12

It is known that $\ce{Mg}$ will reduce ketones. The articles (Ref.1 and 2) in the References section confirm this fact. If you want to form the Grignard reagent of a molecule that also happens to contain a carbonyl group, then the usual practice is to protect it first. A cyclic ketal/acetal, which are stable under basic conditions, is often the first choice ...


9

Unlike other reducing metal hydrides (e.g., $\ce{NaBH4}$ and $\ce{LiAlH4}$), diisobutylaluminum hydride (DIBAL-H) is a liquid at room temperature and dissolve in many hydrocarbons such as toluene and hexanes, which also have very low freezing points. For example, hydrocarbons toluene and hexanes both have freezing points around $\pu{-95 ^\circ C}$. Thus, ...


9

As I noted in a Comment that there should be no significant difference between the use of ethanol or isoamyl alcohol in the reduction of naphthalene with sodium in that both alcohols are primary. The difference lies in the conditions of the reaction. The sequence is a classic case of kinetic vs. thermodynamic conditions. The reaction conditions in the ...


7

It will depend on strict control of the equivalents of reductant, your work-up conditions and your control of temperature during the workup. The intial product of the reduction is the aluminium alkoxide. If you use a proton source such as $\ce{HCl (aq)}$ or $\ce{NH4Cl (aq)}$ to quench the alkoxide and keep the reaction mixture cold then you will get the 4-...


6

The reduction of alkyl halides to the corrosponding alkane by $\ce{LiAlH4}$ is well known reaction: $$\ce{R-X + LiAlH4 -> R-H + AlH3 + LiX}$$ This reaction has been studied in some details (For example, Ref.1). Accordingly, the major product from the reaction of 4-chloropentanal and $\ce{LiAlH4}$ would be pentanol after simultaneous carbonyl and halide ...


5

α,β-Unsaturated carbonyl compounds, for the most stable intermediate with respect to $\ce{H-}$ ion prefer 1,4-addition. However, $\ce{LiAlH4}$ prefers 1,2-addition over 1,4-addition. This can be explained due to the following reasons. It is more reactive so it prefers a kinetically favorable pathway (1,2-addition is often kinetically favored while 1,4-...


5

The major purpose of $\ce{NaBH4}$ in the second step of given reaction sequence is the demercuration, which stops at the hemiacetal as the applied condition is the limit (at $\pu{0 ^\circ C}$ for $\pu{1 min}$). In general under usual conditions (at $\pu{25 ^\circ C}$ in ethanol for for an hour or two), $\ce{NaBH4}$ reacts with hemiacetals and reduce them to ...


4

According to this source here The Na /NH3 reduction of alkynes does not work well on 1-alkynes unless certain modifications are made in the reaction conditions. edit: Thanks to @Buttonwood for providing details on the modifications required. Addition of ammonium sulfate to the reaction mixture gives quantitative yields of alkene according to this edition ...


3

I'll take the opportunity to expand on the Comments and links I provided previously. The intermediate you proposed does not occur during the LiAlH4 reduction of the enol ethers of dihydroresorcinols such as enone 1. The LiAlH4 reduction is a straightforward 1,2-reduction of the carbonyl group. If this reduction is worked up with water, no aqueous acid, ...


3

If you can find another solvent in which to run the reaction, you don't need the ammonia as a solute. The Birch reduction is a reaction with solvated electrons, not with ammonia, and the key is identifying a solvent that can hold the solvated electrons for reaction with the substrate. Many solvents that are liquid at room temperature and atmospheric ...


3

The question you pose is the work of Stork and Dolfini that led to the synthesis of aspidospermine 7 and quebrachamine 9. The aminoketone 1a was condensed with either phenylhydrazine or o-methoxyphenylhydrazine to form the respective arylhydrazones 2a and 2b. Exposure of either hydrazone to hot acetic acid led to a "Fischer indole synthesis" that ...


3

Fischer Indole for the first step looks possible depending on reaction conditions, but this source states that the indole nucleus is not reduced by LiAlH4, nor do I recall ever seeing it done. Therefore I think what is happening is formation of the phenylhydrazone followed by reduction to the 1,2 di-substituted hydrazine.


2

$\ce{LiAlH4}$ is not formed by $6$ independent ions. But, for the present purpose, it reacts as if it was formed of $\ce{Li^+ + Al^{3+} + 4 H^-}$. The ions $\ce{Li+}$ and $\ce{Al^{3+}}$ do not react with water. The only ion reacting with water is the hydride ion $\ce{H^-}$, and the equation is $$\ce{H^- + H2O -> OH- + 1/2 H2}$$ As each $\ce{LiAlH4}$ &...


2

May I introduce you to the Barton-McCombie reaction for the reduction of alcohols to alkanes. image and further information here


2

As Waylander has pointed out in the comments, the mechanism would proceed as follows in case of $\ce{LiAlH4/AlCl3}$ as a reagent to produce a secondary alcohol (SN1), whereas in case of just $\ce{LiAlH4}$, the reaction proceeds via SN2 to give the tertiary alcohol by attacking the less hindered carbon. Alane has more Lewis acid character than $\ce{LiAlH4}$ ...


2

There is just a silly mistake. As TRC told, $\ce{Pd/C}$ is not a Lindlar catalyst and it behaves as a normal $\ce{Pd/H2}$. Refer to the link for more information.


2

After the synthesis, your crude solid acetaminophen contains dark impurities, which have been formed from oxidation of the starting material, 4-aminophenol that carried along during the synthesis (if you have used 4-aminophenol as it is without purification). These impurities are intensely colored dyes of unknown structures with highly conjugated chromophore,...


2

Dextrose equivalent $(DE)$ is a measure of the amount of reducing sugars present in a sugar product, expressed as a percentage on a dry basis relative to dextrose (glucose). Commercial sugar products are generally characterized by their dextrose equivalent $(DE)$ value (Ref.1). The theoretical definition of $DE$ is given by: $$DE = \frac{\text{Reducing power ...


1

I couldn't find a paper which directly discusses imine reduction with the help of sodium metal and alcohol but I did found its mention in a different paper which discusses the reduction using calcium and ethanol. I believe sodium metal follows similar reaction pathway. The reduction of imines is one of the most significant and useful methods for preparation ...


1

Wolff–Kishner reduction is the method for the reduction of aldehydes and ketones to corresponding alkanes. First, the carbonyl compound condenses with hydrazine to form the corresponding hydrazone. The resulting hydrazone is treated with base to induce the reduction of the $\mathrm{sp^2}$-carbon to $\mathrm{sp^3}$-$\ce{CH2}$, and the oxidation of the ...


1

The reason for this is because OH and OR groups are much more electron-donating. Typically, when you have an aldehyde or ketone, the electronegative electron is slightly more negative and thus the carbonyl carbon is more positive. However, with an OH or OR group attached (i.e in carboxylic acid and esters), the group can push electrons towards the carbon ...


1

The likely reason for this is a steric hindrance. The Clemmensen reduction is very vulnerable to hindrance. See the image below: There was also another reaction conducted that had a 67% yield for a gamma keto acid with resonance. See the image below: Therefore, the main factor that makes α- or β-keto acids not undergoing Clemmensen reduction is likely not ...


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