HW13Key - both enantiomers of the product are formed in...

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Graded Homework Problem #13 – Answer Key This problem is worth a total of 20 raw points. When the diastereomer of 3-methyl-2-hexene shown below is treated with HI, 3-iodo-3-methylhexane is the major product formed. This product is formed as a racemic mixture, meaning both enantiomers are formed in equal amounts. C C CH 3 CH 2 CH 2 CH 3 H CH 3 + H I C CH 2 CH 3 I CH 3 CH 3 CH 2 CH 2 C CH 2 CH 3 I CH 3 CH 3 CH 2 CH 2 + ( E ) ( Z ) ( R ) ( S ) ( R ) ( S ) (a) Indicate the absolute stereochemistry of each compound shown above by circling the appropriate choice below each structure. No explanation is necessary. (b) Using curved arrows to show movement of electron pairs, propose a mechanism for the reaction shown above. Be sure to show structures of all important reaction intermediates that form during the course of the reaction! (c) Briefly explain why
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Unformatted text preview: both enantiomers of the product are formed in equal amounts. In other words, why is this reaction nonstereoselective ? A good explanation should require no more than a sentence or two!!! The carbocation intermediate has trigonal planar geometry at the positively charged carbon atom. The iodide anion can react from either side of this plane with equal likelihood, so both possible stereochemical configurations of the new chiral center are formed in equal amounts. C C CH 3 CH 2 CH 2 CH 3 H CH 3 H I C CH 2 CH 3 I CH 3 CH 3 CH 2 CH 2 C C H H CH 3 CH 3 CH 2 CH 2 CH 3 I trigonal planar geometry at the cationic center the other enantiomer is formed when I-attacks from the other side of the plane...
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