Transcript
FOUNDATION IN MEDICAL STUDIES (JULY 2013 INTAKE) CHM1024 LABORATORY REPORT PRACTICAL 4 : REACTIONS OF ALCOHOLS
NAME
:
AKMAL ADIB BIN FADZIL
MATRIX ID
:
CPM0018_2013C CPM0018_2013C
GROUP
:
A
SEMESTER
:
TWO
DATE
:
4th FEBRUARY 2014
LECTURER
:
MR. MOHD YUSOFF HUSSAIN
OBJECTIVE The objective of this experiment is to classify the different types of alcohols.
INTRODUCTION An alcohol is an organic compound which has a hydroxyl group (-OH) attached to a carbon atom. This carbon atom should be saturated which means that it is single – – bonded to a hydroxyl functional group and other three atoms. An important class of alcohols are the aliphatic alcohols which has a general formula of C nH2n+1OH. Aliphatic alcohols can be considered as a derivative of alkanes with a hydrogen atom replaced by a hydroxyl group. In the IUPAC the IUPAC system, the name of the alkane chain loses the terminal "e" and replaced with "ol” "ol” such as "methanol" as "methanol" and "ethanol". When necessary, the position of the hydroxyl group is indicated by a number between the alkane name and the "ol" such as propan-1-ol for CH3CH2CH2OH and propan-2-ol for CH3CH(OH)CH 3. Sometimes, the position number is written before the IUPAC name: 1-propanol and 2-propanol. If a higher priority group is present such as an aldehyde, an aldehyde, ketone, ketone, or or carboxylic carboxylic acid, then acid, then it is necessary to use the prefix "hydroxy". For example: 1-hydroxy-2-propanone 1-hydroxy-2-propanone (CH 3COCH2OH). Alcohols can be classified classified into primary (1°), secondary (2°) and tertiary (3°) alcohols according to the number of alkyl groups bonded to the carbon atom that has the –OH –OH group attached. In 1° alcohols, the –OH –OH group is bonded to a carbon atom with one or no alkyl groups attached. Ethanol is an example of a primary alcohol. In 2° alcohols, the –OH –OH group is bonded to a carbon atom with two alkyl groups attached. For example, 2-propanol is a secondary alcohol. In 3° alcohols, the – the –OH OH group is bonded to a carbon atom with three alkyl groups attached. For example, 2-methyl-2-propanol 2-methyl-2-propanol is a tertiary alcohol. Most simple aliphatic alcohols and the lower aromatic alcohols such as phenyl methanol, C6H5CH2OH are liquids at room temperature while highly branched alcohols and alcohols with twelve or more carbon atoms are solids. In general, the hydroxyl group makes the alcohol molecule polar. molecule polar. Those Those groups can form hydrogen form hydrogen bonds to one another and to other compounds except in certain large molecules where the hydroxyl is protected by steric by steric hindrance of adjacent groups. This hydrogen bonding means that alcohols can be used as protic as protic solvents. Protic solvents. Protic solvents are solvents that can dissolve salts. Because of hydrogen bonding, the boiling points of alcohols are considerably higher than those of alkanes and chloroalkanes of similar relative molecular mass (RMM). This is because intermolecular hydrogen bonds exist between the – the –OH OH groups in the alcohol molecules. For example:
2
C2H5OH
CH3CH2CH3
CH3Cl
ethanol
propane
chloromethane chlorometh ane
RMM
:
46
44
50.5
Boiling point
:
78°C
-42°C
-24°C
For isomeric alcohols, straight – chain – chain alcohols have higher boiling point than their branched – chain – chain alcohols counterpart. The greater the degree of branching in an isomer, the lower its boiling point. For example:
When it comes to solubility of alcohols, there are two opposing solubility trends in which are: the tendency of the polar OH to promote solubility in water, and the tendency of the carbon chain to resist it. Methanol, ethanol, and propanol are miscible in water because the hydroxyl group wins out over the short carbon chain. Butanol, chain. Butanol, with with a four-carbon chain, is moderately soluble because of a balance between the two trends. Alcohols of five or more carbons (pentanol and higher) are effectively insoluble in water because of the hydrocarbon chain's dominance. All simple alcohols are miscible in organic solvents. Alcohols, like water, can show either acidic or basic properties at the -OH group. With a pK a pKa of around 16-19, they are generally slightly weaker acids acids than water, than water, but they are still able to react with strong bases such as sodium hydride or reactive metals such as sodium. as sodium. The salts The salts that result are called alkoxides, called alkoxides, with the general formula R formula RO O- M+. The oxygen atom has lone has lone pairs of nonbonded electrons that render it weakly basic weakly basic in the presence of strong acids such as sulphuric as sulphuric acid. As acid. As one moves from primary to secondary to tertiary alcohols with the same backbone, the hydrogen bond strength, the boiling point, and the acidity typically decrease. decrease. For this experiment, three unknown alcohols will be classified into 1°, 2° and 3° alcohols by conducting several tests to see its results which include the end – product – product formed, colour of the end – product – product and the time taken for the reaction. Besides that, esterification is carried out to determine the properties of esters by mixing an alcohol with a carboxylic acid.
3
APPARATUS AND MATERIALS 1. 0.04 M sodium dichromate, Na2Cr 2O7 solution. 2. Alcohol A. 3. Alcohol B. 4. Alcohol C. 5. Ethanol. 6. Concentrated sulphuric acid, H2SO4 solution. 7. Droppers. 8. Glacial acetic acid. 9. Lucas’ reagent. 10. Cold Schiff’ Schiff’s s reagent. 11. Distilled water. 12. Rubber stopper. 13. 8 test tubes. 14. Water bath (maintained at 70 ℃).
4
METHOD A. Lucas’ Test. 1. 1 ml of alcohol A is placed into a test tube. 2. 5 – 5 – 10 drops of Lucas’ reagent Lucas’ reagent is added to the test tube. 3. Rubber stopper is inserted onto the test tube. The test tube is is then shaken. shaken. 4. Any observations observations and the time taken for the reaction to occur are recorded. recorded. 5. Steps 1 – 4 – 4 are repeated by replacing alcohol A with alcohol B and C. The test tube is placed in the water bath (maintained at 70 ℃) if no change occurs within 10 minutes. B. Oxidation. 1. 10 drops of of 0.04 M sodium sodium dichromate dichromate (VI), Na2Cr 2O7 solution is placed into a test tube. 2. 2 – 3 – 3 drops of concentrated sulphuric acid, H 2SO4 solution is added into the test tube in a fume f ume hood. 3. 5 drops of alcohol A is added into the mixture. 4. Any changes in the colour of the mixture are are recorded. 5. The mixture is further tested by adding 5 drops of cold Schiff’s reagent and heated in the water bath (maintained at 70 ℃) if the colour of the mixture turns green. 6. Any changes in the colour of the mixture are are recorded. 7. Steps 1 – 6 – 6 are repeated by replacing alcohol A with alcohol B and C. C. Esterification. 1. 2 ml of ethanol is placed in a dry test tube. 2. 1 ml of glacial acetic acetic acid and 3 drops of of concentrated sulphuric sulphuric acid, H2SO4 solution are added into the test tube in a fume hood. 3. The mixture is shaken and then warmed warmed in a water bath at 70 ℃ for 5 – 10 minutes. 4. 3 ml of distilled water is added into the mixture. The vapour released is smelled and described.
5
RESULTS OBSERVATION TEST
LUCAS’ TEST
ALCOHOL A
ALCOHOL B
ALCOHOL C
The
The
The
colourless
solution turns cloudy
solution
immediately.
after a few minutes.
unchanged.
of
solution
the 1. Colour
remains
orange. 1. OXIDATION 2. SCHIFF’S
TEST
colourless
solution turns cloudy
1. Colour
REAGENT
colourless
2. Not
performed
because
the
colour
of
the
solution
remains
of
the 1. Colour
remains
of
the
solution changes
solution changes
from orange to
from orange to
green.
green.
