9

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 ...


5

This answer is years later than the original question, but I add it for anyone who come across it now or in the future... All acid anhydride and acid chlorides are thermodynamically unstable in water. They will eventually hydrolyze, but that time could be seconds, minutes, hours, days, or years. Moreover, in the absence of a catalyst, the rate of hydrolysis ...


3

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


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 ...


2

What's meant by stability is whether the compound "wants" to be in current state. If it wants to change the state (separated ions want to bond) then the system is less stable. In your example you can simply imagine 2 spherical magnets placed close to each other - they would rush towards one another. They had potential energy that turned into ...


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.


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