1. Arrange the following in order of decreasing reactivity toward nitration. Fully explain please I could use some help!!
A. acentanilide, b. bromobenzene, C. acetophenone, D. toluene
2. Explain why p-nitrophenol is a stronger acid than phenol itself? Would p-methoxyphenol be a stronger or weaker acid than phenol? Explain using resonance forms please!
1. To determine the order of decreasing reactivity toward nitration, we need to consider the relative electron-donating or electron-withdrawing effects of the different compounds. Nitration involves the substitution of a nitro group (-NO2) onto an aromatic ring.
a) Acetanilide: This compound contains an amino group (-NH2) which is an electron-donating group. Nitration prefers electron-deficient (electron-withdrawing) compounds. Hence, acetanilide is the least reactive towards nitration among the given choices.
b) Bromobenzene: Bromine is an electron-withdrawing group. It has a higher electron-withdrawing effect than the amino group in acetanilide. Hence, bromobenzene is more reactive towards nitration than acetanilide but less reactive than the remaining choices.
c) Acetophenone: Acetophenone has a carbonyl group (C=O), which is also an electron-withdrawing group. The carbonyl group has a stronger electron-withdrawing effect than the bromine in bromobenzene. So, acetophenone is more reactive than bromobenzene but less reactive than the remaining choice.
d) Toluene: Toluene consists of a methyl (CH3) group attached to the benzene ring. The methyl group is weakly electron-donating and has a lower electron density than the other compounds. Therefore, toluene is the most reactive toward nitration among the given choices.
Hence, the order of decreasing reactivity toward nitration is: D > C > B > A.
2. p-nitrophenol is a stronger acid than phenol due to the presence of the nitro group (-NO2) in the para position (4th carbon position) on the phenyl ring. The nitro group is an electron-withdrawing group and stabilizes the negative charge on the phenoxide ion formed in the deprotonation of p-nitrophenol. This stabilization is achieved through resonance forms.
The resonance forms of the deprotonated p-nitrophenol (phenoxide ion) show that the negative charge can be delocalized over both the oxygen atom and the nitro group. This delocalization of the negative charge leads to greater stability and increased acidity.
Now, let's consider p-methoxyphenol. Methoxy group (-OCH3) is an electron-donating group. It is weaker in electron-donating than the -OH (hydroxyl) group in phenol. Therefore, the negative charge on the phenoxide ion of p-methoxyphenol is less stabilized compared to phenol. In fact, the methoxy group can act as a donor in the resonance forms, destabilizing the phenoxide ion, and making it a weaker acid than phenol.
In summary, p-nitrophenol is a stronger acid than phenol due to the electron-withdrawing effect and resonance stabilization provided by the nitro group. On the other hand, p-methoxyphenol is a weaker acid than phenol due to the electron-donating effect of the methoxy group, which destabilizes the phenoxide ion.