[ashish]

please take a look at this question ...

https://postimg.org/image/hs8cy5rhn/

which double bond will reduce ?? i think the terminal one but then what is the purpose of showing two hydrogen like that

am i correct ?? please help

[pgk]

The purpose of showing two hydrogens like that, is that the extremely basic conditions may promote sequential migrations of the allylic double bond and formation of a conjugated double bond system, which is selectively reduced with Na/NH3.
But remember that your teacher’s opinion counts much more that mine.

[ashish]

hi...thanks for replying ....please take a look at what this book by william carrauthers is saying on allylic double bond

https://postimg.org/image/l1wy3p7cl/

[pgk]

Dear Ashish,
You are welcome but remember that Organic Chemistry is an experimental Science and not a theoretical one; which means that theory follows the prior experimental data and not that the experimental results ought to follow the theoretical rules.
In other words, the experimental results cannot be predicted in advance, except in very simple cases.
Indicatively, Birch reduction (metals + liquid ammonia) at 1/1 molar ratio selectively reduces the endocyclic bonds but it also selectively reduces the conjugated bonds and also selectively, the activated allylic moiety. Besides, the extremely alkaline conditions may also cause isomerization of the vinyl bond that is in α- position of the limonene's ring. That’s why it was previously mentioned that is your teacher’s opinion that counts.
Anyway, the product (Z) is also prepared by another method, as seen in the attached file. This can help you to decide which double bond will selectively be reduced, in the question (iii) and interpret the corresponding reaction mechanism, as well as to decide about the role of the shown hydrogens, whether significant or not. 

[ashish]

no no all the Z's are different...please see this

https://postimg.org/image/ipgohdll1/

actually this question was asked in Civil services exam of India ..and they dont release answer for any question they ask ....we have to prepare by ourselves thats why i was so confused

[pgk]

In case of State exams and not simple school exams, it is better to give the chemically correct answer (regardless if this works in practice, or not), which is:
The endocyclic double bond will be reduced in preference, because the co-present ethanol will form sodium ethoxide (EtONa), which is less basic than sodium amide (NaNH2) and does not favor double bond migrations that could lead to the formation of conjugated double bonds (though, the addition of ethanol is not necessary because the initial product already has a primary alcohol group).

[pgk]

What are the answers that you gave in the other two questions?