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Question #44460

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence starting from 1,1,1-trifluoro-2-methylpropane (Compound \textbf{X}): X→Cl2, hν, 298 KY (Major monochlorinated product)\textbf{X} \xrightarrow{Cl_2,\ h\nu,\ 298\text{ K}} \textbf{Y} \text{ (Major monochlorinated product)} Y→NaCN, DMFZ\textbf{Y} \xrightarrow{NaCN,\ \text{DMF}} \textbf{Z} Z→H3O+, ΔW\textbf{Z} \xrightarrow{H_3O^+,\ \Delta} \textbf{W} Identify the IUPAC name of the major product \textbf{W}.

Options

  • A.

    4,4,4-trifluoro-3-methylbutanoic acid

  • B.

    3,3,3-trifluoro-2,2-dimethylpropanoic acid

  • C.

    4,4,4-trifluoro-2-methylbutanoic acid

  • D.

    3-(trifluoromethyl)butanoic 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

Free Radical Substitution Selectivity|Inductive Effect (I-Effect)|Oxidation State Traps

Why This Question Is Tricky

Common Mistakes

This question tests the competition between standard radical stability (hyperconjugation) and the polar effect in the transition state.

Students typically rely on the heuristic that tertiary radicals are always preferred over primary radicals.

However, the highly electrophilic chlorine radical creates a polar transition state.

The strong -I effect of the adjacent -CF3 group drastically destabilizes the alpha-tertiary position.

The 'hidden variable' is the polar effect overriding Markovnikov-adjacent selectivity, causing a radical shift to the less substituted, but less deactivated, primary carbon.

Failing to recognize this leads to the trap product 3,3,3-trifluoro-2,2-dimethylpropanoic acid.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterHydrocarbons - Class 11 Part 2
ConceptFree Radical Substitution Selectivity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree