Expt 8 b report 2007 til seliwanoff test 1

11,239 views

Published on

Published in: Business, Technology
0 Comments
2 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
11,239
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
141
Comments
0
Likes
2
Embeds 0
No embeds

No notes for slide
  • A chromophore is the part of a molecule responsible for its color. [1] The color arises when a molecule absorbs certain wavelengths ofvisible light and transmits or reflects others. The chromophore is a region in the molecule where the energy difference between two differentmolecularorbitals falls within the range of the visible spectrum. Visible light that hits the chromophore can thus be absorbed by exciting anelectron from its ground state into an excited state.Chromophores almost always arise in one of two forms: conjugated pi systems (also known as resonating systems) and metal complexes.
  • Furfural is an organic compound derived from a variety of agricultural byproducts, includingcorncobs, oat, wheat bran, and sawdust. The name furfural comes from the Latin word furfur, meaning bran, referring to its usual source.
  • Expt 8 b report 2007 til seliwanoff test 1

    1. 1. 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
    2. 2. 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, tert-butyl alcohol
    3. 3. Na Metal• Alkali metal• Soft at room temperature• Silvery white in color• Highly reactive
    4. 4. Na Test• Positive result is indicated by evolution of gas
    5. 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 gastert- butyl alcohol no evolution of gas H2 1. REACTIONS OF ALCOHOLS 1.1 REACTIONS WITH Na METAL
    6. 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 + H2Tert-butyl – slowest to react; least acidic - (CH3)3C-OH + Na  (CH3)3C -O-Na+ + H2
    7. 7. Lucas Reagent• ZnCl2 in concentrated HCl solution• Reagent used for classification of alcohols with low MW.
    8. 8. LUCAS TEST• Test us to differentiate primary, secondary and tertiary alcohols• Uses the differences in reactivity of hydrogen halides and the three classes or types of alcohol
    9. 9. 20 DRPS LUCAS REAGENT 10 DROPS N- BUTYLALCOHOL SHAKE AND Na METAL (SEC-BUTYL, COVER TERT-BUTYL WITH ALCOHOLS) STOPPER Structure formula Test Sample Visible Results responsible for results n- butyl alcohol No layer formation n/asec- butyl alcohol Cloudy formation (CH3)2CHCl + H2Otert- butyl alcohol Fast layer formation (CH3)3CCl + H2O 1. REACTIONS OF ALCOHOLS 1.2 LUCAS TEST
    10. 10. LUCAS TESTReaction:speed of this reaction is proportional to the energy required to form the carbocationThe cloudiness observed (if any) is caused by the carbocation reacting with the chloride ion creating an insoluble chloroalkane.
    11. 11. Reactions:Primary Alcohol:Secondary:Tertiary:
    12. 12. Potassium Dichromate K2Cr2O7• Inorganic chemical• Oxidizing agent• Positive result yields the formation of layers
    13. 13. 20 DROPS N- ACIDIFY WITH 5DROPS 3%BUTYLACLCOHO L 2M H2SO4 K2Cr2O7 Structural Formula responsible for Test Sample Visible Results results 3 layers formed: upper light- n- butyl alcohol orange, middle dark- orange Chromic Ion and bottom clear layer 3 layers formed: upper green, Sec- butyl alcohol middle orange and bottom Chromic Ion blue-green layer 3 layers formed: upper orange, tert- butyl alcohol middle green and bottom No rxn orange layer 1. REACTIONS OF ALCOHOLS 1.3 REACTIONS WITH K2Cr2O7
    14. 14. Rxn with K2Cr2O7oxidation occurs: the orange solution containing the dichromate (VI) ions is reduced to a green solution containing chromium (III) ions.Primary: oxidized to aldehydes/carboxylic acidsAldehydes:Carboxylic:
