Hydrocarbons (Alkane, Alkene and Alkyne) MCQ Questions & Answers in Organic Chemistry | Chemistry

It may be noted that only the terminal alkynes react with ammoniacal silver nitrate. Therefore, this reaction can be used to distinguish between 1-alkynes and others such as alkane, alkenes and non-terminal alkynes.
$$C{H_3} - C{H_2} - C \equiv $$     $$CH + 2\left[ {Ag{{\left( {N{H_3}} \right)}_2}} \right]N{O_3} \to $$       $$\mathop {C{H_3}C{H_2}C \equiv CAg \downarrow }\limits_{{\text{White ppt}}{\text{.}}} $$

Molecular formula

\[B{{r}^{\bullet }}\]  is less reactive and more selective and so the most stable free radical \[\left( {{3}^{\circ }} \right)\]  will be the major product.

In $$neo$$  - pentane all the hydrogen are of same type so it can only give one mono-chloro derivative.
\[\underset{neo\text{-Pentane}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{3}}}}\,\xrightarrow{C{{l}_{2}}}C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{3}}Cl\]

On increasing the number of branches, knocking is decreased and octane number is increased. So, branched chain hydrocarbons have less knocking and is more desirable.