Experiment 1: Solubility Behavior of Organic Compounds Test compound
Legend: + : Soluble - : Insoluble
+ distilled H2O
Solubility by IMFA
Solubility by reaction -
+ + Ether
+ 5% NaOH -
-
+ S1
S2
Water-soluble compounds
+
+ 5% HCl
+ 5% NaHCO3 -
+
A2
B
Basic compounds
+ A1
Acidic compounds + H2SO4
M
Inert compounds
Miscellaneous compounds
+ I
N
Neutral compounds
Experimental results:
I.
S1: Ethanol, Acetone
A2: Phenol
N: -
S2: Sucrose
B: Aniline
I: Hexane, Toluene, Tert-butyl chloride
A1: Benzoic acid
M: Benzamide
Water-soluble compounds have, at least, 1 hydrogen-bonding group per 5 carbon atoms along its path. As such, all possible paths towards the hydrogen-bonding group must be taken to ascertain water solubility.
II. Ether-soluble compounds have 2 or more carbon atoms per hydrogen-bonding group. III. Solubility in water and ether is governed by intermolecular forces of attraction, specifically by hydrogen bonding and Van der Waals forces. Entire dissolution of sample is required for a positive result.
IV. Solubility in acidic and basic reagents is governed by chemical reactions, specifically by neutralization and forced protonation (via sulfuric acid) reactions. Partial dissolution of sample is sufficient for a positive result. V. A1 compounds are strong acids since they were dissolved both by sodium hydroxide, a strong base, and sodium bicarbonate, a weak base. Carboxylic acids and phenols with strong electron-withdrawing groups, such as A2,4-dinitrophenol and B2,4,6-trinitrophenol, belong to this group.
A
2,4-dinitrophenol B
2,4,6-trinitrophenol
The acidic proton of “activated” phenols is more easily cleaved since strong electron-withdrawing groups weaken the hydrogen-oxygen bond. VI. A2 compounds are weak acids since they were not dissolved by sodium bicarbonate, a weak base. All other phenols belong to this category. VII. B compounds are bases since they were dissolved by hydrochloric acid, a strong acid. Amines, primary, secondary, and tertiary, belong to this category. However, electron-withdrawing groups make the basic lone pair less stable and, as a result, such amines belong to the M class. These amines are usually Adiand tri- arylamines, Barylamines with electron-withdrawing groups, and Csimple monosubstituted amides.
A
C
Diphenylamine B
Acetamide
2,4,6-trinitroaniline
VIII. M compounds are miscellaneous. These compounds have nitrogen and/or sulfur atoms and don’t belong to any of the previous classes as well. AAmides belong to this group.
A
Benzamide
Amides are ammonia derivatives wherein the one of the hydrogen atoms has been replaced by an acyl (R-C=O) group.
IX. N compounds are neutral. Aldehydes, ketones, alcohols, ethers, esters, alkynes, and alkenes belong to this category. X. I compounds are inert. Alkanes, alkyl halides, and aromatic compounds, especially molecules with benzene rings, belong to this category.
Experiment 2: Recrystallization and Melting Point Determination of Benzoic Acid I.
The properties of an ideal solvent are as follows: A. B. C. D.
Should readily dissolve solute at elevated temperatures and only sparingly at room temperature Should readily dissolve impurities at any temperature or not at all Should not react with the solute Should be sufficiently volatile to permit easy removal or separation of crystals
II. To prevent recrystallization, the following may be employed during hot filtration: A. Fluted filter paper: Faster filtration occurs due to the larger surface area available. B. Insertion of wire: Faster filtration occurs since escaping vapors need not pass through the funnel, counter-current to the downward flow of the filtrated solution. C. Lifting of funnel during filtration: Faster filtration occurs due to the same reason as above. D. Addition of recystallization solvent to the receiving flask: The volatilizing solvent will heat up the entire system without introducing impurities. The solution is also forced through the filter paper by atmospheric pressure. III. The cooling rate has to be controlled as well. If the filtrate is rapidly cooled, equilibrium conditions will not be met and small crystals will be formed. If the filtrate is slowly cooled, large crystals with extensive mother liquor inclusion will be formed. Therefore, the hot filtrate should be allowed to cool at a rate between these two extremes.
IV. Failure to recrystallize the solute may happen due to supersaturation, unsaturation, or colloidal formation due to an impurity. V. To facilitate recrystallization, one may add a small crystal of the pure substance to the solution, or scratch the container’s inner walls to introduce small sites for recrystallization or seeds into the solution. VI. Animal charcoal or boneblack absorbs colored impurities in most organic substances. VII. For the hot filtration procedure, a wide bore short-stemmed funnel is used. VIII. If the solvent is evaporated after one recrystallization step, the hot step is skipped for the subsequent recrystallization. If the solvent is decreased to half its volume, solubilities are halved as well. IX. Percent purity and percent recovery are to be determined from the final insoluble phase in the final recrystallization step. masscompound × 100 % massfinal insoluble phase masscompound in final insoluble phase Percent recovery = × 100 % masstotal compound used Percent purity =
Percent purity is the main basis for determining the best recrystallization solvent. If the used solvents yield the same percent purity, percent recovery then serves as the tiebreaker. X. Percent recovery is somewhat synonymous to percent yield, though the later is only used for solutions involving chemical reactions.
