Chem 31.1 Finals Notes

Experiment 1: Solubility Behavior of Organic Compounds

Experimental results:

I. 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.

Legend: + : Soluble - : Insoluble

Test compound

+ distilled H2O

+ Ether + 5% NaOH

+ 5% HCl + 5% NaHCO3

+ H2SO4

I N

-

+

A2 A1

S1 S2

+

-

-

+

Water-soluble compounds

Acidic compounds

-

+

Basic compounds

-

+

B

+

-

M Miscellaneous compounds

Neutral compounds Inert compounds

Solubility by IMFA Solubility by reaction

S1: Ethanol, Acetone

I: Hexane, Toluene, Tert-butyl chloride S2: Sucrose A1: Benzoic acid

A2: Phenol

B: Aniline

M: Benzamide

N: -

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.

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 Adi-

and tri- arylamines, Barylamines with electron-withdrawing groups, and Csimple monosubstituted

amides.

VIII. M compounds are miscellaneous. These compounds have nitrogen and/or sulfur atoms and dont

belong to any of the previous classes as well. AAmides belong to this group.

A2,4-dinitrophenol

B2,4,6-trinitrophenol

B2,4,6-trinitroaniline

ADiphenylamine CAcetamide

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. Should readily dissolve solute at elevated temperatures and only sparingly at room temperature

B. Should readily dissolve impurities at any temperature or not at all

C. Should not react with the solute

D. 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.

ABenzamide

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 containers 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.

Percent purity = masscompound

massfinal insoluble phase 100 %

Percent recovery = masscompound in final insoluble phase

masstotal compound used 100 %

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 non-

volatile 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 Sublimation

Description Tedious; many steps Easy; one step

Percent purity Low High

Percent recovery 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:

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:

Distancesolvent

Mobile phase Starting line

Solvent front

1

4

6

10

11

DistanceA

A

A

B

C, D

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.

D

B A

C

Chain isomers refer to the length of the parent carbon chain and the presence of alkyl branches.

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:

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:

Isomers

Structural Stereoisomer

Chain

Positional

Functional

Geometric

Optical

3-dimensional 2-dimensional

Conformational

Propanal Propan-2-one

Propanal

Chair conformation Boat conformation

Butane

2-methylpropane

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.

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.

Anti Gauche

II. The SN1 mechanism for the synthesis reaction, which is the formation of tert-butyl chloride from tert-

butanol 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. Cold hydrochloric acid: Prevents the E1 reaction.

B. Concentrated hydrochloric acid: Practicality, since less volume of acid is needed.

C. Excess hydrochloric acid: Pushes the synthesis reaction forward

D. 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.

E. Sodium bicarbonate: Neutralizes unreacted acid while showing equivalence indication via

effervescence.

F. Solid sodium bicarbonate: Aqueous sodium bicarbonate has water, which is counterproductive for

the experiment.

G. Boiling chips: Promotes even boiling and facilitates bubble formation; prevents superheating,

bumping, and loss of solution.

H. 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 Baeyers test, is only possible for

alcohols with -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) + 2H2O(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.

On the other hand, secondary alcohols are always oxidized to ketones. Finally, phenols are oxidized to

quinones.

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 nitrophenols case.

Pyridium chlorochromate

Quinone

IV. Bromine in water indicates the presence of phenols. The positive result is the formation of

tribrominated phenol, a white precipitate.

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. + = , 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:

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 didnt react because it

lacked true double bonds.

2,4,6-tribromophenol

Limonene

V. Baeyers 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.

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 Crafts 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 3CCl(aq ) AlCl4 (aq ) + CH3 3C(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.

Benzene resonance

4-(1,2-dihydroxypropan-2-yl)-1-methylcyclohexane-1,2-diol

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 N H2 N HR N R2 N HC = OR O H O R O C = OR R CH = CR2

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 NR3 NH3 SO3H 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 ortho-

and 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.

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

Acetanilide

groups, the water present in the solution allowed the carboxylic acid to partially dissociate to form an

ionic solution.

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 acid-

catalyzed reaction is:

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. Schiffs test makes use of Schiffs 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

Resonance and final product

Imine formation

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 ) + H2O(l) 2() + ()

+ 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. Benedicts test makes use of Benedicts 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:

2Cu(aq )2+ + RCOH(aq ) + 2H2O(l) () + ()

+ 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.

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 )

+ () + 3I(aq ) + 3H2O(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

Butan-2-one Butan-2-ol

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.

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 -naphthol, which indicates the presence of carbohydrates. The positive

result of this test is the formation of a purple interface.

IV. Bials 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.

V. Seliwanoffs 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.

5-hydroxymethyl furfural Furfural

-naphthol

Orcinol

VI. Benedicts test confirms the presence of reducing carbohydrates.

VII. Barfoeds 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

molecules 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 Benedicts solution, and

has and 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.

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 -

1,4 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 -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.

Resorcinol

Phenylhydrazine

X. Cellulose is very similar to amylose in structure, except for the fact that -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: CH3COOH(aq) + NaHCO3(aq ) CH3COONa(aq) + CO2(g) + H2O(l)

Benzoic acid: C6H5COOH(aq ) + NaHCO3(aq ) C6H5COONa(aq ) + CO2(g) + H2O(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.

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.

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.

Ferric acetate complex

Ferric benzoate complex

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.

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.

Ferric hydroxamate complex

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 esters

smell. Its sourness overpowers the products 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 Chateliers 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.

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.

As such, after tracing occurs, the reaction mixture will no longer revert back into oil and caustic soda.

Sodium palmitate

Reverse reaction intermediate

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 dont 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: Hinsbergs Method for Characterizing Primary, Secondary, and Tertiary Amines

I. 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

Amine sample

+ Benzenesulfonyl chloride, + Excess sodium hydroxide

Primary amine

Discard.

-

+

-

+

+ Hydrochloric acid

-

+

Tertiary amine Secondary amine

Legend: + : Soluble - : Insoluble

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:

Primary amine Secondary amine Tertiary amine

First reagent Soluble Insoluble No reaction

Second reagent Insoluble 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. Hinsbergs 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 -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. -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 nitrogen-

nitrogen double bond takes place beside the ring that directed the bond correctly or beside the more

activated one.