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The compound \(\mathrm{C}\) in the following is sequence: \(\mathrm{C}_{7} \mathrm{H}_{8} \stackrel{3 \mathrm{Cl}_{2} / \Delta}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{Br}_{2} / \mathrm{Fe}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{Zn} / \mathrm{HCl}}{\rightarrow} \mathrm{C}\) (a) p-bromotoluene (b) o-bromotoluene (c) m-bromotoluene (d) 3 -bromo-2, 4,6 -trichlorotoluene

Short Answer

Expert verified
The compound \( \mathrm{C} \) is 3-bromo-2,4,6-trichlorotoluene (option d).

Step by step solution

01

Analyze the first reaction

The compound \( \text{C}_7\text{H}_8 \) is toluene. The reaction \( \text{3Cl}_2 / \Delta \) indicates the chlorination of the methyl group in toluene under heat. This converts toluene into 2,4,6-trichlorotoluene.
02

Recognize the second reaction role

In the next step, \( \mathrm{Br}_2 / \mathrm{Fe} \) is used, which is a typical bromination reaction in aromatic systems favoring electrophilic aromatic substitution. This reaction will add a bromine atom to the aromatic ring. Given the bulky chlorine groups at 2,4, and 6 positions, bromination is most likely to occur at position 5, resulting in 3-bromo-2,4,6-trichlorotoluene (since the methyl group directs ortho and para, and para is less hindered).
03

Interpret the third reaction

The last step is \( \mathrm{Zn} / \mathrm{HCl} \), which is a reduction reaction, often used to remove halogens from alkyl positions. However, in this context, it can be understood to reduce the non-aromatic halogen if any, but here, it primarily confirms \( \mathrm{C} \) retains its aryl halogen configuration, not affecting the bromination product from step 2.
04

Determine the final compound

Given the reactions, \( \text{C} \) must be 3-bromo-2,4,6-trichlorotoluene, as no changes occur to the aromatic bromine anywhere, and the structure is consistent with the pathways given by the reactions.

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Key Concepts

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

Electrophilic Aromatic Substitution
Electrophilic Aromatic Substitution (EAS) is a fundamental type of organic reaction in which an aromatic hydrogen is replaced by an electrophile. This reaction is crucial in aromatic chemistry. It allows the introduction of a variety of functional groups into an aromatic ring. The process typically involves:
  • Formation of a strong electrophile, which is usually facilitated by a catalyst like iron (Fe) or aluminum chloride (AlCl₃).
  • Attack of the aromatic pi electrons on the electrophile, forming a non-aromatic carbocation intermediate called the arenium ion.
  • Re-aromatization of the ring by the loss of a proton, restoring aromaticity.
In the given exercise, the bromination process using Br extsubscript{2} and Fe demonstrates EAS. The position where the bromine is added is influenced by existing substituents, like the chlorine atoms in 2,4, and 6 positions, which are electron-withdrawing groups. This directs the incoming electrophile, bromine, to the most accessible position, often resulting in a para or ortho substitution if there is limited steric hindrance.
Reduction Reaction
Reduction reactions involve the gain of electrons or the loss of oxygen in a chemical species. In organic chemistry, reduction often refers to the process of adding hydrogen (hydrogenation) or removing halogen atoms. It is a critical step in many synthetic pathways, converting functional groups into more useful or less complex forms. In the equation given, Zn/HCl is used after bromination. Here, it's important to note that common reduction reactions involving Zn/HCl typically target nitro groups or other complex additions, facilitating the reduction of selective functional groups without affecting the aromatic bromine product. However, in this context, Zn/HCl supports the stability of the final compound, confirming the brominated structure and ensuring any unwanted transformations are minimized. Its presence confirms that the structure after reduction remains as 3-bromo-2,4,6-trichlorotoluene.
Chlorination of Toluene
Chlorination, specifically of toluene, is characterized by the addition of chlorine atoms to a substrate, where conditions like heat (94) facilitate reaction progress. In the chlorination of toluene, Cl extsubscript{2} reacts with toluene, a methyl-substituted benzene, to form trichlorotoluene. Here, the methyl group acts as an activating substituent for the aromatic ring, directing chlorination to ortho and para positions, thanks to its electron-donating nature. When reacting to conditions mentioned in the exercise, the methyl group, susceptible to radical substitution under heated conditions, results in a chlorination pattern highlighted by chlorine atoms occupying the 2, 4, and 6 positions on the aromatic ring, forming 2,4,6-trichlorotoluene. The controlled addition process here demonstrates regioselectivity influenced by the existing methyl group.
Bromination Reaction
Bromination is a type of electrophilic aromatic substitution that introduces a bromine atom into an aromatic compound. When using Br extsubscript{2} along with a catalyst like iron (Fe), it facilitates the bromination of aromatic rings. Given the discovery of 3-bromo-2,4,6-trichlorotoluene, bromination targets accessible positions on the aromatic ring. In the exercise context, the bromination is strategically processed on trichlorotoluene. The chlorine atoms at positions 2, 4, and 6 make ortho and para positions less favorable due to steric hindrance. The presence of a methyl group, which typically directs ortho/para addition, influences the choice toward the para position as the least hindered. Ultimately, this allows the formation of 3-bromo-2,4,6-trichlorotoluene, showcasing the precision achievable in selective substitution in aromatic chemistry.

