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Ethyl bromide reacts with lead-sodium alloy to form (a) tetramethyl bromide (b) tetraethyl lead (c) both (a) and (b) (d) none of these

Short Answer

Expert verified
The correct answer is (b) tetraethyl lead.

Step by step solution

01

Understanding the Reaction

Ethyl bromide (C2H5Br) is reacting with a lead-sodium alloy. The chemical reaction involves forming a compound containing lead.
02

Identifying Possible Products

The possible product formed in such a reaction is tetraethyl lead (Pb(C2H5)4), a compound widely known for its use as an anti-knock agent in fuels. This is typically the result when ethyl groups (from ethyl bromide) react with metallic lead.
03

Eliminating Other Options

Tetramethyl bromide or other options like both (a) and (b) do not correspond to the typical outcomes of this specific reaction. Ethyl bromide does not contain methyl groups needed to form tetramethyl bromide.
04

Concluding the Correct Answer

Based on the information and typical chemical knowledge, ethyl bromide reacts with lead-sodium alloy primarily to form tetraethyl lead (Pb(C2H5)4).

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Key Concepts

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

Ethyl Bromide
Ethyl bromide, also known as bromoethane, is an alkyl halide with a chemical formula of \( C_2H_5Br \). It's a clear, colorless liquid with a distinctive sweet smell, and it is commonly used in the synthesis of different chemical compounds. An alkyl halide like ethyl bromide features an alkyl group, ethyl in this case, bonded to a halogen atom, bromine.
In chemical reactions, ethyl bromide is often used to introduce ethyl groups into other molecules. This makes it a versatile reagent in organic synthesis. Its reactivity primarily stems from the presence of the carbon-bromine bond, which is relatively weak and thus capable of participating in various substitution and elimination reactions. During these reactions, the ethyl group is transferred, which is essential for forming larger and more complex molecules in organic chemistry.
Lead-Sodium Alloy
A lead-sodium alloy is a combination of the metals lead (Pb) and sodium (Na). The alloy has unique properties and is used in specific chemical reactions due to its structural and chemical characteristics.
  • Reactivity: The alloy is quite reactive, particularly due to the sodium component, which tends to react with many substances.
  • Utility: In the context of the reaction with ethyl bromide, the lead-sodium alloy plays a critical role in transferring ethyl groups to lead, facilitating the formation of tetraethyl lead.
The combination of lead and sodium in an alloy form is primarily used to enhance specific types of reactivity compared to either metal alone. It provides the metallic medium necessary for unique reactions with organic compounds, like the interaction with ethyl bromide leading to the synthesis of useful organometallic compounds.
Tetraethyl Lead
Tetraethyl lead, commonly referred to as TEL, is a chemical compound with the formula \( Pb(C_2H_5)_4 \). It is an organolead compound, which means it has both organic (carbon-containing) and inorganic (lead) components. Tetraethyl lead was famously used as an anti-knock agent in gasoline, which helped improve engine efficiency and performance by preventing the "knocking" noise commonly associated with the premature ignition of fuel.
This compound is synthesized through the reaction of ethyl bromide with metallic lead, often in the presence of a lead-sodium alloy. Here’s why TEL was so valued:
  • Improved Fuel Efficiency: By reducing engine knocking, TEL allowed engines to run more smoothly, thus preserving their longevity and performance.
  • Historical Significance: Before environmental concerns highlighted the hazards of lead-based compounds, TEL was considered innovative in its ability to enhance fuel quality.
However, due to the dangerous health effects and environmental concerns associated with lead pollution, the use of tetraethyl lead in fuels has significantly decreased, and its production is tightly regulated in many countries around the world.

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Most popular questions from this chapter

Identify the set of reagents/reaction conditions 'X' and ' \(\mathrm{Y}\) ' in the following set of transformations: CCCBr (a) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\) HBr/acetic acid, \(20^{\circ} \mathrm{C}\) (b) \(\mathrm{X}=\) concentrated alcoholic \(\mathrm{NaOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{HBr} /\) acetic acid, \(20^{\circ} \mathrm{C}\) (c) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHCl}_{3}, 0^{\circ} \mathrm{C}\) (d) \(\mathrm{X}=\) concentrated alcoholic \(\mathrm{NaOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHCl}_{3}, 0^{\circ} \mathrm{C}\)

Arrange the following in the order of decreasing tendency towards \(\mathrm{S}_{\mathrm{N}}^{2}\) reaction 1\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) 2\. \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{CHCl}-\mathrm{CH}_{3}\) 3\. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{Cl}\) 4\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Cl}\) (a) \(1>3>2>4\) (b) \(4>4>3>1\) (c) \(2>1>3>4\) (d) \(1>2>4>3\)

Identify the final product (C) in the following sequence of reactions. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \stackrel{\text { Alc. } \mathrm{KOH}, \Delta}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{Br}_{2}}{\longrightarrow}(\mathrm{B}) \stackrel{\mathrm{KCN}}{\longrightarrow}(\mathrm{C})\) (a) \(\mathrm{NCCH}_{2} \mathrm{CH}_{2} \mathrm{CN}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\) (d) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CN}\)

An alkyl chloride produces a single alkene on reaction with sodium ethoxide and ethanol. The alkene further undergoes hydrogenation to yield 2 -methylbutane. Identify the alkyl chloride from amongst the following. (a) \(\mathrm{ClCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{ClCH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{Cl})\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Butanitrile is formed by reaction of \(\mathrm{KCN}\) with (a) butyl chloride (b) propyl chloride (c) butyl alcohol (d) propyl alcohol

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