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:

13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u

6.6 APPLICATIONS OF EQUILIBRIUM in the denominator). This implies that a high value of K is suggestive of a high concentration CONSTANTS of products and vice-versa.Before considering the applications of We can make the following generalisationsequilibrium constants, let us summarise the concerning the composition of equilibriumimportant features of equilibrium constants mixtures:as follows: 1. Expression for equilibrium constant is • If Kc > 103, products predominate over applicable only when concentrations of reactants, i.e., if Kc is very large, the the reactants and products have attained reaction proceeds nearly to completion. constant value at equilibrium state. Consider the following examples: 2. The value of equilibrium constant is (a) The reaction of H2 with O2 at 500 K independent of initial concentrations of has a very large equilibrium constant, the reactants and products. Kc = 2.4 × 1047. 3. Equilibrium constant is temperature (b) H2(g) + Cl2(g) 2HCl(g) at 300K has dependent having one unique value for Kc = 4.0 × 1031. a particular reaction represented by a (c) H2(g) + Br2(g) 2HBr (g) at 300 K, balanced equation at a given temperature. Kc = 5.4 × 1018 4. The equilibrium constant for the reverse • If Kc < 10–3, reactants predominate over reaction is equal to the inverse of the products, i.e., if Kc is very small, the equilibrium constant for the forward reaction proceeds rarely. Consider the reaction. following examples: Reprint 2025-26 182 chemistry (a) The decomposition of H2O into H2 and If Qc = Kc, the reaction mixture is already O2 at 500 K has a very small equilibrium at equilibrium. constant, Kc = 4.1 × 10–48 Consider the gaseous reaction of H2 (b) N2(g) + O2(g) 2NO(g), with I2, at 298 K has Kc = 4.8 ×10–31. H2(g) + I2(g) 2HI(g); Kc = 57.0 at 700 K. • If Kc is in the range of 10 – 3 to 103, Suppose we have molar concentrations appreciable concentrations of both [H2]t=0.10M, [I2]t = 0.20 M and [HI]t = 0.40 M. reactants and products are present. (the subscript t on the concentration symbols Consider the following examples: means that the concentrations were measured at some arbitrary time t, not necessarily at(a) For reaction of H2 with I2 to give HI, equilibrium). Kc = 57.0 at 700K. Thus, the reaction quotient, Qc at this(b) Also, gas phase decomposition of N2O4 stage of the reaction is given by, to NO2 is another reaction with a value 2 –3 Qc = [HI]t / [H2]t [I2]t = (0.40)2/ (0.10)×(0.20) of Kc = 4.64 × 10 at 25°C which is neither too small nor too large. Hence, = 8.0 equilibrium mixtures contain appreciable Now, in this case, Qc (8.0) does not equal concentrations of both N2O4 and NO2. Kc (57.0), so the mixture of H2(g), I2(g) and HI(g) These generarlisations are illustrated in is not at equilibrium; that is, more H2(g) and Fig. 6.6 I2(g) will react to form more HI(g) and their concentrations will decrease till Qc = Kc. The reaction quotient, Qc is useful in predicting the direction of reaction by comparing the values of Qc and Kc. Thus, we can make the following generalisations concerning the direction of the reaction (Fig. 6.7) :Fig.6.6 Dependence of extent of reaction on Kc 6.6.2 Predicting the Direction of the Reaction The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient Q. The reaction quotient, Q (Qc with molar Fig. 6.7 Predicting the direction of the reactionconcentrations and QP with partial pressures) is defined in the same way as the equilibrium • If Qc < Kc, net reaction goes from left to constant Kc except that the concentrations right in Qc are not necessarily equilibrium values. • If Qc > Kc, net reaction goes from right to For a general reaction: left. a A + b B c C + d D (6.19) • If Qc = Kc, no net reaction occurs. Qc = [C]c[D]d / [A]a[B]b (6.20) Problem 6.7 Then, The value of Kc for the reaction If Qc > Kc, the reaction will proceed in the 2A B + C is 2 × 10–3. At a given time, direction of reactants (reverse reaction). the composition of reaction mixture is [A] = [B] = [C] = 3 × 10–4 M. In which direction If Qc < Kc, the reaction will proceed in the the reaction will proceed?direction of the products (forward reaction). Reprint 2025-26 EQUILIBRIUM 183 Solution The total pressure at equilbrium was For the reaction the reaction quotient Qc is found to be 9.15 bar. Calculate Kc, Kp and given by, partial pressure at equilibrium. Qc = [B][C]/ [A]2 Solution as [A] = [B] = [C] = 3 × 10–4M Qc = (3 ×10–4)(3 × 10–4) / (3 ×10–4)2 = 1 We know pV = nRT as Qc > Kc so the reaction will proceed in the Total volume (V ) = 1 L reverse direction. Molecular mass of N2O4 = 92 g 6.6.3 Calculating Equilibrium Number of moles = 13.8g/92 g = 0.15 Concentrations of the gas (n) In case of a problem in which we know the Gas constant (R) = 0.083 bar L mol–1K–1 initial concentrations but do not know any of Temperature (T ) = 400 K the equilibrium concentrations, the following pV = nRTthree steps shall be followed: Step 1. Write the balanced equation for the p × 1L = 0.15 mol × 0.083 bar L mol–1K–1 × 400 Kreaction. Step 2. Under the balanced equation, make p = 4.98 bar a table that lists for each substance involved N2O4 2NO2 in the reaction: Initial pressure: 4.98 bar 0 (a) the initial concentration, At equilibrium: (4.98 – x) bar 2x bar (b) the change in concentration on going to Hence, equilibrium, and ptotal at equilibrium = pN2O4 + pNO2(c) the equilibrium concentration. 9.15 = (4.98 – x) + 2x In constructing the table, define x as the 9.15 = 4.98 + xconcentration (mol/L) of one of the substances that reacts on going to equilibrium, then use x = 9.15 – 4.98 = 4.17 bar the stoichiometry of the reaction to determine Partial pressures at equilibrium are, the concentrations of the other substances in terms of x. pN2O4 = 4.98 – 4.17 = 0.81bar Step 3. Substitute the equilibrium pNO2 = 2x = 22 × 4.17 = 8.34 bar concentrations into the equilibrium equation K p = p N 2O 4  p NO 2 / for the reaction and solve for x. If you are = (8.34)2/0.81 = 85.87to solve a quadratic equation choose the mathematical solution that makes chemical Kp = Kc(RT)∆n sense. 85.87 = Kc(0.083 × 400)1 Step 4. Calculate the equilibrium Kc = 2.586 = 2.6 concentrations from the calculated value of x. Problem 6.9Step 5. Check your results by substituting them into the equilibrium equation. 3.00 mol of PCl5 kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the Problem 6.8 mixture at equilibrium. Kc= 1.80 13.8g of N2O4 was placed in a 1L reaction Solution vessel at 400K and allowed to attain PCl5 PCl3 + Cl2 equilibrium Initial N2O4 (g) 2NO2 (g) concentration: 3.0 0 0 Reprint 2025-26 184 chemistry Taking antilog of both sides, we get, Let x mol per litre of PCl5 be dissociated, K = e–∆G/RT (6.23) At equilibrium: (3-x) x x Hence, using the equation (6.23), the reaction spontaneity can be interpreted in Kc = [PCl3][Cl2]/[PCl5] terms of the value of ∆G . 1.8 = x2/ (3 – x) • If ∆G  < 0, then –∆G /RT is positive, x2 + 1.8x – 5.4 = 0 and e –∆DG /RT>1, making K >1, which x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 implies a spontaneous reaction or the x = [–1.8 ± √3.24 + 21.6]/2 reaction which proceeds in the forward direction to such an extent that the x = [–1.8 ± 4.98]/2 products are present predominantly. x = [–1.8 + 4.98]/2 = 1.59 • If ∆G  > 0, then –∆G /RT is negative, and [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M e –∆G </RT 1, that is , K < 1, which implies [PCl3] = [Cl2] = x = 1.59 M a non-spontaneous reaction or a reaction which proceeds in the forward direction