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The following structure shows Geometrical isomerism (GI for short) because the -Cl and -Br are in same plane

2,8-dibromo-2,8-dichlorodispiro[3.1.36.14]decane

This structure does not show GI because the -Cl and -Br are in perpendicular planes and the distances between the atoms remain same

2,6-dibromo-2,6-dichlorospiro[3.3]heptane

But I have these questions which don't seem to be able to be solved by this logic

Q1)

1,5-dioxaspiro[3.3]heptane

Q2)

2-methyl-1,5-dioxaspiro[3.3]heptane

My immediate thinking is that since one sides has two hydrogens, it cannot show GI

Is my answer, logic correct? I really appreciate any help you can provide.

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  • $\begingroup$ Think again. Are really Br and Cl in same plane with C atoms? $\endgroup$
    – Mathew Mahindaratne
    Commented Mar 19 at 4:24
  • $\begingroup$ are you talking about the first structure? If yes , then arent Br , Cl is same plane because spiro rings are perpendicular to each other , and there are three of them .... so they will be in same plane and hence show GI $\endgroup$
    – Ninjametry
    Commented Mar 19 at 11:59
  • $\begingroup$ All $\mathrm{sp^3}$ hybrids are tetrahedral, which have two planes perpendicular to each other such as $\ce{CH4}$. Thus, all your four structures show in here are not planer. So, could have some mirror images not equal. $\endgroup$
    – Mathew Mahindaratne
    Commented Mar 19 at 19:28
  • $\begingroup$ it ain't planar $\endgroup$
    – Anjankumar
    Commented Mar 20 at 8:03

1 Answer 1

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The second one will definitely exhibit Geometrical Isomerism. Imagine the first ring is in the x-y plane. Then the methyl group attached to it will lie along the z axis.The second ring will be in the x-z plane. Now depending on whether the methyl group is oriented to the +z axis or -z axis, the distance between the oxygen atom of the second ring and the methyl group of the first ring will be different, thus the molecule will show GI.

As for the first problem, I don't know why your book says that it will exhibit GI. All possible configurations seem like the same to me.

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  • $\begingroup$ but according to what I had been taught , we always needed to see the extreme ends of the spiro case and not the intermediary atoms on the ring .... So does this mean that GI in spiro is not only specific to the first 2 cases (the ones with -Cl ,-Br)? $\endgroup$
    – Ninjametry
    Commented Mar 20 at 16:08

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