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3

Density is a macroscopic measurement so we should look for average properties spread over several molecules. The average density of a liquid is not that much smaller than that of its solid ($\approx 0.9$, there are a v. few exceptions, water/ice, silicon, gallium, germanium, bismuth where it is greater) which means that the distance between molecules is only ...


5

The density of a substance depends on its molecular density and packing fraction (where you need to be consistent in definiting the molar volume in calculating both attributes). However you are looking for a simpler explanation that gets the general trends right. Just looking at molecular mass is too simple—larger molecules have greater mass, but take up ...


0

Actually there are commercial fire extinguishers that work by displacing the oxygen in a room. I forget the name but the safety experts put spheres containing an inert gas ( not a noble gas ,not CO2) to auto automatically be released in event of fire . Hydrogen was used in these lab rooms , but I don't know if that is why these extinguishes were chosen.


-1

Interestingly, no exotic compound is required to consume the oxygen in a room, or, for example, a large sewer pipe. In fact, an interesting report of workers who died when drainage occurred in a sewer pipe, resulting in the exposure of fresh iron metal surfaces, likely in the presence of acidic conditions and a complexing agent. A reaction pursed resulting ...


3

Here is a possible route: Start from the commercially available 2,6-dichloro-5-fluoropyridine-3-carboxylic acid (1). Form the Weinreb amide (2) via the acid chloride. React this with vinyl magnesium bromide to form (3). React the product with 1 eq of aniline in the presence of non-nucleophilic base to produce (4). Isolate this product and react with excess ...


0

A single hydrogen atom is just one proton and one electron. However, just to make clear, elemental hydrogen exists as $\ce{H2}$, which is a gas. The atomic number gives the number of protons. The atomic mass is, approximately, the number of protons and neutrons. When a hydrogen atom loses an electron, then it becomes the hydrogen ion ($\ce{H+}$) which is ...


2

Try to get samples of marble (pure calcium carbonate) in a marble mason workshop. Any piece of marble will do. Or try to find samples of ordinary calcareous stone (impure calcium carbonate), that can be found everywhere along the footpaths. Check if it is made of calcium carbonate by dipping them into some vinegar : it must produce bubbles if the stone is ...


2

First, a disclaimer: there is no shortage of opinion about what comprises physical or chemical changes. This page is entitled Identifying Physical and Chemical Changes and is intended to help instructors guide STEM students on this topic. You might find good ideas with low overhead in the Teaching Activities section such as these, quotes from the document: ...


6

This is straightforward enough: The $\ce{NaOMe}$ deprotonates the $\ce{OH}$ group on $\ce{C}$5, the alkoxide then does an intramolecular nucleophilic attack on the ester to give the lactone. This is favoured because it is intramolecular and forming a [6]-ring. It is possible that they may be some ester exchange of the $\ce{tBuO}$ group with the $\ce{MeO}$ ...


3

The comments summarized the reasons well: the diazotization reaction for the initial structure would produce a mixture of $\alpha$-azo and $\beta$-azo compounds, which would give a mixture of products. Also, diazonium salts couple with activated aromatic systems. The diazonium salt shown would not couple with nitrobenzene, which is highly deactivating and ...


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 ...


4

In reality, benzene and acetaldehyde are the minor product of the reaction of benzenediazonium chloride with ethanol. The major product is ethoxybemzene (phenetole). Peter Griess, who had discovered diazonium salts in 1858, had reported benzenediazonium salt (nitrate or sulfate) with ethanol undergoes aforementioned redox reaction in 1864 (Ref.1). However, ...


6

I have commented previously that the success of this reaction depends on the protonation of the free amino group of answer (a). In the absence of acid, heating amino ester (a) would give the starting material, the hydroxyamide because esters are more labile than amides. The reason this reaction can occur is that the amino group is protonated in acid which ...


8

tl;dr A slightly anthropomorphic approach with some basis in general organic chemistry: Step 1: The protonation of the carbonyl (An acid-base reaction) Here, we have three options – we can protonate the nitrogen (deactivated by the resonance with oxygen), the alcohol group ($\mathrm{sp^3}$) or the carbonyl group ($\mathrm{sp^2}$). As you can see this is ...


5

You need to understand something in organic chemistry. Not every question is going to be part of some named reaction that you just have to apply. For example in this question, the basic concepts of organic chemistry is used nothing more. First step, you look at the molecule and think, hmmm what can $\ce{H+}$ do? There are two possible things: The nitrogen ...


3

A mechanism has been proposed herein for the Étard oxidation of ethylbenzene that parallels the proposed oxidation of toluene with chromyl chloride (CrO2Cl2) as a [2,3]-sigmatropic rearrangement that is found on Wikipedia.[1] While such a mechanism is a convenient way to rationalize the formation of benzaldehyde, it does not consider the “Étard complex” that ...


4

Ethylbenzene yields acetophenone when it undergoes Étard reaction. The mechanism is as follows:


1

The partial reaction half-life relates to the reaction speed constant by the same way as for the "normal" reaction half-life. If there are 2 parallel reactions of the first order, $\ce{A -> B}$ and $\ce{A -> C}$, and if there is the reaction rate for the former: $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_{\ce{B}} \cdot [\ce{A}]$$ then the ...


-1

It mainly depends on the reagent. If the attacking reagent has more nucleophilic character ie. tendency to act as a nucleophile, Substitution reactions will be preferred. On the other hand, if the reagent has more tendency to act as a base, elimination reactions will occur. In cases where a reagent falls under both categories, you can go into additional ...


5

This would be the mechanism I propose for the reaction: I have taken the $\ce{pH}$ of the medium to be $4.5$. That should explain the protonation of the $\ce{-OH}$ groups.


1

This looks like a standard Friedel-Crafts reaction, although it is intramolecular rather than intermolecular. The cation undergoes electrophilic substitution to give the product. The choice of terminology used by the authors is strange: a search for "ArSE2" returns very few links, such as this one. There seems to be no commonly known reaction ...


2

During diazonium ion preparation, first, sodium nitrite is mixed with hydrochloric acid to produce nitrous acid. The nitrous acid can be protonated under acidic conditions, which undergoes the loss of water and produce the nitrosonium ion ($\ce{O=N+}$; this reaction is usually performed at or around $\pu{0 ^\circ C}$): An aromatic amine can attack the ...


3

If the reaction is $$\ce{2A -> P},$$ then the rates are defined to be related as $$ -\frac{1}{2}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t} = \frac{\mathrm{d}\ce{P}}{\mathrm{d}t} $$ and this is true throughout the reaction. The rate at which $\ce{P}$ is produced is half that at which $\ce{A}$ is consumed. Normally therefore a factor of 2 is expected with ...


1

The first expression $$\frac{\mathrm{d}[A]}{\mathrm{d}t} = -2k_\mathrm{f}[\ce{A}]^{2}\tag{1}$$ is the correct one for the reaction given. As the OP mentions in the comments, both approaches describe the situation correctly. However, the values of $k_\mathrm{f}$ would be different by a factor of two. You could rewrite the chemical equation by dividing by two (...


1

It is known fact that organolithium compounds react with carboxylic acid to give ketones (e.g., Ref.1). Until recently, however, no report has been published on Grignard reagents on this manner. Now, it is evident that Grignard reagents can also be used to prepare ketones under specific conditions (Ref.2): References: Robert. Levine, Marvin J. Karten, “...


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