Animated Solution for Chemistry - Organic Chemistry: Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light? $C{H}_3-C{H}_2-CH=C{H}_2$ $\delta \gamma \beta \alpha$ (E)

When we encounter an alkene reacting with a halogen like bromine ($B{r}_2$) in the presence of ultraviolet light ($h u$) or high heat, our chemical intuition must immediately pivot. While halogens typically add across the double bond of an alkene at room temperature, the introduction of light fundamentally alters the reaction mechanism.

Light provides the energy required to homolytically cleave the $Br-Br$ bond, generating highly reactive bromine free radicals ($B{r}^{\bullet }$). These radicals initiate a Free Radical Substitution reaction. In alkenes, this substitution overwhelmingly favors the allylic position over the addition reaction.

To predict the outcome, we must map the anatomy of our molecule, but-1-ene ($C{H}_3-C{H}_2-CH=C{H}_2$). The question labels the carbons as $\alpha ,\beta ,\gamma ,$ and $\delta$.

By definition, an allylic carbon is an $s{p}^3$ hybridized carbon atom that is directly bonded to an $s{p}^2$ hybridized carbon of an alkene double bond. - The $\alpha$ and $\beta$ carbons are part of the double bond itself ($s{p}^2$ hybridized). - The $\gamma$ carbon is the $s{p}^3$ carbon immediately adjacent to the double bond. This is our allylic position. - The $\delta$ carbon is further down the chain and is considered an ordinary alkyl carbon.

The ease with which a hydrogen atom is replaced in a free radical substitution depends entirely on the stability of the free radical intermediate formed after the hydrogen is abstracted.

Imagine a bromine radical abstracting a hydrogen from the $\gamma$ position. This leaves behind an unpaired electron on the $\gamma$ carbon, creating an allylic free radical:

This specific structural arrangement is incredibly special. The unpaired electron is separated from the adjacent $\pi$ bond by exactly one single $\sigma$ bond. This allows the $p$-orbital containing the unpaired electron to overlap with the $\pi$ system of the double bond, leading to resonance delocalization:

Because the burden of the unpaired electron is shared across multiple atoms, the allylic radical is exceptionally stable—far more stable than a standard primary or secondary alkyl radical that would form if a hydrogen were abstracted from the $\delta$ position.

In chemical kinetics, a more stable intermediate implies a lower activation energy for its formation. Consequently, the $C-H$ bond that leads to the most stable radical is the weakest and easiest to break.

Since the abstraction of the $\gamma$-hydrogen yields the highly resonance-stabilized allylic radical, the $\gamma$-hydrogen is the most easily replaceable atom during this bromination reaction.

Final Answer: The correct option is (c) $\gamma$-hydrogen.