
What are paraffins and olefins? Give one example of each and write chemical reactions to differentiate them.
Answer
481.2k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

