These questions are somewhat difficult. I believe I am on the right track, but I want to make sure.
1. List these compounds in order of decreasing reactivity towards nitration. Explain your answer.
a) acetanilide
b) bromobenzene
c) toluene
d) acetophenone
2. a) Which is the stronger acid (phenol or methoxyphenol) Explain why (include resonance forms in explanation)?
b) Which is the stronger acid (phenol or p-nitrophenol)? Explain why?
1. To determine the order of decreasing reactivity towards nitration, we need to consider the electron-donating or electron-withdrawing groups present in each compound. Nitration is an electrophilic aromatic substitution reaction, where an electrophile (NO2+) attacks the benzene ring.
a) Acetanilide has an amino group (-NH2) as an electron-donating group. It activates the benzene ring, making it more reactive towards electrophilic substitution. Hence, it is the most reactive compound towards nitration.
b) Bromobenzene does not have any electron-donating or electron-withdrawing groups directly attached to the benzene ring. It is only a weak activator, so it is less reactive than acetanilide.
c) Toluene has a methyl group (-CH3) as an electron-donating group. Although it activates the benzene ring, it is not as effective as the amino group in acetanilide. Hence, toluene is less reactive than acetanilide and bromobenzene.
d) Acetophenone has a ketone group (C=O) attached to the benzene ring. This group is a weak deactivator because it withdraws electron density from the ring through resonance. Therefore, acetophenone is the least reactive towards nitration compared to the other compounds.
So, the overall order of decreasing reactivity towards nitration is: acetanilide > bromobenzene > toluene > acetophenone.
2. a) Methoxyphenol (also known as anisole) is the stronger acid compared to phenol. This is because the methoxy group (-OCH3) in methoxyphenol is an electron-donating group that increases the electron density on the oxygen atom. This electron density is delocalized via resonance, allowing for stabilization of the negative charge on the oxygen atom. This stabilization makes it easier for methoxyphenol to donate a proton, making it a stronger acid than phenol.
b) p-nitrophenol is the stronger acid compared to phenol. The presence of the nitro group (-NO2) in p-nitrophenol is an electron-withdrawing group that withdraws electron density from the phenolic oxygen. This reduces the ability of the oxygen atom to stabilize the negative charge through resonance. As a result, p-nitrophenol is less able to delocalize the negative charge and therefore more likely to donate a proton, making it a stronger acid than phenol.
1. To determine the order of decreasing reactivity towards nitration for these compounds, we need to consider their electron density and stability of the intermediate formed during nitration.
a) Acetanilide: It contains an amino group (-NH2) which is electron-donating, increasing the electron density on the benzene ring. Furthermore, the acetyl group (-COCH3) is electron-withdrawing due to the presence of the carbonyl group. Overall, the electron-density on the benzene ring is slightly increased due to the amino group, making it the most reactive compound towards nitration.
b) Bromobenzene: It does not have any substituents that can donate or withdraw electrons, resulting in a benzene ring with moderate electron density. This moderate electron density makes it less reactive compared to acetanilide.
c) Toluene: It contains a methyl group (-CH3) which is electron-donating. The presence of the methyl group increases the electron density on the benzene ring, making toluene more reactive towards nitration than bromobenzene.
d) Acetophenone: It contains a carbonyl group (-COCH3) which is electron-withdrawing. The presence of the carbonyl group decreases the electron density on the benzene ring, making acetophenone least reactive towards nitration among the given compounds.
Therefore, the order of decreasing reactivity towards nitration is: acetanilide > toluene > bromobenzene > acetophenone.
2. a) To determine the stronger acid between phenol (OH group) and methoxyphenol (OCH3 group attached to benzene ring), we need to compare their stability after losing a proton (H+).
Phenol: In the resonance forms of phenol, the negative charge can be stabilized by delocalizing it onto the benzene ring, forming a stable aromatic system. In one resonance form, the negative charge is on oxygen, and in the other, it is delocalized onto the benzene ring. This resonance stabilization makes phenol a stronger acid than the methoxyphenol.
Methoxyphenol: Although methoxyphenol also possesses resonance forms, the negative charge cannot be delocalized onto the benzene ring as effectively as in phenol. The methoxy group (-OCH3) is electron-donating, increasing the electron density on the benzene ring, which hinders the delocalization of the negative charge. Thus, methoxyphenol is a weaker acid compared to phenol.
b) To determine the stronger acid between phenol (OH group) and p-nitrophenol (NO2 group attached to the para position of the benzene ring), we need to consider the electron-withdrawing effect of the nitro group (-NO2).
Phenol: As mentioned earlier, phenol is stabilized by resonance where the negative charge can be delocalized onto the benzene ring. This resonance stabilization makes phenol a weak acid.
p-Nitrophenol: The nitro group (-NO2) is strongly electron-withdrawing due to the presence of the electronegative nitrogen and oxygen atoms. The withdrawal of electron density from the benzene ring enhances the acidity of p-nitrophenol. Furthermore, resonance forms involving the negative charge on the oxygen atom are stabilized by the electronegative nitro group. Therefore, p-nitrophenol is a stronger acid than phenol.
In summary, the order of acid strength is: p-nitrophenol > phenol > methoxyphenol.