Results:
- Lucas test
- Oxidation test
- Esterification test
Discussion:
- Lucas test
The Lucas reagent is an aqueous solution which consists of strong acid (hydrochloric acid, HCl) and zinc chloride. Lucas reagent is commonly used to determine the class of alcohols. The reactants used to react with Lucas reagent must be soluble in water for the reaction to take place. A compound with more than six carbons is to large to be dissolved in the reagent and therefore will not react in most cases. Therefore only water-soluble alcohols of low molecular weight will provide positive results in this test. Hence only four different low molecular weight alcohols are used, which are the 2-methyl-2-propanol, 2-butanol, n-butanol and an unknown alcohol X.
The reaction that occurs in the Lucas test is a nucleophilic substitution. The reaction is a substitution in which the chlorine replaces the hydroxyl group. Only alcohols that can generate stable carbocation intermediates will undergo the reaction. The acid catalyst activates the hydroxyl group (OH group) of the alcohol by protonating the oxygen atom. The C-OH bond breaks to generate the carbocation, which in turn reacts with the chloride ion (nucleophile) to generate an alkyl halide product. Besides, the hydroxyl ions are protonized with hydrogen ions to form water. Thus the reaction involves replacement of hydroxyl group (-OH) in the alcohol by the chloride ions from the Lucas reagent to form alkyl halide (or haloalkane) and water. Alkyl halide is non-polar compound, while water is polar compound. Hence when these two compounds are formed together, they do not mix with each other, forming a cloudy two layers solution, where the less dense colourless water is on top and the cloudy denser alkyl halide is at the bottom.
Lucas test can differentiate the classes of alcohol by comparing the product formed and the time taken for reaction. No reaction within 15 minutes generally implies that the alcohol used is an primary alcohol. If the product formed turns cloudy and form two layers after 3-5 minutes signifies that the alcohol used is a secondary alcohol. When tertiary alcohol is tested in the other hand, the solution will turn cloudy immediately. The speed of this reaction is proportional to the energy required to form the carbocation. The reaction rate is faster when the carbocation intermediate is more stabilized by greater number of electron donating group (R-) bonded to the positively charged carbon atom while less substituted alcohol reacts much more slowly. Thus, a tertiary alcohol reacts fastest followed by secondary alcohol while primary alcohol generally reacts very slowly.
From the experimental results obtained, we know that 2-methyl-2-propanol is a tertiary alcohol (3° alcohol) because this alcohol reacts immediately when Lucas reagent is added to form 2 layered-solution which is cloudy. 2-butanol is a secondary alcohol (2° alcohol) where it took around 7 minutes to form a cloudy solution while n-butanol and alcohol X are primary alcohols (1° alcohol) as they did not react significantly after 15minutes. The equations for the reactions of different alcohols used are as below:-
2-methyl-2-propanol (3° alcohol):
ZnCl2
(CH3)3COH + HCl (CH3)3CCl + H2O
(2-methyl-2-propanol) (2-chloro-2-methylpropane)
2-butanol (2° alcohol):
ZnCl2
CH3CH(OH)CH2CH3 + HCl CH3CHClCH2CH3 + H2O
(2-butanol) (2-chlorobutane / 2-butyl chloride)
n-butanol (1° alcohol):
ZnCl2
CH3CH2CH2CH2OH + HCl CH3CH2CH2CH2Cl + H2O
(n-butanol) (1-chlorobutane / n-butyl chloride)
Despite the fact that the experimental results obtained are accurate, 2-butanol took a longer time to react as compared to the theoretical results where it should react in within 5 minutes. This may due to slight contamination occurred in the reactants and reagents.
- Oxidation test
Oxidation test is also another way to determine the class of alcohols. There are a variety of oxidizing agent that can be chosen. For example sodium permanganate, potassium permanganate, potassium dichromate and hydrogen peroxide. In this experiment, the reagent used to oxidize the alcohols is sodium dichromate (Na2Cr2O7). Before the sodium dichromate is directly used to oxidize any alcohols, this reagent must be treated with concentrated sulphuric acid (H2SO4) to become acidified sodium dichromate. The acidic conditions keep the chromium in the Cr2O72- state so that it is able to oxidize the alcohols. If oxidation occurs, the yellow solution containing the dichromate (VI) ions, Cr2O72- is reduced to a green solution containing chromium (III) ions, Cr3+. Hence the colour of product formed can be used to identify and class the alcohols. Besides, heat is added to increase the efficiency of the reaction.
Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O
(orange / yellow) (green)
Oxidation reaction involve the removal of the hydroxyl group(-OH group) and a hydrogen group from the alcohol and replace it with oxygen by forming a carbon-oxygen double bond. Theoretically, only primary and secondary alcohol will undergo oxidation reaction. Tertiary alcohol will not undergo any oxidation process. Primary alcohols will be oxidized to form a pale green solution while secondary alcohols will be oxidized to form a greenish solution. The end product of oxidation of alcohols will produce different functional group depends on the type of alcohol used. Oxidation of primary alcohol usually will form carboxylic acid with an aldehyde group as the intermediate. Secondary alcohol will form ketone directly from alcohol when oxidized.
From the experiment, it is observed that 2-methyl-2-propanol remains yellow in colour. This proves that 2-methyl-2-propanol is a tertiary alcohol. This is because tertiary alcohols don't have a hydrogen atom attached to that carbon. Since hydrogen atom is unable to be removed, hence no reaction will occur. Hence, tert-butanol is stable towards oxidation reaction.
