- A$Br_2,\,CCl_4$
- BDil. $KMnO_4 /$ $O{H^\Theta }$
- CConc. $H_2SO_4$
- ✓$AgNO_3$ in $NH_4OH$
Propyne has an acidic hydrogen due to which it readily react with Ammoniacal silver nitrate gives white precipitate while propene does not give any precipitate.
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Assertion $A :$- Carbon forms two important oxides $- CO$ and $CO _2 . CO$ is neutral whereas $CO _2$ is acidic in nature.
Reason $R :$- $CO _2$ can combine with water in a limited way to form carbonic acid, while $CO$ is sparingly soluble in water.
In the light of the above statements, choose the most appropriate answer from the options given below :-
$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH - C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,OH\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
if one mole of each of $X$ and $Y$ with $0.05 mol$ of $Z$ gives compound $XYZ _{3}$. (Given : Atomic masses of $X , Y$ and $Z$ are 10,20 and $30 amu$, respectively). The yield of $XYZ _{3}$ is $.........g$.(Nearest integer)