CHEM 303 Organic Chemistry II Exam II 11-April-2006 Answers

Transcription

1 CEM 303 rganic Chemistry II Exam II 11-April-2006 Answers 1) Compound D, C 8 14, is converted by C 2 =PPh 3 into E, C Treatment of D with LiAl 4 yields two isomeric products, F and G, both of formula C 8 16, in unequal yield. eating either F or G with concentrated 2 S 4 produces, C zonolysis of produces, after normal work-up, a keto-aldehyde, which, upon Cr 3 oxidation, produces 2-methyl-6-oxoheptanoic acid, below. Provide structures for compounds D through. Pay particular attention to the stereochemistry of D. This problem is probably best solved by working backwards. We are given the keto-acid the oxidation yields from the keto-aldehyde. xidation of aldehydes yield acids, so we know the keto-aldehyde is: ow, this keto-aldehyde arises from ozonolysis of ; dicarbonyl compounds often arise from cycloalkenes. (ote that all the carbons we need are present, so we have added evidence for a ring.) Therefore, must be: (the other enantiomer is fine) ow, F and G both react with 2 S 4 to yield ; sulfuric acid and heat are dehydration conditions, so F and G must be:

2 (either way is fine) ote that the stereochemistry of the methyls must be as indicated; the trans arrangement would lead to only one alcohol being formed (verify this for yourselves.) Since F and G arise from reduction of D, D must be a ketone, and have the structure: Finally, E arises from the Wittig reaction on D, and therefore must be: 2a) We know that hydrazines, such as hydrazine itself, 2 2, add to carbonyl compounds to yield hydrazones in the presence of acid. Propose a mechanism for this addition / 3 + heat is: Sort of right out of your notes. MJ does this mechanism clearly in the text. ere it

3 b) We also know that carbonyl compounds, such as alkyl-aryl-ketones, undergo reduction (the Wolff-Kishner eduction) to alkanes in the presence of hydrazine and base. Given what you did above, propose a mechanism to account for this reduction. Your mechanism should clearly show how the pathway changes when the reaction is run in the presence of base. 2 2 / - heat Well, what you wrote in (2a) is also able to be done in base, so getting to the hydrazone should mirror what was done there. (emember MJ s admonition about acid and basic catalysis?) What now needs to be shown clearly is why the reaction goes further in base, to yield the alkane. The answer, in words, is that the - bonds are somewhat acidic, and can be abstracted by extra base. This abstraction starts a cascadelike electron movement which results in the abstraction of + from the acid, 2. ere is the scheme:

4 ) Starting with (S)-(+)-carvone, propose a synthetic route to compound A, a key intermediate in the synthesis of the fungal wheat plant toxin helminthosporal. (int: your first target should be compound B; then consider how to get from B to A.)

5 B C C several steps elminthosporal A There may be several ways of doing this synthesis; I ll evaluate each on its own merits. ere is one: 1)LDA 2) C 2 =CC()C 3 3) 3 + / 2 2 /Pd/C 1) a 2 2) heat A B

6 The most troubling part here is the last intramolecular aldol condensation; I would have to say that the reaction is driven by ring size. Also, the yields are not specified (nor do we have access to any specialized reagents which may have been used in the original synthesis. You were asked to provide a reasonable synthetic route, and the above is certainly that, even if the yields might arguably be low.) 4) Provide reagents for the given reactions: a) a) b) c) d) The key here is (was) to realize that, as written, the reagent(s) had to bring the synthesis to an isolable compound. If we treat it this way, then here are a set of reagents which work: (a): C 2 =PPh 3 ; (b): Br, 2 2 (or similar); (c): Mg/Et 2 ; (d) C 2, 3 +. b) a) b) Kind of straightforward: (a): Cr 3 /pyridine (PCC also works); (b): C 3, 3 +. c) a) b) Again, sort of quick: (a) ab 4, 3 + ; (b) C 3 CCl

7 5a) The 1 -M spectrum shown is that of a compound with M + = 134; highresolution mass spectrometry (MS) shows the M + = ow many degrees of unsaturation does the unknown have? If the unknown has an I absorption at 1690 cm -1, what is a likely structure? For both of these compounds, the precise mass gives you the molecular formula C 9 10, with 5 degrees of unsaturation. The peak in the I at 1690 cm -1 implies a conjugated carbonyl; the triplet at δ1.2 shows a relatively normal methyl group next to a methylene; the quartet at δ2.9 shows a methylene next to a methyl group, and also near an electron-withdrawing substitutent. The aromatic protons give the most difficulty why the separated peaks? ne answers is the shielding/deshielding cone of the carbonyl next to the ring. Given all this, the structure which seems to fit the very best (which is correct) is propiophenone,

8 b) The 1 -M shown is that of a compound isomeric with the one given above. This isomer has an I absorption at 1730 cm -1. Propose a structure. ere we have an isomer of (a). The carbonyl at 1730 cm -1 is non-conjugated (and is about where we would expect an aldehyde). The presence of an aldehyde is confirmed by the M signal at δ9.8. Aromatic protons are again present (and not affected by the carbonyl) as a pseudo-singlet. The mulitplet at δ2.8 is two overlapping triplets (although this is difficult to see). Given this, our structure is 3-phenylpropionaldehyde (also called hydrocinnanaldehyde