Identifying the Electrophile in But-2-ene and Bromine Reaction

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Discussion Overview

The discussion revolves around the reaction of but-2-ene with bromine in the presence of concentrated aqueous sodium nitrate, focusing on identifying the electrophile involved, the products formed, and the mechanism of the reaction. Participants explore various statements regarding the electrophile, the nature of the products, and the optical activity of the resultant compound.

Discussion Character

  • Homework-related
  • Debate/contested
  • Exploratory

Main Points Raised

  • Some participants propose that the electrophile could be NO2+, while others argue that Br+ may also form as an electrophile due to polarization of the halide molecule.
  • One participant suggests that only 2,3-dibromobutane is not the sole product, as the reaction can yield compounds with both -Br and -OH groups.
  • There is a contention regarding the optical activity of the resultant compound, with some asserting that it cannot show optical activity due to the possibility of attack from both sides, leading to a racemic mixture.
  • Participants discuss the mechanism involving an initial electrophilic attack followed by a nucleophilic attack, with some expressing uncertainty about the formation of NO2+ from the reaction.
  • One participant expresses concern about potentially bending rules in their reasoning regarding the formation of the compound and the involvement of nitrate ions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the identity of the electrophile, the products formed, or the optical activity of the resultant compound. Multiple competing views remain regarding these aspects of the reaction.

Contextual Notes

There are limitations regarding the assumptions made about the formation of NO2+ and the nature of the electrophile, as well as the conditions under which the reaction occurs. The discussion also reflects uncertainty about the mechanism and the resulting stereochemistry of the products.

chewchun
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Homework Statement


But-2-ene reacts with bromine in the presence of concentrated aqueous sodium nitrate to give the following compound
CH3-C(H)(ONO2)-C(H)(Br)-CH3

Which of the following statement is correct.
a.the electrophile is NO2+
b.Only 2,3-dibromobutane is formed.
c.Resultant solution shows optical activity.
d.Mechanism involves an initial electrophilic attack followed by nucleophilic attack

Homework Equations





The Attempt at a Solution


I know that B is out cause you can have -Br and -OH instead of just 2-Br.
C is also out because the compound can be attacked from both side(Trigonal planar)
A or D is correct.D makes sense to me but isn't NO2+ a electrophile. In this case i would presume NO2+ is somehow formed which attacks OH group...If not what is the case?
 
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See, the halide molecule also polarize and form halide electrophile, in this case, Br+ can form. NaNO3 cannot give NO2+ ion, as it is obtained by "protonation" of nitric acid, which is not a simple task, considering nitric acid itself is very strong acid. That's what I feel.
 
Name these ions... NO3- and NO2+

Which one is present in your proposed mixture?
 
CH3-C(H)(ONO2)-C(H)(Br)-CH3

Care to elaborate, as chemisttree asked?

UPDATE: I have formed on paper the compound H3C-CH(ONO2)-CHBr-CH3, considering Nitrate group, albeit I am afraid I may have have bent some rules (which looks like the case, forming acid out of aqueous nitrate ions) And I have reached to following answers: C, D.
 
Last edited:
AGNuke said:
CH3-C(H)(ONO2)-C(H)(Br)-CH3

Care to elaborate, as chemisttree asked?

UPDATE: I have formed on paper the compound H3C-CH(ONO2)-CHBr-CH3, considering Nitrate group, albeit I am afraid I may have have bent some rules (which looks like the case, forming acid out of aqueous nitrate ions) And I have reached to following answers: C, D.

The answer was given to be D only.
The carbon is directly bonded to oxygen,and the oxygen directly bonded to NO2.So it seems to be O- and NO2+?And i presume the O- to be created from a alcohol group and the H+ is removed...
Answer is not C as the compound is trigonal planar and can be attacked from both sides,hence a racemic mixture..
But i don't understand how C(O)(NO2) is formed as it seems impossible to form NO2+ from this reaction...
 
Esterification I guess.
 

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