expt 8 b report 2007 til seliwanoff test 1

EXPERIM

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

Sec- butyl alcohol Moderate evolution of gas

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

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

BROMINE WATE

RTest Sample Visible Results Structure formula responsible for results

Phenol Turns pinkish, pptbromination of benzene

α – napthol Turns to dark green, ppt

bromination of aromatic ring

Cathechol Dark brown, no pptbromination of benzene

Resorcinol Dark brown, no pptbromination of benzene

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

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

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

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

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

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

Acetone blurry yellowish precipitate

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

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

INCLINE TUBE

AND RUN

THROUGH THE

SIDES 2 Ml

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 +++

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.

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.

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

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

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

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

+ 20 DROPS

25% NaOH

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

20 drps 10%

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

2.2 HYDROLYSIS OF AN ESTER

20 DRPS Ethyl

acetate

5 drps 25%

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

2.3 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

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

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.

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