The given reaction is an elimination reaction using alcoholic KOH as the base. Under such conditions, the reaction favors the formation of an alkene via the E2 mechanism. The elimination occurs by the removal of the ff-hydrogen, leading to the formation of the most stable (more substituted) alkene as the major product. In this case, the product formed is a highly substituted alkene, making option (2) the correct answer.
The major product (A) is:
Given below are two statements:
Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction.
Statement (II): In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the $ \beta $-carbon.
In the light of the above statements, choose the most appropriate answer from the options given below:
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: