
Double bond equivalent of the following is:
A. 7
B. 11
C. 6
D. None of these

Answer
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Hint: The double bond equivalent or level of unsaturation is the number of unsaturation present in an organic molecule. We can find this by the formula:
\[DBE=C+\dfrac{H}{2}-\dfrac{X}{2}+\dfrac{N}{2}\]
Complete step by step solution:
The term unsaturation here means a double bond or a ring system. For example, in benzene there are 3 double bonds and 1 ring which gives us 4 DBE.
Moreover, a triple bond can be regarded as DBE=2.
It must be noted that in the formula
\[DBE=C+\dfrac{H}{2}-\dfrac{X}{2}+\dfrac{N}{2}\]
Presence of an oxygen atom does not affect the DBE calculation.
Now, in this compound the number of Carbon atoms are 14.
Therefore, from the above solution we can conclude that the correct answer is (b).
Note: Remember, the degrees of unsaturation only give the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings and one double bond, 1 ring and two double bonds, 1 ring and triple bond, 1 double bond and one triple bond, or 3 double bonds.
\[DBE=C+\dfrac{H}{2}-\dfrac{X}{2}+\dfrac{N}{2}\]
Complete step by step solution:
The term unsaturation here means a double bond or a ring system. For example, in benzene there are 3 double bonds and 1 ring which gives us 4 DBE.
Moreover, a triple bond can be regarded as DBE=2.
It must be noted that in the formula
\[DBE=C+\dfrac{H}{2}-\dfrac{X}{2}+\dfrac{N}{2}\]
where, X is the total number of halogens, namely Cl, Br, F and I present in the structure.
C = number of carbon atoms.
H = number of Hydrogen atoms.
N = number of nitrogen atoms.
Presence of an oxygen atom does not affect the DBE calculation.
Now, in this compound the number of Carbon atoms are 14.
Number of Hydrogen atoms is 9.
Number of Nitrogen atoms is 1.
Number of halogen atoms is 0
On applying the DBE formula:
\[=14+1-\dfrac{9}{2}+\dfrac{1}{2}\]
= 15 - 4
= 11
Therefore, from the above solution we can conclude that the correct answer is (b).
Note: Remember, the degrees of unsaturation only give the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings and one double bond, 1 ring and two double bonds, 1 ring and triple bond, 1 double bond and one triple bond, or 3 double bonds.
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