Classification Tests for Hydroxyl- and Carbonyl-containing Compounds
Miguel, K.D., Moron, R.S.S., Pazon, A.D., Ramirez, C.V., Raquepo, T.M.R., Razon, D.N.A.Jr.
2B-PH, Group No. 6, Department of Pharmacy, Faculty of Pharmacy,
University of Santo Tomas, España Boulevard, 1015 Manila, Philippines

Abstract
In organic chemistry it is very common to see molecules comprised mainly of a carbon backbone with functional groups attached to the chain. The functional group gives the molecule its properties, regardless of what molecule contains it.[1] Examples of functional groups are that of hydroxyl (-OH) which is usually seen in alcohols, and carbonyl (C=0) which is seen in aldehydes and ketones. In this experiment, several tests were conducted to distinguish and differentiate various sample compounds such as ethanol, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, benzyl alcohol, n-butyraldehyde, benzaldehyde, acetone, acetophenone, isopropyl alcohol, and acetaldehyde. The tests are the following: solubility test of alcohols in water, which gave a soluble result in ethanol, sec-butyl alcohol and tert-butyl alcohol. Next is the Lucas test, which is used to differentiate 1°, 2° & 3° alcohols. In Lucas test, tert-butyl alcohol gave an immediate turbid result; the rate of reaction was noted. Chromic Acid test (Jones Oxidation) which gave a positive result by producing a blue-green solution with the sample n-butyl alcohol, acetaldehyde, benzaldehyde, and isopropyl alcohol. 2,4-Dinitrophenylhydrazone (2,4-DNP) was used to detect presence of carbonyl groups. In this test, only acetophenone gave a red-orange precipitate. The tests to identify between aldehydes and ketones are Fehling’s test and Tollen’s Silver Mirror test. In Fehling’s test, it gave a brick-red precipitate in n-butyraldehyde and benzaldehyde; while in the Tollen’s Silver Mirror test it produced silver mirror precipitate in acetaldehyde. The Iodoform test is a test for the detection of ketones and aldehydes carrying an alpha methyl group. In this test acetaldehyde, acetone, acetophenone and isopropyl alcohol gave a result of yellow precipitate. In conclusion, acetaldehyde, n-butyraldehyde and benzaldehyde are aldehydes; while acetone and acetophenone are ketones. Both ketones and aldehydes contain the carbonyl group. Furthermore, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, and isopropyl alcohol are all alcohols containing the hydroxyl functional group.

Introduction Alcohols are compounds in which one or more hydrogen atoms in an alkane have been replaced by an –OH (hydroxyl) functional group. Alcohols fall into different classes depending on how the –OH group is positioned on the chain of the carbon atoms. [2] Alcohols are classified as primary (1°), secondary (2°), or tertiary (3°), depending on the number of carbon atoms bonded to the carbon bearing the –OH group. [3] In a primary alcohol, the carbon which carries the –OH group is only attached to one alkyl group. In a secondary alcohol, it is directly joined to two alkyl groups. In a tertiary alcohol, the carbon containing the –OH is attached directly to three alkyl groups. [2] Alcohols also have the ability to hydrogen bond with water; however, their solubility decreases as the size of the R group and their hydrocarbon-like character increases. [4] Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group (C=O). The carbon atom of this group has two remaining bonds that may be occupied by hydrogen or alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde. If neither is hydrogen, the compound is a ketone.[5] An aldehyde differs from a ketone by having a hydrogen atom attached to the carbonyl group. This makes the aldehydes very easy to oxidize. For example, ethanal, CH3CHO, is very easily oxidized to either ethanoic acid, CH3COOH, or ethanoate ions, CH3COO-. Ketones don't have that hydrogen atom and are resistant to oxidation. They are only oxidized by powerful oxidizing agents which have the ability to break carbon-carbon bonds. [6] There are various differentiating tests to determine whether a sample is an alcohol, aldehyde or a ketone. Among the commonly used tests are Solubility test, Lucas test, Chromic acid test, 2,4-Dinitrophenylhydrazone (2,4 DNP), Fehling’s test, Tollen’s Silver Mirror test, and Iodoform Test. The objectives of this experiment are: 1) to distinguish whether a compound is hydroxyl- or carbonyl-containing 2) to differentiate the three types of alcohols 3) to differentiate aldehydes from ketones, and finally 4) to explain the mechanism involved in the differentiating tests. [7]

