EQUIVALENT (DBE) OR INDEX OF HYDROGEN DEFICIENCY (IHD)
1.9.3 Effect of Intermolecular Forces on Different Properties
1.9.3.5 Acidic character of organic compounds
(1) Acidity of diastereoiosmeric dicarboxylic acids
Maleic acid, cis —HOOC—CH== CH—COOH (pKa1 = 1.92, pKa2 = 6.04) is stronger for the fi rst ionization but weaker for the second ionization than the isomeric fumaric acid, trans–
HOOC—CH== CH—COOH (pKa1= 3.03, pKa2 = 4.44). In maleic acid, the two carboxyl (—COOH) groups are placed on the same side of the double bond and so, the maleate monoanion is stabilized by intramolecular hydrogen bonding. In fumaric acid, on the other hand, the two —COOH groups are placed on the different sides of the double bond
and consequently, intramolecular hydrogen bonding is not possible for stabilization of the fumarate monoanion. Because of greater stability of the monoanion, maleic acid has a greater tendency to ionize than fumaric acid, i.e., for the fi rst ionization it is a stronger acid than fumaric acid.
Due to stabilization of the maleate monoanion by intramolecular hydrogen bonding, it is very diffi cult to remove a proton from it. Proton release is also disfavoured because the fi eld effect generated by the —COO① group on the nearby —COOH group is free and there is no signifi cant fi eld effect caused by the —COO① group. On the other hand, in fumarate monoanion, the —COOH group is free and there is no signifi cant fi eld effect caused by the
—COO① group. So, it easily ionizes to form the fumarate dianion, i.e., for second ionization fumaric acid is stronger than maleic acid.
The compound I is more acidic than the compound II can also be explained in terms of hydrogen bonding and fi eld effect.
Each of these two diastereoisomeric acids (I and II) is locked in that particular chair conformation in which the bulky —C(CD3)3 group is placed in an equatorial position. In
such conformation, the —COO① and —COOH groups are both axial in I and equatorial in II.
Because of close proximity of these two groups, intramolecular hydrogen bonding occurs in II, but not in I in which they are held farther apart. Therefore, ionization of II is relatively more disfavoured compared to I. Proton release from II is further disfavoured because of fi eld effect generated by the —COO① group, on the —COOH group. No signifi cant fi eld effect operates in I. Hence, the compound I is more acidic than the compound II.
(2) Acidity of hydroxybenzoic acids Hydroxybenzoic acids have two acidic functional groups; one is —COOH and the other is —OH.
The fi rst ionization (i.e., dissociation of carboxyl proton) of o-hydroxybenzoic acid (salicylic acid) is more favourable than that of the p-hydroxybenzoic acid, whereas the second ionization (i.e., dissociation of hydroxyl proton) of the latter acid is more favourable than that of the former acid. This observation can be explained by hydrogen bonding and fi eld effect. Since the —OH and the —COOH groups in o-hydroxybenzoic acid (salicylic acid) are placed in adjacent ring carbons, therefore, the conjugate base, i.e., the monoanion, is stabilized by intramolecular hydrogen bonding leading to the formation of a stable six- membered ring (chelation). On the other hand, the two groups in the p-isomer are held farther apart and no such intramolecular H-bonding is possible. o-Hydroxybenzoate ion is, therefore, more stable than the p-hydroxybenzoate ion and because of this, ionization of the —COOH group of o-hydroxbenzoic acid is relatively more favourable as compared to that of p-hydroxybenzoic acid. Since the o-hydroxybenzoate ion is stabilized by internal H-bonding, it is diffi cult to remove the —OH proton from this negatively charged cyclic system, i.e., the second ionization does not take place easily. Also, the ionization is disfavoured because of the fi eld effect generated by —COO① group on the nearby —OH group. In p-hydroxybenzoate ion, on the other hand, the —OH group is free and so the second ionization occurs easily in this case.
[Although intramolecular H-bonding occurs in undissociated o-hydroxybenzoic acid, it is less effective than its anion because a negative charge on oxygen leads to a stronger H-bonding.]
