- ✓${C_3}{H_6}$
- B${C_3}{H_8}$
- C${C_8}{H_{10}}$
- D${C_8}{H_{12}}$

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$A.$ $KF > KI ; LiF > KF$
$B.$ $KF < KI ; LiF > KF$
$C.$ $SnCl _4 > SnCl _2 ; CuCl > NaCl$
$D.$ $LiF > KF ; \quad CuCl < NaCl$
$E.$ $KF < KI ; CuCl > NaCl$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} O{\mkern 1mu} } \\
{\,\,\,|{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} ||} \\
{C{H_3} - CH - C - OH}
\end{array}$ $+ C{H_3} - N{H_2} \to 'A'\xrightarrow[\Delta ]{}'B'\xrightarrow{\begin{subarray}{l}
{\text{LiAl}}{{\text{H}}_4} \\
{\text{(excess)}}
\end{subarray} }'C'$
The final product $‘C’$ will be