-3
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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 '19 at 8:26
  • 1
    $\begingroup$ I agree, this is not a useful teaching tool. $\endgroup$ – Waylander Jul 7 '19 at 10:11
  • 1
    $\begingroup$ 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 '19 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 '19 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 '19 at 16:52
1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.