1. How can o-nitrophenol be separated from its m- or p-isomer ? Give reasons.
Solution A compound can be steam-distilled if it has an appreciable vapour pressure at the boiling point of water. Due to intramolecular H-bonding, o-nitrophenol is a low-boiling compound and it has appreciable vapour pressure at 100°C. So it can be separated from its m- or p-isomer by steam distillation.
2. Explain why in the vapour state acetic acid (CH3COOH) has molecular weight of 120.
Solution In the vapour phase, acetic acid (CH3COOH) exists as dimmers in which a pair of acid molecules are held together by intermolecular hydrogen bonds. Because of this, acetic acid in the vapour state has a molecular mass of 120.
3. The compound II is more basic than the compound I. Explain.
Solution The conjugate acid (monoprotonated from) of the cis-diamine II is stabilized by intramolecular hydrogen bonding while the conjugate acid of the trans-diamine I is not stabilized by intramolecular hydrogen bonding because the groups are held further apart.
For this reason, the diamine II is relatively more basic than the diamine I.
4. Nonpolar ethane (bp – 88.2°C) boils at a temperature higher than methane (bp – 162°C) at a pressure of 1 atm — Why?
Solution Heavier molecules require greater thermal energy in order to acquire velocities suffi ciently great to escape the surface of the liquid and since the surface areas of heavier molecules are usually much greater, therefore, intramolecular van der Waals attractions are also much larger. Ethane (CH3CH3) is heavier than methane (CH4) and also its surface area is greater than methane. For these reasons, ethane boils at a temperature higher than methane.
5. The fl uorocarbon C5F12 has a slightly lower boiling point than pentane (C5H12), even though it has a far higher molecular mass. Explain.
Solution Since the molecular mass of C5F12 is much higher than C5H12 and also its size is somewhat larger than C5H12 (F is slightly larger than H), therefore, C5F12 is expected to boil at a temperature higher than C5H12. This can be explained in terms of relative polarizability of the electrons of the atoms involved. The highly electronegative fl uorine atoms show a very low polarizability because their electrons are very tightly held with the nucleus. Because of this, C5F12 molecules are held together by very small van der Waals forces and in fact, it is much smaller than that operates in C5H12. For this reason, C5H12 has a slightly lower boiling point than pentane (C5H12), even though it has a far higher molecular weight.
6. Arrange the following compounds in order of increasing boiling point and explain the order:
Solution To predict relative boiling points, we have to consider the differences in (i) hydrogen bonding and (ii) molecular mass and surface area. 2,2-Dimethylpropane (neopentane) is the lightest among these compounds and it has compact spherical shape for which the van der Waals attractions are minimum. For this reason, 2,2-dimethylpropane is the lowest-boiling compound. The other four compounds have similar molecular masses.
Neither hexane nor 2,3-dimethylbutane is hydrogen bonded, so they will be the next higher in boiling points. Because 2-3-dimethylbutane is more highly branched and has a smaller surface area than hexane, the former will be lower boiling than the latter. The
two remaining compounds are alcohols and they remain associated with intermolecular hydrogen bonding. Therefore, their boiling points are higher than the three hydrocarbons.
Now, 1-pentanol has more surface area for van der Waals forces to operate and hence its boiling point is higher than the relatively compact 2-methyl-2-butanol. Therefore, the increasing order of boiling point is: 2,2-dimethylpropane < 2,3-dimethylbutane < hexane
>2-methyl-2-butanol < 1-pentanol.
7. The cis-isomer of 1,2-dibromethylene boils at a higher temperature than its trans-isomer — Why?
Solution The trans-isomer of 1,2-dibromoethylene is nonpolar because the two C — Br bond moments orienting in opposite directions cancel each other. So relatively weak van der Waals forces hold the molecules together. The cis-isomer, on the other hand, is polar because the two Br atoms lie on the same side of the double bond and the two C — Br bond moments do not cancel each other. A net moment resulting from two individual bond moments operates. As a result, there operates relatively stronger dipole–dipole attractions amongst the molecule along with van der Waals forces. Since more energy is needed to separate the molecules by overcoming these forces, the cis-isomer boils at a higher temperature than the trans-isomer.
