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This question came in one of my school tests. They had given us an unsymmetrical pinacol i.e 1,1-diphenyl-2-methylpropane-1,2-diol and asked us whether it will give only one pinacolone or not.

According to me, I had assumed that carbocation will be formed on the diphenyl-substituted carbon due to which a methyl shift will occur and we will get only one product. But my teacher marked it wrong and said that more than one pinacolone is possible here.

How is it possible?

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    $\begingroup$ Then most probably he is also considering the possibility( minor but can also occur in a very small fraction) of forming the tertiary carbocation and a phenyl shift to give the other Pinnacolone. $\endgroup$ – Soumik Das May 20 '18 at 13:30
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The problem is what is a reaction which gives "only one product", the problem is that very few reactions give a 100 % yield of a substance. For almost all real life reactions there are a small fraction of molecules which react he wrong way.

I think that when more than 99 % of molecules react the "right" way that those which react the "wrong" way can be ignored normally. Unless the wrong product has some very harmful effect on the product when it is used for a purpose.

What I think will happen is that the carbocation formed by protonation of the hydroxyl bonded to the carbon bearing the two phenyl groups, followed by the loss of a water molecule will have a much lower energy than the other carbocation. Thus it will be more easy to form.

But there is a problem, for a rearrangement of a carbocation to occur normally we need to rearrange to a lower energy carbocation. By lowering the energy of the carbocation we can lose the thermodynamic drive for rearrangement. In the ideal world we would consider the rate of rearrangement of the two different carbocations and their formation. I think that the reactions forming the carbocation will be reversable while the pinacol rearrangement will be irreversable.

But in some cases such as many homogenious asymmetric rhodium hydrogenations of prochiral alkenes the lower energy alkene rhodium complex which is present at a far higher concentration than the higher energy isomer, the lower energy isomer has far too little reactivity. As a result while the majority of the complex might be in the R form the tiny trace of the complex in the S form is responsible for the vast majority of the hydrogenation reactivity.

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