Composition, Properties and Uses of Polymer Questions in English

(D) Neoprene is a synthetic rubber formed by the free-radical polymerization of chloroprene ($2$-chloro-$1,3$-butadiene).
It is an addition polymer,not a condensation polymer.
It is highly resistant to petroleum,vegetable oils,and light.
Therefore,the statement that it is a condensation polymer is incorrect.

(B) $PVC$ (Polyvinyl chloride) is widely used in the manufacturing of floor tiles,pipes,and various construction materials due to its durability and chemical resistance.

(C) Buna-$S$ is a copolymer of $1,3-$butadiene and styrene.
It is obtained by the polymerization of $1,3-$butadiene and styrene in the ratio of $3:1$ in the presence of a catalyst (often sodium or free radical initiator).
The reaction is represented as:
$nCH_2=CH-CH=CH_2 + nC_6H_5CH=CH_2 \xrightarrow{\text{Polymerization}} (-CH_2-CH=CH-CH_2-CH(C_6H_5)-CH_2-)_n$

(B) $SBR$ (Styrene-Butadiene Rubber) is a synthetic rubber used extensively in the manufacture of automobile tyres due to its high abrasion resistance and durability.

(B) $LDPE$ (Low Density Polyethylene) is used to obtain shopping bags because of its flexibility and toughness.

(A) The copolymerization of isobutylene ($2$-methylpropene) with a small amount of isoprene ($2$-methyl-$1,3$-butadiene) in the presence of anhydrous $AlCl_{3}$ at low temperatures (around $-100^{\circ} C$) yields Butyl rubber.
The chemical reaction is as follows:
$n \ CH_{2}=C(CH_{3})_{2} + n \ CH_{2}=C(CH_{3})-CH=CH_{2}$ $\xrightarrow{AlCl_{3}} -[CH_{2}-C(CH_{3})_{2}-CH_{2}-C(CH_{3})=CH-CH_{2}]_{n}-$
Thus,the correct option is $A$.

(C) Orlon (polyacrylonitrile) is formed by the addition polymerization of acrylonitrile monomers.
Nylon-$6, 6$,polyester,and urea-formaldehyde polymers are formed by condensation polymerization,which involves the elimination of small molecules like $H_2O$ or $CH_3OH$.

(C) The $-CO-NH-$ linkage is known as the amide linkage or peptide linkage.
$Caprolactam$ is a cyclic amide which,upon ring-opening polymerization,forms $Nylon-6$. The structure of $Caprolactam$ contains the $-CO-NH-$ group within its ring.
Vinyl cyanide $(CH_2=CH-CN)$ contains a nitrile group.
Dimethyl terephthalate contains ester linkages $(-COO-)$.
Hexamethylenediamine $(H_2N-(CH_2)_6-NH_2)$ contains amine groups.

(D) Natural rubber is a linear polymer of isoprene ($2-$methyl$-1,3-$butadiene).
It is formed by the polymerization of isoprene units.
The reaction is as follows:
$n \ CH_2=C(CH_3)-CH=CH_2 \xrightarrow{\text{Polymerization}} [CH_2-C(CH_3)=CH-CH_2]_n$
Thus,the monomer of natural rubber is $2-$methyl$-1,3-$butadiene.

(D) $Dacron$ (also known as $Terylene$) and $PHBV$ $(Poly-\beta-hydroxybutyrate-co-\beta-hydroxyvalerate)$ are polyesters,meaning they contain ester linkages $(-COO-)$.
$Nylon-2-nylon-6$ is a polyamide copolymer formed from glycine and amino caproic acid,which contains amide linkages $(-CONH-)$ instead of ester linkages.

(D) Buna-$S$ or styrene butadiene rubber $(SBR)$ is a copolymer formed by the polymerization of styrene $(C_6H_5CH=CH_2)$ and $1,3-$butadiene $(CH_2=CH-CH=CH_2)$.
The reaction is as follows:
$n(C_6H_5CH=CH_2) + n(CH_2=CH-CH=CH_2) \xrightarrow{Peroxide} [-CH(C_6H_5)-CH_2-CH_2-CH=CH-CH_2-]_n$
Thus,the correct answer is $SBR$.

