Zn attacks Bromine : Kaplan Section Test

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SaintJude

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The question is: What alkene is obtained from this following reaction?

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A. trans-4-methyl-2-pentene Kaplan says: "The bromides must be vicinal leading to a trans product, not cis." What?
B. cis-4-methyl-2-pentene
C. 4-methyl-1-pentene
D. mixture of trans-4-methyl-2-pentene and cis-4-methyl-2-pentene And why is there not a mixture?

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If the Br's were close to each other, they wouldn't be able to react well with the Zn.
 
Another question? This is an e2-type mechanism. So how come we can use an ethanol? Just...cause..? The same reaction (dehalogenation of vicinal dibromides) also proceed with NaI/acetone but acetone is aprotic, so that makes sense.
 
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E2 requires anti-coplanar conformation, aka 180 degrees apart from each other. Spin the Br not getting attacked away from the Br getting attacked. This forces the product to be trans and there will not be any cis product.
 
Still confused about this question... any help would be appreciated

Moto, this is an E2 elimination "in disguise". It would be the same as if one were adding sodium iodide/acetone to a vicinal dibromide. Vicinal simply means that the two bromine are on adjacent carbons. E2 elimination requires anti-coplanar vicinal bromines. Because of this stereochemical requirement, the product must be the TRANS product (if you are not convinced, crack out the old model set ;)).

This mechanism is a one shot wonder and is usually taught once in a first semester orgo class. I would not worry about this too much for the MCAT just know that the mechanism is E2 and requires anti-coplanar arrangement of the two bromines which would lead to a trans product in this case (very unlikely to test this one reaction for what it's worth). This reaction is rarely useful in the lab b/c to generate a vicinal dibromide requires bromination of an alkene which is generally not done in lab.


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