Post Lab Notes

Organic Chemistry Laboratory, University of Santo Tomas

1 msaesmalla©2009

GO FOR EXCELLENCE!

COMPARATIVE INVESTIGATION OF ORGANIC COMPOUNDS

1. Physical state, color and odor - The physical appearance of an unknown will be your first datum in the search to

discover its identity. - Solid (amorphous or crystalline) or liquid.

2. Solubility Properties and Reaction with Litmus paper - Solubility of organic compounds in H2O indicates the polarity of the sample and the

intermolecular forces of attraction that exists between the sample and H2O. - Reaction with litmus paper indicates the acidity/basicity of the H2O-soluble

samples. - Red � Blue (Base) RBB - Blue � Red (Acid) BRA - Litmus paper that retains its color indicates a neutral compound. - Solubility of organic compounds in 5% HCl and/or 5%NaOH also reveals the acidity

and basicity of the sample. - The table below summarizes the solubility of different organic compounds in various

solvents.


2 msaesmalla©2009

GO FOR EXCELLENCE!

- The solubility of the sample in different solvents indicates the possible class of

organic compounds to where it belongs as shown in the figure.

• Try this: In one whole sheet of paper, make a summary table for the solubility of the organic compounds used in the experiment.

Which compounds are acidic? basic? Write the chemical equations of the compounds when reacted with H2O, HCl and NaOH.

REVIEW: Types of Intermolecular forces of attraction (IMFA) *Arranged in decreasing order of IMFA strength a. Ion-dipole – exist between ions and polar molecules. e.g. NaCl and H2O (Na+ and Cl- in H2O) b. H-bonding – exist between two polar molecules; a special type of dipole-dipole

interaction. - occurs when there is an –OH, NH and –F bond c. Dipole-dipole – exist between two polar molecules. d. Dipole-induced dipole – exist between polar and non- polar molecules. e. Van der Waals force – exist between two non- polar molecules.


3 msaesmalla©2009

GO FOR EXCELLENCE!

• Try this: Identify the IMFA exist on the reactions/chemical equations involved you just did in the earlier part.

3. Ignition Test – indicates the presence of unsaturation or high carbon to hydrogen ratio.

- �C/H ratio, � luminosity, � sooty - Degree of luminosity can be assessed by the presence of yellow flame and soot.

4. Infrared (IR) Analysis – it identifies the functional groups present in the sample. - It depends upon the interaction of IR light with the vibrating dipole moments of

molecules. - The molecular vibrations of the organic compound are promoted to higher energy

state (exhibited as bond stretching and/or bending modes) as it absorbs energy (IR radiation).

- The IR energy is measured in wave numbers (cm-1). THINGS TO CONSIDER: 1. The Coordinates

The x-axis indicates the wave numbers and examined most of the time. On the other hand, the y-axis describes the intensity of a given peak.

2. Peaks

Examining the peaks in IR spectra is the most important task at hand. Most of the significant peaks can be seen in the diagnostic region (from 4000 to 1500cm-1).

*The fingerprint region is from 1500 to around 400cm-1

.

3. Peak Quality

Certain functional groups are easy to identify from its peak quality. These include: 1) broad, 2)strong (intense but not wide), 3) weak (tiny), 4) sharp (slim) and 5) multiplet (overlapping peaks).

4. Diagnostic Peaks

Examining diagnostic peaks coupled with their molecular formula is enough to elucidate the gross structure of the molecule.


4 msaesmalla©2009

GO FOR EXCELLENCE!

