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EXPERIM
ENT
8BCOKIE
NG. FALL
ORIA. L
USICA
E. TEST FOR ALCOHOLS AND PHENOLS
2. ALCOHOL AgNO3 TEST
1. REACTIONS OF ALCOHOLS REACTION WITH Na METAL LUCAS TEST REACTIONS WITH K2Cr2O7
FeCl3 TEST BROMINE WATER TEST MILLON’S TEST
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
Na Metal
• Alkali metal• Soft at room temperature• Silvery white in color• Highly reactive
Test Sample Visible Results
Structure or formula of compound
responsible for the visible results
n- butyl alcoholRapid evolution of
gasH2
Sec- butyl alcoholModerate evolution of
gasH2
tert- butyl alcoholSlight to no evolution
of gasH2
1. REACTIONS OF ALCOHOLS1.1 REACTIONS WITH Na METAL
20 DRPS N-BUTYLALCOHOL
(SEC-BUTYL, TERT-BUTYL ALCOHOLS)
Na METAL
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+ + H2
Sec-butyl – CH3CH2CH(OH)CH3 + Na CH3CH2CH(O-Na+)CH3 + H2
Tert-butyl – slowest to react; least acidic - (CH3)3C-OH + Na (CH3)3C -O-Na+ + H2
Lucas Reagent
• ZnCl2 in concentrated HCl solution
• Reagent used for classification of alcohols with low MW.
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
1. REACTIONS OF ALCOHOLS1.2 LUCAS TEST
Test Sample Visible ResultsStructure formula responsible for results
n- butyl alcohol No layer formation n/a
sec- butyl alcoholModerate layer
formation(CH3)2CHCl + H2O
tert- butyl alcohol Fast layer formation (CH3)3CCl + H2O
Na METAL
20 DRPS LUCAS REAGENT
10 DROPS N-BUTYLALCOHOL
(SEC-BUTYL, TERT-BUTYL ALCOHOLS)
SHAKE AND COVER WITH STOPPER
LUCAS TEST
Reaction:
speed of this reaction is proportional to the energy required to form the carbocation
The cloudiness observed (if any) is caused by the carbocation reacting with the chloride ion creating an insoluble chloroalkane.
Reactions:
Primary Alcohol:
Secondary:
Tertiary:
Potassium Dichromate K2Cr2O7
• Inorganic chemical• Oxidizing agent
1. REACTIONS OF ALCOHOLS1.3 REACTIONS WITH K2Cr2O7
Test Sample Visible Results
Structural Formula
responsible for results
n- butyl alcoholBlue- green
solutionChromic Ion
Sec- butyl alcoholBlue- green
solutionChromic Ion
tert- butyl alcohol Blue green Dichromate Ion
Rxn with K2Cr2O7
oxidation 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:
Rxn with K2Cr2O7
Secondary: oxidized to ketoneKetone:
Tertiary: cannot be oxidizedWhy? Because tertiary alcohols don't have a hydrogen atom attached to a carbon
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
FeCl3 Test
• Addition of FeCl3 gives a colored solution
• alcohols do not undergo this reaction
• other functional groups produce color changes:• aliphatic acidsyellow solution; • aromatic acidstan precipitate
Test Sample Visible Results Structural formula responsible for
results
PhenolReddish-brown Iron (III) complex w/
Phenol
α - napthol PurpleIron (III) complex w/
Napthol
Cathechol Dark BlueIron (III) complex w/
Cathechol
ResorcinolMoss Green
Iron (III) complex w/ Resorcinol
2. REACTIONS OF PHENOLS2.1 FeCl3 TEST
FeCl3 Test
An iron-phenol complex is formed.FeCl3 + 6C6H5OH [Fe(OC6H5)6]3- +
3H+ +3Cl-
BR2 IN H2O TEST•used to identify alkenes, alkynes and phenols
•Alkenes & alkynes the reaction occurs through electrophilic addition
•Phenol reacts with sites of unsaturation, even aromatic rings, through a complex addition reaction
BR2 IN H2O TEST•Brominedark brown color
•when it reacts, the color dissipates and the reaction mixture becomes yellow or colorless
•ortho and para positions to the phenol are brominated.
