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Chapter 26: Problem 105

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
Compound X is t-butyl chloride (option a).

Step by step solution

01

Understanding the Reaction with Na/Diethyl Ether

The reaction with Na/diethyl ether is a Wurtz reaction. This reaction involves coupling two alkyl halide molecules to form a larger hydrocarbon. Therefore, when compound  satisfies the Wurtz reaction, it forms a hydrocarbon with twice the number of carbon atoms and H atoms, minus two H atoms.
02

Determine the Hydrocarbon Formed

Given the molecular formula of the organic compound, \(\mathrm{C}_{4} \mathrm{H}_{9}\mathrm{Cl}\), when it undergoes a Wurtz reaction (2\(\mathrm{X}\rightarrow \text{Hydrocarbon} + 2\mathrm{NaCl}\)), it forms \(\mathrm{C}_8\mathrm{H}_{18}\), an alkane which is an octane isomer.
03

Analyzing Monochlorination

The question states that upon monochlorination, the resulting hydrocarbon gives only one possible chloro derivative. This means that all hydrogen atoms in the hydrocarbon formed are equivalent in terms of their reaction to a chlorine radical.
04

Identifying Compound X

Only \(\text{2,2,3,3-tetramethylbutane}\) formed from \(\text{t-butyl chloride}\) (choice (a)) has all its hydrogens equivalent upon monochlorination. This means that t-butyl chloride is the original compound X.

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Key Concepts

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

Wurtz Reaction
The Wurtz reaction is a classic method in organic chemistry for creating larger hydrocarbons from smaller alkyl halides. In general, it involves the use of sodium metal in the presence of dry ether as a solvent. The reaction works by coupling two molecules of an alkyl halide to form a hydrocarbon. For example, two molecules of a halide like \(\text{C}_4\text{H}_9\text{Cl}\) can be combined through a Wurtz reaction to form a hydrocarbon with a greater number of carbon atoms.
  • The reaction involves the \( \text{Na} \) metal reacting with the halide to form an alkyl radical.
  • The radicals then couple to form a new carbon-carbon bond.
  • This process results in a hydrocarbon with a double chain, minus one hydrogen for each halide molecule originally present.
This reaction is particularly useful for forming symmetrical alkanes and is a handy tool in synthetic organic chemistry.
Monochlorination
Monochlorination is a substitution reaction where a single chlorine atom replaces one hydrogen atom in a hydrocarbon. It is typically achieved by introducing chlorine gas to the hydrocarbon in the presence of ultraviolet light or heat. This process produces a chloro derivative of the original hydrocarbon.
  • In the exercise, monochlorination of the hydrocarbon results in only one possible chloro derivative.
  • This indicates that the hydrogen atoms in the formed hydrocarbon are equivalent, meaning any hydrogen in the molecule can be replaced with chlorine resulting in the same product.
  • Monochlorination is useful for understanding the symmetry of hydrocarbons based on their possible chloro derivatives.
By examining which hydrogens are equivalent, chemists can gain insights into the structure and symmetry of the hydrocarbons.
Alkyl Halide
Alkyl halides are organic molecules that consist of an alkyl group covalently bonded to a halogen. The halogen can be any of the elements from the group 17 of the periodic table, such as chlorine, bromine, or iodine. These compounds are typically named by combining the name of the alkyl group with the halogen present, resulting in names like butyl chloride for \(\text{C}_4\text{H}_9\text{Cl}\).
  • Alkyl halides are versatile chemical intermediates in organic synthesis because the halogen can be easily replaced in substitution reactions.
  • They are key substrates in the Wurtz reaction, where they couple to form larger hydrocarbons.
  • The stability of alkyl halides often depends on the type of alkyl group present; tertiary alkyl halides are generally more stable than primary ones.
Understanding alkyl halides is crucial because they act as starting points for the synthesis of various organic compounds.
Hydrocarbon Formation
In organic chemistry, the formation of hydrocarbons through reactions such as the Wurtz reaction is a fundamental aspect. Hydrocarbons are compounds made solely of hydrogen and carbon atoms. They can vary greatly in size and structure, from simple chains to complex branched and cyclic forms. In the context of the given exercise:
  • The original compound, \(\text{C}_4\text{H}_9\text{Cl}\), undergoes the Wurtz reaction to form \(\text{C}_8\text{H}_{18}\), an octane isomer.
  • This new hydrocarbon is formed by coupling two butyl chloride molecules together, effectively doubling the carbon chain length.
  • Hydrocarbons formed by such reactions can be further analyzed and manipulated through reactions like monochlorination, providing insight into their symmetry and chemical properties.
The ability to perform and understand these types of chemical transformations is vital for the creation of various materials and fuels.

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Most popular questions from this chapter

In \(\mathrm{S}_{\mathrm{N}}^{2}\) reactions, the correct order of reactivity for the following compounds: \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) and \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) is: \([2014]\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (c) \(\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (d) \(\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)

\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Cl} \stackrel{\mathrm{Acl}_{\mathrm{K} \mathrm{OH}}}{\Delta} \longrightarrow \mathrm{A} \frac{\mathrm{SO}_{2} \mathrm{Cl}_{2}}{475 \mathrm{~K}}{\longrightarrow} \mathrm{B}\) The compound B in the above sequence is (a) 1,2 -dichloro-2-methyl propene (b) 1 -chloro-2-methyl propene (c) 1,2 -dichloro-2-methyl propene (d) 3 -chloro-2-methyl propene

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\) alc. \(\mathrm{KOH}\) 2\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}+\) alc. \(\mathrm{KOH}\) 3\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\) alc. \(\mathrm{KOH}\) List II (Products) A. cis-but-2-ene B. trans-but-2-ene C. but-1-ene D. 2 -methylprop-l-ene The correct matching is \(\begin{array}{lll}1 & 2 & 3\end{array}\) 123 (a) \(\mathrm{C} \mathrm{B} \mathrm{D}\) (b) \(\mathrm{C} \mathrm{A} \mathrm{D}\) (c) \(\mathrm{C} \mathrm{D} \mathrm{A}\) (d) \(\mathrm{B} \mathrm{A} \mathrm{C}\)

Which of the following will not form a Grignard reagent on treatment with \(\mathrm{Mg}\) in dry ether? (a) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (b) \(\mathrm{Br} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (d) All the three above can form Grignard reagent.

Which one is the correct statement? (a) 2 -butene is Saytzeff's product (b) 1-butene is the Hofmann product (c) both (a) and (b) are correct (d) none is correct
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