Your SlideShare is downloading. ×
Expt 8 b report g&h
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Expt 8 b report g&h

7,110
views

Published on

Published in: Business, Technology

0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
7,110
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
109
Comments
0
Likes
1
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. 8 B CA T N IA LU SI ER . IMALLO R .F P EG X KIENE CO
  • 2. E. TEST FOR ALCOHOLS AND PHENOLS 1. REACTIONS OF ALCOHOLS  REACTION WITH Na METAL  LUCAS TEST  REACTIONS WITH K2Cr2O7 2. ALCOHOL AgNO3 TEST  FeCl3 TEST  BROMINE WATER TEST  MILLON’S TEST
  • 3. Alcohols• only slightly weaker acids than water, with a Ka value of approximately 1 × 10−16• Acidity: 1o>2o>3o• N-butyl alcohol, sec-butyl alcohol, ter t-butyl alcohol
  • 4. Na Metal• Alkali metal• Sof t at room temperature• Silver y white in color• Highly reactive
  • 5. 20 DRPS N- BUTYLALCOHOL (SEC-BUTYL, Na METAL TERT-BUTYL ALCOHOLS) Structure or formula of Test Sample Visible Results compound responsible for the visible results n- butyl alcohol Rapid evolution of gas H2 Moderate evolution ofSec- butyl alcohol H2 gas Slight to no evolution oftert- butyl alcohol H2 gas 1. REACTIONS OF ALCOHOLS 1.1 REACTIONS WITH Na METAL
  • 6. Reaction with Metal NaAlcohol(acid) + Na metal(base)  sodium alcohol-oxide + H2(g)N-butyl – fastest to react; most acidic - CH3CH2CH2CH2OH + Na  CH3CH2CH2CH2 O-Na+ + H2Sec-butyl – CH3CH2CH(OH)CH3 + Na  CH3CH2CH(O-Na+)CH3 + H2Ter t-butyl – slowest to react; least acidic - (CH3)3C-OH + Na  (CH3)3C -O-Na+ + H2
  • 7. Lucas Reagent• ZnCl2 in concentrated HCl solution• Reagent used for classification of alcohols with low MW.
  • 8. LUCAS TEST• Test us to dif ferentiate primar y, secondar y and ter tiar y alcohols• Uses the dif ferences in reactivity of hydrogen halides and the three classes or types of alcohol
  • 9. 20 DRPS LUCAS REAGENT 10 DROPS N- SHAKE AND Na METAL BUTYLALCOHOL COVER WITH (SEC-BUTYL, TERT- BUTYL ALCOHOLS) STOPPER Structure formula Test Sample Visible Results responsible for results n- butyl alcohol No layer formation n/a Moderate layer (CH3)2CHCl + H2Osec- butyl alcohol formationtert- butyl alcohol Fast layer formation (CH3)3CCl + H2O 1. REACTIONS OF ALCOHOLS 1.2 LUCAS TEST
  • 10. LUCAS TESTReaction:speed of this reaction is propor tional to the energy required to form the carbocationThe cloudiness obser ved (if any) is caused by the carbocation reacting with the chloride ion creating an insoluble chloroalkane.
