tiểu-luận-hóa-học-hữu-cơ-Nguyễn-Thị-Dịu-27-02-2021

HUE UNIVERSITY HUE UNIVERSITY OF EDUCATION FACULTY OF CHEMISTRY -?&@ - NGUYỄN THỊ DỊU AMINES AND DIAZONIUM SALTS ESSAY OF ORGANIC CHEMISTRY Lecturer: TRẦN ĐÔNG TIẾN Class: Chemistry 3E Student Code: 18S2011007 Course: 2018 - 2022 Hue, 01/2021 ACKNOWLEDGEMENT I would like to thank Mr Tran Dong Tien for imparting valuable knowledge and experiences to me, helping and creating all favorable conditions so that I can complete this thematic exercise perfectly I would also like to thank my friends who supported me in the writing process of the essay by giving me some useful advices and encouraging me when I am busy and tired Hue, January 2021 Nguyen Thi Diu TABLE OF CONTENTS Page INTRODUCTION Reasons of the topic 2 Purposes of the topic .2 Research methods MAIN CONTENTS OF THE TOPIC .3 Chapter 1: Summary .3 1.1 Amines 1.1.1 Introduction 1.1.2 Synthesis of amines 1.1.3 Reactions of amines 1.2 Diazonium salts 1.2.1 Introduction 1.2.2 Preparation .9 1.2.3 Reaction of diazonium salts 10 Chapter 2: Problems and solutions .12 CONCLUSIONS 25 REFERENCES 25 INTRODUCTION Reasons of the topic Organic chemistry is a science that studies the structure, composition, properties, reaction mechanism, synthesis methods and application of organic compounds and organic materials as well as many different compounds These days, the relevant questions and exercise system of organic chemistry are greatly various The amine is an indispensable part of the curriculum Therefore, I choose the topic "AMINES AND DIAZONIUM SALTS" to study sharply about amine Purposes of the topic - Systematizing knowledge about amines; - Systematizing and solving questions and exercises of chapter amines in the section of organic chemistry Research methods - Synthesizing presentations from references related to amines; - Collecting, classifying and compiling chemical exercises about amines MAIN CONTENTS CHAPTER 1: SUMMARY 1.1 Amines 1.1.1 Introduction: Amines are basic nitrogen-containing compounds that are derivatives of ammonia (NH3), in which one or more alkyl, cycloalkyl, or aryl groups replace hydrogen and bond to the nitrogen atom + Classification of amines: - Amine: organic molecule in which a nitrogen atom is bonded to one or more alkyl group - Amine: An organic base formed by replacing one or more of the hydrogen atoms of ammonia by organic groups - Any of a class of compounds derived from ammonia by replacement of one or more hydrogen atoms with organic groups Amines are classified based on: - Number of nitrogen atom; - Constitutional characteristic of hydrocarbon groups bonded to the nitrogen atom: + Aliphatic amine (alkylamine) + Aromatic amine (arylamine) + Amines of primary, secondary and tertiary: Unlike alcohols and alkyl halides, which are classified as primary, secondary, or tertiary according to the degree of substitution at the carbon that bears the functional group, amines are classified according to their degree of substitution at nitrogen - Primary (1°) amine: one alkyl or aryl group attached to the nitrogen atom - Secondary (2°) amine: two alkyl or aryl groups attached to the nitrogen atom - Tertiary (3°) amine: three alkyl or aryl groups attached to the nitrogen atom - Quaternary (4o) amine: an ion in which nitrogen is bonded to four alkyl or aryl groups and bears a positive charge + Nomenclature of amines The naming of amine depends on the classification of amine They are named using either systematic (IUPAC System) or common names Secondary and tertiary amines having identical alkyl groups are named by using the prefix di- or tri- with the name of the primary amine Amines are named in two main ways in the IUPAC system, either as alkylamines or as alkanamines When primary amines are named as alkylamines, the ending -amine is added to the name of the alkyl group that bears the nitrogen When named as alkanamines, the alkyl group is named as an alkane and the -e ending replaced by -amine Example Primary amine 1.1.2 Synthesis of amines + Nucleophilic Substitution Routes to Amines: Direct