8. o-Xylene boils at a higher temperature than p-xylene — Why?
Solution o-Xylene molecules are weakly polar. So there operates weak dipole–dipole attractions amongst the molecules in addition to the van der Waals forces. On the other hand, p-xylene molecules are nonpolar and so only weak van der Waals forces hold the molecules together. Because of this, o-xylene boils at a higher temperature than p-xylene, but the difference in boiling points is actually very small.
9. The melting point of neopentane (Me4C) is much higher than that of isopentane [CH3CH2CH(CH3)2]. Explain.
Solution The compact symmetrical molecules of neopentane (MC34C) pack well into a crystal lattice while isopentane [CH3CH2CH(CH3)2], which has a —CH3 group dagling from a four-carbon chain, does not. Therefore, intracrystalline forces are much stronger in neopentane compared to isopentane. Consequently, neopentane has a much higher melting point than isopentane.
10. Methyl fl uoride (CH3F) is more soluble in water than methyl chloride (CH3Cl) — Why?
Solution Methyl fl uoride (CH3F) containing highly electronegative and small-sized F atom can form hydrogen bond with water but methyl chloride does not. For this reason, methyl fl uoride is more soluble in water than methyl chloride.
CH —F H3 H F—CH3
O
11. tert-Butyl alcohol is more soluble in water than n-butyl alcohol. Explain.
Solution Alcohols containing branched alkyl groups are more water soluble than nonbranched alkyl groups with the same number of carbon atoms because branching minimizes the contact surface of the nonpolar portion of the molecule and thereby decreases the van der Waals forces of attraction. It is for this reason, the relatively more compact tert-butyl alcohol is more soluble in water than n-butyl alcohol.
∞
∞
3
3 3 2 2 2
3
CH|
H C — C — OH CH CH CH CH OH
| -Butyl alcohol
CH (bp 118 C)
-Butyl alcohol (bp 83 C)
n tert
12. Butane (CH3CH2CH2CH3) is soluble in carbon tetrachloride but not in water — Why ?
Solution The molecules of both butane and carbon tetrachloride (CCl4) are nonpolar and the molecules of each compound are held to each other by weak van der Waals interaction.
Since ‘like dissolves like’ is a rule for dissolution of a solute in a solvent and the dissolution process involves a large increase in entropy, i.e., thermodynamically favourable, butane dissolves in carbon tetrachloride. The highly polar water molecules are held together by strong dipole–dipole interactions and hydrogen bonds. There could be only very weak attractive forces (dipole-induced dipole attractions) between butane molecules on the one hand and water molecules on the other hand. Since solute–solvent interactions cannot outweigh or equal to the sum of solute–solute interactions (van der Waals forces) and solvent–solvent interactions (hydrogen bonding forces), butane does not dissolve in water.
13. Ammonium chloride (N H Cl)≈ 4@ is insoluble in the nonpolar solvent carbon tetrachloride while tetramethylammonium chloride (Me N Cl)4 ≈ @ is appreciably soluble in this solvent. Explain.
Solution In nonpolar solvent CCl4, stabilization of N H≈ 4 and Cl① ions by solvation does not take place and because of this, ammonium chloride is insoluble in CCl4. Tetramethyl- ammonium chloride, Me NCl,4≈ @ on the other hand, is appreciably soluble in CCl4 because the cation in which the positive nitrogen is surrounded by four methyl groups presents a large nonpolar hydrocarbon surface to this solvent and becomes involved with it by van der Waals forces of attraction.
14. Explain why glycerol (HOCH2CHOH CH2OH) is a very viscous liquid.
Solution The degree of molecular association through intermolecular hydrogen bonding is much higher in glycerol containing three —OH groups. For this reason, this triol is a very viscous liquid.
15. The melting point of NaCl (801°C) is very much higher than that of CCl4 (–24°C) — Why ?
Solution The crystal units in NaCl are cations and anions, which are held together by strong electrostatic forces of attraction. Hence it requires a large amount of energy to separate the ions. On the other hand, the crystal units in CCl4 are nonpolar molecules, which are held together by weak van der Waals forces. Therefore, these molecules can be separated by applying a small amount of thermal energy. This explains why the melting point of NaCl is very much higher than that CCl4.
16. In which of the following solvents would cyclohexane have the lowest solubility: diethyl ether, 1-pentanol, hexane and ethanol ?
Solution Cyclohexane is a nonpolar compound which is expected to be more soluble in a nonpolar solvent like hexane or a weakly polar solvent diethyl ether (‘like dissolves like’) than in an H-bonding and polar solvent like 1-pentanol and ethanol. Again, the nonpolar hydrocarbon part of ethanol (C2H5OH) is smaller than that in 1-pentanol (CH3CH2CH2CH2CH2OH). Therefore, cyclohexane is least soluble in ethanol.
17. Salicylic acid, o-HOC6H4COOH, is a stronger acid than o-CH3OC6H4COOH.
Explain why?
Solution Because of the ortho-effect of the bulkier —OCH3 group, o-CH3OC6H4COOH is expected to be more acidic. However, the reverse is true, and this is because, the conjugate base of salicylic acid is stabilized by internal H-bonding (a more dominating factor).
18. 2,6-dihydroxy-4-methylbenzoic acid is a much stronger acid than 2-hydroxy-4-methylbenzoic acid. Explain.
Solution The conjugate base of 2,6-dihydroxy-4-methylbenzoic acid is highly stabilized by intramolecular H-bonding (chelation) involving both the o-hydroxyl groups and for this reason, it is a much stronger acid than 2-hydroxy-4-methylbenzoic acid, the conjugate base of which is stabilized by intramolecular H-bonding involving only one —OH group.
19. Arrange the following compounds in order of increasing boiling point and explain your answer.
3 2 2 3 3 2 3 3 2 2 2 3
CH CH OCH CH CH CH CHCH CH CH CH CH CH
Diethyl ether | Pentane
OH -Butyl alcohol sec
Solution Pentane has no polar groups, so its molecules are held together only by weak van der Walls forces. Diethyl ether has the polar ether group, so it can have dipole–dipole interactions in addition to van der Waals forces. 2-Butanol is a polar bent molecule, so it can have dipole–dipole interactions in addition to van der Waals forces. Because it has an O—H bond 2-butanol molecules are held together by intermolecular hydrogen bonds.
Since hydrogen bonding is stronger than dipole–dipole attractions which in turn stronger than van der Waals forces, therefore, the boiling point of these three compounds increases in the order: pentane < diethyl ether < sec-butyl alcohol.
1. Which compound in each of the following pairs would have the higher boiling point?
Explain your answer.
(a) HOCH2CH2OH or CH3CH2CH2OH (b)
(c) (d)
(e) (f)
2. Arrange the following compounds in order of increasing boiling point and explain the order:
3. Which compound in each pair has the higher melting point ? (a)
(b)
(c) (d)
4. Which compound in each pair is more soluble in water and why?
(a) (b)
(c) (d)
5. Symmetry affects the melting point of a compound but not the boiling point — Why?
6. The boiling point of ethanol (78°C) is more than 100°C higher than that of dimethyl ether (–25°C) while ethylmethylamine has a boiling point (37°C) only 34° higher than that of triethylamine (3.5°C). Explain.
7. 1-Butanol (bp 118°C) has a much higher boiling point than its isomer ethoxyethane (bp 35°C). However, both of them show same solubility (8 g pre 100 g) in water.
Account for these observations.
8. Comment on the solubility of the following compounds in water and in organic solvents (such as CCl4):
NaCl, CH3CH2CH2CH3, CH3CH2CH2OH, CH3(CH2)10OH
9. What types of intermolecular forces are present in each of the following compounds?
(a) (b) (c) (CH3)3N
(d) CH2 == CHBr (e) CH3CH2CH2COOH (f) CH3CH2C ∫∫ C CH2CH3 10. o-Hydroxybenzaldehyde has much lower boiling point and much lower water
solubility than its m- and p-isomers. Explain.
