6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; 175 Reprint 2025-26 Physics (b) A circular loop being deformed into a narrow straight wire. FIGURE 6.16 6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? 6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? 6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. 6.6 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 ´ 10–4 Wb m–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? 6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. 6.8 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? Reprint 2025-26 Chapter Seven ALTERNATING CURRENT 7.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points. Reprint 2025-26 Physics 7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by v = vm sin ωt (7.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. Nicola Tesla (1856 – 1943) Serbian-American scientist, inventor and genius. He conceived the idea of the rotating1943) magnetic field, which is the – basis of practically all alternating current machinery, and which(1856 helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor. electric power. He also invented among other To find the value of current through the resistor, we things the induction motor, ε()t = 0 (refer to Section the polyphase system of ac apply Kirchhoff’s loop rule ∑TESLA power, and the high 3.12), to the circuit shown in Fig. 7.1 to get frequency induction coil v m sin ωt = i R (the Tesla coil) used in radio and television sets and v m i = sin ωtNICOLA other electronic equipment. or R The SI unit of magnetic field is named in his honour. Since R is a constant, we can write this equation as i = i m sin ωt (7.2) where the current amplitude im is given by v m i m = (7.3) R Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with FIGURE 7.2 In a pure resistor, the voltage and each other. current are in phase. The We see that, like the applied voltage, the current varies minima, zero and maxima sinusoidally and has corresponding positive and negative values occur at the same during each cycle. Thus, the sum of the instantaneous current respective times. values over one complete cycle is zero, and the average current 178 is zero. The fact that the average current is zero, however, does Reprint 2025-26 Alternating Current not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i2R and depends on i2 (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is p = i 2 R = i m2 R sin 2 ωt (7.4) The average value of p over a cycle is* p = < i 2 R > = < i m2 R sin 2 ωt > [7.5(a)] where the bar over a letter (here, p) denotes its average George Westinghouse value and <......> denotes taking average of the quantity (1846 – 1914) A leading inside the bracket. Since, i2m and R are constants, proponent of the use of p = i m2 R < sin 2 ωt > [7.5(b)] alternating current overUsing the trigonometric identity, sin2 wt = direct current. Thus, GEORGE he came into conflict 1/2 (1– cos 2wt), we have < sin2 wt > = (1/2) (1– < cos 2wt >) with Thomas Alva Edison, and since < cos2wt > = 0**, we have, an advocate of direct 2 1 current. Westinghouse < sin ωt > = was convinced that the 2 technology of alternating Thus, current was the key to 1 2 the electrical future. p = i m R [7.5(c)] He founded the famous 2 Company named after him WESTINGHOUSE To express ac power in the same form as dc power and enlisted the services (P = I2R), a special value of current is defined and used. of Nicola Tesla and It is called, root mean square (rms) or effective current other inventors in the (1846(Fig. 7.3) and is denoted by Irms or I. development of alternating current motors and – apparatus for the transmission of high tension current, pioneering 1914) in large scale lighting. FIGURE 7.3 The rms current I is related to the peak current im by I = mi / 2 = 0.707 im. 1 T F (t ) d t* The average value of a function F (t) over a period T is given by F (t ) = T ∫0 1 T 1  sin 2ωt  T 1 < cos 2ωt > = ∫ cos 2ω t dt = = [ sin 2ω T − 0 ] = 0** T 0 T  2ω  0 2ωT 179 Reprint 2025-26 Physics It is defined by 2 1 2 i m I = i = i m = 2 2 = 0.707 im (7.6) In terms of I, the average