Expt 8 b report g&h

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



Formaldehyde no reaction

Acetaldehyde clear solution

Acetone clear solution

Benzaldehydewhite precipitate

at the bottom(C5H6)CH(OH)SO3

─Na+




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




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




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



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




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




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


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


2.2 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.


2.3 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



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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Expt 8 b report g&h