Preparation
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Preparation. We have already covered these methods nucleophilic ring opening of epoxides by ammonia and amines. addition of nitrogen nucleophiles to aldehydes and ketones to form imines reduction of imines to amines reduction of amides to amines by LiAlH 4 reduction of nitriles to a 1° amine

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Preparation l.jpg
Preparation

  • We have already covered these methods

    • nucleophilic ring opening of epoxides by ammonia and amines.

    • addition of nitrogen nucleophiles to aldehydes and ketones to form imines

    • reduction of imines to amines

    • reduction of amides to amines by LiAlH4

    • reduction of nitriles to a 1° amine

    • nitration of arenes followed by reduction of the NO2 group to a 1° amine


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Preparation

  • Alkylation of ammonia and amines by SN2 substitution.

    • Unfortunately, such alkylations give mixtures of products through a series of proton transfer and nucleophilic substitution reactions.

polyalkylations


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Preparation via Azides

  • Alkylation of azide ion.

Overall

Alkyl Halide  Alkyl amine


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Example: Preparation via Azides

  • Alkylation of azide ion.

Note retention of configuration, trans  trans


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Reaction with HNO2

  • Nitrous acid, a weak acid, is most commonly prepared by treating NaNO2 with aqueous H2SO4 or HCl.

  • In its reactions with amines, nitrous acid:

    • Participates in proton-transfer reactions.

    • A source of the nitrosyl cation, NO+, a weak electrophile.


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Reaction with HNO2

  • NO+ is formed in the following way.

    • Step 1: Protonation of HONO.

    • Step 2: Loss of H2O.

    • We study the reactions of HNO2 with 1°, 2°, and 3° aliphatic and aromatic amines.


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Tertiary Amines with HNO2

  • 3° Aliphatic amines, whether water-soluble or water-insoluble, are protonated to form water-soluble salts.

  • 3° Aromatic amines: NO+ is a weak electrophile and participates in Electrophilic Aromatic Substitution.


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Secondary Amines with HNO2

  • 2° Aliphatic and aromatic amines react with NO+ to give N-nitrosamines.

carcinogens


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Amines with HNO2

  • Reaction of a 2° amine to give an N-nitrosamine.

    • Step 1: Reaction of the 2° amine (a nucleophile) with the nitrosyl cation (an electrophile).

    • Step 2: Proton transfer.


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RNH2 with HNO2

  • 1° aliphatic amines give a mixture of unrearranged and rearranged substitution and elimination products, all of which are produced by way of a diazonium ion and its loss of N2 to give a carbocation.

  • Diazonium ion: An RN2+ or ArN2+ ion


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1° RNH2 with HNO2

  • Formation of a diazonium ion.

    Step 1: Reaction of a 1° amine with the nitrosyl cation.

    Step 2: Protonation followed by loss of water.


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1° RNH2 with HNO2

  • Aliphatic diazonium ions are unstable and lose N2 to give a carbocation which may:

    1. Lose a proton to give an alkene.

    2. React with a nucleophile to give a substitution product.

    3. Rearrange and then react by Steps 1 and/or 2.


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1° RNH2 with HNO2

  • Tiffeneau-Demjanov reaction: Treatment of a -aminoalcohol with HNO2 gives a ketone and N2.


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Mechanism of Tiffeneau-Demjanov

  • Reaction with NO+ gives a diazonium ion.

  • Concerted loss of N2 and rearrangement followed by proton transfer gives the ketone.

Similar to pinacol rearrangement


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Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent lone pair.

Overall:


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Mechanism carbocation by an adjacent lone pair.

Reversible protonation.

Elimination of water to yield tertiary carbocation.

This is a protonated ketone!

1,2 rearrangement to yield resonance stabilized cation.

Deprotonation.


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carbocation by an adjacent lone pair.ArNH2 with HNO2

  • The -N2+ group of an arenediazonium salt can be replaced in a regioselective manner by these groups.


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1° ArNH carbocation by an adjacent lone pair.2 with HNO2

  • A 1° aromatic amine converted to a phenol.


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1° ArNH carbocation by an adjacent lone pair.2 with HNO2

Problem:What reagents and experimental conditions will bring about this conversion?


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1° ArNH carbocation by an adjacent lone pair.2 with HNO2

  • Problem: Show how to bring about each conversion.


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Hofmann Elimination carbocation by an adjacent lone pair.

  • Hofmann elimination: Thermal decomposition of a quaternary ammonium hydroxide to give an alkene.

    • Step 1: Formation of a 4° ammonium hydroxide.


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Hofmann Elimination carbocation by an adjacent lone pair.

  • Step 2: Thermal decomposition of the 4° ammonium hydroxide.


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Hofmann Elimination carbocation by an adjacent lone pair.

  • Hofmann elimination is regioselective - the major product is the least substituted alkene.

  • Hofmann’s rule:Any -elimination that occurs preferentially to give the least substituted alkene as the major product is said to follow Hofmann’s rule.


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Hofmann Elimination carbocation by an adjacent lone pair.

  • The regioselectivity of Hofmann elimination is determined largely by steric factors, namely the bulk of the -NR3+ group.

  • Hydroxide ion preferentially approaches and removes the least hindered hydrogen and, thus, gives the least substituted alkene.

  • Bulky bases such as (CH3)3CO-K+ give largely Hofmann elimination with haloalkanes.


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Cope Elimination carbocation by an adjacent lone pair.

  • Cope elimination:Thermal decomposition of an amine oxide.

    Step 1: Oxidation of a 3° amine gives an amine oxide.

    Step 2: If the amine oxide has at least one -hydrogen, it undergoes thermal decomposition to give an alkene.


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Cope Elimination carbocation by an adjacent lone pair.

  • Cope elimination shows syn stereoselectivity but little or no regioselectivity.

  • Mechanism: a cyclic flow of electrons in a six-membered transition state.


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