2. Colour
of
the 2. Colour
of
the
green
solution
green
solution
turns
magenta
turns
magenta
slowly.
immediately.
orange. ESTERIFICATION
A pleasant smell is detected from the test tube.
6
DISCUSSION Lucas’ Test. The Lucas’ reagent used in this experiment is an aqueous solution of strong hydrochloric acid, HCl and zinc chloride, ZnCl 2. The reactants used to react with this reagent must be water – water – soluble so that a reaction can take place. An alcohol with more than six carbons is too large to be dissolved in the reagent while water – – soluble alcohol which has low molecular weight can react with this reagent. This gives further insight of the unknown alcohols given in which the alcohols can be methanol, ethanol, propanol or butanol of different classification. classification. For this experiment, each eac h of the unknown alcohols is treated with Lucas’ reagent. If the colour of the solution remains unchanged, it is heated in a water bath at 70°C. Positive indicator of the reaction is the formation of water –insoluble –insoluble alkyl chloride as cloudiness or a precipitate. Theoretically, formation of an alkyl chloride with tertiary alcohol is very rapid, followed by the secondary alcohol that may take from 5 to 20 minutes to form visible cloudiness. Primary alcohols do not react with Lucas reagent or it may show very little result in a very long time. When alcohol A is used, the colour of the solution changes to cloudy immediately. When alcohol B is used, the colour of the solution changes to cloudy after a few minutes. When alcohol C is used, the colour of the t he solution remains unchanged. The results obtained shows that alcohol A is a 3° alcohol, alcohol B is a 2° alcohol and alcohol C is a 1° alcohol. The reaction that occurs in the Lucas test is a nucleophilic substitution. Only alcohols that can generate stable carbocation intermediates will undergo the reaction. The acid catalyst activates the OH group of the alcohol by protonating the oxygen atom. The C – OH2+ bond breaks to generate the carbocation, which in turn reacts with the chloride ion (nucleophile) (nucleophile) to generate an alkyl halide product. A general mechanism for this nucleophilic reaction is as follows:
7
In simpler words, the chemical reaction involves replacing the –OH –OH group of the alcohol with a chloride ion from HCl to form the insoluble alkyl chloride. For alcohol A (3° alcohol):
For alcohol B (2° alcohol):
For alcohol C (1° alcohol):
Zinc chloride (ZnCl2) is a catalyst for this reaction, which makes up for the name of test that is Lucas test. The resultant alkyl chloride is insoluble in water and separates from the Lucas reagent (ZnCl2 in concentrated HCl), forming a cloudy mixture, an emulsion or a new “organic” layer to form in the reaction mixture, or sometimes only a colour change will be observed (red or orange).
8
Oxidation. There are two steps for doing this experiment. The first step is to use sodium dichromate (VI), Na2Cr 2O7 solution acidified with concentrated sulphuric acid, H 2SO4 to differentiate between tertiary and primary/secondary alcohols. If oxidation occurs, the orange solution containing the dichromate (VI) ions are reduced to a green solution containing chromium (III) ions. The second step is to use cold Schiff’s reagent to differentiate between primary and secondary alcohols. If the green colour of the solution turns magenta immediately, it is a primary alcohol. If it remains green or turns magenta at a slow rate, it is a secondary alcohol. alcohol. For the first step, all of the unknown alcohols are added to a mixture of sodium dichromate (VI), Na2Cr 2O7 solution and concentrated sulphuric acid, H 2SO4. The result is that alcohol A is 3° alcohol while alcohol B and C is either 2° alcohol or 1° alcohol. The reason is that tertiary alcohols are not oxidised by the acidified sodium dichromate (VI), Na 2Cr 2O7 solution as they do not have a hydrogen atom attached to the carbon. Secondary and primary alcohols can be oxidised by the acidified sodium dichromate (VI), Na 2Cr 2O7 solution as the oxidising agent (acidified Na2Cr 2O7 solution) removes a hydrogen from the –OH –OH group and a hydrogen atom attached to the carbon that has the – the –OH OH group.
