The product B formed in the following reaction sequence is: (Major product)

Step 1: Identify starting material

"Benzene-propene" likely means allylbenzene (3-phenylpropene) or propenylbenzene:

Structure: $\ce{C6H5-CH2-CH=CH2}$ (allylbenzene)

Or: $\ce{C6H5-CH=CH-CH3}$ (β-methylstyrene)

Step 2: Reaction with HCl

Addition of HCl to the double bond follows Markovnikov's rule:

• H goes to the carbon with more H atoms
• Cl goes to the more substituted carbon

For allylbenzene ($\ce{C6H5-CH2-CH=CH2}$): $\ce{C6H5-CH2-CH=CH2 + HCl -> C6H5-CH2-CHCl-CH3}$

Product A = 1-chloro-1-phenylpropane (secondary chloride)

For β-methylstyrene ($\ce{C6H5-CH=CH-CH3}$): $\ce{C6H5-CH=CH-CH3 + HCl -> C6H5-CHCl-CH2-CH3}$

Product A = 2-chloro-1-phenylpropane (secondary benzylic chloride)

Step 3: Reaction with AgCN

Product A + AgCN → substitution

AgCN is a covalent silver salt → attacks through nitrogen → forms isocyanide ($\ce{R-NC}$)

$\ce{C6H5-CHCl-CH2-CH3 + AgCN -> C6H5-CH(NC)-CH2-CH3}$

Product B = isocyanide at the benzylic position

The NC group is at the secondary benzylic carbon (where Cl was).

Answer: (d)

Key Points:

• HCl addition to alkenes: Markovnikov's rule
• AgCN (covalent) → isocyanide ($\ce{R-NC}$)
• KCN/NaCN (ionic) → nitrile ($\ce{R-CN}$)
• Benzylic position is reactive due to carbocation stability