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TEST/PRINCIPLEREAGENT/LIQUIDADDEDRESULTSREACTION MECHANISM/ REACTION MECHANISM/WHY DID IT HAVE THATREACTION?QUALITATIVE ANALYSIS OF OILS AND FATSSOLUBILITYTESTwater desi ghee vegetable ghee refined oilAll samples are immiscible The solubility of lipids is associated with its chemical structure (&am...

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QUALITATIVE ANALYSIS OF OILS AND FATS

TEST/PRINCIPLE

REAGENT/LIQUID ADDED

RESULTS All samples are

water chloroform

alcohol SOLUBILITY TEST

distilled water diluted NaOH diluted HCl cold ethanol hot ethanol chloroform diethyl ether

TRANSLUCENT SPOT TEST distinguishes volatile/nonvolatil e oils and to a certain extent, saturated/ unsaturated oils

ACROLEIN TEST - used to detect glycerol or fats

KHSO4 (Potassium bisulphate)

coconut oil palmitic oil

Principle: based on the reaction of phenols with furfurals in acid solution

immiscible in both oils miscible in both oils

desi ghee vegetable ghee refined oil glycerol coconut oil paraffin oil beeswax desi ghee vegetable ghee refined oil glycerol coconut oil paraffin oil beeswax

BAUDOIN TEST

immiscible All samples are miscible All samples form a lower layer which dissolves upon heating.

desi ghee vegetable ghee refined oil

translucent

pungent, irritating odor

REACTION MECHANISM/ REACTION MECHANISM/WHY DID IT HAVE THAT REACTION?  The solubility of lipids is associated with its chemical structure ("Like dissolves like")  It consists of long, nonpolar hydrocarbon chains (having no charged areas and hydrophobic) only slightly polar (with a very few charged areas); and are therefore: o highly soluble in chloroform, diethyl ether - lipids are readily miscible in nonpolar and organic solvents o partially soluble in ethanol (hot – soluble, cold – insoluble) - because the solubility of the fatty acids increased with increasing temperature o immiscible in water - Since lipids lack charged groups, H2O molecules have nothing to stick to and don't mix with them as water molecules are polar/hydrophilic (have positive and negative ends) o immiscible in dil. NaOH and dil. HCl - both NaOH and HCl were both polar solvents and they could not mix together  hence, lipids are immiscible in polar, ionic solvents, miscible in nonpolar solvents and in higher temperatures (as ↑ kinetic energy, ↑interaction of molecules)  not all oils will remain translucent at a given amount of time  nonvolatile oils – remain translucent as they have higher boiling points  volatile oils – translucence disappear/evaporated at a certain amount of time as they have lower boiling points  used in the distillation process (easier to extract) – boiling points used to separate liquids according to lower, higher BP Dehydration (formation of double bonds)  the glycerol portion of the molecule is dehydrated to form the unsaturated aldehyde, acrolein CH2=CH-CHO, which is a pungent irritating odor that confirms the presence of fat or oil

No odor was released when the sample was heated.

Conc. HCl Furfural solution

Vanaspa ti ghee

Rose red color was observed when Concentrated HCl and Furfural solution was added.

desi ghee

No change in color / remains yellow

Dehydration, Hydrolysis  Vanaspati ghee contains 5% sesame oil which gives it color upon the addition of reagents  Furfurals are formed by the dehydration of the monosaccharides in concentrated mineral acids.  Hence a solution of sucrose in concentrated hydrochloric acid will be hydrolyzed to an equimolar mixture of glucose and fructose, both of which give rise to hydroxymethyl furfural by dehydration.  The latter reacts with the phenols found in sesame oil (sesamol – derived from sesamolin in sesame oil) to give a red chromogenic product  Pure Desi ghee does not contain sesame oil which is why there were no changes observed in the sample upon the addition of Concentrated HCl and Furfural solution.

QUALITATIVE ANALYSIS OF OILS AND FATS

HUBLE’S TEST Principle: Huble’s reagent reacts with Huble’s reagent alcoholic solution of iodine that contains some mercuric chloride IZABELLE A. CONDEZ, BSMLS 2E FINALS LEZZGETIT - p.1

cotton seed oil

The color of the sample changed into violet.

 Cottonseed oil is saturated because the violet color of the iodine did not fade away.

linseed oil

No change was observed in the color of the sample

 Linseed oil is unsaturated because the violet color of the iodine in Huble’s reagent fades away

Olive oil

UNSATURATION TEST

Methylene fluoride Bromine solution

Corn oil

Coconut oil

MODIFIED FURTER-MEYER TEST

n-Butyl alcohol solution conc. nitric acid

Vitamin E

Bromine is decolorized upon contact with the solution which stayed transparent after around more than 60 drops of bromine. Bromine is decolorized upon contact with the solution which stayed transparent after 19 drops (less than 2o drops) of bromine water. Bromine is decolorized upon contact with the solution which stayed transparent after 4 drops (less than 10 drops) bromine water. The solution’s color changed to bronze/copper red.

