Why do even double bonds in allenes not show geometrical isomers?
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How to calculate the number of geometrical isomers in allenes?
organic-chemistry stereochemistry isomers geometrical-isomerism
Can someone explain how to calculate the number of geometrical isomers in allenes like CH3−(CH=CH)4−CH3 or CH3−(CH=CH)5−CH3
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May 15 '15 at 19:13
user14857
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May 15 '15 at 19:54
ron
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In the linear cases you drew, the answer is always two. They’re just sometimes called differently. (E/Z vs. R/S) – Jan May 15 '15 at 19:23
But suppose one terminal group is in a different plane compared to the other,will still there be 2 geometrical isomers? I don't think so. – user14857 May 15 '15 at 19:25
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Other than searching the net, what have you tried to figure this out? Have you drawn any compounds? – jerepierre May 15 '15 at 19:28
Ah, sorry, I misunderstood ‘geometrical isomer’ But what is the reason to consider axial R and S differently from E and Z? Is ther eany? – Jan May 15 '15 at 19:29
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@SanchayanDutta Put that information in the question. Show some of the compounds you have drawn. You will get better answers, and your question won't be seen as homework. – jerepierre May 15 '15 at 19:38
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If we have one double bond in a hydrocarbon compound we have an olefin or alkene. Ethylene is the simplest example of this class of compounds. The carbons in the double bond and the 4 atoms attached to them lie in the same plane. One pair of cis-trans isomers is possible in compounds with a single double bond.
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If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene. The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another.