





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.
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to:
The output of the circuit is low (zero) for:

(A) \( X = 0, Y = 0 \)
(B) \( X = 0, Y = 1 \)
(C) \( X = 1, Y = 0 \)
(D) \( X = 1, Y = 1 \)
Choose the correct answer from the options given below:
The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are
A. $ Cr^{2+} $
B. $ Fe^{2+} $
C. $ Fe^{3+} $
D. $ Co^{2+} $
E. $ Mn^{2+} $
Choose the correct answer from the options given below
Which of the following circuits has the same output as that of the given circuit?