Experiment 3: Purification of Crude Benzoic Acid by Sublimation I.
For this method of purification to work, the solid must possess a relatively high vapour pressure at a temperature below its melting point. Weak intermolecular forces give way to a large tendency for evaporation or sublimation, which in turn leads to high vapor pressures.
II. Two phase changes occurred in this experiment: sublimation and deposition. III. Purification by sublimation has several limitations. For one, it is only useable for compounds that sublimate at easily attainable temperatures. Also, it can only be used for volatile substances with nonvolatile impurities. Although, this need not be fulfilled; at the very least, there should be a significant difference in the sublimation temperatures of the desired compound and the impurities in the solution.
IV. A comparison between purification by recrystallization and sublimation: Procedure Recystallization Description Tedious; many steps Percent purity Low Percent recovery High
Sublimation Easy; one step High Low
Experiment 4: Liquid Phase Chromatography I.
There are generally two types of chromatography. In the normal phase, a polar stationary phase with a relatively less polar mobile phase is used. In the reverse phase, a non-polar stationary phase with a relatively less non-polar mobile phase is used.
II. Paper chromatography is the separation of components based on polarity. Here, the stationary phase is the water adsorbed on the cellulose in the Whatman #1 filter paper, while the mobile phase is the solvent, which was n-butanol diluted with an aqueous acetic acid solution in the experiment. Evidently, paper chromatography is carried out as normal phase. III. Important concepts about paper chromatography:
11
C, D
10
A Distancesolvent
Solvent front
B
6
A
4
DistanceA 1
Mobile phase
Starting line
The retention factor refers to the degree to which the solvent carried a component of the solution through the filter paper. In paper chromatography, wherein the mobile solvent is dominantly non-polar, the higher the retention factor, the less polar the constituent is. Retention factorA =
DistanceA Distancesolvent
Note that the topmost dot contains two components. In order to separate the two, 2-dimensional paper chromatography should be used. This simply involves redoing the procedure with the same filter paper sideways but with a different solvent. That is:
D C
A
B
IV. The procedure is done in a container to minimize the volatilization of the solvent from the paper. This reduces the possibility of contamination as well. V. Tailing may occur in paper chromatography, wherein dots overlap or even streaks of varying hues are formed. This may be caused by sample overload, the inability of the solvent to distinctly separate constituents, or prolonged exposure to moving air, wherein the solvent is constantly volatilizing. VI. The properties of an ideal solvent are as follows: A. Cheap, for practicality B. Relatively volatile, since solvent vapors aid capillary action in lifting the solvent and solute C. Does not react with the sample VII. Evidently, for colored ones, human eyes are sufficient to determine the presence of the spot. Colorles spots, on the other hand, require certain techniques to allow elucidation. Visualization techniques for these are as follows: A. Ultraviolet light: For conjugated compounds, wherein double bonds alternate with single bonds B. I2 chamber: For unsaturated compounds, such as alkenes and alkynes Dark brown spots that become lighter towards the solvent front become visible. C. KMnO4 washing: For oxidizable compounds, such as alcohols and aldehydes Brown spots become visible.
Experiment 5: Isomerism and Stereochemistry I.
Isomers are compounds that have the same molecular formula, but differ in the position of functional groups, length of the parent carbon chain, geometric orientations of substituents, chirality, etc.
Isomers 3-dimensional
2-dimensional Structural
Stereoisomer Chain
Geometric
Positional
Optical
Functional
Conformational
Chain isomers refer to the length of the parent carbon chain and the presence of alkyl branches.
Butane 2-methylpropane Somewhat similar to chain isomers are positional isomers, which tackle all other functional groups. Functional isomers refer to the presence of a different functional group for each isomer; it is possible that a ketone and an aldehyde may be formed from a single moleculer formula:
Propanal Propanal Propan-2-one Geometric isomers refer to the position of substituents about a double bond. Conformational isomers refer to the possible three-dimensional appearances of a cyclical compound, such as chair and boat isomers for cyclohexane:
Chair conformation
Boat conformation
Another type of conformational isomers deals with the staggered and eclipsed orientation of substituents, with respect to each other, as a result of rotation about a single bond.
