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Names: Bryle Kristiann Camarote Date Performed: January 29, 2014 Nimrod Romelo Date Submitted: February 5, 2014 Sarah Jane Valdon EXPERIMENT #6 Hydrocarbons I. INTRODUCTION Hydrocarbons are family of organic compounds composed entirely of carbon and hydrogen. They are the organic compounds of simple...

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Names: Bryle Kristiann Camarote Nimrod Romelo Sarah Jane Valdon

Date Performed: January 29, 2014 Date Submitted: February 5, 2014

EXPERIMENT #6 Hydrocarbons

I. INTRODUCTION

Hydrocarbons are family of organic compounds composed entirely of carbon and hydrogen. They are the organic compounds of simplest composition and may be considered the parent substances from which all other organic compounds are derived. Compounds in which the carbons are bonded to each other by only single bonds are known as saturated compounds; examples include alkanes and cycloalkanes. Alkanes contain only carbon and hydrogen atoms; they are considered saturated because they have only C-C and C-H. Single bonds and thus contain the maximum possible number of hydrogen per carbon. All carbons are sp3 hybridized and are connected by sigma bonds. On the other hand, cycloalkanes are types of alkanes. One example is cyclohexane. It is a colorless, volatile liquid with pungent odor and has the formula C6H12. It is produced by hydrogenation of benzene and distillation of petroleum. It is used primarily as solvent for paints, varnish and oils.

Unsaturated compounds are compounds that contain one or more double or triple bonds between their carbon atoms. Several unsaturated compounds include alkenes, benzenes and cyclohexenes. Alkenes are hydrocarbon that contain a carbon-carbon double bond, C=C. They have the general formula CnH2n. Because of their double bond, alkenes have lesser number of hydrogens per carbon than alkanes and are therefore referred to as unsaturated. The double bond contains a pi bond while some single bonds have sigma bonds instead. Alkenes include cyclohexene, benzene and toluene. Ringed structures such as cyclohexenes, benzenes and toluenes are considered as alicyclic compounds. Benzene is a colorless liquid with a characteristic odor and burning taste. Its formula is C6H6. The benzene molecule is a closed ring of six carbon atoms connected by bonds that resonate between single and double bonds.

It is insoluble in water but miscible in other elements or compounds. Toluene is a colorless hydrocarbon occurring in coal tar. Toluene is sometimes called methylbenzene. It’s a type of benzene with one methyl group attached. It is used as a solvent and as a source of synthetic compounds. Alkynes are hydrocarbons containing C= C. The carbon-carbon triple bond results from the overlap of two sp-hybridized carbon atoms and consists of one sp-sp sigma bond and two pp pi bonds. Alkynes have the general formula: CnH2n-2. They are very actively chemically and are not found free in nature. Acetylene (C2H2) on the other hand, is a colorless, odorless, flammable gas having HC= CH. It is also known as an ethyne. Acetylene gas is usually stored in metal tanks, under pressure, dissolved in acetone. It burns in air with a hot and brilliant flame. Hydrocarbon solubility follows the rule of like dissolves like. This means that polar compounds like water and alcohols dissolve polar compounds. Nonpolar compounds dissolve other nonpolar compounds but tend not to dissolve polar compounds. Halogenation is the replacement of one or more hydrogen atoms in organic compounds by a halogen. The halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed. The chlorination of methane provides a simple example CH4 + Cl2 + hv and heat CH3Cl + HCl. Bromination of alkanes occurs by similar mechanism, but is slower and more selective because a bromine atom is a less reactive hydrogen abstraction agent than a chloride atom as reflected by the higher bond energy of H-Cl than H-Br. Alkynes also undergo halogenation. Chlorine and bromine add easily to a triple bond, the bromine test is able to determine the presence of unsaturation. Baeyer’s reagent, named after the German organic chemist Adolf von Baeyer, is used as a qualitative test for the presence of unsaturation such as double bonds. It is an alkaline solution of potassium permanganate. Reaction with double or triple bonds in an organic material causes the color to fade from purplish pink to brown. The test with ammoniacal silver nitrate can be used to determine whether the carboncarbon triple bond is found in carbon chain’s end or not. The test involves the release of the terminal proton of the alkyne so that the resulting acetylide ion can form an insoluble precipitate with Ag+. In this experiment, hydrocarbons were examined and investigated. Their solubilitie behaviors were tested, their halogenation, oxidation and reactions with ammoniacal silver nitrate were also observed.

