jazz_xox_
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Please could someone help me with this question. The answer is B, but I am confused as to how you would show the double bonded long molecule as a tetrahedral structure with 4 different groups attached, and how to pick the chiral carbon? Thank you
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MexicanKeith
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(Original post by jazz_xox_)
Please could someone help me with this question. The answer is B, but I am confused as to how you would show the double bonded long molecule as a tetrahedral structure with 4 different groups attached, and how to pick the chiral carbon? Thank you
Geometrical isomers will have a double bond with 2 different groups on each end.

so molecules A, C and D do not display geometrical isomerism.

Optical isomers will have an unsaturated carbon with 4 different groups around it.

so see whether these compounds fit the bill, draw out a displayed or skeletal structure.

(A,B and D are optically active, C is not)
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