24

Bond stability is the answer. The energy required to break a $\ce{C-O}$ bond is $\pu{85.5 kcal/m}$; the energy given off when a $\ce{C=O}$ bond is formed is around $\pu{178 kcal/m}$ (reference). These bond energies suggest that a carbonyl will generally be a few $\pu{kcal/mol}$ more stable than the corresponding gem-diol $[(2 \cdot 85.5) - 178 = \pu{-7 kcal/...


20

This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer. Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions. In general this is so that the spontaneous rate can be compared to the enzyme-...


18

Your analysis is slightly incorrect, actually. If you do a $\mathrm{S_N2}$ on the chloride, hydrolyse the nitrile, and decarboxylate, you end up with the wrong oxidation state: Instead, if you actually get to the cyanohydrin, you could (in theory) simply lose $\ce{CN-}$ to form the ketone: However, a $\mathrm{S_N2}$ reaction at that centre is going to be ...


15

Let me add to iad22agp's excellent answer. Often times in introductory biochemistry classes, the difference between phosphoanhydride and phosphate ester bonds is not adequately explained. ATP (and ADP) contain phosphoanhydride bonds. In these bonds, two phosphoric acid groups are condensed into one: $\ce{R~-H2PO3 + R'~-H2PO3 -> R-HO2P-O-PO2H-R' + ...


15

There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave. $\ce{NCl3}$ The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{...


9

Uh... I think the insight you need is about chemical equilibrium. An aqueous solution is exactly the mixture formed when something dissolved in water. This is easy to understand, but here is a more difficult question: what's the component of this mixure? To answer this is not easy. In a glass of pure water, there are still a little $\ce {H+}$ (to say ...


9

Looking at the following references, it seems that most agree that a water solution of $\ce{MgSO4}$ is slightly acidic. "The pH of an aqueous magnesium sulfate solution is related to the molarity of the MgSO4. Typically, the pH is between 5.5 and 6.5 due to magnesium's affinity for hydroxide ion (OH-). As the sulfate goes into solution, hydroxide anions ...


9

It's a little more complicated than that.. When you dissolved $\ce{Na2CO3}$ in water you had a solution of ions: $\ce{Na+}$, $\ce{CO3^2-}$, $\ce{H3O+}$ and $\ce{OH-}$ (the later are present in water even when there is no salt, since it dissociates too, however they are in equal quantity in pure water so the pH is 7). Now when you mix all those ions, the ...


9

In the typical reaction of hydroxide ion with a methyl ester, the hydroxide attacks the carbonyl carbon of the ester, resulting ultimately in elimination of methoxide. Given the basic conditions, the carboxylic acid formed immediately deprotonates to a carboxylate, and the methoxide is protonated to methanol. In older texts particularly, this reaction ...


8

Both can be formed. It depends on the amount of water added: \begin{align} &&\ce{XeF6 + H2O &-> XeOF4 + 2HF}\\ &\text{or}& \ce{XeF6 + 2H2O &-> XeO2F2 + 4HF}\\ &\text{or even}& \ce{XeF6 + 3H2O &-> XeO3 + 6HF} \end{align}


7

Think of ADP and ATP as substituted phosphoric anhydrides. ATP has a strong tendency to hydrolyze due to the electrophilic nature of the central phosphorus atom which bears two electron-deficient phosphate leaving groups. (note that ADP has only one, and is a poorer source of energy, whereas AMP has only a normal phosphate ester linkage.) ATP is kinetically ...


7

Aqueous solution of magnesium sulfate is going to be slightly acidic, and there is no need to even go to the lab to test it. $\ce{MgSO4}$ is a salt formed by a weak base and a strong acid (both dibasic), so there is a two-step hydrolysis of magnesium cation. First and primary step is the formation of a basic salt: $$ \begin{align} \ce{2 MgSO4 + 2 H2O &&...


7

Pressure does not affect the crystal structure (unless it gets insanely great, so that stishovite formation kicks in; that's not going to happen in an autoclave, though). Pressure, however, does affect the temperature (indirectly), and hence solubility, and hence the rate of crystal formation. There is no lower threshold, that's why you keep getting wildly ...


7

To answer this,Think about how an ester is formed. In the formation of an ester, wherin you react an alcohol with an acid in presence of conc.$\ce{H2SO4}$ $\ce{RCOOH + R'OH -> RCOOR' + H2O}$ Now what we have found by replacing the oxygen with an isotope of oxygen is that $\ce{RCOO'H + R''OH -> RCOOR'' + H2O'}$ What this reveals is that the acid ...


