UPSC MainsChemistry (Optional)Science and TechnologyPractice question

Pinacol Rearrangement of 1,1,2-Triphenylpropane-1,2-Diol

Identify the major products in following reactions and discuss their formation : (C6H5)2C(OH)-C(OH)(C6H5)(CH3) -- H+ -->

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Identify the reaction as an acid-catalyzed Pinacol–Pinacolone rearrangement of 1,1,2-triphenylpropane-1,2-diol. Detail the regioselectivity of initial carbocation formation based on resonance stabilization, followed by analyzing the migratory aptitude of competing groups to determine the major product, supported by a step-by-step mechanism.

Model answer

327 words

Introduction

The reaction of 1,1,2-triphenylpropane-1,2-diol in an acidic medium is a classic example of an acid-catalyzed Pinacol–Pinacolone rearrangement. This transformation converts a vicinal diol into a carbonyl compound through regioselective dehydration followed by a 1,2-rearrangement driven by carbocation stability and migratory aptitude.

Product Identification

Major Product: 1,1,1-Triphenylpropan-2-one [1,1,1-triphenylacetone, (C6H5)3C–COCH3].

Minor Product: 1,2,2-Triphenylpropan-1-one [(C6H5)2C(CH3)–COC6H5].

Mechanism and Discussion of Formation

The formation of the major product proceeds via three distinct, thermodynamically and kinetically controlled steps:

  • Step 1: Regioselective Protonation and Carbocation Generation: The diol possesses two distinct hydroxyl-bearing carbon centers: C1 (bonded to two phenyl groups) and C2 (bonded to one phenyl group and one methyl group). Protonation occurs preferentially at the C1 hydroxyl group. Subsequent loss of water generates a tertiary carbocation that is extensively resonance-stabilized by delocalization across two phenyl rings (a benzhydryl-type carbocation). In contrast, loss of water from C2 would generate a carbocation stabilized by only one phenyl ring and hyperconjugation from the methyl group, which is significantly less stable.
  • Step 2: 1,2-Migration: Once the C1 carbocation is formed, rearrangement occurs via a 1,2-shift from C2 to C1 to generate an oxocarbenium ion intermediate. The competing migrating groups at C2 are phenyl (-C6H5) and methyl (-CH3). Because the phenyl group has a vastly superior migratory aptitude (Ph >> Me) due to its ability to stabilize the transition state via a bridged phenonium ion intermediate, phenyl migration dominates overwhelmingly.
  • Step 3: Deprotonation and Carbonyl Formation: The 1,2-phenyl shift yields a resonance-stabilized protonated ketone (oxocarbenium ion). Rapid loss of a proton (H+) from the oxygen atom affords the neutral major product, 1,1,1-triphenylpropan-2-one.

Conclusion

The outcome of the acid-catalyzed pinacol rearrangement is governed primarily by the initial regioselective ionization favoring the more stable bis-benzylic carbocation at C1, coupled with the intrinsically higher migratory aptitude of the phenyl ring over the methyl substituent.

Key facts to remember

definition
Pinacol–Pinacolone Rearrangement

An acid-catalyzed dehydration of vicinal diols yielding aldehydes or ketones via a carbocation intermediate and a 1,2-migration step.

definition
Migratory Aptitude

The relative ease with which a substituent shifts to an adjacent electron-deficient center, typically following the order aryl > alkyl/hydride depending on electron density.

example
Phenonium Ion Intermediate

During phenyl migration, the aromatic pi-system attacks the adjacent carbocation center to form a transient, bridged cyclopropyl-like intermediate that lowers the activation barrier.

Frequently asked questions

Why does the phenyl group migrate preferentially over the methyl group?

The phenyl ring possesses mobile pi-electrons that can delocalize the emerging positive charge in a bridged phenonium ion transition state, making its activation energy significantly lower than that of methyl migration.