II. RESULTS AND DISCUSSION

I. Investigation of Hydrocarbons 1. Solubility Behavior This part of the experiment involves testing the miscibility of each sample to the CCl 4, H2O, 10% NaOH, and H2SO4 solvents. For cyclohexane, mixing it into the CCl4 solvent resulted in a clear solution. The same goes for the benzene and the toluene. The results signify that each sample is miscible to the carbon tetrachloride solvent. This is due to the principle “like dissolves like”. Our solvent is a nonpolar molecule (with zero dipole moment) because of the four chloride atoms arranged in a tetrahedral arrangement around the carbon, cancelling their electronegative forces which can cause polarity. And each organic sample (cyclohexane, benzene, and toluene) only contain C-H bonds, they are generally nonpolar. Because both components are nonpolar, we can therefore say that they are soluble in the solvent. When water was added to cyclohexane, benzene, and toluene, a clear and colorless solution gave rise. The resulting solution is heterogeneous with two layers which imply that hydrocarbons are immiscible and thus insoluble with water. Furthermore, water is a polar compound due to very polar O-H bonds. In addition, nonpolar hydrocarbons, on the other hand, don’t dissolve in polar water. The same goes for the 10% NaOH solvent. Cyclohexane formed a separation layer when mixed with the solvent indicating that both are immiscible with each other, benzene formed a clumpy layer with the solvent showing minute emulsions, and toluene also formed a separation layer with the solvent but with bubble suspension showing the evolution of gas molecules. We can clearly conclude that each compound is insoluble with the sodium hydroxide solvent. The solvent is polar due to the dipole moment of the –OH while the samples are nonpolar. The same principle applies which is “like dissolves like”. Lastly, we used sulfuric acid H2SO4 as the solvent. Mixing with cyclohexane produced an immiscible mixture with the upper layer clearer than the lower layer which may be due to the gases evolved from the agitation of the concentrated acid. The same result was obtained for the benzene and toluene which formed a translucent upper layer. All samples were immiscible with the solvent because H2SO4 is polar while the samples are nonpolar. Moreover, all were exothermic.

Table 6.1 below summarizes all the remarks being observed in the solubility behavior of the given solvents. SOLVENT

SAMPLE

COLOR

OBSERVATIONS MISCIBILITY

CCl4 (nonpolar)

Cyclohexane Benzene Toluene

Colorless Colorless Colorless

Cyclohexane

Colorless

Benzene

Colorless

Toluene

Colorless

Cyclohexane

Colorless

Very clear clear clear Unequal Separation More curved separation layer Bent separation layer Equal Separation

Benzene

Colorless

H20 (polar)

10 % NaOH

Clumpy, separated layer

Clearer upper layer, bubble Toluene Colorless suspension at lower layer Upper layer more clear, Cyclohexane Colorless unequal separation, exothermic Translucent Concentrated upper layer, H2SO4 Benzene Colorless unequal separation, exothermic Translucent Toluene Colorless upper layer, exothermic Table 6.1 Solubility Behavior of the different Hydrocarbons

Miscible Miscible Miscible Immiscible Immiscible Immiscible Immiscible Immiscible

PHYSICAL STATE Liquid Liquid Liquid Liquid with two layers Liquid with two layers Liquid with two layers Liquid with two layers Liquid with two layers

Immiscible

Liquid with two layers

Immiscible

Liquid with 2 layers

Immiscible

Liquid 2 layers

Immiscible

Liquid with 2 layers

2. Halogenation In this part of the experiment involves testing the reactivity of the organic compound with bromine or specifically, determining whether a particular compound have double bonds or triple bonds. A positive result shows the loss of the red-orange colour of the bromine solution. Bromine test involves electrophilic addition reaction were in alkenes react with Br2 to form a transdibromoalkane. For saturated hydrocarbons (alkanes), they undergo substitution reactions with halogens when exposed to ultraviolet light while unsaturated hydrocarbons (alkenes and alkynes) readily undergo addition reactions with halogens.