CHARACTERISTICS INFRARED ABSORPTION FREQUENCIES

BOND COMPOUND TYPE FREQUENCY (cm-1)

C-H Alkanes 2960 - 2850 (s)

C-H Alkenes 3080 - 3020 (m)

Aromatic Rings 3100 - 3000 (m) C-H

Phenyl Ring Substitution Bands 870 - 675 (s) bend

C-H Alkynes 3333 - 3267 (s)

C=C Alkenes 1680 - 1640 (m,w)

CΞC Alkynes 2260 - 2100 (w,sh)

C=C Aromatic Rings 1600, 1500 (w)

Alcohols, Ethers, Carboxylic acids, C-O

Esters 1260 - 1000 (s)

Aldehydes, Ketones, Carboxylic C=O

acids, Esters 1760 -1670 (s)

Monomeric -- Alcohols, Phenols 3640 - 3160 (s,br)

H-bonded -- Alcohols, Phenols 3600 -3200 (b) O-H

Carboxylic acids 3000 - 2500 (b)

N-H Amines 3500 - 3300 (m)

C-N Amines 1340 - 1020 (m)

CΞN Nitriles 2260 - 2220 (v)

1660 - 1500 (s) NO2 Nitro Compounds

1390 - 1260 (s)

* v – variable, m - medium, s – strong, b r - broad, w - weak


5 msaesmalla©2009

GO FOR EXCELLENCE!

CLASSIFICATION TESTS FOR HYDROCARBONS

1. Physical state, color and odor - The physical appearance of an unknown will be your first datum in the search to

discover its identity. - Solid (amorphous or crystalline) or liquid

- Most hydrocarbons are colorless and odorless. However, many liquid compounds oxidize when they are stored for a long time. Often the oxidation products are intensely colored—yellow, green, red, brown, or black. (e.g. phenol becomes red upon oxidation)

2. Solubility in concentrated H2SO4 - Solubility of organic compounds in H2SO4 indicates whether the sample is a very

weak base (can be protonated) or a neutral compound (can’t be protonated). - The dissolution of compounds in H2SO4 may also produce large amounts of heat

and/or a change in the color of the solution, precipitation or any combination of these. (The reaction can be either violent or slow.)

*H2SO4-soluble (very weak base) Esters Ketones Alkenes Aldehydes Alcohols

*H2SO4-insoluble (neutral compound) Alkanes Aryl halides Alkyl halides most aromatic hydrocarbons

3. Ignition Test – indicates the presence of unsaturation or high carbon to hydrogen ratio.

- �C/H ratio, � luminosity, � sooty - Degree of luminosity can be assessed by the presence of yellow flame and soot.

- Aromatic compounds burn with sooty flame due to the incomplete combustion which causes the formation of an unburned carbon.

- In terms of degree of luminosity, aromatic compound > unsaturated hydrocarbon > saturated hydrocarbon.

*Complete combustion is indicated by a blue flame (non-luminous) and there is more

heat than light; the carbon is completely oxidized.

*Incomplete combustion is indicated by a yellow flame (luminous) and there is much

light than heat; the carbon is not completely oxidized.

x y 2 2 2C H + O CO + H O→

x y 2 2 2C H + O CO + CO + C (soot) + H O →


6 msaesmalla©2009

GO FOR EXCELLENCE!

4. Test for Active Unsaturation

a. Baeyer’s Test – test for double bonds Reagents: 2% KMnO4

(+) result: decolorization of a purple solution; formation of a brown ppt. (MnO2) - involves redox reaction - Mn7+ is reduced to Mn4+; alkene is oxidized to a diol. - Alkenes react with potassium permanganate (KMnO4) to give a diol and MnO2. - Aromatic compounds do not react because of their stability. Cyclohexene + KMnO4 �� 1,2-cyclohexanediol + MnO2

(purple) (colorless) (brown) • Try this: Draw the structure of cyclohexene and 1,2-cyclohexanediol.

b. Bromine Test – test for double bonds Reagents: 0.5% Br2 in CCl4 (+) result: decolorization of an orange solution - Involves electrophilic addition reaction - Alkenes react with Br2 to form a trans-dibromoalkane. - Aromatic compounds do not react because of their stability. - However, aromatic compounds will react slowly upon using FeBr3 or through the

action of UV light. • Challenge: Write the reaction mechanism for the bromine test on cyclohexene.

