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3

Using methyl orange for titration of weak acids is wrong idea. The indicator has range $\mathrm{pH}=3.1~\mathrm{(red)} - 4.4~\mathrm{(yellow)}$. If acetic acid with $\mathrm{p}K_\mathrm{a}=4.75$ is to be determined, the indicator starts to turn toward yellow, when majority of acid is not titrated yet. The concentration ratio $\frac {\ce{[CH3COONa]}}{\...


2

Your proposed mechanism involves a ring expansion from cyclopentane to cyclohexane. This gives very little driving force in terms of relieving strain in the system. Cyclopentane, in a twisted or "envelope" conformation, has some angle strain ($102-106^o$ vs $109.5^o$ $^{ref~1}$) and it does also has some torsional strain down some of its bonds. This is ...


-2


1

Regarding bed-bugs, they are notoriously difficult to remove - I'd highly advise getting a pro involved as soon as possible. For slow/steady carbon dioxide generation in a general sense, however, I'd go with yeast. East to find bakers yeast and flour in any supermarket. Google a basic bread recipe, and mix accordingly. If you do a slow rise, the dough will ...


1

Liquid-Phase Photolysis of Dioxane has been studied in several occasions in the past. One of these studies, several neat aliphatic ethers including dioxane were irradiated for $\ce{24 h}$ with a medium-pressure mercury lamp (Ref.1). As a result, the neat liquid-phase dioxane has been reported to yield gaseous, liquid, and solid products after the irradiation....


3

I think you are correct. The $\mathrm{p}K_\mathrm{a}$ of 4-methylphenylacetic acid is quoted as 4.37 here. The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is quoted as 4.19 by the same source here.


1

I could not find a definitive mechanism for 1,4-dioxane, but this is the mechanism for the formation of THF peroxide is likely very similar diagram from here


0

In this case, the absolute configuration remains same(as R) for SN1 and chages(into S) for SN2. This is because the priority order of non substituted groups(wrt C2) remains the same before and after substitution of Br by OH. Let the initial mole fraction of (R)-2bromobutane be 1. After the reaction, let the mole fractions of R and S 2-bromobutane be x and 1-...


1

Not a complete answer, but too long for a comment. First recognize what is really in solution. Magnesium oxide does not dissolve in water. It reacts to give magnesium hydroxide, which in turn is only sparingly soluble in water alone but could react with the acidic lysine. You should proceed as if your solution had magnesium ions and lysine anion -- ...


1

I don't know how you call it in English but you can apply this approximation $\frac{d[D]}{dt} = 0$ because (3) is fast and $D$ have a really short lifetime, so you can say that $$k_2[B]^2[C]=k_3[D][E]$$ This approximation is why the teacher is always saying that only the slow reaction is important Fast equilibrium means that the equilibrium is quickly ...


1

The keyword you should search is iodometric titration. Triiodide solutions are dark brown, just like the tincture of iodine. If you were to add thiosulfate ion (say from a buret) to the triiodide solution would become colorless. The change is not that sharp. Therefore this is not the way end point is detected. A small drop of starch is added. Starch makes a ...


1

Tunneling can be important for any reaction where there is a proton or electron transfer involved and the electron/proton donor/acceptor are placed in a "sweet spot" that increases the importance of this over other mechanisms. As explained here, the conditions for appreciable contribution of tunneling to a reaction mechanism are that (1) the activation ...


1

One thing the exercise is lacking is the physical state of A, B, C, and D. If they are all gases or all liquids, you can calculate as shown in question and in answers. Otherwise, you would need more information to complete the exercise. For some insight regarding part b) of the exercise, you could rewrite your equations as: $$\ce{A <=> B}$$ $$\ce{A &...


1

You can also solve for $\mathrm{[S]} $ using Lineweaver-Burk plot (or double reciprocal plot). One way of writing the Michaelis Menten kinetic equation given in Wikipedia is: $$v_{\circ} = V_\mathrm{Max} \frac {[S]}{K_\mathrm{M} + [S] } = = k_\text{cat} \mathrm{[E]_{\circ}} \frac {[S]}{K_\mathrm{M} + [S]} $$ If you take the reciprocal of both sides of the ...


