Chapter 6 Intermolecular Forces of Attractions Between Particles

Chapter 6 Intermolecular Forces of Attractions Between Particles Why mixtures mix Consider a glass of wine. Why do alcohol, water, & pigment mix together? There must be attractive forces. Intramolecular forces is the force of attraction between atoms in a molecule Intermolecular forces is the attraction between molecules Intermolecular forces are responsible for many properties of molecular compounds, including crystal structures (e. g. the shapes of snowflakes), melting points, boiling points, Types of Intermolecular Forces of attraction 1)

2) 3) 4) Ionic (Ionic bonds form if the EN is 1.7 or greater) dipole dipole (EN is around 0.5-1.7) H-bonding( is a type of dipole-dipole) London forces or Van der Waals Forces (exist in all molecules, but are especially important in nonpolar covalent molecules, where EN is less than 0.5). Ionic intermolecular forces of attraction are strongest Dipole-dipole are not as strong as ionic Hydrogen bonds are about five times stronger than regular dipole-dipole bonds. London forces are weakest force of attraction. London forces Non-polar molecules do not have dipoles like polar

molecules. How, then, can non-polar compounds form solids or liquids? London forces are named after Fritz London (also called van der Waal forces) London forces are due to small dipoles that exist in non-polar molecules. These forces exist between all molecules Because electrons are moving around in atoms there will be instants when the charge around an atom is not symmetrical The resulting tiny dipoles cause attractions between atoms/molecules Dipole-dipole interaction This happens in all molecules having polar bonds. But Polar molecules are held to each other more strongly than the non-polar molecules of comparable molecular weight. occur when polar molecules are attracted to one another. Electrostatic attraction between (+) and (-) ends of molecules

This accounts for the attraction one water molecule has for another water molecule. H-bonding H O H H O H H-bonding is a special type of dipole - dipole attraction that is very strong It occurs when hydrogen is covalently bonded to a very electronegative atom like N, O, or F because of the small size of hydrogen, its positive charge can get very close to the negative dipole of another molecule. H is so strongly positive that it will sometimes exert a pull on a lone pair in a non-polar compound

The high EN (electronegativity difference) between two atoms like of NH, OH, and HF bonds cause these to be strong forces (about 5x stronger than normal dipole-dipole forces). Calculate the EN for HCl and H2O Soln: HCl = 2.9-2.1 = 0.8 H2O = 3.5-2.1 = 1.4 They are given a special name (H-bonding) because compounds containing these bonds are important in biological systems Molecules with 3 Atoms Even though the C-O bond is polar, the bonds cancel each other out because the molecule is linear the dipole moments are equal and in opposite directions. Therefore CO2 is non-polar. CO2 HCN SO2 The dipole moment between H-C

points in the direction of C. The dipole moment points between C-N points in the direction of the N. Therefore the dipole vectors are additive and HCN is polar SO2 is a polar molecule because the SO dipole Moments dont cancel each other out due to the angle Which of the following molecules are polar (have a dipole moment)?H2O, CO2, SO2, and CH4 O H H dipole moment polar molecule S O O

dipole moment polar molecule H O C O no dipole moment nonpolar molecule H C H H no dipole moment

nonpolar 10.2 Molecules with 4 Atoms CCl4 is non-polar CHCl3 is polar Formation of Intermolecular H- bonds 1.Between non-identical molecules a. alcohol and water H R H O H H

O R O H O H O R H b. chloroform and pyridine Cl Cl C Cl H

N c. acid and water O O R H H C H O H O H 2. Between identical molecules a. NH3 molecule

H H H N H N H H b. hydrocyanic acid H C H N c. alcohols O O

R H H R C N d. acids H O R O C O O

H e. amines H R H N C H H N R H R N

R R Intramolecular H- bond results to ring formation or chelation. Ortho substituted are usually involved in chelation, because meta and para substituents are already distal group as far as effective H- bonding is concerned. A. O H F B. H3C C CH O

C. O C O H O CH H CH O CH2CH3 STRUCTURE AND PHYSICAL PROPERTIES 1. MELTING POINT 2. BOILING POINT 3. SOLUBILITY

BOILING POINT One of the most revealing of all physical properties for a chemical substance is its boiling point. Boiling point reflects the strength of the intermolecular attractive forces that hold the molecules of a substance together in a condensed phase, and as such, it is useful to compare the boiling points for related compounds to see how structural differences account for the differences in intermolecular attractions. The trends in boiling points for various groups of compounds helps in understanding how size, shape, and functional group polarity affect boiling point. The is the temperature Thenormal boiling boiling point ispoint the temperature at whichatthe which a liquid vapor

boils when the external pressure (equilibrium) pressure of a liquid is equalisto1 atm. the external pressure. The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure. 11.8 FACTORS THAT AFFECTS BOILING POINT/ MELTING POINT 1. Molecular Weight/Size The bigger the molecule, the higher its boiling point. Thus, for a series of related compounds, the higher the molecular weight, the higher the boiling point. Note the trend for the first five straight-chain alkanes:

