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

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SUMMARY

The reaction of But-2-ene with bromine in the presence of concentrated aqueous sodium nitrate produces the compound CH3-C(H)(ONO2)-C(H)(Br)-CH3. The correct statement regarding this reaction is that the mechanism involves an initial electrophilic attack followed by a nucleophilic attack (option D). The electrophile in this reaction is identified as NO2+, which is formed through the protonation of nitric acid, despite the complexity of its generation from sodium nitrate. The resultant compound exhibits optical activity due to its trigonal planar structure, allowing for attacks from both sides, leading to a racemic mixture.

PREREQUISITES
  • Understanding of electrophilic aromatic substitution mechanisms
  • Familiarity with the properties of But-2-ene and bromine reactions
  • Knowledge of nitrate ion behavior in organic reactions
  • Concept of optical activity and racemic mixtures
NEXT STEPS
  • Study the mechanism of electrophilic attack in organic chemistry
  • Learn about the formation and role of NO2+ in organic reactions
  • Research the properties and reactions of racemic mixtures
  • Explore the concept of protonation in acid-base chemistry
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Chemistry students, organic chemists, and anyone interested in understanding electrophilic reactions and their mechanisms.

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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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