CHEM 31.1 FINALS REVIEWER 24 October 2018 Post-Lab Discussion Experiment 10: Aldehydes and Ketones : Reactions involving aldehydes and ketones : Standard confirmatory test for the presence of aldehydes and ketones INTRODUCTION : Carbonyl group is a polar group polar group C=O, C is the carbonyl carbon : Carbonyl oxygen atom allows molecules of aldehydes and ketones to form strong hydrogen bonds to molecules of water but it cannot have cannot have strong hydrogen bonds between their molecules : aldehydes are always terminal : ketones are always internal Nucleophilic addition to the carbon-oxygen double bond : nawawalan ng saturation o nawawalan ng double bond : occurs via 2 ways (1) Attack of a strong Nu - 1* (2) Acid catalyst + weak Nu - 2* Catalyst will protonate oxygen then the Nucleophile will attack Carbonyl carbon: trigonal carbon: trigonal planar, susceptible to attack both front and back side : highly reactive : partial positive charge; oxygen has the partial negative charge OBJECTIVE: OBJECTIVE: to examine the reactivities of various aldehydes and ketones 2,4-DNPH Test test for aldehydes and ketones reagent: 2,4-dinitrophenyl hydrazine positive result: formation of yellow to orange precipitate negative result: orange to red solution reaction mechanism – acid catalyzed nucleophilic addition Ethanol - negative Acetone – Acetone – positive – positive – yellow yellow orange ppt Benzaldehyde – Benzaldehyde – positive – positive – yellow yellow orange ppt Formalin – positive N/A – be/cause be/cause it doesn’t have another R group Tollens’ Test Reagent: [Ag(NH3)2]NO3 or ammoniacal silver nitrate Test for aldehudes Positive result: formation of silver mirror Negative result: coloreless solution Reaction: redox reaction
Expected results: Ethanol: N/A Formalin: silver mirror Acetone: colorless solution Benzaldehyde: silver mirror (after heating) Schiff’s Test Reagent: leucofuchsin Test for aldehydes Positive result: formation of violet colored solution Negative result: pink solution Reaction Ethanol: N/A Formalin: indigo soln/ppt Acetone: light pink soln Benzaldehyde: indigo soln/ppt Benedict’s Test Reagent: CuSO4, citrate, NaHCO3 Test for aliphatic aldehydes Does not give a positive result for aromatic aldehydes Positive result: formation of brick red ppt which is Cu 2O Negative result: blue soln because of free Cu2+ Reaction: redox reaction Ethanol: N/A Formalin: brick red ppt Acetone: blue sol’n Benzaldehyde: blue sol’n Fehling’s Test Reagent: cupric tartrate complex ion Test for aldehydes Related to benedict’s test Positive result: formation of a brick red ppt (CU ppt (CU2O) Negative result: blue soln Iodoform Test Based on the production of iodoform (CHI3) Reagent: NaOH, KI, I2 Hypoiodite (IO-) is formed and acts as an oxidizing agent (weak) Test for methyl ketones and 2-alkanols Positive result: formation of bright yellow ppt Negative result: yellow solution Needs only one carbon Ethanol: bright yellow ppt Formalin: N/A Acetone: bright yellow ppt Benzaldehyde: N/A
Effect of Acid Concentration on Hydrazine Formation If the acid concentration is too love the reaction occurs slower o 0.02M CH3COOH If the acid concentration is too high, the reaction would not happen o 0.05M HCl o acid-base reaction with hydrazine Reaction is pH dependent Rate of reaction: 0.5M HCl < 0.02M CH3COOH < 0.02 M HCl Differentiate the following cpds Heptanal and heptanol: any tests for aldehydes : schiff’s test, purple sol’n, heptanal p-ethylbenzaldehyde and ethanol : Schiff’s, purple sol’n, p-ethylbenzaldehyde 3-pentanone and 3-pentanol : 2,4-DNPH, yellow to orange ppt, 3-pentanone o-chlorobenzaldehyde and 3-benzylpropanal : Benedict’s, brick red, 3-benzylpropanal 4-methyl-3-pentanone and 3-methyl-2pentanone : iodoform test, yellow ppt, 3-methyl-2pentanone (methyl ketone) hexanone and hexanal : Schiff’s, purple soln, hexanal
