In the following reaction sequence, \(X\) and \(Z\), respectively, are: \[ {CH3CH2CH2OH →[PCl5] CH3CH2CH2Cl + X + HCl} \] \[ {CH3CH2CH2Cl →[alc.\ KOH][\Delta] Y →[HBr] Z} \]
In the following reaction sequence, \(X\) and \(Z\), respectively, are: \[ {CH3CH2CH2OH →[PCl5] CH3CH2CH2Cl + X + HCl} \] \[ {CH3CH2CH2Cl →[alc.\ KOH][\Delta] Y →[HBr] Z} \]
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\(X=H_3PO_3,\ Z={CH3CH=CH2}\)
\(X=POCl_3,\ Z={CH3-CH(Br)-CH3}\)
\(X=H_3PO_3,\ Z={CH3CH2CH2Br}\)
\(X=POCl_3,\ Z={CH3CH2CH2Br}\)
The Correct Option is B
Solution and Explanation
Step 1: Reaction of alcohol with PCl5.
Alcohol reacts with phosphorus pentachloride to form alkyl chloride.
General reaction: $$ ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl $$ Here: $$ CH_3CH_2CH_2OH $$ gives: $$ CH_3CH_2CH_2Cl $$ Therefore: X = POCl3
Step 2: Reaction of alkyl chloride with alcoholic KOH.
Alcoholic KOH causes dehydrohalogenation. $$ CH_3CH_2CH_2Cl \xrightarrow{alc.\ KOH,\ \Delta} CH_3CH=CH_2 $$ Therefore: Y = CH3CH=CH2 (propene)
Step 3: Addition of HBr to propene.
Propene reacts with HBr according to Markovnikov’s rule. $$ CH_3CH=CH_2 + HBr \rightarrow CH_3CH(Br)CH_3 $$ Therefore: Z = CH3–CH(Br)–CH3
Step 4: Final answer.
X = POCl3
Z = CH3–CH(Br)–CH3
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