An aromatic compound ‘A’ on treatment with CHCl3 and KOH gives twocompounds, both of which give same product ‘B’ when distilled with Zincdust. Oxidation of ‘B’ gives ‘C’ with molecular formula C7H6O2. Sodiumsalt of ‘C’ on heating with soda lime gives ‘D’ which may also be obtainedby distilling ‘A’ with Zinc dust. Identify ‘A’, ‘B’, ‘C’ and ‘D’.
OR
How do you convert the following :
(a) Phenol to Toluene
(b) Ethanol to Ethanal
The compound ‘C’ C7H6O2is an aromatic and as the number of C and O atoms is almost the same, we can say that ‘C’ has a phenyl ring a (-C6H5) along with a substituent group.
If we subtract C6H5 from C7H6O2, we see that the substituent group is -CHO2or –COOH.
So, ‘C’ is benzoic acid (C6H5COOH).
To be more sure, let us see the next reactions done on ‘C’.
If C6H5COONa (sodium salt of benzoic acid) is heated with soda lime, it gives benzene, C6H6, ‘D’.
Also, benzene is produced when phenol is subjected to zinc dust, as reduction reaction occurs. So, ‘A’ is phenol (C6H5OH).
When phenol is treated with CHCl3 and KOH, Riemer-Tiemann reaction takes place, and we know that salicyldehyde (p-hydroxybenzaldehyde) is produced as the major product along with o-hydroxybenzaldehyde as the minor product.
And when they are reduced with zinc dust, both of them give benzaldehyde (as the –OH group is removed only).
So, ‘B’ is benzaldehyde (C6H5CHO).
And we know that oxidation of aldehyde gives carboxylic acid, so oxidation of B, benzaldehyde gives C, benzoic acid.
OR
(a) To convert phenol (C6H5OH) to toluene (C6H5CH3), first we have to remove the –OH group from phenol (done by reaction with zinc dust), so that we can introduce the –CH3 group by Fridel-Craft alkylation method.
The reactions involved are:
(b) The conversion of an alcohol (primary) to its corresponding aldehyde is very tricky as over-oxidation of an alcohol oxidises to its corresponding carboxylic acid.
So, we use a poison, PCC (Pyridiniumchlorochromate) to limit the oxidation to the aldehyde stage only.
