New answers tagged

0

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.


-1

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 ...


4

It looks like you are describing continuous wave (CW) and not pulsed NMR. The latter is generally considered a more modern way of doing NMR as it was developed after the CW technique. The pulsed technique was also accompanied by the development of FT processing which resulted in a Nobel price being awarded to Richard Ernst. In CW NMR you usually "sweep&...


3

Normally the external magnetic field is held constant and an r.f. pulse is applied to generate the nmr signal. The electrons in the molecule allow the external field and the magnetic dipole due to nuclear spin to interact with one another. This interaction causes a local magnetic field that can add to or oppose the external field and so a particular nucleus ...


3

The glycoside amygdalin is a product of enzymatic biosynthesis. Its hydrolysis produces 2 molecules of glucose, benzaldehyde and hydrogen cyanide. Mixing potassium cyanide and sugar ( = sucrose = combined glucose + fructose) does not produce amygdalin. Neither you would get it, mixing glucose, benzaldehyde and hydrogen cyanide.


4

With the $\ce{C=C}$ double bond of the stilbenes as geometric reference, you may think of the phenyl rings as substitutents increasing the electron density of the former. This creates a small, locally constraint, dipole moment. Dipoles like dipoles differ from a scalar (like e.g., temperature) that they have a direction, which is why I drew little arrows ...


1

Reacting with what electrophile? In general the silyl ketene acetal will be more reactive, after all the Mukaiyama reaction can be run in Toluene Mukaiyama Reaction wikipedia page


0

For calculating the positional isomers you can only change the functional group's position and the carbon skeleton has to be kept the same. So for this molecule the following 6 position isomers are possible: Reference: (1) Positional Isomers https://chem.libretexts.org/@go/page/40883 (accessed May 6, 2021).


0

In this case, when the carboxyl group is deprotoneted and we draw the resonating structures the resonating structures are equivalent. But in case of phenolic deprotonation the resonating structures are not equivalent. So the carboxyl anion will be more stable, although it have less number of resonating structures than phenolic anion. So the carboxyl H will ...


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 ...


1

My interpretation of the first spectrum: I would say that the 3517 band belongs to OH group. This is a typical shape OH group. Amines look more like two bands with a shoulder. However secondary amine is also a possibility. Other bands of OH should be the broad band between 1400 and 1300 from which the sharp bands arise. Than probably 1166, which seems to be ...


3

No, aryl halides are not a type of vinyl halides. A compound is called an aryl halide when it contains a halogen group directly attached to an aromatic system. For example, Whereas a compound is called a vinyl halide when it contains a halogen group directly attached to a doubly bonded carbon atom. For example, The double bonds in an aromatic system are ...


3

To understand the major factor we have to consider the mechanism of the reaction first. The reaction you mentioned is popularly know as Riley oxidation. The mechanism is as follows: As you can observe the RDS of the reaction is the tautomerization of the keto form into an enol. Thus the extent of enolization determines the major product in this reaction. As ...


1

This is known as the Riley Oxidation. The Wikipedia entry states that ketones with two alpha methylene positions react more quickly at the least hindered site which references Riley's original paper1. Reference Selenium dioxide, a new oxidising agent. Part I. Its reaction with aldehydes and ketones by Riley et.al. J. Chem. Soc., 1932, 1875-1883, DOI: https:/...


4

@user1055 answered the question pretty much. I am just going to add a complementary answer which has the abstract to the paper which @user1055 mentioned. The abstract of the paper mentions the various products of the decomposition reaction: The thermal decomposition behavior of terephthalic acid (TA) was investigated by thermogravimetry/differential thermal ...


7

Following are the Statement from “On Terephthalic Acid and its derivatives” By Warren De la Rue and Hugo Muller February 7, 1861: On heating, terephthalic acid sublimes without previously fusing(melting). The sublimate, which is indistinctly crystalline, has the same composition and properties as the original acid, and therefore, unlike other bibasic acids, ...


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 ...


2

This question is, at best, flawed. The most acidic protons on the structure provided are not any of the labeled OH protons. Rather, the most acidic position is the one I have drawn a rectangle around below. Deprotonation at this position produces an anion with a resonance contributor in which both rings are aromatic. The circa 150 kJ/mol resonance ...


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 ...


4

This is an exercise in applying the Hammond postulate (multiple times). To accurately compare reaction rates, we need to have the activation energy. We don't have this, and the actual energy of the transition state is unknowable without experimental data. Hammond's postulate states that the structure of the transition state is more similar to the ...


7

In the case of option c, the $\ce{-OCOCH_3}$ group will not be showing $\mathrm{-M}$ effect as the ring is directly attached to the oxygen atom. The group rather shows $\mathrm{+M}$ effect because of the two lone pairs present on the oxygen atom, increasing the electron density in the benzene ring and thus making it a good substrate for a nitration reaction. ...


0

You can't really look at the double bonds in the benzene right using that thought process. In fact, if you draw another resonance form for the benzene ring, you just have both functional groups on one double bond. The reason why the positions are labelled 1 and 2 for the salicylic acid is because the carboxyl group has higher priority than the $\ce{OH}$ ...


