CHEM301 F08 CA10 key

CHEM301 F08 CA10 key - CHEM301, Fall 2008 Dr. Ruder 1 Class...

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CHEM301, Fall 2008 Dr. Ruder 1 Class Activity 10 Stereochemistry of E2 Elimination Two groups removed from must be ANTI. Model 1: Stereochemistry of E2 Elimination In any E2 elimination reaction the two groups that are eliminated must be anti (180 ° apart). The following example shows a dehydrohalogenation from a conformation where the leaving groups (H and Br) are anti. This defines the final geometry of the alkene product. H CH 2 CH 3 H 3 C Br CH 2 CH 3 H CH 3 H Br Ph rotate so Br and H anti H H 3 C CH 2 CH 3 H Br Ph CH 3 H CH 3 CH 2 Br α β CH 3 CH 2 CH 3 - HBr Ph H Ph H B H 3 C CH 2 CH 3 H Ph Ph H Ph = phenyl = C 6 H 5 = Questions 1. For the reaction shown in Model 1: (a). In the compound on the left, if the carbon containing the Br is α , label any β carbons. (b). When converting the wedge dash structure on the left to the Newman structure, which carbon is in the front and which is in the back ( α or β )? Confirm that the Newman and the wedge dash structure on the left are the same conformation. (c). To convert from the Newman structure on the left to that on the right, the groups on the back carbon are rotated. Explain why this is done. Confirm that the Newman on the right is the same conformation as the wedge dash on the right. (d). Elimination of the β -H only occurs when the H and Br are ANTI or 180 apart, which defines the stereochemistry of the alkene product. In the Newman projection on the right, are the Ph and CH 3 groups on the same or opposite sides? _______________ In the product alkene are the Ph and CH 3 groups on the same or opposite sides? ________________ Confirm that the Newman and the
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CHEM301 F08 CA10 key - CHEM301, Fall 2008 Dr. Ruder 1 Class...

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