Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds, is
(a) 2-meth

Q: Why do equivalent hydrogens give the same monochlorinated product?

Because replacing any one of the equivalent hydrogens with a chlorine atom results in the exact same chemical structure, just viewed from a different angle or rotated.

Q: How do I quickly spot equivalent hydrogens?

Look for symmetry! If you can fold the molecule in half and both sides match perfectly, the hydrogens that mirror each other are equivalent. Also, hydrogens on the same carbon (like in a methyl group) are always equivalent.

Q: Does the reaction condition (like UV light) change the number of products?

UV light is required for the free radical chlorination to happen, but it doesn't change the *number of possible structural isomers* formed. It only affects the rate and sometimes the relative yield of each product.

Q: What about stereoisomers? Do we count them?

Usually, in such basic questions, we only count structural (constitutional) isomers unless the question specifically asks for total products including stereoisomers. Here, we are only looking for structural monochlorinated compounds.

Q: Why does 2,2-dimethylbutane have 3 types of hydrogens and not 4?

The three methyl groups attached to the quaternary carbon (C2) are all in the exact same chemical environment. They can freely rotate, making all 9 of their hydrogens completely equivalent.

Question

Of the five isomeric hexanes , the isomer which can give two monochlorinated compounds, is

Options:

(A) 2-methylpentane

(B) 2,2-dimethylbutane

(C) 2,3-dimethylbutane

(D) n-hexane

Sigma Percentile · Accepted Answer

### The Core Principle: Symmetry and Equivalence

When an alkane undergoes free radical halogenation (like chlorination in the presence of UV light), a chlorine atom can theoretically replace any hydrogen atom on the carbon skeleton. However, not all replacements lead to a unique molecule.

If two or more hydrogen atoms are in the exact same chemical environment—meaning they are structurally indistinguishable due to the symmetry of the molecule—replacing any one of them will yield the exact same product. We call these **equivalent hydrogens**.

Therefore, the golden rule for these types of problems is beautifully simple: **The number of possible structural monochlorinated products is exactly equal to the number of distinct (non-equivalent) types of hydrogen atoms in the parent alkane.**

### Analyzing the Contenders

Let's put on our detective glasses and analyze the symmetry of each given isomer of hexane ($C_6H_{14}$).

**1. 2-methylpentane**
Imagine the structure: $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$. 
The two methyl groups attached to the second carbon are identical (Type a). The methine ($CH$) proton is unique (Type b). The next methylene ($CH_2$) is unique (Type c), followed by another unique methylene (Type d), and finally a unique terminal methyl group (Type e). 
Because there is no plane of symmetry cutting through the middle of the chain, this molecule has **5 distinct types of hydrogens**, leading to 5 different monochlorinated products.

**2. 2,2-dimethylbutane**
Structure: $CH_3-C(CH_3)_2-CH_2-CH_3$.
Look at the quaternary carbon (C2). It holds three identical methyl groups. All 9 hydrogens on these three methyls are completely equivalent (Type a). The adjacent $CH_2$ group is unique (Type b), and the terminal $CH_3$ is unique (Type c). 
This gives us **3 distinct types of hydrogens**, resulting in 3 products.

**3. n-hexane**
Structure: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$.
This straight-chain alkane has a clear plane of symmetry right down its center. The two terminal methyls mirror each other (Type a). The two adjacent $CH_2$ groups mirror each other (Type b). The two innermost $CH_2$ groups mirror each other (Type c). 
Thus, n-hexane has **3 distinct types of hydrogens**, yielding 3 products.

### The Winning Isomer: 2,3-dimethylbutane

Now, let's look at **2,3-dimethylbutane**: $CH_3-CH(CH_3)-CH(CH_3)-CH_3$.

This molecule is a masterpiece of symmetry. It possesses a $C_2$ axis of symmetry and a plane of symmetry.

Notice how the left half of the molecule is a perfect mirror image of the right half. Furthermore, on each half, the two methyl groups attached to the methine carbon are identical. Because of this high degree of symmetry, **all four methyl groups** in the entire molecule are in the exact same chemical environment (Type a).

Similarly, the two central methine ($CH$) hydrogens are perfectly identical to each other (Type b).

There are absolutely no other types of hydrogens present. We only have Type 'a' and Type 'b'.

### The Final Verdict

Because 2,3-dimethylbutane possesses only **2 distinct types of non-equivalent hydrogens**, it can only possibly form **2 monochlorinated compounds**.

This elegant application of molecular symmetry leads us directly to the correct answer. Whenever you face a product-counting question in organic chemistry, always let symmetry be your guide!

Animated Solution for Chemistry - Hydrocarbons: Of the five isomeric hexanes , the isomer which can give two monochlorinated compounds, is

Select Answer:

Visualized Solution

\text{Isomerism and Monochlorination}

\text{Analyzing 2-methylpentane}

\text{Analyzing 2,2-dimethylbutane}

\text{Analyzing 2,3-dimethylbutane}

\text{Analyzing n-hexane}

\text{Conclusion}

The Sigma Insight: Alkanes

The Core Principle

Analyzing the Contenders

The Winning Isomer: 2,3-dimethylbutane

The Final Verdict

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