Question

In the following reaction sequence, \(X\) and \(Z\), respectively, are: \[ {CH3CH2CH2OH →[PCl5] CH3CH2CH2Cl + X + HCl} \] \[ {CH3CH2CH2Cl →[alc.\ KOH][\Delta] Y →[HBr] Z} \]

• A. \(X=H_3PO_3,\ Z={CH3CH=CH2}\)
• B. \(X=POCl_3,\ Z={CH3-CH(Br)-CH3}\)
• C. \(X=H_3PO_3,\ Z={CH3CH2CH2Br}\)
• D. \(X=POCl_3,\ Z={CH3CH2CH2Br}\)

Hint:

Alcohol with \(PCl_5\) gives alkyl chloride, \(POCl_3\), and \(HCl\). Propene with HBr gives 2-bromopropane by Markovnikov addition.

Answer 1

The Correct Option is B

Step 1: Understanding the Topic:
This problem covers a sequence of reactions involving "Alcohols," "Haloalkanes," and "Hydrocarbons." It requires knowledge of nucleophilic substitution side-products, dehydrohalogenation (elimination), and regioselective addition to alkenes. Crucially, it tests the understanding of how "Peroxide" changes the outcome of $HBr$ addition to an asymmetrical alkene (the Kharasch effect).
Step 2: Key Formulas and Approach:
The approach involves breaking down the sequence step-by-step:
Step 1: Alcohol + $PCl_5 \rightarrow$ Alkyl Halide + inorganic side products.
Step 2: Alkyl Halide + Alcoholic $KOH$ $\rightarrow$ Alkene.
Step 3: Alkene + $HBr$/Peroxide $\rightarrow$ Anti-Markovnikov Halide.

Step 3: Detailed Explanation:

Reaction 1 (Find X): Propan-1-ol reacts with $PCl_5$. The organic product is 1-chloropropane. The inorganic side products are $HCl$ and $POCl_3$. Thus, $X = POCl_3$. (Note: $PCl_3$ would give $H_3PO_3$).
Reaction 2 (Find Y): 1-chloropropane treated with alcoholic $KOH$ undergoes dehydrohalogenation to form an alkene. $CH_3CH_2CH_2Cl \rightarrow CH_3CH=CH_2$ (Propene, $Y$).
Reaction 3 (Find Z): Propene reacts with $HBr$ in the presence of peroxide. Ordinarily, $HBr$ follows Markovnikov's rule (Br goes to the middle carbon). However, peroxide causes anti-Markovnikov addition.
Result: The Bromine adds to the terminal (primary) carbon. $Z = CH_3CH_2CH_2Br$.
Step 4: Final Answer:
X is $POCl_3$ and Z is 1-bromopropane, matching option (B).