Draw a reaction coordinate diagram for the reaction. (Hint: An alkyl halide is more stable than an alkene.) Draw the structure of the intermediate/s and their location/s in the above reaction coordinate diagram.

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Draw a reaction coordinate diagram for the reaction. (Hint: An alkyl halide is more stable than an alkene.) Draw the structure of the intermediate/s and their location/s in the above reaction coordinate diagram.

 

 

**Question 8: Understanding the Reaction**

**For the following reaction:**

\[
\ce{
H   H
 \ | |
  ><
 /   \
}
+ \ce{H-Br -> 
\quad
H H
\ | |
<      Br:

/   \
}
\]

**Explanation:**

In this question, we are looking at the reaction between an alkene and hydrogen bromide (HBr).

**Reactants:**
- The compound on the left is an alkene, indicated by the double bond between the two carbon atoms.
- The second compound is hydrogen bromide (HBr), where bromine (Br) is attached to a hydrogen (H) atom.

**Products:**
- The product on the right side represents an alkyl bromide, with the double bond changed to a single bond, indicating that the bromine has attached to one of the carbon atoms in place of a hydrogen atom.

**Reaction Mechanism:**
This reaction is a typical example of an electrophilic addition reaction where:
1. The π bond of the alkene attacks the hydrogen (H) from HBr, while the bromide ion (Br-) gets attached to the more stable carbon atom of the temporary carbocation formed.

**Diagram explanation:**
- The reactant alkene is shown with a carbon-carbon double bond, and attached hydrogen atoms.
- The hydrogen bromide is depicted with H and Br, where Br has lone pairs of electrons represented by dots.
- In the product, the alkene has been converted to an alkane (trailing single carbon-carbon bonds), with hydrogen (H) and bromine (Br) atoms attached to the carbon atoms, indicating the electrophilic addition mechanism. The bromine retains its lone pairs in the product.
Transcribed Image Text:**Question 8: Understanding the Reaction** **For the following reaction:** \[ \ce{ H H \ | | >< / \ } + \ce{H-Br -> \quad H H \ | | < Br: / \ } \] **Explanation:** In this question, we are looking at the reaction between an alkene and hydrogen bromide (HBr). **Reactants:** - The compound on the left is an alkene, indicated by the double bond between the two carbon atoms. - The second compound is hydrogen bromide (HBr), where bromine (Br) is attached to a hydrogen (H) atom. **Products:** - The product on the right side represents an alkyl bromide, with the double bond changed to a single bond, indicating that the bromine has attached to one of the carbon atoms in place of a hydrogen atom. **Reaction Mechanism:** This reaction is a typical example of an electrophilic addition reaction where: 1. The π bond of the alkene attacks the hydrogen (H) from HBr, while the bromide ion (Br-) gets attached to the more stable carbon atom of the temporary carbocation formed. **Diagram explanation:** - The reactant alkene is shown with a carbon-carbon double bond, and attached hydrogen atoms. - The hydrogen bromide is depicted with H and Br, where Br has lone pairs of electrons represented by dots. - In the product, the alkene has been converted to an alkane (trailing single carbon-carbon bonds), with hydrogen (H) and bromine (Br) atoms attached to the carbon atoms, indicating the electrophilic addition mechanism. The bromine retains its lone pairs in the product.
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