Step 1: Add HBr to propene. Propene with HBr follows Markovnikov addition, putting bromine on the more substituted carbon: $CH_3CH=CH_2+HBr \rightarrow CH_3CHBrCH_3$ (isopropyl bromide), which is A. Step 2: Form the Grignard reagent B. Isopropyl bromide with $Mg$ in dry ether gives the Grignard reagent $(CH_3)_2CHMgBr$, which is B. Step 3: React the Grignard with benzaldehyde to get C. Adding $(CH_3)_2CHMgBr$ to benzaldehyde $C_6H_5CHO$ and working up gives a secondary alcohol, $C_6H_5CH(OH)CH(CH_3)_2$. So C is this phenyl-bearing secondary alcohol. Step 4: Identify D. The carbinol obtained after addition and protonation (before any further conversion) is the alcohol D, the same $C_6H_5CH(OH)CH(CH_3)_2$ framework. Step 5: Convert to E with HCl. Treating the benzylic secondary alcohol with HCl replaces $-OH$ by $-Cl$, giving the chloro derivative $C_6H_5CHCl\,CH(CH_3)_2$, which is E. Step 6: Choose the option. C is the phenyl substituted secondary alcohol and E is its chloro derivative, matching the figure in option (3). \[ \boxed{C=C_6H_5CH(OH)CH(CH_3)_2,\ E=C_6H_5CHClCH(CH_3)_2\ \text{(Option 3)}} \]
