Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Whi

NCERT Exemplar · Q35

Q. Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.

Yanam BieapShort· 3mImportance★★★★★

82% · 74/90 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Structural Isomerism

Structural Isomerism: The First Meeting

Imagine you have a box of identical Lego bricks — four red, ten blue, and six yellow. You build two different models: a car and a house. Both use exactly the same number of each colour brick, but the structures are completely different. That is the core idea of isomerism: same atoms, different arrangement.

In chemistry, molecules are not just a list of atoms. How those atoms are connected matters enormously. Two molecules can have the exact same molecular formula (same number of each atom) but be connected in different ways. Those are structural isomers (also called constitutional isomers).

The Precise Statement

Structural isomers are compounds that have the same molecular formula but different connectivity of atoms — that is, different structural formulas.

The key word is connectivity. Which atom is bonded to which? If you change that, you get a different substance with different physical and chemical properties.

A Concrete Example: C₄H₁₀

Take butane, C₄H₁₀. There are exactly two ways to connect four carbon atoms and ten hydrogen atoms:

n-Butane — a straight chain: C–C–C–C
Isobutane (2-methylpropane) — a branched chain: a central carbon bonded to three methyl groups

Both have formula C₄H₁₀. But n-butane boils at –0.5 °C, while isobutane boils at –11.7 °C. Same atoms, different connectivity → different substance.

Note

Structural isomers are not the same molecule. They are distinct compounds that happen to share a molecular formula. You cannot rotate or flip one to get the other — you must break and reform bonds.

The Three Main Types

Structural isomerism comes in three flavours:

Type	What changes	Example (C₃H₆O)
Chain isomerism	The carbon skeleton (straight vs. branched)	Butane vs. isobutane
Position isomerism	The location of a functional group or substituent	Propan-1-ol vs. propan-2-ol (OH on carbon 1 vs. carbon 2)
Functional group isomerism	The atoms are rearranged into a different functional group	Propanal (aldehyde) vs. propanone (ketone) — both C₃H₆O

Watch out

Do not confuse structural isomers with stereoisomers. Stereoisomers have the same connectivity but differ in spatial arrangement (like left and right hands). That is a completely different chapter. For now: structural isomers = different bond connections.

Why This Matters …

Why this formula?

Structural Isomerism: Why the Key Ideas Hold

Structural isomerism arises when molecules share the same molecular formula but differ in the connectivity of atoms. There is no single "formula" for structural isomerism — instead, the key is understanding why different arrangements are possible.

The Core Principle: Connectivity ≠ Composition

A molecular formula tells you how many of each atom are present, but not how they are joined. Structural isomers exist because atoms can form bonds in multiple distinct sequences while satisfying valency rules.

Why This Happens: The Valency Constraint

Each atom has a fixed bonding capacity (valency):

Carbon: 4 bonds
Hydrogen: 1 bond
Oxygen: 2 bonds
Nitrogen: 3 bonds

Example: For $C_4H_{10}$ , the formula satisfies $4(4) + 10(1) = 26$ valence electrons. But the carbon atoms can be arranged as:

A straight chain: $CH_3-CH_2-CH_2-CH_3$ (n-butane)
A branched chain: $CH_3-CH(CH_3)-CH_3$ (isobutane)

Both satisfy valency, but the connectivity differs.

The "Formula" for Counting Isomers: Why It's Not Simple

There is no closed-form formula to count structural isomers for a given molecular formula. The number grows rapidly and depends on:

Carbon skeleton branching possibilities
Functional group positions
Ring formation possibilities

Why No Simple Formula Exists

The problem is combinatorial — the number of possible trees (acyclic graphs) with $n$ carbon atoms grows exponentially. For example:

$C_4H_{10}$ : 2 structural isomers
$C_5H_{12}$ : 3 structural isomers
$C_6H_{14}$ : 5 structural isomers
$C_{10}H_{22}$ : 75 structural isomers

The pattern follows Cayley's formula for trees, but even that counts only carbon skeletons — not functional group positions.

