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An organic compound \(\mathrm{X}\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\right)\) on reaction with \(\mathrm{Na} /\) diethylether gives a hydrocarbon which on monochlorination gives only one chloro derivative then, \(\mathrm{X}\) is (a) t-butyl chloride (b) s-butyl chloride (c) iso-butyl chloride (d) \(\mathrm{n}\)-butyl chloride

Short Answer

Expert verified
t-butyl chloride is the compound \( \mathrm{X} \).

Step by step solution

01

Identify Reaction Type

The reaction involves an organic compound \( \mathrm{X} \) which is \( \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl} \). This compound reacts with \( \mathrm{Na} \) in diethyl ether, suggesting a Wurtz reaction. The Wurtz reaction typically forms a hydrocarbon by coupling alkyl halides.
02

Determine the Resultant Compound

Given the Wurtz reaction, the coupling of two molecules of \( \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl} \) will form a hydrocarbon with the formula \( \mathrm{C}_{8} \mathrm{H}_{18} \). We need a hydrocarbon from this reaction that on monochlorination yields only one chloro derivative.
03

Analyze Possible Isomers

The hydrocarbon \( \mathrm{C}_{8} \mathrm{H}_{18} \) formed must be symmetrical to provide only one monochloro derivative. \( \mathrm{Neopentane} \) or methylpropane derivatives are favorable as they can yield symmetrical reaction products.
04

Determine Structure of \( \mathrm{X} \)

Since \( \mathrm{X} \) on reaction forms a hydrocarbon that upon monochlorination gives only one chloro derivative, a symmetrical structure like \( \mathrm{t} \)-butyl chloride (option (a)) must be \( \mathrm{X} \). \( \mathrm{t} \)-butyl chloride, when reacted, results in a symmetrical \( \mathrm{C}_{8} \mathrm{H}_{18} \) (2,2,3,3-tetramethylbutane) that can derive to a single monochlorinated product.

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Key Concepts

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

Chloro Derivatives
Chloro derivatives are organic compounds where one or more chlorine atoms are bonded to carbon atoms. These compounds play a significant role in organic chemistry, especially in reactions that involve substitution or elimination processes.
For instance, in the Wurtz reaction, an alkyl chloride reacts with sodium metal in an ether solvent, leading to the formation of a coupling product and a symmetrical hydrocarbon. The involvement of chloro derivatives is crucial here because the chlorine atoms help form the bond between two organic molecules.
In this exercise, the compound \(X\) is t-butyl chloride (\(\text{C}_4\text{H}_9\text{Cl}\)). When \(X\) undergoes a Wurtz reaction, its chloro derivative nature allows it to partner up and form a larger hydrocarbon (\(\text{C}_8\text{H}_{18}\)). This larger hydrocarbon is crucial for achieving the specific type of product required by the problem's conditions. A single monochlorinated chloro derivative can only be obtained with a very specific, symmetrical arrangement of atoms inherent to \(t\)-butyl chloride.
Symmetrical Hydrocarbons
Symmetrical hydrocarbons are compounds where the molecular structure is mirrored equally from a central point or axis. This symmetry is vital in many chemical reactions because it often leads to unique chemical properties and product formations.
In the context of the exercise, a symmetrical hydrocarbon is necessary so that monochlorination results in only one chloro derivative. When a symmetrical hydrocarbon like 2,2,3,3-tetramethylbutane is monochlorinated, only one kind of chloro derivative is formed because all possible sites for chlorination are equivalent.
  • For example, if we consider different positions on a carbon chain for chlorination and all positions yield the same product, we understand the molecule is symmetrical.
  • This characteristic simplifies predictions about chemical reactions, helping chemists isolate exactly what kind of product can form from a reaction.
Hence, the correct identification of a symmetrical arrangement within the hydrocarbon formed from t-butyl chloride ensures understanding of these concepts for chemical reactions.
Organic Compound Reactions
Organic compound reactions involve transformations where the bonds within and between organic molecules are broken and formed. These reactions are the cornerstone of creating complex molecules from simpler ones, which is often the goal of synthetic chemistry.
The Wurtz reaction is a classic example where two alkyl halides are coupled, resulting in the formation of a longer alkane chain. The reaction operates by removing halogens and coupling the resultant radicals. This step finds particular utility in organic synthesis for extending carbon chains.
  • In the case of this exercise, the reaction of \(t\)-butyl chloride with sodium is a perfect representation of the Wurtz reaction. The reaction proceeds smoothly to form a longer and symmetrical hydrocarbon suitable for monochlorination.
  • Monochlorination refers to substituting one hydrogen atom in a molecule for a chlorine atom. Here, it highlights the creation of exactly one type of product, due to the hydrocarbon's symmetrical structure post-reaction.
The reliability and predictability of such reactions make them principal tools in the chemist's arsenal, delivering a method to progressively build complex structures from basic organic compounds.

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Most popular questions from this chapter

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 \(\mathrm{C} \quad \mathrm{B} \quad \mathrm{A}\)

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Compound (A), \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}\), gives a white precipitate when warmed with alcoholic \(\mathrm{AgNO}_{3} .\) Oxidation of (A) gives an acid (B), \(\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}\). (B) easily forms anhydride on heating. Identify the compound (A). [2013]

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