5. Test for Aromaticity: Nitration – test for aromatic compounds Reagents: HNO3, H2SO4

(+) result: yellow globule/yellow oily layer - involves electrophilic substitution reaction - H2SO4 acts as a catalyst and facilitates the formation of nitronium ion (NO2

+), an electrophile

- One hydrogen atom in the benzene ring is substituted by the nitronium ion.

HNO3 + H2SO4 �� NO2

+ + 2HSO4- + H3O

+

Benzene + NO2+ �� nitrobenzene


7 msaesmalla©2009

GO FOR EXCELLENCE!

• Challenge: Write the reaction mechanism for the nitration test on benzene and toluene.

6. Basic Oxidation – test for alkylated aromatics or arenes

Reagents: 2% KMnO4, 10% NaOH

(+) result: green solution (MnO4)/brown ppt (MnO2) - involves redox reaction - NaOH provides a basic environment. - The alkyl group of the aromatic compound is oxidized to a carboxylic acid. (Reaction

occurs with 10 and 20 alkyl side chain, but not with 30.) - Mn7+ is reduced to Mn6+ or Mn4+ depending on the extent of the reaction.

4 4methylbenzene + KMnO benzoic acid + MnO→


8 msaesmalla©2009

GO FOR EXCELLENCE!

CLASSIFICATION TESTS FOR ORGANIC HALIDES

1. Belstein Test - a quick preliminary check for halogens (+) result: blue-gree flame

- The simplest method for establishing presence of a halogen, but does not positively differentiate between Cl, Br, I

- The blue-green color is due to the emission of light from excited states of copper halide that has vaporized in the burner flame.

- Heating the copper wire before the test is carried out removes traces of sodium chloride that may be present on the wire from handling it with the fingers.

- Reactions in Belstein test - CuX2 is volatile and imparts a blue-green flame.

2. Reaction with Alcoholic AgNO3 - test for SN1 Reactivity Reagents: 2% ethanolic AgNO3

(+) result: white ppt * SN1 - substitution, nucleophilic, unimolecular - The kinetics of the reaction depends only on the alkyl halide. �� Substrate Effect: - The more stable the carbocation intermediate, the faster the SN1 reaction. - 30 > 20 > 10> -CH3

�� Leaving Group: - good leaving group is needed - TosO- > I- > Br- > Cl- ≈ H2O more reactive less reactive

�� Nucleophile:

- hardly affected - The nucleophile does not enter into the reaction until after rate-limiting dissociation

has occurred; thus cannot affect the reaction rate.

�� Solvent: - Polar, protic solvent - Polar solvent stabilize the carbocation intermediate by solvation, thereby increasing

the reaction rate. - H2O > 80% EtOH > 40% EtOH > EtOH more reactive less reactive

(EtOH = ethanol)

[ ]reaction rate = RXk

0

2

2 2 2

Cu + O CuO

loop black solid

RX + CuO CuX + CO + H O

→

→


9 msaesmalla©2009

GO FOR EXCELLENCE!

�� Stereochemistry: - Includes inversion and retention

*The SN1 reaction occurs when the substrate spontaneously dissociates to a carbocation in a slow rate-limiting step, followed by a rapid attack of nucleophile. As a result,SN1 reactions show first-order kinetics and take place with RACEMIZATION of configuration at the carbon atom. They are most favored for tertiary substrates.