Epimers: are carbohydrates that differ in configuration on only 1 stereogenic center : allose and altorse, carbon2 : allose OH at the right, altrose OH at left : glucose and mannose, carbon 2 : glucose OH at right (right left right right) and mannose OH at left : glucose and galactose, carbon 4 : glucose OH at right, galactose OH at left
Experiment 11: Carbohydrates Carbohydrates: broad class of polyhydroxylated aldehydes and ketones : OH group and carbonyl carbon : CH2O, CnH2nOn Classification - Simple sugars: monosaccharides - Complex sugars: disaccharides, oligosaccharides, and polysaccharides Fischer Projection Haworth Projection Chair Conformation Monosaccharides : may be classified as either aldoses or ketoses : glucose - aldohexose : fructose – 6 carbon ketohexose Ketoses: dihydroxyacetone Ketopentose: ribolose Family trees of Ketose and Aldose Aldose
Mutarotation: monosaccharides are reducing sugars because of the presence of hemiacetal group : mutarotation stops when a glycosidic bond to the carbonyl carbon is formed Hemiacetal carbon: carbon atom attached to 2 oxygen atoms and has an OH group attached to it Anomeric carbon: carbonyl carbon capable of having alpha and Beta configuration : goes through open-chain form then, ring closing For aldohexoses – glucopyranose For ketohexoses – fructofuranose Tip: look at OH Alpha – one up one down OH Beta – both up OH and CH2OH group Oligosaccharides : 2-10 monosaccharides : bond between them is a glycosidic bond : glycosidic bond = alkoxy group of one sugar unit attaches itself to the anomeric carbon : sucrose, maltose, lactose : sucrose (non-reducing) = glucose and fructose, alpha 1-2 glycosidic bond : alpha because pababa si oxygen : maltose (reducing) = alpha 1,4-glycosidic bond : lactose (reducing) = beta 1,4-glycosidic bond Polysaccharides : more than 10 monosaccharide units : starch – a polymer of a-D-glucose : yields a mixture of water-soluble amylose and amylopectin upon hydrolysis : amylose – linear : amylopectin – branched alpha 1,6 Cellulose: polymer of B-D-glucose : principal structural component of plants : partial hydrolysis cellobiose, complete hydrolysis D-glucose
QUALITATIVE TESTS FOR CARBOHYDRATES Based on the production of furfural and furfural derivative Based on the reactivity of the aldehyde or potential aldehyde group (i.e. hemiacetal) Phenolic compounds react with these in acid to give colored ppt = positive Molisch’s Test Reagent: a-Naphtol, H2SO4 : all carbs will yield positive result (in free or combined form) : positive result: red violet/ purple ring at the interface Molisch Test Glucose – purple interface Sucrose – purple interface Starch – purple interface Bial’s Test Reagent: Orcinol, HCl, FeCl3 + heat (30-45min) : differentiate pentoses from hexoses : pentoses: blue green solution - ribose : hexoses: yellow – brown solution - glucose Seliwanoff’s Test Reagent: resorcinol, H2SO4 + heat (15 min) : differentiate ketoses from aldoses : positive (ketoses): bright red solution : negative (aldoses): pink solution Rate: fructose > sucrose > glucose Glucose – pink solution Fructose – red solution Sucrose – pink solution OXIDATION OF ALDEHYDES Benedict’s Test : test for reducing sugars : positive result: brick-red ppt : negative result: blue solution : rate: glucose > fructose > maltose > sucrose > starch Barfoed’s Test Reagent: cupric acetate, acetic acid + heat (10 min) : differentiate reducing monosaccharides from reducing disaccharides : positive (monosaccharides): brick red ppt) : negative (reducing disaccharides): blue solution : limitation – heat the sample for too long, possibility of hydrolysis positive: flucose, fructose negative: maltose How can you say that it’s a reducing sugar? You have a hemiacetal
