More than three million carbon compounds have been discovered in the field of chemistry. The diversity of these compounds is due to the capacity of carbon atoms for bonding with one another as well as with other atoms. Most of the carbon compounds are poor conductors of electricity and have low melting and boiling points.
Question: 1
Write the molecular formula of first two members of homologous series having functional group –Br.
Step 1: Understanding the homologous series of alkyl bromides (haloalkanes):
Alkyl bromides, also known as haloalkanes, are organic compounds that contain a carbon-bromine (C–Br) bond. They belong to a homologous series, where each successive compound differs by a -CH₂ unit.
Step 2: The first two members of the homologous series of alkyl bromides:
The first two members of this series are:
1. Bromoethane:
The molecular formula for bromoethane is \(C_2H_5Br\). It is formed by replacing one hydrogen atom in ethane (C₂H₆) with a bromine atom.
2. Bromopropane:
The molecular formula for bromopropane is \(C_3H_7Br\). It is formed by replacing one hydrogen atom in propane (C₃H₈) with a bromine atom.
Step 3: Conclusion:
Both bromoethane and bromopropane belong to the homologous series of alkyl bromides (haloalkanes), with each compound differing by a -CH₂ unit. They contain the functional group -Br, which is characteristic of haloalkanes.
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Question: 2
Given below are the formulae of some functional groups: Write the name of these functional groups.
Step 1: Understanding the aldehyde group:
The aldehyde group is represented by −CHO. It consists of a carbonyl group (C=O) attached to a hydrogen atom. The general formula for an aldehyde is R−CHO, where R is a hydrocarbon group.
- Aldehydes are characterized by the presence of the carbonyl group at the end of the carbon chain.
Step 2: Understanding the carbonyl group:
The carbonyl group is represented by −C=O, where a carbon atom is double-bonded to an oxygen atom.
- The carbonyl group is found in both aldehydes and ketones. However, its position distinguishes these two types of compounds:
- In aldehydes, the carbonyl group is attached to a hydrogen atom, forming the −CHO structure.
- In ketones, the carbonyl group is bonded to two carbon atoms, resulting in the structure −C(C)(C)=O.
Step 3: Conclusion:
(i) The aldehyde group is represented by −CHO, where the carbonyl group is attached to a hydrogen atom.
(ii) The carbonyl group −C=O is a functional group found in both aldehydes and ketones, but the difference lies in their bonding to other atoms.
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Question: 3
What would be observed on adding a 5% alkaline potassium permanganate drop by drop to some warm ethanol taken in a test tube ? State the role of KMnO4 in the reaction and write the chemical equation for the reaction involved.
Step 1: Reaction of potassium permanganate with ethanol:
When 5% alkaline potassium permanganate (KMnO₄) is added drop by drop to warm ethanol, a chemical reaction takes place. KMnO₄ acts as an oxidizing agent and oxidizes ethanol to ethanoic acid (acetic acid).
Step 2: Understanding the role of potassium permanganate:
KMnO₄ is a strong oxidizing agent. In this reaction, potassium permanganate provides oxygen atoms (represented as [O]) that help oxidize ethanol (CH₃CH₂OH) to ethanoic acid (CH₃COOH).
Step 3: Chemical equation:
The chemical equation for this reaction is:
\[
\text{CH}_3\text{CH}_2\text{OH} + 2[O] \xrightarrow{\text{KMnO}_4/\text{heat}} \text{CH}_3\text{COOH} + \text{H}_2\text{O}
\]
This shows that ethanol (CH₃CH₂OH) reacts with oxygen (provided by KMnO₄) to form ethanoic acid (CH₃COOH) and water (H₂O).
Step 4: Observation:
The pink color of KMnO₄ fades as it is consumed in the oxidation process, indicating that potassium permanganate is being reduced as it oxidizes ethanol.
Conclusion:
Potassium permanganate, acting as an oxidizing agent, oxidizes ethanol to ethanoic acid, and its pink color fades during this reaction due to its consumption.
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Question: 4
Write the name of the compound formed when ethanol is heated at 443 K temperature with excess of conc. H2SO4. What is the role of conc. H2SO4 in the reaction ? Write the chemical equation for the reaction involved.
Step 1: Understanding the reaction:
When ethanol (C₂H₅OH) is heated with excess concentrated sulfuric acid (H₂SO₄) at 443 K, ethene (C₂H₄) is formed. This reaction is an example of a dehydration reaction, where a water molecule is removed from ethanol.
Step 2: Role of concentrated sulfuric acid:
Concentrated sulfuric acid (H₂SO₄) acts as a dehydrating agent. It removes a water molecule from ethanol, leading to the formation of ethene (C₂H₄).
- H₂SO₄ also helps in promoting the elimination of water by providing the necessary conditions for the reaction to occur.
Step 3: Chemical equation:
The chemical equation for this reaction is:
\[
\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 443 \, \text{K}} \text{C}_2\text{H}_4 + \text{H}_2\text{O}
\]
In this equation, ethanol (CH₃CH₂OH) undergoes dehydration (removal of water) to form ethene (C₂H₄) and water (H₂O).
Step 4: Conclusion:
The reaction involves heating ethanol with concentrated sulfuric acid at 443 K, where sulfuric acid acts as a dehydrating agent, leading to the formation of ethene and water.
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