CHEM 203 HOMEWORK 4 Chemistry of Alkenes - II
1. The following questions may have occurred to you: (i) do carbocations occur in living systems? (ii) Can an olefin (a Lewis base) react with a carbocation (a Lewis acid)? To address these questions, consider the following. At this very moment, certain enzymes in your liver are producing cholesterol and related steroids. These enzymes take up an endogenous olefinic compound called squalene and convert it into a carbocation that may be represented with structure A below. The enzymes also keep this carbocation in the rigidly defined conformation shown, protecting it from the action of external nucleophiles and promoting or disfavoring various chemical reactions, including rearrangements. Under such conditions, A is rapidly transformed into lanosterol, which is the precursor of cholesterol and of all other steroids.
enzymes
squalene
mechanism?
A HO HO lanosterol
Answer the above questions by writing a detailed mechanism for the conversion of A into lanosterol.
R R
HO HO HO the enzyme keeps this carbocation from rearranging (via H-shift) to a more favorable, HO-stabilized one
R R H R H
H H there are HO HO H's here... HO
the enzyme keeps this carbocation now the enzyme promotes from rearranging (via alkyl shift) to a series of H- and alkyl shifts ... a more favorable, tertiary one
CHEM 203 HOMEWORK 1 p. 2
H H H H R H R R R H H H
H H H H H H H H HO HO HO HO
H H R R H H now, the enzyme promotes loss of a proton from (= deprotonation of) H H H H this carbocation (strong Bronsted acid) HO HO
H H R R H H
H H H Lanosterol ! HO HO H B B basic site within the enzyme ... Enzyme Enzyme
2. Predict the structure of the major product of each of the following reactions and write a detailed mechanism (curved arrows) for its formation.
CH3OH OCH ± H+ OCH3 3 OCH3 OCH3 H–BF 4 O–CH3 OCH3 H BF4 O CH3 H
+ H2SO4 ± H
H O 2 O OH H H O H H HSO4 H H
Br2 Br Br
Br Br Br Br
CHEM 203 HOMEWORK 1 p. 3
(as far as we know ...) H H Cl2 Cl Cl O Cl O O Cl H H Cl Cl
H H O H Br2 ± H+ O OH H H2O Br Br Br Br Br Br
H CH3 O H + OCH Br2 OCH3 ± H 3
CH OH 3 Br Br Br Br Br Br
intenal rotation about C-C bond O O 1. O3 O O O O O + O 2. Zn, H O O O O
H H + Zn, H O O O O = O O (no mech.) O O H H there are H's here
O O O O O O O O 1. O H O (flip 3 + H aldehyde) 2. Zn, H+
CHEM 203 HOMEWORK 1 p. 4
H H O O O O Zn, H+ O O O + O H (no mech.)
O O O O O O O O 1. O H O 3 + H + 2.H2O2, H
H (flip aldehyde) O O O + H O H2O2, H O + O O OH O (no mech.)
O O O Os O O Os OH O OsO , then aq. NaHSO 4 O O 3 H ( + reduced H forms of Os ) aq. NaHSO3 H (no mech.) H OH
3. Draw a clear skeletal structure of:
(a) an olefin that produces the same alcohol when treated with either H2SO4 / H2O or BH3 followed by H2O2 / NaOH
e.g. ; ; ...
(b) an olefin that produces one alcohol when treated with H2SO4 / H2O, but an isomeric alcohol when treated with BH3 followed by H2O2 / NaOH
e.g. ; ...
4. Provide all the reagents / catalysts, in the correct order, that are needed to convert vinylcyclopentane (structure below) into compounds a. - h. If a compound appears to be unavailable as the major product of any reaction known to you, answer "inaccessible".
CHEM 203 HOMEWORK 1 p. 5
Cl O OCH3 Br Cl OH a. b. c. d.
OH OH Cl OH Br e. f. g. h.
vinylcyclopentane O OH Cl Cl OH OH H i. j. k. l.