Nucleophilic Substitution by SN2 reaction of an alkyl halide with NH or an amine (alkylation of NH3 or an amine: Hofmann, 1850 or by amination of alkyl halides) SN2 mechanism: the unhindered alkyl halide = CH3X or RCH2X A very large excess of NH3 ⇒ a mixture of 1°, 2° and 3° amines; A very large excess of alkyl halide ⇒ a mixture of 3° amine and quaternary ammonium salt * The Gabriel Synthesis of 1° Amines (Gabriel, 1887) = Traditional Gabriel synthesis: The Gabriel synthesis converts an alkyl halide into a 1° amine by a twostep process: nucleophilic substitution followed by hydrolysis Disadvantages of Gabriel synthesis: - This method generally fails with secondary alkyl halides - The first technique often produces low yields or side products - Separation of phthalhydrazide can be challenging For these reasons, other methods for liberating the amine from the phthalimide have been developed Even with the use of the hydrazinolysis method, the Gabriel method suffers from relatively harsh conditions + Reduction of Other Functional Groups That Contain Nitrogen - From nitro compounds, oximes: The most frequently used methods employ catalytic hydrogenation, or treatment of the nitro compound with acid and iron Zinc, tin, or a metal salt such as SnCl can also be used: - From nitriles: - From amides: - From alkyl azides: + Reductive Amination of Aldehydes and Ketones (or reductive alkylation of the amine) Aldehydes and ketones can be converted to amines through catalytic or chemical reduction in the presence of ammonia or an amine Primary, secondary, and tertiary amines can be prepared this way: + Degradation reactions: Preparation of Primary Amines through the Hofmann and Curtius Rearrangements Hofmann Rearrangement Amides with no substituent on the nitrogen react with solutions of bromine or chlorine in sodium hydroxide to yield amines through a reaction known as the Hofmann rearrangement or Hofmann degradation: 1.1.3 Reactions of amines + Basicity of Amines Both amine and ammonia produce basic aqueous solutions Amine and ammonia are relatively weak bases due to the acceptor of protons Most are stronger bases than water but are far weaker bases than hydroxide ions, alkoxide ions, and alkanide anions A convenient way to compare the base strengths of amines is to compare the pKa values of their conjugate acids, the corresponding alkylaminium ions When considering the basicity of amines, bear in mind that: The more basic than amine, the weaker its conjugate acid; The more basic than amine, the larger the pKa of its conjugate acid + Amines react as bases with a variety of organic and inorganic acids The equilibrium for an amine that is relatively more basic will lie more toward the left in the above chemical equation than for an amine that is less basic The aminium ion of a more basic amine will have a larger pKa than the aminium ion of a less basic amine When we compare aminium ion acidities in terms of this equilibrium, we see that most primary alkylaminium ions (RNH 3+) are less acidic than ammonium ions (NH 4+) In other words, primary alkylamines (RNH 2) are more basic than ammonia (NH3): + Replacement of hydrogen atom bonded to nitrogen - Alkylation of amines (or NH3): Hofmann's exhaustive alkylation: 10 CHAPTER 2: PROBLEMS AND SOLUTIONS Problem (Probem 25.1, page 950, [1]) Classify each amine in the following compounds as 1°, 2°, 3° Solution: Amines are classified as 1o , 2o , or 3o by the number of alkyl groups bonded to the nitrogen atom Problem (Problem 25.2, page 950, [1]) Draw the structure of a compound of molecular formula C4H11NO that fi ts each description: (a) a compound that contains a 1° amine and a 3° alcohol; (b) a compound that contains a 3° amine and a 1° alcohol Problem ( Problem 25.5, page 953, [1]) Name each amine Solution: a Butan-2-amine b Dibutylamine c N,N-dimethylcyclohexanamine d 2-methylnonan-5-amine e N-ethylhexan-3-amine f 2-methyl-N-propylcyclopentan-1-amine 14 Problem (Problem 25.6, page 954, [1]) Draw a structure corresponding to each name a 2,4-dimethyl-3-hexanamine b N-methylpentylamine