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EXPERIM
ENT
8BCO
KIENG
. FA
LLORIA
.
LUSICA
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
Na Test• Positive result is indicated by
evolution of gas
Test Sample Visible Results
Structure or formula of compound
responsible for the visible results
n- butyl alcohol Rapid evolution of gas
H2
Sec- butyl alcohol Moderate evolution of gas
H2
tert- butyl alcohol no evolution of gas H2
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 alcohol Cloudy 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
• Positive result yields the formation of layers
1. REACTIONS OF ALCOHOLS1.3 REACTIONS WITH K2Cr2O7
Test Sample Visible Results
Structural Formula
responsible for results
n- butyl alcohol3 layers formed: upper light-orange, middle dark- orange and bottom clear layer
Chromic Ion
Sec- butyl alcohol3 layers formed: upper green, middle orange and bottom blue-green layer
Chromic Ion
tert- butyl alcohol3 layers formed: upper orange, middle green and bottom orange layer
No rxn
20 DROPS N-BUTYLACLC
OHOL
ACIDIFY WITH 2M H2SO4
5DROPS 3%
K2Cr2O7
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 solution
Test Sample Visible Results Structural formula responsible for
results
Phenol Reddish-brown Iron (III) complex w/ Phenol
α - napthol Purple Iron (III) complex w/ Napthol
Cathechol Dark Blue Iron (III) complex w/ Cathechol
Resorcinol Moss Green Iron (III) complex w/ Resorcinol
2. REACTIONS OF PHENOLS2.1 FeCl3 TEST
20 DROPS 95%
ETHANOL
2 DROPS LIQUID SAMPLE
5 DROPS 3%FeCl3
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
20 DROPS 95% ETHANOL
20 DROP
S PHEN
OL
BROMINE WATE
RTest Sample Visible Results Structure formula responsible for results
Phenol Turns pinkish, pptbromination of benzene
ring
α – napthol Turns to dark green, ppt
bromination of aromatic ring
Cathechol Dark brown, no pptbromination of benzene
ring
Resorcinol Dark brown, no pptbromination 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
• Hg(NO3)2 in HNO3
• Only EGG ALBUMIN will give a positive result
2. REACTIONS OF PHENOLS2.3 MILLON’STEST
5 DROPS MILLON’
S REAGEN
T
5 DROPS
OF SAMPLE
• SHAKE• HEAT IN
WATER BATH (2MINS)
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 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.
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
2,4- DNPH Test
2,4-Dinitrophenylhydrazine (Brady's reagent)
used to qualitatively test for carbonyl groups
2,4- DNPH Test
A positive test is signaled by a yellow or orange precipitate (dinitrophenylhydrazone)
RR'C=O + C6H3(NO2)2NHNH2 → C6H3(NO2)2NHNCRR' + H2O
A condensation reaction
5 DROPS 2,4-DPNH
3 DROPS SAMPLE
•HEAT IN WATER BATH(5 MINS)
1. 2,4-DPNH TEST
F Ad Ac B
Test samples Visible result Structure or formula of compound responsible for the visible results
Formaldehyde Solid yellow precipitate
Acetaldehyde Clear yellow solution with orange precipitate
AcetoneYellow orange solution with orange precipitate
Benzaldehyde Orange precipitate
1. 2,4-DPNH TEST
2,4- DNPH Test
• formaldehyde, acetaldehyde, and benzaldehyde aldehydes
• Acetone ketone
• Produced a POSITIVE RESULT (orange coloration)
Bisulfite Test (NaHSO3)
• Also a test for aldehydes and ketones
• A positive result is indicated by the formation of precipitate
2. BISULFITE TEST
20 DROPS SODIUM BISULFIDE
5 DROPS SAMPLE
MIX. COOL IN AN ICE BATH
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Formaldehyde no reaction
