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 ...


7

The answer varies slightly depending on whether the migration is one-step or two-step via an intermediate. In either case, the respective transition state or intermediate looks like this: You can see that the middle structure would be stabilized most by a group that is able to stabilize the ring via resonance, for example, an alkoxy substituent. If the ...


7

I think the third option as greatest number of resonance structures can be alluded to it. The following may suffice: In the first one, the carbocation is isolated except for the presence of a single double bond in conjugation beside it. The oxygen and the double bond beside it play no role in stablizing it. In the second one the lone pair on oxygen, the ...


6

It isn't just both bromine atoms that are lost, although that might be expected in mass spec given the relative weakness of the carbon-bromine bond. You also lose a hydrogen atom. And that is key. We might suppose that a single bromide ion (or bromine atom plus electron) comes off first, forming a $\ce{C4H8Br^+}$ ion. That is the peaks at 135 and 137, ...


6

Kornblum1 demonstrated that nitrocyclohexane 3 was the major isolable product from the reaction of iodocyclohexane 1 and silver nitrite. Earlier work by Rosanow2 (1915) claimed the additional formation of tertiary nitro compound 7, a result that Kornblum did not confirm. If indeed nitrocyclopentane 7 did form, a ring contraction is required. Ring contraction ...


4

Your answer is the correct one. Acid-catalysed hydrolysis of epoxides proceeds exactly as shown. In the specific case of styrene oxide, nucleophiles attach themselves to the benzylic carbon1,2, which means that A should be the major product. I'm attaching an image from (1) which gives the products of reaction of methanol with styrene oxide: If you want to, ...


4

These would be the arguments I propose under the assumption that the RDS approximation is valid within the scope of the question: The order (B > A > C > D) was said to be correct for the conc. $\ce{H2SO4}$/ $\Delta$ case since there would be no effect of $\ce{CH-}$ acidity since the protonation of the $\ce{-OH}$ group takes place and so the ...


3

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


3

Brønsted Acid has been used in Friedel-Crafts alkylation before (for example, Ref.1). According to this reference, when toluene use as the solvent, the F-C alkylation reaction of toluene is fruitful for tertiary and secondary alkyl bromides, tosylates, and alkenes (which give both carbocations upon protonation). However, the only primary bromide, which ...


2

Your basic assumptions are correct. It can be observed as such: In Q the carbocation is stablised by resonance, inductive and hyperconjugative effects so it is quite stable and will be formed fastest. Comparing P and R: In P, a secondary carbocation is formed and there are 4+1 or 5 hyperconjugative structures possible. Moreover, the inductive effect is also ...


2

How to decide whether +M effect or -I effect will operate in this case? The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$. In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ ...


1

I believe that in the examples where a 1,2 hydride shift was mentioned, the hydrogen atom would have 'shifted' between two adjacent carbon atoms. I think the 1,2 refers to the fact that the carbon atoms involved in the shift are adjacent. A 1,3 shift would mean a the hydrogen atom shifts to the second carbon from the original carbon atom it was bonded to, if ...


1

This seems to be true only because SN1 reactions prefer polar protic solvents which stabilize both the leaving group and the carbocation. But such solvents also surround strong nucleophile and hinder it from the carbocation. Weak nucleophiles on the other hand don't interact with protic solvents as much and thus stand a better chance of getting close to ...


1

An acid catalysed dehydration reaction of alcohols will always proceed with the removal of -OH as H2O molecule and therefore, will be dependent on the formed carbocation's stability because the rate deternining step for this process is formation of carbocation. Carbocation Carbocation can be stabilized by inductive effect, field effect, mesomeric effect (or) ...


1

I would say that many books suggest that +m effect overshines the -I effect but I feel it depend upon the reaction where thee intermediates are formed and type of reaction is happening and stability of intermediate involves the thermodynamics for a general exam like situation you can mark A is more stable than B.


1

If hetro atom with lone pair is neighboring to carbocation, then that lone pair can be donated to empty orbital on carbocation (see figure below). These orbitals are said to be in conjugation. Conjugation leads to delocalization of electrons resulting in resonance structures. Quoting from "Organic Chemistry" by T.W. GRAHAM SOLOMONS , CRAIG B. FRYHLE .SCOTT ...


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