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In the following reaction sequence, X and Z, respectively are: $CH_3CH_2CH_2-OH + PCl_5 \rightarrow CH_3CH_2CH_2Cl + X + HCl$, then $CH_3CH_2CH_2Cl \xrightarrow[\Delta]{Alc.\ KOH} Y \xrightarrow[(C_6H_5CO)_2O_2]{HBr} Z$
A
$X = POCl_3$; $Z = CH_3-CH(Br)-CH_3$
B
$X = H_3PO_3$; $Z = CH_3CH_2CH_2-Br$
C
$X = H_3PO_3$; $Z = CH_3-CH(Br)-CH_3$
D
$X = POCl_3$; $Z = CH_3CH_2CH_2-Br$
Detailed Solution
$CH_3CH_2CH_2OH \xrightarrow{PCl_5} CH_3CH_2CH_2Cl + POCl_3 + HCl$, so X = $POCl_3$
$CH_3CH_2CH_2Cl \xrightarrow[\Delta]{alc.\ KOH} CH_3-CH=CH_2$ (Y)
$CH_3-CH=CH_2 \xrightarrow[(C_6H_5CO)_2O_2]{HBr} CH_3CH_2CH_2Br$ (Z, anti-Markovnikov/peroxide effect)
So X = $POCl_3$ and Z = $CH_3CH_2CH_2-Br$.