  • 11. Reactions:Primary Alcohol:Secondary:Tertiary:
  • 12. Potassium Dichromate K2Cr2O7• Inorganic chemical• Oxidizing agent
  • 13. Structural Formula responsible for Test Sample Visible Results results n- butyl alcohol Blue- green solution Chromic IonSec- butyl alcohol Blue- green solution Chromic Iontert- butyl alcohol Blue green Dichromate Ion 1. REACTIONS OF ALCOHOLS 1.3 REACTIONS WITH K2Cr2O7
  • 14. Rxn with K2Cr2O7oxidation occurs: the orange solution containing the dichromate (VI) ions is reduced to a green solution containing chromium (III) ions.Primar y: oxidized to aldehydes/carboxylic acidsAldehydes:Carboxylic:
  • 15. Rxn with K2Cr2O7Secondary: oxidized to ketoneKetone:Tertiary: cannot be oxidizedWhy? Because tertiary alcohols dont have a hydrogen atom attached to a carbon
  • 16. Phenols• consist of a hydroxyl group (-O H) attached to an aromatic hydrocarbon group.• relatively higher acidities, hydrogen atom is easily removed• acidity: carboxylic acids > OH group in phenols > aliphatic alcohols• pKa is usually between 10 and 12
  • 17. FeCl3 Test• Addition of FeCl3 gives a colored solution • alcohols do not undergo this reaction• other functional groups produce color changes: • aliphatic acidsyellow solution; • aromatic acidstan precipitate
  • 18. Test Sample Visible Results Structural formula responsible for results Reddish-brown Iron (III) complex w/ Phenol Phenol Iron (III) complex w/α - napthol Purple Napthol Iron (III) complex w/ Cathechol Dark Blue Cathechol Iron (III) complex w/ Moss GreenResorcinol Resorcinol 2. REACTIONS OF PHENOLS 2.1 FeCl3 TEST
  • 19. FeCl3 TestAn iron-phenol complex is formed.FeCl3 + 6C6H5OH  [Fe(OC6H5)6] 3- + 3H + +3Cl -
  • 20. BR2 IN H2O TEST•used to identify alkenes, alkynes and phenols•Alkenes & alkynes  the reaction occurs throughelectrophilic addition•Phenol  reacts with sites of unsaturation, evenaromatic rings, through a complex additionreaction
  • 21. BR2 IN H2O TEST•Brominedark brown color•when it reacts, the color dissipates and thereaction mixture becomes yellow or colorless•ortho and para positions to the phenol arebrominated.
  • 22. Structure formulaTest Sample Visible Results responsible for results bromination of benzene Phenol Turns pinkish, ppt ring Turns to dark green, bromination of aromaticα – napthol ppt ring bromination of benzene Cathechol Dark brown, no ppt ring bromination of benzene Dark brown, no pptResorcinol rings 2. REACTIONS OF PHENOLS 2.2 BROMINE WATER TEST
  • 23. Br2 in H2O Test• to detect any phenol or phenolic groupspresent in the unknown.• The positive test is the decoloration ofbromine and the presence of precipitate.• test is able to detect phenol but not benzeneis because of the increased reactivity of thephenol.• The increase in density of phenol makes itmore susceptible to attack by bromine.
  • 24. Millon’s Reagent• Used for determination of the presence ofproteins• Dissolved mercury in concentrated nitric acid,diluted with water and when heated with phenoliccompounds gives a red coloration• Only EGG ALBUMIN will give a positive result
  • 25. Structure formula responsibl Visible e for Test Sample Results results Mercuric complex Phenol pink with phenolic group Mercuric complex Catechol brown with phenolic group Mercuric complex brown with phenolic Resorcinol group Mercuric complex Green,dark with phenolic A-napthol orange group2. REACTIONS OF PHENOLS 2.3 MILLON’STEST
  • 26. Millon’s Test • to detect any phenol or phenolic groups present in the unknown. • A positive test is a pink to red colored solution or precipitate. • The coloration is due to the mercury present in Millon’s reagent reacting with the phenolic OH group to form a complex and/or a precipitate.
  • 27. F. TEST OF ALDEHYDES AND KETONES1. 2,4-DNPH TEST 6. FEHLING’S TEST2. BISULFITE TEST 7. MOLISCH TEST3. SCHIFF’S TEST 8. BENEDICT’S TEST4. TOLLEN’S TEST 9. BARFOED’S TEST5. IODOFORM TEST 10. SELIWANOFF’STEST
  • 28. F Ad Ac B1. 2,4-DPNH TEST
  • 29. Structure or formula ofTest samples Visible result compound responsible for the visible resultsFormaldehyde Solid yellow precipitate Clear yellow solution withAcetaldehyde orange precipitate Yellow orange solution Acetone with orange precipitateBenzaldehyde Orange precipitate 1. 2,4-DPNH TEST
  • 30. Structure or formula ofTest samples Visible result compound responsible for the visible resultsFormaldehyde no reactionAcetaldehyde clear solution Acetone clear solution white precipitate atBenzaldehyde (C5H6)CH(OH)SO3─N the bottom a+ 2. BISULFIDE TEST
  • 31. Structure or formula of compoundTest samples Visible result responsible for the visible results dark violet solution Schiff’s reagentFormaldehyde with metallic complex with appearance methanol Schiff’s reagentAcetaldehyde violet solution complex with ethanol Unconjugated Schiff’s Acetone light pink reagent complex Schiff’s reagentBenzaldehyde royal blue solution complex with methylphenol 3. SCHIFF’S TEST
  • 32. F Ad Ac B 3. SCHIFF’S TEST
  • 33. Structure or formula of compoundTest samples Visible result responsible for the visible results Black solution withFormaldehyde silver substance Silver metal (silver mirror) With silverAcetaldehyde Silver metal substance clear solution [no Acetone - reaction] brown precipitate at Silver metalBenzaldehyde the top 4. TOLLEN’S TEST
  • 34. Before heating After heating 4. TOLLEN’S TEST
  • 35. 5. IODOFORM TEST
  • 36. Structure or formula ofTest samples Visible result compound responsible for the visible results no reaction (clearFormaldehyde NaOH [-] solution) yellow precipitate withAcetaldehyde CHI3 strong odor Acetone blurry precipitate CHI3Benzaldehyde brown precipitate NaOH [-] 5. IODOFORM TEST
  • 37. Test Samples Visible ResultFormaldehyde negativeAcetaldehyde blue to blue green color, layer formation Acetone Fehling’s reagent’s color, electric blueBenzaldehyde blue color, oil layer 6. FEHLING’S TEST
  • 38. 6. FEHLING’S TEST
  • 39. Structure or formula ofTest samples Visible result compound responsible for the visible results Glucose blue violet ring α-naphthol Maltose blue violet ring α-naphthol Sucrose blue violet ring α-naphtholBoiled Starch blue violet ring α-naphthol 7. MOLISCH TEST
  • 40. 7. MOLISCH TEST
  • 41. Structure or formula ofTest samples Visible result compound responsible for the visible results red precipitate Glucose over yellow cuprous oxide solution Maltose green blue solution cuprous oxide Blurry precipitate copper complex Sucrose over blue solution with water [-] darker blue copper complexBoiled Starch solution with water [-] 8. BENEDICT’S TEST
  • 42. 8. BENEDICT’S TEST
  • 43. Structure or formula ofTest samples Visible result compound responsible for the visible results Glucose 2 layers: blue over red cuprous oxide Maltose clear blue solution cuprous oxide clear top over blue Sucrose Cuprous oxide solutionBoiled Starch Aqua blue in color Cuprous oxide 9. BARFOED’S TEST
  • 44. Structure or formula ofTest samples Visible result compound responsible for the visible results Colored complex of Glucose very, very light orange furfural with resorcinol Colore complex of Maltose clear, light brown orange furfural with recorcinol colored complex of Sucrose pink orange furfural with resorcinol colored complex ofBoiled Starch pink orange furfural with resorcinol 10. SELIWANOFF’STEST
  • 45. G. TEST FOR AMINES 1. HINSBERG TEST 2. NITROUS ACID TEST
  • 46. + + 5 DROPS 20 DROPS 10% 5 DROPS benzenesulfonyl NaOH chloride sample cover tube with cork & shake for about 5mins.if not basic + 10% NaOH DROPWISE if precipitate forms + 40 DROPS water then shake + 3M HCl DROPWISE 1. HINSBERG TEST
  • 47. Test samples Visible result Structure or formula of compound responsible for the visible results Clear light orange C6H5SO2NR─Na+ → Methylamine with brown C6H5SO2NRH precipitateDimethylamine No change C6H5SO2NR2 Clear light yellow;Trimethylamine NR3 → 3RNH + Cl- gel Precipitate formed; Aniline - release of heat Evolution of white C6H5SO2NR─Na+ →N-methylaniline smoke C6H5SO2NRH 1. HINSBERG TEST
  • 48. • Differentiate primary amines, secondary amines, tertiary amines and aniline from each other• Involves formation of sulfonamides and shaking with excess sodium hydroxide in the first step. The second step requires acidification of the mixture. The results for the different types of amines allow a determination to be made.
  • 49. Primary amines: substance dissolves in a base and precipitates in an acidSecondary amines: substance precipitates in a base and will have no change in the acidTertiary amines: substance precipitates in a base and dissolves in an acid
  • 50. Primary amines give sulfonamides that are soluble in basic solution
  • 51. Secondary amines give sulfonamides that are insolublein basic solution
  • 52. Tertiary amines do not form stable sulfonamides
  • 53. • Anilines aromaticity prevents ef ficient reaction with the reagent thus it results in a negative test.• N-methylaniline gives a positive test because of the methyl group present which allows the N to react with the reagent because of its electron-repelling ef fects.