Acetaldehyde Little white ppt H3C(OH)SO3- Na+
Acetone Yellow ppt H3CC(OH)SO3- Na+
Benzaldehyde white precipitate at the bottom
(C6H6)CH(OH)SO3- Na+
Bisulfite Test
General equation:
Formaldehyde: no reaction due to steric effects
Acetaldehyde: CH3CH=O + NaHSO3 (CH3)CH(OH)SO3-Na+
Acetone: CH3C=OCH3 + NaHSO3 (CH3)2C(OH)SO3-Na+
Benzaldehyde: + NaHSO3 (CH)5C(OH) SO3-Na+
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
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 purple solution
Schiff’s reagent complex with methylphenol
3. SCHIFF’S TEST
20 DROPS SCHIFF’S REAGENT
ADD
5 DROPS SAMPLE
F Ad Ac B
3. SCHIFF’S TEST
Tollen’s Test• Used to ascertain if compound is a ketone or an
aldehyde
• Tollen’s reagent ammoniacal silver nitrate solution
• Positive resutl is the depositon of silver metal to form the so called “silver mirror”
• Aldehyde silver mirror• Ketone no reaction
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Formaldehyde
Black solution with silver
substance (silver mirror)
Silver metal
Acetaldehyde With silver substance Silver metal
Acetone clear solution [no reaction] -
Benzaldehyde brown precipitate at the top
Silver metal
4. TOLLEN’S TEST
20 DROPS TOLLEN’S REAGENT
5 DROPS SAMPLE
•HEAT IN WATER BATH AND OBSERVE CHANGES
Before heating After heating
4. TOLLEN’S TEST
Iodoform Test
• Iodoform reagent is • a mixture of iodine (I2) and potassium
iodide (KI) and 10% NaOH
• Used to test for methylketones, ketones where one of the two alkyl groups bonded to the carbonyl carbon is a methyl group
•Positive result is a yellow iodoform formation
5. IODOFORM TEST
20 DROPS 10% NaOH +5 DROPS SAMPLE
ADD I2/KI W/ MIXING UNTIL BROWN COLOR
PERSIST
HEAT IN A WATER BATH @ 6O C (2 MINS)
ADD 10% NaOH (DROPWISE) MIX UNTIL BROWN COLOR DISAPPEARS -> YELLOW
ADD 5 DROPS DISTILLED WATER
SHAKE VIGOROUSLY AND STAND (15 MINS)
5. IODOFORM TEST
Test samples Visible resultStructure or formula of compound responsible for the visible results
Formaldehyde no reaction (clear solution) NaOH [-]
Acetaldehyde yellow precipitate with strong odor
CHI3
Acetone blurry yellowish precipitate
CHI3
Benzaldehyde Clear solution NaOH [-]
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 aldehydes and aldoses to carboxylic acids.
• The Cupric ion is reduced to cuprous oxide which then gives the red precipitate.
6. FEHLING’S TEST
20 DROPS
FEHLING’S
REAGENT
5 DROPS
OF SAMPLE
HEAT IN
WATER BATH
Test Samples Visible Result
Formaldehyde Red precipitate
Acetaldehyde Red precipitate
Acetone No reaction
Benzaldehyde No reaction
Molisch Test• For the presence of carbohydrates• based on the dehydration of the carbohydrates
by sulfuric acid
• Molisch Reagent is α-naphthol and conc. H2SO4
• A positive result yields a formation of blue violet ring
Molisch Test• Sugar is hydrolized to yield furfural
• Alpha-naphthol reacts with cyclic aldehyde to form a purple ring, a positive indication
• C6H12O6 + (conc.) H2SO4 → C5H4O2 + 3 H2O • C5H4O2 + 2 C10H8OH (α-naphthol) → colored
product
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
20 DROPS
MOLISCH
REAGENT
5 DROPS
OF SAMPL
E
INCLINE TUBE
AND RUN
THROUGH THE
SIDES 2 Ml
H2SO4
7. MOLISCH TEST
Benedict’s Test• Detect the presence of reducing sugars (sugars
with a free aldehyde or ketone group)
• Monosaccharides and some disaccharides are reducing sugarsfree reactive carbonyl group
• Other disaccharides such as sucrose and starchesnon-reducing sugars and will not react with Benedict's solution.
Benedict’s Test• Benedict’s Reagent is made up of
Sodium citrate; CuSO4; sodium bicarbonate
• Copper sulfate (CuSO4) present in Benedict's solution reacts with electrons from the aldehyde or ketone group of the reducing sugar to form cuprous oxide (Cu2O), a red-brown precipitate.
Benedict’s Test• Results:
Result What it means
No precipitate -
Green A trace
Yellow +
Orange ++
Red +++
Test samples Visible result
Structure or formula of compound
responsible for the visible
results
Glucosered precipitate
over yellow solution
cuprous oxide
Maltose green blue solution cuprous oxide
SucroseBlurry
precipitate over blue solution
copper complex with water [-]
Boiled Starch darker blue solution
copper complex with water [-]
8. BENEDICT’S TEST
20 DROPS Benedict’s
reagent
5 drops sample
• Heat in water bath
8. BENEDICT’S TEST
Barfoed’s test• Similar to Benedict's test, but determines if a
carbohydrate is a monosaccharide or a disaccharide
• Reacts with monosaccharides to produce cuprous oxide at a faster rate than disaccharides do
• Barfoed’s reagent is made up of • Cu(OAc)2 in HOAc
• Positive Result shows a formation of a red precipitate.
Barfoed’s test
• Reducing monosaccharides are oxidized by the copper ion in solution to form a carboxylic acid and a reddish precipitate of copper (I).
• Copper(II) acetate is reduced to copper(I) oxide (Cu2O), which forms a brick-red precipitate. The aldehyde group of the monosaccharide is oxidized to the carboxylate.
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Glucose Red precipitate cuprous oxide
Maltose clear blue solution
Sucrose clear top over blue solution
Boiled Starch Aqua blue in color
9. BARFOED’S TEST
20 DROPS Barfoed’s reagent
5 drops sample
•Heat in water bath
Seliwanoff’s Test• To differentiate between aldose and ketose sugars
• Seliwanoff’ reagent consist of resorcinol and conc. HCl
• Ketose sugar containing a ketone group; if the mixture turns red
• Aldose sugar containing an aldehyde group; mixture turns pink
Seliwanoff’s Test
• Positive result indicates the formation of a red precipitate and a pink solution.
Test samples Visible result
Structure or formula of compound
responsible for the visible results
Glucose very, very light orange
Maltose clear, light brown orange
Sucrose pink orangecolored complex of
furfural with resorcinol
Boiled Starch Light orange
10. SELIWANOFF’STEST
20 DROPS Seliwanoff’s reagent
5 drops sample
•Heat in water bath
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
Trimethylamine Clear light yellow; gel
NR3 → 3RNH + Cl-
AnilinePrecipitate
formed; release of heat
-
N-methylaniline Evolution of white smoke
C6H5SO2NR─Na+ → C6H5SO2NRH
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 result Structure or formula of compound
responsible for the visible results
Methylamineevolution of colorless gas
bubblesN2
Dimethylamine light 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-methylaniline light brown orange solution with gas
C6H5CH3N─N=O
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 white-yellow
solid precipitate
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
2. HYDROLYSIS OF ACID DERIVATIVES
2.1 HYDROYSIS OF BENZAMIDE
Pinch of Benzam
ide
20 drps 10%
NaOH
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
2. HYDROLYSIS OF ACID DERIVATIVES
2.2 HYDROLYSIS OF AN ESTER
20 DRPS Ethyl
acetate
5 drps 25%
NaOH
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
2. HYDROLYSIS OF ACID DERIVATIVES
2.3 HYDROLYSIS OF ANHYDRIDE
20 DRPS water
Red and blue litmus paper
20 drps Acetic
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
2o Amine
3o Amine
1o Amine
Reaction mechanism
3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES
5 drps Ethyl Acetate
20 drps 7% Methanolic
Hydroxylamine Hydrochloride
sol’n
+ KOHRed litmus
PaperBlue
Heat sol’n in water bath
10 drps 3% FeCl3
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
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.