Question

For the following formula, C4H9Cl, calculate the IHD and select all the types of unsaturation that might be present in the molecule based on the IHD.

Answer 1

Answer:

The IHD = 0

There is no unsaturation in the compound C4H9Cl

Explanation:

IHD = Index for Hydrogen Deficiency .It determines total number of unsaturation or Pi- bond present in the molecule.It is calculated using the formula:

For Molecule:

$C_{c}H_{h}O_{o}N_{n}X_{x}$

Here . C = carbon

H = Hydrogen

O = Oxygen

N = Nitrogen

X = Halogen

The IHD is calculated as :

$\frac{2c-h-x+n+2}{2}$...............(1)

For example :

C4H8O2

Here c = 4 , h = 8 , o = 2 .

n = x=0

Put the value in equation (1) and find IHD

$\frac{2(4)-8-0+0+2}{2}$

$\frac{8-8+2}{2}$

$\frac{2}{2}$

IHD = 1

For C4H9Cl

c = 4 , h = 9 and  x = 1

IHD for this molecule can be calculated as :

$\frac{2c-h-x+n+2}{2}$

$\frac{2(4)-9-1+2}{2}$

$\frac{8-9-1+2}{2}$

= 0

This means there is no unsaturation in the molecule.

Each bond is sigma bond and properly saturated.