Van der Waals attractive forces increases as the hydrocarbon chain increases so that the boiling point is high. Compound Formula (mol. wt.) B.P. methane CH4(16) -164C ethane C2H6 (30) -88C propane C3 H8 (44) -42C

n-butane C4 H10 (58) 0C n-pentane C5 H12 (72) 36C 2. Branching in the hydrocarbon lowers the boiling point In the straight chain hydrocarbon molecules can approach each other so stronger Van der Waals attractive force. In branched chain hydrocarbon, there will be no close approach of molecules so that there is weak van der waals attractive force. Example:

CH3 CH3 C CH3 < CH3 CH2 CH2 CH3 neopentene n-pentene CH2 CH3

3. Dipole moment measured in Debye Units (D) = e X d ( magnitude of charge X distance) differences in electronegativity The more polar the higher is the boiling point Molecular nitrogen (N2) and carbon monoxide (CO) have identical molecular weights: 28g/mol. Which has the higher boiling point? 4. Association Associated liquids - liquids whose molecules are held together by H- bonds. Boiling points are abnormally high. The more associated the liquid, the higher the boiling point. Example: methanol CH3OH can form 1 H- bond CH3COOH can form 2 H- bonds

5. CH3 CH CH3 CH3 CH2 CH3 CH3 C CH In propene, diffused pi electron cloud would not allow the molecules to be closer as there will be repulsion. Propane molecule interact with each other more strongly than propene molecule. In propyne, the pi electron cloud is tighter so that

molecule can approach each other closer, so stronger and therefore higher boiling point than propane. b. ) compounds. Determine which The following are set of organic one has a higher boiling point or melting point. Why? a.) OH benzene van der waals b.) O H phenol van der waals & H-bonding O H

O H NO2 NO2 The polar nitro group provides additional attractive forces between molecule O H O H NO2 NO2 intramolecular H-bonding is invoked in the ortho position

When OH and NO2 groups are tied up internally by a hydrogen bond, their effectivity in forming intermolecular hydrogen bonding is lessened. O O H O H H O N O O N O

H intramolecular H-bond O intermolecular H-bonding c.) O H NH2 Lone pair from N can easily be delocalized to the ring, whereas, lone pair in phenol is not as easily delocalized because of the strong attracting power of this atom from the lone pair. d.) H3C CH2 CH2

CH2 NH2 H3C CH2 CH2 CH2 OH Butanol forms stronger H-bonding because O is more electronegative than N. e.) H3C CH2 O CH2

CH3 very volatile (low boilingpoint), molecules cannot form Hbonding, only van der waals f.) O H3C CH2 CH2 C H butanal doest not form H-bonding H3C CH2

CH2 CH2 butanol forms H-bonding OH g.) H3C CH2 CH2 CH2 H3C OH MW= 74

Forms only one H-bond molecules h.) MW= 74 forms 2 H-bond between O R O H O R H Forms only one H-bond molecules CH2

R H O C C O H R O forms 2 H-bond between COOH MELTING POINT Melting a change from a highly ordered arrangement of the particles to the more random arrangement that characterizes liquid. Melting occurs when the temperature is reached at which the thermal energy is

great enough to overcome the intracrystalline forces that hold particles in position. To melt ionic solid enough energy is needed to break the ionic bond. Example: NaCl ( mpt = 801oC) To melt molecular solid enough energy is needed to break the intermolecular forces Example: CH4 ( mpt= -183oC) Electrostatic Example: CH3OH hydrogen bonding and dipole-dipole mpt less than 801oC but higher than -183oC H 2O (l) The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium Freezing

(s) Melting H2O 11.8 11.8 a. Which is of higher melting point? C O H O H C

O H OH OH o-hydroxybenzalhehyde - intramolecular H-bond - less associated with each other b. C O H O C H p-hydroxybenzaldehyde

- intermolecular H-bond - more associated with each other C O H NH2 o-aminobenzaldehyde NH2 p-aminobenzaldehyde C. C O H C O

O H CH3 O CH3 o-methoxybenzaldehyde p-methoxybenzaldehyde more defined negativity less defined negativity stronger interaction weaker interaction Both will not form H- bonding d.) carboxylic acids with increasing MW exhibit increasing melting point. SOLUBILITY - is dissolution of the structural units ( ions or molecules) -are separated using up the energy coming from the attraction

between the solute and the solvent molecules -the energy required to break the bonds between solute particles is supplied by the formation of bonds between the solute particles and the solvent molecules - The solubility of a solute is the maximum quantity of solute that can dissolve in a certain quantity of solvent or quantity of solution at a specified temperature. - The factors that determine solubility are the strength of IMFs and speed of molecules. The main factors that have an effect on solubility are: 1.the nature of the solute and solvent -- While only 1 gram of lead (II) chloride can be dissolved in 100 grams of water at room temperature, 200 grams of zinc chloride can be dissolved. The great difference in the solubilities of the of these two substances is the result of differences in their natures. 2.temperature -- Generally, an increase in the temperature of the solution increases the solubility of a solid solute. A few solid solutes, however, are less soluble in warmer solutions. For all gases, solubility decreases as the temperature of the solution rises. 3. pressure -- For solids and liquid solutes, changes in pressure have practically no effect on solubility. For gaseous solutes, an increase in pressure increases solubility and a decrease in pressure decreases solubility. (When the cap on a bottle of soda pop is removed, pressure is released, and the gaseous solute

