what will be the major organic products for the following reaction: (Ph)2(OH)C-CH2OH under H+ cat reaction?

Answer 1

Lancenigo di Villorba (TV), Italy

I CONFIRM THE LATTER AMONG THE MEN PRECEEDING ME.

Your question interests the ASSISTED REARRANGEMENT which runs strting from the a paticular ELIMINATION EVENT. In an acidic medium, the Organic Substrate follows
an Elimination's Event, e.g. DE-HYDRATATION in Alpha Placement as I named it like i) Step

i) (C6H5)2C(OH)-CH2OH(solv.) + H+(solv.) <--->
<---> [(C6H5)2C-CH2OH]+(solv.) + H2O(solv.)

hence it takes place an Acid-Base's Event,
e.g. DE-PROTONATION as I named it like ii) Step

ii) [(C6H5)2C-CH2OH]+(solv.) <--->
<---> [(C6H5)2C=CHOH](solv.) + H+(solv.)

finally there is an Ibridization Rearrangement among the Side-Chain's Carbon Atoms, e.g. KETO-ENOLIC TAUTOMERISM as I named it like iii) Step

iii) [(C6H5)2C=CHOH](solv.) <--->
<---> [(C6H5)2CH-CHO](solv.)

Thus, the mechanism carry out to the Final as Most Stable Product, e.g. (C6H5)2CH-CHO itself.

IBRIDIZATION ROLE
YOU HAVE TO SEE THAT Side-Chain Carbon Atoms MAY CHANGE ITS IBRIDIZATION ALONG THE STEPs SHOWN HERE ABOVE.
The Ibridization's Changes I cited result the Driving Force employed by Mechanism to join at its own ends.
The STEP i) acts as an Elimination One which interests the BREAKING OF C-O's BOND as the Inner Carbon Atom instead the Terminal One.
I reasoned upon the "Intermediate or Carbocation's Energy", so I SEEK FOR THE CARBOCATION MOST STABLE ONEs, e.g. Carbocation placing Electrical Charge in INNER CARBON ATOM WHICH IS THE LONELY ONE ABLE TO DISPLACE THE UNFAVOURABLE CHARGE TOGETHER TWO AROMATIC RINGs.
The STEP ii) acts as an Acid-Base One which interests the FORMING OF C-=C's BOND between the two Side-Chain's Carbon Atoms.
I reasoned upon the "Lewis's Acid-Base Theory", where Terminal Carbon Atom gives an ELECTRONIC DUET (e.g. as a Lewis's Base) to form the C=C's Bond.
On its own part, Inner Carbon Atom stands in "sp2 Ibridization" and it permits the continuous Orbital's Overlapping as warranted by means of the Two Aromatic Rings.
In this fashion, the C=C's Bond which is forming maintain low the Activation Energy of the Step ii), as it contributes to diminute the "Intermediate Energy" of the Unsaturated Compound which is forming.
The STEP iii) acts as an Keto-Enolic Tautomerism which interests the HYDROGEN ION's EXCHANGE between the two Side-Chain's Carbon Atoms.
I reasoned upon the "Bronsted-Lowry's Acid-Base Theory", where Terminal Carbon Atom gives an HYDROGEN ION (e.g. as a Bronsted-Lowry's Acid) toward Inner Carbon Atom.
As you understood, Hydrogen Ion starting from the Oxydril Group results in a MOST ACTIVE PLACEMENT OR MOST ACIDIC ONE INSTEAD THE POSITION AS BOUND TO THE INNER CARBON ATOM.

I hope this helps you.

Answer 2

(Ph)2CH-CHO by pinacol-pinacolone mechanism.

Answer 3

(Ph)2(OH)CH2OOH