3. Reaction with NaI in Acetone – test for SN2 Reactivity

Reagents: 15% NaI in acetone

(+) result: white ppt (insol. In acetone) * SN2 - substitution, nucleophilic, bimolecular - The kinetics of the reaction depends only on the alkyl halide and nucleophile. �� Substrate Effect: - The more stable the carbocation intermediate, the faster the SN1 reaction. - -CH3 > 10 > 20 > 30 (due to steric effect)

�� Leaving Group: - good leaving group is needed - TosO- > I- > Br- > Cl- > F- > HO- , H2N

- , RO- more reactive less reactive

�� Nucleophile:

- Strong nucleophile works best. - Nucleophlicity parallels basicity. - Nucleophilicity usually increases going down a column of the periodic table.

e.g. HS- is more nucleophilic than HO-

I- > Br- > Cl-

�� Solvent:

- Polar, aprotic solvent - HMPA > CH3CN > DMF > DMSO > H2O > CH3OH

more reactive less reactive

�� Stereochemistry:

- Involves Walden inversion of configuration

*The SN2 reaction occurs as the entering nucleophile attacks the halide 1800 away from the leaving group, resulting in an umbrella-like Walden inversion of configuration at the carbon atom. The reaction shows second-order kinetics and is strongly inhibited by increasing steric bulk of the reagents. Thus, SN2 reactions are favored for primary and secondary substrates.

[ ] -reaction rate = RX Nu:k


10 msaesmalla©2009

GO FOR EXCELLENCE!

CLASSIFICATION TESTS FOR HYDROXYL- AND CARBONYL-CONTAINING COMPOUNDS

1. Solubility of alcohols in water

a. Alcohols are insoluble in water except under C6. b. Factors affecting solubility:

i. number of carbon atoms c. the higher the number of carbon atom, the more insoluble or less soluble d. e.g. ethanol (CH3CH2OH) is more soluble than butanol (CH3CH2CH2CH2OH)

b. branching of carbon chain - the more branching present, the more soluble (with the same number of

carbons) - e.g. tert-butanol > sec-butanol > n-butanol c. presence of polar functional groups ( -OH, -NH2, -CO2H) - compound with polar functional group is more soluble e.g. butanol > butane; 1,3-butanediol > butanol

2. Lucas Test – differentiates 10, 20, 30 alcohols Reagents: anhydrous ZnCl2, HCl (+) result: based on turbidity (alkyl chloride formation); the rate of the reaction was observed (time is noted) - based on SN1 reaction (30 > 20 > 10 alcohols); depends on the formation of a stable

carbocations - 30 alcohols form the second layer in less than a minute. - 20 alcohols require 5-10 minutes. - 10 alcohols are usually unreactive. - The presence of ZnCl2 (a good LEWIS ACID) makes the reaction mixture even more

acidic, thus enhances the formation of carbocations.

3. Chromic Acid Test/Dichromate Test/Jones Test – test for oxidizables or any compounds that possess reducing property (has an alpha acidic hydrogen); 10, 20 alcohols and aldehydes give a (+) visible result. Reagents: 10% K2Cr2O7, 6M H2SO4 (+) result: green or blue-green solution - involves redox reaction; 10, 20 alcohols and aldehydes undergoes oxidation,

chromium undergoes reduction (from Cr+6 to Cr+3). - A 10, 20 alcohols and aldehydes will reduce the orange-red chromic acid/sulfuric acid

reagent to an opaque green or blue suspension of Cr(III) salts in 2-5 s.


11 msaesmalla©2009

GO FOR EXCELLENCE!

- A 10 alcohol reacts with chromic acid to yield aldehydes, which further oxidized to carboxylic acid.

- A 20 alcohol react with chromic acid to yield ketones, which do not oxidize further. - A 30 alcohols are usually unreactive. - Aldehydes are oxidized to carboxylic acid.

4. DNPH Test/2,4-DNP– test for carbonyl groups; positive for aldehydes and ketones Reagents: 2,4-dinitrophenylhydrazine, ethanol, H2SO4 (+) result: red-orange ppt (conjugated carbonyl compounds) or yellow ppt (nonconjugated carbonyl compounds) - Some high molecular weight ketones may fail to react or may yield oils. - Mechanism: condensation or addition/elimination - Involves nucleophilic addition of NH2 to C=O and elimination of H2O - Most aromatic aldehydes and ketones produce red dinitrophenylhydrazone. - Many nonaromatic aldehydes and ketones produce yellow products. - The reaction of 2,4-DNPH with aldehydes and ketones in an acidic solution is a

dependable and sensitive test.