2. REACTIONS OF PHENOLS2.2 BROMINE WATER TEST
Test Sample Visible Results Structure formula responsible for results
Phenol Turns pinkish, ppt bromination of benzene ring
α – naptholTurns to dark
green, pptbromination of aromatic
ring
Cathechol Dark brown, no ppt bromination of benzene ring
ResorcinolDark brown, no ppt bromination of benzene
rings
Br2 in H2O Test
• to detect any phenol or phenolic groups present in the unknown.
• The positive test is the decoloration of bromine and the presence of precipitate.
• test is able to detect phenol but not benzene is because of the increased reactivity of the phenol.
• The increase in density of phenol makes it more susceptible to attack by bromine.
Millon’s Reagent• Used for determination of the presence of proteins
• Dissolved mercury in concentrated nitric acid, diluted with water and when heated with phenolic compounds gives a red coloration
• Only EGG ALBUMIN will give a positive result
2. REACTIONS OF PHENOLS2.3 MILLON’STEST
Test Sample
Visible Results
Structure formula responsible for results
Phenol pink Mercuric complex with phenolic group
Catechol
brown Mercuric complex with phenolic group
Resorcinol
brown Mercuric complex with phenolic group
A-naptholGreen,dark
orange
Mercuric complex with phenolic group
• 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.
Millon’s Test
F. TEST OF ALDEHYDES AND KETONES
2. BISULFITE TEST
1. 2,4-DNPH TEST
3. SCHIFF’S TEST4. TOLLEN’S TEST5. IODOFORM TEST
6. FEHLING’S TEST7. MOLISCH TEST8. BENEDICT’S TEST9. BARFOED’S TEST10. SELIWANOFF’STEST
1. 2,4-DPNH TEST
F Ad Ac B
Test samples Visible resultStructure or formula of compound responsible for the visible results
Formaldehyde Solid yellow precipitate
AcetaldehydeClear yellow solution
with orange precipitate
AcetoneYellow orange solution with orange precipitate
Benzaldehyde Orange precipitate
1. 2,4-DPNH TEST
2. BISULFIDE TEST
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Formaldehyde no reaction
Acetaldehyde clear solution
Acetone clear solution
Benzaldehydewhite precipitate
at the bottom(C5H6)CH(OH)SO3
─Na+
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Formaldehydedark violet solution
with metallic appearance
Schiff’s reagent complex with
methanol
Acetaldehyde violet solutionSchiff’s reagent complex with
ethanol
Acetone light pinkUnconjugated
Schiff’s reagent complex
Benzaldehyde royal blue solutionSchiff’s reagent complex with methylphenol
3. SCHIFF’S TEST
F Ad Ac B
3. SCHIFF’S TEST
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Formaldehyde
Black solution with silver
substance (silver mirror)
Silver metal
AcetaldehydeWith silver substance
Silver metal
Acetoneclear solution [no
reaction]-
Benzaldehydebrown precipitate
at the topSilver metal
4. TOLLEN’S TEST
Before heating After heating
4. TOLLEN’S TEST
5. IODOFORM TEST
5. IODOFORM TEST
Test samples Visible resultStructure or formula of compound responsible for the visible results
Formaldehydeno reaction (clear
solution)NaOH [-]
Acetaldehydeyellow precipitate with
strong odorCHI3
Acetone blurry precipitate CHI3
Benzaldehyde brown precipitate NaOH [-]
6. FEHLING’S TEST
Test Samples Visible Result
Formaldehyde negativeAcetaldehyde blue to blue green color, layer
formation
Acetone Fehling’s reagent’s color, electric blue
Benzaldehyde blue color, oil layer
6. FEHLING’S TEST
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Glucose blue violet ring α-naphtholMaltose blue violet ring α-naphtholSucrose blue violet ring α-naphthol
Boiled Starch blue violet ring α-naphthol
7. MOLISCH TEST
7. MOLISCH TEST
Test samples Visible result
Structure or formula of compound
responsible for the visible
results
Glucosered precipitate
over yellow solution
cuprous oxide
Maltosegreen blue
solutioncuprous oxide
SucroseBlurry
precipitate over blue solution
copper complex with water [-]
Boiled Starchdarker blue
solutioncopper complex with water [-]
8. BENEDICT’S TEST
8. BENEDICT’S TEST
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Glucose2 layers: blue over
redcuprous oxide
Maltose clear blue solution cuprous oxide
Sucroseclear top over blue
solutionCuprous oxide
Boiled Starch Aqua blue in color Cuprous oxide
9. BARFOED’S TEST
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Glucosevery, very light
orange
Colored complex of furfural with resorcinol
Maltoseclear, light brown
orange
Colore complex of furfural with recorcinol
Sucrose pink orangecolored complex of
furfural with resorcinol
Boiled Starch pink orangecolored complex of
furfural with resorcinol
10. SELIWANOFF’STEST
G. TEST FOR AMINES
1. HINSBERG TEST2. NITROUS ACID TEST
+ +20 DROPS 10% NaOH
5 DROPS sample
5 DROPS benzenesulfon
yl chloride
cover tube with cork & shake for about 5mins.
