Question
$$36.4\,g$$ of 1, 1, 2, 2-tetrachloropropane was heated with zinc dust and the product was bubbled through ammoniacal $$AgN{O_3}.$$ What is the weight of precipitate obtained?
A.
30.0 $$g$$
B.
29.4 $$g$$
C.
28.0 $$g$$
D.
25.7 $$g$$
Answer :
29.4 $$g$$
Solution :
The dehalogenation of vicinal tetrahaloalkane gives alkynes. It is done by heating the compound with $$Zn$$ dust.
\[\underset{\begin{smallmatrix}
\text{1, 1, 2, 2-tetrachloropropane} \\
\text{Molecular weight = 182 }
\end{smallmatrix}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix}
| \\
Cl\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
\,Cl \\
|
\end{smallmatrix}}{\mathop{-\,\,C\,\,-}}}\,\underset{\begin{smallmatrix}
|\,\,\,\,\,\, \\
Cl\,\,\,\,\,\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
Cl\,\,\, \\
|\,\,\,\,\,
\end{smallmatrix}}{\mathop{CH}}}\,\xrightarrow{Zn/\Delta }}}\,\] \[\underset{\text{Propyne}}{\mathop{C{{H}_{3}}-C\equiv C-H}}\,\]
\[C{{H}_{3}}C\equiv C-H+AgN{{O}_{3}}\xrightarrow[\Delta ]{N{{H}_{4}}OH}\] \[\underset{\begin{smallmatrix}
\text{Silver propynide (ppt}\text{.)} \\
\left( Mol.\,wt\text{ }=\text{ }147 \right)
\end{smallmatrix}}{\mathop{C{{H}_{3}}-C\equiv C-Ag}}\,+N{{H}_{4}}N{{O}_{3}}+{{H}_{2}}O\]
\[\because \,182\,g\] of tetrachloro alkane gives \[147\,g\] of ppt.
\[\therefore \,36.4\,g\] tetrachloro alkane gives \[\frac{147\times 36.4}{182}=29.4\,g\] of precipitate