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

will not show geometrical isomerism as it has two \[-C{{H}_{3}}\]  groups on the same carbon atom of the double bond.

$$2-methyl-2-butene$$      ( molecular formula $${C_5}{H_{10}}$$  ) yields acetone on ozonolysis.

Anti Markovnikov's addition

\[C{{H}_{3}}C\equiv CH\xrightarrow{HBr}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{C}}\,\]      \[=C{{H}_{2}}\xrightarrow{HBr}\]    \[\underset{\text{2, 2-Dibromopropa}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,Br \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,Br \\ | \end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{3}}}}\,\]

Oxymercuration - demercuration reaction addition of $${H_2}O$$  molecule is according to Markovnikoff's rule.

(i) Chlorination at $$C{\text{ - }}2$$   and $$C{\text{ - }}4$$   produces no chiral compounds
(ii) Chlorination at $$C{\text{ - }}3$$  produces achiral carbon marked with star $$\left( {d\,{\text{and}}\,l\,{\text{form}}} \right).$$
(iii) Chlorination at $$C{\text{ - }}1$$  also produces a chiral carbon marked with star $$\left( {d\,{\text{and}}\,l\,{\text{form}}} \right).$$

\[C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow[{{H}_{2}}{{O}_{2}},\,NaOH]{{{B}_{2}}{{H}_{6}}}\]       \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\]

Geometrical isomerism arisesdue to restricted rotation around $$C = C$$  bond.

From the reaction of benzene with ethylene in the presence of $$anhy.$$  $$AlC{l_3},$$  ethylbenzene is produced.

\[\left( \text{i} \right)C{{H}_{3}}\underset{\begin{smallmatrix} | \\ H\, \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,Br \\ | \end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{2}}C{{H}_{3}}\xrightarrow{alc.\,KOH}\underset{\begin{smallmatrix} \text{2-Butene} \\ \text{(main product}) \\ 'X' \end{smallmatrix}}{\mathop{C{{H}_{3}}CH=CHC{{H}_{3}}}}\,\]