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Show how the following compounds may be prepared from ammonia and the given starting materials: a. 1,2 -ethanediamine from ethene b. 2 -aminoethanol from ethene c. benzenamine from chlorobenzene

Short Answer

Expert verified
Convert ethene via dehydration to ethanol, then amination to ethylamine, 1,2-dichloroethane, and finally 1,2-ethanediamine. For 2-aminoethanol, epoxidize and react ethylene oxide with ammonia. Convert benzene to chlorobenzene, then benzenamine.

Step by step solution

01

Convert Ethene to Ethylamine

First, convert ethene (\(CH_2=CH_2\)) to ethylamine (\(CH_3CH_2NH_2\)) using the following steps:1. React ethene with water in the presence of sulfuric acid to form ethanol (\(CH_3CH_2OH\)), a hydration reaction.2. Convert ethanol to ethylamine by reacting ethanol with ammonia in the presence of an appropriate catalyst or through nucleophilic substitution using ethanol chloride followed by ammonolysis (\(CH_3CH_2Cl + NH_3\rightarrow CH_3CH_2NH_2\)).
02

Prepare 1,2-Ethanediamine

To prepare 1,2-ethanediamine from ethylamine:1. First convert ethylamine into 1,2-dichloroethane. Start by halogenating ethylene to form 1,2-dichloroethane (\(CH_2ClCH_2Cl\)) via chlorination.2. Conduct a nucleophilic substitution reaction by mixing 1,2-dichloroethane with excess ammonia to replace both chlorine atoms with amino groups, resulting in 1,2-ethanediamine (\(H_2NCH_2CH_2NH_2\)).
03

Convert Ethene to Ethylene Oxide

To synthesize 2-aminoethanol, start by converting ethene to ethylene oxide:1. Perform an epoxidation of ethene with oxygen in the presence of a silver catalyst to form ethylene oxide (\(C_2H_4O\)). This involves the oxidation of ethene.2. By controlling the reaction conditions, ensure no over-oxidation occurs.
04

Synthesize 2-Aminoethanol

React ethylene oxide with ammonia:1. Conduct a ring-opening reaction of ethylene oxide in the presence of ammonia, which will give 2-aminoethanol (\(C_2H_7NO\)).2. This reaction involves a nucleophilic attack by ammonia on the epoxide ring, resulting in 2-aminoethanol formation in a single step.
05

Prepare Chlorobenzene from Benzene

To obtain benzenamine from chlorobenzene:1. Start by converting benzene to chlorobenzene via electrophilic substitution using chlorine and a metal catalyst (iron or aluminum chloride), resulting in chlorobenzene (\(C_6H_5Cl\)).
06

