# Most favorable position of attack of an electrophile on 4-trifluoromethyl-1,2-diphenylbenzene

### Problem

This problem is from a FIITJEE AITS paper

Which position is most favorable for the attack of an electrophile?

Position 2

### My attempt

The more favorable position for attack should be the on the para ring (considering ring with $$\ce{-CF3}$$ group as the reference ring), because of the lower inductive effect as compared to meta position.

If the electrophile attacks at position $$1$$ or $$6$$ (which means the carbocation will be formed at the carbon connected via a double bond), the carbocation would be equally stabilized by resonance, also the charge can move to the other 2 rings forming 9 Resonating structures.

If carbocation is formed at position $$6$$ by the electrophilic attack on 2nd position, the carbocation won't be able to move out to other rings for resonance implying that there would be only 3 resonating structures.

Position 1 or 6

Why is my answer is wrong? What is the correct explanation?

• btw which year's paper? – B.Anshuman Jul 29 '20 at 11:25
• 2018 year AITS ppr – the.eleventh.letter Jul 29 '20 at 15:54

TL;DR - The ring with positions 1, 2, and 6 on it is more deactivated than the ring with positions 3, 4, and 5, so the correct answer cannot be position 2 (as marked) or positions 1 and 6 (as you determined). The correct answer must be on the other ring. Since the central ring imposes a -I effect, the correct answer is position 4.

There are two factors to consider in this question. The first is that the trifluoromethyl group ($$\ce{CF3}$$) is electron-withdrawing by induction. Fluorine is the most electronegative element, and the $$\ce{C-F}$$ bond is polarized toward $$\ce{F}$$. The $$\ce{CF3}$$ group has a -I effect on the ring, and that effect is strongest at ortho and para.

The second is that the three rings in this structure are not coplanar. The steric interactions between hydrogen atoms on the two rings with labeled positions in your image causes the rings to twist out of coplanarity. This feature of the structure means that the central ring cannot exert a resonance effect on either of the other rings. The central ring will is electron poor thanks to the $$\ce{CF3}$$ group, and so it will provide a -I effect to the other two rings. Since the -I effect on the central ring is stronger at the position para to the $$\ce{CF3}$$ group than at the position meta, the ring with positions 1, 2, and 6 will be more deactivated than the ring with positions 3, 4, and 5.

Since the lower ring is experiencing a -I effect, electrophilic attack at meta is most likely, so position 4 is the true correct answer.

• One query, how did you get the fact that the -I effect is stronger at ortho and para when it is distance dependant? – Safdar Jul 27 '20 at 18:51
• -H effect would be more true here. – B.Anshuman Jul 29 '20 at 11:23
• @Safdar Try to form the arenium ion by attack of an electrophile on tert-Butylbenzene. You will see that when the attack is on ortho and para position, the +ve charge will appear on the carbon where the tert-butyl is present(through resonance). – Robin Singh Jul 31 '20 at 12:22

This is an another way to answer the question: On the way, I also want to show OP that why central ring bearing $$\ce{CF3}$$ group has the highest electron deficiency at ortho- and para-positions (meaning, the highest electron density is at meta-position in this ring). Let's look at the nitration of trifluoromethylbenzene:

According to this University of Columbia website, the highest percentage of nitration is at meta-position. If the rate of nitration of benzene is considered to be $$1$$, that for trifluoromethylbenzene is $$6.7 \times 10^{-5}$$ for meta-substitution while that for ortho- and para-positions are about $$4.5 \times 10^{-6}$$. That means, $$\ce{CF3}$$ group is a strong deactivator (c.f., rate of nitration of toluene is 58, 42, and 2.5 for para-, ortho-, and meta-substitution, respectively compared to benzene being 1).

The reason for preference for meta-substitution of trifluoromethylbenzene is depicted in the scheme. At least one of resonance contributors of the relevant Wheland intermediate during each of ortho- and para-substitution is destabilized by the neighboring $$\ce{CF3}$$ group ($$\ce{F3C-C^+}$$ destabilization). Meantime, none of resonance contributors of Wheland intermediate corresponding to meta-substitution do not deactivated and enjoys lower activation energy. All of these directed to the conclusion that the highest electron density is at meta-position when the aromatic nucleus is deactivated by +I effect (Ref.1-4). However, keep in mind that this highest electron density at meta-position is still way lower than the electron density of benzene.

Based on these facts, the substitution happens in the aromatic ring with 3, 4, and 5 positions (Keep in mind that there are no resonance contribution what so ever due to aromatic nuclei being non co-planner). When you consider the aromatic ring with 3, 4, and 5 positions, it has a substitution next to 3-position, which is an electron deficient biphenyl with $$\ce{CF3}$$ group on one of the phenyl ring. This substituted biphenyl is, therefore, meta-directing (a similar way as described for the substitution of trifluoromethylbenzene in the scheme). Thus, up coming substitution on the sought ring would be at meta-position, which is marked as position 4.