FORMATION OF OSAZONE Based on the reactivity of the aldehyde and potential aldehyde group Reagent: phenylhydrazine, HCl + heat 20 min : identify monosaccharides and reducing disaccharides : positive: yellow to yellow-orange ppt : negative: red soln Osazone formation : monosaccharides that are epimeric at C-2 give the same osazone : phenylhydrazine targets C1 and C2 : disaccharides with a free anomeric carbon gives a positive result Expected results Glucose yellow-orange ppt Maltose yellow-orange ppt Fructose yellow-orange ppt Sucrose – red soln HYDROLYSIS OF OLIGOSACCHARIDES AND POLYSACCHARIDES In the presence of acid disaccharides can be hydrolyzed back into its monosaccharide units Sucrose : reagent: HCl + heat, 10% NaOH, benedict’s reagent unhydrolyzed: blue solution hydrolyzed: brick red ppt Starch : components 20-30% amylose linear a1 4 and 70-80% amylopectin branched a1 4 and a16 : amylose: soluble in water, amylopectin: insoluble in water : kapag nawala ang helix structure ni starch, nawawala yung blue complex and forms a yellow solution Amylodextrin : short-chained amylose produced from debranching amylopectin : I2 in KI: wine red : benedict’s test: green : polysaccharides take longer because they have more glycosidic bonds to break benedict’s test – I2 in KI test unhydrolyzed sucrose – blue soln – n/a hy sucrose – brick red ppt – n/a unhy starch – blue soln - blue-violet soln hy starch – brick red ppt - yellow soln
Experiment 12: Carboxylic Acid and Derivatives Reactivity Basicity Trend NR2 amide > RO- (ester) > OH- carboxylic acid > RCOO- anhydrided > Cl- acyl halide
Reactivity Acyl halide > anhydride > carboxylic acid > ester > amide RESULTS AND DISCUSSIONS Solubility of NaHCO3 Solvent: 5% NaHCO3 Observation: miscibility and efferevescence Reactions : * Test for Acetic Acid Reagents: NaOH + CH3COOH + FeCl3 Addition of CH3COOH is controlled until solution is slightly acidic Observation: formation of red-colored sol’n If basic, false positive -> FeOH 3 Reaction: * Test for Benzoic Acid Benzoic Acid + heat + NH 4OH until slightly basic boil of xs NH3 FeCl3 Observation: formation of flesh-colored ppt Reaction: * Boil bc we don’t want NH 3 giving false positive Formation of Esters General Observation: production of fruity odor Reaction pathways Carboxylic acids and amides not reactive enough to synthesize an ester requires a catalyst Acyl halides and anhydrides ester synthesis does not require catalysts From caraboxylic acid (catalyst required) CH3COOH + EtOH + H2SO4 (catalyst) heat (warm) Observation: production of “plastic balloon smell” From Acyl Halide (no catalyst required ) Benzoyl Chloride + EtOH Observation: wintergreen-like or medicinal-like smell Mas maganda kasi hindi harsh yung conditions to react Acyl chloride to ester is more favored rather than from ester to acyl chloride
Hydroxamic Test: General test for esters Procedure: EtOAc or ether + alcoholic NH 2OH in HCl hear alcoholic HCl +FeCl3 : basis is the production of hydroxamic acid positive result: purple or magenta colored solution; presence of esters Why alcoholic solutions? Use of alcoholic solutions: ensures miscibility of esters with reagents Addition of alcoholic HCl before FeCl 3We need our pH to be slightly more acidic to prevent the formation of Fe(OH) 3 precipitate Hydrolysis of Acid Derivatives All carboxylic acid derivatives form carboxylic acid (or carboxylate) when hydrolyzed Hyrdolysis reaction can either be baseor acid-catalyzed Base catalyzed – > carboxylate ion Acid catalyzed – > carboxylic acid BASE CATALYZED Ester Ethyl acetate + NaOH reflux + HCl Observation: production of sour smell (formation of acetic acid) Anhydride Water + acetic anhydride litmus Observation: blue litmus paper turns to red Solution is acidic Amide Benzamide + NaOH heat blue litmus Observation: production of pungent odor; vapors turn red litmus paper to blue Production of ammonia will symbolize hydrolyzed benzamide ACID CATALYZED Ester Observation: production of sour smell NOT all acid-catalyzed hydrolysis of ester produces a sour smell Amide Observation: no pungent odor (NH4 + is produced instead of NH 3)