3

Compounds like this are named based on the nitrile from which they are derived. So, in your case, the nitrile is propanenitrile: $$\ce{CH3-CH2-C#N}$$ A protonated nitrile would be called a nitrilium cation, following the general form: amine --> ammonium imine --> iminium nitrile --> nitrilium Thus, the following cation is named propanenitrilium: $$...


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 ...


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 $\...


1

Cloro substituent in an aryl ring has two effects combined: I- effect, owing to its electronegative character and tendency to pull electrons towards itself from the ring and thus deactivating the ring. M+ effect to donate electron to the ring by resonance. Generally the I- effect is stronger than the M+ effect, leading to an overall decrease in ...


-1

The regioselectivity of the overall reaction depends on the temperature. The carbocation intermediate can pass through the two different transition states that lead to the 1,2- and 1,4- products, respectively. If we choose a temperature low enough, then the product distribution will reflect the difference in energy between the two activation energies Ea (1,...


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....


2

I'm not sure if I correctly understand your intent. My assumptions are you have a set of molecules, e.g., as a list of SMILES strings (say FC(c1ccccc1)(F)F, CCCC(F)(F)F, CC(Cl)(Cl)Cl, and CCCO) in a file like probe.smi you want to identify the molecules without the $\ce{-CF3}$ group If so, I would i) search for the molecules with said group in a first ...


1

For alkyl groups attached to the phenol ring, there are two main points that your reasoning does not take account of - Inductive effect is distance dependent. As the distance increases, its effect becomes very small. Even if the group were near, the general trend is that hyperconjugation has stronger effect than inductive effect. You can refer the ...


3

The reason provided by ML cannot explain the difference in acidities since the +I effect doesn't come into play normally for the para positions. A more likely reason is hyperconjugation (or no bond resonance) where the number of hydrogens directly attached to the benzyl carbon decreases as it goes from P to S. This means that the number of destabilizing ...


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 ...


1

After discussion with OP in chat, the issue in logic can be rectified as follows. Simply put: $\ce{A<=>B<=>[slow]C <=> D}$ doesn't imply that $\ce{B <=> C}$ is the RDS in both $\ce{A -> D}$ and $\ce{D -> A}$. Rather than this, we need to consider $\ce{A->D}$ and $\ce{D-> A}$ as two separate reactions. Doing so for the ...


1

Taken from my answer here: Comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one ...


2

Reaction mechanism of aromatic nucleophilic substitution is as given - Source : https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/ The rate determining step is the formation of the C-Nu bond, which causes the development of a negative charge on the carbon atom at ortho position to leaving group (fluorine). Presence of ...


-2

Rules exist in order to make our life easier, but not to make it more complicated or miserable. For example, there are heterocyclic and intermediate organic compounds where different alphabetical order will make your life harder like a lot. In short: it would be very hard otherwise.


5

Recall that Ce(IV) is a pretty strong oxidizing agent. So oxidation of alcohols and fading of the color is expected. The mystery lies in the immediate intense red color. It has been relatively recently studied in great detail and well known long time ago. See these references: V. Briois, D. Lützenkirchen-Hecht, F. Villain, E. Fonda, S. Belin, B. Griesebock, ...


-2

Why be qualitative when with just a little more effort you can be quantitative? Acetylsalicylic acid has a pK$_a$ of 2.97, i.e., it is fairly acidic (Ref 1). Salicylic acid, one of the hydrolysis products, is comparably acidic (pK$_a$ = 2.79 Ref 2) and its sodium salt is fairly neutral (pH = 6.0 - 8.5 for a commercial material Ref 3; calculations suggest ~7....


0

First I was thinking that maybe a chemical indicator could be used, however the pKa values of all three components are relatively close to each other (especially aspyrin and salicylic acid). It might still be a bit of a help to stick a pH electrode in the system and observe the change - however, this is not exactly something you can see easily. Another thing ...


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, ...


-4

You are correct. Na/NH3 is a standard method for reducing internal alkynes to trans (E) alkenes. More reading here and here


1

First up, no '1-ol' doesn't mean there is one OH, the 1 signifies the position of the OH group in your parent chain (here ,hexane) which in this case is the C1 carbon. How many OH groups are present is represented by adding di/tri... as a prefix to '-Ol'. You need to learn how to identify and number a parent chain in organic compounds first in order to make ...


1

What you have witnessed is gelation coupled to phase separation. It depends on the composition of your pudding. The process of preparing a starch gel begins with starch granules swelling and eventually dissolving with uptake of water. This initial step requires disrupting the structure of both amylopectin (more amorphous) and amylose (more crystalline) ...


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{...


4

According to this patent here, N-ethyl 1,8-naphtholactam nitrates in the 4 position i.e. structure a. It seems entirely likely the N-unsubstituted material will do likewise.


1

Wow. This is a very bad question (the exam/assignment, not the OP's). There is no way that you can categorically determine the structure of this based only on the information given. This would have to be open-book providing access to chemical shift tables in the very least. Peaks are poorly labelled and what's more, I believe the molecular mass is incorrect. ...


2

My thoughts are along the lines of @Andrew and @Waylander. Step F doesn't tell us much, @orthocresol! I prefer a cyclization, 1 $\rightarrow$ 2a, followed by hydrolysis of ester 2a to alcohol 2b. Alternatively, N-O bond homolysis with light would probably involve a cage mechanism. .


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