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.
Consider the gas phase reaction: \[ CO + \frac{1}{2} O_2 \rightleftharpoons CO_2 \] At equilibrium for a particular temperature, the partial pressures of \( CO \), \( O_2 \), and \( CO_2 \) are found to be \( 10^{-6} \, {atm} \), \( 10^{-6} \, {atm} \), and \( 16 \, {atm} \), respectively. The equilibrium constant for the reaction is ......... \( \times 10^{10} \) (rounded off to one decimal place).
Molten steel at 1900 K having dissolved hydrogen needs to be vacuum degassed. The equilibrium partial pressure of hydrogen to be maintained to achieve 1 ppm (mass basis) of dissolved hydrogen is ......... Torr (rounded off to two decimal places). Given: For the hydrogen dissolution reaction in molten steel \( \left( \frac{1}{2} {H}_2(g) = [{H}] \right) \), the equilibrium constant (expressed in terms of ppm of dissolved H) is: \[ \log_{10} K_{eq} = \frac{1900}{T} + 2.4 \] 1 atm = 760 Torr.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: