyl propan-2-ol.
References:
- Advanced chemistry by Michael Clugston and Rosalind Flemming
- As and A level chemistry by Erik Lewis and Martyn Berry
- An introduction to the chemistry of carbon compounds by E.O Arene and T.M. Kitwood
- IB chemistry by John Green and Sadru Damji.