Methodology
Apparatus & Sample Compounds Used The materials and reagents used in this experiment were: Lucas reagent, chromic acid reagent, 95% ethanol, Fehling’s A & B, Tollen’s reagent, 5% NaOCl solution, iodoform test reagent, 2,4-DNP, Pasteur pipette, test tubes, vials, and beaker. The sample compounds used were: ethanol, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, benzyl alcohol, n-butyraldehyde, benzaldehyde, acetone, acetophenone, isopropyl alcohol, and acetaldehyde.
Solubility of Alcohols in Water In five test tubes, 10 drops of each sample alcohols (ethanol, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, and benzyl alcohol) were placed separately using a dropper. One mL of water was added dropwise to each of the test tubes, shaking the mixture thoroughly after each addition. When cloudiness resulted, 0.25 mL of water was added at a time, with vigorous shaking until a homogenous dispersion appeared. The total volume of water added was noted. When no cloudiness resulted after the addition of 2.0 mL water, the alcohol was noted down to be soluble in water.
Lucas Test The reagent was prepared by dissolving 16g of anhydrous zing chloride in 10mL of concentrated HCl. The mixture was allowed to cool. Fifty milligrams (2-3 drops) of the sample was added to the reagent prepared in a small vial. It was capped and was shaken vigorously for a few seconds. The mixture was allowed to stand at room temperature. This test was performed on n-butyl alcohol, sec-butyl alcohol, and tert-butyl alcohol. The rate of formation of the cloudy suspension or the formation of two layers was observed.
Chromic Acid Test (Jones Oxidation) A prepared reagent where 20g of chromium trioxide was dissolved in 60mL of cold water with the addition of 20mL of concentrated sulfuric acid was provided for this experiment. One drop of the liquid samples or a small amount of the solid samples (n-butyl alcohol, isopropyl alcohol, tert-butyl alcohol, acetaldehyde, benzaldehyde and acetone) was dissolved separately in 1mL of acetone in a test tube, and 5 drops of the prepared reagent was added one drop at a time while shaking the mixture. The test tubes where placed in a 60°C water bath for 5 minutes. The color of each solution was noted.
2,4-Dinitrophenylhydrazone Test (2,4-DNP) One drop of a liquid sample (acetone, acetaldehyde, benzaldehyde and acetophenone) was placed in a test tube and 5 drops of 95% ethanol were added. After shaking well, 3 drops of 2,4-dinitrophenylhydrazine were added. When no yellow or orange-red precipitate formed, the solution was allowed to stand for at least 15 minutes.
Fehling’s Test Fehling’s A was prepared by dissolving 7g of hydrated copper (II) sulfate in 100mL of water. Fehling B was prepared by dissolving 35g of potassium sodium tartrate and 10g of NaOH in 100mL of water. Both of these reagents were provided. Into each test tube, 1 mL of freshly prepared Fehling’s reagent (made by mixing equal amounts of Fehling’s A and Fehling’s B) was placed. Three drops of the sample to be tested were added and the test tubes were placed in a beaker of boiling water. Changes that occurred within 10-15 minutes were observed. The test was performed on acetaldehyde, acetone, benzaldehyde and acetophenone.
Tollen’s Silver Mirror Test Four test tubes with 1mL each of freshly prepared Tollens’ reagent were prepared. In each of the test tubes, two drops of each of the samples (acetaldehyde, benzaldehyde, acetone, and acetophenone) were added separately. The mixtures were shaken and allowed to stand for 10 minutes. When no reaction occurred, the test tubes were placed in a beaker with warm water (30-50°C) for 5 minutes. Observations were recorded.
Iodoform Test Two drops of each sample (acetaldehyde, acetone, acetophenone, butyraldehyde, benzaldehyde, and isopropyl alcohol) was placed in its own vial. Twenty drops of 10% KI solution was added. Twenty drops of fresh chlorine bleach (5% sodium hypochlorite) was slowly added while shaking to each test tube and was mixed. The formation of a yellow precipitate was noted.