In the experiment, 0.05 M Br2ICCl4 was added to the wrapped and unwrapped test tubes of cyclohexane, benzene and toluene. The tubes were allowed to stand for 10 minutes and later examined. The set of unwrapped samples were found to have changed colour from red-orange to whitish clear: Cyclohexane appeared clearer than toluene and benzene, while both toluene and benzene showed to be translucent white. For the set of wrapped samples, all displayed a translucent light orange colour with the cyclohexane clearer than toluene and benzene. Benzene and toluene are aromatic hydrocarbons. Though benzene and toluene have C=C, they don’t react like alkenes or alkynes. The result showed a negative test as the yelloworange remained. Aromatic rings are less reactive toward electrophiles like bromine than alkenes or alkynes are. For bromination of aromatic hydrocarbons to take place, a catalyst is needed. And this bromination occurs as electrophilic aromatic substitution. Electrophilic addition is not favored by aromatic rings because the stabilities will be lost. Instead, they undergo substitution for stability. Pi-bonds in benzene and toluene contain localised electrons sited above and below the two carbon atoms in the double bonds. This produces a region of high electron density. When bromine approaches benzene/toluene, electrons in pi-bond repel electrons in Br-Br bond, inducing a dipole in the Br2 molecule, it is now polar. Pi-electron pair from the double bond is now attracted to the slightly positive bromine atom, causing the double bond to break, forming a new bond between one carbon atom and one bromine atom, forming a positively charged carbocation. The bond between the two Br atoms breaks by heterolytic fission, forming a bromide ion. The Br- ion is attracted towards the intermediate carbocation, forming a covalent bond. Furthermore, the function of CCl4 is to be the solvent for bromine. Since bromine is

immiscible with nonpolar hydrocarbon compounds, carbon tetrachloride aids by allowing the two compounds easily react. Moreover, by dissolving the Br2, the reactant energy is lowered, lowering the activation energy, and thereby making the reaction go more readily. While for cyclohexane, the ultraviolet light acts as a catalyst in order for the reaction to occur. The energy from the UV light is absorbed by the Bromine gas and breaking it into two bromine radicals. The bromine would then remove and replace one hydrogen from the cyclic ring forming a cyclic halide. HBr is produced as a by-product.

Benzene and toluene comprise of double bonds and bromine readily reacts with them even without the prescence of catalysts. Via electrophilic substitution, the Br2 replaces one of the hydrogen singly bonded to the benzene/toluene ring with HBr produced as a by-product. Table 6.2 below summarizes all the results and observation in halogenation.

Sample

Unwrapped

RESULT

Cyclohexane

Less translucent white

POSITIVE

Benzene

Translucent white

POSITIVE

Toluene

Translucent white

POSITIVE

Wrapped Lighter translucent orange Light translucent orange Light translucent orange

RESULT NEGATIVE NEGATIVE NEGATIVE

Table 6.2 Halogenation

3. Oxidation: Reaction with Baeyer’s Reagent 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 (-C=C- or C≡C-) in an organic material causes the color to fade from purplish-pink to brown. Decolorization of a purple solution and formation of a brown precipitate (MnO2) is a positive result for this kind of test. Bayer’s test involves a redox reaction were in Mn7+ is reduced to Mn4+ and alkene is oxidized to a diol. Alkenes react with KMnO4 to give a diol and MnO2. In the experiment, when cyclohexane reacted with Baeyer’s reagent, there was no color change observed, the solution still remains purple, and no layers were formed. This confirms that potassium permanganate does not react with alkanes because they are saturated (single bonds which are taken and have more amount of hydrogen bonded). For benzene, bubble suspension at the top of the light purple layer was perceived. The same remark also goes with toluene but only has an exception, this time it is only lighter than what we have observed in benzene. Despite the unsaturation of aromatic rings, they do not usually react with strong oxidizing agent such as KMnO4. They are inert to oxidation. Table 6.3 below summarizes the results of this part of the experiment. Table 6.3 Oxidation: Reduction with Baeyer’s Reagent SAMPLE OBSERVATION RESULTS Color change is unobservable Hexane NEGATIVE (Purple color) Benzene Color change is unobservable NEGATIVE

(Purple color), Bubble suspension showed Color change is unobservable NEGATIVE (Purple color) Brown POSITIVE

Toluene Cyclohexene

4. Reaction with Ammoniacal Silver Nitrate Ammoniacal silver nitrate serves as a test to determine the presence of a triple bond (alkyne). The reaction includes the release of terminal proton of the alkyne so that the acetylide ion can form an insoluble precipitate with Ag+. For example, acetylene reacts with ammoniacal solution of AgNO3 forming acetylide of silver. H-C

C-H + 2 AgNO3 + 2 NH4OH AgC CAg + 2NH4NO3 + 2H20 (white precipitate) When ammoniacal AgNO3 was added to cyclohexane, the solution formed two immiscible layers in which bubble suspensions were present. Since the reaction with ammoniacal AgNO3 is exclusive to alkynes, the cyclohexane resulted in a negative test in which there is no reaction. Table 6.4 below summarizes all the observations and results of the experiment. Table 6.4 Reaction with Ammoniacal Silver Nitrate Ammoniacal Silver Nitrate Sample Pre-observation Color Colorless with Clear and Cyclohexane formation of soluble two layers

