- A$35$
- B$37$
- ✓$38$
- D$36$
Iron pentacarbonyl has $EAN$ number of $36= Z - X + Y =(26-0+2 x )$
[ $Z =$ atomic number, $X =$ oxidation state of metal, $Y =$ total electrons donated by ligand]
$\therefore x =5$
So, the formula will be $Fe ( CO )_5$.
Hence it exists in $Fe ( CO )_5$ form.
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$(I)$ $C{H_2} = CH\mathop C\limits^ + HC{H_3}$
$\begin{array}{*{20}{c}} {{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\ {(II)\,\,\,\,\,\,\,\,\,\,C{H_2} = C - \mathop {{\text{ }}C}\limits^ + {H_2}} \end{array}$
$(III)$ $C{H_3}CH = CH\mathop C\limits^ + {H_2}$
$(V=33\, volt)$ $(m_p = 1.672 \times 10^{-27} \,kg)$
$(i)\,CH_3CH_2OH$
$(ii)\ CH_3COCH_3$
$(iii)\ \begin{array}{*{20}{c}}
{C{H_3} - CHOH} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(iv) \,CH_3OH$
Which of the above compound(s), on being warmed with iodine solution and $NaOH,$ will give iodoform ?
Above reaction is an example of $1,4$ -elimination. Predict the product.