11. Which of the following compounds exhibit chelation:
(a) methyl salicylate, (b) o-iodophenol, (c) o-fl uorophenol, (d) o-cyanophenol, and (e) o-cresol?
[Hint: Chelation occurs only in (a) and (c). In (d), although the N atom of —CN is electronegative, the linearity of the —CN group places the N atom too far away from the —OH group to form an H-bond.]
12. Acietic acid dimmer, (CH COOH)3 2 C H6 6 2CH COOH3 DS = +20 e.u.; despite the favourable entropy of dissociation, the equilibrium lies to the left. Explain why?
[Hint: Strength of intramolecular H-bonding in the dimer is more than enough to comprensate for the loss of freedom in the dimer.]
13. 2-methylpyrrolidine boils at temperature higher than the boiling point of pyrrolidine
— Why.
14. Between the anti- and syn-isomers of pyridine-2-carboxaldoxime the anti- form predominates in the isomeric distribution of the compound. Explain this observations.
[Hint: The anti-form gets extra stability due to intramolecular hydrogen bonding (chelation). No such chelating effect is observed in the syn-isomer. For this reason, the anti-form predominates in the isomeric distribution of the compound.
15. Resorcinol has higher boiling point than 2-nitroresorcinol — Why?
16. 8-hydroxyquinoline can be separated from 4-hydroxyquinoline by steam distillation.
Account for this observation.
[Hint: Intramolecular H-bonding occurs in 8-hydroxyquinoline but not in 4-hydroxyquinoline.
17. Draw the H-bonding arrangements in CH3OH — H2O and CH3NH2 — HCHO systems.
[ ]
18. Why is the mp of sulphanilic acid so high?
[Hint: Sulphanilic acid exists as a salt (a dipolar ion or zwitter ion):
H N—3 —SO3–. Because of this, sulphanilic acid has a much higher melting point.]
19. Explain why H2O has a higher boiling point than CH3OH (65°C), NH3 (–33°C) and HF (20°C).
20. Arrange the following compounds in order of increasing boiling point. Explain your answer.
21. Explain why 1-pentanol has solubility of 2.7 g per 100 ml of water, where as ethanol is completely miscible in water.
22. Why does one expect the cis-isomer of an alkene to have a higher boiling point than the trans-isomer?
23. Alcohols with fewer than four carbons are soluble in water, but alcohols with more than four carbons are insoluble in water — Why?
24. The boiling point of propylamine (bp 49°C) is higher than that of ethylmethylamine (bp 37°C) which in turn is higher than that of triethylamine (bp 3.5 °C). Explain.
[Hint: Trimethylamine (MC3N) has no N — H bond and so it cannot form hydrogen bonds with each other. Ethylmethylamine (CH3 CH2NH CH3) has one N — H bond and as it remains associated through intermolecular hydrogen bonding. Propylamine (CH3CH2CH2NH2) with two N — H bonds is more extensively hydrogen bonded.
This explains their boiling points]
25. There are four amides with the molecular formula C3H7 NO. Write their structures.
One of these amides has melting and boiling point that is substantially lower than that of the other three. Identify this amide and explain your answer.
[Hint: CH3CH2CONH2, CH3CONHCH3, HCONHCH2CH3 and HCON (CH3)2. The last one has a melting and boiling point that is substantially lower than that of the other three because it does not have a hydrogen that is covalently bonded to nitrogen and, therefore, its molecules cannot form hydrogen bonds to each other. The other molecules all have a hydrogen covalently bonded to nitrogen, and therefore, hydrogen-bond formation is possible.]
26. Which of the following compounds is expected to volatilize easily and why?
I II
N NH OH
N NH
CH CH
[Hint: Due to intramolecular hydrogen bonding, I is expected to volatilize easily.]