The general equations for the reactions mentioned above are as follows: 3RCH2OH
+
(1° alcohol)
Cr 2O72-
+
(orange)
8H+ → 3RCHO (aldehyde) (aldehyde)
R’
R’
|
|
3R – 3R – CH CH – – OH +
Cr 2O72-
(2° alcohol)
(orange)
+
+
8H+ → 3R – C = O (ketone)
2Cr 3- +
7H2O
(green)
+
2Cr 3+ +
7H2O
(green)
9
Tertiary alcohols do no react with the acidified Na 2Cr 2O7 solution and thus no equation can be made. Also note that if the aldehyde formed from the oxidation of 1° alcohol is allowed to oxidise further, it will produce a carboxylic acid. The diagram of the results obtained is as follows:
Since the colour produced for both mixture C and B is green, it is not sure which one produce an aldehyde (1° alcohol) and which one produces a ketone (2° alcohol). Now that the first f irst step is done, the experiment experiment now proceeds to the second step. The second step involves adding the green-coloured alcohol C and B from the previous step with cold Schiff’s reagent. Schiff's reagent is a fuchsin dye decolorized by passing sulfur dioxide through it. In the presence of even small amounts of an aldehyde, it turns magenta. The reagent is cooled to prevent ketone from reacting with the reagent to give the same colour. Theoretically, if the colour of the mixture containing Schiff’s reagent turns magenta, the alcohol used is a 1° alcohol. This is because 1 ° alcohols are oxidised by the acidified Na2Cr 2O7 solution to form aldehydes which causes the colour of the solution with cold Schiff’s reagent to turn magenta. magenta . If there is no change in colour of the mixture or only a trace of pink colour is detected after a while, it is a 2 ° alcohol as 2° alcohols are oxidised by the acidified Na 2Cr 2O7 solution to form ketones. Ketones have no effect on the colour of the solution with cold Schiff’s reagent. From the experiment, the colour of the mixture containing alcohol C turns magenta while the mixture containing alcohol B remains green in colour when added with c old Schiff’s reagent. reagent . Therefore, alcohol C is a 1° alcohol while alcohol B is a 2° alcohol. Alcohol A has been stated earlier which is a 3° alcohol.
10
Esterification. This experiment is based on the special, classic type of esterification which is the Fischer – Speier esterification. It involves a reflux between a carboxylic acid with an alcohol in the presence of an acid catalyst to produce an ester. It is a nucleophilic acyl substitution reaction. Most carboxylic acids are suitable for the reaction, but the alcohol should generally be a primary or secondary alkyl. Tertiary alcohols are prone to elimination and and phenols are usually too unreactive to give useful yields. Since esterification is highly reversible, the yield of the ester can be improved by using using Le Chatelier's principle which includes: using the alcohol in large excess and using a dehydrating agent such as concentrated sulphuric acid to remove water (one of the end – products). – products). The general equation for this type of esterification is:
For this experiment, ethanol and glacial acetic acid (ethanoic acid) is mixed in a dry test tube and added with concentrated sulphuric acid. The mixture is later shaken and heated in a water bath at 70°C for a few minutes. After that, a few amount of distilled water is added into the mixture before lightly smelling the vapour released. The smell is a pleasant smell which seems to be a bubble – bubble – gum gum like smell. The full equation of the reaction is as follows:
CH3CH2OH + CH3COOH → CH3COOCH2CH3 + H2O
The mechanism of reaction between the ethanol and acetic acid is as follows: f ollows:
11
Step 1: An acid/base reaction. Protonation of the carbonyl carbonyl makes it more electrophilic. electrophilic.
Step 2: The alcohol O functions as the nucleophile attacking the electrophilic electrophilic C in the C=O, with the electrons moving towards the oxonium ion.
Step 3: An acid/base reaction. reaction. Deprotonate Deprotonate the alcoholic oxygen.
Step 4: An acid/base reaction. reaction. Need to make make the -OH leave, leave, so convert it into a good leaving group by protonation.
Step 5: Use the electrons of an adjacent oxygen to help "push out" the leaving group, a water molecule.
Step 6: An acid/base reaction. reaction. Deprotonation Deprotonation of the oxonium oxonium ion reveals the carbonyl in the ester product.
12
Precaution. Throughout the experiment, there are several precautions that were taken and noted in the experiment. One of them is to use concentrated sulphuric acid in a fume hood to avoid inhalation of the fume released from the acid is dangerous as it is toxic and capable of damaging the mucous membrane. Besides that, all alcohols must be kept away from any source of fire throughout the experiment as alcohols are highly flammable. Another one is to use safety gloves and lab coat to avoid contact with corrosive or irritating substances such as concentrated sulphuric acid acid and Lucas’ reagent in case if caution is not fully taken. Wearing safety gloves also protects the hand from staining of substances that are not easily removed such as sodium dichromate solution. In addition, all test tubes used in the experiment should be cleaned and rinsed with distilled water before drying it for use. Last but not least, all chemical wastes of the experiment need to be disposed thoroughly and carefully.
CONCLUSION As a conclusion, the unknown alcohols are able to be classified. By conducting co nducting the Lucas’ test and oxidation test, the alcohols A, B and C are able to be classified into 1°, 2° and 3° alcohol. Alcohol A is a 3° alcohol, alcohol B is a 2° alcohol and alcohol C is a 1° alcohol. 1° alcohol undergoes oxidation but react slowest in halogen substitution reaction. 2° undergoes oxidation and react with a medium speed under halogen substitution reaction. Tertiary alcohols never undergo oxidation but react immediately in halogen substitution reaction. Alcohols react with with carboxylic carboxylic acids to form an ester.
13
QUESTIONS 1. Write the full equation for for esterification esterification reaction in this experiment. experiment. CH3CH2OH + CH3COOH → CH3COOCH2CH3 + H2O 2. What does the formation of two layers in the Lucas’ test indicate? Explain. The formation of two layers in the Lucas’ test indicates a positive result of the experiment. A positive result is the formation of the insoluble alkyl chloride from the reaction of tertiary/secondary alcohols with Lucas’ reagent. The alkyl halide forms a layer at the top of the mixture. 3. Give two examples examples for each each type of alcohol alcohol and draw the structures. For the 1° alcohol:
For the 2° alcohol:
For the 3° alcohol:
14
REFERENCES 1. CHEMISTRY FOR MATRICULATION MATRICULATION SEMESTER SEMESTER 2 FOURTH EDITION, Tan Yin Toon, Oxford Fajar Sdn. Bhd. 2013. 2. http://www.xdocs.com/doc/501518 http://www.xdocs.com/doc/50151853/exp-3-classi 53/exp-3-classification-of-alcoh fication-of-alcohols ols 3. http://www.chemguide.co.uk/orga http://www.chemguide.co.uk/organicprops/alco nicprops/alcohols/esteri hols/esterification.html fication.html 4. http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch15synthesisesters.html 5. http://www.organicchem.org/oc2w http://www.organicchem.org/oc2web/lab/exp/ox eb/lab/exp/oxid/lucas.pdf id/lucas.pdf 6. http://swc2.hccs.edu/pahlavan/2425L4.pdf 7. http://chemwiki.ucdavis.edu/Organ http://chemwiki.ucdavis.edu/Organic_Chemistry ic_Chemistry/Alcohols/Rea /Alcohols/Reactions_of_Alco ctions_of_Alcohols/The hols/The _Oxidation_of_Alcohols _Oxidation_of_Alcohols 8. http://www2.chemistry.msu.edu/facul http://www2.chemistry.msu.edu/faculty/reusch/virttxtjml ty/reusch/virttxtjml/alcohol2.htm /alcohol2.htm
15
