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SN1 vs SN2 Reactivity

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SN1 Reactivity Methyl<primary<secondary<tertiary Why? Carbocation stability Steric congestion does NOT play a role SN2 Reactivity Tertiary<secondary<primary<methyl Steric congestion DOES play a role No carbocation formation

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Good Nucleophiles

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Not electronegative Poorly solvated polarizable Form strong bonds with carbon small impt in SN2

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Bases

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Availability and Stability of electrons: The less stable a lone pair of electrons is, the more basic it will be. Inductive effects: Electron donating groups lead to increased basicity. Ammonia Pkb = 4.75 Methylamine pkb= 3.36

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Leaving Groups

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A good leaving group should be: electronegative and well solvated (to stabilize electrons), polarizable and able to form weak bonds with the carbon (so that less energy is required to undergo the reaction) Large (reduces steric congestion upon leaving) Leaving group ability can be estimated by comparing the pKa of its conjugate acid.

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Substituents SN1

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SN1 reactions also favor a greater degree of substitution and such substituents should be electron donating so as to stabilize the transition state and carbocation through inductive effects and/or resonance effects and/or hyperconjugation

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Substituents SN2

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In SN2 reactions, substituents should be small to reduce steric congestion. Since the transition state of SN2 reactions is trigonal bipyramidal (with the hydroxyl above and the nucleophile below the central carbon) greater substitution entails greater steric strain, and the larger the groups of such substituents, the greater the crowding.

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Solvent SN1

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Should be polar protic (since it stablizes the charge in the TS and carbocation) able to undergo hydrogen bonding, contains an acidic hydrogen, and is able to stabilize ions.

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Solvent SN2

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SN2 Should be polar aprotic (stabilizes charge w/o decreasing the reactivity of nucleophile). Does not have an acidic hydrogens nor exhibits hydrogen bonding, yet, is able to stabilize ions.

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Part 1-Synthesis of n-butyl Bromide

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-n-butyl bromide occurs via the SN2 reaction mechanism. -Excess sulfuric acid serves to shift the equilibrium and thus speeds the reaction by producing a higher concentration of hydrobromic acid. -The sulfuric acid protonates the hydroxyl group of n-butyl alcohol, displacing the water as opposed to the hydroxide ion—the water is a better leaving group, which is important for SN1 reactions. -The acid also serves to protonate the water, which deactivates it as a nucleophile hence preventing conversion of the alkyl halide back to an alcohol.

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Part 1 Problems

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The production of unwanted side products such as di-n-butyl ether (alcohol attacking the butyloxnium ion) and 1-butene (butyloxonium ion by attack of water, a bimolecular elimination reaction)

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Part-2: Synthesis of T-butyl Alcohol

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-t-butyl alcohol, a tertiary alcohol, reacted with hydrochloric acid to produce t-butyl chloride. -Reaction occurs SN2 because of formation of a stable carbocation--the carbocation formed after the dissociation of the tert-butyloxonium ion is trisubstituted, and receives the inductive effects and hyperconjugation benefits from the three adjoining methyl groups

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Part 2 Problems

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Low yield could indicates the possible regeneration of the starting material, t-butyl alcohol, which would occur if water attacked the carbocation.

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SN1 vs SN2 and compounds

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primary: SN2 Secondary/Benzylic: SN1 and SN2 tertiary: SN1 phenol: nucleophilic aromatic subs.

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Silver Nitrate Test in Ethanol

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SN1 reactivity: of silver nitrate in ethanol, ethanol is a moderately powerful ionizing solvent, and nitrate ion is a poor nucleophile. The silver ion will assist the ionization of the alkyl halide, and form a silver halide precipitate.

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Sodium Iodide in Acetone (SN2)

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The iodide ion is an excellent nucleophile, and acetone is a nonpolar solvent. Sodium iodide is soluble in acetone, but a precipitate occurs when bromide ions or chloride ions are produced in the reaction as they are insoluble in acetone.

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Extraction Scheme

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1. H2SO4: Ionic compounds dissolve in concentrated H2SO4. Concentrated H2SO4 protonates all organic compounds containing heteroatoms or a double bond. Alkyl halides are not protonated. 2. Water 3. NaHCO3

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Exp 2: Aspirin Synthesis

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Salicylic Acid + Acetic Anhydride using acid catalyst

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Catalyst

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-increase reaction rate -Not consumed by reaction Provide alternative pathway for mechanism to occur

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3 ways to catalyze Aspirin reactioin

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-Activation of the nucleophile -Activation of the electrophile -Activation of both the nucleophile and electrophile

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Acid will catalyze ___________ while the base will catalyze __________

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electrophile by protonation(acetic anhydride); nucleophile by deprotonation

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Mechanism- Minor route

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phenol attacks anhydride directly. Tetrahedral intermediate collapses and you get your product directly.

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Carbonyl chemistry

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With an anhydride, such as acetic anhydride, an even more potent electrophile is present since the leaving group electronegative (C-O bond) and resonance stabilized (acetate anion).

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Side Products

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Polymer formation hydrolysis(use excess acetic anhydride to react with water and protect product)

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Why use excess acetic anhydride?

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-Drive reaction forward-Le Chatlier -Used as solvent- reacts with water and protects product -Prolonged contact of Aspirin with acid catalysts in presence of excess water results in regeneration of starting material

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Testing for Purity

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-Aspirin tablets: about .32g of acetylsalicylic acid & small amount of starch, which binds the ingredients. -Pure acetylic acid- no starch -Impure acetyl salicylic -acid (mix of phenols and product) -Salicylic acid- just phenols

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Iodine Test

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-Iodine is an orange-red-brown in solution -used to test for the presence of starch, producing a purple black color.

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FeCl3 Test

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-Ferric chloride is light brown in solution. -turns to a deep violet complex in the presence of a phenol due to the formation of an iron phenol complex

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Results of Tests: Aspirin Tablet Pure acetyl salicylic acid Impure acetyl salicylic acid Salicyclic acid

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-FeCl3- Nothing; Iodine: Yes -FeCl3-Nothing; Iodine: Nothing -FeCl3: mild violet color; Iodine: nothing -: deep violet color; Iodine: Nothing

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Exp 3: Synthesis of Lidocaine

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Nitrobenzene→dimethyl aniline→acetanilide→lidocaine

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1. Formation aniline

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Redox: Nitrogen-Reduced(+3 to -3) Tin-oxidized(+2 to +4) we start with 2,6-dimethylnitrobenzene, which is ultimately converted to dimethylaniline by virtue of using stannous chloride as a reducing agent.

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Redox eq.

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6e- + ArNO2 +6H ->ArNH2 + 2H20 3Sn2+ -> 3Sn4+6e- ArNO2 + 6HCl + 3SnCl2-> ArNH2 + 3SnCl4 +2H2O

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Formation of aniline con't

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-The hydrochloric salt is formed upon being reduced, causing it to precipitate from the mixture. -The salt is filtrated to separate it from unreacted starting materials and/or side products that are soluble in the reaction medium. (Use KOH)

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2. Formation of acetanilide

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Nucleophilic acyl substitution aniline + chloroacetyl chloride -aniline product is converted into a-Chloro-2-6-dimethylacetanilide through an esterification reaction. -acyl carbon is more reactive than the alkyl chloride carbon because of the differences in electrophilicity and the steric environment.

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Why Glacial Acetic Acid?

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-reaction is performed in glacial acetic acid (which means that dimethyl aniline is in equilibrium with its salt) to drive the reaction forward -pka of acetic acid is greater than aniline...its a waeker acid so it won't react but it will keep water from interacting with product.

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Why add NaOAc con't?

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-hydrochloric acid is liberated, leading dimethylaniline partially converting to its salt with this acid -Reaction stops at 50% completion. Add NaOAc to regenerate RNH2 and complete reaction Two key things: -reestablishes the reaction shown in as an acid-base equilibrium -NaOAc , serves to remove the HCl that is formed in the reaction and yield acetic acid, and to prevent co-precipitation of the hydrochloride salt. Only acetanilide will precipitate

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3. Formation of Lidocaine

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SN2 Reaction acetanilide + dimethylamine---> Lidocaine with toluene solvent 1. Heat is needed to provide activation energy 2. Excess diethylamine is used: -Reacts with a-chloro-2,6-dimethylaniline -Forms salt with HCl byproduct -Helps drive reaction by Le Chatlier's Principle

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TLC & Polarity of Compounds (complete reaction

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Most Polar →Least Polar Nitrobenzene Nitroaniline Lidocaine acetanilide

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TLC & Polarity of Compounds- formation of Lidocaine

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Most Polar → least polar Lidocaine acetanilide toluene

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Lidocaine Extraction Scheme

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HCl- protonated Lidocaine & diethylamine(moves L & D to aqueous layer 30% NaOH- Deprotonates L & D (moves L to organic layer & trace D) Ethyl acetate- forms aqueous layer 2% NaOH- Removes D MgSO4 and Evap- Lidocaine

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Experiment 4

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Qualitative anaylsis

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Beilestein Test

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Qualitative test for halides, indicated by a blue/green flame caused by the formation of a copper halide (positive result). A negative result would be an unchanged/normal flame color

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Brady Reagent (2,4 Dinitrophenylhydrazine)

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-tests for carbonyl groups associated with aldehydes and ketones, forming a yellow, orange or red precipitate (known as a dintrophenyhrdrazone) as a positive result. -a positive result involves a condensation reaction/addition elimination reaction -negative: no precipitate

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Ferric Chloride

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-Mainly used to detect the presence of phenols, but enols, hydroxamic acids, oximes and sulfinic acids may yield postive results as well. -Invovles the formation of a Ferric-phenol complex which is violent in color. -Negative result is the lack of such intense violet coloration.

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Hydroxamic Acid Test

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-Tests for most nitriles, esters and amides.. -Positive: On heating with hydroxylamine, nitriles, esters and amides form hydroxamic acid, which then may react with ferric chloride to form a ferric hydroxamate complex having a violet color -Negative result would be the lack of such violet ferric complex formation.

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Hinsberg Test

answer

-Hinsberg tests for primary, secondary and tertiary amines. --primary amine: clear solution, which on acidification with HCl, would form a precipitate; -secondary amines form an oil/solid suspension which does not dissolve on acidification. -Tertiary amines initally form an insoluble solid or oil, which, on acidification with HCl, dissolves to give a clear solution of the amine salt.

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Lucas Test

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-identifies the presence of and differention between primary, secondary and tertiary alcohols. -Positive result: precipitate. -The differing reactivity reflects the ease of formation of the corresponding carbocations of the alcohols which ulmitately react in a substitution in which the chloride anion from lucas's reagent (zinc chloride in concentrated HCl) displaces the hydroxyl group. -Based on carbocation stability, it is expected that a positive result for a tertiary alcohol would be reacting immediately, secondary in 3-5 minutes, and primary only on heating.

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Infrared Spectroscopy

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Used to detect functional groups

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Explanation of IR

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used to detect function groups...look at notes

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Hydrocarbon C=C

answer

alkene 1650 cm-1

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Hydrocarbon C-H

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alkane 3000-2700 cm-1

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Arene C-H

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>3000 cm-1

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Arene C=C

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benzene 1500, 1600 cm-1

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Hydroxyl

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3200-3500 cm-1

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Carbonyl

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1715 cm-1

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Alcohol

answer

3600-3200 cm-1

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