1. HINSBERG TEST
if not basic+ 10% NaOH DROPWISE
if precipitate forms +
then shake
40 DROPS water
+ 3M HCl
DROPWISE
1. HINSBERG TEST
Test samples Visible result Structure or formula of compound
responsible for the visible results
MethylamineClear light orange
with brown precipitate
C6H5SO2NR─Na+ → C6H5SO2NRH
Dimethylamine No change C6H5SO2NR2
TrimethylamineClear light yellow;
gelNR3 → 3RNH + Cl-
AnilinePrecipitate
formed; release of heat
-
N-methylanilineEvolution of white
smokeC6H5SO2NR─Na+ →
C6H5SO2NRH
• 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.
Primary amines: substance dissolves in a base and precipitates in an acid
Secondary amines: substance precipitates in a base and will have no change in the acid
Tertiary amines: substance precipitates in a base and dissolves in an acid
Primary amines give sulfonamides that are soluble in basic solution
Secondary amines give sulfonamides that are insoluble in basic solution
Tertiary amines do not form stable sulfonamides
• Aniline's aromaticity prevents efficient 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 effects.
2. NITROUS ACID TEST
+
3 DROPS sample
40 DROPS 2M HCl
cool in ice bath + 5 DROPS cold
20% NaNO2
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
Test samples Visible resultStructure or formula
of compound responsible for the
visible results
Methylamineevolution of colorless gas
bubblesN2
Dimethylaminelight orange, clear
solution(CH3)2N─N=O
Trimethylamine yellow; clear gas (CH3)3N+
Aniline
evolution of gas; yellow, brown
solution; release of heat
N2
N-methylanilinelight brown orange solution with gas
C6H5CH3N─N=O
• primary aliphatic amines give off nitrogen gas and a clear solution
• Primary amines form diazonium salt as a product. This product is very unstable and degrades into a carbocation that tends to react non-selectively with nucleophiles present in the solution.
• Upon reaction with nitrous acid, secondary 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 tertiary 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
H. TEST FOR CARBOXYLIC ACID AND ITS DERIVATIVES
1. FORMATION OF ESTERS2. HYDROLYSIS OF ACID DERIVATIVES
3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES
1. FORMATION OF ESTERS1.1 REACTION OF CARBOXYLIC ACID AND ALCOHOL
pinch salicylic acid
20 DROPS methanol
+ + 5 DROPS conc. H2SO4
shake well
5 mins
Test Sample Visible Result Structure responsible
Salicylic acid mint odor
• 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
1. FORMATION OF ESTERS1.2 SCHOTTEN-BAUMANN REACTION
+
+
20 DROPS water
10 DROPS ethanol
5DROPS benzoylchlorid
e
+ 20 DROPS
25% NaOH
mix
cover tube with cork & gently shake the mixture
Test Sample Visible Result
Benzoylchloride solid white precipitate (bottom)
smells like alcohol
Also known as “Reactions of Acylhalide 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
2. HYDROLYSIS OF ACID DERIVATIVES
2.1 HYDROYSIS OF BENZAMIDE
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 bath
TEST SAMPLES VISIBLE RESULTS
Benzamide red litmus to blue, burnt odor
• benzamide was hydrolyzed with the use of sodium hydroxide
• sodium benzoate and Ammonia was formed
2. HYDROLYSIS OF ACID DERIVATIVES
2.2 HYDROLYSIS OF AN ESTER
loosely cover the test tube with a cork and heat in water bath for 15 minutes
HCl (dropwise)
TEST SAMPLES VISIBLE RESULTS
Ethylacetate strong sour odor
This reaction is reverse of the esterification reaction. A
carboxylic acid and an alcohol are formed resulting to the
odor observed.