Base ang mas maganda for hydrolysis of ester : because of the carboxylate ion intermediate , irreversible. Acid catalyzed is reversible reaction. Hindi siya magrereact via nucleophilic acyl substitution Test Ch3COOH miscible with effervescene C6H5Cooh partial solubility ; bubble form on the solids
Test for acetic acid – NaOH and FeCl3 – red colored solution Test for benzoic acid – NH4OH – flesh-colored ppt Frormation of esters – carboxylic acid – plastic balloon smell ethyl acetate Acyl halide – wintergreen-like smell benzoyl Hydroxamic test – positive for esters – purple or magents colored solution Hydrolysis of acid derivation Anhydride – blue to red Esters – sour smell after addition of HCl Amide – evolution of pungent odor (production of NH3), red litmus paper turns to blue Experiment 13: Nucleophilic Acyl Substitution: The Synthesis of Esters Esters: carboxylic acid derivatives R-COOR or Ar-COOR Known for its aroma and flavor Either natural or synthetic; commonly found in fruits Synthesis; acid catalyzed reactions of RCOOH and ROH ROH serves as the nucleophile Minimize the H2O introduced, use excess amount of one reactant in the synthesis Le Chatelier’s Principle Xs ROH – forward reaction, shift to right Minimize H2O – shift to the right Leaving group is H 2O Methodology ROH + RCOOH RBF with boiling chips + conc. H2SO4 Reaction of ROH and RCOOH Alcohol (xs reactant) and carboxylic acid (limiting reactant Why should alcohol be in xs? Le Chatelier’s Principle What are other way that will favor formation of esters? Minimize introduction of H2O in the reaction More reactive carboxylic acid derivatice. Use acyl halides or anhydrides as starting reagents than RCOOH. Use absolute ROH (ethanol). + concentrated H 2SO4 acid catalyst – to protonate carboxylic acid to make it more reactive Why was H 2SO4 used instead of any other acid? Because it is also a dehydrating agent
Synthesis of Esters Reflux (45-60min) Cool to Room Temp Why are these steps necessary? We reflux to prevent volatilization of ester product Transfer to sep funnel + saturated NaCl Salting out to enhance layer separation organic layer + NaHCO 3 + anhydrous Na2SO4 Separation of layer Organic layer: + NaHCO3 (s) used the effervescence as an indicator of the acid-base neutralization + anhydrous Na2SO4 until it no longer aggregates as drying agent? Why didn’t we use CaCl2? Para matanggal yung alcohol product inside the organic layer. CaCl 2 removes ester Simple Distillation Recap Give at least 3 ways to prevent decrease - Add alcohol in excess - Use absolute alcohol - Use of more activated carboxylic acid derivatives such as anhydrides and acyl chlorides - Use of H2SO4 instead of other acids – concentrated to prevent introduction of water and acts as a dehydrating agent - Use of solid NaHCO 3 instead of aquaeous NaHCO 3 - Use of Na2SO4 instead of CaCl 2 Experiment 14: Saponification: Herbal Soap Making Saponification: formation of soaps : hydrolysis of ester linkages in a fat or oil sample forming glycerol and fatty acid salts (soap) : makes use of hydrolyzing agent aka lye (NaOH or KOH, strong base) : NaOH solid soaps, KOH liquid soaps : Only additives: coloring powder, essential / fragrance oil, herbal extracts Saponification is irreversible Tracing: indication that the process is around 90% complete and the lye will no longer separate form the oil Soap is cured for 5-6 weeks for the NaOH to be completely consumed Hot Process vs. Cold Process HOT : ends as soon as cooking is finished
