Homework 9 - Answers - Homework Problem Set 9 Iverson...

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Unformatted text preview: Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 NAME (Print): _____________________________ SIGNATURE: _____________________________ Chemistry 310N Dr. Brent Iverson 9th Homework April 9, 2008 Please print the first three letters of your last name in the three boxes Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 Score: _________ Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 These are enolate synthesis problems. In each case, all of the carbons of the products must come from the listed starting materials. You may use any reagents we have discussed this semester or last semester. Show all molecules synthesized along the way. For each step, you will only get full credit if the product you list is the major product of that transformation. Remember to work backwards, count carbons, and make sure you know your KRE's. O (7 pts) A) OH H ? E,Z Mixture PCC H3 O Heat (dehydration) H cat. NaOH (aldol) O H O (racemic) OH The key recognition element is the !,"-unsaturated aldehyde, which indicates an aldol followed by dehydration. Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 (10 pts) B) OH ? O O O 1) 0.5 Eq. NaOEt H2 CrO4 2) mild H3 O (Claisen) OH O O SOCl2 Cl OH O O The key recognition element is the !-keto ester product indicating a Claisen as the last step. Alternatively, the ethyl ester could have been made from a reaction of ethanol with sufuric acid directly, avoiding the acid chloride intermediate. Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 (20 pts) C) O O ? O + CO2 1) 0.5 Eq. NaOEt 2) mild H3 O (Claisen) O O O Heat O OH O 1) 1.0 Eq. NaOEt 2) O O Br (See below) H3 O 1) 1.0 Eq. NaOEt O O O 2) Br O (racemic) Following the initial Claisen, this is just an aceto ester synthesis. The key recognition element in the product is the substituted methyl ketone. The ethyl bromide is made according to the scheme below: O 1) LiAlH4 O PBr3 2 OH Br 2) H2 O Alternatively, the ester could have been hydrolyzed in acid base, to give the acetic acid and ethanol, that is then converted into ethyl bromide. Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 (7 pts) O D) O H O O O H ? O * * O 1) 0.5 Eq. NaOEt 2) mild H3 O (Claisen) O O 1) cat. NaOH (aldol) Racemic 2) H3 O heat 1) cat. NaOEt O H (Michael) O 2) mild H3 O The key recognition element is as the product of a Michael reaction between acetoacetic ester and the !,"-unsaturated aldehyde shown. These can be produced by a Claisen and aldol, respectively. This problem demonstrates how multiple reactions can be put together in powerful ways to create very complex molecules from simple starting materials. Homework Problem Set 9 (16 pts) O E) Iverson CH310N Due Wednesday, April 16 O O ? O or * (racemic) O H3O ! 1) 1.0 Eq. HO 2) H3O /H N H H3O N HO O N SOCl 2 O Cl O The key recognition element here is that there was an acylation via the enamine. The acid chloride is made from the starting ester via hydrolysis to the carboxylic acid followed by reaction with SOCl 2. Note that the above could have also been accomplished using a directed enolate reaction in which the enolate of the ketone is created by addition of 1.0 Eq. of LDA, followed by treatment with the ester in a reaction that follows mechanism B. Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 (19 pts) F) O O O ? O HO 1) 1.0 Eq. NaOEt 2) O O HO O O Br (See below) O 1) LiAlH4 2) H2 O 1) 1.0 Eq. NaOEt 2) Br (See below) Heat O O O O O O H3 O HO OH This can be tough to spot, but it is just a reduction following a malonic ester synthesis. The key recognition element in the product is the alkylation pattern with the alcohol on the terminal carbon, indicating reduction of a carboxylic acid. The ethyl bromide is made according to the scheme below: O O 1) NaOH O 2) mild H3 O O O HO + OH O PBr3 2 OH Br Alternatively, the diester could be reduced with LiAlH4 followed by water to give ethanol and 1,3-propanediol, the latter of which is not used further. Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 (13 pts) O G) O O O ? + CO2 HO OH Heat O O 2 PBr3 O O O OH 1) 1.0 Eq. NaOEt 2) Br O O Br O O H3 O O 1.0 Eq. NaOEt Br O This is just plain hard. The methyl ketone is the key recognition element that indicates an acetoester synthesis, a possibility that is confirmed by the presence of aceto acetic ester as a starting material. The insight needed here is that the 1,5-pentanediol can be converted into the 1,5-dibromopropane, which will create the required six-membered ring by sequential double aklkation of the enolate anion. Note that this could have been accomplished in a single step by adding two equivalents of NaOEt, but I wanted to break it up to show you what was happening. Homework Problem Set 9 Iverson CH310N Due Wednesday, April 16 The KRE might be very difficult to spot here. Hint: A substantial portion of this one comes from ones you have done earlier in this homework set. Look at problems B) and D). (22 pts) O * H racemic O H) OH ? + CO2 H2 CrO4 O PCC O # H O O OH OH racemic OH OH / H O * * 1) cat. NaOH 2) H3 O # O O H 1) 0.5 Eq. NaOEt 2) H3 O (mild) O O 1) 1.0 Eq. NaOEt O O H * H3 O O (strong) 2) O O racemic * 3) H3 O (mild) O H This is also just plain hard. The methyl ketone is the key recognition element that indicates an acetoester synthesis. The aldehyde substituted at the !-carbon indicates a Michael reaction. Both the ",!-unsautrated aldehyde and acetoester can be derived from ethanol via an aldol and Claisen reaction, respectively. If you got this, I am VERY impressed! ...
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