2. HYDROLYSIS OF ACID DERIVATIVES
2.3 HYDROLYSIS OF ANHYDRIDE
gently shake and feel the tube
TEST SAMPLES VISIBLE RESULTS STRUCTURE/FORMULA OF COMPOUND RESPONSIBLE FOR RESULT
Acetic anhydride blue litmus to red (CH3CO)2O + H2O → 2CH3COOH
• 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.
2o Amine
3o Amine
1o Amine
Reaction mechanism
3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES
3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES
Test Samples Visible Result Structure/ Formula of Compound
Responsible for Result
Ethylacetate blue litmus; odorless HXBenzamide pink litmus; odorless NH4
Acetic anhydride red litmus; acetic acid odor RCO2HBenzoylchloride blue litmus; alcohol odor ROH
Hydroxamic acid is a compound in which an amine is inserted into a carboxylic acid. In the test for RCOOH derivatives, esters, acid anhydrides and aryl/acyl halides would give positive results.
• purple to red solution and means a positive test, a dark brown solution is uncertain while a yellow solution means negative
When an ester, like ethyl acetate, is reacted with hydroxamic acid, it produces an alcohol.
When an acid anhydride, like acetic anhydride, is reacted with hydroxamic acid, it produces a carboxylic acid.
Acyl Halide
Ferric Hydroxamate Complex Formation
• purple to red solution and means a positive test, a dark brown solution is uncertain 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
What property of alcohol is demonstrated in the reaction with Na metal? What is the formula of the gas liberated?
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 between the alcohols and the Na metal. Why?
Because Na metal reacts with water that may cause ignition.
Why is the Lucas test not used for alcohols containing more than eight carbon atoms?
The Lucas test applies only to alcohols soluble in the Lucas reagent (monofunctional alcohols with less than 6 carbons and some polyfunctional alcohols). The long chains of C-bond atoms act as non-polar makes the hydroxyl group less functional. This results in the insolubility of the alcohol in the reagent and would make the test ineffective.
Explain why the order of reactivity of the alcohols toward Lucas reagent is 3°>2°>1°.
The reaction rate is much faster when the carbocation intermediate is more stabilized by a greater number of electron donating alkyl group bonded to the positive carbon atom.This means that the greater the alkyl groups present in a compound, the faster its reaction would be with the Lucas solution.
What functional group is responsible for the observed result in Millon’s test?
Hydroxyphenyl group or the phenolic –OH
Why is the Schiff’s test considered a general test for aldehydes?
This is because any aldehyde readily reacts with Schiff’s reagent to form positive results.Schiff’s reagent involves a bisulfite ion stuck in the original molecular structure. Aldehydes change this arrangement and thus there is a consequent change as the reaction progresses.
Why is it advantageous to use a strong acid catalyst in the reaction of aldehyde or ketone with 2,4-DNPH?
It is because a strong acid when used as a catalyst reverses the sequence of reactions. In the presence of a relatively weaker acid, the strong nucleophile attacks the substrate then the electrophile follows suit.
Whereas in the presence of a strong acid, the strong hydronium ion is more ready for protonation to the oxygen of the carbonyl group. The weaker nucleophile (which thrives in basic medium) then attacks the carbon to stabilize the forming hemiacetal. Water abstracts the H+ and a hemiacetal is formed. Hemiacetals are relatively less stable products that will form acetals and will not show the visible changes that are expected of the test.
Show the mechanism for the reaction of acetaldehyde with the following reagents:
a. 2,4-DNPH
b. NaHSO3
What structural feature in a compound is required for a positive iodoform test? Will ethanol give a positive iodoform test? Why or why not?
Show the mechanisms for the iodoform reaction using acetaldehyde as the test sample.
What test will you use to differentiate each of the following pairs? Give also the visible result.
a. acetaldehyde and acetone
Schiff’s test – reaction with acetaldehyde will result to a purple solution. Acetone on the other hand will not react.Tollen’s test – acetaldehyde will form a silver mirror. Acetone on the other hand will not have any reaction.
b. acetaldehyde and benzaldehyde
BIsulfite’s test – will differentiate an aliphatic aldehyde from an aromatic aldehyde.Aldehyde will react faster than benzaldehyde. Both will form a re precipitate due to cuprous oxide.
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