9.4 Alkynes 1 2 3 4 5 Like alkenes, alkynes are also unsaturated II. H3C–C≡ C– CH2– CH3 Pent–2-yne hydrocarbons. They contain at least one triple 4 3 2 1 bond between two carbon atoms. The number III. H3C–CH–C≡ CH 3-Methyl but–1-yne |of hydrogen atoms is still less in alkynes as CH3compared to alkenes or alkanes. Their general Structures I and II are position isomers formula is CnH2n–2. and structures I and III or II and III are chain The first stable member of alkyne series isomers. is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Problem 9.13 purposes in the form of oxyacetylene flame Write structures of different isomers obtained by mixing acetylene with oxygen corresponding to the 5 th member of gas. Alkynes are starting materials for a large alkyne series. Also write IUPAC names of number of organic compounds. Hence, it all the isomers. What type of isomerism is interesting to study this class of organic is exhibited by different pairs of isomers? compounds. Solution 9.4.1 Nomenclature and Isomerism th 5 member of alkyne has the molecular In common system, alkynes are named as formula C6H10. The possible isomers are: derivatives of acetylene. In IUPAC system, they Table 9.2 Common and IUPAC Names of Alkynes (CnH2n–2) Value of n Formula Structure Common name IUPAC name 2 C2H2 H-C≡CH Acetylene Ethyne 3 C3H4 CH3-C≡CH Methylacetylene Propyne 4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne 4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne Reprint 2025-26 Hydrocarbons 315 (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne (c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne 3-Methylpent-1-yne 4-Methylpent-1-yne 4-Methylpent-2-yne Fig. 9.6 Orbital picture of ethyne showing (a) sigma overlaps (b) pi overlaps. orbitals of the other carbon atom, which undergo lateral or sideways overlapping to 3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C different pairs. π bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol -1) is more than those 9.4.2 Structure of Triple Bond of C=C bond (bond enthalpy 681 kJ mol –1) Ethyne is the simplest molecule of alkyne and C–C bond (bond enthalpy 348 kJ mol–1). series. Structure of ethyne is shown in The C≡C bond length is shorter (120 pm) Fig. 9.6. than those of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon Each carbon atom of ethyne has two sp atoms is cylindrically symmetrical about thehybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear bond is obtained by the head-on overlapping molecule. of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised 9.4.3 Preparation orbital of each carbon atom undergoes 1. From calcium carbide: On industrial overlapping along the internuclear axis with scale, ethyne is prepared by treating the 1s orbital of each of the two hydrogen calcium carbide with water. Calcium atoms forming two C-H sigma bonds. carbide is prepared by heating quick lime H-C-C bond angle is of 180°. Each carbon with coke. Quick lime can be obtained byhas two unhybridised p orbitals which are heating limestone as shown in the followingperpendicular to each other as well as to the reactions:plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p CaCO3 ∆ CaO + O2 (9.55) Reprint 2025-26 316 chemistry CaO + 3C CaC2 + CO (9.56) the sp hybridised carbon2 atoms whereas they are attached to sp hybridised carbon Calcium 3 atoms in ethene and sp hybridised carbons carbide in ethane. Due to the maximum percentage of CaC2 + 2H2O Ca(OH)2 + C2H2 (9.57) s character (50%), the sp hybridised orbitals of carbon atoms in ethyne molecules have2. From vicinal dihalides : Vicinal dihalides highest electronegativity; hence, these attract on treatment with alcoholic potassium the shared electron pair of the C-H bond of hydroxide undergo dehydrohalogenation. ethyne to a greater extent than that of the One molecule of hydrogen halide is 2 sp hybridised orbitals of carbon in ethene eliminated to form alkenyl halide which 3 and the sp hybridised orbital of carbon in on treatment with sodamide gives alkyne. ethane. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes. HC ≡ CH + Na → HC ≡ C–Na++ 1/2 H2 9.4.4 Properties Monosodium Physical properties ethynide Physical properties of alkynes follow the same (9.59) trend of alkenes and alkanes. First three HC ≡ C– Na + Na → Na+ C–Na+ ≡ C–Na++ 1/2 H2members are gases, the next eight are liquids and the higher ones are solids. All alkynes Disodium ethynide are colourless. Ethyene has characteristic (9.60)odour. Other members are odourless. Alkynes are weakly polar in nature. They are lighter CH3 – C ≡ C – H + Na+ NH–2 than water and immiscible with water but ↓ soluble in organic solvents like ethers, carbon CH3 – C ≡ C– Na+ + NH3 tetrachloride and benzene. Their melting Sodium propynide (9.61) point, boiling point and density increase with These reactions are not shown by alkenesincrease in molar mass. and alkanes, hence used for distinction Chemical properties between alkynes, alkenes and alkanes. What Alkynes show acidic nature, addition reactions about the above reactions with but-1-yne and and polymerisation reactions as follows : but-2-yne ? Alkanes, alkenes and alkynes A. Acidic character of alkyne: Sodium follow the following trend in their acidic metal and sodamide (NaNH2) are strong behaviour : bases. They react with ethyne to form sodium i) CH ≡ CH > H2C – CH2 > CH3 –CH3acetylide with the liberation of dihydrogen gas. These reactions have not been observed ii) HC ≡ CH > CH3 –C≡ CH >> CH3 –C≡C–CH3in case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison B. Addition reactions: Alkynes contain a to ethene and ethane. Why is it so ? Has triple bond, so they add up, two molecules of it something to do with their structures dihydrogen, halogen, hydrogen halides etc. and the hybridisation ? You have read that Formation of the addition product takes place hydrogen atoms in ethyne are attached to according to the following steps. Reprint 2025-26 Hydrocarbons 317 The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic addition reactions. A few addition reactions (9.66)are given below: (i) Addition of dihydrogen (iv) Addition of water Pt/Pd/Ni H2 Like alkanes and alkenes, alkynes are alsoHC≡CH+H2 [H2C = CH2] CH3–CH3 immiscible and do not react with water. (9.62) However, one molecule of water adds to alkynes on warming with mercuric sulphate CH3–C≡CH + H2 Pt/Pd/Ni [CH3–CH=CH2] and dilute sulphuric acid at 333 K to form Propyne Propene carbonyl compounds. ↓H2 CH3–CH2–CH3 Propane (9.63) (ii) Addition of halogens (9.67) (9.64) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation. (iii) Addition of hydrogen halides (9.68) Two molecules of hydrogen halides (HCl, HBr, (v) Polymerisation HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable which two halogens are attached to the same conditions, linear polymerisation of ethyne carbon atom) takes place to produce polyacetylene or H–C≡C–H+H–Br [CH2 = CH–Br]→ CHBr2 polyethyne which is a high molecular Bromoethene weight polyene containing repeating units of CH3 (CH = CH – CH = CH ) and can be represented 1,1-Dibromoethane as —(CH = CH – CH = CH)n— Under special (9.65) conditions, this polymer conducts electricity. Reprint 2025-26 318 chemistry Thin film of polyacetylene can be used as but in a majority of reactions of aromatic electrodes in batteries. These films are good compounds, the unsaturation of benzene ring conductors, lighter and cheaper than the is retained. However, there are examples of metal conductors. aromatic hydrocarbons which do not contain a (b) Cyclic polymerisation: Ethyne on benzene ring but instead contain other highly unsaturated ring. Aromatic compoundspassing through red hot iron tube at 873K containing benzene ring are known asundergoes cyclic polymerization. Three benzenoids and those not containing amolecules polymerise to form benzene, which benzene ring are known as non-benzenoids.is the starting molecule for the preparation of Some examples of arenes are givenderivatives of benzene, dyes, drugs and large below:number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (9.69) Biphenyl Problem 9.14 How will you convert ethanoic acid into 9.5.1 Nomenclature and Isomerism benzene? The nomenclature and isomerism of aromatic Solution hydrocarbons has already been discussed in Unit 8. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below:

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

3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).