 There are more unsaturated hydrocarbon chains of the olive oil present than the other two experimented lipids.

 There are less unsaturated hydrocarbon chains of the corn oil present than the olive oil.  Coconut oil has the least amount of unsaturated hydrocarbon chains present (with about 90% of saturated fat and only 9 % unsaturated fat) compared to the other two experimented lipids.

 Oils that react with Huble’s reagent fades the violet color of iodine, hence it is unsaturated and if the color persists, then the given fat or oil is saturated

Addition Reaction  Upon contact of the bromine with the unsaturated hydrocarbon chain of the lipid, an addition reaction will occur which breaks the double bond  the Br2 will split and the bromine atoms will be added across the double bond which give a colorless product, and since the bromine is used up, there will be no brown color present unless there is an excess of bromine  The saturated hydrocarbon chain cannot have an addition reaction with the bromine since there is no room for them to bond with the carbon atoms of the hydrocarbon chain

Ammonia Oxidation  There is presence of tocopherol; the tocopherol will be deprotonated which leaves the oxygen with a negative charge  This made that oxygen’s electron move to form a double bond with carbon which also made the other double bonds move  A water molecule attacks the carbon, and since water can be a potential leaving group, it is deprotonated  Further exchange of electrons will lead to the product ion of orthotocopherol quinone/tocopherol-o-quinone (???) that gives off the bronze red color

SAPONIFICATION

SALKOWSKY TEST

SAPONIFICATIO N

conc. Sulfuric acid

20% Sodium Hydroxide

Cholest e-rol

The two liquids are not immiscible. The upper layer turns red upon adding.

Making of soap

 There is cholesterol present.  sulphuric acid concentrate separates two water molecules from two cholesterol molecules in this reaction and bicholestadien is formed  Sulphuric acid sulphonates are thus formed by bicholestadiene and red color bi-sulphonic acid is produced since there is cholesterol present  Bicholestadiene – conc. H2SO4 dehydration of 2 molecules from 2 molecules of cholesterol Base Promoted Ester Hydrolysis  first reaction: exothermic reaction when NaOH contacted the fatty acid  hydrolyze the lipid using the base, which is the sodium hydroxide, to separate the lipid into glycerol and fatty acids  The nucleophile will attack the carbonyl carbon due to its partial positive nature; this attack occurs at all three positions in the triglyceride: o the OH group attacks the carbon of the ester from the fatty acid, which has a double bond with the oxygen o oxygen of the ester now has a single bond with the carbon; hence, it carries a partially negative charge which causes the reform of the carbon-oxygen double bond in order to gain more stability

IZABELLE A. CONDEZ, BSMLS 2E FINALS LEZZGETIT - p.2 (cont. saponification)  it kicks off the OR group which became the leaving group  this forms a carboxylic group which will be deprotonated by a strong base  results to the formation of three fatty acid salts and a glycerol molecule  sodium from the NaOH will attack the now anion oxygen creating soap; common salt is added to precipitate the salts as thick curds of soap wherein the water layer is the Glycerol and the upper layer is the crude soap.

NUCLEOPROTEIN, RNA PREPARATION

PREPARATION OF YEAST RNA

10% NaOH 6N acetic acid 30 mL of 95% ethyl alcohol with 8 drops of concentrated HCl

