16

Neither $\ce{LiAlH_4}$ nor $\ce{NaBH4}$ are able to reduce an isolated $\ce{C=C}$ bond. But if you have an enal (a conjugated aldehyde) it can react (as an electrophile) either at the $\beta$-carbon or at the carbonyle group's carbon. According to the HSAB Principle the $\beta$-carbon is a "soft" center and would react preferably with "soft" nucleophiles ...


13

The first reaction is called a catalytic reduction. Here is a picture of how it works with an olefin like ethylene. You can see that the first step involves adsorption of hydrogen onto the catalyst surface followed by dissociation of the hydrogen molecule into its catalytically active form. The same things happen with nitrobenzene. As the following ...


12

You've got it correct in your comment with the $\ce{N}$ lone pair assisting. The reason why the $\ce{N}$ lone pair kicks oxygen out, and not the other way round, is because oxygen-based groups are better leaving groups than nitrogen-based ones under the reaction conditions. (Just as a rough guide, you could compare $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ ...


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 ...


12

First of all, I should clarify that hydrogenolysis of epoxides does not always proceed with such predictable regioselectivity simply based on a switch in catalyst. Often it depends on the substituents present, as well as their relative configuration (for example, cis/trans for 2,3-disubstituted epoxides).[1] In Reaxys I find no records of the epoxide in the ...


11

I don't buy that mechanism. Like you noted, breaking a C-H bond in methanol seems unlikely, especially since the reaction is run in the presence of acid, so a ready source of protons is available. Most of these reactions that involve a metal surface proceed through what is known as a single electron transfer (SET) mechanism. Reductions of nitro compounds, ...


10

The Clemmensen reduction is really only effective at reducing ketones to alkenes when one group is aromatic. Additionally, the Clemmensen reduction uses concentrated HCl, which might cause problems. Another functional group in your molecule, the tertiary alcohol, will be removed by HCl via an elimination reaction. Choice (b) is the Wolff–Kishner reaction, ...


10

Cinnamaldehyde (1) does not undergo 1,4-reduction but rather 1,2-reduction with LiAlH4. The product of the reduction is 3-phenylpropan-1-ol (2). If conjugate addition were to occur, enolate 3 would be produced (stable to the reductant; enolates can be prepared to protect ketones from this reductant) and aldehyde 4 would be expected, as long as the aqueous ...


9

@Halake: The protection of adipaldehyde as the monoacetal was solved by Schreiber and Claus some years ago [Also see, Schreiber, S.L.; Claus, R.E.; Reagan, J. Tetrahedron Letters, 1982, 23, 3867.] Ozonolysis of cyclohexene (1) in the presence of methanol leads to the molozonide 2, that opens to the zwitterion 3 which is trapped as the methoxy hydroperoxide 4....


9

Carbamates are usually reduced to N-methyl groups. There are numerous examples: J. Am. Chem. Soc. 2012, 134 (16), 6936–6939 Org. Lett. 2012, 14 (18), 4834–4837 But it is not always a given. In this next example, the nitrogen is part of a three-membered ring (aziridine). These nitrogens are better leaving groups than usual, cf. Ketone/aldehyde synthesis ...


7

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

Sodium in ammonia ($\ce{Na/NH3}$) and sodium amide ($\ce{NaNH2}$) are very different sets of conditions, despite superficially looking similar! Sodium metal itself, $\ce{Na}$, is a one-electron reducing agent – it is oxidised to $\ce{Na+}$ in the process. When it reduces neutral hydrocarbons, negatively charged species are formed, which pick up a proton ...


7

Steps: Use $\ce{Br2/FeBr3}$ to make bromobenzene. Nitrate with $\ce{HNO3/H2SO4}$ to make 4-bromo-nitrobenzene. Treat with $\ce{NaOH}$ to displace the $\ce{Br}$ making 4-nitrophenol. Reduce with $\ce{Sn/HCl}$ to make the required p-aminophenol.


7

The "Lindlar catalyst" (Pd/CaCO3) is probably the most common way of effecting stereoselective reduction of alkynes to (Z)-alkenes. It is fairly accessible and the reaction is easy to run. But it is not the only way: there are other catalyst systems. One of the more notable ones is probably the Brown catalyst "P2-Ni" (Wikipedia). There is a nice review on ...


