Which of the following show `G.I.`?
`a.`
`b.`
`c.`
`d.`

Explanation

`a.` It shows `G.I.`
`i.` Two atoms on the dotted line `(------------)` means that the groups are below the plane of the ring.
`ii.` Two atoms on the bold lines
`iii.` One atome on the dotted line `(------------)` and another atoms on the bold line means that one group is below and another is above the plane of the ring.
`b.` It shows `G.I.`

`c.` It does not show `G.I.`

`d.` It shows `G.I.`
Two higher priority groups on the same side , on two different `C` atoms of the double bond
Priority of `C_(2)H_(3)gt CH_(3)` and `OH gt LP` of `bar(e)s`
Tow higher priority groups on oppostie sides on two different `C` atoms of the double bond.
`e.` It does not show `G.I.`

`f.` It does not show `G.I.`
`g.` It shows `G.I.`

`h.` It shows `G.I.`

`i.` It does not show `G.I.`

`j`. It shows `G.I.` It shows three G.I.[Terminal groups `(CH_(3)`group) are same.]
where `n` is the number of double bonds.
`1.`
In the above case, both the structures reperesent the same compound.
2.
3.
`k`.
In case of aldo `-` oximes, the naming of syn and anti is different from that of keto `-` oximes. In aldo `-` oximes, syn is named in which the position `(-H)` and `(-OH)` groups are on the same side and anti in which position of `(H)` and `(OH)` groups are on the opposite sides.
Whereas in case of keto `-` oximes, the names syn and anti are carried out according to priority rule, just like the namiing of `E` and `Z`.

In `(I)` , the position of `(H)` and `(OH)` groups are in the opposite (anti) position, but according to the priority rule, `CH_(3)gt H, OH gt LP` of `bar(e)s,` two higher priority groupss `(CH_(3)` and `OH)` are on the same side , so it is Z.
In `(II)` , the position of `(H)` and `(OH)` groups are in the same `(syn)` position, and accroding to the priority rule, two higher priority groups are on the opposite sides, so it is `E`.
`1.` It shows `G.I.`