PREPARATION OF NUCLEOPROTEI NS

0.4% NaOH toluene one drop of 10%HCl

TEST FOR SOLUBILITY cold water

hot water

Precipitation  NaOH was added as it enhances the permeability of yeast cell wall; it disrupts the cell membrane and lyses the cell to extract nucleic acids  Acetic Acid ensured that desired RNA was not degraded, RNA is easily hydrolyzed by acid compared to base, lowers pH levels  During the method of removing the RNA precipitate, water makes it difficult for the Na+ to interact with the PO3- because of the high dielectric constant of the water, even if the positively charged sodium ions and the negatively charged phosphate groups are electrostatically attracted to each other.  8 drops of conc. HCl was used to protonate the phosphate groups in nucleic acid backbone, used to ionize making it more soluble in RNA  In order for the salt to dissociate to a degree and has a weaker dielectric constant which is enough to allow the Na+ and PO3-to come together, ethyl alcohol is used; Molecules of ethanol can form hydrogen bonds with water molecules, reducing the amount of water molecules necessary to hydrate DNA  Ethyl Alcohol caused RNA to precipitate from solution; Since ethanol increases the salt-RNA relationship, it allows the neutralized RNA to remain less polar and hence less soluble Precipitation  use of sand particles for mechanical elimination of the cell wall does not need chemicals  ether was mixed in with the yeast as it destroys the yeast by lysis and dividing the cell into smaller pieces (removing lipid substance), causing yeast to open  Trituration was done to help purify the impure yeast by taking advantage of the solubility differences of the compound  use of 0.4% NaOH was to hydrolyze the compound as the said reagent enhances the permeability of the yeast’s cell walls; decomposition of a nucleoprotein into its nucleic acid and protein constituent is more readily caused by alkaline hydrolysis  toluene prevents bacterial growth and permeabilizes the yeast membrane  Hydrochloric acid is added to precipitate the nucleoprotein;  filtration precipitates water-soluble proteins  The yeast RNA is insoluble in cold water because the phosphodiester bonds in the sugar-phosphate backbone are not broken in cold water.  When salt breaks down into positive and negative ions, the negative phosphate group Yeast RNA settled at the bottom is neutralized by the positive ions which makes nucleic acids less polar and therefore, less soluble in water The Yeast RNA dissolved within the  The yeast RNA dissolves in hot water because of the presence of purines. Hot water also inactivates the RNase enzyme. The phosphodiester bonds in the sugar-phosphate solution but left a few sediments at backbone are broken in hot water. the bottom.  Also, the molecules gain kinetic energy as the solution rise in temperature that helps the solvent molecules break apart the solute molecules that are bound together by intermolecular attractions more efficiently

TEST FOR THE CHEMICAL COMPONENTS OF NUCLEIC ACIDS

ethanol

Yeast RNA precipitate settled at the bottom.

Precipitation  ethanol can dissociate the salt to an extent and at the same time has a weaker dielectric constant, making it easier for Na+ and PO3- to come together  Ethanol molecules can form hydrogen bonds with water molecules, they decrease the amount of water molecules needed to hydrate DNA  Since the Ethanol improves the interaction of salt with the RNA, it helps the neutralized RNA stay less polar and therefore, less soluble  Ethanol causes the RNA to aggregate with positive ions in the solution, forming a solid or precipitate (ammonium phosphomolybdate) at the bottom of the tube.

Yeast RNA settled at the bottom.

 Diluted HCl cannot generate a chemical species that can deprotonate hydroxyl at carbon-2 of ribose moiety.  nitrogen lone pairs in amines are less closely held than oxygen lone pairs, due to oxygen being more electronegative, and therefore the nitrogen’s lone pairs are more available for donation since it is more basic  This makes the nitrogen-carbon double bond’s resonance form more significant and more resistant to acid hydrolysis

IZABELLE A. CONDEZ, BSMLS 2E FINALS LEZZGETIT – p. 3

TEST FOR SOLUBILITY

diluted HCl

TEST FOR THE CHEMICAL COMPONENTS OF NUCLEIC ACIDS

diluted NaOH

HYDROLYSIS

10% sulfuric acid solution

TEST FOR PURINE - used to detect the presence of purine bases.

ammoniacal silver nitrate (AgNO3) concentrated ammonium hydroxide (NH4OH)

TEST FOR PENTOSE SUGAR - used to detect carbohydrates and carbohydratecontaining compounds.

Molisch reagent concentrated sulfuric acid (H2SO4)

IZABELLE A. CONDEZ, BSMLS 2E FINALS LEZZGETIT - p.4

Solution turned to a yellowish color as the Yeast RNA dissolved in the solution.

Formation of hydrolysate

Formation of white precipitate

Formation of a purple ring

Base Hydrolysis - RNA hydrolysis occurs as the ribose's deprotonated 2' position OH, due to the alkaline environment, acts as a nucleophile that attacks the neighboring phosphorus in the RNA's sugar-phosphate backbone phosphodiester bond  The five oxygen atoms will be temporarily bound to the phosphorus, a transition state  then, the phosphorus detaches from the oxygen that binds it to the adjacent sugar, resulting in the RNA backbone's ester cleavage  This creates a 2', 3'-cyclic phosphate that when hydrolyzed, will produce a 2'- or a 3'nucleotide  Interaction of RNA molecules with the positive charges in an alkaline environment results in the formation of soluble nucleates. Also, the disruption of the hydrogen bonding between the RNA bases makes the yeast RNA soluble. Hydrolysis  done in order for the polymers (nucleic acids – DNA and RNA) to be broken down into monomers (nucleotides and its parts: inorganic phosphate, deoxy/ribose, 4 diff. heterocyclic nitrogenous bases), in which a bond is broken, or lysed, by addition of a water molecule  During a hydrolysis reaction, a molecule composed of multiple subunits is split in two: one of the new molecules gains a hydrogen atom, while the other gains a hydroxyl (-OH) group, both of which are donated by water Hydrolysis  Hydrolysis of N-β-glycosidic bonds between purine bases and ribose results in a release of purine base (A and G) caused by NH4OH  Ag+ precipitate caused the formation of the foamy gelatinous substance. Dehydration  Dehydration of the carbohydrate upon the addition of sulfuric acid resulting in the formation of a furfural.  When monosaccharide is treated with conc H2SO4 or conc HCl, -OH group of sugar are removed in the form of water and furfural is formed from pentose sugar  The furfural undergoes condensation with the α-naphthol resulting in the formation of a purple colored complex in the interface.

