MCQ
Which will undergo reaction with ammoniacal $AgN{O_3}$
- A
- ✓$C{H_3} - CH = CH - C \equiv CH$
- C$C{H_3} - C{H_2} - CH = CH - C{H_2} - C{H_3}$
- D$C{H_2} = CH - C{H_2} - C{H_3}$
($H$ atom attached to triple bond) is present therefore it gives reaction with ammoniacal $AgN{O_3}$.
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$1.$ The compound ${X}$ is
$(A)$ $\mathrm{NaNO}_3$ $(B)$ $\mathrm{NaCl}$ $(C)$ $\mathrm{Na}_2 \mathrm{SO}_4$ $(D)$ $\mathrm{Na}_2 \mathrm{~S}$
$2.$ The compound ${Y}$ is
$(A)$ $\mathrm{MgCl}_2$ $(B)$ $\mathrm{FeCl}_2$ $(C)$ $\mathrm{FeCl}_3$ $(D)$ $\mathrm{ZnCl}_2$
$3.$ The compound ${Z}$ is
$(A)$ $\left.\mathrm{Mg}_2 \mid \mathrm{Fe}(\mathrm{CN})_6\right]$ $(B)$ $\operatorname{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
$(C)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$ $(D)$ $\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
Give the answer question $1,2$ and $3.$
$C{H_3} - \mathop {\mathop {C = }\limits_{|\,\,\,\,} }\limits_{C{l_{}}\,} \mathop {\mathop {C\, - }\limits_{|\,\,\,\,\,\,} }\limits_{C{H_3}\,} \mathop {\mathop {CH - \,}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{{C_2}{H_5}\,} C{H_2} - C \equiv CH$