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In this question, I am not quite sure what will be the major product in the reaction and on reacting (2-methoxyethyl)benzene with $\ce{NO2-O-NO2}$ ($\ce{N2O5}$). I know $\ce{NO2}$ will act as electrophile and gets attached to ortho or para. In this will it be ortho position instead of para due to hydrogen bonding with $\ce{CH2CH2OCH3}$?

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  • $\begingroup$ And what "the reaction" really is? You don't provide the context. $\endgroup$
    – Mithoron
    Commented Dec 3, 2022 at 16:37
  • $\begingroup$ I edited the ques can you pls review it again. $\endgroup$
    – Vansh Bhardwaj
    Commented Dec 3, 2022 at 16:38
  • $\begingroup$ @Poutnik I know it would be a nitration reaction but what should be the major product. $\endgroup$
    – Vansh Bhardwaj
    Commented Dec 3, 2022 at 16:47
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    $\begingroup$ @VanshBhardwaj Compare to the nitration of toluene. 2-Nitrotoluene It is made by nitrating toluene at above -10 °C. This reaction affords a 2:1 mixture of 2-nitro and 4-nitro isomers $\endgroup$
    – Poutnik
    Commented Dec 3, 2022 at 16:55

2 Answers 2

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This is a case of the nitration of alkylbenzene by dinitrogen pentoxide (DNP), which can be considered as an anhydride of nitric acid. The substrate in the question is (2-methoxyethyl)benzene, which is closely related to toluene except for the extra bulkiness of the substitution. Thus, one can except nitration mechanism would be closely related to that of toluene.

What's lacking in this question is the conditions for the nitration. The alkyl group is basically a o,p-director. However, DNP-based nitration reactions are dramatically influenced by the medium (solvent) used (Ref.1 & 2). That means, the ortho to para products ratio is different from condition to condition (e.g., solvent). For example, it was 1:2 if the reaction performed at $\pu{-10 ^\circ C}$ as Poutnik pointed out in his comment above (in this case, I assumed the reaction is in toluene). When the nitration of toluene by $\ce{N2O5}$ is done in an organic solvent other than toluene such as liquefied 1,1,1,2-tetrafluoroethane (TFE), the product ratio varies (Ref.2):

Nitration of toluene by N2O5

If the reaction is conducted in the presence of a solid Zeolite with small pores and with DNP as the nitrating agent and dichloromethane as the solvent, the reaction products have a very high proportion of para-nitro substituted isomer (Ref.3).

References:

  1. R. W. Millar, M. E. Colclough, P. Golding, P. J. Honey, N. C. Paul, A. J. Sanderson, and M. J. Stewart, "New synthesis routes for energetic materials using dinitrogen pentoxide," Philos. Trans. R. Soc., A 1992, 339(1654), 305-319 (DOI: https://doi.org/10.1098/rsta.1992.0037).
  2. Alexandr K. Kharchenko, Ruslan V. Fauziev, Mikhail N. Zharkov, Ilya V. Kuchurov, and Sergei G. Zlotin, "Nitration of aromatics with dinitrogen pentoxide in a liquefied 1,1,1,2-tetrafluoroethane medium," RSC Adv. 2021, 11(42), 25841-25847 (DOI: https://doi.org/10.1039/D1RA04536A).
  3. Reddy Damavarapu, Keerthi Jayasuriya, and Thomas J. Kwok, "Regioselective nitration of aromatic compounds by dinitrogen pentoxide and the reaction products thereof," US Patent 1999, 5,977,418 (PDF).
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    $\begingroup$ Interesting, for o- and p- isomers, I would expect the preference difference would be small and 2 : 1 was the result of statistical availlability ( NO2+ meeting o-C 2x more frequently than p- carbon. ). So for 1 : 1, preference of p-C must be 2x stronger, what seems suspicious. Perhaps the steric effect of methyl are more important than I have thought. /// The solvent influence must be really tricky. // But I have just residual knowledge of organic synthesis. $\endgroup$
    – Poutnik
    Commented Dec 4, 2022 at 14:31
  • $\begingroup$ I think the trick is DNP stays as ion pair in aqueous solvents and as molecular form in organic solvents. $\endgroup$
    – Mathew Mahindaratne
    Commented Dec 4, 2022 at 14:53
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    $\begingroup$ And as a neutral molecule, it is then weaker electrofil and therefore more selective and possibly more sterically picky (my guess) $\endgroup$
    – Poutnik
    Commented Dec 4, 2022 at 17:07
  • $\begingroup$ I see it must vary according to solvent in which reaction is carried out.Thanks a lot. $\endgroup$
    – Vansh Bhardwaj
    Commented Dec 5, 2022 at 2:42
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Nitrogen pentoxide is a good source of $\ce{NO2^+}$ (in fact the solid form directly contains $\ce{NO2^+}$). So, as you imply, we would expect a nitration reaction especially with an activating, ortho-para directing substituent such as the alkoxy group shown.

The ortho product would be favored by hydrogen bonding, but here there is no good mechanisms for that. Hydrogen bonding with hydrogen attached to carbon may actually be possible, but it is weak, and the nitrogen group will not provide more strongly protection hydrogen. Therefore the more sterically favored para product would be major.

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  • $\begingroup$ Or, you might want to ask the professor why it would be ortho, as it would not be evident to me. $\endgroup$
    – Oscar Lanzi
    Commented Dec 3, 2022 at 16:53

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