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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 ...


-1

Per a source, to quote: Like ozone, the predominant oxidation reaction mechanism for chlorine dioxide proceeds through a process known as free radical electrophilic (i.e. electron-attracting) abstraction rather than by oxidative substitution or addition (as in chlorinating agents such as chlorine or hypochlorite). So, the reaction: $\ce{.ClO2 + e- -> ...


1

This is an example of Cope Elimination where the base takes up hydrogen from the least sterically hindered Carbon atom to avoid crowding in the Transition state. Hence, Hoffman Alkene is obtained and ethene is formed. The reaction is not an example of E1cb since the Leaving Group is a stable alkene whereas the basic requirement for a reaction to be ...


0

The OH group of phenol is so activating that the nitration mixture rapidly overheats even with dilute nitric acid. The result is oxidation with the evolution of copious amounts of nitrogen oxides. Instead, you heat the phenol with sulfuric acid to get mostly phenol-2,4-disulfonic acid.The sulfonic acid groups are so electron-withdrawing that nitration can ...


1

Higher temperature has several effects: The rate of chemical reactions increases empirically and typically 2-4 times by every increase be 10° C Heating toward boiling or intensive gasing turbulently mixes solution, what brings fresh solution to limescale. Faster diffusion in hot solution due lower viscosity and higher ion/molecule mobility. I never had ...


1

Look for equation solvers in Numpy/ Scipy and load at the top of your script scipy.integrate import odeint and import numpy as np . There are examples of how to numerically solve differential equations in the examples and on line. The Master equation approach does not work for second order steps. There is also the Gillespie method which is a sort of Monte-...


1

The use of dry NaHCO3 to extinguish a fire works in two ways: First, the smothering of the flame with CO2 and water per the decomposition reaction: $\ce{2NaHCO3 ⟶ Na2CO3 + CO2 + H2O}$ Second, the decomposition reaction is endothermic, so it lowers the temperature of the fire. To quote a source: In an endothermic reaction heat must be continuously ...


6

Reaction rates can be calculated from the Arrhenius equation $k=Ae^{E_a/RT}$ where $E_a$ is the activation energy found at the transition state. Please check out the Reference 1 for calculating transition states under force. Following is an excerpt from the paper (Ref.1): As a force is delivered to the mechanophore, the energy provided by the work can ...


1

Here is a short simple available answer provided online by Prachi Sawant in 2016 to quote: Formic acid has both aldehydic (-CHO) and carboxylic (-COOH) functional groups. Hence, it gives both Tollen's and Fehling's test positive. Reaction of formic acid with Tollen's reagent: $\ce{HCOOH + 2[Ag(NH3)2]+ +2OH- → 2Ag + CO2 + 2H2O + 4NH3 }$ ...


1

When heated, $\ce{NaHCO_3}$ is easily decomposed and produces much $\ce{CO_2}$ in the chemical reaction : $$\ce{2 NaHCO_3 -> Na_2CO_3 + CO_2 + H_2O}$$ and $\ce{CO_2}$ is a gas that tends to prevent Oxygen from maintaining the fire. As to your question 2, I have no idea how to solve it. And as to your question 3, $\ce{NaHCO_3}$ is by far the best ...


-2

2NaHCO3 --> Na2CO3+H2O+CO2 at temperatures above 50 degrees Celsius: TRUE. However, Na2CO3+H2O --> NaHCO3+NaOH is a hydrolysis reaction: it only goes to a small extent, maybe 2-5%, and on evaporating the water, it goes back to Na2CO3. Na2CO3 melts at 851C without decomposition. So, if you take a solution of NaHCO3 (pH ~8.5) and boil it, you will generate ...


1

The problems with reacting pure nylon salt are as follows: The molecules of hexamethylene diamine (HMD) and adipic acid will have a very hard time accessing each other's reacting groups, because of limited mobility within the solid structures. The reaction is a condensation, so it will produce a substantial amount of water anyway, so pretty much any reason ...


0

I think the cinnamaldehyde gives fecl3 test due the similar reason why acetic acid can give fecl3 test as Fe3+ can act as oxidizing agent. As in case of acetic acid Fe(Ch3COO)3 (red colour) is formed a similar complex may be formed. (This is my opinion corrections are accepted)


-2

Actually, if one adds your two reactions: $\ce{2 NaHCO3 <=> Na2CO3 + H2O + CO2}$ $\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$ Upon cancelling, to derive the reaction: $\ce{ NaHCO3 <=> NaOH + CO2}$ which implies that heated Sodium bicarbonate is a chemically active source of Lye, if one allows the CO2 gas to escape (which answers (a) of your ...


