Organic Chemistry - Oklahoma City Community … Chemistry Review Information for Unit 1...

Organic Chemistry

Review Information for Unit 1Intermolecular ForcesEmpirical and Molecular Formulas

Intermolecular Forces

Attractive forces between molecules are very important in “condensed” phases (solids and liquids) molecules are constantly in contact with each other

important in determining the BP, MP, and solubility of compounds


3 major kinds of intermolecular (attractive) forces in solids and liquids:

dipole-dipole interactionspolar molecules

hydrogen bondingmolecules with -OH or -NH groups

London dispersion forcesall molecules


Polar molecules have permanent dipoles positive end negative end

The most stable arrangement of polar molecules is the one in which the positive end of one molecule is oriented toward the negative end of the other molecule.

- -++

Dipole-dipole interaction (force)


Dipole-Dipole Interaction (Force) an attractive intermolecular force resulting from the attraction of the positive and negative ends of the dipole moments of polar molecules

When a polar liquid vaporizes, the dipole-dipole interactions must be overcome. More polar molecules have stronger dipole-dipole interactions. large DHvap

high BP


Organic compounds containing O-H and/or N-H bonds exhibit hydrogen bonding: a strong dipole-dipole interaction between a hydrogen atom that is covalently bonded to either O, N, or F and a lone pair of electrons on a different O, N, or F atom


Due to the large difference in electronegativity, O-H and N-H bonds are highly polar H has a partial positive charge in such bonds

The H atom is an electrophile and has a strong affinity for nonbonding electrons on a N or O atom (other than the one it is covalently bonded to).

O - H N - Hd- d+d-d+


H

H

O

O

Hydrogen bonding

between 2 ethyl

alcohol molecules:

Hydrogen bonding

between 2 methyl

amine molecules:

Hydrogen

bond

H

H

N

N

CH3

HCH

3

Hd-

d+

d+

d- d-

d+

d+

d+

d+

d-


Although H-bonding is a strong attractive force, it is weaker than covalent bonds: H-bond

~20 kJ/mol to break

C-H, N-H, O-H covalent bonds:~400 kJ/mol to break

H-bonding has a large effect on physical properties.


Impact of H-bonding on BP: H-bonding leads to higher BP

Alcohols have stronger H-bonds than aminesalcohol BP > amine BP (when # of C is the same)

As the number of H’s on a N increases, H-bonding increases:amine BP increases with increasing # of H’s on the N


In nonpolar molecules, London dispersion forcesare the principal attractive force. Arise from temporary dipole moments induced in one molecule by other nearby moleculeselectrons in molecules are displaced from symmetrical arrangement

temporary dipole established last for a fraction of a secondchange continuously


London dispersion forces in nonpolar

molecules:


London dispersion forces require close surface contact between two (or more) molecules. The strength of the London dispersion forces is roughly proportional to surface area.As surface area increases, LDF increase and BP increases


For simple hydrocarbons with the same total number of carbons: Unbranched (linear) has greater surface areahigher BP

branched has less surface area (more spherical) lower BP

98.4oC

81oC


Example: Which of the following should have the highest BP?

C

C

Cl

Cl

ClCl

Cl

H

ClCl

C

C

Cl

Cl

ClCl

Cl

H

ClCl

C

C

Cl

Cl

ClCl

Cl

H

ClCl

N-pentane 2-methyl butane neopentane


Example: Which of the following should have the highest BP?

C

C

Cl

Cl

ClCl

Cl

H

ClCl

C

C

Cl

Cl

ClCl

Cl

H

ClCl

C

C

Cl

Cl

ClCl

Cl

H

ClCl

N-pentane 2-methyl butane neopentane

Since all of the molecules are nonpolar, LDF are the

most important attractive force.

As surface area increases, BP increases.

For simple hydrocarbons, surface area is greatest

for unbranched hydrocarbons.

C

C

Cl

Cl

ClCl

Cl

H

ClCl

Answer:


Given a series of compounds, you should be able to rank them in order of their BP based on these intermolecular forces and explain why.

To predict relative boiling points, look for differences in: Molecular weight Hydrogen bonding Polarity Surface area


If two compounds have similar molecularweights, then the IMFs with the greatest influence on BP are:

hydrogen bonding > dipole-dipole> LDF

OH O

60 g/mol 58 g/mol60 g/mol 58 g/mol

89oC 56.5oC -11.7oC


If two compounds have drastically different molecular weights, then molecular weight maybe more important than hydrogen bonding or polarity in determining their relative BPs.

64.7oC 174.1oC69oC

CH3OH

32 g/mol 86 g/mol 142 g/mol


Example: Arrange the following in order of increasing boiling point. Explain why.

