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

A carbonyl group normally destabilizes carbocations but stabilizes carbanions. In this particular case, the rate of dehydration isn't guarded by carbocation stability, but by CH- acidity. In acidic conditions, OH group in alcohols is relatively easily protonated. Then, if there is a relatively acidic hydrogen in $\beta$-position it immediately dissociates, ...


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 ...


5

Resonance energy is a fictitious value, it cannot be measured, it is always only estimated. The definition in the gold book[1] clearly states this: The difference in potential energy between the actual molecular entity and the contributing structure of lowest potential energy. The resonance energy cannot be measured, but only estimated, since contributing ...


5

If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet. Well … you get an average oxidation state. This calculation arguably implicitly assumes that all nitrogen atoms be equivalent. In some cases ...


4

When the cases of carboxilic acids are considered, we must note that the conjugation of $\ce{COO-}$ group is so strong in itself, that it's conjugation with the rest of the compound is rendered relatively weak. In such cases, the inductive effect is dominant. Let us take the example of benzoic acid. The carbon to which $\ce{COOH}$ is attached shall be ...


4

Your reasoning is not quite correct. Since electron density on $\ce{N=O}$ bond is more around $\ce{O}$ than $\ce{N}$ (electronegativity of $\ce{O}$ is greater than that of $\ce{N}$), $\ce{N}$ should have a slight $\delta+$ charge, which demands electrons from adjacent $\pi$-system. As a result, the electron density on the phenyl nucleus of nitrozobenzene is ...


4

Considering that you're referring to the difference in the number of resonating structures, you're right in that aspect. 4‐Nitroaniline does have more resonating structures than piperidin‐2‐one. But in organic chemistry, you should be knowing that for a phenomenon to happen, there have to be several contributing factors. While this includes the number of ...


3

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


3

It's most certainly true that proton transfer between the two acid groups can occur and is facile (particularly in an intermolecular manner). However, that only means that over a sufficiently long period of time, the four C–O equilibrium bond lengths will all average out to be the same. It does not mean that at any one instant in time, they are all the same.*...


3

For naphtol both the acidic and the basic forms have many resonance structures along the two fused benzene cycles. However, in the case of 2-hydroxy-1,4-naphthoquinone, the basic conjugated form creates novel resonance structures that engage the double bond of the quinone cycle as seen in the written resonance structure in the question. The novel resonance ...


3

I am not sure the 3D conformation stills as reported on PubChem are all that accurate. Although crystal structures are sometimes referenced for specific molecules, some of the displays are in fact models, some of which do not make sense to me. Here are some observations regarding the three molecules you noted. 2-Nitrobenzoic acid The PubChem model shows both ...


3

A short answer is that we mostly use resonance structures to represent delocalization when the bond orders or formal charges predicted by VBT are not approximately integers, simply because non-integer bond orders and charges are more cumbersome to draw. A typical example is a carboxylate, which we could draw with 1.5 bonds between both C-O pairs and -0.5 ...


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 $\ce{e-}$ ...


2

Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect. But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation . In ...


2

No atom has to obey the octet rule. It is a rule. It has no theoretical justification. It works often. But that's all that could be stated. And a lot of compounds are known that do not follow this rule like $O_2, NO, B_2H_6$, etc. And Nitrogen does not form 5 bonds. It cannot. $4$ bonds is the maximum for atoms of the second period (Li, Be, B, C, N, O, F).


2

Some background knowledge You need to be aware of electronegativities, assignement of s and p character, relation of structure to the donating or accepting nature of the substituent, and most importantly, the overlap efficiency of the donor and acceptor positions (roughly estimated by overlap integral , although for the sake of this discussion, most of the ...


2

What you have seen is not an accurate depiction of the bonding situation according to the currently accepted theory. The correct structure of sulphate, shown below, has exactly zero double bonds. Martin performed a calculation on the closely related sulphite ion (in which there is one less oxygen leading to a lone pair on sulphur) which shows zero π-type ...


