Complete Question:
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Answer:
Value of IHD = 2
Types of unsaturation that might be present is :Three double bonds or three ring closings or two double bonds and one ring closing or one double bond and two ring closings or one triple bond and one double bond or one triple bond and one ring closing
Explanation:
In order to get a better understanding of this answer we need to know that this question is based on the concept of Hydrogen Deficiency (IHD).IHD can be defined as the degree of unsaturation in a molecule,it is common to double equivalent.
IHD is simply a measurement of unsaturation in a hydrocarbon and this tells us that the more the unsaturation in a hydrocarbon the more the deficiency of hydrogen.if the IHD is one it means the presence of double bond or one cycle in the structure.
The formula for calculating IHD is given below,
IHD = [tex] (2c+2-h-z+n)÷2[/tex]
Here , c is the number of carbon atoms,h is the number of hydrogen atoms,x is the number of halogen atoms and n is the number of nitrogen atom
Considering the molecule C₇H₁₀NBr
from the molecule
c = 7 , h = 10 ,Br = halogen = 1 and N = 1
Substituting these values into the equation we have
IHD = [tex] (2(7)+2-(10)-(1)+(1))÷2[/tex]
= 3
From our calculation IHD is 3 it means that the there are unsaturation or ring closings(i.e cycles).One unsaturation or one ring closing is not possible because the value of IHD is higher than one and also it cannot be benzene because the IHD value of benzene is 4.
From our calculation the value of IHD is 3. Hence, the types of bond that might be present in the molecule include three bonds or three ring closings or double bond and two ring closings or one triple bond and one ring closings or one triple bond and one ring closing.
