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23-1 Preparation  We have already covered these methods nucleophilic ring opening of epoxides by ammonia and amines. addition of nitrogen nucleophiles.

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Presentation on theme: "23-1 Preparation  We have already covered these methods nucleophilic ring opening of epoxides by ammonia and amines. addition of nitrogen nucleophiles."— Presentation transcript:

1 23-1 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 nitration of arenes followed by reduction of the NO 2 group to a 1° amine

2 23-2 Preparation  Alkylation of ammonia and amines by S N 2 substitution. Unfortunately, such alkylations give mixtures of products through a series of proton transfer and nucleophilic substitution reactions. polyalkylations

3 23-3 Preparation via Azides  Alkylation of azide ion. Overall Alkyl Halide  Alkyl amine

4 23-4 Example: Preparation via Azides Alkylation of azide ion. Note retention of configuration, trans  trans

5 23-5 Reaction with HNO 2  Nitrous acid, a weak acid, is most commonly prepared by treating NaNO 2 with aqueous H 2 SO 4 or HCl.  In its reactions with amines, nitrous acid: Participates in proton-transfer reactions. A source of the nitrosyl cation, NO +, a weak electrophile.

6 23-6 Reaction with HNO 2  NO + is formed in the following way. Step 1: Protonation of HONO. Step 2: Loss of H 2 O. We study the reactions of HNO 2 with 1°, 2°, and 3° aliphatic and aromatic amines.

7 23-7 Tertiary Amines with HNO 2 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.

8 23-8 Secondary Amines with HNO 2 2° Aliphatic and aromatic amines react with NO + to give N-nitrosamines. carcinogens

9 23-9 Amines with HNO 2  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.

10 23-10 RNH 2 with HNO 2  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 N 2 to give a carbocation.  Diazonium ion:  Diazonium ion: An RN 2 + or ArN 2 + ion

11 ° RNH 2 with HNO 2  Formation of a diazonium ion. Step 1: Reaction of a 1° amine with the nitrosyl cation. Step 2: Protonation followed by loss of water.

12 ° RNH 2 with HNO 2  Aliphatic diazonium ions are unstable and lose N 2 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.

13 ° RNH 2 with HNO 2  Tiffeneau-Demjanov reaction:.  Tiffeneau-Demjanov reaction: Treatment of a  - aminoalcohol with HNO 2 gives a ketone and N 2.

14 23-14 Mechanism of Tiffeneau-Demjanov Reaction with NO + gives a diazonium ion. Concerted loss of N 2 and rearrangement followed by proton transfer gives the ketone. Similar to pinacol rearrangement

15 23-15 Pinacol Rearrangement: an example of stabilization of a carbocation by an adjacent lone pair. Overall:

16 23-16 Mechanism Reversible protonation. Elimination of water to yield tertiary carbocation. 1,2 rearrangement to yield resonance stabilized cation. Deprotonation. This is a protonated ketone!

17 ° ArNH 2 with HNO 2  The -N 2 + group of an arenediazonium salt can be replaced in a regioselective manner by these groups.

18 ° ArNH 2 with HNO 2  A 1° aromatic amine converted to a phenol.

19 ° ArNH 2 with HNO 2 Problem: Problem: What reagents and experimental conditions will bring about this conversion?

20 ° ArNH 2 with HNO 2  Problem:  Problem: Show how to bring about each conversion.

21 23-21 Hofmann Elimination  Hofmann elimination:  Hofmann elimination: Thermal decomposition of a quaternary ammonium hydroxide to give an alkene. Step 1: Formation of a 4° ammonium hydroxide.

22 23-22 Hofmann Elimination Step 2: Thermal decomposition of the 4° ammonium hydroxide.

23 23-23 Hofmann Elimination  Hofmann elimination is regioselective - the major product is the least substituted alkene.  Hofmann’s rule:  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.

24 23-24 Hofmann Elimination The regioselectivity of Hofmann elimination is determined largely by steric factors, namely the bulk of the -NR 3 + group. Hydroxide ion preferentially approaches and removes the least hindered hydrogen and, thus, gives the least substituted alkene. Bulky bases such as (CH 3 ) 3 CO - K + give largely Hofmann elimination with haloalkanes.

25 23-25 Cope Elimination  Cope elimination:  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.

26 23-26 Cope Elimination 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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