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How to know if a nucleophile will attack from front-side(retention of configuration) or back-side(inversion of configuration) in sn2 reaction?

https://www.youtube.com/watch?v=L01vv1Mj64I

In this video same reaction is showing both frontside and backside.

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closed as too broad by Todd Minehardt, Mathew Mahindaratne, Mithoron, Zhe, Tyberius Jul 8 at 18:16

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ That "frontside $S_N2$" isn't an SN2 reaction, and it is extremely unfavourable. Stupid video. Retention occurs with the $S_Ni$ mechanism, e.g. chlorination of alcohols via thionyl chloride. $\endgroup$ – Karl Jul 7 at 8:26
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    $\begingroup$ I agree, this is not a useful teaching tool. $\endgroup$ – Waylander Jul 7 at 10:11
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    $\begingroup$ The video is an animation ,without any commetry with respect to which type of attack is energitically more feasible. Without that ,this video is not usefull $\endgroup$ – Chakravarthy Kalyan Jul 7 at 10:39
  • $\begingroup$ I feel like the so-called “frontside Sn2” looks more like the mechanism for electrophillic Attack to the carbon, though. $\endgroup$ – ANZGC FlyingFalcon Jul 7 at 15:05
  • $\begingroup$ Due to the leaving group and the nature of the compound in which the nucleophile is reacted with I.e either primary or secondary haloalkane as the case may be $\endgroup$ – Bin badaweeyyerh Jul 7 at 16:52
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The video you linked to mentioned a particular research paper which you should read if interested (either just ask your institution for a copy or use you-know-what-website). The conclusion writes:

“...backside SN2-b barriers increase along the nucleophiles F-, Cl-, Br-, and I- and decrease along the substrates CH3F, CH3Cl, CH3Br, and CH3I. Frontside SN2-f barriers show the same trends but are in all cases much higher (ca. 10-60 kcal mol-1) because of more steric repulsion as a result of the proximity between the nucleophile and leaving group.”

In this paper, they used SN2-b and SN2-f to represent backside/frontside reactions. I think that should answer your question.

There’s something that you should take note of, though: the “front-side Attack” mechanism would be the optimal one for the electrophillic Attack of methane. You might want to read that at J. Am. Chem. SOC. 1995,117, 1336-1343 (“Electrophilic Substitution of Methane Revisited”). This can be explained in molecular orbital terms (you might want to read up on the isolobal analogy if interested).

In a concerted nucleophillic substitution, the transition state is “electron-rich” and tends to adopt the most “open” geometry possible. This is anagolous to the linear geometry of the trihydrogen anion.

Meanwhile, in a concerted electrophillic substitution, the transition state is “electron-deficient” and tends to adopt a “closed” geometry, à là the trigonal planar shape of the trihydrogen cation.

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