Problem 110 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The hydrocarbon is 2,2,3,3-tetramethylbutane (option c).

Step 1: Understand the Molecule

The problem states that a hydrocarbon with the formula \(\mathrm{C}_{8} \mathrm{H}_{18}\) gives only one monochloro derivative. This implies that all hydrogen atoms in the molecule must be equivalent. When a single hydrogen is replaced with a chlorine, only one type of product should be possible.

Step 2: Analyze the Options

Let's first analyze each option individually and determine if it fulfills the condition of having all hydrogen atoms equivalent.(a) 2-methyl heptane: It is a branched alkane so hydrogen atoms are not equivalent.(b) Octane: Straight-chain alkanes have equivalent hydrogens only if they are symmetrical, which octane is not.(c) \( \text{2,2,3,3-tetramethylbutane} \): This is a fully symmetrical molecule where all hydrogen atoms are equivalent.(d) \(2,2,4\)-trimethyl pentane: This is a branched alkane, meaning not all hydrogen atoms are equivalent due to different environments.

Step 3: Identify the Correct Hydrocarbon

Based on the analysis, the correct hydrocarbon must be \( \text{2,2,3,3-tetramethylbutane} \) because it is fully symmetrical. Due to this symmetry, all hydrogen atoms in this molecule are in identical environments, resulting in only one type of monochloro derivative being possible.

Key Concepts

Hydrocarbons

Hydrocarbons are the simplest organic compounds composed solely of carbon (C) and hydrogen (H) atoms. They form the foundational structure for organic chemistry and are classified into various types based on the types of bonds between the carbon atoms:

• Alkanes: Contain single bonds only, following the formula \(C_nH_{2n+2}\).
• Alkenes: Include one or more double bonds, with the general formula \(C_nH_{2n}\).
• Alkynes: Feature one or more triple bonds, and are represented by \(C_nH_{2n-2}\).

Hydrocarbons can be straight-chained, branched, or cyclic, which impacts their physical and chemical properties. The focus here is on a hydrocarbon with the formula \(C_8H_{18}\), known as octane. This molecule can be found as different isomers, including straight and branched forms.
Each form may have different reactivity and properties, and understanding these variations is crucial when predicting the behavior in chemical reactions, such as chlorination.

Symmetry in Molecules

Symmetry in molecules refers to the uniformity and arrangement of parts within a molecule. This is crucial when predicting molecular properties and reactivity, including interactions with light and rearrangements with substituents like chlorine in monochloro derivatives.

• Molecular symmetry involves elements such as planes, axes, and centers of symmetry.
• A symmetric molecule appears identical upon certain rotations, reflections, or inversions.

In the context of the exercise, symmetry explains why only one monochloro derivative can be formed. For \(2,2,3,3\)-tetramethylbutane, symmetry means all hydrogens are equivalent. This is because its branches are evenly distributed around the carbon framework, making each hydrogen atom react identically with chlorine, thus resulting in only one type of monochlorinated product.
Identifying symmetrical hydrocarbons within options requires assessing molecular geometry, which plays a direct role in determining the equivalence of hydrogen atoms.

Monochloro Derivatives

Monochloro derivatives are products formed when one hydrogen atom in a hydrocarbon is substituted with a chlorine atom. This simple reaction is essential in organic synthesis and industrial applications. Here’s how it unfolds:

• The reaction requires a hydrocarbon and a chlorine source, typically involving ultraviolet light or heat to initiate.
• The substitution happens at a specific hydrogen atom, forming a new compound with a chlorine atom attached.

Significantly, the formation of monochloro derivatives depends on the equivalency of hydrogen atoms in the original hydrocarbon. A hydrocarbon like \(2,2,3,3\)-tetramethylbutane, which is fully symmetrical, permits only one unique monochloro compound because all hydrogens are equivalent.
Recognizing which hydrocarbons will yield a single monochloro derivative requires understanding both their structure and symmetry. This knowledge aids in predicting reaction outcomes and optimizing chemical processes.