Anti
Gauche
II. Optical isomers possess chiral carbon atoms. These carbon atoms have four different substituents and as a result, these molecules are asymmetric with respect to the chiral carbon centers. III. Enantiomers are non-superimposable mirror images of a molecule. They have the same properties, except for the rotation of plane-polarized light. Being mirror images, their absolute configurations are switched, with one being R and the other being S. For each compound, there are 2n enantiomers, where n is the number of chiral centers. IV. To determine the absolute configuration of a chiral center, the least prioritized group should be positioned the farthest. This can be achieved by performing an even number of ligand switches. The three other substituents are then assigned their priorities based on atomic weight. The absolute configuration is determined by the circular position of the three substituents according to their priorities. V. An R configuration is assigned for centers that have clockwise-positioned priority while an S configuration is assigned for centers that have counterclockwise-positioned priority. VI. Diastereomers are non-superimposable non-mirror images of a molecule. This is formed by performing only one ligand switch. As a result, only one absolute configuration, given multiple chiral centers¸ is different. VII. Mesomers, or meso compounds, are superimposable onto its mirror image. These are symmetrical molecules, which reduce the number of enantiomers by one.
Experiment 6: Synthesis of an Alkyl Halide I.
SN1 reactions form carbocation intermediates, while SN2 reactions form pentavalent carbon transition states.
II. The SN1 mechanism for the synthesis reaction, which is the formation of tert-butyl chloride from tertbutanol and hydrochloric acid, is as follows:
III. The E1 mechanism for the side reaction, which is the formation of 2-methylpropene from tert-butanol and hydrochloric acid, is as follows:
IV. Good leaving groups are weak bases, such as halides, which are conjugate bases of hydrohalogenic acids, water, which is the conjugate base of hydronium, ammonia, ethanol, etc. V. Rationalization of steps in the experiment: A. B. C. D.
E. F. G. H.
Cold hydrochloric acid: Prevents the E1 reaction. Concentrated hydrochloric acid: Practicality, since less volume of acid is needed. Excess hydrochloric acid: Pushes the synthesis reaction forward Saturated sodium chloride: Squeezes out organic molecules from the aqueous layer and pulls water molecules into the aqueous layer; slightly increases the purity of the organic layer by removing some water molecules. Sodium bicarbonate: Neutralizes unreacted acid while showing equivalence indication via effervescence. Solid sodium bicarbonate: Aqueous sodium bicarbonate has water, which is counterproductive for the experiment. Boiling chips: Promotes even boiling and facilitates bubble formation; prevents superheating, bumping, and loss of solution. Anhydrous calcium chloride: Complexes with oxygen-containing compounds, such as water and unreacted alcohols. Clumped drying agent: wet solution Free-floating drying agent (grains): dry solution
Experiment 7: Alcohols, Phenols, and Ethers
I.
Lucas reagent (Zinc chloride in concentrated hydrochloric acid) differentiates primary, secondary, and tertiary alcohols through a modified SN1 reaction. The positive result of this test is layer formation, due to the presence of the aqueous zinc, chloride, and hydronium layer, and the organic alkyl halide layer. Benzyl alcohol is a false positive since it readily forms an organic layer on its own. However, it should be noted that a reaction still took place. Benzylic and allylic alcohols still undergo the SN1 reaction as they more reactive than tertiary alcohols given the said mechanism.
II. Oxidation by neutral potassium permanganate, also known as Baeyer’s test, is only possible for alcohols with α-hydrogen atoms. As such, tertiary alcohols give a negative result, which is a purple solution. The positive result for this test is the decolorization of the purple solution accompanied by the formation a brown precipitate. This occurs to the reduction of manganese in the reaction: − + − 4H(aq ) + MnO4 (aq ) + 3e → MnO2 (s) + 2H2 O(l)
Primary alcohols are oxidized to carboxylic acids by strong oxidants such as potassium permanganate, chromic anhydride, and potassium dichromate. Weak oxidants, such as pyridium chlorochromate, reduce the alcohols to aldehydes instead.
Pyridium chlorochromate On the other hand, secondary alcohols are always oxidized to ketones. Finally, phenols are oxidized to quinones.
Quinone III. Ferric chloride confirms the presence of phenols via complex formation, the colors of which differ depending on the structure of the phenol. Most complexes are greenish or yellowish, but some may exhibit totally different colors, such as red in nitrophenol’s case.
IV. Bromine in water indicates the presence of phenols. The positive result is the formation of tribrominated phenol, a white precipitate.
2,4,6-tribromophenol
Experiment 8: Aliphatic and Aromatic Hydrocarbons I.
Alkanes, alkenes, and alkynes are aliphatic hydrocarbons. Hydrocarbons with benzene rings are aromatic.