  • 54. 3 DROPS sample + cool in ice bath + 5 DROPS cold 20% NaNO240 DROPS 2M HCl if no evolution of colorless gas nor formation of yellow to orange color is obtained, warm half of the sol’n at room temp.  + ice cold sol’n (dropwise) of about 50mg of β-naphthol in 2ml of 2M NaOH 2. NITROUS ACID TEST
  • 55. Test samples Visible result Structure or formula of compound responsible for the visible results evolution of colorless Methylamine N2 gas bubbles light orange, clearDimethylamine (CH3)2N─N=O solutionTrimethylamine yellow; clear gas (CH3)3N+ evolution of gas; yellow, brown Aniline N2 solution; release of heat light brown orangeN-methylaniline C6H5CH3N─N=O solution with gas 2. NITROUS ACID TEST
  • 56. • primar y aliphatic amines give of f nitrogen gas and a clear solution• Primar y amines form diazonium salt as a product. This product is ver y unstable and degrades into a carbocation that tends to react non-selectively with nucleophiles present in the solution.
  • 57. • Upon reaction with nitrous acid, secondar y aliphatic and aromatic amines form n-nitrosoamine which appears to be a yellow oily liquid.• When the mixture becomes acidic, all the amines present in the mixture tend to undergo reversible salt formation. This happens with ter tiar y amines. The ammonium salts that are formed are usually soluble in water.• an orange coloration may probably have come from the n- nitrosoamine produced in the reaction
  • 58. H. TEST FOR CARBOXYLICACID AND ITS DERIVATIVES1. FORMATION OF ESTERS2. HYDROLYSIS OF ACIDDERIVATIVES3. HYDROXAMIC ACID TESTFOR ACID DERIVATIVES
  • 59. pinch salicylic acid + + 5 DROPS conc. H2SO4 5 mins shake well20 DROPS methanol Test Sample Visible Result Structure responsible Salicylic acid mint odor 1. FORMATION OF ESTERS 1.1 REACTION OF CARBOXYLIC ACID AND ALCOHOL
  • 60. • The ester methyl salicylate was produced when the salicylic acid was heated with methanol in the presence of an acid catalyst (H2SO4). The esterification reaction is both slow and reversible.• Sweet fruity smell was produced aka oil of wintergreen
  • 61. 20 DROPS water cover tube with cork & gently shake the mixture + 20 DROPS 25% NaOH + mix 10 DROPS ethanol Test Sample Visible Result + Benzoylchloride solid white precipitate (bottom) smells like alcohol 5DROPSbenzoylchloride 1. FORMATION OF ESTERS 1.2 SCHOTTEN-BAUMANN REACTION
  • 62. Also known as “Reactions ofAcylhalide and Alcohol”• The acyl halides will undergo a reaction with alcohols under basic conditions to form esters. Esters are both insoluble in water and less dense than water and thus will form a layer on top of the water
  • 63. With a stirring rod, hold a piece of moist red litmus paper over the mouth of the test tube while heating the mixture to boiling in a H2O bathTEST SAMPLES VISIBLE RESULTSBenzamide red litmus to blue, burnt odor 2. HYDROLYSIS OF ACID DERIVATIVES 2.1 HYDROYSIS OF BENZAMIDE
  • 64. • benzamide was hydrolyzed with the use of sodium hydroxide• sodium benzoate and Ammonia was formed
  • 65. loosely cover the test tube with a cork and heat in water bath for 15 minutes HCl (dropwise)TEST SAMPLES VISIBLE RESULTSEthylacetate strong sour odor 2. HYDROLYSIS OF ACID DERIVATIVES 2.2 HYDROLYSIS OF AN ESTER
  • 66. This reaction is reverse of the esterification reaction. Acarboxylic acid and an alcohol are formed resulting to the odor obser ved.
  • 67. gently shake and feel the tubeTEST SAMPLES VISIBLE RESULTSSTRUCTURE/FORMULA OF COMPOUND RESPONSIBLE FOR RESULTAcetic anhydride blue litmus to red (CH3CO)2O + H2O → 2CH3COOH 2. HYDROLYSIS OF ACID DERIVATIVES 2.3 HYDROLYSIS OF ANHYDRIDE
  • 68. • In the hydrolysis of acetic anhydride, acetic acid was formed.• In the litmus paper test, the blue litmus paper turned red because an acid was formed.
