Step 1. To determine the structure of compound (C).
(i) Since (C) on treatment with HBr in presence of peroxide (anti-Markovnikov's addition) gives 1-bromo-3-phenylpropane, therefore, (C) must be 3-phenyl-1-propene.
C6H5CH2CH=CH2HBr/Peroxide−−−−−−−−−−−−→(Anti-Mark. Addn.)C6H5CH2CH2CH2Br
3-Phenyl -1-propene (C) 1-Bromo-3-phenylpropane
Step 2 To determine the structures of compounds (D) and (E).
Now (C), i.e., 3-phenyl-1-propene being a terminal alkene does not show geometrical isomerism. However, when (C) is heated with KOH, it gives a mixture of two geometrical isomers, (D and E). Since (C) has the same M.F. as (E) , therefore, (D) and (E) are isomers of (C) , i.e., on heating with KOH (C) undergoes isomerization to give the more stable 1-phenyl-1-propene.
Further since out of (D) and (E) , (D) predominates, therefore (D) must be the more stable trans isomer, i.e., trans-1-phenyl-1-propene and (E) must be the less stable cis-isomer, i.e., cis-1-phenyl-1-propene isomer. (Refer Image)
Step 3: To determine the structures of compounds (A) and (B).
Since (C), i.e., 3-phenyl-1-propene is obtained by the reaction between a Grignard reagent (A) and a haloalkane (B), therefore (A) must be phenyl magnesium bromide and (B) must be 3-bromo-1- propene.
C6H5MgBr + BrCH2CH=CH2Δ−→C6H5CH2CH=CH2+MgBr2
Phenyl mag. bromide (A) 3-Bromo-1-propene (B) 3-phenyl-1-propene (C)
Thus, (A) = phenyl magnesium bromide , (B) = 3-bromo-1-propene, (C) = 3-phenyl-1-propene, (D) = trans-1-phenyl-1-propene and (E) = cis-1-phenyl-1-propene.