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As has been demonstrated in the text, when the starting alkene has \(\mathrm{CH}_{2}\) as its terminal group, the Heck reaction is highly stereoselective for formation of the \(E\) isomer. Here, the benzene ring is abbreviated \(\mathrm{C}_{6} \mathrm{H}_{5}\)-. Show how the mechanism proposed in the text allows you to account for this stereoselectivity.

Short Answer

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Answer: The Heck reaction with an alkene containing a terminal methyl group shows stereoselectivity towards the E-isomer due to steric interactions between the large groups during the migratory insertion step. The syn-insertion transition state has more steric interaction between the alkene and the aryl group, while the anti-insertion transition state minimizes steric interactions due to the arrangement of these groups. Consequently, the anti-insertion transition state, which leads to the E-isomer, is favored as it has a lower energy.

Step by step solution

01

Recall the general mechanism of the Heck reaction

The Heck reaction involves the coupling of an alkene and an aryl halide to form a substituted alkene. In the initial step, the palladium(0) catalyst forms a complex with the aryl halide, undergoing oxidative addition to create a palladium(II) complex. Next, the alkene coordinates to the palladium(II) complex, and migratory insertion takes place to form the carbon-carbon bond. Finally, the product is released, and the palladium(0) is regenerated through a reductive elimination process.
02

Mechanism with Terminal Methyl Group

Consider an alkene with a terminal methyl group (\(\mathrm{CH}_{2}\)) reacting with an aryl halide (\(\mathrm{C}_{6} \mathrm{H}_{5}\)-X) through the Heck reaction. The alkene coordinates to the activated aryl-palladium(II) complex; once the migratory insertion occurs, two possible transition states can be considered: syn-insertion and anti-insertion. The syn-insertion leads to the Z-isomer, while the anti-insertion leads to the E-isomer.
03

Evaluate Steric Interactions

Compare the steric interactions between the syn and anti transition states. Remember that the transition state with less steric interaction is favored as it has lower energy. In the case of syn-insertion, there is more steric interaction between the alkene and the aryl group (large groups are close to each other), whereas for anti-insertion, there is minimal steric interaction between the alkene and the aryl group. This difference in steric interactions leads to a preference for the formation of the E-isomer, thereby explaining the observed stereoselectivity. In conclusion, the mechanism proposed in the text adequately accounts for the observed stereoselectivity towards the E-isomer in the Heck reaction with an alkene containing a terminal methyl group. The key factor is the steric interaction between the large groups during the migratory insertion step, which is minimized in the case of anti-insertion.

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Most popular questions from this chapter

Show how you might prepare each compound by a Heck reaction using methyl 2-propenoate as the starting alkene.

Show the sequence of Heck reactions by which the following conversion takes place. Note from the molecular formula given under each structural formula that this conversion corresponds to a loss of \(\mathrm{H}\) and I from the starting material.

Treatment of cyclohexene with iodobenzene under the conditions of the Heck reaction might be expected to give 1-phenylcyclohexene. The exclusive product, however, is 3-phenylcyclohexene. Account for the formation of this product.

Over the past several decades, chemists have developed a number of synthetic methodologies for the synthesis of steroid hormones. One of these, developed by Lutz Tietze at the Institut für Organische Chemie der Georg-August- Universität, Göttingen, Germany, used a double Heck reaction to create ring B of the steroid nucleus. As shown in the following retrosynthetic analysis, a key intermediate in his synthesis is compound (1). Two Heck reaction disconnects of this intermediate give compounds (2) and (3). Compound (2) contains the aromatic ring that becomes ring A of estrone. Compound (3) contains the fused five- and six-membered rings that become rings \(C\) and \(D\) of estrone. (a) Name the types of functional groups in estrone. (b) How many chiral centers are present in estrone? (c) Propose structural formulas for compounds (2) and (3). (d) Show how your proposals for compounds (2) and (3) can be converted to compound (1). (Note: In the course of developing this synthesis, Tietze discovered that vinylic bromides and iodides are more reactive in Heck reactions than are aryl bromides and iodides.) (e) In the course of the double Heck reactions, two new chiral centers are created. Assume in compound (3), the precursor to rings \(C\) and \(D\) of estrone, that the fusion of rings \(C\) and \(D\) is trans and that the angular methyl group is above the plane of the ring. Given this stereochemistry, predict the stereochemistry of compound (1) formed by the double Heck reaction. (f) To convert (1) to estrone, the tert-butyl ether on ring D must be converted to a ketone. How might this transformation be accomplished?

Vancomycin is an important antibiotic. It is isolated from the bacterium Streptomyces orientalis and functions by inhibiting bacterial mucopeptide synthesis. It is a last line of defense against the resistant Staph organisms that are now common in hospitals. In 1999, Professor Dale Boger (The Scripps Research Institute) reported a synthesis of vancomycin aglycon (aglycon = lacking a sugar) involving the following steps, among others. Compound (I) was prepared from simple starting materials by a series of steps involving forming amide bonds. (a) Suggest reasonable precursors and show how the bonds could be formed (the actual reagents used have not been introduced, but they work in a similar way to those you know). (b) Give reagents for this reaction and suggest the mechanism. One of the interesting features of this synthesis is that ring \(C\) in compound (II) (and subsequent compounds in this synthesis) has extremely hindered rotation. As a result, compound (II) exists as two atropisomers (Section 3.2) that are interconverted only at \(140^{\circ} \mathrm{C}\). (c) Show these two isomers. (II) was then converted to (III). (d) Suggest reagents to accomplish this transformation. Compound (III) was then converted to (IV). (e) Suggest reagents and the ring A fragment that could be used for this reaction. Closure of an amide link between the amine on ring A (after removal of the protecting group) and the carbomethoxy group above it led to a precursor of vancomycin. (f) Show the ring closure reaction of the deprotected free amino group and its mechanism. Another interesting feature of this synthesis is that rings \(A\) and \(B\) also form atropisomers. These can be converted into a \(3: 1\) mixture of the desired and undesired atropisomers on heating at \(120^{\circ} \mathrm{C}\). (g) Draw these atropisomers and show that only one can be converted to vancomycin. The synthesis of the aglycon was completed by functional manipulation and addition of ring \(\mathrm{E}\) by chemistry similar to that detailed earlier. Yet, another set of atropisomers (this time of ring E) was formed! However, this one was more easily equilibrated than the others; model studies had shown that the activation barrier for this set of atropisomers should be lower than that of the others.

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