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6

This was going to be a comment but it got too long. TL;DR: - This is not an answer, rather a justification for why the question is (probably) wrong. This paper (linked by @Rishi) gives us experimental evidence that under action of concentrated $\ce{H2SO4}$ hydrogens are exchanged from paraffins in the following fashion: $$ \begin{align} \ce{(CH3)2CHCH3 + ...


2

It is true that normally in the Schmidt reaction of ketones, amides are formed as the major product (Ref.1). However, the products from the Schmidt reaction is highly depeends on the conditions used. The other byproducts include tetrazole and urea derivatives (Ref.2). The mechanism for tetrazole formation given by OP is acceptable but it need harsh ...


3

As napstablook pointed out in the comment section, I didn't understand where you got this erronous information about Schmidt reaction. For example, it definitely didn't give a amide with the reaction with carboxylic acid. The product is a corresponding amine with one less carbon. Basically, the Schmidt reaction is the reaction of hydrazoic acid $(\ce{HN3})$ ...


0

This reaction called Electrophilic addition to α,β-unsaturated carbonyl compounds. Electrophilic addition to α,β-unsaturated carbonyl compounds is analogous to electrophilic addition to isolated double bonds, except that the electrophile adds to the carbonyl oxygen, the most basic atom in the molecule. After that, the nucleophile adds to the β carbon, and ...


0

The enthalpy of solution is equal to the difference in the enthalpy of hydration minus the lattice energy of the material. Enthalpy of solution = Enthalpy of hydration− Lattice energy. Maybe you should look up the KI hydration energy value somewhere, like the NIST website.


8

It depends on the reaction conditions. I suspect the answer the question setter wants to see is phthalic acid, but the true answer is none of these. I have found multiple references (such as this US Patent and this Org. Syn prep) that refer to the production of naphthoquinones by oxidation of naphthalenes with high valent Chromium reagents under acid ...


2

It appears that your teacher is wrong! The synthesis of pyridine from acetylene and hydrogen cyanide over red hot iron was originally reported by William Ramsay in 1876 wikipedia. Further examples have been reported JACS paper here, patent with Rhodium and Cobalt catalysis here The second reaction is the Pechmann Pyrazole synthesis article here. The ...


4

No. Example 1 - neopentyl halide: slow SN2 as it has considerable steric hindrance; slow SN1 as it is a primary carbon, so it does not form stable carbocation intermediate. Example 2 - benzyl halide: fast SN2 being a primary carbon; fast SN1 due to a resonance-stabilized benzyl carbocation intermediate.


4

Short answer: no, it is not correct. You can have compounds that are thermodynamically/kinetically impeded for a variety of reasons and present both slow SN1 and SN2 reactivities. The exceptions would be carefully handcrafted “everything else being equal” lists.


0

Actually one can't always find answers through logical formulations to these JEE problems on organic chemistry , but we can understand and compare logical constraints. I remember this question was asked in two different ways in my examinations - one considering only $\mathrm{S_N2}$ among the reactions and the other one is same as yours. $$\ce{CH3Cl ->[OH-]...


9

This is a rather unusual and interesting case of aromaticity, which has been given a special name: homoaromaticity. The Wikipedia page does a quite nice job of explaining what's going on. As you state, protonation of cyclooctatetraene generates a $\ce{C8H9^{+}}$ cation containing an $\mathrm{sp^3}$-hybridised carbon atom between all the other $\mathrm{sp^2}$-...


3

It is entirely possible you form aniline, but it gets consumed. During the reduction nitrosobenzene would also be formed, and Delmagro et al. [1] report that in basic solution the aniline can couple with the nitrosobenzene to give azobenzene. This coupling depends on the nitrogen in aniline acting as a nucleophile, so under acidic conditions as in a tin/...


5

According to this site here citing Carey & Sundberg and Morrison & Boyd "In case of 2-methyl cyclohexanone, this planarity would cause steric clashes between the methyl group and the pyrrolidine hydrogen atoms. Due to this, formation of the less stable 6-position carbanion is preferred and hence forces reactions to proceed at this position."...


6

The Von-Richter reaction's accepted mechanism is given by- As you can see, this is not a simple case of say, removal of $\ce{NO2-}$ ion, instead it involves removal of $\ce{N2}$ which provides thermodynamic stability. the tables you use have been calculated on the basis of $\ce{pK_a}$ values which don't give the right answers in many cases due to solvent ...


3

I think you are in correct track in preparing Polylactide (PLA)-starch bioplastic, in a short. The question is actually an opinion-based but interesting one. So, even though there is high possibility that I may get down-voted, I'd like to give few of my opinions. I think you may have to tryout few steps. I agree with the crushing PLA pellets into powder ...


4

Truly collisionless unimolecular reaction are rare, as even highly unstable molecules must be triggered somehow to gain the reaction activation energy. This is frequently obtained as an energy gain by thermal collision with other molecule, or by a photon absorption. Collisions are usually much more frequent, but photons are usually much more energetic than ...


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