Question

Explain why the addition of $ ~HI $ to $ 3,3-dimethylbut-1-ene $ gives $ 2-iodo-2,3-dimethyl $ butane as the major product and not the $ 1-iodo-3,3-dimethyl $ butane.

Answer 1

Hint : We know that the addition of $ HI $ to an alkyl halide followed by heating the substance leads to the formation of an alkene and this takes place through an elimination reaction. The hydroxyl atom of the halogen acts as a string base that helps in the extraction of a beta hydrogen atom from the alkyl halide.

Complete Step By Step Answer:
Butane indicates that the parent alkane is butane and contains four carbon atoms. $ 1-iodo $ indicates that the iodine atom is attached to carbon number one of the parent alkane. The chemical symbol for iodine is $ 2,3- $ dimethyl indicates that two methyl groups are attached. One is attached to the carbon number two of the parent alkane and the other is attached at carbon number three. The representation of a molecule in such a way that the bonds are represented by lines, carbon atoms are represented by line ends and intersections and atoms other than carbon and hydrogen are represented by their chemical symbols is known as a bond line formula. We know that the representation of a molecule in such a way that the bonds are represented by lines, carbon atoms is represented by line ends and intersections and atoms other than carbon and hydrogen.
 $ \underset{3,3-Dimethylbut-1-ene}{\mathop{C{{H}_{3}}-C\left( 2C{{H}_{3}} \right)-CH=C{{H}_{2}}}}\,\xrightarrow[+{{H}^{+}}\text{ }]{\text{ }\left( HI \right)}\underset{2{}^\circ Carbocation}{\mathop{C{{H}_{3}}-\left( 2C{{H}_{3}} \right)-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}}}\, $
 $ \underset{2{}^\circ Carbocation\text{ }}{\mathop{C{{H}_{3}}-\left( 2C{{H}_{3}} \right)-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}}}\,\xrightarrow{\text{ }1,2-Methyl\text{ }shift\text{ }}\underset{\begin{smallmatrix}
 3{}^\circ Carbocation\text{ } \\
 \left( More\text{ }Stable \right)
\end{smallmatrix}}{\mathop{C{{H}_{3}}-\overset{+}{\mathop{C}}\,\left( C{{H}_{3}} \right)-CH\left( C{{H}_{3}} \right)-C{{H}_{3}}}}\, $ $ \underset{\begin{smallmatrix}
 3{}^\circ Carbocation\text{ } \\
 \left( More\text{ }Stable \right)
\end{smallmatrix}}{\mathop{C{{H}_{3}}-\overset{+}{\mathop{C}}\,\left( C{{H}_{3}} \right)-CH\left( C{{H}_{3}} \right)-C{{H}_{3}}}}\,\xrightarrow{+{{I}^{-}}}\underset{2-Iodo-2,3-di-methylbutane}{\mathop{C{{H}_{3}}-\left( I \right)C\left( C{{H}_{3}} \right)-CH\left( C{{H}_{3}} \right)-C{{H}_{3}}}}\, $ .

Note :
Remember that the bromination of the alkenes or the alkynes is a non-stereo selective reaction as the bromine molecule being non polar, it adds onto the alkene or the alkyne non-selectively. $ ~HI $ when dissolved in water and that aqueous solution is added to any alkene or alkyne then it behaves as a strong base that adds to the double bond and forms the alcohol from the alkene and alkynes.