Due: Monday, November 15, 2010

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The alcohol module in ORGO will give you a good review of some of the fundamental reactions discussed in class and in Chapters 8 and 9. As you master the chemistry of alcohols, you should try the Web of Reactions. 1. How many grams of KMnO4 in aqueous KOH are required to oxidize 20 grams of 1,2-cyclohexanediol to adipic acid? [Note: MnO2 is the reduction propuct of permanganate. This is a redox reaction from Gen. Chem. Go here for help. Derive the balanced equation and show your work.]. First, MnO4- ---> MnO2 then add two oxygens on the right by using two hydroxyls and one water for each oxygen. 2H2O + MnO4- ---> MnO2 + 4HO - (+ 3-electrons; eq. 1) Now, C6H12O2 ---> C6H10O4 requires two oxygen on the left and two hydrogens on the right. First, the oxygen: 4HO - + C6H12O2 ---> C6H10O4 + 2H2O Now the hydrogens: 2HO - + 4HO - + C6H12O2 ---> C6H10O4 + 2H2O + 2H2O This equation simplifies to 6HO - + C6H12O2 ---> C6H10O4 + 4H2O (- 6-electrons; eq. 2) Multiplying eq. 1 by a factor of two to balance the electrons gained and lost, we have 4H2O + 2MnO4- ---> 2MnO2 + 8HO - (+ 6-electrons; eq. 3) Adding eqs. 2 and 3, we have 6HO - + C6H12O2 + 4H2O + 2MnO4- ---> C6H10O4 + 4H2O + 2MnO2 + 8HO - Simplifying yields C6H12O2 + 2MnO4- ---> C6H10O4 + MnO2 + 2HO - (eq. 4; Note: the adipic acid will undergo deprotonation by the hydroxide at this point. This does not affect the solution to the problem.) ------------------------------------- KMnO4 MW. = 154; C6H12O2 MW = 116 Thus, 2 x 154/116 = x/20; x = 54.5 g KMnO4

Victor Grignard (1871-1935) Co-Nobel Prize in Chemistry (1912)

2. Optically-active compound A (C10H20O2) reacts with LiAlH4 in ether to form a single optically-inactive compound B (C5H12O). Bromide C is converted into its Grignard reagent D. Reagent D reacts with A to form optically-active E (C9H20O) and (S)-B. What are the structures A-E? Explain and illustrate. 1 DU. A is C10, 1 DU, two oxygens, reacts with LiAlH4 and a Grignard reagent and forms a single C5 compound after reduction. What is A but an ester whose carboxylic acid and alcohol portions are both C5 units. The two fragments of A must be branched to allow for optical activity. The racemic alcohol B must be primary because of the LiAlH4 reduction. There is only one possibility for a primary alcohol that is C5 and chiral. At this point the gross structure of A is known. The lack of optical activity in B is due to the presence of a racemate. So the two asymmetric carbons must have opposite handedness. Esters undergo double Grignard addition. Since E is C9 (0 DU), the Grignard reagent D is ethyl magnesium bromide and C is ethyl bromide [C5 + (2 x C2) = C9]. The production of (S)-B in the Grignard addition means that the alcohol portion of ester A is (S) and the carboxylic acid portion is (R).

3. Predict the products and/or reagents in each of the following examples. Justify your answers.

4. Two bottles on a shelf have had their labels fall off. Both of the labels read "C5H11Br". A student decides to run some reactions on the contents of bottle A and B to determine the structures of the two compounds. From the flow chart determine the structure of A and B and identify C-F. Show your reasoning. [Hint: Draw all of the structures of C5H11Br. Eliminate non-contenders? Only the major product in the formation of C should be considered.] A has no DU. It reacts rapidly with water (SN1). A is a tertiary bromide. One possibility: 2-bromo-2-methylbutane. B is 2-methyl-2-butanol. Tertiary alcohols are not oxidized by PCC. KOH causes E2 elimination of A to form C, 2-methyl-2-butene. Hydroboration of C gives anti-Markovnikov addition of water to the alkene. E is 3-methyl-2-butanol. Chromic acid oxidation of E, a secondary alcohol, provides F, 3-methyl-2-butanone. HBr treatment of secondary alcohol E afford hydride migation product A in addition to direct substitution product B. KOH E2 elimnation of B gives C.

5. Neosporol (1), which is shown in two views, was successfully synthesized from racemic ketone 2, whose synthesis is well beyond the scope of this question. The immediate problem was to convert ketodiol 2 into triol 3. [The fact-oid-s have been altered slighted to facilitate the question. (J. Am. Chem. Soc., 1993, 115, 2581) ] When an excess of methyllithium was used to convert the ketone function of 2 into the tertiary alcohol of 3, only ketodiol 2 was recovered upon aqueous workup. A Jmol structure of neosporol is provided. Move the structure around to compare it with the two views of neosporol 1.

a) What is the minimum amount of methyllithium required in this reaction? Explain? Methyllithium reacts with each of the alcohols and the ketone. Three equivalents of methyllithium.

b) What events occurred prior to aqueous work up? [Hint: Generally, organolithium and Grignard reagents undergo addition but they are also the conjugate bases of weak acids.] What was the fate of the ketone group? Rather than undergoing addition to the ketone, the methyllithium acted as a base, abstracting a hydrogen atom adjacent to the ketone forming a ketone enolate. The enolate is stable until it is protonated in the aqueous workup.

When methyl magnesium bromide was employed, both 2 and a mixture of the diastereomers of 3 were obtained. Complete conversion of 2 to 3 (5/1 mixture of diastereomeric tertiary alcohols) was effected cleanly with the cerium reagent, CH₃CeCl₂.

c) Draw the structures of the two diastereomers of 3, i.e., provide stereochemistry in structure 3. See 3a and 3b below from addition of the organometallic reagent to either face of the ketone.

d) Provide conditions and a mechanism for the conversion of 3 to 1. Is it necessary to separate the diastereomers of 3 prior to forming 1? See above. A proton can protonate any of the oxygen atoms of 3a and 3b. The only productive event is protonation of the tertiary alcohol, which leads to a tertiary carbocation. The carbocation is captured by an intramolecular S_N1 reaction followed by loss of a proton to form 1. No separation of 3a and 3b is required.

Neosporol (How to manipulate Jmol structures)  (Larger Version)