Orgoman Ch4, Q7
Started by babowc
that my friend is a good question
The product produces a racemic mixture of enantiomers. Both the products by itself will rotate light to be optically active; however, when they are together, they will cancel each other out to be optically inactive.
"An equal mixture of two enantiomers is called a racemic mixture or racemate.
If two enantiomers rotate plane-polarized light in opposite directions, a racemate will not rotate light at all. The effects of the two enantiomers will cancel out."
"An equal mixture of two enantiomers is called a racemic mixture or racemate.
If two enantiomers rotate plane-polarized light in opposite directions, a racemate will not rotate light at all. The effects of the two enantiomers will cancel out."
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You beat me to it!
Edit: Fixed.
Br2 can undergo free radical halogentation @ either hydrogen forming 2 products which are enantiomers of each other. We can assume this is a 50/50 racemic mixture because each product has an equal chance of forming. In a racemic mixture, one product will rotate light to the left, and the other will rotate light the same amount to the right. In this case, the rotation cancel each other out and light rotates 0 degrees (not at all). Therefore, the products will not be optically active.
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It doesn't. The original compound is chiral, it has an enantiomer pair. He probably drew the pair on the same skeleton, or he simply made a mistake. Either way, both the reactant and the products are violating the octet.
lol he just edited and reposted it after my post haha
Yup. I drew an extra bond in there somehow 🙄 Fixed my original post. Thanks for pointing out funkychicken!