II. Criteria for determining aromaticity: A. 𝟒𝐧 + 𝟐 = 𝛑 𝐞− , where n should be a whole number B. Conjugated molecule with a cyclic structure C. Planar structure via sp2 bond hybridization III. Bromination in light is a free radical substitution reaction. As such, only alkanes can undergo this reaction. The positive result is decolorization of the yellow solution to a colorless one. In the experiment, hexane and limonene gave positive results. The latter was able to undergo free radical bromination due to the presence of alkane substituents in its structure:
Limonene IV. Bromination in dark is a halogenations substitution reaction. Here, pi bonds are broken and bromide ions attach themselves to the (previously) vinylic carbon atoms. As with the previous bromination reaction, the positive result is decolorization of the yellow solution to a colorless one. In the experiment, limonene was the only one that gave a positive result. Benzene didn’t react because it lacked “true” double bonds.
Benzene resonance V. Baeyer’s test indicates the presence of double bonds. The pi bonds are broken and diol formation ensues. The positive result for this test is the decolorization of the purple solution accompanied by the formation a brown precipitate. In the experiment, only limonene gave a positive result since it was the only compound of the three that possessed double bonds.
4-(1,2-dihydroxypropan-2-yl)-1-methylcyclohexane-1,2-diol VI. Combustion distinguishes between aliphatic and aromatic compounds based on the sootiness of the produced flame. Carbon double bonds make the flame somewhat sooty and aromatic rings increase the sootiness all the more. As such, in the experiment, hexane showed a clear, yellow flame, limonene gave a sooty, orange flame, while benzene exhibited a sootier, dark orange flame. VII. Friedel Craft’s Alkylation, with tert-butyl chloride and aluminium chloride, is an electrophilic aromatic substitution reaction. The said reactants are needed, since a Lewis acid must catalyze the reaction. Aluminum choride forms a complex by abstracting the halide of the tert-butyl chloride: AlCl3 (s) + CH3 3 CCl(aq ) → AlCl4
− (aq )
+ + CH3 3 C(aq )
The negative charge of the complex allows to it act as a strong nucleophile for abstracting a hydrogen atom from benzene rings. As a result, alkylation involving the hydrogen-deficient ring and the tertiary carbocation proceeds faster. The positive result of this test is the colorization of white anhydrous aluminium chloride crystals to orange or red. VIII. Powdery aluminium chloride is no longer anhydrous.
IX. Steam distillation is used when the boiling points of the solute and solvent are very near each other. Otherwise, fractional distillation is used. In the experiment, this was used since limonene had to boil at a lower temperature since it degrades at a temperature that is a few degrees past its normal boiling point. Steam distillation accomplishes the lowering of the boiling point since the vapor pressure of the solvent contributes to the total pressure within the set-up. As a result, atmospheric pressure is more easily reached.
Experiment 9: Electrophilic Aromatic Substitution Reaction I.
Electron-donating groups increase electron density on the aromatic ring. They stabilize the carbocation intermediate through resonance and inductive effects. As such, the aromatic molecule becomes more reactive towards electrophilic attack. Examples are, arranged from most activating to least activating: O
NH2
NHR
NR 2
NHC = OR
OH
OR
O − C = OR
R
CH = CR 2
Electron-donating groups are ortho- and para-directing. II. Electron-withdrawing groups decrease the electron density on the ring. As a result, they destabilize the carbocation intermediate. The aromatic molecule becomes less reactive towards electrophilic attack. Examples are, arranged from most deactivating to least deactivating: NO2
NR 3
NH3
SO3 H
CN
CF3
ClC = O
HC = O
RC = O C = OOH
C = OOR X
Electron-withdrawing groups are meta-directing. Exceptions to this are the halogens, which are orthoand para-directing. III. In the experiment, the reaction was bromination in dark. A possible error in the experiment involves acetanilide. Since it has a methyl substituent, it can undergo bromination in light and, as a result, it may seem to react faster than normal.
Acetanilide IV. A polarizing solvent increases the rate of reaction since it imparts partial charges, or dipoles, to the halogen reactant. The carbocation intermediate is more readily formed. In the experiment, diluted glacial acetic acid was used. Besides the presence of the strongly polarizing carbonyl and hydroxyl
groups, the water present in the solution allowed the carboxylic acid to partially dissociate to form an ionic solution.
Resonance and final product Experiment 10: Aldehydes and Ketones I.