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Most popular questions from this chapter

Unpleasant smell of carbylamine is produced when chloroform and alcoholic KOH are heated with (a) any amine (b) any aliphatic amine (c) any aromatic amine (d) any primary amine

\(\mathrm{CH}_{3} \mathrm{Br}+\mathrm{Nu} \longrightarrow \mathrm{CH}_{3}-\mathrm{Nu}+\mathrm{Br}^{-}\) The decreasing order of the rate of the above reaction with nucleophiles \(\left(\mathrm{Nu}^{-}\right)\)(a) to \((\mathrm{d})\) is [2006] \(\left[\mathrm{Nu}^{-}=\left(\right.\right.\)a) \(\mathrm{Ph} \mathrm{O}^{-}\)(b) \(\mathrm{AcO}^{-}\)(c) \(\mathrm{HO}^{-}\)(d) \(\left.\mathrm{CH}_{3} \mathrm{O}^{-}\right]\) (a) \(4>3>1>2\) (b) \(4>3>2>1\) (c) \(1>2>3>4\) (d) \(2>4>3>1\)

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{OH} \frac{\mathrm{NaBr}, \mathrm{H}_{2} \mathrm{SO}_{4}, \Delta}{\longrightarrow}\) 3\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \stackrel{\mathrm{PBr}_{3}}{\longrightarrow}\) 4\. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{OH} \stackrel{\mathrm{sOC}_{3}}{\longrightarrow}\) List II (Alkyl halides) A. \(\mathrm{CH}_{3} \mathrm{CHBr}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}\) B. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{Cl}\) C. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) D. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{Br}\) The correct matching is \(\begin{array}{rrrrr}1 & 2 & 3 & 4 \\ \text { (a) } \mathrm{C} & \mathrm{D} & \mathrm{B} & \mathrm{A}\end{array}\) (b) \(\mathrm{C} \quad \mathrm{D} \quad \mathrm{A} \quad \mathrm{B}\) (c) \(\mathrm{D} \quad \mathrm{C} \quad \mathrm{A} \quad \mathrm{B}\) (d) D \(\mathrm{C} \quad \mathrm{B} \quad \mathrm{A}\)

The decreasing order of reactivity of I. m-nitrobromobenzene II. \(2,4,6\)-trinitrobromobenzene III. p-nitrobromobenzene IV. and 2,4 -dinitrobromobenzene towards \(\mathrm{OH}^{-}\)ions is (a) \(\mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV}\) (b) \(\mathrm{II}>\mathrm{IV}>\mathrm{I}>\mathrm{III}\) (c) \(\mathrm{II}>\mathrm{IV}>\mathrm{III}>\mathrm{I}\) (d) IV \(>\mathrm{II}>\mathrm{III}>\mathrm{I}\)

How many elimination products are possible when 1chloromethyl cyclopentane is treated with alcoholic \(\mathrm{KOH} ?\)

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