bubbles out of solution. This escape of a gas from solution is called effervescence.) The rate of solution is a measure of how fast a substance dissolves. Some of the factors determining the rate of solution are: 1.size of the particles -- When a solute dissolves, the action takes place only at the surface of each particle. When the total surface area of the solute particles is increased, the solute dissolves more rapidly. Breaking a solute into smaller pieces increases its surface area and hence its rate of solution. (Sample problem: a cube with sides 1.0 cm long is cut in half, producing two pieces with dimensions of 1.0 cm x 1.0 cm x 0.50 cm. How much greater than the surface area of the original cube is the combined surface areas of the two pieces? 2.0 cm2 2. stirring -- With liquid and solid solutes, stirring brings fresh portions of the solvent in contact with the solute, thereby increasing the rate of solution. 3. amount of solute already dissolved -- When there is little solute already in solution, dissolving takes place relatively rapidly. As the solution approaches the point where no solute can be dissolved, dissolving takes place more slowly. 4. temperature -- For liquids and solid solutes, increasing the temperature not only increases the amount of solute that will dissolve but also increases the rate at which the solute will dissolve. For gases, the reverse is true. An increase in temperature decreases both solubility and rate of solution. Solvation can be done in 2 ways: 1.Use of lonepairs -------aprotic solvents

2. Use of H-bonding ------protic solvents Protic Solvents - solvent containing H that is attracted to O and N and appreciably acidic. Example: CH3OH and H2O ( can solvate both + and ion) Solvation of NaCl using H2O Aprotic solvent polar solvent of moderately high dielectric constants, which do not contain acidic hydrogen. Example: Dimethylsulfoxide CH3 S CH3 O O N,N-dimethylformamide(DMF) H C O

N CH3 CH3 Sulfolane O O Solubility of Acids and Bases General Rule: Strong acids are soluble in strong bases and vise versa. Weak bases are soluble strong acids and vise versa.

in Solubility of amines in dilute HCl is associated with the tendency of the lone pair of electron of N to bond with the proton. H 2N + HCl H 3 N+ Cl - salt formation In general aliphatic amines are soluble in dilute HCl. When alkyl groups are bulky in secondary and tertiary amines, solubility in

dilute HCl decreases. This is a consequence of steric inhibition of the approach of the acid to bond with the lone pair or the instability of the salt formed as a result of steric overcrowding. CH3 H3C H3C NH2 CH3 CH3 CH3 Triphenylamine is not soluble in dilute HCl because of unavailability of the lone pair for coordination with the acid as result of resonance effects or effects due to electron delocalization. N

Amides are not soluble in dilute HCl as simple amines because of lesser availability of lone pair of electron of the N of amides compared with simple amines. The lone pair of amides is delocalized towards the carbonyl carbon. O R C NH 2 Amide R amine NH2

Disubstituted amides, however, in contrast to the simple amides are more soluble in dilute HCl. In these disubstituted amides the alkyl groups increases the availability of the lone pair of electrons on the nitrogen for the acid to coordinate with. O R C N R R Solubility

of unsaturated noncyclic hydrocarbons and some aromatic hydrocarbons in cold concentrated H2SO4 is a consequence of the availability of pi electron for coordination with proton. CH3CH=CH2 + H2SO4 CH3CH2-CH3 H + H2SO4 + H

H RO R + H2SO 4 R -O -R + H RO H + H 2S O 4 R -O -H +

O R-C-OR + OH H2SO 4 R-C OR Solubility of organic compounds in dilute sodium hydroxide is a consequence of the presence of acidic hydrogen. OH O Na + S- H + CH3 - N H

+ aqNaOH N Na + + HOH O + O + aq NaO H

O O OH HO H O O O + S Na aqNaO H O N HO H CH2

N CH2 O aq NaO H N O O + O O Na + +

HO H In chelated phenols, acidic H is tied up as a hydrogen bond, thus insoluble in dilute NaOH. C H O O H Ketones and aldehydes are insoluble in dilute NaOH even though they posses acidic H, because the acidity of H is too weak to allow dissolution in dilute NaOH. Solubility of organic compounds in dilute sodium bicarbonate(NaHCO3) can reflect strength in acidity of these systems. Only carboxylic acid, sulfonic acids, and

sulfinic acids are soluble in dilute sodium bicarbonate solution(NaHCO3). Phenols and aliphatic alcohols which are regarded as weak acids do not dissolved in NaHCO3, but 2, 4, 6-trinitrophenol is an exception, because the 3 NO2 increases the acidity of phenol. Solubility of chloroform in organic bases such as pyridine and trimethylamine consequence of H- bonding. Solubility of is

a unsaturated noncyclic hydrocarbons and some aromatic hydrocarbons in cold concentrated H2SO4 is a consequence of the availability of pi electron for coordination with proton.

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