power, denoted by P is 1 2 2 P = p = i m R = I R (7.7) 2 Similarly, we define the rms voltage or effective voltage by v m V = = 0.707 vm (7.8) 2 From Eq. (7.3), we have vm = imR v m i m or, = R 2 2 or, V = IR (7.9) Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of vm = 2 V = (1.414)(220 V) = 311 V In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation (7.7) can also be written as P = V2 / R = I V (since V = I R) Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. Solution (a) We are given P = 100 W and V = 220 V. The resistance of the bulb is 2 V 2 ( 220 V ) R = = = 484 Ω P 100 W (b) The peak voltage of the source is V 7.1 v m = 2V = 311 (c) Since, P = I V P 100 W I 0.454A EXAMPLE V 220 V Reprint 2025-26 Alternating Current

4.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10–19 C, me = 9.1×10–31 kg)

9.3 Alkenes Alkenes are unsaturated hydrocarbons Fig. 9.4 Orbital picture of ethene depictingcontaining at least one double bond. What σ bonds only should be the general formula of alkenes? If there is one double bond between two carbon 9.3.2 Nomenclature atoms in alkenes, they must possess two For nomenclature of alkenes in IUPAC system, hydrogen atoms less than alkanes. Hence, the longest chain of carbon atoms containing general formula for alkenes is CnH2n. Alkenes the double bond is selected. Numbering of the are also known as olefins (oil forming) since chain is done from the end which is nearer to Reprint 2025-26 Hydrocarbons 307 Fig. 9.5 Orbital picture of ethene showing formation of (a) π-bond, (b) π-cloud and (c) bond angles and bond lengths the double bond. The suffix ‘ene’ replaces ‘ane’ Solutionof alkanes. It may be remembered that first member of alkene series is: CH2 (replacing (i) 2,8-Dimethyl-3, 6-decadiene; n by 1 in CnH2n) known as methene but has (ii) 1,3,5,7 Octatetraene; a very short life. As already mentioned, first (iii) 2-n-Propylpent-1-ene; stable member of alkene series is C2H4 known (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene;as ethylene (common) or ethene (IUPAC). IUPAC names of a few members of alkenes Problem 9.8 are given below : Calculate number of sigma (σ) and pi (π) Structure IUPAC name bonds in the above structures (i-iv). CH3 – CH = CH2 Propene SolutionCH3 – CH2 – CH = CH2 But – l - ene σ bonds : 33, π bonds : 2CH3 – CH = CH–CH3 But-2-ene σ bonds : 17, π bonds : 4CH2 = CH – CH = CH2 Buta – 1,3 - diene σ bonds : 23, π bond : 1CH2 = C – CH3 2-Methylprop-1-ene | σ bonds : 41, π bond : 1 CH3 CH2 = CH – CH – CH3 3-Methylbut-1-ene 9.3.3 Isomerism | Alkenes show both structural isomerism and CH3 geometrical isomerism. Structural isomerism : As in alkanes, ethene Problem 9.7 (C2H4) and propene (C3H6) can have only one Write IUPAC names of the following structure but alkenes higher than propene compounds: have different structures. Alkenes possessing (i) (CH3)2CH – CH = CH – CH2 – CH C4H8 as molecular formula can be written in  the following three ways: CH3 – CH – CH | I. 1 2 3 4 C2H5 CH2 = CH – CH2 – CH3 (ii) But-1-ene (C4H8) (iii) CH2 = C (CH2CH2CH3)2 (iv) CH3 CH2 CH2 CH2 CH2CH3 II. 1 2 3 4 | | CH3 – CH = CH – CH3 CH3 – CHCH = C – CH2 – CHCH3 | But-2-ene CH3 (C4H8) Reprint 2025-26 308 chemistry III. 1 2 3 In (a), the two identical atoms i.e., both CH2 = C – CH3 the X or both the Y lie on the same side | of the double bond but in (b) the two X or CH3 two Y lie across the double bond or on the 2-Methylprop-1-ene opposite sides of the double bond. This (C4H8) results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in Structures I and III, and II and III are the two arrangements is different. Therefore,the examples of chain isomerism whereas they are stereoisomers. They would have thestructures I and II are position isomers. same geometry if atoms or groups around C=C bond can be rotated but rotation around Problem 9.9 C=C bond is not free. It is restricted. For Write structures and IUPAC names of understanding this concept, take two pieces different structural isomers of alkenes of strong cardboards and join them with the corresponding to C5H10. help of two nails. Hold one cardboard in your Solution