  • 69. Reaction mechanism 1o Amine 2o Amine 3o Amine 3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES
  • 70. Test Samples Visible Result Structure/ Formula of Compound Responsible for ResultEthylacetate blue litmus; odorless HXBenzamide pink litmus; odorless NH4Acetic anhydride red litmus; acetic acid odor RCO2HBenzoylchloride blue litmus; alcohol odor ROH 3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES
  • 71. Hydroxamic acid is a compound in which an amine is inser ted into a carboxylic acid. In the test for RCOOH derivatives, esters, acid anhydrides and ar yl/acyl halides would give positive results.• purple to red solution and means a positive test, a dark brown solution is uncer tain while a yellow solution means negative
  • 72. When an ester, like ethyl acetate, is reacted with hydroxamic acid, it produces an alcohol.When an acid anhydride, like aceticanhydride, is reacted with hydroxamicacid, it produces a carboxylic acid.
  • 73. Acyl HalideFerric Hydroxamate Complex Formation
  • 74. • purple to red solution and means a positive test, a dark brown solution is uncer tain while a yellow solution means negative• The resulting ferric hydroxamate has a distinct burgundy or magenta color.• esters, anhydrides, amides and acyl chlorides give positive tests because all solutions were colored dark brown of brownish red
  • 75. What property of alcohol is demonstrated in thereaction with Na metal? What is the formula of the gasliberated? The acidity of alcohol is demonstrated in the reaction w/Na metal. The gas liberated is H2.Dry test tube should be used in the reaction betweenthe alcohols and the Na metal. Why? Because Na metal reacts with water that may causeignition.
  • 76. Why is the Lucas test not used for alcoholscontaining more than eight carbon atoms? The Lucas test applies only to alcohols soluble in theLucas reagent (monofunctional alcohols with less than 6carbons and some polyfunctional alcohols). The long chainsof C-bond atoms act as non-polar makes the hydroxyl groupless functional. This results in the insolubility of the alcoholin the reagent and would make the test ineffective.Explain why the order of reactivity of the alcoholstoward Lucas reagent is 3°>2°>1°. The reaction rate is much faster when the carbocationintermediate is more stabilized by a greater number ofelectron donating alkyl group bonded to the positive carbonatom.This means that the greater the alkyl groups presentin a compound, the faster its reaction would be with theLucas solution.
  • 77. What functional group is responsible for the observedresult in Millon’s test? Hydroxyphenyl group or the phenolic –OH Why is the Schiff’s test considered a general test foraldehydes? This is because any aldehyde readily reacts with Schiff’sreagent to form positive results.Schiff’s reagent involves abisulfite ion stuck in the original molecular structure. Aldehydeschange this arrangement and thus there is a consequentchange as the reaction progresses.
  • 78. Why is it advantageous to use a strong acid catalystin the reaction of aldehyde or ketone with 2,4-DNPH? It is because a strong acid when used as a catalystreverses the sequence of reactions. In the presence of arelatively weaker acid, the strong nucleophile attacks thesubstrate then the electrophile follows suit.
  • 79. Whereas in the presence of a strong acid, the stronghydronium ion is more ready for protonation to the oxygenof the carbonyl group. The weaker nucleophile (whichthrives in basic medium) then attacks the carbon tostabilize the forming hemiacetal. Water abstracts the H+and a hemiacetal is formed. Hemiacetals are relatively lessstable products that will form acetals and will not show thevisible changes that are expected of the test.
  • 80. Show the mechanism for the reaction of acetaldehyde with the following reagents: a. 2,4-DNPH b. NaHSO3
  • 81. What structural feature in a compound is requiredfor a positive iodoform test? Will ethanol give apositive iodoform test? Why or why not?
  • 82. Show the mechanisms for the iodoform reaction using acetaldehyde as the test sample.
  • 83. What test will you use to differentiate each of thefollowing pairs? Give also the visible result.a. acetaldehyde and acetoneSchiff’s test – reaction with acetaldehyde will result to apurple solution. Acetone onthe other hand will not react.Tollen’s test – acetaldehyde will form a silver mirror. Acetoneon the other hand will not have any reaction.b. acetaldehyde and benzaldehydeBIsulfite’s test – will differentiate an aliphatic aldehyde from anaromatic aldehyde.Aldehyde will react faster than benzaldehyde. Both will form a reprecipitate due to cuprous oxide.