Organic Chemistry Experiment 9

Rogeni Misaelle M. Petinglay B.S. in Chemistry II

Date performed: April 6, 2017 Date submitted: April 7, 2017

Experiment No. 9 ORGANIC DERIVATIVES OF WATER I.

Summary of Results

Table 1. Solubility Behavior Sample

1-butanol

Solubility Test 10% H20 H2SO4  NaOH S+

-

Observations H20

10% NaOH

+

Colorless with emulsions, oily layer

2 layers, emulsions

2 layers, bubble suspennsions upon shaking

2-butanol

+

-

+

Colorless -> turbid white (upon shaking) -> colorless

t-butyl alcohol

+

-

+

Colorless w/ small white particles

2 layers,

+

2 layers, yellow  bubbly upper layer, coloress bottom layer

2 layers, yellow upper layer, colorless  bottom layer

Turbid white on top, dirty white on  bottom

Gelatinous cloudy white solution w/ small white  precipitates

Diethyl ether

S+

Phenol

Legend:

-

+

-

-

= soluble

n/a

S+ = slightly soluble

H2SO4 Colorless, soluble, exothermi c  Neon yellow, soluble, exothermi c Yellow, soluble, exothermi c Pale  brown, soluble, exothermi c

n/a

- = insoluble

Table 2. Reaction with Potassium Permanganate Sample 1-butanol 2-butanol

Before heating Brown violet solution with dark violet precipitate Brown dark violet solution with

KMnO4 After heating Light brown solution with dark violet precipitate Dark brown solution with

Result + +

violet precipitate t-butyl alcohol Diethyl ether Phenol

Dark violet solution Dark violet solution with violet  precipitate Light brown solution with dark  brown precipitate

dark brown precipitate Dark violet solution, no changes Dark brown solution with dark brown precipitate Pale yellow -> brown solution with dark red  precipitate, light red  precipitate on the surface

Table 3. Reaction with Ammoniacal Silver Nitrate Ammoniacal AgNO3 Sample Before heating After heating Colorless liquid, 1-butanol Colorless liquid emulsions 2-butanol Colorless liquid Colorless liquid t-butyl alcohol Colorless liquid Colorless liquid Diethyl ether Colorless liquid Colorless liquid Colorless bottom Pale yellow solution layer with pale yellow Phenol with emulsions, dark upper layer  brown precipitate

Table 4. Lucas Test: Reaction with HCl –  HCl –  ZnCl  ZnCl2 Mixture HCl –  HCl –  ZnCl  ZnCl2 Mixture Sample Observations 1-butanol Colorless solution 2-butanol Colorless solution -> Cloudy white solution t-butyl alcohol Colorless solution -> Cloudy white solution Diethyl ether Colorless liquid Phenol Colorless solution

+

+

Result -

Results + + -

Table 5. Reaction with Ferric Chloride Sample 1-butanol 2-butanol t-butyl alcohol Diethyl ether Phenol

Ferric Chloride Observations Colorless solution -> Pale yellow solution Colorless solution -> Pale yellow solution Colorless solution -> Pale yellow solution Colorless solution -> Pale yellow solution Colorless solution -> light violet with emulsion

Result +

Table 6. Iodoform Test: Reaction with I2, NaOH Sample Before addition of NaOH 1-butanol 2-butanol t-butyl alcohol Diethyl ether Phenol

II.

Orange solution, 2 layers Orange solution, 2 layers, oily upper layer 2 layers, yellow orange upper layer, orange bottom layer 2 layers, orange upper layer, yellow orange bottom layer 2 layers, orange upper layer, gelatinous pink bottom layer

After addition of NaOH 2 layers, pale yellow upper layer, la yer, colorless lower layer 2 layers, pale yellow upper layer, la yer, colorless lower layer

Result

Soluble, pale yellow solution

-

2 layers, yellow upper layer with some yellow immiscible liquid

-

2 layers, colorless upper layer, oily white turbid lower layer

-

-

Discussion

Solubility Behavior This part of the experiment examined the solubility of the each sample in water, 10%  NaOH and H2SO4 if insoluble in the basic solvent. solvent. The samples used were 1-butanol, 2-butanol, t-butyl alcohol, diethyl ether, and phenol. Solubility in Water. The presence of a polar functional group in a sample will be soluble in water following the rule “like dissolves like”. Alcohols are insoluble in water except if they are in Carbon-6. The more branching is present in a compound, the more it will be soluble  because branching lowers intermolecular forces and deceases intermolecular attraction. The higher the number of carbon atoms, the more nonpolar the compound making it insoluble in water. Diethyl ether is a diprotic solvent which means that it has limited solubility in water. In the experiment, results showed that phenol was insoluble in water while 1-butanol and diethyl ether are slightly soluble. Samples soluble in water are 2 -butanol and t-butyl alcohol. Solubility in 10% NaOH. A compound is deemed soluble in NaOH if it is significantly soluble in water. Insoluble compounds in this solvent are some sodium salts with highly substituted phenols. In In the experiment, all of the samples were were insoluble in this solvent. Alcohols are insoluble in NaOH solution because it is the hydroxide reacting with the alcohol, not the sodium ions (Rxn: NaOH + R-OH -> R-ONa + H2O). The free energy difference between the products and reactants side of the equation is not very large, however, so there will be some equilibrium between hydroxide ions (OH-) and alkoxide ions (R-O-) in solution, which will depend on the pKa of the alcohol (Net ionic equation: OH- + R-OH -> R-O - + H2O). The reaction of phenol with NaOH gave a gelatinous cloudy white solution w/ small white precipitates. Theoretically, it should give a colorless solution containing sodium phenoxide. Phenol behaves as an acid and gives up its proton to the OH- which is a base that is why it dissolves in NaOH solution. Solubility in H2SO4. Sulfuric acid protonates all organic compounds that contains oxygen and/or nitrogen, as well as few aromatic carbons and alkenes. The reaction of the compound with