one hand and try to rotate the other. Can (a) CH2 = CH – CH2 – CH2 – CH3 you really rotate the other cardboard ? The answer is no. The rotation is restricted. This Pent-1-ene illustrates that the restricted rotation of atoms (b) CH3 – CH=CH – CH2 – CH3 or groups around the doubly bonded carbon Pent-2-ene atoms gives rise to different geometries of such compounds. The stereoisomers of this (c) CH3 – C = CH – CH3 type are called geometrical isomers. The | isomer of the type (a), in which two identical CH3 atoms or groups lie on the same side of the 2-Methylbut-2-ene double bond is called cis isomer and the (d) CH3 – CH – CH = CH2 other isomer of the type (b), in which identical | atoms or groups lie on the opposite sides of CH3 the double bond is called trans isomer . Thus 3-Methylbut-1-ene cis and trans isomers have the same structure but have different configuration (arrangement (e) CH2 = C – CH2 – CH3 of atoms or groups in space). Due to different | arrangement of atoms or groups in space, CH3 these isomers differ in their properties like 2-Methylbut-1-ene melting point, boiling point, dipole moment, solubility etc. Geometrical or cis-trans isomersGeometrical isomerism: Doubly bonded of but-2-ene are represented below :carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways : Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of cis-but-2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that Reprint 2025-26 Hydrocarbons 309 trans-but-2-ene is non-polar. This can be (ii) CH2 = CBr2 understood by drawing geometries of the two (iii) C6H5CH = CH – CH3forms as given below from which it is clear that in the trans-but-2-ene, the two methyl (iv) CH3CH = CCl CH3 groups are in opposite directions, Threfore, dipole moments of C-CH3 bonds cancel, thus Solution making the trans form non-polar. (iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom. 9.3.4 Preparation 1. From alkynes: Alkynes on partial reduction with calculated amount of cis-But-2-ene trans-But-2-ene dihydrogen in the presence of palladised (µ = 0.33D) (µ = 0) charcoal partially deactivated with poisons like sulphur compounds or quinoline give In the case of solids, it is observed that the alkenes. Partially deactivated palladisedtrans isomer has higher melting point than charcoal is known as Lindlar’s catalyst.the cis form. Alkenes thus obtained are having cis Geometrical or cis-trans isomerism geometry. However, alkynes on reductionis also shown by alkenes of the types with sodium in liquid ammonia form transXYC = CXZ and XYC = CZW alkenes. Problem 9.10 Draw cis and trans isomers of the following compounds. Also write their IUPAC names : (i) CHCl = CHCl (9.30) (ii) C2H5CCH3 = CCH3C2H5 Solution (9.31) iii) CH≡ CH+H2 Pd/C CH2 =CH2 (9.32) Ethyne Ethene CH3–C≡ CH+H2 Pd/C CH3–CH =CH2 iv) Propyne Propene (9.33) Will propene thus obtained show Problem 9.11 geometrical isomerism? Think for the reason in support of your answer. Which of the following compounds will show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X) on (i) (CH3)2C = CH – C2H5 heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, Reprint 2025-26 310 chemistry ethanol) eliminate one molecule of halogen takes out one hydrogen atom from the acid to form alkenes. This reaction is β-carbon atom. known as dehydrohalogenation i.e., removal of halogen acid. This is example of β-elimination reaction, since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached). (9.37) 9.3.5 Properties Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. (9.34) The first three members are gases, the next fourteen are liquids and the higher ones are Nature of halogen atom and the alkyl group solids. Ethene is a colourless gas with a faint determine rate of the reaction. It is observed sweet smell. All other alkenes are colourless that for halogens, the rate is: iodine > and odourless, insoluble in water but fairly bromine > chlorine, while for alkyl groups soluble in non-polar solvents like benzene, it is : tertiary > secondary > primary. petroleum ether. They show a regular increase 3. From vicinal dihalides: Dihalides in in boiling point