: since lye is used up already, any additives added will not be affected by the base : “rough” appearance bc of heat and form bubbles as it heats and cools COLD : takes 24-36 hours : adding additives may react with the base : less dense and smoother in texture STAGES OF HOT PROCESS Separation – separation of solids from liquids, occasionally stirred Coming together and rinsing – stirred vigorously to release hot air “Mashed potato” like consistency – look less bubbly and greasier; pH should be around 8-10 saponification : process of soap making: combination of fats or oils with a lye solution : saponification value, the amount of lye needed for a certain kind of oil : qualities of soap hardness / softness amount of lather harshness / mildness : these qualities are determined by types of fats / oils used solubility of soap in water : the sodium salts of long chain carboxylic acids (soaps) are almost completely miscible in water : micelle formation hydrophobic tail and hydrophilic head : soap allows water to remove normallyinsoluble matter by emulsification hydrophobic part traps the dirt, hydrophilic part is soluble in H 2O Synthetic detergents : with sulfates or sulfonates as their polar groups : offer an advantage over soaps as they can function in hard water (i.e. watercontaining Ca2+, Mg2+, Fe2+, and Fe3+) ion rich commonly, Ca2+ and Mg2+ mahirap magbula : water-softening soap additive … compounds that bind with metal cations to prevent precipitation with soap : NaHCO3 Borax (sodium borate or sodium tetraborate), EDTA compounds, citric acid : sodium lauryl sulfate Saponification vs. Transeseterification Saponification : reagent: triglyceride, NaOH, H2O : Nu: OH
: product carboxylate salt : application soap making Transesterification : Reagents: triglyceride, NaOH, ROH : Nu: RO: product esters : application biodiesel production Experiment 15: Hinsberg’s Method for Characterizing Amines Amines: contain an amine group : relatively weak bases due to lone pair on N : can be classified as 1, 2, 3. Characterization of 1,2,3 Reagent: benzenesulfonyl chloride
sulfonamides: products of benzenesulfonyl chloride and amines : density is greater than water Hinsberg’s method: based on solubility of sulfonamide products Primary Amines: produce acidic sulfonamides when reacted with benzenesulfonyl chloride Acidic sulfonamide soluble in NaOH , insoluble in acid : white ppt Secondary Amines: produce non-acidic sulfonamide : insoluble in aquaeous NaOH layer formation : layer formation in NaOH, and dense liquid residue in HCl Tertiary Amines: it does not react with the benzenesulfonyl chloride : no sulfonamide formed; layer separation dure to polarity difference : Addition of HCl reacts with tertiary amine to form quarternary ammonium salt : solubility observed is due to acid-base reaction of tertiary amine with HCl Expected results Aniline - sulfonamide is dissolved – white/cream ppt Diethylamine – dense, organic residue (layer of globule containing sulfonamide – layer formation Triethlyamine – layer formation (upper layer is just the tertiary amine) – miscible Primary – secondary – tertiary B.S.Cl Soluble – insoluble – NVR HCl Insoluble – NVR – soluble
Limitation: amphiprotic amine (i.e. amines containing acidic and basic properties) Example: N-methylaminobenzoic acid : rxn false positive Other methods of characterization NITROUS ACID TEST: Liebermann nitroso test Reagent: Nitrous acid (HO-N=O) : reacts will all classes of amines, and differentiates them based on the products Primary Amines: evolution of N2 (g) Aromatic primary amines: yield orange or red precipitate, stability of diazonium salt is resonance stabilized Secondary Amines: yellow organic layer (low MW) or low melting soids (for aromatic amines of higher aliphatic amines) Tertiary amines: no reaction; except for tertiary alkylamins which form green para-nitroso derivatives CYCLIC IMIDE FORMATION 1 and 2 amines react with cyclic anhydrides to form crystalline amides 3 amines do not react REACTION WITH ISOCYANATES 1 and 2 amines react with isocyanates to form crystalline substituted urea 3 amines do not react
