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

To solve the given reaction sequence, we need to analyze each step and determine the structures of the intermediate and final compounds.

1. First Reaction:

The reaction given is: \(CH_3CH_2CH_2OH \xrightarrow{PCl_5} CH_3CH_2CH_2Cl + X + HCl\)

In this reaction, 1-propanol reacts with phosphorus pentachloride (\(PCl_5\)) to form 1-chloropropane (\(CH_3CH_2CH_2Cl\)), hydrogen chloride (\(HCl\)), and another by-product \(X\). With \(PCl_5\), the by-product formed is typically phosphorus oxychloride (\(POCl_3\)). Hence, \(X = POCl_3\).

1. Second Reaction:

The sequence proceeds with: \(CH_3CH_2CH_2Cl \xrightarrow{alc. \ KOH, \ \Delta} Y\)

This step involves dehydrohalogenation, where alkali (alcoholic KOH) is used to eliminate hydrogen halide (HCl) from the alkyl halide. The reaction results in the formation of an alkene. From 1-chloropropane, 1-propene (\(CH_3CH=CH_2\)) is formed.

1. Third Reaction:

The sequence continues with: \(Y \xrightarrow{HBr} Z\)

When 1-propene reacts with \(HBr\), Markovnikov's rule is applicable here, which states that the hydrogen atom will attach to the carbon with more hydrogen atoms. Therefore, the major product is 2-bromopropane (\(CH_3-CH(Br)-CH_3\)).

1. Conclusion:

To summarize, the correct choices for \(X\) and \(Z\) are:

• \(X = POCl_3\)
• \(Z = CH_3-CH(Br)-CH_3\) (2-Bromopropane)

Therefore, the correct answer to the reaction sequence is:

\(X=POCl_3,\ Z={CH3-CH(Br)-CH3}\)