this solvent may produce color changes and heat. In the experiment all the samples tested as observed to have an exothermic reaction. Some color changes were observed in 2-butanol (colorless to neon yellow solution), t-butyl alcohol (colorless to yellow solution), diethyl ether (yellowish to pale brown solution). In 1-butanol, no color change was observed.

Reaction with Potassium Permanganate Baeyer's reagent is an alkaline solution of cold potassium permanganate, which is a  powerful oxidant making this a redox reaction. Reaction with double or triple bonds in an organic material gives a positive test which includes the disappearance of purple color and appearance of brown precipitate. In the experiment, all of the samples turned into a brown colored solution with the appearance of precipitates except for t-butyl alcohol which had no changes. Tertiary alcohols cannot be oxidized for the reason that all the hydrogen atoms are already filled or occupied by the remaining 3 spaces needed and there are no hydrogen atoms attached to the alcohol carbon. KMnO4 oxidizes phenol to para-benzoquinone while alcohols will be oxidized to carbonyls.

R eaction eaction with Tollen’s reagent (Ammoniacal (Ammoniacal Silver Nitrate) Tollens' reagent is a chemical reagent used to determine the presence of an aldehyde, aromatic aldehyde and alpha-hydroxy ketone functional groups and not for alcohols. The reagent consists of a solution of silver nitrate and ammonia. It can oxidize aldehydes into corresponding carboxylate ion, while it reduces itself to metallic silver. The aldehyde itself is oxidised to a salt of the corresponding carboxylic acid  because the solution is alkaline. Only phenol reacted to the reagent but it was not a positive result. The solution turned pale yellow instead of a silver mirror” solution. The rest of the samples remained colorless.

Lucas Test: Reaction with HCl –  HCl –  ZnCl  ZnCl2 Mixture Lucas test distinguishes tertiary, secondary and primary alcohols from each other by using zinc chloride as the reagent in concentrated hydrochloric acid. It is based on the rate of formation of insoluble alkyl chloride. An emulsion is formed. This test is reliable only for alcohols that are fairly soluble in water. The rate of reaction of the alcohols is determined by the ease of formation of the carbocation intermediate. Tertiary alcohols react quickly because they form a relatively stable tertiary carbocation while secondary alcohols react more slowly because the secondary carbocation is less stabilized than the tertiary carbocation. Primary alcohols generally take to react, about an hour. The reaction of a tertiary alcohol, involves a Lewis acid- base reaction between zinc

chloride and a lone pair of electrons on the alcohol oxygen. In a slow, rate determining step, the HO-ZnCl complex leaves and the carbocation is then attacked by a chloride anion from the HCl solvent to form the insoluble alkyl chloride which forms the cloudy emulsion indicative of a  positive test. Again, the overall rate of the reaction is determined by the slow step, which is the  breaking of the C-OH-ZnCl bond. This rate is determined by the stability of the carbocation  being formed. Experimental results show that only 2-butanol and t-butyl alcohol gave positive results while t-butyl alcohol, diethyl ether and phenol gave negative results. Experimental results slightly deviated from the expected result because of t-butyl alcohol. It should have given a  positive result. The experimenters might have thought that no more reaction will take place, immediately after 30 minutes, the samples were discarded because the test tubes will be used for the next part of the experiment.

Reaction with Ferric Chloride The ferric chloride test is used to determine the presence of phenols in a given sample or compound. This test is traditional calorimetric test for phenols which uses a 1% iron (III) chloride solution that has been neutralized with sodium hydroxide until a slight precipitate of FeO(OH) is formed. Enols, hydroxamic acids, oximes, and sulfinic acids give positive results as well. A positive result forms a violet complex with iron (III), which is intensely colored. In the experiment, only the phenol setup obviously turned into a violet solution while the rest turned into pale yellow solution which indicates a negative result. Phenol is acidic, and can form the  phenoxide ion. This ion can complex with iron (III) and form a coloured substance. Alcohols are much less likely to form ions than phenol is, meaning alcohols cannot complex the iron(III) ion and therefore cannot form coloured compounds.