2,4-dinitrophenylhydrazine tests for the presence of carbonyl compounds, namely ketones and aldehydes. The positive result for this test is the formation of a yellow or orange precipitate. The acidcatalyzed reaction is:
Imine formation The negative result is the formation of a yellow solution. Note that the 2,4-DNPH solution is already yellow. Such a result would signify that no reaction took place. This test is affected by acid concentration. In a very acidic environment, the 2,4-DNPH molecules become protonated, which greatly hinders the continuation of the reaction. In a very basic environment, not all of carbonyl oxygen atoms are protonated and, as a result, activated. The pH of the solution must be kept between these two extremes. II. Schiff’s test makes use of Schiff’s reagent, which contains acidified leucofuchsin dye, which is colorless or light purple. This test indicates the presence of aldehydes. The positive result is the formation of a (dark) purple solution. The negative result is the formation of a light purple or pink solution. III. Tollens’ test makes use of Tollens’ reagent, which is ammoniacal silver nitrate. In an aqueous solution, this dissociates to form diamminesilver(I) ions. This test indicates the presence of aldehydes only via a reduction-oxidation reaction. The positive result of this test is the formation of an elemental silver
mirror. Due to the basicity of the solution, as a result of the presence of ammonia, a carboxylate ion was formed instead of a carboxylic acid. The reaction is: 2 Ag NH3
+ 2 (aq )
+ + RCOH(aq ) + H2 O(l) → 2𝐀𝐠 (𝐬) + 𝐑𝐂𝐎𝐎− (𝐚𝐪) + 4NH3 (aq ) + 3H(aq )
The negative result of this test is the formation of a turbid solution. Occasionally, a colorless solution may be formed as well. IV. Benedict’s test makes use of Benedict’s reagent, which is a basic cupric solution. This test indicates the presence of aliphatic aldehydes, which have their carbonyl functional groups directly attached to a non-aromatic carbon atom, via a reduction-oxidation reaction. The positive result of this test is the formation of a brick red precipitate. The reaction is: + − 2Cu2+ (aq ) + RCOH(aq ) + 2H2 O(l) → 𝐂𝐮𝟐 𝐎(𝐬) + 𝐑𝐂𝐎𝐎(𝐚𝐪) + 5H(aq )
The negative result is the formation of a blue solution, which signifies that no reaction took place. V. The iodoform test makes use of iodine in basic potassium iodide solution, which indicates the presence of methyl-ketones and 2-alkanols.
Butan-2-one
Butan-2-ol
The positive result is the formation of a bright yellow iodoform precipitate, which is CH3I. The precipitation is base-catalyzed. The hydrogen atoms of the methyl group are first replaced by iodine atoms. Then, an additional reaction with a hydroxide ion releases the iodinated carbon and a hydrogen atom attaches to it. In reality, only methyl-ketones facilitate the formation of iodoform but the hypoiodite ions (IO-) oxidize the secondary alcohols to methyl-ketones. The overall reaction is: − − − RCOCH3 (aq ) + 3I2 (aq ) + 4OH(aq ) → RCOO(aq ) + 𝐂𝐇𝐈𝟑 (𝐬) + 3I(aq ) + 3H2 O(l)
An exception to this test is ethanol. Hypoiodite oxidizes ethanol to ethanal, which is an aldehyde with a terminal methyl group. This methyl group undergoes the said reaction above. The negative result for this test is the formation of a yellow solution.
Experiment 11: Carbohydrates
I.
Six-membered carbohydrate rings are termed as pyranoses, while five-membered ones are furanoses.
II. Hexoses are dehydrated by strong acids to form 5-hydroxymethyl furfural, while pentoses form furfural. Evidently, this test is limited to these two types of carbohydrates; carbohydrates with four or less carbon atoms will always yield a negative result.
Furfural
5-hydroxymethyl furfural
Carbohydrates with more than six carbon atoms, such as disaccharides, oligosaccharides, and polysaccharides, are first hydrolyzed by the acid into their monosaccharide constituents before forming the corresponding furfural products. III. Molisch test makes use of α-naphthol, which indicates the presence of carbohydrates. The positive result of this test is the formation of a purple interface.
α-naphthol IV. Bial’s test makes use of orcinol, which indicates the presence of pentoses or hexoses. Pentoses form a blue to green solution or compound while hexoses form a yellow brown solution condensation product.
Orcinol V. Seliwanoff’s test makes use of resorcinol, which differentiates between ketoses and aldoses. Ketoses are more easily dehydrated by a hot strong acid than aldoses and, as such, for a given interval of time before the completion of the dehydration and resorcinol reaction, ketoses will form bright red condensation products or solutions while hexoses will only yield a pink coloration.
Resorcinol VI. Benedict’s test confirms the presence of reducing carbohydrates. VII. Barfoed’s reagent confirms the presence of monosaccharides. Though it has cupric ions, which are readily reduced by reducing sugars, the reaction rate decreases for each carbohydrate ring in the molecule’s structure. As such, the reduction-oxidation reaction with disaccharides proceeds much, much slower than one involving a monosaccharide; oxidation of oligosaccharides proceeds even slower. The positive result of this test is the formation of a brick red precipitate, Cu2O. VIII. Osazone formation occurs in sugars that are capable of mutarotating, reduces Benedict’s solution, and has α and β forms. In short, only reducing sugars, ones which have free anomeric carbons or hemiacetal structures, are capable of forming osazones. Phenylhydrazine replaces the attached groups to the first two carbon atoms of the reacting carbohydrate and a double bond is formed between the carbon and the terminal nitrogen atoms. The positive result for this test is the formation of yellow precipitate or crystals.
