Question

Polymerisation of buta-1,3-diene by free radical mechanism gives _________.
A. trans-1,4-Polybutadiene
B. cis-1,4-Polybutadiene
C. Polyvinyl polythene
D. Polyallyl polythene

Answer 1

Hint: Polymerisation occurs through centres of radical formation. There are four radical centres in buta-1,3-diene as it has two double bonds present. The structure of buta-1,3-diene is

Just form the radicals and make the bonds between the radicals to produce three polymers of buta-1,3-diene.
Complete answer:
Free radical mechanism is a chain terminating type of polymerization. It occurs through formation of radicals at high temperature and pressure. Let us start forming the polymers.

S. No.	Centres through which polymerization will take place	Mechanism of polymerization	Structures of polymers	Name of polymers
1	Carbon -1,2 or ${{\text{C}}_{1}}-{{\text{C}}_{2}}$	In this, the double bond between ${{\text{C}}_{1}}-{{\text{C}}_{2}}$ will break into radicals and other molecules of buta-1,3-diene will also break from the same centre into radicals and then radicals of every other molecule will form single bond with each other and there is no breaking of another double bond.		The name of the polymer is polyvinyl polythene.
2	Carbon -1,4 or ${{\text{C}}_{1}}-{{\text{C}}_{4}}$	In this, all the double bonds breaks into radicals and radicals between ${{\text{C}}_{1}}-{{\text{C}}_{4}}$ will form single bond with other molecules of buta-1,3-diene will also break from the same centre into radicals. But here, the radicals at ${{\text{C}}_{2}}-{{\text{C}}_{3}}$ join together to form a bond and single bond was already present so, double bond is made. The hydrogen atoms are on the same side on the double bond, so ‘cis’ polymer is formed.		The name of the polymer is cis-1,4-Polybutadiene.
3	Carbon -1,4 or ${{\text{C}}_{1}}-{{\text{C}}_{4}}$	In this, all the double bonds breaks into radicals and radicals between ${{\text{C}}_{1}}-{{\text{C}}_{4}}$ will form single bond with other molecules of buta-1,3-diene will also break from the same centre into radicals. But here, the radicals at ${{\text{C}}_{2}}-{{\text{C}}_{3}}$ join together to form a bond and single bond was already present so, double bond is made. The hydrogen atoms are on the opposite side of the double bond, so a ‘trans’ polymer is formed.		The name of the polymer is trans-1,4-Polybutadiene.

The correct answer to this question is option ‘a’, ‘b’ and ‘c’.

Note: Do not form same type of polymers again and again by joining radicals formed at ${{\text{C}}_{1}}-{{\text{C}}_{2}}$ centre from the either side as the structure is symmetrical at ${{\text{C}}_{1}}-{{\text{C}}_{2}}$. So, be careful while forming polymers. The compounds formed will not be called as isomers as they are identical.