A Hydrocarbon $ [A] $ (molecular formula $ C_3H_6 $) on reaction with $ \text{Br}_2/\text{CCl}_4 $ gave $ [B] $. When $ [B] $ is heated with 2 moles of alcoholic KOH, it gave compound $ [C] $. 3 moles of Compound $ [C] $ when passed through red hot Iron tube forms $ [D] $. Identify $ [D] $.
Show Hint
To identify the final product in such reactions, consider the possibility of dehydrohalogenation and dehydrogenation reactions. These are common reactions involving halogenated hydrocarbons and alkenes, leading to the formation of aromatic compounds like benzene.
The molecular formula of the hydrocarbon \( [A] \) is \( C_3H_6 \), which indicates that it is propene. Let's analyze the reactions step by step: - Step 1: \( A \) (C₃H₆) reacts with \( \text{Br}_2/\text{CCl}_4 \), a halogenation reaction that leads to the formation of 1,2-dibromopropane. Therefore, compound \( [B] \) is 1,2-dibromopropane. - Step 2: When \( [B] \) (1,2-dibromopropane) is heated with alcoholic KOH, a dehydrohalogenation occurs, which leads to the formation of propene. Hence, compound \( [C] \) is propene. - Step 3: When 3 moles of \( [C] \) (propene) are passed through red-hot iron tubes, a benzene ring is formed through dehydrogenation. In this process, benzene is formed by the elimination of hydrogen atoms from propene.
Now, the reaction proceeds to form a specific substituted benzene. The correct substitution pattern on the benzene ring is 1, 3, 5-trimethylbenzene, where the methyl groups are located at the 1st, 3rd, and 5th positions of the benzene ring.
Thus, the correct compound \( [D] \) is 1, 3, 5-Trimethylbenzene.
