Question

C6H14, calculate the IHD and select allthe types of unsaturation that might be present in the molecule based on the IHD.

Answer 1

Concepts and reason

The molecular formula represents the total number of atoms presented in the compound. In Chemistry, IHD refers to index of hydrogen deficiency which is also called as the degree of unsaturation. IHD (index of hydrogen deficiency) is a formula which is used to determine the chemical structure from the molecular formula.

Fundamentals

The molecular formula gives the information about the total number of atoms present in the compound. The structure of the compound can be determined with the help of an index of hydrogen deficiency (IHD) formula.

${\rm{IHD}}\,\,{\rm{ = }}\,\,\left( {{\rm{number of C atoms}}\,\,{\rm{ - }}\,\frac{{{\rm{number of}}\,{\rm{H atoms}}}}{{\rm{2}}}\,{\rm{ + 1}}} \right)$

C = total number of carbons present in the molecular formula.

H = total number of hydrogen present in the molecular formula.

A cyclic structure, double bond and triple bond in a structure can be determined with the help of the degree of unsaturation values.

IHD = 0, no unsaturation in the molecule. The structure is saturated hydrocarbon. All bonds are single bonds in the compound.

IHD = 1, the structure may contain 1 double bond or 1 ring.

IHD = 2, the structure may contain 2 double bond or 1 ring with 1 double bond.

IHD = 3, the structure may contain 3 double bond or 1 ring with 2 double bond.

IHD = 4, the structure may contain 4 double bond or 2 triple bonda or 1 ring with three double bonds or 2 rings with 2 double bonds or 3 rings with 1 double bond.

The given molecular formula is ${{\rm{C}}_6}{{\rm{H}}_{{\rm{14}}}}$

Number of individual atoms present in the molecule

$\begin{array}{l}\\{\rm{C}}\,\,\,{\rm{ = }}\,\,{\rm{6}}\\\\{\rm{H}}\,\,{\rm{ = }}\,\,{\rm{14}}\\\\{\rm{IHD}}\,\,{\rm{ = }}\,\,\left( {{\rm{number of C}}\, - \frac{{{\rm{number of}}\,{\rm{H}}}}{{\rm{2}}}\,{\rm{ + 1}}} \right)\\\end{array}$

$\begin{array}{l}\\{\rm{IHD}}\,\,{\rm{ = }}\,\,\,\left( {{\rm{6}} - \,\,\frac{{{\rm{14}}}}{{\rm{2}}}\,{\rm{ + 1}}} \right)\\\\{\rm{IHD}}\,\,{\rm{ = }}\,\,\,{\rm{7}} - {\rm{7}}\\\\{\rm{IHD}}\,{\rm{ = }}\,{\rm{0}}\\\end{array}$

IHD = 0

Degree of unsaturation is zero; therefore, the given compound is saturated hydrocarbon. Double bonds, triple bonds and rings will not appear in the structure of the compound. All bonds present in the structure are single bonds only.

Ans:

IHD = 0

• Degree of unsaturation is zero

• Represents saturated hydrocarbon

• All bonds are single bonds in the structure.