(B) $PVC$ (Polyvinyl chloride) is widely used in the manufacturing of synthetic leather,often referred to as artificial leather or rexine. It is durable,water-resistant,and cost-effective compared to natural leather.

(B) Bakelite is a thermosetting polymer formed by the condensation reaction of phenol and formaldehyde.
It is a poor conductor of heat and electricity,which makes it ideal for manufacturing electrical switches,plugs,and handles of cooking utensils like cookers.

(D) Novolac is a linear polymer of phenol and formaldehyde. It is used as a constituent in the preparation of paints and as a binder in laminated wooden planks.

(B) The copolymerization of lactic acid $(CH_3CH(OH)COOH)$ and glycolic acid $(HOCH_2COOH)$ leads to the formation of a biodegradable polymer known as Dexon (also referred to as polyglycolide-co-lactide or $PLGA$). The reaction involves the condensation of these hydroxy acids with the elimination of water molecules. The structure of the resulting polymer is: $[ -O-CH(CH_3)-CO-O-CH_2-CO- ]_n$.

(A) Bakelite is a thermosetting polymer formed by the condensation polymerization of phenol and formaldehyde. Because it is a poor conductor of heat and electricity and is resistant to heat,it is widely used for making handles of cookers,frying pans,and electrical switches.

(A) Lexan is a type of polycarbonate polymer known for its high impact resistance and durability,which is why it is used for making bulletproof helmets.

(B) $Nylon-6,6$ is a condensation polymer formed by the reaction between hexamethylene diamine $(nH_2N-(CH_2)_6-NH_2)$ and adipic acid $(nHOOC-(CH_2)_4-COOH)$.
Since the repeating units are linked by amide bonds $(-CONH-)$,it is classified as a polyamide.

(C) Natural rubber is a linear polymer of isoprene ($2-$methyl-$1,3-$butadiene).
It is formed by the polymerization of isoprene units.
The reaction is as follows:
$nCH_2=C(CH_3)-CH=CH_2 \xrightarrow{\text{Polymerization}} [CH_2-C(CH_3)=CH-CH_2]_n$

(C) Nylon-$66$ is a condensation polymer formed by the reaction between hexamethylenediamine and adipic acid.
$n HOOC(CH_2)_4COOH + n H_2N(CH_2)_6NH_2 \rightarrow [NH(CH_2)_6NHCO(CH_2)_4CO]_n + 2n H_2O$

(D) Terylene (Dacron) is a polyester fibre which is prepared by the condensation polymerisation of ethylene glycol and terephthalic acid with the elimination of water molecules.
The chemical reaction is as follows:
$n(HO-CH_2-CH_2-OH) + n(HOOC-C_6H_4-COOH) \xrightarrow{\Delta, -nH_2O} [-O-CH_2-CH_2-O-CO-C_6H_4-CO-]_n$
This reaction takes place at approximately $425-475 \ K$.

(D) Nylon$-66$ is a polyamide fibre which is manufactured by the condensation polymerisation of adipic acid and hexamethylene diamine. Since it is both a condensation polymer and a polyamide,the statement 'Nylon$-66$ is not a' is false for both options $(a)$ and $(b)$. Therefore,the correct answer is $(d)$.

(C) Fibres are thread-forming solids that possess high tensile strength and high modulus. These characteristics are due to strong intermolecular forces like hydrogen bonds or dipole-dipole interactions.
Among the given polymers:
$(i)$ Teflon is an elastomer/plastic.
$(ii)$ Starch is a natural polymer (carbohydrate).
$(iii)$ Terylene is a polyester fibre.
$(iv)$ Orlon (polyacrylonitrile) is a synthetic fibre.
Therefore,the correct set is $(iii)$ and $(iv)$.