    15. 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. 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. 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 solution
    18. 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. 19. FeCl3 TestAn iron-phenol complex is formed.FeCl3 + 6C6H5OH  [Fe(OC6H5)6]3- + 3H+ +3Cl-
    20. 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. 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. 22. 20 DROPS 20 DROPS BROMINE 95% PHENOL WATERETHANOL Structure formulaTest Sample Visible Results responsible for results bromination of benzene Phenol Turns pinkish, ppt ring Turns to dark green, bromination of aromaticα – napthol ring ppt 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. 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. 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 •Hg(NO3)2 in HNO3• Only EGG ALBUMIN will give a positive result
    25. 25. Structure formula responsibl Visible e for5 DROPS Test Sample Results resultsMILLON’SREAGENT • SHAKE pink Mercuric complex • HEAT IN Phenol with phenolic group WATER BATH Mercuric complex (2MINS) Catechol brown with phenolic5 DROPS group OF brown Mercuric complex with phenolicSAMPLE Resorcinol group Mercuric complex Green,dark with phenolic A-napthol orange group 2. REACTIONS OF PHENOLS 2.3 MILLON’STEST
    26. 26. Millon’s Test • to detect any phenol or phenolic groups present in the unknown. • A positive test is a red to brown 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. 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. 28. 2,4- DNPH Test2,4-Dinitrophenylhydrazine (Bradys reagent)used to qualitatively test for carbonyl groups
    29. 29. 2,4- DNPH TestA positive test is signaled by a yellow or orangeprecipitate (dinitrophenylhydrazone)RRC=O + C6H3(NO2)2NHNH2 →C6H3(NO2)2NHNCRR + H2OA condensation reaction
    30. 30. 5 DROPS2,4-DPNH • HEAT IN WATER BATH(5 MINS)3 DROPSSAMPLE F Ad Ac B 1. 2,4-DPNH TEST
    31. 31. 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 with Acetone orange precipitateBenzaldehyde Orange precipitate 1. 2,4-DPNH TEST
    32. 32. 2,4- DNPH Test• formaldehyde, acetaldehyde, and benzaldehyde aldehydes• Acetone  ketone• Produced a POSITIVE RESULT (orangecoloration)
    33. 33. Bisulfite Test (NaHSO3)• Also a test for aldehydes and ketones• A positive result is indicated by the formation ofprecipitate
    34. 34. 20 DROPS SODIUMBISULFIDE Structure or formula of compound Test samples Visible result responsible for the visible results5 DROPSSAMPLE Formaldehyde no reaction Acetaldehyde Little white ppt H3C(OH)SO3- Na+MIX. COOL Acetone Yellow ppt H3CC(OH)SO3- Na+IN AN ICE BATH (C6H6)CH(OH)SO3- white precipitate at Benzaldehyde Na+ the bottom 2. BISULFITE TEST
    35. 35. Bisulfite TestGeneral equation:Formaldehyde: no reaction due to steric effectsAcetaldehyde: CH3CH=O + NaHSO3 (CH3)CH(OH)SO3-Na+Acetone: CH3C=OCH3 + NaHSO3 (CH3)2C(OH)SO3-Na+Benzaldehyde: + NaHSO3  (CH)5C(OH) SO3-Na+
    36. 36. Schiff’s Test• For the detection of aldehydes• Schiff’s reagent •reaction product of Fuchsine or the closely related Pararosaniline (lacks a methyl group) and sodium bisulfite (NaHSO3 )….Fuschine in NaHSO3 solution• Schiff’s reagent reacts with aldehydes,regenerating the chromophoric system• Positive result: magenta/purple colored solution