7

As M. Farooq pointed in the comments, the concept of $n$-factor is applied to the reactions a given compound participates in, namely of the following types: acid-base, neutralization (in a context of Arrhenius approach, i.e. the amount of transported hydronium ions); redox (total change in oxidation state per mole of the substance); precipitation and double ...


6

We miss some data to make a good hypothesis. The concentration of the $\ce{H_2SO_4}$ solution. The temperatures. These are the main two factors that affect cellulose hydrolysis. Usually are used temperature above 200 °C and pressure above 25 MPa. For me its strange that the exothermic reaction between SA and $\ce{NaOH}$ could lead to an increase of ...


6

Both molecules can theoretically act as a Lewis acid. This would involve the approach of a Lewis base (i.e. a lone pair-carrying group), its interaction with the central atom and the formation of a four-electron-three-centre bond. These bonds with a bond order of $0.5$ can be understood by invoking the following two resonance structures: $$\ce{\overset{-}{X}...


6

The reaction will stop after first methyl rearrangement as it leads to tertiary allylic carbocation which is highly stable, so there will be no further rearrangement, as it does not lead to more stable carbocation intermediate. Thus one will get tertiary allylic alcohol.


5

In principle, it is difficult to distill salts (ionic compounds). Therefore you first make sure that you have a free base in the solution. This is guaranteed in alkaline solution. For the steam distillation the compounds should be immiscible, which is another reason for having a free base. Let’s compare boiling points and solubility for aniline and its salt: ...


5

The mechanism really isn't much more complicated than what you describe. The $\ce{H+}$ binds to the acetal linkage. Water then binds to the fructose ring, breaking its bond with the acetal linking $\ce{O}$. Finally, the water molecule on the fructose loses a proton, restoring the catalyst. As Jan mentioned in the comments, the specific acid used is ...


5

The overall potential of an electrochemical cell is equal to the reduction potential for the substance being reduced plus the oxidation potential for the substance being oxidized. In other words, $ΔE_{rxn} = E°_{red} + E°_{ox}$. Now, all that we need to do is find a list of reduction potentials. From here, I found the following reduction potentials: $$\ce{...


5

Sulfuric acid is an excellent dessicant and is also very non-volatile itself (it boils at 337 °C). In other words it would likely still have been there in yet another month. Your instructor was correct to suggest that you rinse the sulfuric acid with some cold water. Benzoic acid does have some non-negligible solubility in water (1.7 g/L at ...


5

Your issue is hardly with the Wittig reaction, although one could very well nitpick. The addition of phosphonium ylid to carbonyl is generally thought to be concerted: see Which is the currently accepted mechanism of a Wittig reaction? for more details. From what I can tell, you are having more trouble with the acidic hydrolysis of an enol ether to an ...


5

Going by the mechanism of hydrolysis under basic medium, the lone pair of hydroxide ion will attack on C=O carbon. Due to the negative mesomeric (-M/-R) effect and negative inductive effect of the nitro group, the C=O will act as a more electrophilic substrate as compared to the one which is has ether substituent, (which due to its +M and +I effect ...


5

I'd say, yes it is possible. The pioneering work on interconversion reactions of cyclobutyl, cyclopropylcarbinyl, and allylcarbinyl derivatives has been done by John D. Roberts and coworkers (Ref.1). In these works, they have checked reactions under both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ conditions. They have observed the reactions expected to involve ...


4

From the comments: I used 50 grams of cellulose, then I added 55 ml $\ce{H2SO4}$ solution (99%) and it was conducted at normal room temperature and normal air pressure, and in the end I added about 10 sodium hydroxide pellets (1 pellet being around 1cm long) So unless I am missing something, in an attempt to hydrolyze cellulose, no water was added to ...


4

So G is the expected cyanohydrim, but the second step looks very vigorous (concentrated sulfuric acid and heat). I would expect two things to happen under these conditions, 1) hydrolysis of the nitrile group to the corresponding carboxylic acid (the reaction conditions are too strong to stop at the intermediate amide, rather the amide is further converted ...


4

As per Nicolau's suggestion, : $$\{(1-2)\times\color\red{4}\}-2$$ gives $$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$ Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$ $2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3} $ molecules. Wikipedia also ...


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