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A hydrocarbon of molecular formula, \(\mathrm{C}_{8} \mathrm{H}_{18}\) gives only one monochloro derivative. The hydrocarbon is: (a) 2-methyl heptane (b) Octane (c) \(2,2,3,3\) -tetramethyl butane (d) \(2,2,4\) -trimethyl pentane

Short Answer

Expert verified
The correct answer is (c) 2,2,3,3-tetramethylbutane.

Step by step solution

01

Analyze the Problem

To solve this problem, we need to determine which hydrocarbon structure given in the options would produce only one monochloro derivative. A monochloro derivative is formed when one hydrogen atom is replaced by one chlorine atom. This can occur only if all the hydrogen atoms in the molecule are equivalent, meaning that the hydrocarbon should lack distinct types of hydrogen atoms.
02

Option Analysis

Each option must be evaluated to see how substituting a single hydrogen atom with a chlorine atom would affect the structure of the compound. This hinges on the symmetry and type of hydrogens present in the molecule. (a) In 2-methyl heptane, hydrogens are not equivalent. (b) Octane is a linear chain, but hydrogens are not all equivalent. (c) In 2,2,3,3-tetramethylbutane, all carbon atoms are linked such that all hydrogens are equivalent due to symmetry. (d) In 2,2,4-trimethylpentane, the hydrogens are not equivalent.
03

Identify the Correct Hydrocarbon

The hydrocarbon that produces only one structural isomer when a chlorine atom replaces any hydrogen atom is the one where all hydrogens are equivalent. For 2,2,3,3-tetramethylbutane, each carbon atom is bonded in a symmetric manner, making all hydrogens equivalent and resulting in only one monochloro derivative.

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Key Concepts

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

Monochloro Derivative
A monochloro derivative occurs when a chlorine atom replaces a single hydrogen atom in a hydrocarbon molecule. This process results in a new compound where only one of the many hydrogen atoms has been substituted by chlorine. Forming just one monochloro derivative from a hydrocarbon requires a special condition: all of its hydrogen atoms must be equivalent. This means any hydrogen can be swapped out with chlorine, and the resulting chemical structure remains the same no matter which one is replaced. Only molecules with such symmetry can form a single type of monochloro derivative. This concept plays a crucial role in understanding the exercise, as it helps identify which hydrocarbon structures are capable of generating just one derivative.
Symmetrical Hydrocarbons
Symmetrical hydrocarbons are molecules where the structure is evenly balanced around a center or axis. This symmetry results in the equivalence of all hydrogen atoms attached to the carbon backbone.
  • Such symmetry means that these compounds have identical environments for all hydrogens.
  • This also ensures that no matter which hydrogen is replaced with a chlorine atom, the resulting compound will be identical to others where a different hydrogen is replaced.
One great example is the symmetrical structure of 2,2,3,3-tetramethylbutane. In this molecule, symmetry allows for only one possible monochloro derivative to form, because all hydrogens are structurally identical. Recognizing this symmetry is key to identifying which hydrocarbons can produce only one derivative.
Hydrocarbon Structure
In organic chemistry, understanding hydrocarbon structure is fundamental for predicting chemical behavior. A hydrocarbon's structure determines how its atoms are organized, which affects physical and chemical properties, such as its ability to form specific derivatives.
  • Linear hydrocarbons have a straightforward chain-like structure, like octane, that may or may not have equivalent hydrogens.
  • Branched hydrocarbons, like 2-methyl heptane, often lack symmetry, resulting in non-equivalent hydrogen atoms.
The structure involving multiple branches, as seen in 2,2,3,3-tetramethylbutane, provides a perfectly symmetrical arrangement. This symmetry ensures that each hydrogen is equivalent, allowing only one monochloro derivative to form. Understanding the placement and connections of carbon atoms is thus essential for grasping how the molecule behaves during chlorination.
Equivalent Hydrogen Atoms
"Equivalent hydrogen atoms" refers to the concept where all hydrogens in a molecule can be interchanged without altering the overall structure of the molecule. This equivalence is often a result of structural symmetry in the hydrocarbon. When every hydrogen atom in the molecule shares the same type of environment, they are considered equivalent.
  • Equivalency of hydrogen atoms is crucial in determining if a hydrocarbon can create only one monochloro derivative.
  • It's a unique scenario because any substitution of these hydrogens results in the same molecular structure.
In the case of the exercise example, 2,2,3,3-tetramethylbutane has its hydrogens in an equivalent arrangement due to its symmetrical framework. This allows for the creation of just one monochloro derivative. Recognizing equivalent hydrogens helps in predicting the outcomes of substitution reactions in organic chemistry.

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Most popular questions from this chapter

The number of structural and configurational isomers of a bromo compound, \(\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{Br}\), formed by the addition of \(\mathrm{HBr}\) to 2 -pentyne respectively are: (a) 1 and 2 (b) 2 and 4 (c) 4 and 2 (d) 2 and 1

Reductive ozonolysis of the alkene, \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}_{2}\) will give: (a) Only \(\mathrm{CO}_{2}\) (b) Only \(\mathrm{CH}_{2}^{2} \mathrm{O}\) (c) Only \(\mathrm{CH}_{3}^{2}-\mathrm{CHO}\) (d) Mixture of \(\mathrm{CH}_{3} \mathrm{CHO}, \mathrm{HCHO}\) and \(\mathrm{CO}_{2}\)

Compare the \(\mathrm{C}-\mathrm{H}\) bond lengths of the following hydrocarbons:1. \(\mathrm{CH}_{3}-\mathrm{CH}_{3}\) 2\. \(\mathrm{CH}_{2}=\mathrm{CH}_{2}\) 3\. \(\mathrm{CH} \equiv \mathrm{CH}\) The decreasing order of \(\mathrm{C}-\mathrm{H}\) bond lengths is: (a) \(3,2,1\) (b) \(1,2,3\) (c) \(2,1,3\) (d) \(3,1,2\)

Which is the decreasing order of strength of bases: \(\mathrm{OH}^{-}, \mathrm{NH}_{2}^{-}, \mathrm{HC} \equiv \mathrm{C}^{-}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2}^{-} ?\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{OH}^{-}\) (b) \(\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}\) (c) \(\mathrm{OH}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}\) (d) \(\mathrm{NH}_{2}^{-}>\mathrm{HC} \equiv \mathrm{C}^{-}>\mathrm{OH}^{-}>\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}\)

The products formed by the ozonolysis-hydrolysis of a compound of formula, \(\mathrm{C}_{5} \mathrm{H}_{8}\) are \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) and \(\mathrm{CO}_{2} .\) The compound is: (a) Pent-2-yne (b) Pent-1-yne (c) Penta-1, 3 -diene (d) Penta-1, 4-diene

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