(C) The biodegradable polymer nylon-$2$-nylon-$6$ is an alternating polyamide copolymer of glycine $[H_2N-CH_2-COOH]$ and aminocaproic acid $[H_2N-(CH_2)_5-COOH]$.

(D) $PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a biodegradable polymer that belongs to the family of polyhydroxyalkanoates $(PHAs)$.
It is synthesized by the copolymerization of two monomeric units,namely $3-$hydroxybutanoic acid and $3-$hydroxypentanoic acid.

(C) Low density polythene $(LDPE)$ is prepared by the polymerization of ethene under high pressure and temperature in the presence of traces of dioxygen or a peroxide initiator.
This process involves free radical addition and $H$-atom abstraction,which results in a highly branched structure.

(D) The correct matches are:
$1$. Caprolactam polymerizes to form Nylon-$6$,which has the structure $(-CO-(CH_2)_5-NH-)_n$ $(d)$.
$2$. Vinyl chloride polymerizes to form Polyvinyl chloride $(PVC)$,which has the structure $(-CH_2-CH(Cl)-)_n$ $(c)$.
$3$. Styrene polymerizes to form Polystyrene,which has the structure $(-CH_2-CH(C_6H_5)-)_n$ $(b)$.
$4$. Propene polymerizes to form Polypropylene,which has the structure $(-CH_2-CH(CH_3)-)_n$ $(a)$.
Thus,the correct match is $1-d, 2-c, 3-b, 4-a$.

(B) The correct pair is $Bakelite$ and $Novolac$.
$Bakelite$ is formed by the polymerization of $phenol$ and $formaldehyde$ monomers.
$Teflon$ is the polymer of $tetrafluoroethene$.
$Polyester$ is the polymer of $ethylene glycol$ and $terephthalic acid$.
$Nylon 6$ is the polymer of $caprolactum$.

(D) The correct answer is $(D)$.
Nylon-$6$ is prepared by the polymerization of caprolactam.
The monomer,caprolactam,contains a seven-membered ring with one nitrogen atom and six carbon atoms.
In the repeating unit of Nylon-$6$,which is $-[NH-(CH_2)_5-CO]-$,there are $5$ methylene $(-CH_2-)$ groups.

(B) $Cis-1,4-polyisoprene$ is the chemical name for natural rubber. It is a polymer of $2-methyl-1,3-butadiene$ (isoprene). The structure is as follows:
$[-CH_2-C(CH_3)=CH-CH_2-]_n$
Due to the $cis$ configuration of the double bonds,the polymer chains are coiled and exhibit elastic properties,making it natural rubber.

(B) $Nylon-6,6$ is a polyamide polymer because it contains amide linkages $(-CONH-)$ formed by the condensation polymerization of hexamethylenediamine and adipic acid.

(B) The statement in option $B$ is incorrect because natural rubber (cis$-1,4-$polyisoprene) has a 'cis' configuration at every double bond,not a 'trans' configuration. Gutta-percha is the 'trans' isomer of natural rubber.
Buna-$S$ is a co-polymer of $1,3$-butadiene and styrene,which is correct.
Vulcanisation makes rubber harder and stronger by forming cross-links between polymer chains.
Natural rubber is indeed a $1,4$-polymer of isoprene ($2$-methyl-$1,3$-butadiene).

(C) Orlon is a polymer of acrylonitrile,which is also known as vinyl cyanide.
The polymerization reaction is as follows:
$nCH_2=CH(CN) \xrightarrow{\text{Peroxide}} -(CH_2-CH(CN))_n-$
Thus,the monomeric unit of Orlon is vinyl cyanide.

(D) When a mixture of $1,3-$butadiene and styrene in the ratio of $3:1$ undergoes copolymerisation in the presence of sodium,it forms a styrene-butadiene copolymer,commonly known as styrene-butadiene rubber $(SBR)$ or Buna-$S$.
In the name Buna-$S$,'$Bu$' stands for butadiene,'$Na$' stands for sodium (which acts as the polymerising agent),and '$S$' stands for styrene.

(D) Biodegradable polymers are polymers that decompose by bacterial action.
Polyhydroxy butyrate-co-$\beta$-hydroxyvalerate $(PHBV)$ and Nylon $2$-Nylon-$6$ are examples of biodegradable polymers.
Glyptal is a condensation polymer of ethylene glycol and phthalic acid,which is not biodegradable.

(A) The correct matching is:
$A$. $Nylon$ $6,6$ is a polyamide formed from $Adipic$ $acid$ and $Hexamethylenediamine$ $(v)$.
$B$. $Terylene$ (also known as $Dacron$) is a polyester formed from $Ethylene$ $glycol$ and $Terephthalic$ $acid$ $(i)$.
$C$. $Buna-S$ is a copolymer of $1,3-Butadiene$ and $Styrene$ $(ii)$.
$D$. $Neoprene$ is a polymer of $Chloroprene$ $(iii)$.
Therefore,the correct sequence for $A, B, C, D$ is $(v, i, ii, iii)$.

(B) Dextron - Biodegradable polymer of polyester class.
$(b)$ Nylon-$2$-nylon-$6$ - Biodegradable polymer of polyamide class.
$(c)$ Nylon-$6, 6$ - Non-biodegradable polymer of polyamide class.
$(d)$ $PHBV$ - Biodegradable polymer of polyester class.
Therefore,option $(b)$ is correct.

(B) The correct matches are as follows:
$1$. Electrodes in batteries are made from conducting polymers like $Polyacetylene$ $(A-II)$.
$2$. Welding of metals is performed using $Oxyacetylene$ flame $(B-III)$.
$3$. Toys are commonly manufactured using $Polypropylene$ $(C-I)$.
Therefore,the correct sequence is $A-II, B-III, C-I$.

(D) Statement-$I$ is incorrect because Nylon $6$ is a condensation homopolymer formed from the monomer caprolactam,not a copolymer.
Statement-$II$ is incorrect because Nylon $6, 6$ is formed from adipic acid and hexamethylenediamine,not tetramethylenediamine.
Therefore,both statements are incorrect.

(A) The number average molecular mass $(\overline{M}_n)$ is calculated using the formula: $\overline{M}_n = \frac{\sum N_i M_i}{\sum N_i}$
Here,$N_1 = 10$,$M_1 = 5,000$ and $N_2 = 5$,$M_2 = 50,000$.
$\sum N_i M_i = (10 \times 5,000) + (5 \times 50,000) = 50,000 + 250,000 = 300,000$.
$\sum N_i = 10 + 5 = 15$.
$\overline{M}_n = \frac{300,000}{15} = 20,000 = 2 \times 10^4$.

(D) $1$. Linear polymers: These are long chain polymers. Examples from the list are: Nylon $6,6$,$PVC$,and Novolac (which is a linear polymer). So,there are $3$ linear polymers.
$2$. Crosslinked polymers: These are polymers with crosslinks between chains. Examples from the list are: Bakelite and melamine-formaldehyde resin. So,there are $2$ crosslinked polymers.
$3$. Therefore,the number of linear and crosslinked polymers are $3$ and $2$ respectively.
The correct option is $D$.

(C) The biodegradable polymer $PHBV$ (poly $\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is obtained by the copolymerization of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.
The monomers are:
$Y = CH_3-CH(OH)-CH_2-COOH$ ($3$-hydroxybutanoic acid)
$Z = CH_3-CH_2-CH(OH)-CH_2-COOH$ ($3$-hydroxypentanoic acid)
Thus,option $C$ is the correct answer.

(B) $PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a copolymer formed by the condensation polymerization of two hydroxy acids:
$1$. $3$-Hydroxybutanoic acid $(CH_3-CH(OH)-CH_2-COOH)$
$2$. $3$-Hydroxypentanoic acid $(CH_3-CH_2-CH(OH)-CH_2-COOH)$
Thus,$X$ is $3$-hydroxybutanoic acid and $Y$ is $3$-hydroxypentanoic acid.
Comparing this with the given options,option $B$ matches these structures.

