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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 ...

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 ...

0

Considering the intermediate or the transition state which bears a negative charge, electron donating groups will destabilize it thereby decreasing the rate of the reaction whereas electron withdrawing groups will increase the reaction rate.

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.*...

1

Yes it is there due to the increased size of nucleus. But, the difference in the electronegativity of the two isotopes would be so small so you can easily neglect it. The resonating structure will be different in C14 and C12. The reason is C14 is radioactive (very less) so it may cause difference between the chemical properties of the two resonating ...

2

As OP suggested there are three resonance structures of diazomethane contributing to the structure: If you want to neutralize the formal charges of last resonance structure, you can put a double bond on negatively charged carbon and adjacent positively charged nitrogen to get the fourt resonance structure. However, traditionally we don't do that because it ...

-1

In comparing acidic strength between molecules firstly make its conjugate base , by -H+. Remember the fact Stability of conjugate base is directly proportional to acidic strength of its corresponding molecule. Also while comparing acidic strength check factors that are uncommon to all molecules and increase stability (check electronic effect and also steric ...

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, ...

1

In (b) part negative charge is on less electronegative atom carbon so it will hold negative charge loosely in comparison to oxygen. So resonance can happen more betterly in second one. In (g) part according to me a should me more stable as in a part there is linear conjugation but in b part there is cross conjugation. Sorry for the English.

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 ...

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