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Draw the three alkenes, each of formula C5H10, that will form 2-methylbutane upon hydrogenation?
Unsaturated hydrocarbons undergo reduction reactions in order to create saturated hydrocarbons. In the example above an alkene goes through a reduction to yield an alkane. The reduction reaction is often referred by the name catalytic hydrogenation since the catalyst must be a metal.
Hydrogenation refers to the reaction where hydrogen is added in the presence of unsaturated molecules. This reaction occurs when molecular hydrogen is added to unsaturated molecules with a metal catalyst.
Hydrogenation describes the chemical reaction between unsaturated molecules, and molecular hydrogen. H2 any unsaturated compounds in the presence a catalyst like nickel, palladium, or platinum to produce the corresponding saturated compound.
Four steps can be taken to find the unsaturated reactionant from the saturated product.
1. Use DBE formulae to determine degree of unsaturation for a given molecular formula.
2. Drawing isomers
3. The unsaturated isomers that have been obtained to be hydrogenated
4.isolation and synthesis of unsaturated isomers that are responsible for 2-methylbutane
The DBE (Double bonds equivalence formulae) can be used to determine the number of unsaturation in a molecule.
Below is the formula for calculating DBE
[katex]{\rm{DBE = C – }}\frac{{\rm{H}}}{{\rm{2}}}{\rm{ – }}\frac{{\rm{X}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{N}}}{{\rm{2}}}{\rm{ + 1}}[/katex]
The number of hydrogen atoms in the area is H
The number of halogenatoms in a halogenatom is X
N is the number of nitrogenatoms
The number of carbonatoms in a given carbonatom is C
Isomers are compounds that have the same chemical formula but different order of arrangement. You can draw many isomers from the unsaturated reactants given chemical formulae by using double bond equivalence.
Hydrogenation can be used to reduce alkenes, and other higher unsaturated substances, in order to produce their alkanes. The isomers are obtained by hydrogenation. The target of interest is 2-methyl butane, which is an isomer.
The chemical formula [katex]{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{10}}}}[/katex] is given.
[katex]\begin{array}{l}\\{\rm{Number of carbons,\;\;\;\;}}\,\,\,\,\,{\rm{C}}\,\,{\rm{ = }}\,\,{\rm{5}}\\\\{\rm{Number of hydrogens,\;\;H }}\,{\rm{ = 10}}\\\\{\rm{Number of halogens,\;\;\;}}\,\,{\rm{X}}\,\,\,{\rm{ = }}\,\,{\rm{0}}\\\\{\rm{Number of nitrogens, }}\,\,{\rm{N = 0}}\\\end{array}[/katex]
The double bond equivalent is:
[katex]\begin{array}{l}\\{\rm{DBU = C – }}\frac{{\rm{H}}}{{\rm{2}}}{\rm{ – }}\frac{{\rm{X}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{N}}}{{\rm{2}}}{\rm{ + 1}}\\\\{\rm{ = 5 – }}\frac{{{\rm{10}}}}{{\rm{2}}}{\rm{ – }}\frac{{\rm{0}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{0}}}{{\rm{2}}}{\rm{ + 1}}\\\\{\rm{ = 1}}\\\end{array}[/katex]
All possible isomers for [katex]{{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{10}}}}[/katex]
Hydrogenation 1 pentene
Hydrogenation 2 pentene
Hydrogenation 2M-1-Butene
Hydrogenation 2M2B-Butene
Hydrogenation 3-methyl-1-butene
These are the isomers that give 2-methylbutane
2-methyl-1, 2-methyl-2, and 3-methyl-1 butene. Ans:
These are the three alkenes that give 2-methylbutane after hydrogenation: