Question

What are paraffins and olefins? Give one example of each and write chemical reactions to differentiate them.

Answer 1

We know that the paraffin refers to many chemicals that are white soft solids and combustible liquid that consists of saturated hydrocarbons, mainly alkane series, obtained from a petroleum refinery and olefin is a very common term used in organometallic compound chemistry.

Complete answer:
As we know that the Paraffin hydrocarbons are known as alkanes. Additional carbon atoms are joined to each other in a long chain alkane molecule by a single covalent bond. Each atom is connected to four hydrogen atoms in order to form four single covalent connections. It is known to you that alkane is the modern IUPAC name for the class of hydrocarbons formerly known as paraffin hydrocarbons or paraffins. Paraffin is a truncation of Latin ‘parum affinum’ meaning less affinity i.e., less reactivity. Alkanes have only single bonds which are sigma covalent. These bonds are pretty strong and hard to break and to break one requires a large amount of energy. Basically, alkanes are pretty stable because of this hence, they are inert substances. +
Meanwhile, the common olefins examples are propane, ethene, butene, and pentene. In the IUPAC naming system, the name of the olefins suffixes with the ene. Olefins are unstable compounds. These are saturated hydrocarbons, which are also called alkanes. They contain carbon-carbon single bonds. There general formula is ${{C}_{n}}{{H}_{2n}}_{+2}.$ Example: Methane $\left( C{{H}_{4}} \right)$ Olefins: Olefins means oil forming. These are unsaturated hydrocarbons containing at least one double bond. These are also called alkenes. Their general formula is ${{C}_{n}}{{H}_{2n}}$ ​ Example: Ethylene $\left( {{C}_{2}}{{H}_{4}} \right).$ The chemical reaction is given by: $\underset{Methane}{\mathop{C{{H}_{4}}}}\,+C{{l}_{2}}\to C{{H}_{3}}Cl+HCl$ and $\underset{Ethylene}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,+C{{l}_{2}}\to \left( C{{H}_{2}}Cl \right)C{{H}_{2}}-Cl$

Note:
Remember that the boiling point of the olefins depends on the number of a carbon atom. It increases with the increase in the number of a carbon atom. The boiling point of the olefins is inversely proportional to the surface area of the compound.