with increase in size i.e., every which two halogen atoms are attached – CH2 group added increases boiling point by to two adjacent carbon atoms are known 20–30 K. Like alkanes, straight chain alkenes as vicinal dihalides. Vicinal dihalides on have higher boiling point than isomeric treatment with zinc metal lose a molecule branched chain compounds. of ZnX2 to form an alkene. This reaction Chemical properties is known as dehalogenation. Alkenes are the rich source of loosely held CH2Br–CH2Br + Zn CH2=CH2+ ZnBr2 pi (π) electrons, due to which they show (9.35) addition reactions in which the electrophiles add on to the carbon-carbon double bond toCH3CHBr–CH2Br + Zn CH3CH=CH2 form the addition products. Some reagents +ZnBr2 also add by free radical mechanism. There (9.36) are cases when under special conditions, alkenes also undergo free radical substitution4. From alcohols by acidic dehydration: reactions. Oxidation and ozonolysis reactions You have read during nomenclature of are also quite prominent in alkenes. A brief different homologous series in Unit 12 description of different reactions of alkenes that alcohols are the hydroxy derivatives is given below: of alkanes. They are represented by R–OH where, R is CnH2n+1. Alcohols on heating 1. Addition of dihydrogen: Alkenes add with concentrated sulphuric acid form up one molecule of dihydrogen gas in alkenes with the elimination of one water the presence of finely divided nickel, molecule. Since a water molecule is palladium or platinum to form alkanes eliminated from the alcohol molecule in (Section 9.2.2) the presence of an acid, this reaction is 2. Addition of halogens : Halogens like known as acidic dehydration of alcohols. bromine or chlorine add up to alkene to This reaction is also the example of form vicinal dihalides. However, iodine β-elimination reaction since –OH group does not show addition reaction under Reprint 2025-26 Hydrocarbons 311 normal conditions. The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation. Addition of halogens to alkenes is an example of electrophilic addition reaction involving cyclic halonium ion formation (9.42) which you will study in higher classes. Markovnikov, a Russian chemist made a generalisation in 1869 after studying such reactions in detail. These generalisations led Markovnikov to frame a rule called Markovnikov rule. The rule states that negative part of the addendum (adding (9.38) molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. Thus according to this rule, product I i.e., 2-bromopropane is expected. In actual practice, this is the principal product of the reaction. This generalisation of Markovnikov (9.39) rule can be better understood in terms of mechanism of the reaction. 3. Addition of hydrogen halides: Hydrogen halides (HCl, HBr,HI) add up to alkenes Mechanism to form alkyl halides. The order of Hydrogen bromide provides an electrophile, reactivity of the hydrogen halides is H +, which attacks the double bond to form HI > HBr > HCl. Like addition of halogens carbocation as shown below : to alkenes, addition of hydrogen halides is also an example of electrophilic addition reaction. Let us illustrate this by taking addition of HBr to symmetrical and unsymmetrical alkenes Addition reaction of HBr to symmetrical alkenes (a) less stable (b) more stableAddition reactions of HBr to symmetrical primary carbocation secondary carbocationalkenes (similar groups attached to double bond) take place by electrophilic addition (i) The secondary carbocation (b) is more mechanism. stable than the primary carbocation (a), therefore, the former predominates CH2=CH2+H–Br CH3–CH2–Br (9.40) because it is formed at a faster rate. (ii) The carbocation (b) is attacked by Br– ionCH3–CH=CH–CH3+HBr CH3–CH–CHCH3 to form the product as follows : Br (9.41) Addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H – Br add to propene ? The two 2-Bromopropane possible products are I and II. (major product) Reprint 2025-26 312 chemistry Anti Markovnikov addition or peroxide effect or Kharash effect In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place contrary to