A short fragment of a polymer chain (three repeat units, H at both ends) is drawn. It is made by 1,4-addition of three molecules of a diene. Using degree of unsaturation = rings + pi bonds, find (sum of the degrees of unsaturation of the three separate monomer molecules) minus (degree of unsaturation of the fragment shown). Give the answer as an integer.

Numerical answer type, tolerance 0.5.

Answer: 3

The fragment is cis-1,4-polybutadiene, made from buta-1,3-diene, which has two C=C bonds (degree of unsaturation 2). Three monomers give

3 \times 2 = 6

. The fragment

C X_{12} H X_{20}

has three C=C and no rings, so its degree of unsaturation is 3 (checked with RDKit). The difference is

6 - 3 = 3