





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.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
