2-methyl butane on reacting with bromine in the presence of sunlight gives mainly
(a) 1-bromo-3-meth

Q: Why does the reaction need sunlight?

Sunlight (UV light) provides the exact quantum of energy required to undergo homolytic cleavage of the weak Br-Br bond. This initiates the free radical chain reaction by generating the first bromine radicals.

Q: Why is the tertiary free radical more stable than the others?

The tertiary free radical is highly stabilized by hyperconjugation. The adjacent C-H sigma bonds can overlap with the half-filled p-orbital of the radical carbon, effectively delocalizing the unpaired electron and lowering the energy of the system.

Q: Would chlorination give the exact same major product?

Chlorination is much less selective than bromination. While the $3^\circ$ product might still be formed, chlorine radicals are so reactive that they attack $1^\circ$ and $2^\circ$ hydrogens almost as easily, resulting in a significant mixture of different chlorinated products.

Q: What is the reactivity-selectivity principle?

It states that a highly reactive species (like a chlorine radical) is less selective in its attacks, grabbing whatever is closest. Conversely, a less reactive species (like a bromine radical) is highly selective and will wait to form the most thermodynamically stable intermediate.

Q: Can we use iodine for this free radical substitution?

Iodination is generally too slow and highly reversible to be practical under these standard conditions. The HI formed reduces the alkyl iodide back to the alkane. It requires an oxidizing agent (like nitric acid) to drive the reaction forward.

Question

2-methyl butane on reacting with bromine in the presence of sunlight gives mainly

Options:

(A) 1-bromo-3-methylbutane

(B) 2-bromo-3-methylbutane

(C) 2-bromo-2-methylbutane

(D) 1-bromo-2-methylbutane

Sigma Percentile · Accepted Answer

### The Setup: Sunlight and Halogens

Imagine you are in a laboratory, and you mix an alkane like **2-methylbutane** with bromine water. If you keep the flask in the dark, absolutely nothing happens. But the moment you expose it to sunlight or UV light ($h
u$), a vigorous reaction kicks off. Why? Because the energy from the light is just enough to break the weak $	ext{Br}-	ext{Br}$ bond symmetrically, creating highly reactive bromine free radicals.

This is the hallmark of a **free radical substitution reaction**. The bromine radical is hungry for an electron and will snatch a hydrogen atom from the alkane to form $	ext{HBr}$, leaving behind an alkyl free radical.

### The Hunt for Stability: Radical Intermediates

Now, 2-methylbutane isn't just a simple straight chain; it has different types of hydrogen atoms. We have primary ($1^\circ$) hydrogens at the ends, secondary ($2^\circ$) hydrogens in the middle, and exactly one tertiary ($3^\circ$) hydrogen attached to the branched carbon.

When the bromine radical approaches, it has a choice. Which hydrogen should it abstract? This is where the concept of **radical stability** comes into play. Free radicals are electron-deficient species. They desperately need electron density to stabilize themselves. Alkyl groups are electron-donating through the $+I$ (inductive) effect and, more importantly, through **hyperconjugation**.

The stability order of free radicals is strictly:
$$3^\circ > 2^\circ > 1^\circ$$

The tertiary carbon in 2-methylbutane is surrounded by three other carbon atoms, providing massive hyperconjugative stabilization to the resulting radical.

### The Final Strike: Bromine's Choice

Here is a crucial fact about bromination: it is **highly selective**. According to the Reactivity-Selectivity Principle, because the bromine radical is relatively stable and less reactive (compared to a chlorine radical), it is very picky. It will patiently wait to abstract the hydrogen that yields the most stable intermediate.

Therefore, the bromine radical almost exclusively attacks the $3^\circ$ hydrogen. Once the $3^\circ$ alkyl radical is formed, it quickly reacts with another $	ext{Br}_2$ molecule to form the final product. The tertiary hydrogen is replaced by a bromine atom, yielding **2-bromo-2-methylbutane** as the overwhelming major product.

*A quick tip for your exams:* If the reagent was chlorine instead of bromine, the reaction would be much less selective, and you would end up with a messy mixture of $1^\circ$, $2^\circ$, and $3^\circ$ chlorinated products. Always pay attention to the specific halogen being used!

Animated Solution for Chemistry - Hydrocarbons: 2-methyl butane on reacting with bromine in the presence of sunlight gives mainly

Select Answer:

Visualized Solution

\text{Reaction Setup}

\text{Regioselectivity of Bromination}

\text{Identifying Hydrogen Types}

\text{Free Radical Stability}

\text{Major Product Formation}

\text{Conclusion \& Extension}

The Sigma Insight: Haloalkanes & Haloarenes

The Setup

The Hunt for Stability

The Final Strike

Similar Questions

3-methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

How many chiral compounds are possible on monochlorination of 2-methyl butane?

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields I. $C_{2} H_{5} C H_{2} C H_{3} ∣ C ∣ C H_{3} - O C H_{3}$ II. $C_{2} H_{5} C H_{2} C H_{3} ∣ C = C H_{2}$ III. $C_{2} H_{5} C H = C H_{3} ∣ C - C H_{3}$

trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces

HBr reacts with $CH_{2} = CH - OCH_{3}$ under anhydrous conditions at room temperature to give

The major product in the following conversion is

Heating of 2-chloro-1-phenyl butane with EtOK/EtOH gives $X$ as the major product. Reaction of $X$ with $Hg(OAc)_{2} / H_{2} O$ followed by $NaBH_{4}$ gives $Y$ as the major product. $Y$ is

In the following monobromination reaction, the number of possible chiral products is

Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light? $C H_{3} - C H_{2} - C H = C H_{2}$ $δ γ β α$ (E)

The product formed in the first step of the reaction of $CH_{3} - CH_{2} - Br ∣ CH - CH_{2} ∣ CH - CH_{3} ∣ Br$ with excess $Mg/Et_{2} O$ ( $Et = C_{2} H_{5}$ ) is