OH

OHHO

NH2

N

H

OH

OHHO

NH2

N

HOH

OHHO

NH2

N

H

OH

OHHO

NH2

N

H

OH

OHHO

NH2

N

H

hexane 2,2-dimethylbutane

1-hexanoldipropylamine

n-hexylamine1,6-hexanediol

OHHO


Neither hexane or 2,2-dimethylbutane are capable of hydrogen bonding so they will have the two lowest BP’s. 2,2-dimethylbutane is more branched

lower BP hexane is linear

higher BPOH

OHHO

NH2

N

H

OH

OHHO

NH2

N

H

hexane 2,2-dimethylbutane


1-hexanol, 1,6-hexanediol, dipropylamine, and n-hexylamine all have groups capable of forming hydrogen bonds. amines form weaker hydrogen bonds than alcoholsdipropylamine and n-hexylamine will have lower BP’s than the alcohols

Dipropylamine has one N-H bond (less H-bonding) while n-hexylamine has 2 N-H bonds (more H-bonding)


So far:

2,2-dimethylbutane < hexane < dipropylamine < n-hexylamine

1,6-hexanediol has two OH groups while 1-hexanol has one OH group 1,6-hexanediol has higher BP than 1-hexanol


So:

2,2-dimethylbutane < hexane < dipropylamine <

n-hexylamine < 1-hexanol < 1,6-hexanediol

49.7oC 68.95oC 109.4oC

250oC158oC130oC


Intermolecular forces also determine the solubility of organic compounds. “Like dissolves like”

polar compounds dissolve in polar solventsnonpolar compounds dissolve in nonpolar solvents

Ionic compounds generally dissolve readily in water because water hydrates or solvates the individual ions


Polar organic compounds dissolve in polar solvents due to: dipole-dipole interactions H-bonding

Hydrogen bonding

between methyl

amine and water. H

H

N

OH

HCH

3


Nonpolar compounds do not dissolve in water because they cannot break the H-bonding network that exists in a sample of water

Calculating Empirical & Molecular Formulas

The empirical formula for an unknown compound can be determined using quantitative elemental analysis.

Often, the % composition for all elements except oxygen is reportedExample: 40.0% C and 6.67% H

Remainder (unless it adds up to very close to 100%) is assumed to be oxygenIn previous example: 53.3% O


Calculate empirical formula by: % to mass mass to mole divide by smallest multiply ‘til whole


Example: Calculate the empirical formula for a compound containing 81.8% C and 18.2% H.

Cmol6.82C g 12.0

C mol 1 Cg81.8

Steps 1 & 2: % to Mass; Mass to Moles

Hmol18.0H g 1.01

H mol 1 Hg18.2


Step 3: Divide by Smallest

C: 6.82 = 1.00 = 1

6.82

H: 18.0 = 2.64

Step 4: Multiply ‘til Whole

Since CH2.64 doesn’t make any sense, you

must multiply to get a whole number ratio.

2.64 ~ 2.67 ~ 2 2/3 or 8/3. So multiply both

numbers by 3.

C3H8


Molecular formulas are always some whole number multiple of the empirical formula:

C2H4O (acetic acid)

CH2O

C6H12O6 (glucose)

X 2

X 6


Molecular weight is the same whole number multiple of the empirical formula’s formula weight (FW).

C2H4O (acetic acid)

CH2O

C6H12O6 (glucose)

X 2

X 6FW = 30.0 amu

MW = 30.0 amu x 2 = 60.0 amu

MW = 30.0 amu x 6 = 180.0 amu


Molecular weights can be determined experimentally in several ways: colligative properties ideal gas law mass spectrometry

To determine the molecular formula for an unknown: calculate the empirical formula calculate ratio of MW/empirical FW multiply subscripts of empirical formula by previous ratio


Example: Determine the molecular formula for a compound that contains 39.8 % C, 3.32% H, and 39.2% Cl if its molecular weight is about 181 amu.

Step 1: Find Empirical Formula

33.021.10Cmol3.32Cg12.0Cmol1Cg39.8

11.001.10Clmol1.10Clg 35.5

Clmol1Clg39.2

39921.10Hmol3.29Hg 1.01Hmol1Hg3.32 .

11.011.10Omol1.11Og16.0Omol1Og17.7


Empirical Formula = C3H3ClO

Step 2: Find MW/FW (empirical)

MW = 181 (given)FW (empirical) = 3(12.0) + 3 (1.01) + 35.5 +

16.0= 90.5

MW = 181 = 2FW 90.5

Step 3: Multiply subscripts

Molecular Formula = C(3x2)H(3 x 2)Cl(1x2)O(1x2)

= C6H6Cl2O2

Organic Chemistry - Oklahoma City Community … Chemistry Review Information for Unit 1...

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