- ✓$2E_1 -E_2$
- B$E_1 -E_2$
- C$E_1 -2E_2$
- D$(E_1 + E_2)/2$
hence $\mathrm{EA}=2 \mathrm{E}_{1}-\mathrm{E}_{2}$
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$(A) i)$ Lindlar's catalyst, $H _2$; $ii$) $SnCl _2 / HCl$; iii) $NaBH _4$; iv) $H _3 O ^{+}$
$(B) i)$ Lindlar's catalyst, $H _2$; $ii$) $H _3 O ^{+}; iii)$ $SnCl _2 / HCl$; iv) $NaBH _4$
$(C) i)$ $NaBH _4; ii)$ $SnCl _2 / HCl; $iii$)$ $H _3 O ^{+}; iv)$ Lindlar's catalyst, $H _2$
$(D) i)$ Lindlar's catalyst, $H _2$; $ii$) $NaBH _4; iii)$ $SnCl _2 / HCl; iv)$ $H _3 O ^{+}$
$(1)$ $\begin{array}{*{20}{c}}
{C{H_3}CH = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(2)$ $C{H_3}CH = CHC{H_3}$
$(3)$ $\mathop {C{H_3}C}\limits^{\begin{subarray}{l}
\,\,\,\,\,\,\,\,{\begin{array}{*{20}{c}}
{}&H
\end{array}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,|
\end{subarray} } = CHC{H_2}C{H_3}$
$(4)$ $\begin{array}{*{20}{c}}
{C{H_3}C = C - C{H_3}} \\
{\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_{3\,\,}}\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$

