Post Lab Notes

Organic Chemistry Laboratory, University of Santo Tomas

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 ident i dentity. ity. - Solid (amorphous or crystalline) or liquid.

2. Solubility Properties and Reaction with Litmus paper - Solubility of organic compounds compounds in H 2O indicates the polarity of the sample and the intermolecular forces of attraction that exists between the sample and H 2O. - Reaction with 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 indicate s a neutral compound. neutral compound. - Solubility Solubilit y of organic organic compounds compounds in 5% HCl HCl and/or and/or 5%NaOH also reveals the acidity and basicity of the sample. - The table table below summarizes the solubility of different organic compounds in various solvents.

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The solubility of the sample in different solvents indicates the possible class of organic compounds to where it belongs as shown in the figure.

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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 H 2O, HCl and NaOH.

REVIEW: Types of Intermolecular forces of attraction (IMFA) *Arranged in decreasing order of IMFA strength

a. Ion-dipole – Ion-dipole – exist between ions and polar molecules. e.g. NaCl and H 2O (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 – Dipole-dipole – exist between two polar molecules. d. Dipole-induced dipole – exist between polar and non- polar molecules. e. Van der Waals force – force – exist between two non- polar molecules. 2


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

3. Ignition Test – Test – indicates the presence of unsaturation or 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 – Analysis – it identifies i dentifies the functional groups present groups present in the sample. - It depends upon upon the interaction of IR light with the vibrating dipole dipole moments of molecules. - The molecular 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 ). -1

*The fingerprint region is from 1500 to around 400cm .

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.

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CHARACTERISTICS INFRARED ABSORPTION FREQUENCIES -1 BOND COMPOUND TYPE FREQUENCY (cm ) 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) Alkynes 2260 - 2100 (w,sh) CΞC C=C Aromatic Rings 1600, 1500 (w) Alcohols, Ethers, Carboxylic acids, C-O 1260 - 1000 (s) Esters Aldehydes, Ketones, Carboxylic C=O 1760 -1670 (s) acids, Esters Monomeric -- Alcohols, Phenols 3640 - 3160 (s,br) O-H H-bonded -- Alcohols, Phenols 3600 -3200 (b) Carboxylic acids 3000 - 2500 (b) N-H Amines 3500 - 3300 (m) C-N Amines 1340 - 1020 (m) CΞN Nitriles 2260 - 2220 (v) NO2

* v – variable, m -

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Nitro Compounds

1660 - 1500 (s) 1390 - 1260 (s)

medium, s – strong, strong, b r - broad, w - weak


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CLASSIFICATION 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 ident i dentity. ity. - Solid (amorphous or crystalline) or liquid - Most hydrocarbons hydrocarbon s 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 H 2SO4 - Solubility Solubilit y of organic compounds in H 2SO4 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 (The reaction can be either violent or slow .) 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 – Test – indicates the presence of unsaturation or 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 flame due to the incomplete combustion which causes the formation of an unburned carbon. - In terms of degree of luminosity, luminosi ty, aromatic compound 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. C xHy + O 2

→

CO 2 + H2O

*Incomplete combustion is indicated by a yellow flame (luminous) and there is much light than heat; the carbon is not completely oxidized. C xHy + O 2

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→

CO 2 + CO + C (soot) + H 2O


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4. Test for Active Unsaturation a. Baeyer’s Test – test for double bonds Reagents: 2% KMnO 4 (+) result: decolorization of a purple solution; formation of a brown ppt. (MnO 2) - involves redox reaction - Mn7+ is reduced to Mn 4+; alkene is oxidized to a diol. - Alkenes react with potassium permanganate (KMnO 4) to give a diol and MnO 2. - Aromatic compounds do not react because of their stability. Cyclohexene + KMnO4  1,2-cyclohexanediol + MnO2 (purple) (colorless) (brown)

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Try this: Draw the structure of cyclohexene and 1,2-cyclohexanediol.

b. Bromine Test – test for double bonds Reagents: 0.5% Br 2 in CCl 4 (+) result: decolorization of an orange solution - Involves electrophilic addition reaction - Alkenes react with Br 2 to form a trans -dibromoalkane. -dibromoalkane. - Aromatic compounds do not react because of their stability. - However, aromatic compounds will react slowly upon using FeBr 3 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: HNO 3, H2SO4 (+) result: yellow globule/yellow oily layer - involves electrophilic substitution reaction - H2SO4 acts as a catalyst and facilitates the formation of nitronium ion (NO 2+), an electrophile - One hydrogen atom atom in the benzene ring is substituted substituted by the nitronium nitronium ion. HNO3 + H2SO4  NO2+ + 2HSO 4- + H3O+ Benzene + NO2+  nitrobenzene 6


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

6. Basic Oxidation – Oxidation – test for alkylated aromatics or arenes Reagents: 2% KMnO 4, 10% NaOH (+) result: green solution (MnO 4)/brown ppt (MnO 2) - involves redox reaction - NaOH provides a basic environment. - The alkyl group of the aromatic compound is oxidized oxidized to a carboxylic acid. (Reaction 0 0 occurs with 1 and 2 alkyl side chain, but not with 3 0 .) - Mn7+ is reduced to Mn 6+ or Mn4+ depending on the extent of the reaction. methylb methylbenz enzene ene + KMnO KMnO 4

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→

benzoic benzoic acid + MnO 4


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CLASSIFICATION CLASSIFICATION TESTS FOR ORGANIC HALIDES 1. Belstein Test - a quick preliminary check for halogens (+) result: blue-gree flame - The simplest simplest method for establishing presence of a halogen, 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 Cu0 + O2 loop RX + CuO

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CuO

→

black solid →

CuX 2 + CO 2 + H2O

CuX2 is volatile and imparts a blue-green flame.

