Question

The decreasing order of priority for the following functional groups is:
I. $-C\equiv N$ II. $-CON{H}_2$
III. $-OH$ IV. $-CHO$
(A)- (II) > (I) > (IV) > (III)
(B)- (III) > (IV) > (I) > (II)
(C)- (I) > (II) > (IV) > (III)
(D)- (I) > (II) > (III) > (IV)

Answer 1

Hint: Carboxylic acid derivatives get priority over other functional groups, like nitriles, aldehydes, ketones, alcohols, amines, etc. In a compound containing more than one functional, the priority order has to be followed for writing the IUPAC name of the compound.

Complete answer:
International Union of Pure and Applied Chemistry (IUPAC) have developed an order of precedence of functional groups which is to be followed while naming a compound containing more than one functional group. The higher priority functional group is the principal functional group of a compound and is identified by the suffix in the IUPAC name of the compound.
Highest in the functional group priority list are carboxylic acids, sulphonic acids and derivatives of carboxylic acids.
     $$-COOH>-S{O}_{3}H>-COOR>-COCl>-CON{H}_2$$
These groups are followed by nitrile, aldehyde, ketones, alcohols, thiols, amines and imines.
     $$-C\equiv N>-CHO>-COR>-OH>-SH>-N{H}_2>=NH$$
Following the above groups are carbon-carbon multiple bonds.
Based on the above functional group priority sequence, we can infer that amide ($-CON{H}_2$) gets the highest priority among the functional groups given, which are
I. $-C\equiv N$
II. $-CON{H}_2$
III. $-OH$
IV. $-CHO$
After amide, nitrile ($-C\equiv N$) gets preference over aldehyde ($-CHO$) and alcohol ($$-OH$$).
If both aldehyde and alcohol are present in a compound, the aldehyde group gets the lower number, hence, higher priority.
Therefore, the decreasing order of priority for the given functional groups is:
     $$-CON{H}_2>-C\equiv N>-CHO>-OH$$

Hence, the correct answer is option (A), i.e. (II) > (I) > (IV) > (III).

Note:
We may get confused between $-C\equiv N$ and $-CHO$ for priority order. To overcome such confusions, keep in mind the most oxidized carbon, i.e. carbon containing less or no hydrogen always gets preference over the carbon containing hydrogen. Therefore, $-C\equiv N$ gets priority over $-CHO$.