Chemistry 1

Chemistry Level 2

The smallest alkane which can show optical isomerism possesses:

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8 carbon 6 carbon 7 carbon 5 carbon

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Sahil Bansal
Oct 2, 2015

For the alkane to be optically active i.e to show optical isomerism it is necessary that all the 4 groups attached to the carbon are different and we want the smallest alkane so this can happen if Hydrogen,Methyl,Ethyl and Propyl groups are attahced to a Carbon. So in total 7 carbons are required(1=Methyl,2=Ethyl,3=Propyl,1=carbon to which these groups are attached,hence total = 1+2+3+1=7)

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http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Lecture Notes 104/lect15c.html

Someone else suggested 7 as answer as well. But 1,3-dibromocyclopentane was also suggeted 5 as answer; cyclo for 5.

Lu Chee Ket - 5 years, 4 months ago

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Cyclo: cyclic (forming a "ring") Pentane: simple alkane with 5 carbon atoms Cyclopentane: a ring of pentane

A Former Brilliant Member - 5 years, 3 months ago

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