The major products A and B in the following reactions are

Step 1: Hydrogen Abstraction (Formation of Radical A)

When the starting nitrile, 2,3-dimethylbutanenitrile ($$(\text{CH}_3)_2\text{CH--CH(CH}_3\text{)--CN}$$), is heated with a peroxide initiator, the peroxide decomposes to form free radicals. These radicals abstract the most reactive hydrogen atom from the nitrile:

$$\text{Product A} = (\text{CH}_3)_2\text{CH--}\dot{\text{C}}(\text{CH}_3)\text{--CN}$$

• The hydrogen atom at the $$\alpha$$-position (adjacent to the cyano group) has a significantly lower bond dissociation energy because the resulting radical is stabilized by resonance delocalization into the $$\pi$$-system of the nitrile group ($$\text{--C}\equiv\text{N}$$).
• Therefore, abstraction happens exclusively at the $$\alpha$$-position to yield the stable $$\alpha$$-cyano radical intermediate:



• Step 2: Radical Addition to Alkene (Formation of B)

  The highly stable carbon radical A acts as a nucleophilic radical and adds to the terminal, less sterically hindered carbon of pent-1-ene ($$\text{CH}_2=\text{CH--CH}_2\text{--CH}_2\text{--CH}_3$$). This forms a new carbon-carbon bond and generates a secondary alkyl radical:

  $$\text{Intermediate Radical} = (\text{CH}_3)_2\text{CH--C(CH}_3\text{)(CN)--CH}_2\text{--}\dot{\text{C}}\text{H--CH}_2\text{--CH}_2\text{--CH}_3$$



• Step 3: Chain Propagation

  This newly formed secondary alkyl radical then abstracts a hydrogen atom from another molecule of the starting nitrile compound, completing the radical chain mechanism and yielding the final addition product B:

  $$\text{Product B} = (\text{CH}_3)_2\text{CH--C(CH}_3\text{)(CN)--CH}_2\text{--CH}_2\text{--CH}_2\text{--CH}_2\text{--CH}_3$$

Conclusion:

The reaction proceeds via preferential formation of the resonance-stabilized alpha-cyano radical intermediate followed by anti-Markovnikov-like radical addition across the terminal alkene, matching the structures given in Option B.

Answer: Option B