Results and Discussions
Solubility of Alcohols in Water In case of alcohols, just as it happens in case of many other biological molecules, the basic solubility rule that “like dissolves like” is a bit more complexed. Each alcohol consists of a carbon chain (always nonpolar) and a -OH group (which is polar). For ethanol for example the chemical formula looks like this: C2H5OH. Ethanol has a two carbon chain and a -OH group. As water is polar, it attracts OH group. Carbon chain on the other hand, as nonpolar is repelled. Solubility of alcohols is therefore determined by the stronger of the two forces. Because of the strength of the attraction of the -OH group, first three alcohols (methanol, ethanol and propanol) are completely miscible or soluble. They dissolve in water in any amount. Starting with the four-carbon butanol the solubility of alcohols is starting to decrease. After the 7-carbon heptanol, alcohols are considered immiscible. [8] For the test of solubility of alcohols in water, turbidity was observed in the solution. A cloudy appearance meant that the alcohol is not soluble in water. The amount of water needed to produce a homogenous dispersion was noted. Results are seen in Table1.
Table 1 Solubility of Alcohols in Water
Alcohol Amount of water (in mL) needed to produce a homogenous dispersion Solubility in water
Ethanol 0.25mL Soluble n-butyl alcohol 2mL Insoluble sec-butyl alcohol 1.25mL Soluble tert-butyl alcohol 0.25mL Soluble
Benzyl alcohol 2mL insoluble It can be concluded that the compounds that dissolved in water are polar since water is a polar compound. These compounds include ethanol, sec-butyl alcohol and tert-butyl alcohol. Both ethanol and tert-butyl required only 0.25mL for it to dissolve. There are certain factors that affect solubility; one of these is the presence and number of carbon atoms in the compound. The lower the number of carbon atoms, the more soluble the compound becomes. Branching of atoms also affects solubility. The more branching present, the more soluble a compound is (tert-butyl alcohol is an example). This is only true for organic compounds that have the same number of carbon atoms present.
Lucas Test Lucas test on alcohols is a way to differentiate between primary, secondary and tertiary alcohols. It is based on the reactivity of each alcohol with the hydrogen halide. [9] The reaction is a substitution reaction where the chloride of the zinc chloride gets replaced by the hydroxyl group of the alcohol. The reactivity of the alcohol with Lucas reagent is measured by the degree of turbidity which may vary from colorless to turbid. The formation of turbid solution happens due to the formation of chloroalkane. [10] Table2 shows the results of the Lucas test.
Table 2 Lucas Test Results
Substance Reaction Observed n-butyl alcohol Clear, colorless sec-butyl alcohol Clear, colorless at first then became turbid tert-butyl alcohol Turbid solution

In Lucas test, Zinc Chloride acts as catalyst. The classification of the alcohols is usually done based on the difference in reaction with concentrated hydrochloric acid. The reaction which normally occurs is a SN1 nucleophilic substitution is a two-step reaction. Alcohols which have a capability to form carbocation intermediates exhibit this reaction. Only secondary and tertiary alcohols exhibit SN1 nucleophilic mechanism. [10] Figure1 shows the Sn1 Reaction with Lucas reagent.

In the first step the proton (H+) from Hydrochloric acid (HCl) will protonate the OH– group of the alcohol. Water (H2O) attached to the carbon is a weaker nucleophile than Cl (Chloride). Thus nucleophile Cl– replaces the H2O group forming a carbocation as its present in excess. In the second step the Cl– attacks the carbocation and thus forms alkyl chloride. Here the first step is generally the slowest step and is the rate determining step. As the tertiary carbocation is much stabilized, they are the ones which undergo reaction and form a turbid solution. The opposite of it happens in the case of primary alcohols. [10] Tertiary alcohols such as tert-butyl alcohol took its reaction immediately which produced a turbid solution. Secondary alcohols, such as sec-butyl alcohol require 2-3 minutes to undergo reaction. It took 2 minutes and 38 seconds to become turbid. N-butyl alcohol, which is a primary alcohol, did not produce turbidity.

