Question

Out of the following compounds
-Pent-1-ene and Pent-2-ene
-2-methyl but-1-ene and 2-methyl but-2-ene
Which pair has the lowest and the highest heats of combustion, respectively?

Answer 1

Hint: The heat of combustion is defined as the amount as the heat released for complete combustion of a compound in its standard state to form products that are in their standard states.

Complete step by step solution:
The heat of combustion of a compound is dependent upon the stability of the compound. In the first pair of compounds, i.e., Pent-1-ene and Pent-2-ene, the first compound is more substituted than the second one due to the higher number of substituents. Hence the amount of energy required for the combustion of Pent-1-ene will be more than that for Pent-2-ene.
Similarly the heat of combustion required for the conversion of 2-methyl but-2-ene will be less than that for 2-methyl but-1-ene. Because the double bond substitution in the former has more substituents than the later.
Hence the order is: 2-methyl but-1-ene < 2-methyl but-2-ene < Pent-2-ene < Pent-1-ene.

Note:
The calorific value is the total amount of energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. Combustion is always accompanied by the release of heat and hence different organic compounds are burnt for the release of fuels. Some of them are butane, iso-butane, etc. these gases are major components of the natural cooking gas or the compressed natural gas. When methane burns in air it releases about $ -418\text{kJ/ mol} $ of energy.