26

To my knowledge the mechanism of the Wittig reaction isn't fully resolved yet. But maybe I can give you some ideas about why the Wittig reaction with unstabilized ylides is (Z)-selective (well, with the exception of the Schlosser modification) instead of (E)-selective. In the excellent book by Clayden, Warren, and Greeves, there is a section beginning p. ...


25

Racemization isn't "exact," but rather very very close to equality. It is just simple probability. Think of flipping a coin, p=probability for heads, and q=probability of tails. Now for a fair flip p=q=0.5. From binomial theory the standard deviation is $\sqrt{n\cdot p \cdot q}$ where n is the number of flips. Now let's assume 2 standard deviations ...


20

Great question! It turns out that the rate of formation of the "expected" endo product is actually ~500 times faster than the rate of formation of the exo product. However, the Diels–Alder is a reversible reaction. In this case, the exo product is thermodynamically favored over the endo product by about $\pu{1.9 kcal/mol}$. So, even though the predicted ...


17

It is well known that SN1 reactions often give incomplete racemisation: Although many first-order substitutions do give complete racemization, many others do not. Typically there is 5–20% inversion, although in a few cases, a small amount of retention of configuration has been found. These and other results have led to the conclusion that in many SN1 ...


15

As we know, $\ce{S_N1}$ reaction is stereochemically non specific. This means, that it does not force the formation of one stereoisomer, as $\ce{S_N2}$ does. In $\ce{S_N1}$ reaction, presence of carbenium ion (=carbocation) as intermediate, results in the feasibility of attack from both sides of the ion, resulting in the formation of both stereoisomers(R and ...


13

Dihydroxylation responds to steric effects and proceeds at less hindered positions. What will be the major product in the following reaction if the reagent used is osmium tetroxide followed by sodium hydrogen sulfite? The crux of the original question seems to be why standard Upjohn dihydroxylation conditions (cat. osmium tetroxide w/ NMO as the ...


10

Key The reaction is run in heptane, a non-polar solvent. A non-polar solvent will not stabilize or support ions as well as a polar solvent. Background When the $\ce{Cl2}$ begins to interact with the indene an initial ion pair is formed. In the case of chlorine, it is not always clear whether a cyclic halonium ion (the 3-membered structure on the ...


10

TL;DR: With stabilised ylids, the trans oxaphosphetane is formed, under kinetic control. This is due to minimisation of dipole-dipole repulsions in the transition state for oxaphosphetane formation: There is no equilibrium established between cis and trans oxaphosphetanes. The formation of the trans oxaphosphetane is irreversible, and its collapse leads to ...


10

Dave Evans taught me that you should generally analyze these types of problems by drawing out allyl conformations. I've drawn the nitronate to minimize 1,3-allylic strain. Now, the top face of the nitronate is largely blocked by the t-butyl carboxylate. I strongly suspect that having a bulking alkyl group there is important to getting good selectivity. ...


9

The addition of Grignards to propargyl alcohols occurs via the alkoxide salt, after deprotonation:[1] Here: the reactivity may be explained by the stabilising influence of complexation; the regioselectivity may be explained by the preference of a five-membered chelate ring over a four-membered one; and the stereoselectivity arises because syn addition ...


8

You haven't mixed up your terms: a stereoselective reaction is defined as one which preferentially forms one stereoisomer over another. Likewise, all stereospecific reactions are also stereoselective, also stereospecificity has stricter criteria then plain old stereoselectivity. However, you have to remember that the term stereoisomer encapsulates both ...


8

1. The acetate aldol reaction Before looking at selective variants, its worth pointing out what an acetate aldol is. An acetate aldol (below), is the simplest kind of aldol in which the β-hydroxyketone product has no methyl group at the α-position to the carbonyl. This is in contrast to the (more usual?) propionate aldol reaction in which the α-position ...


8

1. Origins of the Houk model The initial idea behind the Houk model was infact provided by Kishi, who proposed that alkenes with a chiral centre on the adjacent carbon adopted a 'reactive conformation' in which the small group eclipsed the alkene in order to minimise unfavourable steric interactions.[1] Fig1: Kishi's "reactant conformer" model, ...


8

It's just theory vs. real life. When mixing components, you always have limitations with the purity of chemicals and the accuracy of the balance available. When you look at chemical reactions which yield chiral compounds, starting from achirals and there is no bias towards d or l, then you will end up with a true racemate.


