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WITCHER [35]
1 year ago
7

Why is 15 m acetic acid an appropriate solvent in which to perform rate studies of electrophilic bromintations?.

Chemistry
2 answers:
Alekssandra [29.7K]1 year ago
8 0

Acetic acid activates the bromine and makes it a better electrophile.

<h3>What is bromination?</h3>

When a substance undergoes bromination, bromine is added to the compound as a result of the chemical reaction. After bromination, the result will have different properties from the initial reactant.

<h3>Why is 15M acetic acid used as a solvent for bromination?</h3>

DCM (dichloromethane) requires more time. Acetic acid has protons that can give one of the Br (bromine) a positive charge and activate it. There is a brief loss of aromaticity that calls for high energy activation.

Refer to the attached image for bromination reaction.

Learn more about bromination here:

brainly.com/question/26428023

#SPJ4

Download docx
Ann [662]1 year ago
3 0

The bromine is activated by acetic acid, which improves its electrophilic properties.

<h3>What is meant by bromination?</h3>
  • When a material gets brominated, the chemical reaction causes bromine to be added to the compound.
  • The outcome will differ from the initial reactant in terms of its properties after bromination.

<h3>Why is 15M acetic acid employed as a bromination solvent?</h3>
  • DCM (dichloromethane) takes longer to process.
  • One of the Br (bromine) atoms can be activated by the protons in acetic acid, which can give it a positive charge.
  • Short-lived loss of aromaticity necessitates high-energy activation.

<h3>For the reaction of bromination, refer to the attached file.</h3>

To learn more about electrophilic bromination visit:

brainly.com/question/4589491

#SPJ4

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A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so
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Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

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Formula used :

\text{Molarity}=\frac{\text{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

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