secondary alcohol (X) reacts with PCL5 to give alkyl halide (Y) which on dehydrohalogenation yield an alkene(Z).The alkene upon ozonolysis gives the mixture of ethanal and methanal. Identify X ,Y ,Z

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

        |                                     |

       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

        |

       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  

                                                 ||     ||                     ||        ||

                                                O    O                    O      O

The final products as seen above are methanal and ethanal

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;