Start Learning Free

Before You Begin

Question #44404

Difficulty: 8/10 | 🟢 Free

The Question

Consider the reaction of phenyl neopentyl ether (\cePh−O−CH2−C(CH3)3\ce{Ph-O-CH2-C(CH3)3}) with excess concentrated \ceHI\ce{HI} at 373 K373\text{ K}. Identify the major organic products formed.

Options

  • A.

    Phenol and 2-iodo-2-methylbutane

  • B.

    Phenol and 1-iodo-2,2-dimethylpropane

  • C.

    Iodobenzene and 2,2-dimethylpropan-1-ol

  • D.

    Iodobenzene and 2-iodo-2-methylbutane

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

Ether Cleavage Patterns|Steric Hindrance in SN2|Pinacol-Pinacolone Rearrangement

Why This Question Is Tricky

Common Mistakes

This question tests a critical exception to the standard ether cleavage rules.

Students are taught the heuristic: 'Ether cleavage with HI proceeds via SN2 for primary/secondary alkyl groups and SN1 for tertiary.' Applying this blindly leads to the assumption of an SN2 attack on the primary neopentyl carbon, yielding 1-iodo-2,2-dimethylpropane (Option B).

The 'Hidden Variable' is the extreme steric hindrance of the neopentyl group, which makes SN2 kinetically unviable despite being a primary center.

This forces the mechanism into an SN1 regime accompanied by a 1,2-alkyl shift, testing the student's ability to prioritize steric and electronic factors over simplified mnemonic rules.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAlcohols, Phenols, and Ethers - Class 12
ConceptEther Cleavage Patterns
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree