Answer:
Compound A and compound B are constitutional isomers with molecular formula C3H7Cl.
When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates.
Explanation:
Constitutional isomers are the one which differs in the structural formula.
When compound A is treated with sodium methoxide, a substitution reaction predominates.
That means sodium methoxide is a strong base and a strong nucleophile.
But when it reacts with primary alkyl halides it forms a substitution product and when it reacts with secondary alkyl halide it forms mostly elimination product.
The reaction and the structures of A and B are shown below:
Water is neutral with ph 7. when we add water to acidic solution, it will be less acidic. so the pH of acid increases
To change only one variable which is very important than to test the experiment to match the hypothesis again, I think. It’s been a while since I was on that lesson♀️
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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.
<span>Consider two solutions: solution X has a pH of 4; solution Y has a pH of 7. From this information, we can reasonably conclude that </span>the concentration of hydrogen ions (H⁺) or hydronium ions (H₃O⁺) in solution X is thousand times as great as the concentration of hydrogen ions or hydronium ions in solution Y.
Solution X: c(H⁺) = 10∧-pH = 10⁻⁴ mol/L = 0,0001 mol/L.
Solution Y: c(H⁺) = 10⁻⁷ mol/L = 0,0000001 mol/L.
0,0001 mol/L / 0,0000001 mol/L = 1000.
