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Question #44503
The Question
An alkene , 2-methyl-1-phenylprop-1-ene, is subjected to the following reaction sequence: \begin{enumerate} \item \item \item \end{enumerate} Identify the correct IUPAC name of product .
Options
- A.
2,2-dimethyl-3-phenylpropanoic acid
- B.
3-methyl-2-phenylbutanoic acid
- C.
3-methyl-3-phenylbutanoic acid
- D.
2-methyl-2-phenylbutanoic acid
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.
Concepts Behind This Question
Peroxide Effect (Kharasch)|Resonance Energy and Stability|Hydrocarbons
Why This Question Is Tricky
Common Mistakes
This question tests the student's reliance on the simplified 'Anti-Markovnikov = addition to the less substituted carbon' heuristic.
In 2-methyl-1-phenylprop-1-ene, the less substituted carbon is C1.
A student blindly applying the rule will place the bromine at C1, leading to the incorrect product 3-methyl-2-phenylbutanoic acid.
However, radical addition is governed by the thermodynamic stability of the intermediate radical.
The secondary benzylic radical formed by addition to C2 is stabilized by resonance (making it significantly more stable than the tertiary alkyl radical formed at C1).
This resonance stabilization completely overrides the steric preference, inverting the expected regioselectivity and trapping students who memorize rules without understanding the underlying radical stability principles.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Hydrocarbons - Class 11 Part 2 |
| Concept | Peroxide Effect (Kharasch) |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |