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Predict the products of the monobromination of (a) \(m\) -dinitrobenzene; (b) aniline; (c) \(p\) -bromoanisole.

Short Answer

Expert verified
The products of the monobromination will be 1-bromo-3,5-dinitrobenzene, ortho-bromoaniline and 2,4-dibromoanisole, respectively.

Step by step solution

01

Monobromination of m-Dinitrobenzene

m-Dinitrobenzene has two nitro groups which are meta to each other. Nitro group is a meta-directing group (meaning that it directs incoming groups to the position meta to it) and is electron withdrawing by inductive and resonance effect. As such, the position it leaves the most electron density at is the position meta to it, which is where bromine will add. This would give 1-bromo-3,5-dinitrobenzene as the product.
02

Monobromination of Aniline

Aniline, which is a benzene ring with an amino (-NH2) group, will react differently. The amino group donates electrons through resonance to the benzene ring (due to the unshared pair on the nitrogen), activating the ortho and para positions. So bromine will preferably add to either available ortho position (since para position is occupied), giving ortho-bromoaniline as the final product.
03

Monobromination of p-Bromoanisole

p-Bromoanisole is a benzene ring with a methoxy (-OCH3) and bromine (-Br) group on the para positions. The methoxy group is an activating group and will enhance reactivity of the aromatic ring by donating electron density. It will direct incoming bromine to the ortho or para positions (relative to it). The para positions are occupied, the available ortho positions will accept bromine to give 2,4-dibromoanisole.

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Key Concepts

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

Bromination
Bromination is a chemical reaction where a bromine atom is introduced into a molecule. When discussing organic chemistry, bromination often targets specific positions in aromatic compounds, influenced by substituents already present on the molecule. This involves the use of bromine (\(\text{Br}_{2}\)) or a bromine-containing reagent.

During bromination, an existing compound reacts with bromine atoms, typically on an aromatic ring. It is a vital process in laboratories given how it can modify chemical properties without changing the fundamental structure too much.
  • Monobromination: Adding only one bromine atom to a compound.
  • Ortho, meta, para positions: Describe potential substitution points on an aromatic ring, controlled by substituents already present.
Understanding the details of bromination allows chemists to predict which positions on the ring are most likely to be affected by incoming bromine atoms.
Electrophilic Aromatic Substitution
Electrophilic Aromatic Substitution (EAS) is the cornerstone reaction for modifying aromatic compounds. This complex process allows for substitution of hydrogen atoms in aromatic rings with electrophilic species, like bromine. Unlike other substitution reactions, EAS retains the aromatic character of the ring.

The process generally involves these steps:
  • Formation of an Electrophile: A bromine molecule interacts with a catalyst, becoming more reactive.
  • Formation of a Sigma Complex: The electrophile briefly disrupts the aromaticity as it bonds with the ring.
  • Dissociation: Aromaticity is restored as the hydrogen initially displaced by bromine is expelled.
Controlling the positioning and rate of substitution requires understanding the substituents on the ring, as they dictate the reaction's pathway and speed.
Aromatic Compounds
Aromatic compounds are organic molecules characterized by their ring structure, consisting of alternating double and single bonds. This special configuration grants them unique stability, known as aromaticity.

In the context of bromination and EAS, understanding aromatic compounds requires knowledge of how different substituents affect reactivity:
  • Electron Withdrawing Groups (EWGs): Groups like nitro (\(-\text{NO}_{2}\)) decrease electron density, making rings less reactive.
  • Electron Donating Groups (EDGs): Groups like amino (\(-\text{NH}_{2}\)) and methoxy (\(-\text{OCH}_{3}\)) increase reactivity by adding electron density.
These groups influence where and how bromination occurs, by attracting electrons or repelling them, thus determining the preferred sites for reaction.
Chemical Reactions
Understanding chemical reactions involves grasping how substances interact to form new materials. In organic chemistry, specifically with aromatic compounds, these reactions follow particular patterns based on the principles of reactivity.
  • Reagents: Substances like reagents in bromination cause aromatic compounds to undergo specific transformations.
  • Conditions: Reaction conditions such as temperature, solvent, and pressure also influence the outcomes.
  • Substituents: Each substituent on an aromatic ring impacts how and where reactions happen.
In bromination, how efficiently a bromine atom attaches and which positions it targets hinges upon these factors. By mastering the interactions among these elements, you can better predict and direct the desired outcomes in aromatic chemistry.

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Most popular questions from this chapter

The reduction of aldehydes and ketones with a suitable hydride-containing reducing agent is a good way of synthesizing alcohols. This approach would be even more effective if, instead of a hydride, we could use a source of nucleophilic carbon. Attack by a carbon atom on a carbonyl group would give an alcohol and simultaneously form a carbon-to-carbon bond. How can we make a C atom in an alkane nucleophilic? This was achieved by Victor Grignard, who created the organometallic reagent \(\mathrm{R}-\mathrm{MgBr},\) with the following reaction in diethyl ether: $$\mathrm{R}-\mathrm{Br}+\mathrm{Mg} \longrightarrow \mathrm{R}-\mathrm{MgBr}$$ The Grignard reagent is rarely isolated. It is formed in solution and used immediately in the desired reaction. The alkylmetal bond is highly polar, with the partial negative charge on the \(\mathrm{C}\) atom, which makes the C atom highly nucleophilic. The Grignard reagent \((\mathrm{R}-\mathrm{MgBr})\) can attack a carbonyl group in an aldehyde or ketone as follows: Addition of dilute aqueous acid solution to the metal alkoxide furnishes the alcohol. The important synthetic consequence of this procedure is that we have prepared a product with more carbon atoms than present in the starting material. A simple starting material can be transformed into a more complex molecule. (a) What is the product of the reaction between methanal and the Grignard reagent formed from 1-bromobutane after the addition of dilute acid? (b) By using a Grignard reagent, devise a synthesis for 2-hexanol. (c) By using a Grignard reagent, devise a synthesis for 2 -methyl- 2 -hexanol. (d) Grignard reagents can also be formed with aryl halides, such as chlorobenzene. What would be the product of the reaction between the Grignard reagent of chlorobenzene and propanone? Can you think of an alternative synthesis of this product, again using a Grignard reagent? (e) The basicity of the \(C\) atom bound to the magnesium in the Grignard reagent can be used to make Grignard reagents of terminal alkynes. Write the equation of the reaction between ethylmagnesium bromide and 1-hexyne. [Hint: Ethane is evolved.] (f) By using a Grignard reagent, suggest a synthesis for 2 -heptyn-1-ol.

In referring to the molecular mass of a polymer, we can speak only of the average molecular mass. Explain why the molecular mass of a polymer is not a unique quantity, as it is for a substance like benzene.

Starting with the compounds chloromethane, chloroethane, sodium azide, sodium cyanide, and a reducing agent, suggest how the following compounds could be synthesized. (a) \(N\) -methylpropanamide (b) ethylethanoate (c) methylethylamine (d) tetramethylammonium chloride

When \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}\) is added to \(\mathrm{CH}_{3} \mathrm{OH}\) at room temperature, the major product is \(\mathrm{CH}_{3} \mathrm{O}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{3}\) and a minor product is \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{2} .\) Write out the mechanisms for the reactions leading to these products and use curved arrows to show the movement of electrons.

Predict the main product(s) of (a) the mononitration of benzoic acid; (b) the monosulfonation of phenol; (c) the monobromination of 2 -nitrobenzaldehyde.

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