Key Reasoning: The Branching Principle

The fundamental reason structural isomers exist is that carbon chains can branch. Consider $C_5H_{12}$ :

Straight chain: $C-C-C-C-C$ (n-pentane)
One branch: $C-C-C(C)-C$ (isopentane) — the branch can be at position 2 or 3, but these are identical due to symmetry
Two branches: $C-C(C)(C)-C$ (neopentane) — a quaternary carbon

Why position matters: The branch location changes the carbon's environment, altering physical and chemical properties.

The Functional Group Position Rule

For compounds with functional groups (e.g., alcohols $C_nH_{2n+2}O$ ), the position of the -OH group creates isomers:

$CH_3CH_2CH_2OH$ (propan-1-ol) — OH at end
$CH_3CH(OH)CH_3$ (propan-2-ol) — OH in middle …

Start your 14-day free trial to unlock the full solution →

Monochlorination of 2-methylpropane proceeds via free-radical intermediates. The two possible hydrocarbon radicals are the primary (1°) radical (from a methyl C–H) and the tertiary (3°) radical (from the methine C–H). The tertiary radical is more stable due to hyperconjugation and inductive effects.

The question asks about the radical intermediates formed during monochlorination of 2-methylpropane (isobutane). This is a classic free-radical substitution reaction, where a chlorine atom abstracts a hydrogen atom from the alkane, leaving behind an alkyl radical. The stability of these radicals determines the major product.

Why radicals? In the propagation step of chlorination, a chlorine radical ( $\text{Cl}\cdot$ ) pulls off a hydrogen atom. The carbon that loses the hydrogen becomes a carbon radical ( $\text{R}\cdot$ ). The type of carbon (primary, secondary, tertiary) from which hydrogen is removed dictates the radical's structure and stability.

Let’s break it down step by step.

Structure of 2-methylpropane 2-methylpropane has the formula $\text{C}_4\text{H}_{10}$ and a branched structure:

\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_3

The central carbon (C-2) is bonded to three methyl groups and one hydrogen — it is a tertiary carbon. The three methyl carbons are each primary carbons (each bonded to one other carbon and three hydrogens).

Possible hydrogen abstractions A chlorine radical can abstract a hydrogen from two distinct positions:
- From a primary carbon (any of the three methyl groups):

\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_3 + \text{Cl}\cdot \rightarrow \text{CH}_2\cdot-\text{CH}(\text{CH}_3)-\text{CH}_3 + \text{HCl}

 This gives a **primary radical** (1°), where the unpaired electron is on a carbon that is attached to only one other carbon.

From the tertiary carbon (the central carbon):

\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_3 + \text{Cl}\cdot \rightarrow \text{CH}_3-\text{C}\cdot(\text{CH}_3)-\text{CH}_3 + \text{HCl}

 This gives a **tertiary radical** (3°), where the unpaired electron is on a carbon attached to three other carbons.

No secondary radical is possible here because there is no secondary carbon in the molecule.

Stability comparison: Why is the tertiary radical more stable? Radical stability follows the order: tertiary > secondary > primary > methyl. The reason lies in two effects:
- Hyperconjugation: The unpaired electron in a radical can be delocalised by overlap with adjacent C–H sigma bonds. A tertiary radical has three alkyl groups attached, each providing C–H bonds that can donate electron density into the half-filled p-orbital. More alkyl groups mean more hyperconjugative structures, which stabilises the radical. For the tertiary radical here, there are 9 C–H bonds on the three methyl groups adjacent to the radical centre, offering extensive delocalisation.
- Inductive effect: Alkyl groups are electron-donating (by hyperconjugation and induction). They push electron density toward the electron-deficient radical centre, stabilising it. A tertiary carbon has three such donating groups, while a primary has only one. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ34 — Write the structures and names of products obtained in the reactions o…→Next questionQ36 — An alkane C8H18 is obtained as the only product on subjecting a primar…

Q.Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days