    37. 37. 5 DROPS SAMPLE ADD Structure or formula of compound Test samples Visible result responsible for the20 DROPS visible resultsSCHIFF’S dark violet solution Schiff’s reagentREAGENT Formaldehyde with metallic complex with appearance methanol Schiff’s reagent Acetaldehyde violet solution complex with ethanol Unconjugated Schiff’s Acetone light pink reagent complex Schiff’s reagent royal blue purple Benzaldehyde complex with solution methylphenol 3. SCHIFF’S TEST
    38. 38. F Ad Ac B 3. SCHIFF’S TEST
    39. 39. Tollen’s Test• Used to ascertain if compound is a ketone or analdehyde• Tollen’s reagent ammoniacal silver nitratesolution• Positive resutl is the depositon of silver metal toform the so called “silver mirror”• Aldehyde silver mirror• Ketone no reaction
    40. 40. 20 DROPSTOLLEN’SREAGENT • HEAT IN WATER BATH AND OBSERVE CHANGES Structure or formula5 DROPS of compoundSAMPLE Test samples Visible result responsible for the visible results Black solution with Formaldehyde silver substance Silver metal (silver mirror) With silver Acetaldehyde Silver metal substance clear solution [no Acetone - reaction] brown precipitate at Silver metal Benzaldehyde the top 4. TOLLEN’S TEST
    41. 41. Before heating After heating 4. TOLLEN’S TEST
    42. 42. Iodoform Test• Iodoform reagent is • a mixture of iodine (I2) and potassium iodide (KI) and 10% NaOH• Used to test for methylketones, ketoneswhere one of the two alkyl groups bonded tothe carbonyl carbon is a methyl group•Positive result is a yellow iodoform formation
    43. 43. 20 DROPS ADD I2/KI W/10% NaOH + MIXING UNTIL HEAT IN A WATER BATH @ BROWN COLOR 5 DROPS PERSIST 6O C (2 MINS) SAMPLE ADD 10% MIX UNTIL BROWN COLOR NaOH DISAPPEARS ->(DROPWISE) YELLOWADD 5 DROPS SHAKE VIGOROUSLY DISTILLED AND STAND (15 WATER MINS) 5. IODOFORM TEST
    44. 44. Structure or formula ofTest samples Visible result compound responsible for the visible results no reaction (clearFormaldehyde NaOH [-] solution) yellow precipitate withAcetaldehyde CHI3 strong odor blurry yellowish Acetone CHI3 precipitateBenzaldehyde Clear solution NaOH [-] 5. IODOFORM TEST
    45. 45. Fehling’s Test• Fehling’s Reagent is made up of • Sodium tartrate; NaOH; CuSO4• A positive result yields the formation of a red precipitate(Cu2O)• This coloration is caused by the oxidation of aldehydesand aldoses to carboxylic acids.•The Cupric ion is reduced to cuprous oxide which thengives the red precipitate.
    46. 46. Test Samples Visible ResultFormaldehyde Red precipitateAcetaldehyde Red precipitate Acetone No reactionBenzaldehyde No reaction 6. FEHLING’S TEST
    47. 47. Molisch Test• For the presence of carbohydrates• based on the dehydration of the carbohydratesby sulfuric acid•Molisch Reagent is α-naphthol and conc. H2SO4•A positive result yields a formation of blue violetring
    48. 48. Molisch Test• Sugar is hydrolized to yield furfural• Alpha-naphthol reacts with cyclic aldehyde toform a purple ring, a positive indication•C6H12O6 + (conc.) H2SO4 → C5H4O2 + 3 H2O• C5H4O2 + 2 C10H8OH (α-naphthol) → coloredproduct
    49. 49. 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
    50. 50. 7. MOLISCH TEST
    51. 51. Benedict’s Test• Detect the presence of reducing sugars (sugarswith a free aldehyde or ketone group)• Monosaccharides and some disaccharides arereducing sugarsfree reactive carbonyl group• Other disaccharides such as sucrose andstarchesnon-reducing sugars and will not reactwith Benedicts solution.
    52. 52. Benedict’s Test• Benedict’s Reagent is made up of Sodium citrate; CuSO4; sodium bicarbonate• Copper sulfate (CuSO4) present in Benedictssolution reacts with electrons from the aldehyde orketone group of the reducing sugar to formcuprous oxide (Cu2O), a red-brown precipitate.
