New answers tagged

0

A few references into the rabbit hole and we have an answer. The major reference was given in the link mentioned by OP which was Chem. Ber. 1994, 127 (2), 409–425[1], in which the reaction mechanism of the synthesis is given as follows: [...] Reaction of 33Chem. Ber. 1994, 127 (2), 409–425 with tetrathiophene-S,S-dioxide is performed by heating the ...


1

The image shows the complete mechanism


3

I appreciate Waylander's answer. However, although the answer gives you the basic mechanisms for Reimer–Tiemann and Kolbe-Schmidt reactions, it does not address some question and OP's misconceptions such as: The $\ce{CO2}$ liberated will react with the phenol. Thus, I'd like to address these by using the same starting materials used in the question. ...


3

This is the mechanism of the Reimer-Tiemann reaction (source Wikipedia here) The polyhaloform reacts with strong base to create a carbene, a highly reactive species which reacts with the phenoxide anion through the alpha carbon to give an anionic intermediate with the -ve charge on the halogenated carbon. It is this species that picks up the deuterium. In ...


2

No. It is not a precipitation reaction. It is physical reaction. The pores are made of the rest of the membranes of the cells. When wood is overheated in air, it burns. When this is done without too much contact with the air, the cellular water gets evaporated. The content of the cell (cytoplasm, proteins, etc.) is pyrolyzed, destroyed and mostly transformed ...


-1

Here is the small scale manufacturing process we are giving to our students in their first organic chemistry laboratory, for their synthetizing and destroying picric acid. Take $1$ mL concentrated nitric acid $\ce{HNO3}$ ($65$%) in a $\pu{20 mL}$ test tube. Add with care, and in several portions, a tiny amount of phenol (maximum $0.02$ g). The reaction is ...


1

You should know that the molecularity is not relevant in your case and has no significance for multi-step reactions, it is just applicable to elementary reactions. The relevant parameter for multi-step reactions is the, reaction order wich is originally an experimental parameter indirectly related to the molecularity of all elementary reactions that makeup ...


2

$1$) Let's admit first that the reaction rate is proportional to the amount of reactants, in the usual way. Whatever the order $1$ or $2$ of the reaction, the rate constant ${k_{1} }$ or ${k_2}$ will not change if the amount of reactant is increased. But the rate of reaction ${r = - dn/dt}$ does increase when the amount of reactant ${n}$ increases, whatever ...


2

I think your mechanism is reasonable. Coming to rate of dehydration, In first case Alpha hydroxy ketone will have the least rate of dehydration because a carbocation is formed which is destabilized by Ketone group and even notice there is hydrogen bonding in reactant making reactant more stable and thus rate of dehydration is decreased in first case. In ...


1

Product of all the three compounds would be same. You have missed space tautomerism in the third one. Besides,charge beside a carbonyl is considered very unstable. Thus, first one is lowest in rate. Third one is placed second because there is an extra step of tautomerism.


4

This does not answer your question exactly, but I hope this can shed some light on the anomalous behaviour of anilinium cations. First of all, it is not true that the para-product is always favoured with the $\ce{NH3+}$ ion, it depends on the reaction conditions. There are two effects to consider here—1) kinetic/thermodynamic control, and 2) rate determining ...


1

Your reasoning is correct and the problem is with the publisher as they made a mistake while compiling these previous year questions. I found out the original 2003 question paper(pdf). The question you mentioned had the following options: It is clear form the options that the answer is C(which is the product you expected) which can also be confirmed by the ...


0

Yes, the reasons you have given for eliminating options seem correct...I think the answer should be the two Ph groups both with COOH and CH2OH (since it has to have carboxylic acid and alcohol groups in the product) so the structure will be similar to option (C) but with CH2OH in place of the OH groups.


-3

yeah so basically no reaction is purely SN1 or purely Sn2 its always a mixture of both and the overall order which we see is nothing but what that majority of reaction follows so now going by ur question, we do have to check about the substrate to inspect SN2 mechanism but in this case, since the reaction conditions are made in such a way that the overall ...


5

The Hunsdiecker reaction is a halogenation-decarboxylation reaction of silver salts of carboxylic acids. To understand what causes the yield to change, first we have to consider the mechanism of the reaction. As seen from the mechanism the homolytic cleavage of the $\ce{O-X}$ bond is the slowest step in the reaction and thus is the rate determining step. ...


6

You may notice that B is an intermediate to eventually reach D. Indeed, it is quite possible, performing the reaction in the lab, B would spontaneously yield D with only little chance to isolate this intermediate as such. $\ce{NaBH4}$ reduces the aldehyde to an alcohol. Then, mediated by the acid, the actetal is converted into an aldehyde to yield B. The ...


6

Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer ...


3

An electrophilic aromatic substitution reaction is one in which the aromatic ring reacts (as a nucleophile) with an electrophilic reagent. In all of the reactions that follow this naming convention, the adjectives "nucleophilic" and "electrophilic" describe the reagents, not the substrate. For example, the following reaction is a ...


2

Here is a mechanism for the cleavage. After proton transfer, the carbocation has a resonance contributor. Nucleophilic attack can occur on the ethyl group because there is a good leaving group. The electrophilic addition can happen, with iodide attacking the carbocation. However, it is reversible. Iodide is a good leaving group, and the carbocation is ...


-2

Yes, since nitrogen is more electronegative than carbon, and also since it has the lone pair available for taking, the proton is likelier to go to it.


7

I appreciate Waylander's answer, but it did not address the OP's curiocity of why the base abstracts the proton from only ortho-position. Yet, Waylander correctly pointed out that since the aromatic $\pi$-system is at right angle to the triple bond (the second $\pi$-bond of the triple bond), the ability to donate a lone pair to the ring by a substituent (...


6

Short answer: The $\ce{OMe}$ stabilizes the negative charge in the ortho position From Master Organic Chemistry.com here see Case#2 below So how do substituents on the ring affect addition to the triple bond? Since the aromatic pi-system is at right angles to the triple bond, what’s NOT relevant is the ability to donate a lone pair to the ring (like $\...


4

@Harry Holmes: Here is a simplified example of how the formation of Bakelite may occur under base-catalyzed conditions to incorporate formaldehyde as -CH2- at the ortho and para positions of phenol. There are a myriad of permutations as to the order of condensations. I have selected one that illustrates the Michael-like addition that seems to have eluded you....


5

The above depicts a concept, rather than an equation for that there is no balance of atoms on the left and right of the arrow. Because if $\ce{Mo2O3}$ is said to be used as catalyst rather than a (stoichiometric) reagent, then the oxygen in formaldehyde must be of different origin than $\ce{Mo2O3}$. On industrial scale, formaldehyde is prepared from ...


3

In 1928, Malaprade demonstrated that periodic acid reacted with ethylene glycol to produce iodic acid and formaldehyde (Ref.1). Hence, the oxidation of adjacent diols with periodic acid or its salt in aqueous solution is now generally known as the Malaprade reaction, the mechanism of which is depicted below: The reaction proceeds faster under acidic ...


3

The research paper sighted in the question (Ref.1) is about the extraction of boron (as boric acid: $\ce{H3BO3}$) from salt lake brine using 2-ethylhexanol/kerosene as organic phase. It is true that the given diagram in this paper (as shown in the question) is showing an esterification, but the authors refer it to two references, one of them is Ref.2. ...


9

Unlike other reducing metal hydrides (e.g., $\ce{NaBH4}$ and $\ce{LiAlH4}$), diisobutylaluminum hydride (DIBAL-H) is a liquid at room temperature and dissolve in many hydrocarbons such as toluene and hexanes, which also have very low freezing points. For example, hydrocarbons toluene and hexanes both have freezing points around $\pu{-95 ^\circ C}$. Thus, ...


5

I already made a comment about some of what I am about to say but I will provide a partial answer. I say partially because I could not find any mechanism for the second product. However, from literature, what I found was that in acidic conditions, $\ce{KMnO4}$ will oxidize naphthalene into the first product, phthalic acid1. In basic/alkaline conditions, $\ce{...


3

Your suggested mechanism should be acceptable (need two more steps, protonation and deprotonation, see last two steps of my scheme below), if you have isolated the carboxylic acid product (compound $\bf{1}$ in my scheme). If compound $\bf{1}$ is not isolable, I suggest following path would be taken to decarboxylation: Your reaction is an analog to Pictet ...


1

Note that hydrogen peroxide is slow to oxidize DPD except at high concentrations but using peroxidase cleaves at the oxygens to form hydroxyl radicals. As noted in the cited paper [1], this can be used to quantify hydrogen peroxide as it becomes activated to oxidize DPD. So the hydroxyl radical from hypochlorous acid can form water from the hydrogen on DPD ...


1

Since electron-withdrawing groups (EWGs) lower orbital energy (we can rationalise this in terms of reduced electron density leading to reduced shielding), they will stabilise multiple bonds - you see this in the Diels-Alder reaction, where EWGs on the dienophile reduce the energy of its LUMO and so accelerate the reaction by reducing the gap between the ...


3

The rate equation is $$\ln[\ce{Fe^{3+}}] = \ln[\ce{Fe^{3+}}]_0 − kt$$ So, $a-x=[A]$. Don't subtract the next value. Put it as it is. That's just for your understanding, however. Don't try to use the integrated rate form to plot the graph. $[\ce{Fe^{3+}}]=238 \text{ where } t=10 \text{ and } [\ce{Fe^{3+}}]=227 \text{ where } t=20$. Using the above mentioned ...


0

Wolff–Kishner reduction is the method for the reduction of aldehydes and ketones to corresponding alkanes. First, the carbonyl compound condenses with hydrazine to form the corresponding hydrazone. The resulting hydrazone is treated with base to induce the reduction of the $\mathrm{sp^2}$-carbon to $\mathrm{sp^3}$-$\ce{CH2}$, and the oxidation of the ...


1

This reaction is mostly ignored in basic organic texts and the mechanism you give is straight out of Wikipedia with some modification. I am sure some serious study exists some where but not on the web. Please bear with me because I can't[or just don't know how to] do drawings. The first step is probably as written and is most likely slow because it involves ...


3

[OP] The equilibrium constant would not include the solid $\ce{I2}$, but why is this? Let me explain this with a different example. If you have a saturated solution (e.g. lemonade with too much sugar in it) it is at equilibrium. If you add more sugar, the lemonade does not get sweeter. That tells you that the amount of solid does not matter (as long as ...


Top 50 recent answers are included