Phenylhydrazine IX. Iodine in potassium iodide solution test for the presence of starch. The linear amylose can form a helical tube-like or coil structure, which is made up of six D-glucose sub-units, linked together with α1,4 glycosidic bonds, per turn, that can encase triiodide ions, which were formed via the reaction: − − I2 (aq ) + I(aq ) → I3 (aq )
As a result, a blue-black solution is formed. The other component of starch, which is branched amylopectin, is incapable of yielding the same physical result since it cannot form a helical structure. On a side note, it has with α-1,6 glycosidic bonds between some of its glucose sub-units. (Acid-catalyzed) hydrolysis breaks down the linear amylose chains into its glucose sub-units, and maltose units if the hydrolysis is not extensive enough.
X. Cellulose is very similar to amylose in structure, except for the fact that β-1,4 glycosidic bonds are the ones that link sub-units. This results to a two-dimensional polymer structure that is much more rigid than that of starch. Amylopectin branching and helical coiling does not occur in cellulose as well; rather, a stiff chain is formed. Hydrogen bonding, brought about by the hydroxyl groups and hydrogen atoms between two chains, also increases the stability of the material. As such, cellulose is harder to break down than starch.
Experiment 12: Carboxylic Acid and Derivatives I.
Carboxylic acids are soluble in sodium bicarbonate via a neutralization reaction. Effervescence is observed as the formed carbonic acid degrades to water and carbon dioxide: Acetic acid: CH3 COOH(aq ) + NaHCO3 (aq ) → CH3 COONa(aq ) + CO2 (g) + H2 O(l) Benzoic acid: C6 H5 COOH(aq ) + NaHCO3 (aq ) → C6 H5 COONa(aq ) + CO2 (g) + H2 O(l)
II. The positive result for the acetic acid test is the formation of a red complex or solution. Sodium hydroxide neutralizes acetic acid to form acetate, which then complexes with ferric ions to form the said result.
Ferric acetate complex III. The positive result for the benzoic acid test is the formation of a flesh precipitate. Ammonium hydroxide neutralizes benzoic acid to form benzoate, which then complexes with ferric ions to form the said result.
Ferric benzoate complex IV. In the two previous tests, the addition of excess hydroxide ions leads to the formation of ferric hydroxide, a brown precipitate, which could be mistaken for a positive result.
V. The formation of esters proceeds faster with acyl halides than carboxylic acids since the latter is more reactive due to the presence of the electron-withdrawing halide. For acyl halides, addition of the alcohol is enough for the reaction to take place. However, for carboxylic acids, heat and an acid catalyst are both required. VI. The hydroxamic test indicates the presence of esters. The “uncommon” reagents that were used were alcoholic hydrochloric acid, alcoholic potassium hydroxide, and alcoholic acidic hydroxylamine. These reagents were simply diluted with alcohol in order to dissolve the ester sample. The positive result of this test is the formation of a purple complex or solution.
Ferric hydroxamate complex Hydrochloric acid was added to the solution in order to neutralize unreacted hydroxide ions from the dissociation of potassium hydroxide. This is to prevent the formation of ferric hydroxide. VII. Carboxylic acid derivatives hydrolyze in water to reform the parent acid. However, their reactivities determine the ease of accomplishing the said reaction. Anhydrides easily hydrolyze in water. Esters require the presence of a base-catalyst during the reflux procedure to form the corresponding carboxylate anion upon the expulsion of an alcohol, followed by a base catalyst to reform the carboxylic acid. Amides require the presence of an acid catalyst for the hydrolysis reaction. Ammonium cations are also formed in this reaction. VIII. The hydrolysis of amides can also produce a carboxylate product, instead of the acid, if a base catalyst is used. In this case, ammonia is formed and this serves as the indicator for the completeness of the reaction instead of the presence of a sour smell. This can be kept track of with a red litmus paper.
Experiment 13: Nucleophilic Acyl Substitution: The Synthesis of Esters I.
In the reaction mechanism, Athe nucleophile attaches to the carbonyl group, producing a tetrahedral intermediate. BThe electron pair from oxygen then displaces the leaving group to form a new carbonyl compound product.