the Markovnikov rule. This happens only with HBr but not with HCl The secondary free radical obtained in the and Hl. This addition reaction was observed above mechanism (step iii) is more stable than by M.S. Kharash and F.R. Mayo in 1933 the primary. This explains the formation of at the University of Chicago. This reaction 1-bromopropane as the major product. It may is known as peroxide or Kharash effect be noted that the peroxide effect is not observed or addition reaction anti to Markovnikov in addition of HCl and HI. This may be due rule. to the fact that the H–Cl bond being (C6H5CO)2O2 stronger (430.5 kJ mol –1) than H–Br bond CH3 – CH=CH2+HBr CH3–CH2 (363.7 kJ mol –1), is not cleaved by the free radical, whereas the H–I bond is weaker CH2Br (296.8 kJ mol –1) and iodine free radicals 1–Bromopropane combine to form iodine molecules instead of adding to the double bond. (9.43) Mechanism : Peroxide effect proceeds via Problem 9.12 free radical chain mechanism as given below: Write IUPAC names of the products obtained by addition reactions of HBr to(i) hex-1-ene (i) in the absence of peroxide and (ii) in the presence of peroxide. Solution Homolysis C. 6H5+H–Br C6H3+ B. r(ii) 4. Addition of sulphuric acid : Cold concentrated sulphuric acid adds to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate by the electrophilic addition reaction. Reprint 2025-26 Hydrocarbons 313 ketones and/or acids depending upon the nature of the alkene and the experimental conditions (9.49) KMnO4/H+ CH3 – CH=CH–CH3 2CH3COOH (9.44) But-2-ene Ethanoic acid (9.50) 7. Ozonolysis : Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide, and then cleavage of the ozonide by Zn-H2O to smaller molecules. This reaction is highly useful in detecting the position of the double (9.45) bond in alkenes or other unsaturated compounds.5. Addition of water : In the presence of a few drops of concentrated sulphuric acid alkenes react with water to form alcohols, in accordance with the Markovnikov rule. (9.51) (9.46) 6. Oxidation: Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent) produce vicinal glycols. Decolorisation of KMnO4 solution is used as a test for unsaturation. (9.52) 8. Polymerisation: You are familiar with (9.47) polythene bags and polythene sheets. Polythene is obtained by the combination of large number of ethene molecules at high temperature, high pressure and in the presence of a catalyst. The large molecules thus obtained are called (9.48) polymers. This reaction is known as b) Acidic potassium permanganate or acidic polymerisation. The simple compounds potassium dichromate oxidises alkenes to from which polymers are made are called Reprint 2025-26 314 chemistry monomers. Other alkenes also undergo are named as derivatives of the corresponding polymerisation. alkanes replacing ‘ane’ by the suffix ‘yne’. n(CH2 =CH2) High temp./pressureCatalyst —( CH2–CH2 )— The position of the triple bond is indicated by the first triply bonded carbon. Common Polythene and IUPAC names of a few members of alkyne (9.53) series are given in Table 9.2. High temp./pressure You have already learnt that ethyne and n(CH3 –CH=CH2) Catalyst —( CH–CH2 )—n propyne have got only one structure but there are two possible structures for butyne – CH3 (i) but-1-yne and (ii) but-2-yne. Since these Polypropene two compounds differ in their structures (9.54) due to the position of the triple bond, they Polymers are used for the manufacture of plastic are known as position isomers. In how bags, squeeze bottles, refrigerator dishes, toys, many ways, you can construct the structure pipes, radio and T.V. cabinets etc. Polypropene for the next homologue i.e., the next alkyne is used for the manufacture of milk crates, with molecular formula C5H8? Let us try to plastic buckets and other moulded articles. arrange five carbon atoms with a continuous Though these materials have now become chain and with a side chain. Following are the common, excessive use of polythene and possible structures : polypropylene is a matter of great concern for Structure IUPAC name all of us. 1 2 3 4 5 I. HC≡ C– CH2– CH2– CH3 Pent–1-yne