# Tag Info

## Hot answers tagged chemoselectivity

21

Your assertions are correct, for the most part. $\ce{BH3}$ is a Lewis acid, and it does not reduce carbonyl groups by directly donating a hydride group like $\ce{NaBH4}$ does. The $\ce{NaBH4}$ reduction mechanism is fairly short and involves a direct transfer of the a hydride ion to an electron deficient carbonyl carbon: $\hspace{2.4cm}$ $\ce{NaBH4}$ is ...

12

You will most likely be successful if you convert the ketone to the corresponding tosylhydrazone and reduce that with $\ce{NaCNBH3}$ (DOI). A previous study (Bull. Chem. Soc. Jpn., 1974, 47, 2323-2324) has shown that the tosylhydrazone of cyclohexanone can be reduced to cyclohexane with $\ce{NaBH4}$ in polar aprotic solvents. $\ce{NaCNBH3}$ is an even ...

10

The interaction of the alkene with m-CPBA is primarily nucleophilic/electrophilic in nature, i.e. the π MO of the alkene attacks the O–O σ* orbital. Therefore, the more electron-rich alkene (i.e. more highly substituted alkene) preferentially reacts. With methyl groups there probably isn't too much steric hindrance. m-CPBA approaches above/below the alkene ...

9

Ketones have two alkyl groups attached to the carbonyl moiety, which are electron-donating by hyperconjugation. That is, the overlap of $\ce{C-C}$ and $\ce{C-H}$ $\sigma$-bonds with the $\pi$-type molecular orbitals of the carbonyl increases the electron density around the carbon of the carbonyl group, thus making it less electrophilic. Alternatively, I ...

8

The answer appears to be "it depends" (perhaps this should not surprise an experienced organic chemist). According to the abstract of the paper by Mendelovici and Glotter [1] in steroid chemistry: Treatment of 3β- and 3α-hydroxy-(acetoxy-)cholest-4-en-6-one and of 6β-and 6α-hydroxy-(acetoxy-)cholest-4-en-3-one with MCPBA gives two types of product, ...

8

In general, the Rosenmund reduction isn't actually used very often, mainly since the catalyst is also very good at reducing other functional groups which may be present in the molecule (the catalyst is similar to that used for a Lindlar reduction). To quote from Comprehensive Organic Name Reactions and Reagents: [...] the Rosenmund reduction is an ...

7

Using the haloform reaction to convert the methyl ketone 1 to the corresponding carboxylic acid 2 is a neat idea. According to an old review by R. C. Fuson and B. A. Bull, published in Chem. Rev., 1934, 15, 275-309 (DOI), this seems to work for 4-hydroxyacetophenone. Your starting material 1 is a phenyl homolog of that. In 2, you have two acidic protons ($\... 7 The best strategy is probably to chemoselectively protect the aldehyde before the borohydride reduction. Protecting groups are derivatives of certain moieties in a molecule that are far less reactive under most conditions, but which can be easily removed under very specific conditions to regenerate the original moiety. For carbonyl compounds -- this ... 6 There are quite a few reasons why the episulfide might open before the epoxide: Sulphur is less electronegative than oxygen. Therefore, by Bent's rule , more atomic s-character will be directed by the ring towards sulphur in the episulfide as compared to the oxygen in the epoxide. An increased s-character will lead to a bigger increase in the$\ce{C-S-C}$... 6 An easy and convenient method to achieve this is the use of the Luche reduction.[1,2] The ‘organic chemist’s bible’, Strategic Applications of Named Reactions in Organic Synthesis by L. Kürti and B. Czakó notes on the page of the Luche reduction (p. 268): 8) the combination of$\ce{CeCl3/NaBH4}$is excellent for the chemoselective reduction of ketones in ... 6 It’s not really a dichotomy, it’s just two very different reaction mechanisms. For the cuprate addition, which also applies to general Grignard-type additions, in principle the reaction is reversible (figure 1). Figure 1: Reversibility of Michael additions. However, the reverse direction is only really possible if the anion released is somewhat stable. In ... 5 The key question is how the reduction by the two reagents proceeds. For tetraborohydride the reduction proceeds as a standard nucleophilic addition of hydride ion to carbonyl group. For tin/HCl the reduction proceeds as a series of one-electron transfers from metal surface mixed with protonation. It seems that the lowest free orbital in the compound ... 5 The$\ce{N-H}$protons are relatively weakly acidic and this makes deprotonation kinetically less favored if an alternative nucleophilic attack mechanism is available. With$\ce{NH3}$no such alternative is there, due to lack of a decent leaving group, so deprotonation ultimately prevails; but$\ce{NH2Cl}$allows displacement of a chloride ion and thus ... 5 The selective Baeyer-Villiger Oxidation of unsaturated ketones is definitely depends on the conditions used. I do not want to eloborate this fact any further because it was well documented in Waylander's answer. Howevr, so far into the literature suggest that use of peracids in presence of unsaturation in the substrate ketone is not a good idea. The quote "... 4 I'm not sure I buy the strain argument either. Is there enough difference between the reactivities that you would get selectivity? In the real world I think you would get a mixture, possible complex enough to be called a mess. However according to this reference here Grignard reagents attack the least hindered carbon atom of thiiranes. So the sequence ... 4 Let the carbonyl compound under consideration be acetone. Case I: Hydrogen Abstraction The species formed are NaH and an allyl-type radical. The resonating system includes an electronegative oxygen atom, which also has an incomplete octet in one of the canonical forms. In essence, products are not really stable. Case II: Attack on Carbonyl The species ... 4 For simplicity, I will consider the reaction of acetic acid with ammonia, because the same principles apply. The two alternatives are: $$\underbrace{\ce{CH3C(O)NH2}}_{\mathrm{amide}} + \ce{H2O} \leftarrow \ce{CH3C(O)OH} + \ce{NH3}\rightarrow \underbrace{\ce{CH3C(NH)OH}}_{\mathrm{imidic \ acid}} + \ce{H2O}$$ where the group connected via a double bond has ... 3 As noted earlier, the determining factor is the greater strength of the carbon-chlorine bond versus carbon-bromine, which leads to a higher activation energy for the magnesium to react with the former. The strength of a carbon-halogen bond, and therefore the activation energy required for magnesium to form a Grignard reagent, is greatest with fluorine and ... 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 There are certain ethers which are somewhat easier to remove than others. If you stipulate that the ether must have a CH2 group immediately adjacent to oxygen, then a couple of possibilities, off the top of my head, would be: Benzyl ethers (R' = Ph), or substituted derivatives thereof (e.g. PMB, R' = 4-methoxyphenyl). These are typically removed with ... 3 I'm going to take "more reactive" as in the alkene undergoes bromination at a faster rate than the alkyne, which seems to imply that the activation energy of the transition state is greater. A Kinetic Perspective The formation of the bromonium ion involves a cyclic transition state where the the Br-Br bond is attacked by the alkene/alkyne pi-bond, and the ... 2 According to Bordwell$\mathrm{p}K_\mathrm{a}$Table,$\mathrm{p}K_\mathrm{a}$of$\ce{PhCH2COSPh}$is$16.9$, which is compatible with alcoholic$\ce{OH}$. We all know that Grignard reagents and organolithium reagents are very susceptible to acidic hydrogens because they are strongly basic as well as nucleophilic. Between$\ce{t-BuLi}$and$\ce{n-BuLi}$,$\...

2

Your given answer to the textbook question is seemingly correct. I said seemingly, because this reaction is not well understood to date, even though its first introduction was given at time as far as 1929 (Ref.1). The reaction was mostly hidden without getting attention in literature until to the middle of 20th century. This is mostly because the reactive ...

2

In Wutz's book on protecting groups,[1] I found an example of almost instantaneous reductive debenzylation which leaves an ester untouched:[2] The use of BCl3 is also possible:[3] References: Wuts, P. G. M. Greene's Protective Groups in Organic Synthesis, 5th ed.; Wiley: Hoboken, NJ, 2014, p 131. Sulikowski, M. M.; Davies, G. E. R. E.; Smith, A. B., III. ...

2

A discussion about alternative reduction methods was here Clemmensen reduction vs. Wolff-Kishner reduction. I've not seen nitro groups reduced by the acid-triethylsilane or tetramethyldisiloxane treatment. One used on plant that I've seen (and used) is TFA - triethylsilane. The carbonyl is protonated and the first hydride delivery gives the alcohol. A ...

1

It has to do with hybridization. The nitrogen in a nitrile is sp hybridized; imine sp2 and ammonia sp3. The more s-character in hybrid orbitals, the less basic the electron pair (protonated nitrile pKa ~ -10; pKb ~24). Ammonia has the lowest percent s-character (25%, NH4+ pKa 9.25). Imines lie in between although most examples you will find are measured in ...

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