Experiment 16: Synthesis of 1-phenylazo-2-naphthol (Sudan 1) Objectives: To synthesize a red-orange azo dye called SUDAN I or 1-phenylazo-2naphthol using the two-pot synthesis To purify the synthesized dye using recrystallization To utilize ingrain dyeing to color cotton fabric using the synthesized dye Aniline + 12 M HCl + d.H2O -> phenyldiazonium solution Cool to below 5C – NaNO2 – ice bath B-naphthol + 5% aq. NaOH – ice bath -> Bnaphthoxide solution
Crude Sudan I + unreacted phenyldiaoznium + B-naphthoxide + water Filtrate – discard Residue – recrystallize in ethanol pure Sudan I. melting point 132C Ingrain dyeing dyed slight fade MAIN REACTION Diazotization and coupling reaction EAS – electrophilic aromatic substitution
Preparation of B-naphthoxide ion B-naphthol + OH - resonance-stabilized napthoxide ion Addition of NaOH enhances electron donating capability of hydroxyl group Do not let the solution get too acidic If acidic env’t, protonate, leaving you with a carbocation that is susceptible to forming other side products Preparation of phenyldiazonium ion (diazotization) Primary aromatic amines Only formed in strongly acidic solutions with sodium nitrite High pH will cause side reactions Coupling reaction: electrophilic aromatic substitution Electrophile: phenyldiazonium ion Attachment ortho or para? Number of resonance structures intermediates o-7, p-6. O4 intermediates remained aromaticity, p-2 remained aromaticity
urea CONVERSION TO AMIDE 3 amines do not produce amides PICRATE SALTS OR MOLECULAR COMPLEXES addition of 2,4,6-trinitrophenol precipitates with distinct melting points CHLOROANIL TEST amines dissolved in dioxane produces colored adducts when mixed with chloroanil (2,3,5,6-tetrachloro-pbenzoquinone)
Properties of dyes : dyes are colored compounds
chromophore: whole extended conjugated system; color-bearer, responsible for light absorption auxochrome: parts of the molecule that increase/decrease the wavelength absorbed by the whole molecule TYPES OF DYEING Direct dying: simply putting together the fabric and dye moleculres : similar IMFA between dye and fabric Mordant dyeing: involves use of a mordant (glue) that reacts with both the fxl groups of the dye and fabric, making them stick together : most mordants are heavy metal salts; limited application dure to increasing toxicity : example: chromium Disperse dying: dye simply being dispered on the fabric as it is : for hydrophobic synthetic fiber (acetate rayon and polyester) Ingrain dyeing: synthesize the dye within the fabric : employed when dye and cloth do not have very similar IMFAs : cotton is very polar, and Sudan I is non-polar : dyes used for ingrain dyeing are usually to big to be able to anchor to the cloth itself Properties fastness and levelness Fastness: retentiveness of the cloth for the dye Levelness: uniformity of the color on the fabric Property Synthesis and Staining
IMFA
Attachment of dye to fabric
Direct Dying Ingrain Dying Dye is Dye is synthesized synthesized before after staining staining Similar IMFA IMFA between dye between dye and fabric and fabrics is not very similar Dye is not Dye usually big molecule is and is too large and attached to is trapped fabric via inside grains IMFA of fabric
Application of Sudan Dyes : mainly used for coloring solvents, oils, waxes, petrol, shoe, and floor polishes (hydrophobic interactions)
: lysochrome – dye used for biochemical staining of triglycerides, fatty acids, lipoproteins : generally, they are considered carcinogenic compounds DETERMINATION OF AMINE AND COUPLING COMPONENT OF AN AZO DYE : identify precursors of azo dyes 1 locate azo group (N=N) 2 look at the fxl groups attached to the aromatic rings, identify wheter they EWG or EDG 3 mark areas susceptible to EAS in each aromatic ring 4 the markings that coincide with the azo group then that is the coupling component and the other is the amine component 5. If, on the other hand, the markings coincide with – N=N- group on both aromatic rings, then the coupling components is the ring that is more activated toward EAS.