5. Fehling’s Test - test for aldehydes

Reagents: CuSO4, NaOH ( Cu2+ in alkaline solution) (+) result: brick-red ppt (Cu2O/cuprous oxide) - Involves redox reaction. - Aldehyde is oxidized to carboxylic acid; ketones do not undergo oxidation. - Copper is reduced (from Cu2+ to Cu1+)

RCOH + 2Cu2+ + 5-OH � RCOO- + Cu2O + 3H2O

6. Tollens Test/Silver Mirror Test – test for aldehydes

Reagents: AgNO3, NH3 (+) result: silver mirror


12 msaesmalla©2009

GO FOR EXCELLENCE!

- The preparation of Tollens reagent is based on the formation of a silver diamine complex that is water soluble in basic solution.

- Involves redox reaction. - Aldehyde is oxidized to carboxylic acid; ketones do not undergo oxidation except

alpha-hydroxyketone. - Silver is reduced (Ag1+ to Ag0). - Formic acid and hydroxylamine will also give a (+) result.

7. Iodoform Test – test for methyl carbinol (20 alcohol with adjacent methyl group) and methyl carbonyl groups Reagents: 10% KI, NaClO (+) result: yellow crystals or ppt (CHI3 m.pt. 119-1210C) - An alkaline solution of sodium hypoiodite, formed from sodium hydroxide and iodine,

will convert acetaldehyde and aliphatic methyl ketones into iodoform (haloform reaction).

- Since the reagent is also an oxidizing agent, alcohols which are readily oxidized to acetaldehyde or methyl ketones also give a (+) reaction.

- The mechanism of iodoform synthesis occurs through a series of enolate anions, which are iodinated; hydroxide diplaces the Cl3

- anion though an addition/elimination pathway.


13 msaesmalla©2009

GO FOR EXCELLENCE!

CLASSIFICATION TESTS FOR CARBOXYLIC ACIDS AND DERIVATIVES

1. Hydrolysis of Acid Derivatives a. Acyl Halides

- Reagent: H2O - Acyl halides react with water to yield carboxylic acids. - This hydrolysis reaction is a typical nucleophilic acyl substitution process, initiated by

the attack of water on the acyl halide carbonyl group. - The tetrahedral intermediate undergoes elimination of Cl- and loss of H+ to give the

product carboxylic acid and HCl. - Sample Compounds: acetyl chloride (+)

CH3COCl + H2O �� CH3COOH + HCl

- The above reaction is accompanied by a warming effect.* - Other reactions performed on the resulting mixture of the above reaction: 1. Hydrolysis with Aqueous Silver Nitrate

- Reagent: 2% AgNO3 - (+) result is indicated by the immediate formation of the ppt of silver halide - Acid halides, halide salts and 2,4-dinitroaromatic halides give immediate ppt.

2. Solubilty in Bicarbonate

- Reagent: saturated NaHCO3 - Effervescence due to the evolution of CO2 (formation of bubbles) indicated the

presence of carboxylic acid. - All water-soluble compounds will dissolve in the bicarbonate solution, but only acids

will give bubbles.

b. Acid Anhydrides - The chemistry of acid anhydrides is similar to that of acyl halides. - The kinds of the reactions the two groups undergo are the same, though acid

anhydrides react more slowly. - Acid anhydrides also undergo hydrolysis to form a carboxylic acid. - CH3COOCOCH3 + H2O �� 2 CH3COOH - For other reactions, just refer to the acyl halides. - Sample compound: acetic anhydride (+)

c. Esters - Reagents: 25% NaOH, 10% HCl

- Basic hydrolysis of an ester converts an ester into the carboxylate salt of the parent acid and the alcohol from which the ester was formed.

- Acidification of the carboxylate solution with HCl leads to the recovery of the parent acid.


14 msaesmalla©2009

GO FOR EXCELLENCE!