Iodoform Test: Reaction with I2, NaOH This test does not distinguish 1°, 2°, 3° alcohol but is specific for only one class of alcohol. This is the secondary methyl alcohol. If the alcohol contains a methyl group attached to a carbon that also has a hydrogen and an OH group then it will give a positive iodoform test. The formation of a yellow precipitate indicates a positive test. Apart from its colour, this can be recognised by its faintly "medical" smell. In the reaction, there is a formation of a mild oxidizing agent sodium hypoiodite, NaOI, from sodium hydroxide and iodine. NaOI oxidizes the alcohol to a carbonyl. The alcohol is oxidized to to a ketone and iodine gains two electrons. In the third step, which really consists of several individual steps, the methyl ketone is converted to the anion of a carboxylic acid (with one carbon less than the alcohol) and iodoform (yellow precipitate). Experimental data shows that all the samples were negative. There were color changes to a yellow solution, however, no yellow precipitate formations were observed.

III.

Conclusion

The organic derivatives of water, such as alcohols, phenols and ethers have different chemical properties. To be acquainted with these properties, certain tests can be made. The solubility behavior follows the general rule like dissolves like”. Reaction with potassium  permanganate involve redox reaction with double or triple bonds in an organic material. A  positive test which includes the disappearance of purple color and appearance of brown  precipitate. The Tollen’s test is used to determine the presence of an aldehyde, aromatic aldehyde and alpha-hydroxy ketone functional groups. The formation of black precipitates indicates a  positive result. Lucas test distinguishes tertiary, secondary second ary and primary alcohols from each other  by using zinc chloride as the reagent in concentrated hydrochloric acid. A positive test turns a solution into cloudy white. The ferric chloride test is used to determine the presence of phenols in a given sample or compound. A positive result forms a violet complex with Fe3+, which is intensely colored. Iodoform test is used to determine the presence of secondary methyl alcohol. If the alcohol contains a methyl group attached to a carbon that also has a hydrogen and an OH group then it will give a positive po sitive iodoform test. Some errors were observed based on the experimental results. This may be due to human errors, contaminations, and inappropriate proportions.

IV.

Answers to Questions

1. Explain the solubility behavior of the representative compounds in water as a function of: a) branching in the molecule The more branching is present in a compound, the more it will be soluble in water  because branching lowers intermolecular forces and decreases intermolecular attraction  b) relative proportions of hydrophilic bonds to hydrophobic bonds Hydrophilic bonds are water soluble while hydrophobic bonds are water insoluble. Hydrophobic bonds are those without any functional groups or C-C bonds and have change in the cluster or position of water and solute. Hydrophilic Hydrophilic bonds have O, N- very electronegative atoms and involves H-bonding. 2. Explain the acidity differences observed for the compounds used in this exercise in terms of the stability of the corresponding conjugate bases. K-OH bond gives alkoxide anion. It is a strong base which can easily attract p roton. ROH acts as lewis acids. All samples except phenol are insolubkke on the 10% NaOH. Phenol is a strong base. 3. Based on the results of the oxidation test, classify the compounds tested according to the following categories: easily oxidizable, oxidizable, and resistant to ox idation. The appearance of black particles indicates that all all compounds except phenol were easily oxidizable according to the students’ experimentation. Phenol can be b e oxidizable to easily oxidizable due to change with solutions color. 4. Discuss the reactivity differences of alcohols towards the Lucas rea gent. Possible error could have occurred despite the formation of a cloudy white solution in in the experiment. 

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Tertiary alcohols react immediately, secondary alcohols react for about 8 -10 minutes and  primary alcohol react the longest for about an hour, and sometimes it gives no reaction. 1-butanol- primary alcohol (no reaction when time was stopped) 2-butanol- secondary alcohol t-butyl alcohol –  alcohol –  tertiary  tertiary alcohol

5. Suggest simple chemical test that will differentiate between the following pairs o f compounds. Write the equations for the reactions involved. a. phenol and isopentyl alcohol -> dilute NaOH solution C6H5OH(s) + OH-(aq) -> C6H5O-(aq) + H2O(g)  b. tert-butyl alcohol and isobutyl alcohol -> water oxidation C4H9OH(s) + H2O(l) -> C4H10O(aq) c. neopentyl alcohol and ether -> evaporation C5H912O + H2O -> C4H10O d. sec-butyl alcohol and neopentyl alcohol -> Lucas Test C5H912O + HCl -> C5H12O e. propene and butanol -> Acid hydrolysis C3H6 + H2O -> C4H9OH 

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V.

References

Experiment 6: Qualitative Tests for Alcohols, Alcohol Unknown, IR of Unknown. Retrieved April 8, 2017 from http://myweb.brooklyn.liu.edu/swatson/Site/Laboratory_Manuals_ files/Exp6.pdf  Organic Derivatives of Water. Retrieved April 8, 2017 from https://www.scribd.com /doc/48026330/Organic-Derivatives-of-Water The Iodoform Test. Retrieved April 8, 2017 from http://www.harpercollege.edu/tmps/chm/ 100/dgodambe/thedisk/qual/iodo.htm Physics Forums. Retrieved April 9, 2017 from https://www.physicsforums.com/ https://www.physicsforums.com/threads/whythreads/whycant-alcohol-react-with-aqueous-naoh.410255/