    53. 53. Benedict’s Test•Results: Result What it means No precipitate - Green A trace Yellow + Orange ++ Red +++
    54. 54. 20 DROPSBenedict’s reagent • Heat in water bath Structure or 5 drops formula of sample compound Test samples Visible result 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 complex Boiled Starch solution with water [-] 8. BENEDICT’S TEST
    55. 55. 8. BENEDICT’S TEST
    56. 56. Barfoed’s test• Similar to Benedicts test, but determines if acarbohydrate is a monosaccharide or adisaccharide• Reacts with monosaccharides to producecuprous oxide at a faster rate than disaccharidesdo•Barfoed’s reagent is made up of •Cu(OAc)2 in HOAc•Positive Result shows a formation of a redprecipitate.
    57. 57. Barfoed’s test• Reducing monosaccharides are oxidized by thecopper ion in solution to form a carboxylic acid anda reddish precipitate of copper (I).• Copper(II) acetate is reduced to copper(I) oxide(Cu2O), which forms a brick-red precipitate. Thealdehyde group of the monosaccharide is oxidizedto the carboxylate.
    58. 58. Structure or formula ofTest samples Visible result compound responsible for the visible results Glucose Red precipitate cuprous oxide Maltose clear blue solution clear top over blue Sucrose solutionBoiled Starch Aqua blue in color 9. BARFOED’S TEST
    59. 59. Seliwanoff’s Test• To differentiate between aldose and ketosesugars• Seliwanoff’ reagent  consist of resorcinol and conc. HCl• Ketose  sugar containing a ketonegroup; if the mixture turns red• Aldose  sugar containing an aldehydegroup; mixture turns pink
    60. 60. Seliwanoff’s Test• Positive result indicates the formation of ared precipitate and a pink solution.
    61. 61. Structure or formula ofTest samples Visible result compound responsible for the visible results Glucose very, very light orange clear, light brown Maltose orange colored complex of Sucrose pink orange furfural with resorcinolBoiled Starch Light orange 10. SELIWANOFF’STEST
    62. 62. G. TEST FOR AMINES 1. HINSBERG TEST 2. NITROUS ACID TEST
    63. 63. + + 5 DROPS 20 DROPS 10% 5 DROPS benzenesulfonyl NaOH sample chloride 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
    64. 64. 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
    65. 65. 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
    66. 66. 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
    67. 67. H. TEST FOR CARBOXYLICACID AND ITS DERIVATIVES1. FORMATION OF ESTERS2. HYDROLYSIS OF ACIDDERIVATIVES3. HYDROXAMIC ACID TESTFOR ACID DERIVATIVES
    68. 68. pinch salicylic acid + + 5 DROPS conc. H2SO4 5 mins shake well20 DROPS methanol Test Sample Visible Result Structure responsible Salicylic acid white-yellow solid precipitate 1. FORMATION OF ESTERS 1.1 REACTION OF CARBOXYLIC ACID AND ALCOHOL
    69. 69. 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
    70. 70. 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
    71. 71. 20 DRPS 5 drps loosely cover the test tube with a Ethyl 25% cork and heat in acetate NaOH water bath for 15 minutes HCl (dropwise) TEST SAMPLES VISIBLE RESULTS Ethylacetate strong sour odor 2. HYDROLYSIS OF ACID DERIVATIVES 2.2 HYDROLYSIS OF AN ESTER
    72. 72. 20 DRPS Red and 20 drps gently blue litmus Acetic shake water paper Anhydride and feel the tubeTEST SAMPLES VISIBLE RESULTS STRUCTURE/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
    73. 73. Reaction mechanism 1o Amine 2o Amine 3o Amine 3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES
    74. 74. 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
    75. 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/ Nametal. 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. 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. 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. 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. 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 to stabilizethe forming hemiacetal. Water abstracts the H+ and ahemiacetal is formed. Hemiacetals are relatively less stableproducts that will form acetals and will not show the visiblechanges that are expected of the test.
    80. 80. Show the mechanism for the reaction of acetaldehyde with the following reagents: a. 2,4-DNPH b. NaHSO3
    81. 81. What structural feature in a compound is requiredfor a positive iodoform test? Will ethanol give apositive iodoform test? Why or why not?
    82. 82. Show the mechanisms for the iodoform reaction using acetaldehyde as the test sample.
    83. 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.

    ×