Cl OCH H SO 3 2 4 Cl2 a. b. CH OH Cl 3
O Br O , then HBr 3 d. c. OH H2O2, acid O + H OH
OH inaccessible BH3, then e. f. H O , aq. OH ( why ...? ) 2 2 NaOH
OH Cl Br2 inaccessible g. h. H O 2 Br ( why ...? )
OH O OsO , then O3, then 4 i. H j. aq. NaHSO OH Zn, acid O 3 + H H
Cl Cl HCl inaccessible k. l. ( why ...? ) OH CHEM 203 HOMEWORK 1 p. 6
4. Propose a method for the preparation of compounds a. – h. below starting from appropriate alkenes. Draw a clear structure of your proposed starting olefin and list all reagents / catalysts, in the correct order, that are required to induce the desired transformation. Your method must be a good one, i.e., the desired compound must be the major product of your reaction(s). Note: it is understood that chiral compounds will be obtained as racemic mixtures.
CH3 OH OCH3 OH H H a. b. c. Br d. O O OH H Cl OH OH e. f. g. h. OH Cl OH
• this is the anti-diastereomer of a vicinal diol OH • the only method known to us for the creation of vicinal a. diols is the dihydroxylation of alkenes with OsO4 • to make this compound, one must treat an alkene with OH OsO4, followed by aqueous NaHSO3
BUT: which alkene do we need to obtain the desired product?
HO OH 1. OsO4 • A general diagram for the dihydroxylation reaction is: C C C C 2. aq.
NaHSO3 therefore, product a. must result from dihydroxylation CH CH CH–CH –CH of a molecule of 2-pentene: 3 2 3
BUT: which isomer of 2-pentene will produce the desired product? • The dihydroxylation of alkenes is a syn-addition reaction, so the OH groups must enter from the same face of the π bond
• If we were to attempt making a. by dihydroxylation of trans-2-pentene, we would get: HO H OH H OH C H syn diastereomer: C C C = wrong product H H H OH
but if we were to dihydroxylate cis-2-pentene . . . : HO H OH H OH C H H anti diastereomer: C C H C = desired product ! H OH
CHEM 203 HOMEWORK 1 p. 7
so, the correct answer is:
1. OsO OH notice that this molecule is chiral: it will be 4 H obtained as the racemate. For simplicity, we draw only one enantiomer, with the 2. aq. H understanding that both will actually be OH NaHSO3 formed during the reaction.
A quicker way to obtain the correct answer:
OH groups now point in OH anti diastereomer: OH the same direction, as H H imagine a rotation H OH required for a syn addition. C H C H C H about the internal C Imagine forming a π bond OH C–C σ bond ... : between the OH-bearing C atoms ... cis-2-pentene
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• this is an alcohol OH b. • the only methods known to us for the creation of alcohols are the acid- catalyzed hydration of alkenes and the hydroboration-oxidation of alkenes "H–OH" H OH • A general diagram for either reaction is: C C C C
Therefore, we can make b. starting from:
BUT: which method will furnish the desired product?
• If we were to hydrate the above alkene under acid-catalyzed conditions, we would obtain:
H2SO4 OH "Markownikov" alcohol: wrong product H2O • If we were to subject the above alkene to hydroboration-oxidation, we would obtain: R 1. BH3 B R OH "anti-Markownikov" alcohol: desired 2. H2O2 product aq. NaOH
so, this is the correct answer
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CH • this is a bromohydrin-like compound that — in principle — one can 3 OCH c. 3 create by treating an alkene with Br2 and CH3OH Br Br H 2 CH3O Br • A general diagram for either reaction is: C C C C CH3OH
CHEM 203 HOMEWORK 1 p. 8
Therefore, one could make c. starting from:
BUT: the above reaction is an anti addition, meaning that Br and OCH3 will add from opposite faces of the alkene.
Will this produce the correct stereoisomer of the product ?
• Because the Br atom and the OCH3 groups in c. are trans to each other, one will indeed obtain the desired product from the alkene shown above (1-methylcycloheptene): so the correct answer is:
CH3 OCH3 notice that this molecule is chiral: it Br2 will be obtained as the racemate. Br For simplicity, we draw only one CH OH enantiomer, with the understanding 3 H that both will actually be formed
* * *
• this is a carbonyl compound; a dialdehyde, to be precise. H H d. • the only method known to us to make aldehydes is the ozonolysis of alkenes followed by treatment of the ozonide with Zn and acid O O 1. O3 • A general diagram for either reaction is: C C C O + O C 2. Zn, H+
H H 1. O H H O O Therefore, one could 3 + + certainly make d. from 1 2 3 4 + O O R R R R an alkene like: R1 R2 R3 R4 2. Zn, H
But there is a better option . . .
• If the two carbonyl carbons were initially doubly bonded to each other, then ... H H H reconnect = O cyclopentene ! O O O H
1. O3 H H so: sole product 2. Zn, H+ O O
• Whereas the first method also yields two additional carbonyl compounds of no interest to us, the second one would produce only the desired d.
• The second method is better, and it represents the best answer
* * *
CHEM 203 HOMEWORK 1 p. 9
• this is an alcohol e. • the only methods known to us for the creation of alcohols are the acid- catalyzed hydration of alkenes and the hydroboration-oxidation of alkenes OH "H–OH" H OH • A general diagram for either reaction is: C C C C
• If one were to make e. by acid-catalyzed hydration of an alkene, the latter would have to undergo protonation to form the following carbocation:
• protonation of either 1-butene, cis-2-butene, or trans-2-butene would afford the above carbocation; therefore, reasonable answers are: H SO H2SO4 2 4 or H O H2O OH 2
• If one were to make e. by hydroboration-oxidation of an alkene, one could not use 1-butene as the substrate: H O BH3 R 2 2 wrong B OH product R aq. NaOH
• however, one could employ either cis or trans-2-butene:
BH H2O2 or 3 desired product B aq. NaOH OH R R
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Cl f. • this is the syn-diastereomer of a vicinal dichloride • the only method known to us for the creation of vicinal dichlorides is the chlorination of alkenes Cl Cl2 Cl Cl • A general diagram for either reaction is: C C C C
CH3 • to prepare f. we need the following CH CH CH–CH3 alkene (4-methyl-2-pentene): CH3
BUT: which which geometric isomer of the above alkene is required to obtain the desired product?
• The halogenation (chlorination) of alkenes is an anti-addition reaction, so the halogen atoms must enter from opposite faces of the π bond
• If one were to attempt making f. by chlorination of trans-4-methyl-2-pentene, one would get: Cl H H Cl H C C anti diastereomer: C C = wrong product H H H Cl Cl
CHEM 203 HOMEWORK 1 p. 10
but if one were to chlorinate cis-4-methyl-2-pentene . . . : Cl H H Cl H C C H syn diastereomer: C C = desired product H H Cl Cl Cl Cl H so, the correct answer is: 2 H Cl cis-4-methyl-2-pentene
A quicker way to obtain the correct answer: Cl atoms now point in opposite directions, syn diastereomer: as required for an Cl Cl H imagine a rotation anti addition. H C C Imagine forming a H about the internal H cis-4-methyl-2-pentene Cl C–C σ bond ... : Cl π bond between the Cl-bearing C atoms ...
* * *
g. • this is an alcohol • the only methods known to us for the creation of alcohols are the acid- catalyzed hydration of alkenes and the hydroboration-oxidation of alkenes OH
• If one were to make g. by acid-catalyzed hydration of an alkene, the latter would have to undergo protonation to form the following carbocation:
• the only way to create the above carbocation is to protonate 3-methyl1-butene:
H–OSO3H BUT . . .
• the secondary cation would CH3 CH3 rapidly rearrange via a H shift = C CH3 C CH to furnish a better stabilized, CH 3 3 C CH3 C tertiary one: H H H H from which an incorrect product would be obtained: OH so, one cannot make g. by acid-catalyzed hydration of an alkene.
• If one were to make g. by hydroboration-oxidation of an alkene — a reaction that produces an "anti-Markownikov" alcohol — the required alkene would be:
• Then . . . H O BH3 2 2 desired product B aq. NaOH OH R R CHEM 203 HOMEWORK 1 p. 11
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OH • this is a vicinal diol of cis-configuration h. • the only method known to us for the creation of vicinal diols is OH the dihydroxylation of alkenes with OsO4 OsO then 4 HO OH • A general diagram for either reaction is: C C C C aq. NaHSO3 • to prepare f. we could dihydroxylate 1-methyl-cyclopentene:
BUT: will we get the correct diastereomer of the product ?
• The dihydroxylation of alkenes is a syn-addition reaction, so the OH groups must enter from the same face of the π bond
• If one were to dihydroxylate 1-methyl-cyclopentene, one would get: OH OH H H OH C cis isomer: C C = OH C desired product
OH OsO4 then so, the correct answer is: OH aq. NaHSO3