Convert Chlorobenzene to Benzenamine

Finally, convert chlorobenzene to benzenamine:1. Perform a nucleophilic substitution reaction of chlorobenzene with an excess of ammonia.2. Use a copper(I) catalyst under elevated temperature and pressure to facilitate the reaction, which substitutes the chlorine atom with an amino group to form benzenamine (\(C_6H_5NH_2\)).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ammonia Reactions
Ammonia ( H_3 ) is a versatile compound in organic synthesis. It acts as a reagent in various chemical reactions, allowing for the insertion of an amino group. One of the key roles ammonia plays in synthesis is through nucleophilic substitution, where it replaces leaving groups like halide ions.
This reaction is crucial for transforming compounds into amines. Ammonia's small size and lone pair of electrons make it a powerful nucleophile, attacking electrophilic sites in other molecules.
In the context of the given exercise, ammonia facilitates the transformation of precursors, such as converting 1,2-dichloroethane to 1,2-ethanediamine. Similarly, in the presence of ammonia, epoxides can undergo ring-opening to result in valuable amine-alcohols.
  • An essential feature of ammonia reactions is the need for suitable reaction conditions, including solvents and temperature, to drive the desired synthesis forward.
  • Using ammonia often requires controlling the stoichiometry to avoid overreaction or unwanted byproducts.
Nucleophilic Substitution
Nucleophilic substitution reactions are fundamental in organic chemistry. They involve the replacement of a leaving group in a molecule by a nucleophile. During such reactions, the nucleophile attacks an electrophilic carbon atom, displacing a weaker nucleophile, typically a halogen in the case of alkyl halides.
For example, when chlorobenzene reacts with ammonia, the chlorine atom, acting as a leaving group, is replaced by an amino group. This step is facilitated by the presence of a copper catalyst.
This category of reactions is significant in the synthesis of amines. It accommodates different types of solvents and conditions, depending on the reactivity of substrates involved.
  • This type of reaction is called SN2 when it involves a coordinated mechanism where both bond breaking and making happen simultaneously.
  • The rate of nucleophilic substitution is influenced by the nature of the nucleophile, the leaving group, and the substrate.
Epoxide Ring-Opening
Epoxides are three-membered cyclic ethers known for their high reactivity due to strain in the ring. In synthesis, they undergo reactions that open the ring and allow incorporation of different nucleophiles.
When ethylene oxide, an epoxide, reacts with ammonia, a nucleophilic ring-opening occurs, resulting in 2-aminoethanol. This process involves the attack of the ammonia on the less hindered carbon atom of the epoxide, leading to the formation of an amine-alcohol.
  • Ring-opening reactions are essential in creating multi-functional compounds from simple and reactive substrates.
  • They often proceed under mild conditions, making them useful in industrial synthesis.
  • Control over reaction conditions helps avoid side reactions such as polymerization or overreaction.
Electrophilic Substitution
Electrophilic substitution is a reaction where an electrophile replaces a hydrogen atom in an aromatic compound. This is significant in converting benzene into other useful derivatives. In electrophilic substitution, a positively charged electrophile attacks an electron-rich aromatic ring.
For instance, when benzene is converted to chlorobenzene, a chlorine electrophile replaces a hydrogen atom. This reaction is catalyzed by Lewis acids like aluminum chloride, which makes the chlorine a potent electrophile.
These reactions are crucial for introducing functional groups onto aromatic rings, allowing for further transformations.
  • Electrophilic aromatic substitution is usually characterized by the temporary loss of aromaticity, followed by rapid rearrangement to restore it.
  • The reaction conditions, such as temperature and catalyst type, can significantly affect the rate and outcome.

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Most popular questions from this chapter

Show how the following transformations may be achieved. List reagents and approximate reaction conditions. a. 3-bromopropene to 3-butenamine b. cyclohexanone to cyclohexamine c. benzenecarboxylic acid to phenylmethanamine (not N-phenylmethanamine) d. benzenecarbaldehyde to N-methylphenylmethanamine \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NHCH}_{3}\right)\)

Offer plausible explanations of the following facts: a. Aza-2,4-cyclopentadiene (pyrrole) is unstable in acid solution and polymerizes. (Consider the effect of adding a proton to this molecule at the nitrogen and at carbon.) b. 1,3-Diaza-2,4-cyclopentadiene (imidazole) is a much stronger base than 1,3-diazabenzene (pyrimidine). c. The triaminomethyl cation, \(\left(\mathrm{NH}_{2}\right)_{3} \mathrm{C}\), is an exceptionally weak acid.

Assess the possibility of O-alkylation in the reaction of \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NH}^{\ominus} \mathrm{Na}^{\oplus} .\) Give your reasoning.

a. Explain why 1,3-diazacyclopentadiene (imidazole) is a much stronger acid than azacyclopentadiene (pyrrole). b. Would you expect benzenamine to be a stronger or weaker acid than cyclohexanamine? Give your reasoning.

Write a structural formula (one for each part) that fits the following descriptions. (These descriptions can apply to more than one structural formula.) a. A liquid basic nitrogen compound of formula \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{~N}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}\) and excess \(\mathrm{NaOH}\) solution gives a clear solution. This solution when acidified gives a solid product of formula \(\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{O}_{2} \mathrm{NS}\). b. A liquid diamine of formula \(\mathrm{C}_{5} \mathrm{H}_{14} \mathrm{~N}_{2}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}\) and \(\mathrm{NaOH}\) gives an insoluble solid. This solid dissolves when the mixture is acidified with dilute hydrochloric acid.

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