In the experiment, the carbonyl oxygen atom is protonated by the acid catalyst before the formation of the tetrahedral intermediate through the attack of the alcohol on the carbonyl carbon and the breaking of the carbonyl pi bond. The hydroxyl hydrogen atom of the attached alcohol transfers to the hydroxyl group of the carboxylic acid, forming water. A free water molecule then abstracts the proton catalyst from the carbonyl oxygen and the carbonyl double bond is reformed. II. Rationalization of steps in the experiment: A. Excess alcohol: Pushes the synthesis reaction forward. B. Carboxylic acid as the limiting reagent: It is the substrate for the reaction, while the alcohol is the reactant that bonds to the substrate. Furthermore, excess acid hinders the observation of ester’s smell. Its sourness overpowers the product’s scent. C. Boiling chips: Promotes even boiling and facilitates bubble formation; prevents superheating, bumping, and loss of solution. D. Concentrated sulfuric acid: Catalyst for the reaction by protonating the carbonyl oxygen atom. The nucleophilic attack of the alcohol proceeds faster due to the full positive charge of the molecule. Also acts as a dehydrating agent; removes water from the solution which prevents the reverse reaction according to Le Chatelier’s principle. E. Cooling to room temperature: Prevents the volatilization of the ester as it is being transferred to the separatory funnel F. Cold water: Prevents the volatilization of the ester G. Rinsing the round bottom flask: Collects all of the formed products H. Saturated sodium chloride: Squeezes out organic molecules from the aqueous layer and pulls water molecules into the aqueous layer; slightly increases the purity of the organic layer by removing some water molecules. I. Sodium bicarbonate: Neutralizes unreacted acid (both from sulphuric acid and the used carboxylic acid) while showing equivalence indication via effervescence. J. Solid sodium bicarbonate: Aqueous sodium bicarbonate has water, which is counterproductive for the experiment. K. Anhydrous sodium sulfate: Dries the solution. Calcium chloride cannot be used since the product contains oxygen atoms.
III. Distillation increases the percent yield of the procedure. There are two cases for this, regarding the boiling points of the formed ester and used (unreacted) alcohol: A. BPester > BPalcohol: The ester is in the distilling flask. B. BPester < BPalcohol: The ester is in the receiving flask. IV. The use of excess alcohol and cold water, the addition of saturated sodium chloride, concentrated sulphuric acid, and anhydrous sodium sulfate, and refluxing the reaction solution all increased the percent yield of the reaction. V. The addition of saturated sodium chloride, solid sodium bicarbonate, and anhydrous sodium sulfate, and distillation all increased the percent purity of the ester product.
Experiment 14: Saponification: Herbal Soap-Making I.
Saponification is the exothermic base hydrolysis of triesters or triglycerides. Triglycerides are broken down into glycerols and fatty acid salts, which is soap.
II. These fatty acid salts have very long carbon chains, termed usually as the hydrophobic tail, and an ionic carboxylate end, also known as the hydrophilic head.
Sodium palmitate
III. Soaps solidify due to the ionic and Van der Waals forces. IV. The reverse reaction of saponification process does not take place due to the lack of any good leaving groups in the resulting intermediate from the bonding of the glycerol and the carboxylate anions.
Reverse reaction “intermediate” As such, after tracing occurs, the reaction mixture will no longer revert back into oil and caustic soda.
V. Soap can function as an emulsifier, capable of mixing a liquid into another immiscible one. When hydrophobic matter, such as oil and grease, is added into a solution containing soap molecules, layers are not formed. Rather, a solution containing small globules, called micelles, is the result. The nonpolar side of the soap molecules allows them to trap dirt and oily matter while their polar side allows them to stay soluble in, and be easily washed away by, water. This property is also handy when it comes to washing dirt, especially by hand. Since human skin has a natural thin coat of oil, dirt particles easily adheres to it and consequently becomes harder to remove due to their newly-acquired oily coating. This becomes more troublesome due to the fact that oil is not volatile and as a result, these particles will just stay there. However, upon the introduction of soap molecules to the skin surface, they are evidently washed away with ease as the nonpolar tails of the soap molecules adhere to dirt particles’ oil coating. VI. Soaps don’t work as efficiently when used with hard water. This is because metal cations in the water bond to the ionic head of the soap molecules to form insoluble precipitates. Certain additives have to be added to the water, such as sulfite compounds, to prevent the precipitation reaction by reacting with the metal ions themselves. VII. In transesterification, water is replaced by an alcohol; the reaction is still base-catalyzed. Regarding the final product, the sodium cation is replaced by the alkyl group of the alcohol that was used for the reaction. In performing this reaction, the environment has to be completely dry or else soap formation will occur.
Experiment 15: Hinsberg’s Method for Characterizing Primary, Secondary, and Tertiary Amines Amine sample
+ Benzenesulfonyl chloride, + Excess sodium hydroxide -
+ Primary amine
-
+
+ Hydrochloric acid Legend: + : Soluble - : Insoluble
+ Tertiary amine
I.
