
To determine the correct IUPAC nomenclature, follow these steps:
Step 1: Identify the parent chain The parent chain consists of 7 carbon atoms (hept-) with a double bond at the 6th carbon (hept-6-en-).
Step 2: Functional groups and priority The compound contains the following functional groups: 1. A formyl group (−CHO) at the 2nd carbon, 2. A hydroxyl group (−OH) at the 4th carbon, 3. A carboxylic acid group (−COOH) at the end of the chain. The carboxylic acid group has the highest priority, so the chain is named as a derivative of ”hept-6-enoic acid”.
Step 3: Naming the substituents 1. The −CHO group is named as ”formyl” since it is a substituent and not the main functional group. 2. The −OH group is named as ”hydroxy”.
Step 4: Combine the name The substituents and parent chain are combined in the order
of their positions:
2-formyl-4-hydroxyhept-6-enoic acid.
Step 5: Validate the given options From the options provided, the correct name matches:
(3) 2-formyl-4-hydroxyhept-6-enoic acid.
Final Answer: (3)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
IUPAC name of \(K_2MnO_4\) is
is

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,