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Likurg_2 [28]
3 years ago
10

An alkene reacts with a strong protic acid to form a carbocation. In Part 1 draw the curved arrow notation for the reaction betw

een an alkene and HBr. However, an alkene will react with a halogen electrophile to form a cyclic intermediate. In Part 2 draw the curved arrow notation for the reaction between an alkene and Br2.
Chemistry
1 answer:
lozanna [386]3 years ago
6 0

Answer:

See explanation and image attached

Explanation:

Now, we have chosen the alkene 1-propene in our example.

In the first reaction of  1-propene  with HBr, the reaction proceeds by ionic mechanism leading to the formation of 2-bromo propane.

In the second reaction of  1-propene  with the bromine molecule, the first step is the formation of the brominium cation which is a cyclic intermidiate followed by the addition of Br^- yielding the 1,2- dibromopropane product

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Successive ionization energy for Cu²+​
kogti [31]

Answer:

Here's what I get  

Explanation:

The ionization energy (I) is the energy required to remove an electron from an atom in the gaseous phase.

You can remove electrons in succession and measure the energies required as I₁, I₂, I₃, etc.

Thus, the removal of two electrons from Cu gives you Cu²⁺.

I found the ionization energies of Cu and used them to create those of Cu²⁺ (see table and graph below).

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You can remove the nine 3d electrons and then there is a sudden jump from I₉ to I₁₀ as you break into the filled [Ar] configuration.

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8 0
3 years ago
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AleksandrR [38]

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OleMash [197]

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Have a nice day!

6 0
4 years ago
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DIA [1.3K]

b. a chemical that cannot be broken down or separated into other chemicals
5 0
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