This question asks for the number of monochloro derivatives (monochlorinated products) formed when the given compound undergoes free radical chlorination. The compound is 2,2,3-trimethylpentane. To determine the number of distinct monochloro products, we must identify all the unique types of hydrogen atoms in the molecule that can be replaced by a chlorine atom.

Step 1: Draw the Structure

The compound is 2,2,3-trimethylpentane. Let's draw its carbon skeleton.

The parent chain is pentane: C1-C2-C3-C4-C5.
A methyl group is attached to carbon 2: so carbon 2 is a tertiary carbon with three alkyl groups.
A methyl group is attached to carbon 3: so carbon 3 is also a tertiary carbon.

The structure can be represented as: CH₃-C(CH₃)₂-CH(CH₃)-CH₂-CH₃.

Step 2: Identify Different Types of Hydrogens

We need to find all carbon atoms that have hydrogens and are chemically distinct. Let's label the carbons for clarity:

Carbon 1 (CH₃-): This is a primary carbon. However, it is attached to a tertiary carbon (C2). This makes its hydrogens distinct.
Carbon 2 (C(CH₃)₂): This is a tertiary carbon with no hydrogens.
The two methyl groups on Carbon 2: These are identical primary carbons. All 6 hydrogens on these two methyl groups are equivalent.
Carbon 3 (CH(CH₃)-): This is a tertiary carbon with one hydrogen.
The methyl group on Carbon 3: This is a primary carbon. Its 3 hydrogens are equivalent but distinct from others.
Carbon 4 (CH₂-): This is a secondary carbon. Its 2 hydrogens are equivalent.
Carbon 5 (CH₃): This is a primary carbon. Its 3 hydrogens are equivalent.

Step 3: List the Distinct Hydrogen Environments

From the analysis, we find five distinct types of hydrogens:

Type a: The 3 hydrogens on the methyl group attached to Carbon 3.
Type b: The 6 hydrogens on the two identical methyl groups attached to Carbon 2.
Type c: The 1 hydrogen on the tertiary Carbon 3.
Type d: The 2 hydrogens on the secondary Carbon 4.
Type e: The 3 hydrogens on the terminal methyl group, Carbon 5.
Type f: The 3 hydrogens on the methyl group attached to Carbon 1. (This is often a point of confusion. Carbon 1 is a CH₃ group attached to the highly substituted Carbon 2. Its environment is different from the methyl on Carbon 3 and Carbon 5, making it a sixth distinct type).

Let's double-check Carbon 1. Its structure is H₃C-C(CH₃)₂-. The carbon of this methyl group is attached to a carbon that has three methyl groups. None of the other primary carbons (on C3 or C5) are attached to such a crowded carbon. Therefore, the hydrogens on Carbon 1 are in a unique environment.

Step 4: Count the Monochloro Products

Chlorination can occur at any of these six distinct carbon atoms, each leading to a different structural isomer (monochloro derivative).

Therefore, the number of monochloro derivatives formed is 6.

Final Answer

The number of monochloro derivatives formed from 2,2,3-trimethylpentane is $6$ .

Related Topics & Formulae

Topic: Structural Isomerism (513)

Structural isomers are compounds that have the same molecular formula but different connectivity of atoms. In this problem, each chlorination at a different type of carbon creates a new structural isomer of chlorinated pentane.

Topic: Free Radical Halogenation (814, 1434)

This reaction mechanism involves three steps: initiation, propagation, and termination. The reactivity of different hydrogen atoms follows the order: tertiary > secondary > primary. However, the number of products depends on the number of distinct hydrogen types, not just their reactivity.

Key Theory: The number of monohalogenated structural isomers is equal to the number of distinct types of hydrogen atoms in the alkane. Equivalent hydrogens (those in identical chemical environments) will yield the same product upon substitution.