Problem 70
Question
On monochlorination of \(n\)-pentane, the number of isomers formed is are (a) 4 (b) 3 (c) 2 (d) 1
Step-by-Step Solution
Verified Answer
The number of isomers formed is 3.
1Step 1: Understanding the Question
The problem asks us about the number of isomers formed when monochlorination occurs on \(n\)-pentane. This means we need to find how many unique structures can form when one hydrogen atom in \(n\)-pentane is replaced by a chlorine atom.
2Step 2: Structuring \(n\)-Pentane
\(n\)-Pentane is a straight-chain alkane with the structure \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \). Visualize this as a chain of five carbon atoms, each surrounded by hydrogen atoms.
3Step 3: Identifying Distinct Hydrogen Types
In \(n\)-pentane, we must examine each hydrogen atom to see how many distinct replacements with chlorine could occur. There are three differentiated positions based on the uniqueness of their adjacent carbons: hydrogen atoms on the terminal (end) carbons and hydrogen atoms on the internal (middle) carbons.
4Step 4: Counting Terminal Hydrogen Sites
Both ends of \(n\)-pentane have \(\text{CH}_3\) groups. Replacing a hydrogen from either \(\text{CH}_3\) yields the same molecule, therefore these collectively contribute only one unique isomer.
5Step 5: Counting Secondary Carbon Site Hydrogens
The two internal carbon atoms each connect to other carbons, forming a \(\text{CH}_2\) group. Replacing a hydrogen from any internal \(\text{CH}_2\) group gives rise to two more unique isomers, one from each \(\text{CH}_2\) group.
6Step 6: Summing the Total Isomers
Adding the contributions from terminal sites (1 isomer) and secondary carbon sites (2 isomers), we get a total of 3 unique isomeric forms.
Key Concepts
n-Pentane StructureChemical IsomersChlorination Reaction
n-Pentane Structure
To understand the n-pentane structure, imagine a simple chain of carbon atoms. This molecule, scientifically called pentane, consists of five carbon atoms bonded in a straight line, forming what's known as a straight-chain or linear alkane. The chemical formula of n-pentane is \( \text{C}_5\text{H}_{12} \). Each carbon atom is saturated with enough hydrogen atoms to make it stable. When you visualize n-pentane, think of it as:
The structural formula of n-pentane can be represented as \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \). Notice how the first and the last carbon atoms, known as terminal carbons, are different from the middle or internal carbons in their number of hydrogen atoms. Terminal carbons each bond with three hydrogen atoms to form \( \text{CH}_3 \) groups, while internal carbons bond with two hydrogen atoms forming \( \text{CH}_2 \) groups. This differentiation will play a key role when discussing isomers.
- A central spine of five carbon atoms
- Hydrogen atoms branching out from each carbon
The structural formula of n-pentane can be represented as \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \). Notice how the first and the last carbon atoms, known as terminal carbons, are different from the middle or internal carbons in their number of hydrogen atoms. Terminal carbons each bond with three hydrogen atoms to form \( \text{CH}_3 \) groups, while internal carbons bond with two hydrogen atoms forming \( \text{CH}_2 \) groups. This differentiation will play a key role when discussing isomers.
Chemical Isomers
Chemical isomers are fascinating phenomena. They occur when molecules with the same formula have different structural arrangements. For the case of n-pentane, mono-chlorinated isomers are formed when one hydrogen atom is replaced by a chlorine atom. Because of this single substitution, the resulting molecules will still have the same molecular formula, but vary in their structure due to where the substitution occurs.
The number of these distinct arrangements is what is counted as the total number of isomers in such a chemical reaction.
- Three forms of these isomers can be generated based on their different substitution sites in n-pentane.
- When a hydrogen atom from either \( \text{CH}_3 \) end is substituted, it leads to the same isomer. This explains why there is only one outcome for chlorine attaching to any terminal carbon.
- The internal carbons can lead to two different isomers, with each \( \text{CH}_2 \) group offering a distinct structure upon chlorination.
The number of these distinct arrangements is what is counted as the total number of isomers in such a chemical reaction.
Chlorination Reaction
The chlorination reaction involves substituting a chlorine atom for a hydrogen atom within an organic molecule. In the context of n-pentane, monochlorination is a process where one chlorine atom replaces a single hydrogen atom. This reaction requires the presence of ultraviolet light, which provides the energy necessary for breaking the \( \text{C-H} \) bond and forming the \( \text{C-Cl} \) bond.
Such reactions are crucial in organic chemistry. They allow chemists to create halogenated derivatives, which have various industrial and pharmaceutical applications. The efficiency and selectivity of the chlorination can be influenced by factors like the site of reaction, the type of starting material, and reaction conditions.
Through careful analysis of the chlorination reaction on n-pentane, three distinct isomers can be identified, originating from hydrogen atoms being replaced at varying positions in the carbon chain. Understanding these molecular interactions enhances our grasp of organic reactions, showcasing their importance in synthetic chemistry.
Such reactions are crucial in organic chemistry. They allow chemists to create halogenated derivatives, which have various industrial and pharmaceutical applications. The efficiency and selectivity of the chlorination can be influenced by factors like the site of reaction, the type of starting material, and reaction conditions.
Through careful analysis of the chlorination reaction on n-pentane, three distinct isomers can be identified, originating from hydrogen atoms being replaced at varying positions in the carbon chain. Understanding these molecular interactions enhances our grasp of organic reactions, showcasing their importance in synthetic chemistry.
Other exercises in this chapter
Problem 67
The total number of optically active isomers for \(\mathrm{CH}_{2} \mathrm{OH}(\mathrm{CHOH})_{3} \mathrm{CHO}\) are (a) 16 (b) 8 (c) 4 (d) 2
View solution Problem 69
Total number of isomers and optically active compounds in the isomers of \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{Br}\) are (a) 6,4 (b) 5,3 (c) 3,5 (d) 8,3
View solution Problem 71
The number of possible open chain (acyclic) isomeric compounds for molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10}\) would be (a) 5 (b) 6 (c) 7 (d) 4
View solution Problem 75
Assign the IUPAC name for the following compound. (a) 3,4 -dichloroazobenzene (b) (4-chlorophenyl) (3-chlorophenyl) diazene (c) 3,4 -bis(chlorophenyl) diazene (
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