2

As according to Molecular Orbital theory , when anti- aromatic compound is formed, the electron goes to anti bonding, and is unpaired even, which reduces it's stability., Or we can say that in anti- aromatic,the plane of electron cloud twists there bonds between carbons, making it unstable, thus resonance is not possible. Antiaromaticity is a characteristic ...


2

(a) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Even though $\ce{O}$ is more electronegative than $\ce{C}$, its octet is fulfilled while $\ce{C}$'s is not (only 6 $e^-$s on positively charged $\ce{C}$). Therefore, second structure is more favored over the first one. (b) Negative charge on $\ce{O}$ versus negative charge on $\ce{C}$: Here, ...


2

You don't have to do anything for benzene because it occurs in every choice. You have to compare $\ce{SO2}$ and $\ce{SO3-}$ because they distinguish A) from B) and C) from D). You have to compare $\ce{NH3}$ and $\ce{NO3-}$ because the distinguish A) and B) from C) and D). If you use this strategy, you miss out on the chance to figure out $\ce{SO3}$, ethanol ...


2

Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy. I would argue the phenolate ion has five mesomeric structures, two with the charge assigned to oxygen (analogous to the two resonance structures of phenol), and three with the ...


2

I'm not sure if G is an activating or deactivating compound. You may want to look at the relative order of activation/deactivation of some substituents, (source: www.chem.ucalgary.ca) As per the image, $\ce{-NR2}$ is a strongly activating group, while $\ce{-CONH2}$ is a moderately deactivating group (bcoz, it lies somewhere in between $\ce{-COOH}$ and $\ce{...


2

In figure B, nitrogen is $\mathrm{sp^{2}}$ hybridized which means that the lone pair sticking out from the ring is contained in a $\mathrm p$ orbital. This $\mathrm p$ orbital is geometrically restricted from interacting with the $\mathrm p$ orbital of the adjacent carbon. The $\mathrm p$ orbital containing the lone pair on nitrogen is in a different plane ...


2

But that doesn't make sense, since the same thing is true in A! No, it isn’t; and identifying which of the two cases is true is precisely the key point to understanding when resonance can happen and when it cannot. First and foremost: we usually draw in two dimensions to represent three-dimensional molecules. Obviously, this is inadequate and often we need ...


2

Perhaps the words in the picture are explained at greater length in the text, or perhaps actual numbers are given for the dissociation constants (https://www.quora.com/Why-is-o-flurophenol-is-more-acidic-than-p-flurophenol). As it turns out, the dissociation constants are 8.7, 9.3. 9.9 and 10.0 for ortho, meta, para and unsubstituted phenol. Ortho-...


2

You might be confused but not missing anything big or fundamental. A pi-conjugated system is one in which delocalisation of pi-electrons can take place. It is all. When effective, pi-conjugation is a subset, or a type, of delocalisation than can be described by resonance. However: to be strict, conjugation refers to bonds, and specifically to an alternating ...


1

The other answer by permeakra is sufficient and complete. But, I'd like to add an explanation on how "A carbonyl group normally destabilizes carbocations but stabilizes carbanions" For instance, someone can think that the carbocation would resonate with the carbonyl group as what we've learnt earlier, I'd also thought in that way, initially. But, ...


1

The first acid dissociation constant for genistein ($\mathrm{p}K_\mathrm{a1}$) is $6.51 \pm 0.20$ (Ref.1), mainly because of hydrogen bonding between $\ce{C_{(4)}}$ carbonyl oxygen and $\ce{C_{(5)}-OH}$, similar to that in acetylacetone (see the diagram): Therefore, in physiological $\mathrm{pH}$ ($\mathrm{pH} \approx 7.3$) it should be partially ionized. ...


1

Chloro group is considered a deactivating group according to Hammett plots of each possible reaction. The $\sigma_\mathrm{para}$ of $\ce{Cl}$-substituent is listed as $+0.227$ while $\sigma_\mathrm{meta}$ of $\ce{Cl}$ is listed as $+0.373$. It is known fact that $\sigma_\mathrm{meta}$ is an indicative of how much inductive effect contribute to the reaction, ...


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