Discard. -
Secondary amine
Primary amines form a sulfonamide precipitate with benzensulfonyl chloride. A hydrogen atom is abstracted from the amine, while the halide is removed from the reagent. Upon the addition the of
excess sodium hydroxide, the final hydrogen atom is cleaved from the molecule by sodium cations, giving it a full negative charge. II. Secondary amines form a sulfonamide precipitate with benzenesulfonyl chloride as well. However, due to the lack of hydrogen atoms bonded to the nitrogen atom, the precipitate stays insoluble after the addition of sodium hydroxide. III. Tertiary amines are incapable of forming sulphonamides and, as such, no reaction takes place between one and benzensulfonyl chloride. However, hydrochloric acid, the second reagent, protonates the amine, making it soluble. IV. A summary of the solubilities in the first and second reagent:
First reagent Second reagent
Primary amine Soluble Insoluble
Secondary amine Insoluble No reaction
Tertiary amine No reaction Soluble
V. The discarded layer contains sodium hydroxide and benzenesulfonate. The latter is a result of the reaction between benzensulfonyl chloride and hydroxide ions to form benzenesulfonic acid, which was neutralized by excess hydroxide ions to form the said compound. VI. The residue, may it be solid or liquid, may contain either the amine sample, sulfonamide precipitate, or both. Sulfonamides are usually denser than water but, occasionally, the opposite occurs. VII. Hinsberg’s test cannot be used to classify amphoteric amines, which possess (carboxylic) acid and amine structures. They are soluble both in the first, due to the neutralization reaction of acid substituent, and second, due to the neutralization reaction of the amine substituent, reagents.
Experiment 16: Synthesis of 1-phenylazo-2-naphthol I.
Azo compounds are organic compounds that contain an azo group (–N=N–). The azo group is attached to aryl groups on both sides. Azo dyes also exhibit resonance of electrons which stabilizes the compound.
II. Dyes are colored for they contain at least one color-bearing group or chromophore. In azo dyes, the extended conjugated pi electron system found in the azo group allows the dyes to absorb light in the visible region. The conjugated structure also adds to the color of the dye. Furthermore, the delocalization of electrons and conjugation are both extended due to the nitrogen-nitrogen double bond linkage and this further deepens the color of the dye.
Aside from chromophores, azo dyes also contain auxochromes or color helpers in the form of –NHR, – NR2, –OH, and –OR. The presence of auxochromes can deepen the color of the dye. Auxochromes are also used to influence dye solubility. III. The synthesis of azo dyes involves two reactions: diazotization reaction and coupling reaction. In diazotization, a primary aryl amine is made to react with nitrous acid, which was generated from the reaction between sodium nitrite and a concentrated acid, to form diazonium ions. This reaction is carried out under freezing conditions. The diazonium ion is among the most versatile intermediates in organic synthesis.
The produced diazonium ion then reacts with highly activated aromatic substrates in the coupling reaction to give the azo dye. The coupling reaction is an electrophilic aromatic substitution wherein the diazonium ion serves as the electrophile. Since the diazonium ion is a relatively weak electrophile, the aromatic ring it attacks must be highly activated with groups such as –OH and –NH2. IV. Dyes are made with the following properties in mind: fastness and levelness. Fastness refers to the ability of the dye to strongly bond to the fibers of the material being dyed. The dye must not fade after repeated washings (water fast) and exposure to light (light fast). On the other hand, levelness refers to the uniformity of the dye on the fiber when applied. The dye must evenly color the fabric upon application. V. When dying fabric, it is important to know the structure of both the dye and the fabric to be used to determine the appropriate dyeing method. Dyes can be classified according to the method by which they were applied to a fabric. Dyes are commonly classified as follows: direct dyes, ingrain dyes, vat dyes, and disperse dyes. Direct dyes attach themselves to a fabric via a direct chemical interaction. Ingrain dyes, such as simple dyes, are synthesized within the fabric itself. Smaller molecules that can be used to synthesize the ingrain dye are allowed to penetrate the fabric. These two smaller molecules then react to form the larger ingrain dye which gets trapped in the fabric. Ingrain dyeing is commonly used for cotton fabric. Mordant dyes make use of substances known as mordants that bond with the dye itself to form lakes, which are dye-mordant complexes. The same principle when the dye bites onto the fabric. Coordinate covalent bonds are formed between the lakes and the fabric molecules. Water-insoluble disperse dyes are applied to fabric such as polyesters by allowing them to penetrate a fabric by using an organic solvent to carry the dye. The fabric is first suspended in an aqueous solution with a little amount of the organic solvent. The organic solvent then dissolves the dye and disperses it into the fabric. The dye gets trapped in the fabric because of water-insolubility.
VI. The synthesis is a 2-pot reaction. In the first pot, phenyldiazonium chloride is formed, while in the second pot, naphthalen-2-olate or β-naphthoxide is produced. The two are then reacted later to form the crude Sudan-1 solution, which consists of the synthesized dye, unreacted reactants, and water. VII. Nitrous acid was generated in-situ in the first pot since it is unstable. A prepared solution cannot be used. VIII. β-naphthol was reacted with sodium hydroxide to give it a negative charge, which activates it toward electrophilic aromatic substitution. It also increases reaction rate since naphthalen-2-olate is soluble in water. IX. The dye was recrystallized to remove any occluded molecules or ions. X. To determine the reactants needed to form an azo dye, the aromatic rings on both sides of the nitrogen-nitrogen double bond have to be inspected. The breaking of the linkage to the nitrogennitrogen double bond takes place beside the ring that directed the bond correctly or beside the more activated one.
