Write down the steps which correctly describe the steps required to decide whether or not a precipitate forms when two aqueous s
1 answer:
The correct question is as follows.
place the steps required to determine whether or not a precipitate forms when two solutions are mixed in the correct order.
-consider possible cation-anion combinations
-note the ions present in the reactions
-use the solubility rules to determine whether or not either of the combinations gives an insoluble salt
Explanation:
A precipitate is an insoluble substance formed due to chemical reaction between two aqueous solutions. But it is not necessary that two aqueous solutions will always give an insoluble solid substance on chemical reaction with each other.
For e.g.
Here, AgCl is the precipitate formed.
Hence, correct steps that are required to decide whether or not a precipitate forms when two aqueous solutions are mixed:
1). note the ions present in the reactants
2). consider possible cation-anion combinations
3). use the solubility rules to determine whether or not either of the combinations gives an insoluble salt
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<h2>Answer</h2>
Bromination:
Any reaction or process in which bromine (and no other elements) are introduced into a molecule.
Bromonium Ion:
The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)
Mechanism:
Step 1:
In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".
Step 2:
When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.
In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.