Orgoman Ch4, Q7

[babowc]

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The compound is optically active, and has an enantiomer.. But why is the solution stating that the answer is D?

 

[funkyCHICKEN]

View attachment 23536

The compound is optically active, and has an enantiomer.. But why is the solution stating that the answer is D?

that my friend is a good question

 

[J777]

Forms a racemic mixture, which is optically inactive.

 

[KyoPhan]

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."

 

[cmF]

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."

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.

 

[KyoPhan]

Yes, but your explanation is better 🙂

 

[funkyCHICKEN]

You beat me to it!

why does the middle carbon have 5 bonds :S? KILL ME NOW MY DAT IS IN 2 DAYS

 

[Slumber]

why does the middle carbon have 5 bonds :S? KILL ME NOW MY DAT IS IN 2 DAYS

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.

 

[funkyCHICKEN]

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

 

[cmF]

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.

Yup. I drew an extra bond in there somehow 🙄 Fixed my original post. Thanks for pointing out funkychicken!

 

[babowc]

why does the middle carbon have 5 bonds :S? KILL ME NOW MY DAT IS IN 2 DAYS

I drew it on there when I was working on it, to see that the Br attaches there.
I didn't think about erasing it, sorry.

and I got it! Thanks for the clarification👍😀