Clarity Physical Sate Clear with bubble Liquid suspensions

B. Preparation and Testing of Acetylene Gas The acetylene gas was prepared by reacting water to calcium carbide. The slowly addition of water to calcium carbide resulted to the release of acetylene gas. The chemical reaction for the preparation of acetylene is: CaC2(s) + 2H2O (l) → Ca(OH)2(s) + C2H2 (g) Acetylene gas was then released upon the slow addition of water. Afterwards, the collected gas was then tested for its solubility behaviour with cyclohexane, toluene, and benzene. For cyclohexane, it mixed well with no sign of layer formation. The same goes for toluene and benzene. Acetylene was next tested with halogenation using 0.05 M Br2ICCl4.

The unwrapped mixture appeared less clear or slightly foggy than the wrapped mixture. For oxidation using Baeyer’s reagent, acetylene caused the solution to change in color with bubble formation. And lastly, acetylene was tested with its reaction with ammoniacal silver nitrate and resulted in the formation of precipitates. Table 6.5 below summarizes the results of this part of the experiment. Solvent Baeyer’s reagent (violet) Ammonium silver Br2 in CCl4 (red orange)

Observation Change in color of the solution with bubble formation Precipitate formed Wrapped Unwrapped Less clear, slightly Clearer foggy

Table 6.5 Preparation and Testing of Acetylene Gas

Terminal alkynes are acidic because of its sp hybridization. In sp hybridized orbitals, the electron pair is nearest to the nucleus. This causes the alkyne to lose its Hydrogen and form an H+ ion. Since the electron pair is near the nucleus, the anionic compound becomes stable. And a stable conjugate base means that it is an acidic compound. But they acidity is still greatly less than water. Given below are the reagents and the reaction conditions that would distinguish between the following compounds. Written also are the equations for the reactions involved: a). Benzene and Ethylbenzene Baeyer’s Reagent  Benzene does not get oxidation with baeyer’s reagent, but when ethyl benzene is oxidised with baeyer’s reagent (alkaline KMnO4) it gives Benzoic acid which gives effervescence of CO2 with bicarbonate b.) 1-butyne and 2-butyne Reaction with ammoniacal AgNO3  With the reaction of Ammoniacal AgNO3, it’s possible to determine whether the C≡C is found at the end of the carbon chain or not. Upon reaction, the terminal end of 1butyne releases a proton so that the resulting acetylide ion can form an insoluble precipitate with Ag+. The reaction for 2-butyne is different.

c. 2-methylpentane and 2-methyl-2-pentene Reaction with bromine  2-methylpentane is an alkane and 2-methyl-2pentene is an alkene, to distinguish one from the other, bromine test can be used. 2-methylpentane react with bromine to produce H-Br gas as a by product, while 2-methyl-2pentene does not produce H-Br. d. toluene and 1-methylcyclohexene  Baeyer’s Test III. CONCLUSION Hydrocarbons are compounds that contain only hydrogen and carbon atoms and because of the cancellation of C-H, they become a non-polar compound. Since they are nonpolar compounds they will be only dissolves in non-polar compounds like CCl4 as the rule of solubility states that like dissolves like. Hydrocarbons are immiscible in a polar compound such as water, NaOH , and H2SO4. If Br2 is used as a halogen, the reaction is said to be positive to the free radical substitution if it loses red-orange color of the molecular bromine and the evolution of gas, hydrogen bromine has occurred. The Baeyer’s test is a test for double bonds or alkenes. For positive results using KMnO4 it must show decolorization of a purple solution and formation of a brown precipitate (MnO2). Bayer’s test involves a redox reaction were in Mn7+ is reduced to Mn4+ and alkene is oxidized to a diol. Alkenes react with KMnO4 to give a diol and MnO2. Different errors can be encountered in this experiment like the calibration of equipment, the personal biases of the individual especially on the color change of the compound and the describing the compound in the looks, physical state, miscibility etc. Errors can be eradicated by the proper calibration of the equipment and by the and by taking care of what is being noted. Various test in determining the specific functional groups by observing its behaviour when it reacts with certain reagent. These chemical reactions such as halogenation and oxidation help us predict what type of hydrocarbon is being tested (saturated or unsaturated). In the experiment, halogenation and oxidation reaction more likely took place for unsaturated hydrocarbon due to electron dense areas making substitution of the hydrogen by a halogen easier. We also observed that most of the chemical reactions accompany change in color, change in temperature of the given solution and formation of precipitates which signals that reaction really happened. Careful handling of the solutions should be done...