NEET-UG exam Preparation on acidity of phenols and carboxylic acids, exceptions #neetchemistry

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[00:00:01]Hello students, welcome back. In this video, I'm going to solve a question based on acidic strength of the molecules. If you solve this one question, you will be revising many concepts from the organic chemistry. So, let's get started.

[00:00:16]You have to choose the correct order of acidic strength for the given set of molecules. First, let's analyze the molecules given here. They are phenols.

[00:00:26]So, there is a ortho nitrophenol, para nitrophenol. So, first of all, why phenol is acidic in nature? Because it can donate the proton. It's a proton donor. What are the factors which will influence the acidic strength of the organic molecules? There are different phenomena. There is inductive effect, there is a resonance, then there is a hybridization, and aromaticity is there, and there is ortho effect, and many more.

[00:00:54]So, you have to apply all these concepts to solve such question.

[00:01:01]The nitro group present here is exhibiting what effect? It is exhibiting -I and -R. Both the nitro groups, right?

[00:01:11]But, the positions of the nitro group is different. Will it matter here?

[00:01:16]Definitely, it matters. How so? When the nitro group is at the ortho position, it will exhibit intramolecular hydrogen bonding like this.

[00:01:29]Because of this intramolecular hydrogen bonding, the hydrogen is held in the molecule. That means what? It is not donating the proton easily because of this stable six-membered ring.

[00:01:45]Hence, what happens to its acidic strength? The acidic strength decreases.

[00:01:50]Therefore, the order given here is wrong.

[00:01:53]Let me revise this once again. Whenever there is a nitro group at the ortho position, it exhibits intramolecular hydrogen bond with the hydrogen to form a six-membered ring. Therefore, it is not donating the proton easily, hence it will be less acidic in nature.

[00:02:12]Many students, what they think is, "Okay, nitro is ortho."

[00:02:16]That means it's closer to the OH, so it withdraws more electrons. So, you are applying only minus I effect, then you are wrong. In case of para, there is no intramolecular because the groups are away, right? Therefore, this will donate the proton easily, hence this will be more acidic than ortho. Therefore, you can eliminate option A here. Where else we have A in A and D. You can eliminate these two. Now, let's check for the second set of molecule. So, which concept do we have to use here? See, benzoic acids are acidic in nature, and electron donating group decreases, electron withdrawing group increases. If we use this concept, then methyl should donate, therefore it should be less. So, you might have eliminated this. Then, you are in trouble. This is not correct.

[00:03:08]Why so? Because along with this concept, you should also know that it is exhibiting ortho effect.

[00:03:16]So, what is ortho effect in benzoic acid? See, it's applicable for acids, not phenols, okay? So, when there is a group like methyl, ethyl, or any other group, okay? It doesn't matter whether it is electron donating or electron withdrawing. So, if there is a group at the ortho position, due to steric hindrance, it will kick out this hydrogen easily as a proton. Hence, ortho isomer is always a better acid for benzoic acid. Therefore, this is correct. Where do we have B? This one.

[00:03:50]So, this is the correct option. B is the correct option. However, let's verify why para-chlorophenol is more acidic than para-fluorophenol.

[00:04:02]Many students assume that fluorine is more electron withdrawing. Therefore, it withdraws more electron. Therefore, the acidic strength should be more. Hence, para-fluorophenol must be more acidic than the para-chlorophenol. No, this is not correct. There is one more additional concept here. What is that?

[00:04:21]Fluorine 2p orbital and and the carbon 2p orbital will undergo overlapping, which is more effective.

[00:04:31]That means it is donating electron through plus R effect. The plus R effect is predominating over minus I effect.

[00:04:40]The overlapping of 2p 2p orbital is more effective as compared to the chlorine is 3p and carbon is 2p.

[00:04:51]Because the energy of the orbitals will be different. Therefore, here the overlapping of the orbital is not as effective as this. If it is not donating electron through resonance easily, that means what? This is a better acid. Let me repeat this once again. You should not consider inductive effect here because resonance predominates over inductive effect. And which is exhibiting more resonance here? Fluorine as compared to chlorine because the 2p 2p overlapping is more effective than 3p 2p overlapping. That is the reason fluorine donate more electron. If it is donating more electron, then it is a weak acid. Hence, C is also correct.

[00:05:35]The last one is for phenol. There are two nitro groups. So, both of them withdraw electrons. So, since there are two electron withdrawing group, they withdraw more and more electron making this phenol more acidic in nature.

[00:05:48]Hence, this is wrong.

[00:05:51]So, finally, what is the correct answer here? Option B.

[00:05:55]Solve the question given at the end of this video to master the concept. Thank you for watching this video. Until next time.