- Often you will find that either the carboxylic acid or the alcohol formed from the hydrolysis of an ester is extremely soluble in H2O.

- Sample compound: ethyl acetate (+)

CH3COOCH2CH3 + NaOH �� CH3COO-Na+ + CH3CH2OH CH3COO-Na+ + HCl �� CH3COOH + NaCl

- Evidence for an ester is indicated by: 1) the disappearance of organic layer |(if any), 2) odor of the sample (during heating), and 3) appearance of a ppt or odor of a carboxylic acid.

d. Amides - Reagent: 10% NaOH

- A red litmus paper turned to blue accompanied by an ammonia or amine-like odor, indicates an amide.

- Amides of higher amines that do not turn the litmus paper blue may nevertheless give an amine-like odor.

- Some amides will yield a precipitate or a separated liquid phase (the carboxylic acid).

RCONR’2 + NaOH �� RCOO-Na+ + R2’NH RCOO-Na+ + H+ �� RCOOH + Na+

- Sample compound: benzamide (+) • Try this: Write the chemical equation for the hydrolysis of benzamide.

2. Alcoholysis: Schotten-Baumann Reaction a. Carboxylic acid

- Reagents: ethanol, conc’d H2SO4

- Involves Fischer esterification reaction; this is a nucleophilic acyl substitution reaction carried out under acidic conditions

- Strong mineral acid such as H2SO4 or HCl make the carboxylic acid more reactive enough to be attacked by alcohol.

- The net effect of Fischer esterification is substitution of an –OH group by OR’. - All steps are reversible. - Sample compound: acetic acid (+)

- Reaction: 2 4

3 2

H SO

3 3 2 3 3CH CH OHCH COOH CH COOCH CH + H O+

←→

- Evidence of reaction: heat, HCl gas, phase separation, ester-like odor


15 msaesmalla©2009

GO FOR EXCELLENCE!

b. Acyl halide/Acid anhydride - Reagents: ethanol, H2O, 20% NaOH - The esterificaion reaction of an alcohol with acid chloride is strongly affected by

steric hindrance. - Bulky groups on either partner slow down the reaction (reactivity: 10> 20> 30 ROH) - Alcoholysis reactions are usually carried out in the presence of NaOH or pyridine to

react with the HCl formed and prevent it from causing side reactions. - On the other hand, only half of the acid anhydride molecule is used; the other half

acts as the leaving group during the nucleophilic acyl substitution reaction. - Thus, acid chlorides are preferred for introducing acyl substituents. - Sample compounds: acetyl chloride (+)

acetic anhydride (+)

- Reaction: CH3COCl + CH3CH2OH �� CH3COOCH2CH3 + HCl CH3COOCOCH3 + CH3CH2OH �� CH3COOCH2CH3 + CH3COOH - Evidence of reaction: heat, HCl gas, phase separation, ester-like odor

3. Aminolysis: Anilide Formation - Reagents: aniline, H2O - Acyl halides react rapidly with ammonia or amines to give amides in good yield - Acid anhydride also react but it takes a longer time - Both mono- and disubstiuted amines can be used but not trisubstituted amines. - Since HCl is formed durng the reaction, 2 equivalents of the amine must be used (1

eq. reacts with the acid chlorides and another eq. reacts with the HCl by-product) - Sample compounds:

a. Acetyl chloride (+) b. Acetic anhydride (+)

- A (+) test is indicated by the formation of a ppt. • Challenge: Write the reaction mechanism for the aminolysis of acetyl chloride and

acetic anhydride.

4. Hydroxamic Acid/Ferric Hydroxamate Test – test for esters

- Reagents: hydroxylamine hydrochloride (NH2OH•HCl), 1M KOH, 5% FeCl3

- Preliminary test is done to eliminate those phenols and enols that give colors with ferric chloride in acidic solution and that would therefore give a false-positive result in the ferric hydroxamate test.

- If a color other than yellow results, the ferric hydroxamate test cannot be used. - A (+) test is indicated by the formation of a blue-red (burgundy or magenta) color.


16 msaesmalla©2009

GO FOR EXCELLENCE!

- Esters react with hydroxylamine in basic solution to form hydroxamic acids, which in turn react with ferric chloride in acidic solution to form bluish-red ferric hydroxamates.

- Sample Compounds:

a. Ethyl acetate (+) b. Acetamide (-)

• Try this: Write the chemical equation for the hydroxamic acid test of ethyl acetate.

*IMPORTANT POINTS TO CONSIDER

- The chemistry of all acid derivatives is similar and dominated by a nucleophilic acyl substitution reaction:

RCOY + :Nu- �� RCONu + :Y-

- Nucleophilic acyl substitution reactions take place in two steps: 1) addition of the

nucleophile, and 2) elimination of a leaving group. - Any factors that make the carbonyl group more easily attacked by nucleophile favor

the reaction. - Steric and electronic factors affect the reactivity of the acid derivative towards

nucleophilic acyl substitution reactions. - Steric factors: Unhindered, accessible carbonyl groups react with nucleophiles more

readily than do sterically hindered groups. - e.g. CH3COCl > (CH3)3CCOCl - Electronic factors: strongly polarized acid derivatives are attacked more readily than

less polar ones. - Reactivity order: RCOCl (Acid halide) > RCO2COR (Acid anhydride) > RCOOR

(Ester) > RCONH2 (Amide) - RULE: A more reactive derivative will be transformed to a less reactive one. - The same kind of reactions occur on acid derivatives: 1) Hydrolysis – reaction with water to yield a carboxylic acid. 2) Alcoholysis – reaction with alcohol to yield an ester.

3) Aminolysis – reaction with ammonia or amine to yield an amide. *Other reactions include reduction and Grignard reaction.


17 msaesmalla©2009

GO FOR EXCELLENCE!

CLASSIFICATION TEST FOR AMINES

1. Hinsberg’s Test – differentiates 10, 20, and 30 amines - Reagents: 10% NaOH, benzene sulfonyl chloride, 6M HCl - Formation of a white ppt (p-toluenesulfonamide) when the reaction mixture is

acidified indicates a 10 amine. - *Most 10 amine yield a clear solution after the initial reaction, but some form sodium

salts or disulfonyl derivatives that precipitate during the reaction. RNH2 + ArSO2Cl + 2NaOH �� ArSO2NR-Na+ + NaCl + 2H2O ArSO2NR-Na+ + HCl �� ArSO2NHR + NaCl - Most 20 amine yield a white solid that does not dissolve in H2O or 6M HCl. - *A liquid residue that is more dense than H2O and insoluble in 6M HCl may be a 20

amine’s arenesulfonamide that has failed to crystallize. R2NH + ArSO2Cl + NaOH �� ArSO2NR2 + NaCl - 30 amines should not react. - *Water-soluble 30 amines yield a clear solution that does not form a separate phase

on acidification. R3N + ArSO2Cl �� no reaction R3N + HCl �� R3NH+Cl-

- Sample compounds: Aniline N-methylaniline N,N-dimethylaniline

2. Nitrous acid test – differentiates 10 from 20 and 30 amines; also distinguish alkyl amines from aromatic amines

*The chemistry of amines is dominated by the lone pair of electrons in nitrogen. The presence of this lone pair makes amine both basic and nucleophilic. - They react with acids to form acid-base salts. Also, they react with electrophiles in many of the polar reactions.


18 msaesmalla©2009

GO FOR EXCELLENCE!

References: Mayo, D., Pike, R., and Trumper, P. (2000). Microscale Organic Laboratory with Multistep Synthesis. John Wiley and Sons. McMurry, J. (2000). Organic Chemistry. Brooks/Cole: NY Garcia, C. (2005), Organic Chemistry Laboratory Manual

Post Lab Notes

Documents

Transcript of Post Lab Notes