Question

${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{CH}}_3-{\mathrm{CH}}^{+}-{\mathrm{CH}}_2-{\mathrm{CH}}_3\to$Major product. Identify the major product formed in the following reactions.

• A. ${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3$
• B. ${\mathrm{CH}}_3-{\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_3$
• C. ${\mathrm{CH}}_3-{\mathrm{CH}}_3=\mathrm{CH}-{\mathrm{CH}}_3$
• D. ${\mathrm{CH}}_3-\underset{\underset{{\mathrm{OSO}}_3\mathrm{H}}{|}}{\mathrm{CH}}={\mathrm{CH}}_2-{\mathrm{CH}}_3$

Answer 1

The correct option is D

${\mathrm{CH}}_3-\underset{\underset{{\mathrm{OSO}}_3\mathrm{H}}{|}}{\mathrm{CH}}={\mathrm{CH}}_2-{\mathrm{CH}}_3$

The explanation for the correct option:

Option (D): ${\mathrm{CH}}_3-\underset{\underset{{\mathrm{OSO}}_3\mathrm{H}}{|}}{\mathrm{CH}}={\mathrm{CH}}_2-{\mathrm{CH}}_3$

1. In organic chemistry, an electrophilic addition reaction occurs when a double or triple bond in a chemical molecule is broken, resulting in the production of two new bonds.
2. When alkenes react with concentrated sulfuric acid Alkyl hydrogen sulfates are formed.
3. For a symmetrical alkene result is the same, thus it doesn't matter which end of the double bond is used for the new Hydrogen bond.
4. The probability of hydrogen contributing to one side or the other is equal.
5. When Sulfuric acid reacts with an unsymmetrical alkene, it might react in two ways.
6. Depending on which carbon atom the hydrogen connects to, we will end up with one of two products.
7. Otherwise for symmetrical alkene reaction takes place normally
8. However, the primary product adheres to Markovnikov's Rule, which states that when a compound $\mathrm{HX}$ is added to an unsymmetrical alkene, the hydrogen attaches to the carbon that already has the most hydrogens connected to it.
9. Given alkene is a symmetrical alkene so the results will be the same from both sides.
10. Protonation of a carbocation in butene is as follows i.e. A conjugate acid is formed by adding a proton (or hydron, or hydrogen cation) to an atom, molecule, or ion ${\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{H}}{|}}{\mathrm{C}}=\stackrel{\stackrel{\mathrm{H}}{|}}{\mathrm{C}}-{\mathrm{CH}}_3\rightleftharpoons {\mathrm{CH}}_3-\underset{\underset{\mathrm{H}}{|}}{\stackrel{\stackrel{\mathrm{H}}{|}}{\mathrm{C}}}-\underset{+}{\stackrel{\stackrel{\mathrm{H}}{|}}{\mathrm{C}}}-{\mathrm{CH}}_3$
11. Sulfonation is the process of substituting hydrogen with sulfonic acid (${\mathrm{SO}}_3$) and forming an electrophile.
12. The electrophile in sulfuric acid is the slightly positive hydrogen atom, which is strongly attracted to the electrons in the pi bond.
13. The pi bond's electrons travel down to the somewhat positive hydrogen atom.
14. The Hydrogen-Oxygen bond's electrons are pulled down until they are fully on the Oxygen atom, resulting in a negative ion.
15. The lone pair of electrons on the Oxygen atom is strongly attracted to the positive Carbon and advances towards it until a bond is formed.
16. Hence it is the correct option.

Therefore, the product formed is ${\mathrm{CH}}_3-\underset{\underset{{\mathrm{OSO}}_3\mathrm{H}}{|}}{\mathrm{CH}}={\mathrm{CH}}_2-{\mathrm{CH}}_3$