6

The torquoselectivity of the above reactions is dictated by stereoelectronic effects, which (in this case) are significant enough to take precedence over steric effects. In general, electron-donating substituents (e.g. OMe, OTMS, ...) prefer to move outwards, and electron-accepting substituents (e.g. CHO) prefer to move inwards. These can be understood by ...


5

Having the methyl group psuedo-axial means that when the nucleophile approaches, the transitiation state is chair-like. The energy penalty for having the methyl group initially in the less favourable position is compensated for the low energy TS. The ring-flipped version, in which the methyl group is psuedo-equatorial, would open via a high energy twisted ...


5

Cram's rule, proposed by Donald Cram in 1952,[1] was the earliest model proposed for rationalising the stereoselectivity of nucleophilic additions to carbonyl groups with α-stereocentres. The model involved assigning each α-substituent a relative "size" (small, medium, and large), then placing the carbonyl oxygen antiperiplanar to the largest of these three ...


5

I think Jan did an excellent job of answering your question, however I am writing a slightly longer answer largely with more details concerning the Felkin–Anh rule, just so that someone who is not already familiar with them can learn something too. Assuming that there isn’t an unusually electronegative atom on the carbon next to the carbonyl, the largest ...


5

The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by $\ce{I-}$ on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the ...


5

The results of the iodolactonization example presented in this question were published in 1982 by Bartlett, et al.1 While @Aniruddha Deb has captured the essence of the reaction, there is more than one product formed in the reaction. A more detailed mechanism is offered, an issue not addressed by the original authors. Carbonate 1 is a racemate. No products ...


4

Problem #1 is that if you interchange the methyl and hydrogen groups, you get diastereomers, not enantiomers. But this is a relatively minor point. I assume your question is about the diastereoselectivity, not enantioselectivity (as ron rightly pointed out there is no enantioselectivity in this reaction). The analysis you have drawn is somewhat similar to a ...


4

The $\ce{TBDPS}$ group (tert-butyldiphenylsilyl) is a very bulky protecting group. A silicon atom is bonded to the oxygen, which already acts as a larger carbon atom. That is then substituted by three alkyl or aryl groups, one of which is the very bulky tert-butyl group. There is a lot of congestion going on next to that oxygen atom, so its ability to ...


4

If there is no source of chirality present in the reaction mixture, then there is no reason for the reaction to be enantioselective. In this case there are two possible sources of chirality: firstly the light, which could be circularly polarised (not a common thing to do), and secondly the catalyst, which can be chiral. Although the exact catalyst is not ...


4

Yes. Right after Clayden mentions this stereoselectivity, he proceeds to state that this is due to "coordination between the zinc atom and the hydroxyl group in the transition state" (pp. 1067-8, 1st ed):


4

Without further details, I would say that both products are possible. The reaction that you have given here is an Iodolactonization reaction. You have correctly identified that the tert-butyl group would be removed and that the oxygen atom would prefer to form a 6-member ring. In terms of stereochemistry, however, it is tough to make a decision, as: The ...


3

Enyzmatic reduction (as would occur in yeast) could easily be a stereoselective process - probably via selective delivery of a hydride from NADH. As the other respondent notes, there is simply no way that combining achiral reagents (pyruvic acid and NaBH4) can yield anything other than a racemic product.


3

For most aldol reactions, the relative 1,2-diastereoselectivity is a consequence of the enolate geometry.* Both tin and titanium enolates are easily formed, and have been shown to exist predominantly in the (Z) configuration. When reacting a cyclic (Zimmerman Traxler) TS, its commonly observed that (Z) enolates afford 1,2-syn aldol adducts, due to the ...


3

I have had great results for enantioselective ketone reduction using the Noyori Ru-TsDPEN system. This has advantages through being catalytic with low stoichiometry and doesn't require high pressure, or molecular hydrogen for that matter. The downside might be the initial cost of the catalyst / chiral ligand but you don't need very much. The typical ...


3

While @orthocresol was busy finding the Carreira paper, I was busy constructing my own thoughts on the stereochemical consequences of the reactions of 1 (iii) and 5 (ii). Yes, it is an issue of conformation. It is far easier to rationalize the results after the fact. To predict the most reactive conformation of epoxide 1 the Newman projection 2, viewed along ...


3

This figure from this reference (p. 170). might help you visualize things. As you can see, the stereochemistry is given by electronic repulsion between the carbonyl group and the incoming aldehyde.


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