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I have went through each option: A) Freon is $\ce{CCl2F2},$ which is derived from methane B) Teflon is a polymer of $\ce{-F2C-CF2 -}$ (or) $\ce{(C2F4)_n},$ derivative of 1,2-difloroethene. Here, carbon is saturated, which means no double or triple bond is present. C) Iodoform is $\ce{CHI3},$ which again is a derivative of methane. D) Vinyl chloride is ...

10

As Oscar Lanzi suggested, both $+M$ and $-I$ applies here, but $\ce{Cl}$ stabilizes carbocation, meaning $+M$ is more effective than $-I$. This fact was confirmed by this peer-reviewed paper (Ref.1): The lowering of $\ce{C_\beta–H}$ stretching frequencies in carbocations 1a–d and 2a–c induced by hyperconjugation was tested as a possible probe for ...

9

There is indeed very little to be found in the literature. A recent Russian paper (Ref.1) mentions benzenediazonium bromide as a reactant. This thesis from 1975 (Ref.2) contains a preparation of benzenediazonium bromide and an attempted preparation of benzenediazonium iodide; it notes that the iodide is unstable as does this reference here. The discussion ...

8

TL;DR - The ring with positions 1, 2, and 6 on it is more deactivated than the ring with positions 3, 4, and 5, so the correct answer cannot be position 2 (as marked) or positions 1 and 6 (as you determined). The correct answer must be on the other ring. Since the central ring imposes a -I effect, the correct answer is position 4. There are two factors to ...

8

This is a question based on two reactions. The first reaction is the formation of benzyne using $\ce{Li/Hg}$. After this a Diels–Alder reaction takes place between furan and benzyne. The reaction mechanism would be as follows [1]: A more generalised form of this reaction, dealing with the conditions needed and the yield of reaction would be [2]: Reference ...

8

Nothing against Safdar, but this is technically not a two-step reaction. When lithium amalgam is added to 2-bromofluorobenzene in presence of furan, the reaction goes directly to the product, a Deals-Alder adduct: However, in the absence of furan, the reaction proceeds to give biphenylene and triphenylene (Ref.1&2). The tentative reaction mechanism can ...

7

Steric hindrance does indeed interfere with the stability of carbon tetraiodide, as does poor covalent bonding overlap between the very differently sized carbon and iodine atoms. Even so, carbon tetraiodide is definitely established and in use as an iodinating agent. Storage below normal room temperature is recommended. Carbon tetraiodide is also very red....

7

Reaction-1 is a simple example of an $\mathrm{S_N2}$ reaction, where the acetate group ($\ce{AcO-}$) substitutes the tosyl group ($\ce{OTs}$) forming the product (1) as follows: But for Reaction-2, the lone pairs present on the $\ce{NMe2}$ provide anchimeric assistance in the removal of the tosyl group, leading to the formation of a bicyclic intermediate (...

7

This is an another way to answer the question: On the way, I also want to show OP that why central ring bearing $\ce{CF3}$ group has the highest electron deficiency at ortho- and para-positions (meaning, the highest electron density is at meta-position in this ring). Let's look at the nitration of trifluoromethylbenzene: According to this University of ...

6

The density of a substance depends on its molecular density and packing fraction (where you need to be consistent in definiting the molar volume in calculating both attributes). However you are looking for a simpler explanation that gets the general trends right. Just looking at molecular mass is too simple—larger molecules have greater mass, but take up ...

5

Mostly it's a matter of economy. In most diazonium salt synthesis the counterion is not involved in the organic reactions; it's just a spectator. So we choose chloride because hydrochloric acid is a cheap and effective chemical to acidify the aromatic amine+nitrite mixture and make the diazonium ion in the first place. Ergo an aryldiazonium chloride. The ...

5

This reaction is a special case of Darzen's halogenation. I think you are aware of mechanism of Darzen's reaction or else you can check it here. It occurs through a 6 membererd cyclic intermediate. I'll show you the first reaction you would be able to answer second one in succession.

5

You need to reckon with how strong a base the nucleophile is compared with what you are trying to displace. Only if the nucleophile is the stronger base can you expect the reaction to go based on thermodynamics. Thus, both of the following reactions can go: $\ce{RCl + OCH3^- -> ROCH3 +Cl^-}$ $\ce{RCl + CN^- -> RCN +Cl^-}$ because both ...

4

The reaction proceeds by SN2, which is highly accelerated at the benzylic position, due to stabilisation of the transition state by conjugation with the aromatic ring. So you'll have to replace the benzylic bromine. References: Benzyl chloride: chemtube3d

4

In this question, we need to find the most electron-rich carbon out of positions $1,2,3,4,5,6$ in the compound given below: We proceed using the process of elimination. Note: Not sure how scientific this is, however an attempt I made in finding the effect of the inductive effect was to run a DFT B3LYP/3-21 G on trifluoromethylbenzene. The result showed the ...

4

Before going into details, let me state Wikipedia: $\ce{^{208}PoF6}$ was probably successfully synthesised via the same reaction (passing fluorine gas over $\ce{^{208}Po}$) in 1960 where a volatile polonium fluoride was produced, but it was not fully characterized before it underwent radiolysis and decomposed to polonium tetrafluoride. Even attempts were ...

4

Good answer by @Waylander and @OscarLanzi. They provided some literature evidence of the synthesis procedure of Benzenediazonium bromide and iodide and discussed its usage and worth and explained why they are not that important as compared to benzenediazonium chloride. My answer just revolves on the actual synthesis procedure(slightly abridged in my answer) ...

4

Density is a macroscopic measurement so we should look for average properties spread over several molecules. The average density of a liquid is not that much smaller than that of its solid ($\approx 0.9$, there are a v. few exceptions, water/ice, silicon, gallium, germanium, bismuth where it is greater) which means that the distance between molecules is only ...

3

You get both, but the +M effect wins. See the discussion of the effect of atoms with lone pairs over here.

3

Simple answer is consider the stability of carbocations. You already got the tertiary carbocation in hand that is much stable than secondary carbocation, which would be resulted by ring expantion. Sure, a methide migration would give you another tertiary carbocation, but before you get that, the reaction progress has to climb another actvation barrier, which ...

3

Cleavage of epoxides can happen in 2 ways, resulting from the cleavage with the 2 carbons to which the oxygen atom is bonded. If the medium is acidic, it will follow the SN1 pathway, proceeding via the most stable carbocation. This makes sense, since in acidic medium, the alcohol can get protonated and break away from one of the carbon atoms. If the medium ...

3

The answer is (b); majority elimination by an E2 mechanism. Quoting from chem.libretext.org: Acetylide anions are strong bases and strong nucleophiles[...] Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.

3

The mechanism involving the chlorination of $2^\circ$-alcohols has been studied tremendously. Based on these studies, it was revealed that if nucleophilic solvents such as 1,4-dioxane or ether has been used in the reaction the stereochemistry of the original alcohol would be retained in the resultant chloroalkane product (retention). However, if a base such ...

3

Although atomic bond in a compound such as $\ce{M+X-}$ (e.g., $\ce{Na+Cl-}$) is considered to be 100% ionic, in reality, it also has some covalent character. An explanation for the partial covalent character of an ionic bond has been given by Kazimierz Fajans in 1923 (Ref.1). According to Fajans, if two oppositely charged ions (say $\ce{M+}$ and $\ce{X-}$) ...

3

The strength of Lewis acids relative to your Lewis base has been commented on. A fuller answer to your question needs to include the Lewis acid you are considering. The original Bronsted-Lowry theory of comparing acidity and basicity, or acids and bases, involved $\ce{H^+}$ and $\ce{OH^-}$ and the transfer of a proton. The broader Lewis theory allowed ...

2

The solvent used is a polar protic solvent which heavily solvates the OH- radical thereby decreasing it's nucleophilicity as the solvent forms a cage around the radical anion .Even though the basic strength of the anion is also decreased by solvation still the decrease in nucleophilicity is dominant as it is more difficult for a large solvated anion to ...

2

Protic solvents affect nucleophilicity because they form a shell around the nucleophile via their hydrogen which has a positive partial charge and therefore interacts with it, thus the nucleophilicity will decrease. This effect is stronger for smaller nucleophiles as they can be "trapped" more easily. So taking the classical nucleophiles you obtain this ...

2

There are three mechanisms to discuss here: $S_N2$, $S_N1$ for $RX$ and nucleophilic aromatic substitution by addition-elimination for $ArX$: $S_N2$ and $S_N1$: (Image from https://www.quora.com/Why-steric-hindrance-doesnt-affect-Sn1-reaction) $S_N2$ is concerted (one step), so this one step is rate determining, or the slowest of the mechanism. In $S_N1$, ...

2

Be careful! Ethanol is not a strong acid ($\mathrm{p}K_\mathrm{a} \approx 16$). So its conversion to ethoxide anion is not so easy. Rather, it would attack the $\ce{C}$ center by its lone pair and then it would release proton to form the product. Actually here what I think is, the hetero atom $\ce{O}$ just near to the $\ce{C}$ undergoing substitution will ...

2

If we consider $\ce{HX}$ as the common formula for hydrogen halogenides, then there is an equilibrium $$\ce{HX(aq) + H2O <=> X-(aq) + H3O+(aq)}$$ that can be expressed also as: $$\ce{acid1 + base2 <=> base1 + acid2}$$ The equilibrium is competition of 2 acids, $\ce{HX}$ ansd $\ce{H3O+}$. $\ce{H3O+}$ is the strongest acid stable in aquaeous ...

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