2

What you do is write $[D]=[C]K_{CD}$, $[C]=[B]K_{BC}$, and $[B]=[A]K_{AB}$. So, $$[C]=K_{BC}K_{AB}[A]$$ and $$[D]=K_{CD}K_{BC}K_{AB}[A]$$So, $$[A]+[B]+[C]+[D]=(1+K_{AB}+K_{BC}K_{AB}+K_{CD}K_{BC}K_{AB})[A]=10$$


1

You can solve for [S]. There are two common ways to write the equation, like you have it: $$v_0 = k_\text{cat} [E_0] \frac{[S]}{K_\text{m} + [S]}$$ or after dividing both numerator and denominator by [S]: $$v_0 = k_\text{cat} [E_0] \frac{1}{\frac{K_\text{m}}{[S]} + 1}$$ In this form, it is easier to see that [S] occurs once and you can solve for it. Get ...


1

Different compounds always have to have the same number of each type of atom in them. For example, in respiration: $$\ce{C6H12O6 + O2 -> CO2 + H2O}$$ The chemical formulas for these compounds must always remain the same else it is no longer that compound. With this in mind balancing increases the number of molecules of each compound you have to ensure ...


3

How is it that a chemical reaction produces an equation, that is essentially an impossibility, according to the law of conservation of mass? You find unbalanced equations as exercises and as a step in writing a balanced equation when there is insufficient information that still needs to be discovered. Balancing equations as an exercise The observable ...


5

First, provided the ball is an atom, you are not writing a chemical equation for a dozen of balls, rather a dozen times at least $10^{16}$ (to speak of equilibrium), usually a dozen times $10^{23}$ (macroscale that normal people operate on). That's not the level you want to use addition for; smaller numbers are easier to operate with. Second, it's not ...


2

AS @Karsten Theis pointed out that this question is too broad. Yet, if you are willing to study, here is a review article (Ref.1), which included more than 250 references to go with: Abstract: The skeletal rearrangement of α-halogeno-ketones, which is known as the Favorskii rearrangement, is met most frequently in aliphatic monocyclic, and polycyclic ...


2

I would comment however I am new and have yet to unlock that feature. Do you have an exact substrate of interest? I have studied the Favorskii rearrangement a tiny bit in a graduate mechanisms course and if your seeking a very general answer to a general question; The favorskii seems to be favorable when your substrate is containing rings that are ...


7

You are not too far off. It is somewhat of a mixture of the two mechanisms you proposed. The carbonyl oxygen is much more basic than the non-carbonyl oxygen and will be protonated preferentially. Then hydrogen peroxide can attack, and once the tetrahedral intermediate collapses, deprotonation yields peracetic acid.


5

This is a compilation from the comments (some of them deleted by now) showing the thought process in finding the reaction product and eliminating some of the steps that might seem plausible. None of the insight is mine. Formation of a cation I think the H+ will first attack the oxygen of the epoxide Yes, that is the first step. No attack by a ...


3

$\ce{NaNO2/HCl}$ is as you suggest a source of $\ce{HONO}$. This will be protonated by $\ce{HCl}$ to give the nitrosonium ion $\ce{NO+}$ (in equilibrium with nitrosyl chloride). This strongly electrophilic species will react with the enol of the ketone to nitrosate the alpha position. This nitroso compound then rearranges to give the alpha oxime. Image from:...


1

It is more probable like $$\begin{align} \ce{NO2Br &-> NO2 + Br} \\ \ce{NO2Br + Br &-> NO2 + Br2} \\ \ce{2 Br &-> Br2} \\ \end{align}$$ The last reaction is a minor one in case concentration of $\ce{Br}$ is low. The reaction rate order can be concentration dependent and need not be the integer. In fact, it is rather mathematical ...


8

The conventional mechanism is as follows: The phenol displaces one acetate group on iodine – this makes the iodine itself act as what is essentially a fancy leaving group. Nucleophilic attack at either the ortho or para position allows the loss of iodobenzene and another molecule of acetic acid. What I can't explain right now is why it chooses to go para ...


4

Explanation In case 1. You can see it is a 2° carbocation (or secondary carbonation) with angle strain of ~25° on cyclopropane In case 2. You can see it is 3° carbocation (or tertiary carbocation) with angle strain of ~10° on cyclopropane It is obvious that case 2 will be major intermediate yielding $\ce{1-chlorobicyclic [3,2,0] heptane}$ as main product ...


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