Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light?

Step 1: Identify compound E and its hydrogens

Compound E: $\ce{CH2=CH-CH2-CH3}$ (but-1-ene)

Labeling positions relative to the double bond:

• C1 ($\ce{CH2=}$): vinyl position
• C2 ($\ce{=CH-}$): vinyl position
• C3 ($\ce{-CH2-}$): allylic position ($\alpha$ to double bond in some notations, or the allylic carbon)
• C4 ($\ce{-CH3}$): homoallylic

Step 2: Apply allylic free radical stability

In free radical bromination (light), the allylic position is most reactive due to:

• Allylic radical stabilization by resonance: $\ce{CH2=CH-CH* <-> *CH2-CH=CH}$
• Bond dissociation energy of allylic C–H is lower (~356 kJ/mol) vs. secondary (~397 kJ/mol)

For but-1-ene: the C3 hydrogen is allylic and most easily replaced.

Answer: Option (b) — $\gamma$-hydrogen

Key Points to Remember:

• Allylic position = most reactive in free radical halogenation
• Allylic radical is stabilized by resonance with the $\pi$ system
• Reactivity: allylic > tertiary > secondary > primary > vinylic