
Which one of the following is a polyamide polymer?
A. Terylene
B. Nylon – 6
C. Buna – S
D. Bakelite
Answer
517.2k+ views
Hint: Amide is a carbonyl group attached beside an amine group, with the formula, \[CON{{H}_{2}}\]. Repetitive units of this amide group result in a polyamide. Polymers are substances having continuous units of a single small unit called as a monomer. A polyamide polymer will have an amide linkage in the monomer.
Complete answer:
Polyamide polymers are the type of polymers whose monomers consist of an amide linkage. These types of polymers are an example of condensation polymerization. It is that type of polymerization in which two monomers condense together with a loss of simple molecules.
We have been given several polymers, among them we have to find that polymer that has a polyamide linkage.
Terylene is a polymer called Dacron, that has ethylene glycol and terephthalic acid as monomers, so they do not consist of an amide linkage. While Buna – S and Bakelite also does not consist of monomers having amide linkage. So, Nylon – 6 has amide linkage.
Nylon – 6 is a polymer formed from the monomer of Caprolactum. Caprolactum is a 6 – carbon atom ring structure, consisting of an amine and carbonyl group, it has formula, ${{C}_{6}}{{H}_{11}}NO$ , which has an amide linkage. Various units of this caprolactam join together and the reaction of homo condensation carries out at high temperature in presence of water that result in removal of water molecules to form Nylon – 6 polymer, which is also called as perlon. The reaction is:
${{C}_{6}}{{H}_{11}}NO\xrightarrow[{{H}_{2}}O]{533-543K}{{H}_{3}}\overset{+}{\mathop{N}}\,-{{(C{{H}_{2}})}_{5}}-CO{{O}^{-}}\xrightarrow{-{{H}_{2}}O}{{(-HN-{{(C{{H}_{2}})}_{5}}-\overset{\overset{O}{\mathop{||}}\,}{\mathop{C}}\,-)}_{n}}$
Hence, Nylon – 6 is a polyamide polymer, so option B is correct.
Note:
Other polymers Buna – S has buta-1,3-diene, and styrene as monomers that does not contain an amide linkage, while Bakelite is the urea – formaldehyde resin, that has urea and formaldehyde as monomers, so no amide linkage. Terylene is a type of polyester, consisting of ester linkages.
Complete answer:
Polyamide polymers are the type of polymers whose monomers consist of an amide linkage. These types of polymers are an example of condensation polymerization. It is that type of polymerization in which two monomers condense together with a loss of simple molecules.
We have been given several polymers, among them we have to find that polymer that has a polyamide linkage.
Terylene is a polymer called Dacron, that has ethylene glycol and terephthalic acid as monomers, so they do not consist of an amide linkage. While Buna – S and Bakelite also does not consist of monomers having amide linkage. So, Nylon – 6 has amide linkage.
Nylon – 6 is a polymer formed from the monomer of Caprolactum. Caprolactum is a 6 – carbon atom ring structure, consisting of an amine and carbonyl group, it has formula, ${{C}_{6}}{{H}_{11}}NO$ , which has an amide linkage. Various units of this caprolactam join together and the reaction of homo condensation carries out at high temperature in presence of water that result in removal of water molecules to form Nylon – 6 polymer, which is also called as perlon. The reaction is:
${{C}_{6}}{{H}_{11}}NO\xrightarrow[{{H}_{2}}O]{533-543K}{{H}_{3}}\overset{+}{\mathop{N}}\,-{{(C{{H}_{2}})}_{5}}-CO{{O}^{-}}\xrightarrow{-{{H}_{2}}O}{{(-HN-{{(C{{H}_{2}})}_{5}}-\overset{\overset{O}{\mathop{||}}\,}{\mathop{C}}\,-)}_{n}}$
Hence, Nylon – 6 is a polyamide polymer, so option B is correct.
Note:
Other polymers Buna – S has buta-1,3-diene, and styrene as monomers that does not contain an amide linkage, while Bakelite is the urea – formaldehyde resin, that has urea and formaldehyde as monomers, so no amide linkage. Terylene is a type of polyester, consisting of ester linkages.
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