Chromic Acid Test (Jones Oxidation)

The Chromic acid test distinguishes primary and secondary alcohols from tertiary. Chromic acid will oxidize a primary alcohol first to an aldehyde and then to a carboxylic acid and it will oxidize a secondary alcohol to a ketone. Tertiary alcohols do not react. The –OH bearing carbon must have a hydrogen atom attached. Since the carbon atom is being oxidized in primary and secondary, the orange chromi-um Cr 6+ ion is being reduced to the blue-green Cr 3+ion. [12] Aldehydes can also be differentiated from ketones using this test. [13] Table3 shows the results of this test.
Table 3 Chromic Acid Test Results
Substance Results Observed n-butyl alcohol Blue-green solution tert-butyl alcohol Orange solution
Acetaldehyde Blue-green solution
Benzaldehyde Blue-green solution
Acetone Orange solution
Acetophenone Orange solution
Isopropyl Alcohol Blue-green solution

Based on the results, n-butyl alcohol, acetaldehyde, benzaldehyde, and isopropyl alcohol gave a positive result of a blue-green solution. Chromic Acid test involved reduction-oxidation or redox reaction. 1˚ and 2˚alcohols and aldehydes underwent oxidation (lost electrons) and chromium underwent reduction (gain electrons) from Cr 6+ to Cr 3+. 1˚ and 2˚ alcohols and aldehydes reduced the orange-red chromic acid/sulfuric acid reagent to an blue green solution of Cr (III) salts. As seen on the results, n-butyl alcohol and isopropyl alcohol which are a primary alcohol and a secondary alcohol respectively, gave a positive result, as well as the aldehydes; while the ketones and tertiary alcohol did not. The chemical reaction for this test is seen in Figure2. Figure 2 Chromic Acid Test Reaction Mechanism

2,4-Dinitrophenylhydrazone (2,4-DNP) 2,4-Dinitrophenylhydrazone is a test for the carbon-oxygen double bond; meaning, aldehydes and ketones react to this test to form a solid 2,4-dinitrophenlyhydrazone derivative. [14] It is used to qualitatively test the carbonyl functionality of a ketone or aldehyde. The resulting color after conducting the test provides structural information. If the color of the solid is yellow, this means that the carbonyl group in the unknown is unconjugated. The resulting color will be red-orange for conjugated carbonyl compounds (having alternate single and double bonds). [13] Results from the experiment can be seen in Table4.
Table 4 2,4-Dinitrophenylhydrazone Test Results
Substance Reactions Observed
Acetaldehyde Yellow precipitate
Benzaldehyde Yellow precipitate
Acetone Yellow precipitate
Acetophenone Red-orange precipitate

As seen on the table, only acetaphenone gave a red-orange result while the rest of the sample compounds tested gave a yellow precipitate. This means that acetaldehyde, benzaldehyde, acetone and acetaphenone have carbon-oxygen double bond in their compound, and that acetaphenone is a conjugated carbon compound. The reaction mechanism for this test is seen in Figure3. R and R’ can be any combination of hydrogen or hydrocarbon groups (such as alkyl groups). If at least one of them is hydrogen, then the original compound is an aldehyde. If both are hydrocarbon groups, then it is a ketone. The reaction is known as a condensation reaction, in which two molecules join together with the loss of a small molecule in the process. In this case, the small molecule is water. In terms of mechanisms, this is a nucleophilic addition-elimination reaction. The 2,4-dinitrophenylhydrazine (seen in the reactant) first adds across the carbon-oxygen double bond (the addition stage) to give an intermediate compound which then loses a molecule of water (the elimination stage). [14] Figure 3 2,4-Dinitrophenylhydrazone Reaction Mechanism

Fehling’s Test In Fehling’s test the presence of aldehydes is detected. Most simple ketones do not react with this test, although certain hydroxyketones and carbohydrates do. [13] The reactions observed for this experiment is listed in Table5.
Table 5 Fehling's Test Results
Substance Reactions Observed
Acetaldehyde Brick-red precipitate
Benzaldehyde Brick-red precipitate
Acetone Blue solution
Acetophenone Blue solution

It is seen in the results that both acetaldehyde and benzaldehyde gave a positive result while acetone and acetophenone did not. The reaction mechanism for this reaction is seen in Figure4. The bistartratocuprate (II) complex in Fehling’s solution is an oxidizing agent and the active ingredient in the test. The compound to be tested is added to the Fehling’s solution and the mixture is heated. Aldehydes such as benzaldehyde and acetaldehyde are oxidized, giving a positive result (precipitation of red copper (I) oxide), but ketones (like acetone and acetophenone) do not react, unless they are alpha-hydroxy-ketones. [15]
Figure 4 Fehling's Test Reaction Mechanism
Tollen’s Silver Mirror Test Tollen’s Silver Mirror test is a qualitative laboratory test used to distinguish between aldehydes and ketones. It shows the fact that aldehydes are readily oxidized, whereas ketones are not. This test uses a Tollen’s reagent which is an ammoniacal solution of silver ion prepared by dissolving silver oxide in ammonia. The reagent is reduced to metallic silver by aldehydes, which in turn are oxidized into corresponding acids. Silver ions in the presence of hydroxide ions come out of solution as a brown precipitate of silver(I) oxide, Ag2O(s). This precipitate dissolves in aqueous ammonia, forming the diamminesilver(I) ion, [Ag(NH3)2]+. Ketones do not react with Tollens' reagent [16] Reactions observed in regards with the Tollen’s Silver Mirror test is seen in Table6.
Table 6 Tollen's Silver Mirror Test Results
Substance Reactions Observed
Acetaldehyde Silver mirror precipitate
Benzaldehyde Silver precipitate
Acetone Clear, colorless solution
Acetophenone Turbid solution with globules

It can be concluded that acetaldehyde and benzaldehyde are aldehydes because they gave a positive result in this test. The reaction mechanism for this test can be seen in Figure5.

Figure 5 Tollen's Silver Mirror Test

Iodoform Test Iodoform test or haloform reaction is a chemical reaction where a haloform (CHX3, where X is a halogen) is produced by the exhaustive halogenation of a methyl ketone (a molecule containing the R – CO – CH3 group) in the presence of a base. R may be H, alkyl or aryl. The reaction can be used to produce chloroform(CHCl3), bromoform (CHBr3), or iodoform(CHI3). This reaction was traditionally used to determine the presence of a secondary alcohol oxidizable to a methyl ketone. When iodine and sodium hydroxide are used as the reagents, a positive reaction gives iodoform. Iodoform (CHI3) is a pale-yellow substance. Due to its high molar mass caused by the three iodine atoms, it is solid at room temperature (chloroform and bromoform). It is insoluble in water and has an antiseptic smell.Avisible precipitate of this compound will form from a sample only when either a methyl ketone, ethanal, ethanol, or a methyl secondary alcohol is present. [17] The results for this experiment is seen in Table7.
Table 7 Iodoform Test Results
Substance Reactions observed
Acetaldehyde Red solution n-butyraldehyde Red solution
Benzaldehyde Red solution
Acetone Yellow precipitate
Acetophenone Yellow solution with precipitate
Isopropyl alcohol Yellow solution

As seen on the results, acetone, acetophenone and isopropyl alcohol gave a positive result, confirming that acetone and acetophenone are ketones, and that isopropyl alcohol is a secondary alcohol. The reaction mechanism of the iodoform test can be seen in Figure6. In the presence of a base, most methyl ketones are converted to an enolate ion by the removal of one of the protons adjacent to the carbonyl group. In the presence of a halogen like iodine, the enolate anion and halogen react to form a monohaloketone which forms an anion more readily than the original ketone because of the electronegativity of the halogen. The process rapidly repeated until the methyl group has become a trihalomethyl group. Cleavage of trihalo compound by a base then produces the salt of a carboxylic acid and a haloform. In the case of iodine, the water-insoluble iodoform is formed. It is a bright yellow solid. [18]

Figure 6 Iodoform Test Reaction Mechanism

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[18] Bathan, G., Bayquen, A., Crisostomo, A., Cruz, C., de Guia, R., Farrow, F., . . . Torres, P. (2014). Laboratory Manual Organic Chemistry (Revised ed.). Quezon City, NCR: C & E Publishing Inc. p.109. Retrieved November 18, 2015