Answer:
X = 2-propanol
Y = 2-chloropropane
Z = propene
Explanation:
Secondary alcohol is an alcohol in which the carbon that the functional group (OH group) is attached to is also attached to two alkyl group. Examples are 2-propanol and 2-butanol.
The secondary alcohol referred to in the question here is 2-propanol and it reacts with PCl₅ as seen below
H₃C-CH-CH₃ + PCl₅ ⇒ H₃C-CH-CH₃ + HCl + POCl₃
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OH Cl
The alkylhalide produced there is 2-chloropropane. Thus, we can see that X is 2-propanol and Y is 2-chloropropane.
When this 2-chloropropane undergoes dehydrohalogenation (removal of hydrogen and halogen) an alkene is formed. The reaction is seen below
H₃C-CH-CH₃ (-HCl) ⇒ H₃C-CH=CH₂
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Cl
The alkene produced (as seen above) is propene.
Confirmation: When propene undergoes ozonolysis, the compounds produced are ethanal and methanal as seen in the equation below
O
/ \
H₃C-CH=CH₂ + O₃ ⇒ H₃C - CH CH₂ [-O] ⇒ CH + CH₂
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O O O O
The final products as seen above are methanal and ethanal
