If \( \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc \), then the value of \( k \) is:
We are given the following determinant and are tasked with finding its value: \[ \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc \]. Taking \(a\), \(b\), and \(c\) out of the matrix from columns \(C_1\), \(C_2\), and \(C_3\), respectively: \[ abc \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = kabc \] Dividing both sides by \(abc\), we get: \[ \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = k \] Using column operations \(C_2 \to C_2 + C_1\) and \(C_3 \to C_3 + C_1\), the determinant simplifies to: \[ \begin{vmatrix} -1 & 0 & 0 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{vmatrix} = k \] Expanding the determinant along the first row: \[ -1(0 \times 0 - 2 \times 2) = k \] .
Simplifying further: \[ -1(-4) = k \] \[ k = 4 \] \(\therefore k = 4\) Given the problem setup, the value of \(k\) is 4, and thus the correct option is <strong>(D) 4</strong>.
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.