2. Reaction with Alcoholic AgNO 3 - test for SN1 Reactivity Reagents: 2% ethanolic AgNO 3 (+) result: white ppt * SN1 - substitution, substitution , nucleophilic, unimolecular - The kinetics of the reaction depends only on the alkyl halide. reaction rate = k [ RX]  Substrate Effect:

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The more stable the carbocation intermediate, intermediat e, the faster the S N1 reaction. 0 0 0 3 > 2 > 1 > -CH3

 Leaving Group:

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good leaving group is needed TosO- > I- > Br- > Cl- ≈ H2O more reactive

less reactive

 Nucleophile:

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hardly affected The nucleophile nucleophile does not enter into the reaction until after rate-limiting dissociation has occurred; thus cannot affect cannot affect the reaction rate.

 Solvent:

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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) 8


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 Stereochemistry:

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Includes inversion and inversion and retention

* The The S N N 1 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,S N N1 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 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. reacti reaction on rate rate = k [RX] Nu:-   Substrate Effect:

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The more stable the carbocation intermediate, intermediat e, the faster the S N1 reaction. 0 0 0 -CH3 > 1 > 2 > 3 (due to steric effect)

 Leaving Group:

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good leaving group is needed TosO- > I- > Br- > Cl- > F- > HO- , H2N- , ROmore reactive

less reactive

 Nucleophile:

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Strong nucleophile works best. Nucleophlicity Nucleophlicit y parallels basicity. Nucleophilicity Nucleophili city usually increases going going down down a column of the periodic table. table. - - e.g. HS is more nucleophilic than HO I - > Br - > Cl -

 Solvent:

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Polar, aprotic solvent HMPA > CH3CN > DMF > DMSO > H2O > CH3OH more reactive

less reactive

 Stereochemistry:

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Involves Walden inversion of inversion of configuration

0 * The The S N N 2 reaction occurs as the entering nucleophile attacks the halide 180 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, S N 2 reactions are favored for primary and secondary substrates.

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CLASSIFICATION CLASSIFICATION TESTS T ESTS FOR HYDROXYL- AND CARBONYL-CONTAINING COMPOUNDS 1. Solubility of alcohols in water a. Alcohols are insoluble in water water except under C 6. b. Factors affecting solubility: i. number of carbon atoms c. the higher the number of carbon atom, the more insoluble insolubl e or less soluble soluble d. e.g. ethanol ethanol (CH 3CH2OH) is more soluble than butanol (CH 3CH2CH2CH2OH) b. branching of carbon chain - the more more branching present, the more soluble (with the same same number number of carbons) - e.g. tert-butanol > sec-butanol > n-butanol c. presence of polar functional groups ( -OH, -NH 2, -CO2H) - compound with polar functional group is more soluble e.g. butanol > butane; 1,3-butanediol > butanol 2. Lucas Test – differentiates 1 0, 20, 30 alcohols Reagents: anhydrous ZnCl 2, HCl (+) result: based on turbidity (alkyl chloride formation); the rate of the reaction was observed (time is noted) - based on S N1 reaction (3 0 > 20 > 1 0 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. unreactive. - The presence of ZnCl 2 (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); 1 0, 20 alcohols and aldehydes give a (+) visible result. Reagents: 10% K 2Cr2O7, 6M H2SO4 (+) result: green or blue-green solution - involves redox reaction; 1 0, 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. 10


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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, H 2SO4 (+) result: red-orange ppt (conjugated carbonyl compounds) or yellow ppt (nonconjugated carbonyl compounds) - Some high molecular molecular weight ketones may fail to to react or may yield oils. - Mechanism: condensation or addition/elimination addition/elimina tion - Involves nucleophilic addition of NH 2 to C=O and elimination of H 2O - Most aromatic aldehydes and ketones produce red dinitrophenylhydrazon dinitrophenylhydrazone. e. - 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: CuSO 4, NaOH ( Cu 2+ in alkaline solution) (+) result: brick-red ppt (Cu 2O/cuprous oxide) - Involves redox reaction. - Aldehyde is oxidized to to carboxylic acid; ketones do not undergo oxidation. oxidation. 2+ 1+ - Copper is reduced (from Cu to Cu ) RCOH + 2Cu 2+ + 5-OH  RCOO- + Cu2O + 3H2O 6. Tollens Test/Silver Test/Sil ver Mirror Test – test for aldehydes Reagents: AgNO 3, NH3 (+) result: silver mirror 11


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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 (Ag 1+ 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 (CHI 3 m.pt. 119-121 0C) - An alkaline solution of sodium hypoiodite, hypoiodi te, formed from sodium hydroxide and iodine, will convert acetaldehyde and aliphatic methyl ketones into iodoform (haloform reaction). - Since the reagent is also also an oxidizing agent, alcohols which are readily oxidized to acetaldehyde or methyl ketones also give a (+) reaction. - The mechanism mechanism of iodoform iodofo rm synthesis occurs through a series of enolate anions, anions, which are iodinated; hydroxide diplaces the Cl 3 anion though an addition/elimination pathway.

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CLASSIFICATION TESTS FOR CARBOXYLIC ACIDS AND DERIVATIVES 1. Hydrolysis of Acid Derivatives a. Acyl Halides - Reagent: H 2O - Acyl halides react with water to yield carboxylic acids. - This hydrolysis reaction 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 acid and HCl. - Sample Compounds: acetyl chloride (+) CH3COCl + H2O  CH3COOH + HCl -

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

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Reagent: 2% AgNO 3 (+) result result is indicated by the immediate formation formation of the pp pptt of silver silver halide halide Acid halides, halide salts and 2,4-dinitroaromatic 2,4-dinitroaro matic halides give immediate ppt. 2. Solubilty in Bicarbonate

-

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Reagent: saturated NaHCO 3 Effervescence due to the evolution of CO 2 (formation of bubbles) indicated the presence of carboxylic acid. All water-soluble compounds compounds will dissolve in the the bicarbonate solution, but only acids will give bubbles.

b. Acid Anhydrides The chemistry chemistry of acid anhydrides is similar to that of acyl halides. The kinds of the reactions the two groups undergo are the same, same, though acid anhydrides react more slowly. Acid anhydrides also undergo hydrolysis to to form a carboxylic acid. acid.

CH3COOCOCH3 + H2O  2 CH3COOH -

-

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For other reactions, just 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 Acidificati on of the carboxylate carboxylate solution with HCl leads to the recovery of the parent parent acid. msaesmalla © 2009 2009

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Often you will find that either the carboxylic acid or the alcohol formed formed from the hydrolysis of an ester is extremely soluble in H 2O. 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 b blue lue accompanied by an ammonia ammonia or amine-like odor, indicates an amide. Amides of higher amines that do not turn the litmus paper blue may 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 H 2SO4 - Involves Fischer esterification reaction; this is a nucleophilic acyl substitution reaction carried out under acidic conditions acidic conditions - Strong mineral acid such as H 2 S 2SO O 4 4 or HCl make the carboxylic acid more reactive enough to be attacked by alcohol. - The net effect of Fischer esterification esterificatio n is substitution of an –OH group by OR’. - All steps are reversible. - Sample compound: acetic acid (+)

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H2SO4 ←    → CH3CH2OH

-

Reaction: CH3COOH

-

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

+

CH3 COOCH2 CH3 + H3 O


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b. Acyl halide/Acid anhydride Reagents: ethanol, H 2O, 20% NaOH The esterificaion reaction of an alcohol with with acid chloride chloride is strongly affected by steric hindrance. Bulky groups groups on either partner slow slow down the reaction (reactivity: (reactivi ty: 1 0> 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 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 (+)

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Reaction: CH3COCl + CH3CH2OH  CH3COOCH2CH3 + HCl CH3COOCOCH3 + CH3CH2OH  CH3COOCH2CH3 + CH3COOH

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

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

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Reagents: hydroxylamine hydrochloride (NH 2OH•HCl), 1M KOH, 5% FeCl 3 Preliminary test is done to eliminate those eliminate those phenols and enols that give colors with ferric chloride in acidic solution and that would therefore give a false-positive result 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. color. msaesmalla © 2009 2009

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

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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-

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Nucleophilic Nucleophili c acyl substitution reactions reactions take take place in two steps: 1) addition of the nucleophile , nucleophile , and 2) elimination of a leaving group . group . Any factors that make 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) > RCO 2COR (Acid anhydride) > RCOOR (Ester) > RCONH 2 (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 – Hydrolysis – reaction with water to yield a carboxylic acid. 2) Alcoholysis – Alcoholysis – reaction with alcohol to yield an ester. 3) Aminolysis – Aminolysis – reaction with ammonia or amine to yield an amide. *Other reactions include reduction and Grignard reaction.


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CLASSIFICATION TEST FOR AMINES 1. Hinsberg’s Test – differentiates 1 0, 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 2 0 amine yield a white solid that does not dissolve in H 2O or 6M HCl. *A liquid residue that is more dense than H 2O and insoluble in 6M HCl may be a 2 0 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. acidification. R3N + ArSO2Cl  no reaction R3N + HCl  R3NH+Cl-

-

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

2. Nitrous acid test – differentiates 1 0 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 electrons in nitrogen. The presence of this lone pair makes amine both basic and basic and nucleophilic . - They react with acids to form acid-base salts. Also, they react with electrophiles in many of the polar reactions. 17


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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

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