Degree of Unsaturation Formula: Equation, Calculation, Solved Examples

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Degree of Unsaturation Formula is also termed as Index of Hydrogen Deficiency (IHD). This is an easier way of calculating the number of multiple bonds or rings in an unknown chemical structure. There are multiple styles for determining the structure of an organic emulsion and for knowing about the structural parcels of the emulsion. Determining the degree of achromatism is one way to identify the structure of the organic emulsion. It helps you also to double-check the number of \(\pi\) bonds and/ or presence of any cyclic rings. 

What is the Degree of Unsaturation Formula?

Impregnated composites are bones having a single bond. This single bond in a logged emulsion shouldn't be part of any cyclic or ring structure. On the other hand, an unsaturated emulsion is a bone that has a double bond (s), triadic bond (s), and/ or ring (s). The alkanes having single bonds are classified as impregnated whereas the alkenes and alkynes having double and triadic bonds independently are classified as unsaturated hydrocarbons. 

Degree of Unsaturation Formula

Degree of Unsaturation Formula

Calculating the degree of unsaturation gives you information about the total number of pi bonds and rings that can be present within a patch which can help you in determining the molecular structure rather than using precious spectroscopic ways. The degree of unsaturation formula is given below in simplest terms as -

DU = Degree of Unsaturation = number of pi bonds number of rings 

How to Find a Degree of Unsaturation?

The degree of unsaturation formula takes the number of carbon titles (C), number of nitrogen titles (N), number of halogen titles (X), and the number of hydrogen titles (H) as input values. These values can be fluently obtained from the molecular formula of the emulsion. For how to calculate the degree of unsaturation, the formula is given as-

DU = ( (2C 2) N-X-H)/ 2 

The input values are a suggestion of the number of hydrogen titles that should be present for the given emulsion to be classified as impregnated. An impregnated patch contains only single bonds and no rings. In other words, an impregnated form of a hydrocarbon will have the maximum number of hydrogen titles in an acyclic alkane form. 

Hence, the 2C 2, in the formula is the total number of hydrogen titles that should be bound to the logged carbon titles and H is the number of hydrogen titles that are present in the emulsion. The reason that the value of halogen titles (X) is abated is that it replaces an equal number of hydrogen i.e. one halogen snippet for one hydrogen snippet. Piecemeal from that, nitrogen brings in further hydrogen titles and because of that, the value is added. Oxygen and Sulphur don't feel to have any effect in changing the achromatism status of an emulsion. 

The information obtained from the degree of unsaturation can be characterized as 1 value of DU gives 1 king or 1 double bond. For a triadic bond, 2 is the minimal DU value. The following two simple exemplifications will illustrate the use and understanding of how to find the degree of unsaturation and the degree of unsaturation formula. 

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Things to Remember

  • The input values are a suggestion of the number of hydrogen titles that should be present for the given emulsion to be classified as impregnated. An impregnated patch contains only single bonds and no rings.
  • To know how to find the degree of unsaturation of a given emulsion formula is therefore given which includes these characteristics as part of the formula. Hence, the degree of unsaturation formula helps you in determining whether an emulsion is impregnated or unsaturated.
  • Generally, saturation is considered as the point when a result can dissolve no further of the substance which is added to it. In terms of the degree of unsaturation, as mentioned over, it gives information about the cling of motes i.e. it tells you whether a patch has a single bond, double bond, triadic bond, or a cyclic ring structure. 

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Sample Questions

Ques. What's the Degree of Unsaturation of Benzene? (1 Mark)

Ans. The molecular formula of benzene is C6H6. Therefore, it has six carbon titles and it needs 8 further hydrogen titles to be classified as impregnated. Thus, the degree of unsaturation of benzene is 4, which gives one ring and three double bonds. 

Ques. How to use the Degree of Unsaturation Calculator? (3 Marks)

Ans.  The procedure to use the degree of unsaturation calculator is as follows 

  1. Enter the molecular formula in the input field 
  2. Now click the button “ Calculate Degree of Unsaturation” to get the result 
  3. Eventually, the degree of unsaturation for the given molecular formula will be displayed in the affair field 

Ques. What's Meant by Degree of Unsaturation? (3 Marks)

Ans. The Degree of Unsaturation is defined as the sum of the number of the rings involved plus the total number of multiple bonds present in an organic emulsion. The molecular formula is abridged to CnHm and the degree of unsaturation formula is given by -

Degree of Unsaturation = ( Total Number of Double Bonds) (2 x Total Number of Triple Bonds) ( Total Number of Rings) 

The notion of the degree of unsaturation of an organic emulsion is deduced simply from the tetravalency of carbon. Any emulsion whose chemical formula has two hydrogens lower than the maximum number possible (2n 2) must have one ring or one double bond. 

Ques. How to Determine the Degrees of Unsaturation of a Patch? (3 Marks)

Ans. The degrees of unsaturation in a patch are cumulative — a patch with one double bond has one degree of unsaturation, a patch with two double bonds has two degrees of unsaturation, and so forth. 

Just as the conformation of a double bond causes two hydrogens to be lost, the conformation of a ring also results in the loss of two hydrogens, so every ring in the patch also adds one degree of unsaturation. 

For every triadic bond, two degrees of unsaturation are added to a patch, because a patch must lose four hydrogens to make a triadic bond. 

CBSE CLASS XII Related Questions

1.

Give the IUPAC names of the following compounds:

(i)CH3CH(Cl)CH(Br)CH3

(ii)CHF2CBrClF

(iii)ClCH2C≡CCH2Br

(iv)(CCl3)3CCl

(v)CH3C(p-ClC6H4)2CH(Br)CH3

(vi)(CH3)3CCH=CClC6H4I-p

      2.
      Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: 
      (i) Fe3+ (aq) and I- (aq) 
      (ii) Ag+ (aq) and Cu(s) 
      (iii) Fe3+(aq) and Br-(aq) 
      (iv) Ag(s) and Fe3+(aq) 
      (v) Br2 (aq) and Fe2+(aq).

          3.
          A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

              4.

              Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

                  5.

                  Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. 
                  \((i) Methanal \)
                  \((ii) 2-Methylpentanal \)
                  \((iii) Benzaldehyde \)
                  \((iv) Benzophenone \)
                  \((v) Cyclohexanone \)
                  \((vi) 1-Phenylpropanone \)
                  \((vii) Phenylacetaldehyde \)
                  \((viii) Butan-1-ol \)
                  \((ix) 2, 2-Dimethylbutanal\)

                      6.

                      Write down the electronic configuration of:
                      (i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+ 
                      (ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+

                          CBSE CLASS XII Previous Year Papers

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