DETECTION OF SUGAR IN URINE ACIDS TEST FOR THE CHEMICAL COMPONENTS OF NUCLEIC

 Chemical test performed to detect the presence of pentoses and its derivatives, pentosans  When pentoses are dehydrated, it forms a furfural which condenses with orcinol, resulting to the formation of blue-green precipitate  Can be used for qualitative analysis by measuring the absorbance in a spectrophotometer

BIAL’S TEST

TEST FOR INORGANIC PHOSPHATE GROUP - used to indicate the presence of a phosphate group

ammonium hydroxide (NH4OH) solution 10% nitric acid (HNO3) solution ammonium molybdate solution

BENEDICT’S TEST - used to detect the presence of simple carbohydrates such as reducing sugars or monoand disaccharides that have free aldehyde or ketone groups

Benedict’s reagent: sodium citrate sodium carbonate pentahydrate of copper (II) sulfate

FEHLING’S TEST - used to differentiate water-soluble aldehyde from

Fehling’s solution A: blue aqueous copper (II) sulphate solution

BIAL’S ORCINOL TEST – derivation of Bial’s Test used to detect the presence of RNA in solutions Hydrolysis  Hydrolysis of pyrophosphate to phosphate results to a formation of yellow precipitate  when a solution containing phosphate ions is heated with a solution of ammonium molybdate [(NH4)2MoO4] and dilute nitric acid a bright yellow precipitate of ammonium phosphomolybdate [(NH4)3PO4 • 12MoO4] is Formation of yellow precipitate formed  The yellow precipitate which was the ammonium phosphomolybdate, is extremely insoluble in nitric acid

0 (blue) - no reducing sugar / glucose present 0.5-1% (green) - traceable amounts of glucose present 1-1.5% (yellow) – low (RESULT FROM EXPERIMENT) 1.5-2% (orange) - moderate

2% up (brick red) - high

0 (blue) - no reducing sugar / glucose present 0.5-1% (green) - traceable amounts of glucose present

Reduction Reaction (COPPER SULFATE => CUPROUS OXIDE)  glucose as a reducing sugar is capable of transferring hydrogens (electrons to other compounds)  The resulting color of the precipitate varied depending on the amount of reducing sugars present in the sample.  Benedict’s reagent allows for glucose to react and produce a complex (due to the reducing property of simple carbohydrates) that is an indicative result for the presence of sugar in the analyte.  Copper (II) ions in Benedicts solution are reduced to copper (I) ions, which causes the color change.  Complex carbs, such as starches do not react positive with the test unless they are broken down through heating or digestion  Table sugar (disaccharide)is a non-reducing sugar and does not react to the test  Results (pic under Fehling’s test): positive – reddish brown ppt (simple sugars: glucose, fructose, lactose) negative – no red ppt (complex sugars: sucrose, starch)

1-1.5% (yellow) – low (RESULT FROM EXPERIMENT) 1.5-2% (orange) - moderate

ketone functional groups

Fehling’s solution B: clear, colorless alkaline solution of Sodium potassium tartrate, also known as Rochelle Salt

DETECTION OF UREA IN URINE

IZABELLE A. CONDEZ, BSMLS 2E FINALS LEZZGETIT - p.5 SODIUM HYPOBROMITE TEST alkaline sodium hypobromite Principle: urea solution decomposes when reacted with NaBrO phenol red solution (phenolphthalein) weak acid; indicator 2% of Na2CO3 UREASE TEST alkalizes the solution 1% of acetic acid neutralizes the mixture

SULPHOSALICYL IC ACID TEST

sulphosalicylic acid solution

2% up (brick red) - high

 aqueous tartrate ions from the dissolved alkaline Sodium potassium tartrate chelates to copper (II) ions from the dissolved copper sulfate crystals as bidentate ligands which yields a bistartratocuprate (II) complex  bistartratocuprate (II) complex - works as an oxidizing agent; active reagent in the test - oxidizes the aldehyde to a carboxylate anion which causes copper (II) ions to reduce into copper (I) ions  The resulting color of the precipitate varied depending on the amount of reducing sugars present in the sample.  on heating an aldehyde or reducing sugar with Fehling’s solution give reddish brown precipitate; formation of red precipitate of cupr...