





This is an elimination reaction (E2 mechanism) where the hydroxide ion acts as a strong base. The steps of the reaction are as follows:
Step 1: Identify the base and leaving group
In the presence of \(\text{OH}^-\) and EtOH, the bromine atom (\(\text{Br}\)) acts as the leaving group. The hydroxide ion abstracts a proton (\(\text{H}^+\)) from the adjacent carbon atom, resulting in the formation of a double bond.
Step 2: Formation of the double bond
Since the reaction proceeds via the E2 mechanism, the elimination occurs in a single step. The base removes a \(\beta\)-hydrogen atom (attached to the carbon next to the carbon with \(\text{Br}\)), and the \(\text{Br}^-\) leaves, resulting in the formation of a double bond between the \(\alpha\)- and \(\beta\)-carbons.
Step 3: Determine the major product
The double bond forms between the \(\alpha\)-carbon (the one originally bonded to the bromine atom) and the \(\beta\)-carbon. Due to Zaitsev's rule, the most substituted alkene is the major product. Hence, the major product is cyclopentene with a methyl (\(\text{CH}_3\)) substituent.
Final Answer: (3) Cyclopentene with a \(\text{CH}_3\) substituent.
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



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,