References:

1. C. Sandorfy, "Note on the Inductive and Mesomeric effects in Substituted Benzene Derivatives," Can. J. Chem. 1958, 36(12), 1739-1741 (https://doi.org/10.1139/v58-251).
2. F. A. Matsen, "Molecular Orbital Theory and Spectra of Monosubstituted Benzenes. I. The Resonance Effect," J. Am. Chem. Soc. 1950, 72(11), 5243–5248 (https://doi.org/10.1021/ja01167a127).
3. W. W. Robertson, F. A. Matsen, "Molecular Orbital Theory and the Near Ultraviolet Absorption Spectrum of Monosubstituted Benzenes. IV. The Phenyl Halides and the Inductive Effect," J. Am. Chem. Soc. 1950, 72(11), 5252–5256 (https://doi.org/10.1021/ja01167a130).
4. Christopher K. Ingold, "Principles of an Electronic Theory of Organic Reactions," Chem. Rev. 1934, 15(2), 225–274 (https://doi.org/10.1021/cr60051a003).
• So is the method used by the OP correct? – Safdar Jul 29 '20 at 5:26
• @ Safdar: No, OP chose para-position of central nucleus as the position with the highest electron density (relatively speaking). That is not correct at the first place. – Mathew Mahindaratne Jul 29 '20 at 7:26
• I mean the part where he attacked the ring and then checked for stability. I was asking if that was correct? – Safdar Jul 29 '20 at 7:32
• @ Safdar: OP's method is not correct because of thinking of mesomeric effect. The rings are not co-planner so no resonance contribution what so ever. The resonance contributor with a positive charge on ipso-carbon in both cases (substitution on 1 and 6 positions) will be destabilized by +I effect of central ring. – Mathew Mahindaratne Jul 29 '20 at 8:02
• Great answer Mathew! Also, good find on the UoC lecture slides: they explain what is happening perfectly. +1 – Aniruddha Deb Aug 11 '20 at 16:26

In this question, we need to find the most electron-rich carbon out of positions $$1,2,3,4,5,6$$ in the compound given below:

We proceed using the process of elimination.

Note: Not sure how scientific this is, however an attempt I made in finding the effect of the inductive effect was to run a DFT B3LYP/3-21 G on trifluoromethylbenzene. The result showed the m-carbon to have a charge of $$-0.173$$ and the p-carbon had a charge of $$-0.172$$. The original value in the benzene molecule is $$-0.180$$ and so inductive effect is present due to $$\ce{-CF3}$$. This meant that the m-carbon was more electron rich compared to the p-carbon. I do not have a reason for this however.

### Elimination $$1$$

We reject positions $$1,2,6$$ since as stated above the meta-carbon is more electron rich than the para-carbon.

### Elimination $$2$$

Now, we are left with the ring containing positions $$3,4,5$$. In order to proceed further, we assume a $$\delta +$$ charge on the carbon para to the $$\ce{-CF3}$$ group. This would mean that the phenyl ring resonates with the $$\delta +$$ to give five different possible resonance structures. In order to simplify the resonance structures, I am considering only this ring.

Due to the two eliminations made, we can say that position $$4$$ would be the most electron rich carbon and so the most likely to be attacked by an electrophile.

• How did you assume a +ve charge para to CF3? What's the effect making up the charge? – the.eleventh.letter Jul 27 '20 at 11:59
• It isn't a full positive charge. Due to the $\ce{-CF3}$ group there would be a delta positive charge formed since it is highly EW. So, to show the delocalisation of the charge i took it to be a positive charge. – Safdar Jul 27 '20 at 12:01
• CF3 will cause a positive charge. But then since CF3 doesn't show -M / hyperconjucation, the positive charge should only be due to inductive effect? But as far as I know, it becomes almost insignificant at the 3rd carbon. – the.eleventh.letter Jul 27 '20 at 12:08
• I suggested eliminating 3,4,5 because of inductive effect because when no other effects (M/ hyperconjucation) are differtiating, it's the last choice. But the positive charge you suggested, is so minute, I think won't be able to participate in resonance as such. We usually see cations doing such stuff. And given that it is question at +2 level (not research stuff) , I somewhere think that it must some concept which hasn't been pointed out yet. :) – the.eleventh.letter Jul 27 '20 at 12:33
• The funny thing is if you try to calculate the values of the actual charges on the trifluoro methyl benzene C-4 is more positive than C-3.. minute difference so may be due to approximations but still.. Also, the nucleophilic centre would be the most electron rich carbon.. – Safdar Jul 27 '20 at 12:36

$$\ce{CF3}$$ loves to pull electrons, hence it is easier to polarize the molecule in this way (which is inaccurate because you cannot polarize it completely, but considering partial effects). This polarized structure is made considering the requirement of negative charge at the carbon positioned adjacent to CF3.

position 3 and 5 maybe a dispute but, as a known fact, $$-\ce{I}$$ is distance dependent, 3,5 will be lesser favourable that 2.

Your reason is apparently wrong because you considered an attack first and stabilised the product of your attack.

Optimised geometry{tinker:force field mm3} Then exported as xyz form and performed Hartree Fock using TeraChem, the following conclusions were made after the chat in comments.

1. The inductive effect is dependent on distance and has nothing to do with indivisual bond polarities as in the first diagram.

2. The distance between CF3 , (2,3,5) is maximum for 2.

• How do you decide the dispute is between 3,4 and 2??? – Safdar Jul 28 '20 at 17:10
• I remarked the polarisation, please read the first line – user96208 Jul 28 '20 at 17:10
• This is more of speculation than a proper answer. First, you have compared 3,5 and 2 which are on different rings. Second, the most your answer has done is to say that what OP did is wrong . And haven't properly explained your reasoning.. – Safdar Jul 28 '20 at 17:12
• if so you wanted not to compare any positions you might proceed by leaving the question. OP has not asked for reasoning yet. If it's your question, ask it as your own – user96208 Jul 28 '20 at 17:14
• I think explanation and reasoning mean the same thing. Last line in Q asks for the explanation – Safdar Jul 28 '20 at 17:20