As for 2-butanol, a greenish solution is formed after heating. One hydrogen from the hydroxyl group (-OH) and one hydrogen atom is removed from the carbon where hydroxyl functional group is attached and replaced by a carbon-hydrogen double bond. Since the carbon atom (where the hydrogen from –OH group and the hydrogen atom is removed) is attached to two alkyl group, hence the product formed is a ketone group, 2-butanone. Besides, the two removed hydrogen atom will react with oxygen to form a water molecule.
For n-butanol and alcohol X, a pale green solution is formed. If excess alcohol is used, an aldehyde group will be obtained; n-butanol will form butanal and water. However, because excess oxidising agent is used, n-butanol is further oxidised by adding oxygen to the hydrogen atom to from a carboxylic acid. Hence the end product is butanoic acid. As for alcohol X, it is determined as an primary alcohol from both the Lucas test and oxidation test. The equations of the chemical reactions of n-butanol (1° alcohol) and 2-butanol (2° alcohol) are shown as follows:
n-butanol (1° alcohol):
Na2Cr2O7 / H+
CH3CH2CH2CH2OH + [O] CH3CH2CH2CHO + H2O
(n-butanol) Δ (butanal)
Na2Cr2O7 / H+
CH3CH2CH2CHO + [O] CH3CH2CH2COOH
(butanal) Δ (butanoic acid)
Overall:
Na2Cr2O7 / H+
CH3CH2CH2CH2OH + 2[O] CH3CH2CH2COOH + H2O
(n-butanol) Δ (butanoic acid)
2-butanol (2° alcohol):
Na2Cr2O7 / H+
CH3CH(OH)CH2CH3 + [O] CH3COCH2CH3 + H2O
(2-butanol) Δ (2-butanone)
2-methyl-2-propanol (3° alcohol):
Na2Cr2O7 / H+
(CH3)3COH + [O] No reaction
(2-methyl-2-propanol) Δ
Overall, primary alcohols are oxidized to carboxylic acids or aldehydes, where in the experiment n-butanol was oxidised to butanoic acid due to the usage of the strong oxidising agent like Na2Cr2O7 in excess. On the other hand, secondary alcohols are oxidised to ketones, where 2-butanol was oxidised to 2-butanone. Tertiary alcohols show no reaction due to their stable structure where the carbon atom is already substituted with three stable alkyl groups.
- Esterification test
Esters are derived from carboxylic acids and alcohols. A carboxylic acid contains the -COOH group while alcohol contains the –OH group. Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid just like in this experiment. The esterification reaction is both slow and reversible. Hydroxyl group from the acid and hydrogen atom from the alcohol is removed to form water and ester. Due to the reaction of removal of water, esterification is also known as dehydration reaction. The general equation is shown as below:
In this test, ethanol, CH3CH2OH react with ethanoic acid, CH3COOH, in the presence of a few drops of concentrated H2SO4 as catalyst and water bath, to form an ester called ethyl ethanoate or ethyl acetate. The chemical composition of acids, alcohols and catalyst used can affect the rate of reaction. In general, ester produces pleasant scent. However in this experiment, a rubbery scent is detected. This is because esterification reaction is slow and reversible; hence not much ester is produced. The smell might be slightly distorted or covered by the carboxylic acid. Besides, addictive need to be added to the ester to produce a stronger pleasant, often fruit smell. Esters are widely used in the fragrance and flavour industry. It is also used in the organic chemistry for the test of alcohols and carboxylic acid.
The equation of the chemical reaction that occurred is shown as follows:
Concentrated HCl
CH3CH2OH + CH3COOH CH3COOCH2CH3 + H2O
(Ethanol) (Ethanoic acid) Δ (Ethyl ethanoate)
Precautionary steps:
- Keep all the alcohols from any source of fire as alcohols are highly flammable.
- All the unused or used waste materials must be poured into a specific waste container prepared in the fume chamber to prevent pollution to the environment.
- Cautions must be taken when pouring concentrated sulphuric acid as it is toxic and is destructive to mucus membrane.
- 2-methyl-2-propanol is harmful if inhaled. May cause skin and respiratory irritant. Severe eye irritant.
- All transfer of chemicals should be done in the fume chamber to avoid inhalation of any chemicals vapour as it may cause severe sickness such as cancer.
Conclusions:
From the experiment, it can be concluded that n-butanol and the unknown alcohol X are primary alcohol, 2-butanol is a secondary alcohol and 2-methyl-2-propanol is a tertiary alcohol. Primary alcohol will undergo oxidation but react slowest in halogen substitution reaction. Secondary alcohol in the other hand undergoes oxidation reaction but will react with a medium speed under halogen substitution reaction. Tertiary alcohol will not undergo oxidation reaction but will react almost immediately in halogen substitution. Alcohols will react with carboxylic acid to form an ester.
References:
- From the internet:
-
“boiling point of alcohols and phenols”, retrieved on 1 February 2011 from
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“Lucas test for alcohols”, retrieved on 1 February 2011 from
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“oxidation of alcohols”, retrieved on 1 February 2011 from
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“introducing alcohols”, retrieved on 2 February 2011 from
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“reactin of alcohols”, retrieved on 2 February 2011 from
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“chapter 15, alcohols, diols and triols”, retrieved on 2 February 2011 from
- From the books:
- McMurry – Fay, Chemistry Fifth Edition, Pearson Prentice Hall, 2008.
- E.N. Ramsden, A-level Chemistry Fourth Edition, Nelson Thornes, 2000.