3

The vicinal-glycols can be cleaved to corresponding aldehydes and/or ketones in high yield by the action of periodic acid ($\ce{HIO4}$) or lead tetraacetate ($\ce{Pb(OAc)4}$). This oxidative cleavage of a carbon-carbon single bond provides a two-step involving cyclic intermediate reaction mechanism with high-yield. A generally accepted equation for these ...


2

You're absolutely right: cyclopropane's maximum ring strain makes it highly prone to addition reactions. Keep the following definition in mind: An addition reaction, in organic chemistry, is in its simplest terms an organic reaction where two or more molecules combine to form a larger one (the adduct). In your case: Thus, it breaks down the ring ...


0

Both the mechanisms you've mentioned exist, but they both have some kinetic difficulties. Your first mechanism (called $\ce{ArSN2}$) involves breaking the aromatic character of the ring, while your second mechanism (a form of $\ce{E1cB}$ leading to a benzyne intermediate) involves removal of a minimally acidic hydrogen. So a good, strong base favours ...


0

I would say (B). The mechanism involves the breaking of the Halogen–Halogen bond. The strength of this bond decreases down the group because atomic radius increases. The attraction of the halogen nuclei on the electrons in the covalent bond decreases as the valence electrons are further away from the nucleus. Therefore, the I–I bond is weakest while F–F ...


0

Ans is (D) Since F+ is not possible and in case. of I2 it will be reversible so In I2 &F2 the order is I2>F2 So ans is (D)


0

Per a source, the chemical reactions in a zinc–carbon battery can be detailed as follows, to quote: In a zinc–carbon dry cell, the outer zinc container is the negatively charged terminal. The zinc is oxidised by the charge carrier, chloride (Cl−) via the following half reactions: Anode (oxidation reaction, marked −) $\ce{Zn + 2 Cl− → ZnCl2 + 2 e−...


2

No, it is not a good disinfectant. To quote from the CDC on chemical disinfectants 'Guideline for Disinfection and Sterilization in Healthcare Facilities (2008)' available here: Methyl alcohol (methanol) has the weakest bactericidal action of the alcohols and thus seldom is used in healthcare 488.


1

It's a tricky example, but the $S_N1$ process is favored because the solvent is protic and the resulting 6-member ring is more stable than a 5-member ring. If ring expansion weren't possible, I believe your intuition is correct that the $S_N2$ mechanism would dominate. See this discussion. You're not alone!


2

You didn't give a mechanism for your expected product, so I can only speculate as to how you intended to form the final product. If you thought to use an Sn2 mechanism, I'd say that the reaction is highly improbable as the halide is attached to a trisubstituted carbon, i.e. a carbon with a lot steric hinderance, and therefore, it would be rather difficult ...


1

I think Bard's is more general than Newman's, Newman's assumes Cx=Cx* where Bard does not. Bard says the current is proportional to the forward rate, and if there are metal ions near the electrode, the forward reaction removes electrons from the electrode to reduce the ions. That means the current flows into the electrode. Bard then says the applied ...


1

The first step (reaction with acetic anhydride) creates a mixed anhydride, this is a good electrophile with acetate as a leaving group. This is nucleophilically attacked by the oxygen of acetone (drawn as its enol in the scheme, but it is easier to visualise it as acetone) to give a cationic intermediate which is immediately attacked by the -OH of the ...


0

The second pathway looks to be correct because : 1. The carbanion formed is considerably stable 2. $\ce{S_{N^2}}$ dosen't look favourable in the first pathway(in the second step in the first reaction) The first pathway on the other hand dosen't look to be energetically favorable neither by $\ce{S_{N^1}}$ nor by $\ce{S_{N^2}}$ as the carbon is way too ...


1

Energy is needed for separating positive from negative ions in the dissolution process. This energy is taken in the surrounding water. Water is loosing energy in the dissolution process. That is why the temperature of the water decreases. There is nothing special in using $NH_4Cl$. The same phenomena happens when dissolving a salt like $NaCl$ or any other ...


2

An exothermic reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released into the surroundings, resulting in an overall negative quantity for the heat of reaction (-ΔE ). An endothermic reaction occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system ...


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