6

The acyloin condensation of diethyl adipate (1) on your Wikipedia citation generates ethoxide that can cause a Dieckmann condensation (intramolecular Claisen condensation, structure 6). TMSCl scavenges ethoxide and silylates the enediolate 3 formed in the acyloin condensation. The process is a 4-electron reduction (see details below). Mild hydrolysis of 4 ...


6

Yes, that is used as a pathway for synthesis of amines from amides, albeit rarely. I quote from the abstract of a communication that reports the same: (emphasis mine) The reduction of amides to their corresponding amines has been used extensively in the synthesis of many drugs and their intermediates. This reaction has usually been carried out by means ...


5

Hydrogenation of the alkene introduces a stereo centre, thus the products are:


5

You are correct that ammoniacal silver nitrate is Tollen's reagent. Tollen's reagent is most notable for its reaction with aldehydes to produce a silver mirror, but it also reacts with several other types of compounds, notably alpha-hydroxy ketones and terminal alkynes. With terminal alkynes, the alkyne reacts as a acid to form a silver acetylide which ...


5

I would look at it in the following way: The oxidation state of the carbonyl carbon is +II, while the oxidation state of the nitrogen in the nitro group is +III. So nitrogen is in a higher oxidation state and it is also more electronegative. Thus, isn't it sensible that it gets reduced first before the carbonyl group? In fact, reduction of a nitro group is ...


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

You guessed it right: a ketone will be formed. This reaction involves migration of alkyl groups from boron to the carbon atom of CO. There will be three migrations after which you will get your alcohol: Image taken from National Programme on Technology Enhanced Learning (NPTEL). Lecture 26 : Boron Containing Compounds. Image link: Scheme 9 See also ...


5

There are two replaceable $\ce{H}$s in the starting material, but only 1 equiv of $\ce{TsCl}$ is given. Thus, primary $\ce{-OH}$ will win the battle due to steric reasons, and therefore, B represent the intermediate product X. The second reaction is firstly acid-base reaction between secondary $\ce{-OH}$ and $\ce{H-}$, and then, nucleophilic substitution ($\...


5

Hydrogenation of aromatic nitro groups (over, for example, Pd/C) usually results in reduction to the corresponding aniline and is probably easier to carry out and work up as compared to Sn/HCl. Wikipedia has a list of several conditions for reduction of ArNO2 to ArNH2: https://en.wikipedia.org/wiki/Reduction_of_nitro_compounds#Aromatic_nitro_compounds It ...


5

Acetone and sodamide give acetone sodium enolate A, nothing further happens until acetylene is introduced. The pKas of acetone and acetylene are both around 25 so an exchange occurs reforming acetone and creating sodium acetylide. This seems an unusual way of making sodium acetylide, but I don't think this is a real world example (unless someone can find a ...


4

Chemistry is an experimental science supported by computational methods. Qualitative explanations for reaction mechanisms follow as a result. More on that issue later.The mechanism of the Birch reduction has been investigated by Zimmerman and Wang.1 What are the experimental results? The Birch reduction is first order in substrate, electrons and alcohol. ...


4

In the mechanism you are considering, there are two options for reduction of R-$\ce{N^{.}H}$ (and similar for other neutral radical species) $\ce{CH3OH + R-N^{.}H -> RNH2 + C^{.}H_2OH }$ $\ce{CH3OH + R-N^{.}H -> RNH2 + CH3O^{.}} $ In the first option (1) nitrogen oxidizes carbon - rare, but possible, as carbon has lower electronegativity. In option (...


4

Does diisobutylaluminium hydride (DIBAL-H) reduce only carboxylic acids? or only carboxylic acid esters? or reduce both both types of compounds? The answer is yes, it does reduce both. Not only them, it also reduces nitriles to aldehydes, and is a more selective reagent than lithium aluminum hydride (LAH) in the reduction of nitriles (Ref.1). About ...


4

According to the article on Birch Reduction on the Organic Chemistry portal: The question of why the 1,3-diene is not formed, even though it would be more stable through conjugation, can be rationalized with a simple mnemonic. When viewed in valence bond terms, electron-electron repulsions in the radical anion will preferentially have the nonbonding ...


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

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 ...


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