Quiz 4 Answer key S2009

Quiz 4 Answer key - Reagent 1 is formed by mixing 2 equivalents of Me 3 Al with H 2 NCH 3(OCH 3)Cl 1 Balance the equation by adding the missing

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Name:_____________________ Chem 41c Quiz 4 Stoltz, Spring 2009 May 8, 2009 You have 25 min to take this quiz. It is closed note, closed book, and no collaboration is allowed. Please do not discuss the quiz with anyone until you receive it back graded. Place a box around your answers. There is no partial credit. Predict the products (if any) of the following reactions: (5 points each) 1. OEt O OH O 1. NaOH H 2 O, EtOH 60 °C 2. HCl, H 2 O 2. H 2 O–THF (1:1) Cl O OH O 3. CH 3 N O OCH 3 MgBr O THF, 0 °C Provide reagents for the following transformations. (5 points each) 4.
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Cl O N H N H aromatic or trialkyl amine base N O N O
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Bonus (5 points) In class we learned that reagent 1 was useful for the conversion of esters to Weinreb amide derivatives.
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Unformatted text preview: Reagent 1 is formed by mixing 2 equivalents of Me 3 Al with H 2 NCH 3 (OCH 3 )Cl. 1. Balance the equation by adding the missing other products. 2. Draw a mechanism for the formation of 1 from these reagents. Al CH 3 H 3 C CH 3 (2 equiv) N H OCH 3 H 3 C H Cl Al N OCH 3 H 3 C H 3 C CH 3 other products 1 CH 3 N H OCH 3 H 3 C H (CH 3 ) 2 Al CH 4 + (CH 3 ) 2 AlCl N OCH 3 CH 3 H 2 CH 4 + (CH 3 ) 2 AlCl CH 3 (CH 3 ) 2 Al 1 In a very basic way. ..this is the mechanism:-you can think of (CH 3 ) 3 Al like other organometallics you have learned about (like Grignard reagents) CH 4...
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This note was uploaded on 01/03/2012 for the course CH 41c taught by Professor List during the Fall '10 term at Caltech.

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Quiz 4 Answer key - Reagent 1 is formed by mixing 2 equivalents of Me 3 Al with H 2 NCH 3(OCH 3)Cl 1 Balance the equation by adding the missing

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