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H(3)C-underset(CH(3))underset(|)(CH)-CH=...

`H_(3)C-underset(CH_(3))underset(|)(CH)-CH=CH_(2)+HBr to A`.
A is predominantly

A

`CH_(3)-underset(Br)underset("| ")"CH"-underset(CH_(3))underset("| ")"CH"-CH_(3)`

B

`CH_(3)-underset(CH_(3))underset("| ")"CH"-underset(Br)underset("| ")"CH"-CH_(3)`

C

`CH_(3)-underset(CH_(3))underset("| ")"CH"-CH_(2)-CH_(2)Br`

D

`CH_(3)-overset(Br)overset("| ")underset(CH_(3))underset("| ")"C "-CH_(2)CH_(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: 1. **Identify the Reactants**: The given compound is 3-methyl-2-buten-1 (H₃C-CH(CH₃)-CH=CH₂) and it reacts with HBr. 2. **Understand the Reaction Type**: This is an addition reaction where HBr adds across the double bond of the alkene. 3. **Apply Markovnikov's Rule**: According to Markovnikov's rule, when HX (where X is a halogen) adds to an alkene, the hydrogen atom (H⁺) will attach to the carbon with the greater number of hydrogen atoms, and the halide (Br⁻) will attach to the carbon with fewer hydrogen atoms. 4. **Identify the Double Bond Carbons**: In the structure of 3-methyl-2-buten-1, the double bond is between the second and third carbons. The second carbon (C2) has one hydrogen atom, and the third carbon (C3) has two hydrogen atoms. 5. **Determine the Product**: According to Markovnikov's rule, Br will attach to C2 (the carbon with fewer hydrogens), and H will attach to C3. This results in the formation of 2-bromo-3-methylbutane. 6. **Write the Final Product**: The final product of the reaction is 2-bromo-3-methylbutane. ### Summary of the Solution: The product A formed from the reaction of 3-methyl-2-buten-1 with HBr is predominantly **2-bromo-3-methylbutane**.
To solve the problem, we will follow these steps: 1. **Identify the Reactants**: The given compound is 3-methyl-2-buten-1 (H₃C-CH(CH₃)-CH=CH₂) and it reacts with HBr. 2. **Understand the Reaction Type**: This is an addition reaction where HBr adds across the double bond of the alkene. 3. **Apply Markovnikov's Rule**: According to Markovnikov's rule, when HX (where X is a halogen) adds to an alkene, the hydrogen atom (H⁺) will attach to the carbon with the greater number of hydrogen atoms, and the halide (Br⁻) will attach to the carbon with fewer hydrogen atoms. ...
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