Expt 8 b report g&h

8 B CA
T
N IA LU
SI
ER .
IMALLO
R .F
P EG
X KIEN
E CO

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

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

Na Metal
• Alkali metal
• Sof t at room temperature
• Silver y white in color
• Highly reactive

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 of
Sec- butyl alcohol H2
gas
Slight to no evolution of
tert- butyl alcohol H2
gas

1.1 REACTIONS WITH Na METAL

Reaction with Metal Na
Alcohol(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

Ter t-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 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

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 + H2O
sec- butyl alcohol
formation
tert- butyl alcohol Fast layer formation (CH3)3CCl + H2O

1.2 LUCAS TEST

LUCAS TEST
Reaction:

speed of this reaction is propor tional to
the energy required to form the
carbocation

The cloudiness obser ved (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

Structural Formula
responsible for
Test Sample Visible Results results

n- butyl alcohol Blue- green solution Chromic Ion
Sec- butyl alcohol Blue- green solution Chromic Ion
tert- butyl alcohol Blue green Dichromate Ion

1.3 REACTIONS WITH K2Cr2O7

Rxn with K2Cr2O7
oxidation 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:

Rxn with K2Cr2O7

Secondary: oxidized to ketone
Ketone:

Tertiary: cannot be oxidized
Why? 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 acidsyellow solution;
• aromatic acidstan precipitate

Structural formula
Reddish-brown Iron (III) complex w/
Phenol
Phenol
Iron (III) complex w/
α - napthol Purple
Napthol
Cathechol Dark Blue
Cathechol
Moss Green
Resorcinol Resorcinol

2. REACTIONS OF PHENOLS
2.1 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
•Brominedark 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.

Structure formula

bromination of benzene
Phenol Turns pinkish, ppt ring
Turns to dark green, bromination of aromatic
α – napthol
ppt ring
Cathechol Dark brown, no ppt ring
Dark brown, no ppt
Resorcinol rings

2.2 BROMINE WATER TEST

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

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 group

2.3 MILLON’STEST

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.

F. TEST OF ALDEHYDES AND KETONES
1. 2,4-DNPH TEST 6. FEHLING’S TEST
2. BISULFITE TEST 7. MOLISCH TEST
3. SCHIFF’S TEST 8. BENEDICT’S TEST
4. TOLLEN’S TEST 9. BARFOED’S TEST

5. IODOFORM TEST 10. SELIWANOFF’STEST

F Ad Ac B

1. 2,4-DPNH TEST

Test samples Visible result
compound responsible for the
visible results

Formaldehyde Solid yellow precipitate

Clear yellow solution with
Acetaldehyde
orange precipitate

Yellow orange solution
Acetone with orange precipitate

Benzaldehyde Orange precipitate

1. 2,4-DPNH TEST

Structure or
formula of
Test samples Visible result compound
responsible for the
visible results

Formaldehyde no reaction

Acetaldehyde clear solution

Acetone clear solution

white precipitate at
Benzaldehyde (C5H6)CH(OH)SO3─N
the bottom
a+

2. BISULFIDE TEST

Structure or formula
of compound
responsible for the
visible results
dark violet solution Schiff’s reagent
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
Benzaldehyde royal blue solution complex with
methylphenol

3. SCHIFF’S TEST

F Ad Ac B

3. SCHIFF’S TEST

of compound
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

Before heating After heating

4. TOLLEN’S TEST

Test samples Visible result compound responsible for
the visible results

no reaction (clear
Formaldehyde NaOH [-]
solution)

yellow precipitate with
Acetaldehyde CHI3
strong odor

Acetone blurry precipitate CHI3
Benzaldehyde brown precipitate NaOH [-]

5. IODOFORM TEST

Test Samples Visible Result

Formaldehyde negative
Acetaldehyde 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 compound responsible for
the visible results
Glucose blue violet ring α-naphthol
Maltose blue violet ring α-naphthol
Sucrose blue violet ring α-naphthol
Boiled Starch blue violet ring α-naphthol

7. MOLISCH TEST

Structure or
formula of
Test 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 complex
Boiled Starch
solution with water [-]

8. BENEDICT’S TEST

Test samples Visible result compound responsible

Glucose 2 layers: blue over red cuprous oxide
Maltose clear blue solution cuprous oxide
clear top over blue
Sucrose Cuprous oxide
solution
Boiled Starch Aqua blue in color Cuprous oxide

9. BARFOED’S TEST

Test samples Visible result compound responsible

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
colored complex of
Boiled Starch pink orange

10. SELIWANOFF’STEST

G. TEST FOR AMINES

1. HINSBERG TEST

2. NITROUS ACID TEST

+ +
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

of compound
responsible for the
visible results
Clear light orange C6H5SO2NR─Na+ →
Methylamine with brown
C6H5SO2NRH
precipitate
Dimethylamine 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

• 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
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.

3 DROPS
sample

+ cool in ice bath
+ 5 DROPS cold
20% NaNO2

40 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


compound responsible
evolution of colorless
Methylamine N2
gas bubbles
light orange, clear
Dimethylamine (CH3)2N─N=O
solution
Trimethylamine yellow; clear gas (CH3)3N+
evolution of gas;
yellow, brown
Aniline N2
solution; release of
heat
light brown orange
N-methylaniline C6H5CH3N─N=O
solution with gas


• 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.

• 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

H. TEST FOR CARBOXYLIC
ACID AND ITS DERIVATIVES

1. FORMATION OF ESTERS

2. HYDROLYSIS OF ACID
DERIVATIVES

3. HYDROXAMIC ACID TEST
FOR ACID DERIVATIVES

pinch salicylic acid

+ + 5 DROPS conc. H2SO4
5 mins

shake well
20 DROPS methanol

Test Sample Visible Result Structure responsible

Salicylic acid mint odor

1.1 REACTION OF CARBOXYLIC ACID AND ALCOHOL

• 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

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
5DROPS
benzoylchloride

1.2 SCHOTTEN-BAUMANN REACTION

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

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
2.1 HYDROYSIS OF BENZAMIDE

• benzamide was hydrolyzed with
the use of sodium hydroxide
• sodium benzoate and Ammonia
was formed

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

2.2 HYDROLYSIS OF AN ESTER

This reaction is reverse of the
esterification reaction. A
carboxylic acid and an alcohol are
formed resulting to the odor
obser ved.

gently
shake
and feel
the tube

TEST SAMPLES VISIBLE RESULTSSTRUCTURE/FORMULA
OF COMPOUND
RESPONSIBLE FOR RESULT

Acetic anhydride blue litmus to red (CH3CO)2O + H2O → 2CH3COOH

2.3 HYDROLYSIS OF ANHYDRIDE

• 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.

Reaction mechanism

1o Amine

2o Amine

3o Amine

3. HYDROXAMIC ACID TEST FOR ACID DERIVATIVES

Test Samples Visible Result Structure/ Formula of Compound
Responsible for Result

Ethylacetate blue litmus; odorless HX
Benzamide pink litmus; odorless NH4
Acetic anhydride red litmus; acetic acid odor RCO2H
Benzoylchloride blue litmus; alcohol odor ROH

3. HYDROXAMIC ACID TEST FOR ACID
DERIVATIVES

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

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 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

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.

Expt 8 b report g&h

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