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Which of the following compounds is not chiral? [2004] (a) 2 -chloropentane (b) 1 -chloropentane (c) 3-chloro-2-methylpentane (d) 1-chloro-2-methylpentane

Short Answer

Expert verified
The non-chiral compound is (b) 1-chloropentane.

Step by step solution

01

Understanding Chirality

A compound is chiral if it has a carbon atom with four different groups attached, making it asymmetric. This carbon is known as a chiral center or stereocenter.
02

Analyzing Option (a) 2-chloropentane

2-Chloropentane has the structure: CH3-CHCl-CH2-CH2-CH3. The second carbon atom is bonded to four different groups: CH3, Cl, H, and CH2-CH2-CH3, making it a chiral center.
03

Analyzing Option (b) 1-chloropentane

1-Chloropentane is structured as: Cl-CH2-CH2-CH2-CH2-CH3. The first carbon atom bonded to chlorine is also attached to two hydrogen atoms and one alkyl group (butyl chain), meaning it doesn't have four different groups attached. This compound lacks any chiral centers.
04

Analyzing Option (c) 3-chloro-2-methylpentane

In 3-chloro-2-methylpentane, the third carbon is bonded to Cl, H, CH3, and CH(CH3)2 (isopropyl group). With four different groups attached, this carbon is a chiral center.
05

Analyzing Option (d) 1-chloro-2-methylpentane

For 1-chloro-2-methylpentane, the first carbon atom is bonded to Cl, two hydrogen atoms, and a carbon chain, lacking chirality as it does not have four different substituents.
06

Conclusion: Identify the Non-Chiral Compound

Upon examination, (b) 1-chloropentane has no chiral center because its first carbon is bonded to two identical hydrogen atoms. Therefore, it is the non-chiral compound.

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Key Concepts

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

Chiral Center Identification
Chiral center identification is a crucial step in determining the chirality of organic compounds. A chiral center, also known as a stereocenter, is typically a carbon atom bonded to four different groups. This configuration causes the molecule to exist in two non-superimposable mirror images, known as enantiomers. To identify chiral centers in a molecule, follow these steps:
  • Look at each carbon atom and identify the groups attached to it.
  • If exactly four distinct groups exist, you've found a chiral center.
  • If any two groups are identical, then that carbon cannot be a chiral center.
The presence of a chiral center in a molecule often leads to unique physical and chemical properties, crucial in fields like pharmaceuticals and materials science.
Stereochemistry
Stereochemistry refers to the study of the spatial arrangement of atoms within a molecule. This branch of chemistry is essential for understanding how different molecular shapes affect the properties and reactions of substances. The focus is on isomers—molecules that have the same molecular formula but differ in the spatial arrangement of atoms.

There are several key terms in stereochemistry to be familiar with:
  • Enantiomers: These are pairs of molecules that are non-superimposable mirror images of each other.
  • Diastomers: Not mirror images and have different physical properties.
  • Cis-Trans Isomerism: A form of stereoisomerism where the same atoms are connected but differ in the spatial positioning around double bonds or ring structures.
Stereochemistry is significant because the spatial arrangement of atoms can influence how a compound interacts with biological molecules. This affects the compound’s functionality, especially in drugs, where the "wrong" isomer might not bind as intended in biological systems.
Asymmetric Carbon Atom
An asymmetric carbon atom is the key to defining chiral compounds. It is a carbon atom attached to four different atoms or groups. This asymmetry makes the carbon atom a chiral center. Asymmetric carbon atoms are central to the concept of chirality and stereochemistry because they lead to optical activity, meaning the compound can rotate plane-polarized light.

To understand asymmetric carbon atoms, consider these points:
  • An asymmetric carbon is usually denoted by an asterisk (*) in chemical structures.
  • They give rise to enantiomers, each of which is a mirror image of the other that cannot be superimposed.
  • Identifying asymmetric carbons aids in determining the specific stereochemistry of a compound.
When analyzing complex organic compounds, noting all asymmetric carbon atoms helps in predicting behavior and reactivity, which is vital for applications in chemical synthesis and pharmaceuticals.

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Most popular questions from this chapter

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{OH} \frac{\mathrm{NaBr}, \mathrm{H}_{2} \mathrm{SO}_{4}, \Delta}{\longrightarrow}\) 3\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \stackrel{\mathrm{PBr}_{3}}{\longrightarrow}\) 4\. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{OH} \stackrel{\mathrm{sOC}_{3}}{\longrightarrow}\) List II (Alkyl halides) A. \(\mathrm{CH}_{3} \mathrm{CHBr}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}\) B. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{Cl}\) C. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) D. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{Br}\) The correct matching is \(\begin{array}{rrrrr}1 & 2 & 3 & 4 \\ \text { (a) } \mathrm{C} & \mathrm{D} & \mathrm{B} & \mathrm{A}\end{array}\) (b) \(\mathrm{C} \quad \mathrm{D} \quad \mathrm{A} \quad \mathrm{B}\) (c) \(\mathrm{D} \quad \mathrm{C} \quad \mathrm{A} \quad \mathrm{B}\) (d) D \(\mathrm{C} \quad \mathrm{B} \quad \mathrm{A}\)

Identify \(Z\) in the following series: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \stackrel{\mathrm{Alc} . \mathrm{KOH}}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{Br}_{2}}{\longrightarrow \mathrm{Y}} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{Z}\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\) (b) \(\mathrm{NCCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}\) (c) \(\mathrm{BrCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}\) (d) \(\mathrm{BrCH}=\mathrm{CHCN}\)

Which of the following represents the correct order of densities? (a) \(\mathrm{CCl}_{4}>\mathrm{CHCl}_{3}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CCl}_{4}>\mathrm{CHCl}_{3}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{3} \mathrm{Cl}\) (c) \(\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{CHCl}_{3}>\mathrm{CCl}_{4}\) (d) \(\mathrm{CCl}_{4}>\mathrm{CHCl}_{3}>\mathrm{H}_{2} \mathrm{O}>\mathrm{CH}_{2} \mathrm{Cl}_{2}>\mathrm{CH}_{3} \mathrm{Cl}\)

1-chlorobutane on reaction with alcoholic potash gives (a) 1 -butene (b) 1 -butanol (c) 2 -butene (d) 2-butanol

Consider the following sequence of reactions \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\text { Ethanolic } \mathrm{KOH}}{\longrightarrow}(\mathrm{X}) \stackrel{\mathrm{Br}_{2}}{\longrightarrow}(\mathrm{Y})\) \(\frac{\mathrm{NaNH}_{2}}{\mathrm{liq} \cdot \mathrm{NH}_{3}}(\mathrm{Z})\) The end product \((Z)\) is (a) propane (b) propyne (c) propene (d) propan-2-al

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