Problem 70 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

(a) Show by chemical equations only, how you would prepare the following from the indicated starting materials. Specify the reagents in each step of the synthesis. (i) Hexachlorethane, \(\mathrm{C}_{2} \mathrm{Cl}_{6}\), from calcium carbide. (ii) Chloroform from carbon disulphide. (b) Give one chemical test which would distinguish between \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) from \(\mathrm{CHCl}_{3}\).

Step-by-Step Solution

Answer

Hexachloroethane: prepared from calcium carbide by chlorination. Chloroform: from carbon disulphide via carbon tetrachloride. Distinguish ethanol by the iodoform test.

Step 1: Preparing Hexachloroethane from Calcium Carbide

To prepare hexachloroethane (\(\mathrm{C}_2\mathrm{Cl}_6\)) from calcium carbide, we start by reacting calcium carbide (\(\mathrm{CaC}_2\)) with water to produce ethylene (\(\mathrm{C}_2\mathrm{H}_2\)) and calcium hydroxide:\[\mathrm{CaC}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{C}_2\mathrm{H}_2 + \mathrm{Ca(OH)}_2\]Next, we chlorinate the ethylene in the presence of a catalyst such as iron (III) chloride, to obtain hexachloroethane:\[\mathrm{C}_2\mathrm{H}_2 + 6\mathrm{Cl}_2 \xrightarrow{\text{FeCl}_3} \mathrm{C}_2\mathrm{Cl}_6\]

Step 2: Preparing Chloroform from Carbon Disulfide

To obtain chloroform (\(\mathrm{CHCl}_3\)) from carbon disulphide (\(\mathrm{CS}_2\)), first react carbon disulphide with chlorine:\[\mathrm{CS}_2 + 6\mathrm{Cl}_2 \rightarrow \mathrm{CCl}_4 + \mathrm{S}_2\mathrm{Cl}_2\]The product, carbon tetrachloride (\(\mathrm{CCl}_4\)), can then be reacted with hydrogen in the presence of light:\[\mathrm{CCl}_4 + \mathrm{H}_2 \rightarrow \mathrm{CHCl}_3 + \mathrm{HCl}\]

Step 3: Chemical Test to Distinguish Ethanol from Chloroform

Ethanol (\(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\)) can be differentiated from chloroform (\(\mathrm{CHCl}_3\)) using an Iodoform Test. Add iodine and aqueous sodium hydroxide to both solutions. Ethanol will produce a yellow precipitate of iodoform (\(\text{CHI}_3\)), indicating a positive result:\[3\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 4\mathrm{I}_2 + 6\mathrm{NaOH} \rightarrow \mathrm{CHI}_3 + 5\mathrm{H}_2\mathrm{O} + 6\mathrm{NaI}\]Chloroform will not undergo this reaction and will remain unchanged.

Key Concepts

Chemical Equations

In organic chemistry synthesis, chemical equations play a crucial role. They represent the series of chemical reactions involved in transforming starting materials into desired products. For example, the preparation of hexachloroethane involves converting calcium carbide into ethylene, then chlorinating it to get hexachloroethane. Here's the breakdown:

• The first reaction involves water and calcium carbide to produce acetylene and calcium hydroxide: \(\mathrm{CaC}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{C}_2\mathrm{H}_2 + \mathrm{Ca(OH)}_2\).
• The second reaction is chlorination of acetylene to form hexachloroethane: \(\mathrm{C}_2\mathrm{H}_2 + 6\mathrm{Cl}_2 \xrightarrow{\text{FeCl}_3} \mathrm{C}_2\mathrm{Cl}_6\).

These equations demonstrate how starting materials transform step-by-step, relying on reagents and conditions for successful reactions.

Reagent Selection

Choosing the correct reagents is vital in chemical synthesis to ensure each step proceeds smoothly and efficiently. In the preparation of chloroform from carbon disulphide, careful choice of reagents greatly impacts the reaction's outcome.

• Initially, chlorine reacts with carbon disulphide, leading to the formation of carbon tetrachloride and disulfur dichloride: \(\mathrm{CS}_2 + 6\mathrm{Cl}_2 \rightarrow \mathrm{CCl}_4 + \mathrm{S}_2\mathrm{Cl}_2\).
• Next, hydrogen reduces carbon tetrachloride in the presence of light to produce chloroform: \(\mathrm{CCl}_4 + \mathrm{H}_2 \rightarrow \mathrm{CHCl}_3 + \mathrm{HCl}\).

The reagents serve as essential components, driving reactions forward effectively.

Chemical Tests

Chemical tests help identify compounds by examining their reactions with specific indicators. In distinguishing ethanol from chloroform, the Iodoform Test is a reliable method.

• In this test, iodine and sodium hydroxide are used.
• Ethanol reacts, producing a yellow precipitate known as iodoform: \(3\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 4\mathrm{I}_2 + 6\mathrm{NaOH} \rightarrow \mathrm{CHI}_3 + 5\mathrm{H}_2\mathrm{O} + 6\mathrm{NaI}\).
• Chloroform, however, does not react, indicating a negative test result.

This simple chemical test is widely used for identifying alcohols capable of forming iodoform.

Hexachloroethane

Hexachloroethane, a highly chlorinated organic compound, stems from chlorinating ethylene. Understanding its preparation involves appreciating both the initial formation of ethylene from calcium carbide and its subsequent reactions.

• Calcium carbide reacts with water, forming acetylene, a key intermediate.
• The presence of a catalyst like \(\text{FeCl}_3\) expedites the chlorination process, converting acetylene to hexachloroethane.

Hexachloroethane has applications in different industries, including pyrotechnics and as a smoke agent, demonstrating its versatility as a chemical compound.

Chloroform Preparation

Chloroform, an organic solvent, has a unique preparation method involving carbon disulphide transformation through chlorination and reduction.

• The initial step involves chlorinating carbon disulphide to yield carbon tetrachloride, an intermediate.
• Subsequently, through reduction with hydrogen, chloroform is produced along with hydrochloric acid as a byproduct.

Chloroform has historically been used as an anesthetic, though its usage has diminished due to safety concerns. Its synthesis showcases the complexity and precision required in chemical transformations.