Question

The total number of possible monochloro derivative of 3-methyl pentane on reaction with \(\mathrm{Cl}_{2} / \mathrm{h} v\) is (A) 4 (B) 6 (C) 7 (D) 8

Short answer

The total number of possible monochloro derivatives of 3-methyl pentane when reacting with $\mathrm{Cl}_{2} / \mathrm{h} v$ is 6. This is because there are 6 unique positions within the molecule where a hydrogen atom can be replaced by a chlorine atom, resulting in different monochloro derivatives.

Step-by-step solution

Draw the structure of 3-methyl pentane

The structure of 3-methyl pentane is given by the following IUPAC name: pentane with a methyl group (CH3) attached to carbon atom 3. This molecule consists of five carbon atoms in a straight chain (pentane), and a methyl group bonded to carbon 3. Thus, its structure is: CH3 | CH3-CH2-C-CH2-CH3

Identify the reaction mechanism

The reaction of 3-methyl pentane with Cl2 under the influence of light (hv) proceeds via a radical mechanism, in which the Cl-Cl bond is homolytically cleaved into separate radicals. These chlorine radicals then abstract the hydrogen atom from various positions in the 3-methyl pentane molecule to form HCl and a (semi-stable) alkyl radical. In the next step, another chlorine radical reacts with this alkyl radical to form the monochloro-substituted product.

Identify unique hydrogen atom positions for substitution

To find out the total number of possible monochloro derivatives, we need to identify unique positions in the 3-methyl pentane molecule where hydrogen atoms could be replaced by a chlorine atom, leading to different derivative structures. There are the following groups of equivalent hydrogen atoms (meaning that if any hydrogen atom is replaced with chlorine atom, the resulting compound will be the same): 1. Hydrogens on C1 (primary) - 3 H atoms 2. Hydrogens on C2 (secondary) - 2 H atoms 3. Hydrogens on C5 (primary) - 3 H atoms 4. Hydrogen on C4 (secondary) - 1 H atom 5. Hydrogens on the methyl group attached to C3 (primary) - 3 H atoms 6. Hydrogen on C3 (tertiary) - 1 H atom This gives a total of 6 different unique positions where the hydrogen atom can be replaced with a chlorine atom to form monochloro derivatives. So the correct answer is (B) 6.

Key concepts

Organic Chemistry JEE

Organic chemistry is a crucial part of the Joint Entrance Examination (JEE), an engineering entrance assessment conducted in India. It explores the structure, properties, composition, reactions, and synthesis of carbon-containing compounds. In JEE, organic chemistry encompasses various reactions such as substitution, elimination, addition and more. A clear understanding of concepts like the configuration of organic molecules, their functional groups, and reaction mechanisms is vital. Concepts like monochloro derivatives become especially important, as they build the foundation for understanding complex organic reactions and synthesis.

Monochloro derivatives are organic compounds that contain a single chlorine atom in their molecular structure. These derivatives can be created from hydrocarbons through halogenation reactions, a topic often covered in JEE organic chemistry syllabus. Mastery over these reactions and understanding how to identify the number of possible isomers, as in the case of 3-methyl pentane, is an essential skill for students preparing for the exam.

Halogenation Reaction Mechanism

Halogenation is a type of substitution reaction widely discussed in organic chemistry, especially in the context of preparing for exams like the JEE. It involves the substitution of one or more hydrogen atoms in an organic compound with halogen atoms. This reaction occurs through a radical mechanism, which proceeds in three steps: initiation, propagation, and termination.

In the initiation step, a halogen molecule, such as Cl₂, is split into two halogen radicals, often with the help of ultraviolet light (represented as hv). During propagation, these halogen radicals react with the organic substrate, abstracting hydrogen atoms to form a new radical species. Finally, the new radical can react with another halogen radical to form the product while regenerating the reactive halogen radical, thus continuing the chain process. In termination, two radicals combine to form a stable molecule, thus ending the reaction chain. Understanding this mechanism is key to identifying the products formed in halogenation reactions, like the monochloro derivatives of 3-methyl pentane.

Unique Hydrogen Positions in Alkanes

Identifying unique hydrogen positions in alkanes is fundamental to predicting the number of possible isomers formed during reactions like halogenation. In organic chemistry, hydrogens attached to the same carbon, or to equivalent carbon atoms, are considered equivalent or 'chemically identical'. This means replacing any one of them with a halogen results in the same compound.

For example, in the case of 3-methyl pentane, the hydrogen atoms on the 1st and 5th carbon are all primary and equivalent; therefore, substitution on any one of these hydrogens yields the same product. Similarly, hydrogen atoms on carbons with the same substituents and connectivity are considered equivalent. It is essential to recognize the primary, secondary, and tertiary nature of carbon atoms to understand the replacement patterns correctly. Knowing how to identify these unique positions allows one to determine the number of different monochloro derivatives that can be formed from an alkane, which is a type of question frequently encountered in JEE organic chemistry problems.