c N-isopropyl-p-nitroaniline d N-methylpiperidine e N,N-dimethylethylamine f 2-aminocyclohexanone g N-methylaniline h N-ethylaniline Solutions: Problem (Problem 25.7, page 954, [1]) Arrange each group of compounds in order of increasing boiling point Solutions: Primary (1°) and 2° amines have higher bp’s than similar compounds (like ethers) incapable of hydrogen bonding, but lower bp’s than alcohols that have stronger intermolecular hydrogen bonds Tertiary amines (3°) have lower boiling points than o and 2o amines of comparable molecular weight because they have no N–H bonds 15 Problem (Probem 25.41, page 994, [1]) Give a systematic or common name for each compound Solution: Problem (Problem 25.42, page 994, [1]) Draw the structure that corresponds to each name a cyclobutylamine f N-methylcyclopentylamine b N-isobutylcyclopentylamine g cis-2-aminocyclohexanol c tri-tert-butylamine h 3-methyl-2-hexanamine d N,N-diisopropylaniline i 2-sec-butylpiperidine e N-methylpyrrole j (2S)-2-heptanamine 16 Solution: Problem8 (Problem 25.44, page 994, [1]) How many stereogenic centers are present in each compound? Draw all possible stereoisomers Solution: 17 Problem (Problem 25.11, page 961, [1]) Draw the product of each reaction Solutions: SN2 reaction of an alkyl halide with NH3 or an amine forms an amine or an ammonium salt Problem 10 (Problem 25.45, page 954, [1]) Which compound in each pair is the stronger base? Solution: Problem 11 (Problem 25.58, page 996, [1]) How would you prepare benzylamine (C6H5CH2NH2) from each compound? In some cases, more than one step is required a C6H5Cl e C6H5CH3 b C6H5CN f C6H5COOH c C6H5CONH2 g C6H5NH2 d C6H5CHO h benzene Solution: 18 Problem 12 (Problem 25.60, page 996, [1]) What products are formed when Nethylaniline (C6H5NHCH2CH3) is treated with each reagent? a HCl b CH3COOH f CH3I (excess), followed by Ag2O and ∆ g CH3CH2COCl c (CH3)2C =O h The product in (g), then HNO3, H2SO4 d CH2O, NaBH3CN i The product in (g), then [1] LiAlH4; [2] H2O e CH3I (excess) j The product in (h), then H2, Pd-C Solution: 19 Problem 13 (Problem 25.61, page 996, [1]) Draw the products formed when pmethylaniline (p-CH3C6H4NH2) is treated with each reagent a HCl f CH3COCl, AlCl3 b CH3COCl g CH3COOH c (CH3CO)2O h NaNO2, HCl d CH3I i Step (b), then CH3COCl.AlCl3 e (CH3)2C=O j CH3CHO, NaBH3CN Solution: 20 Problem 14 (Problem 25.62, page 998, [1]) How would you convert CH3CH2CH2CH2NH2 into each compound? a CH3CH2CH2CH2NHCOC6H5 b CH3CH2CH2CH2N=C(CH2CH3)2 c CH3CH2CH=CH2 d CH3CH2CH2CH2NHCH2C6H5 e CH3CH2CH2CH2NHCH2CH3 f [CH3CH2CH2CH2N(CH3)3]+ I– Solution: Problem 15 (Problem 25.72, page 998, [1]) Draw a stepwise mechanism for the following reaction Solution: 21 Problem 16 (Problem 25.77, page 998, [1]) Devise a synthesis of each compound from benzene You may use any other organic or inorganic reagents Solution: 22 Problem 17 (Problem 25.80, page 998, [1]) Safrole, which is isolated from sassafras, can be converted to the illegal stimulant MDMA (3,4methylenedioxymethamphetamineMDMA, “Ecstasy”) by a variety of methods (a) Devise a synthesis that begins with safrole and uses a nucleophilic substitution reaction to introduce the amine (b) Devise a synthesis that begins with safrole and uses reductive amination to introduce the amine Solution: 23 Problem 18 (Problem 25.89, page 1001, [1]) Draw the product Y of the following reaction sequence Y was an intermediate in the remarkable synthesis of cyclooctatetraene by Wilstatter in 1911 Solution: Problem 19: Write the chemical equation according to the following conversion diagram: Solution: 24 Problem 20 Give the structures of A,B and C in the following reaction: Solution: Problem 21 Compound A (C5H11O2N) is a photoactive liquid Reducing A by H2 with Ni catalyst will be photoactive B (C5H13N) Let B react with HNO2 acid to obitan a mixture of photosensitive alcohol C and tert-amylic alcohol (2-methyl-2-butanol) Determine the structural formula of A Using the structural formula, write down the equations for the formation of B, C and tert-amylic alcohol from A Solution: 25