(C) Let us analyze the given polymers based on their type and classification:
$1$. $Polystyrene$ is a homopolymer of styrene and is a thermoplastic.
$2$. $Bakelite$ is a condensation polymer (phenol-formaldehyde resin) and is thermosetting.
$3$. $Nylon$ $6$ is a homopolymer of caprolactam and is classified as a fibre due to strong intermolecular forces like hydrogen bonding.
$4$. $Buna-N$ is a copolymer of $1,3-butadiene$ and $acrylonitrile$ and is an elastomer.
Comparing these with the options:
- Option $A$: $Polystyrene$ is a homopolymer,not a copolymer.
- Option $B$: $Bakelite$ is a condensation polymer,not an addition polymer.
- Option $C$: $Nylon$ $6$ is a homopolymer and is a fibre. This is correct.
- Option $D$: $Buna-N$ is a copolymer,not a homopolymer.
Therefore,the correct set is $Nylon$ $6$ - Homopolymer - fibre.

(A) The classification of polymers based on intermolecular forces is as follows:
$1$. Neoprene is an elastomer,which possesses weak intermolecular forces. Thus,$A$ is correct.
$2$. Terylene is a fibre,which possesses strong intermolecular forces (dipole-dipole or hydrogen bonding). Thus,$B$ is correct.
$3$. Polystyrene is a thermoplastic,which possesses weak intermolecular forces,not 'very weak'.
$4$. Polythene is a thermoplastic,which possesses weak van der Waals forces,not hydrogen bonding.
Therefore,the correctly matched pairs are $A$ and $B$.

(D) Condensation polymers are formed by the repeated condensation reaction between two different bi-functional or tri-functional monomeric units,usually involving the elimination of small molecules like $H_2O$,$HCl$,etc.
$1$. $PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a condensation polymer.
$2$. $Nylon-6$ is a condensation polymer (formed from caprolactam via ring-opening polymerization,which is a type of condensation).
$3$. $Glyptal$ is a condensation polymer (polyester).
$4$. $Buna-N$ and $Neoprene$ are addition polymers.
Therefore,the condensation polymers are $A, D$,and $E$.

(NONE) The correct matches are:
$(A)$ $-(NH-(CH_2)_6-NH-CO-(CH_2)_4-CO)_n-$ is Nylon-$6,6$,which is formed from Hexamethylene diamine and Adipic acid. So,$A \rightarrow IV$.
$(B)$ $-(CO-(CH_2)_5-NH)_n-$ is Nylon-$6$,which is formed from Caprolactam. So,$B \rightarrow III$.
$(C)$ $-(CF_2-CF_2)_n-$ is Teflon,which is formed from Tetrafluoroethene. So,$C \rightarrow V$.
$(D)$ $-(O-CH_2-CH_2-OOC-C_6H_4-CO)_n-$ is Dacron (Terylene),which is formed from Ethylene glycol and Terephthalic acid. So,$D \rightarrow I$.
Thus,the correct sequence is $A-IV, B-III, C-V, D-I$.

(A) The matching is as follows:
$A$. Addition polymer is $V$. Polythene.
$B$. Condensation polymer is $I$. Bakelite.
$C$. Acrilan is $IV$. Vinyl cyanide (Acrylonitrile).
$D$. Rubber (Natural rubber) is $II$. $2$-methyl-$1,3$-butadiene (Isoprene).
Therefore,the correct sequence is $A-V, B-I, C-IV, D-II$.

(D) The number average molecular mass $(M_n)$ of a polymer is calculated using the formula:
$M_n = \frac{\Sigma N_i M_i}{\Sigma N_i}$
Given:
$N_1 = 15, M_1 = 8,000$
$N_2 = 15, M_2 = 80,000$
Substituting the values:
$M_n = \frac{15 \times 8,000 + 15 \times 80,000}{15 + 15}$
$M_n = \frac{120,000 + 1,200,000}{30}$
$M_n = \frac{1,320,000}{30} = 44,000$