9

I think the reason for inverting the the round flask is to capture sunlight very efficiently. A round bottomed flask filled with a fluid is like a ball lens. Sun rays can be considered parallel and see how they focus. If light is coming from all directions, every part will be illuminated well as the Sun moves. See ball lenses here ball lenses The clue comes ...


5

The reducing power of alkali metal such as $\ce{Na}$ and $\ce{K}$ in liquid $\ce{NH3}$ is duee to the solvated electrons ($e^-$), which can reduced certain compounds into unusual oxidation states. One such example is reduction of oxygen ($\ce{O2}$). The the solvated electrons can be reduced $\ce{O2}$ to $\ce{O2^{.-}}$ (superoxide ion) first and then to $\ce{...


4

The course of the reaction depends on whether or not a proton source other than ammonia is present. In the absence of an alcohol (ethanol, tert-butanol) the α-dione 1 adds an electron from sodium to give a resonance stabilized radical anion 2. A second electron is added to afford enediolate 3. [This is the same species formed with sodium in the acyloin ...


4

This reaction was conducted using a 3- to 5-fold excess of ethyl cyanoacetate (ECA) and KOH over the arylnitro compounds in dimethylformamide (DMF). It is also important to realize that most KOH is 15% water, which allows for a source of protons in this basic medium. Steps 1 --> 3 (Scheme 1) illustrate a possible route to nitrone 4. In so doing the ECA ...


4

Imines survive, that's for sure. Some 20 years ago junkies developed a way of turning a relatively innocent drug, which was then legal (and which has since ceased to be legal in many jurisdictions for this very reason) into a much more potent substance. I'd rather not identify it precisely; let's just say it is prominently featured in a TV series with name ...


3

The $\ce{LiAlH4}$ will reduce the cyclic anhydride to the diol. $\ce{KMnO4}$ re-oxidises both the alcohol groups (diol formed) to carboxylic acids and also oxidises the double bond to the diacid so the product you have drawn is correct, but it is an intermediate. You then have to consider the effect of heating this (to a pretty high temperature, I would ...


3

In comparision to Clemmensen reduction, both $\ce{Na-Hg}$ and $\ce{Al-Hg}$ cannot be used as direct substitutes for $\ce{Zn-Hg}$. Reaction with $\ce{Na-Hg}$ amalgam: $\ce{Na-Hg}$ amalgam is also a good reducing agent and is capable of reducing ketones/aldehydes to alcohols1. However, Do not use conc. $\ce{HCl}$ with $\ce{Na-Hg}$ amalgam, as there would be a ...


3

The correct question (taken from the 2016 JEE Paper is): Lithium Aluminium Hydride reduces epoxides1, hence (A) would not give the required product. LAH also reduces ethers to primary alcohols, which is an easier way of eliminating the option (A). $\ce{BH3}$ in THF can reduce the acid, as explained in this question, hence the requisite product would not be ...


3

The only case I know of where activated carbon causes reduction at room temperature would be filtration of ozone, $\ce{O3}$. Even fluorine will not oxidize plain carbon: Carbon is stable in a fluorine atmosphere up to about $\pu{400 ^\circ C}$ . Oh, well perhaps dioxygen difluoride, $\ce{FOOF}$, could also oxidize carbon (or could be reduced by carbon, ...


3

It is not so much the glass, it is the contact between the substrate and the reflecting layer. Some of the problems with the silver/glucose approach, like their optical imperfections is related to the bad adhesion of elemental silver to glass. The «normal glass mirror» in the bathroom thus uses a tin layer between the cleansed glass, and the subsequently ...


2

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


2

My organic chemistry is rusty by now, but the full mechanism for the reduction itself is likely something along these lines: In a "typical" lactone / ester, the alkyl oxygen can't be lost so easily, but in this case there's a nitrogen lone pair that can assist in that. The resulting iminium ion is reduced by the silane. The acid at the end will ...


2

You can do a normal hydrogenation reaction(H2,Pd) under low pressures as suggested by @Waylander. Alternatively you can use the Wilkinson's catalyst which also is a way to do catalytic hydrogenation but is capable of only reducing the double bond.


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 ...


1

One example of a reaction like this is the reaction of Benzaldehyde and Formaldehyde in the presence of a basic hydroxide (OH-). This reaction proceeds as follows: Thus, using Sodium Hydroxide (NaOH), The reaction would yield Phenylmethanol (C6H5CH2OH) and Sodium Formate (HCOONa, sodium salt of formic acid) If you want a better answer, or if this was not ...


1

The book's reasoning is correct. Fe/HCl gives $\ce{FeCl2}$ on reaction and it gets hydrolyzed by steam vapors to produces more hydrochloric acid and hydrogen to push the reaction forward thus making the reaction self-sustaining to produce more aniline from nitrobenzene. $$\ce{Fe + 2HCl ->[160°C,CH3OH] FeCl2 + H2}$$ $$\ce{3FeCl2 + 4H2O(steam) ->[...


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