Question

Acetone can be converted into pinacol by:
(A) \[Mg/Hg/{H_2}O\]
(B) $Zn/Hg/HCl$
(C) $Na/Hg/{H_2}S{O_4}$
(D) All of the above

Answer 1

Hint:In order to solve this question we should know that in the stereochemistry of the reduction of acetone is done with the help of magnesium amalgam. Reaction takes place at the metal surface and the necessary is only the surface of magnesium of amalgam. The metal salts other than mercuric salts can also activate magnesium in this reaction but are less effective.

Complete answer:For solving this question we need to know first that what is acetone and pinacol so first we will study these of the two:So acetone is an organic compound with the formula ${(C{H_3})_2}CO$ that contains 3 carbon atoms, 6 hydrogen atoms and an oxygen atom. This is the simplest and smallest ketone. It is a colourless , highly volatile and flammable liquid with a characteristic pungent smell.Now, the pinacol is a white solid organic compound that has a hydroxyl group on the vicinal carbon atoms. It has the formula ${C_6}{H_{14}}{O_2}$ which contains 6 carbon atoms 14 hydrogen atoms and 2 oxygen atoms whose orientation is such as the two –OH groups are present.First of all when the double bond between C and O breaks for a while and becomes the lone pair then the two electrons which are present in the outermost orbit attack similarly one such intermediate is created after it mutually combines to form another intermediate.
According to equation:
${(C{H_3})_2}C = O\xrightarrow{{M\ddot g}}{(C{H_3})_2} - O - Mg$
Now the another step:
${(C{H_3})_2}C - O - Mg\xrightarrow{{{{(C{H_3})}_2}C = O}}{(C{H_3})_2}C - O - Mg - O - C{(C{H_3})_2}$
Now after the bonding between two carbons connected by oxygen and hydrolysis the pinacol is formed.
${(C{H_3})_2}C - O - Mg - O - C{(C{H_3})_2}\xrightarrow{{{H_2}O}}{C_6}{H_{14}}{O_2}$
Hence the pinacol is formed.

So, the correct option will be (A).

Note:Pinacol itself is produced by magnesium reduction of acetone, probably by the way of ketyl intermediate. Since the diol is symmetrical, protonation and loss of water takes place with equal probability at either hydroxyl group.