Basic Organic Chemistry
1. Brief introduction
1) Degree of unsaturation (double bond equivalents, DBE)
① CnHm
( )
I. DBE=
×
II. e.g. prismane C6H6, DBE= = 4 (4 rings)
② heteroatoms
I. oxygen & sulphur
a. not included in the formula
×
b. e.g. ethanol C2H6O, DBE= =0
II. nitrogen
a. DBE=
×
b. e.g. CH3CH2N(CH3)2 C4H11N, DBE= =0
III. Halogen
a. Treat halogens as hydrogen
2. Alkanes
1) Structure and bonding
① Carbon atoms form four σ-bonds (fully saturated). So tetrahedral
② Linear Combination of Atomic Orbitals (LCAO)
I.
II. Bond angle = 109.5°
2) σ-framework (only -bonds)
①
② For methane, all equal electron pairs are equally repulsive. So methane adopts a
tetrahedral shape
, ③ Lone pairs of electrons are held closer to the nucleus of the parent atoms; therefore,
exhibit a greater repulsive effect than those in a covalent bond
3) conformational analysis
① conformation equilibrium
I. particularly stable conformations are called conformers
II. Boltzmann distribution of conformations is governed by the energy of that state
(i.e. the most stable conformer is the most populated)
② ethane
I. ethane can adopt infinite conformations by differing degree of rotation around C-C
bond
II. there are two extreme conformations: staggered & eclipsed
III.
IV. torsional angles (τ)
,V.
VI. Not to be confused with average bond angles, which are still 109°
VII. Staggered is lower in energy than eclipsed, because:
a. Minimise the steric interactions of the hydrogen atoms (not the main reason,
as the hydrogen atoms are too small to get in each other’s way)
b. there are some stabilising σ to σ* interaction between the C–H σ bonding
orbital on one carbon and the C–H σ*antibonding orbital on the other carbon
(maximised when exactly parallel, in staggered conformation)
c. the repulsion between the electrons in the bonds is at a minimum in the
staggered conformation
VIII. potential energy diagram
, ③ propane
I.
II. Methyl is electron donating, so has more steric effect. But steric factor is still not
important
III. The repulsion (van der Waals repulsion) is greater in propane than ethane as methyl
group is larger than H.
IV. Van der Waals repulsion is caused by the additional internal energy in the molecule
caused by the valance electron shells of two atoms competing for the same space
V. The larger the groups the greater this interaction (a large group requires more
space)
VI. The energy graph for bond rotation in propane would look exactly the same as that
for ethane except that the barrier is now 14 kJ mol–1.
① n-butane
I. the two methyl groups are large enough to get in the way of each other so steric
factor is important
1. Brief introduction
1) Degree of unsaturation (double bond equivalents, DBE)
① CnHm
( )
I. DBE=
×
II. e.g. prismane C6H6, DBE= = 4 (4 rings)
② heteroatoms
I. oxygen & sulphur
a. not included in the formula
×
b. e.g. ethanol C2H6O, DBE= =0
II. nitrogen
a. DBE=
×
b. e.g. CH3CH2N(CH3)2 C4H11N, DBE= =0
III. Halogen
a. Treat halogens as hydrogen
2. Alkanes
1) Structure and bonding
① Carbon atoms form four σ-bonds (fully saturated). So tetrahedral
② Linear Combination of Atomic Orbitals (LCAO)
I.
II. Bond angle = 109.5°
2) σ-framework (only -bonds)
①
② For methane, all equal electron pairs are equally repulsive. So methane adopts a
tetrahedral shape
, ③ Lone pairs of electrons are held closer to the nucleus of the parent atoms; therefore,
exhibit a greater repulsive effect than those in a covalent bond
3) conformational analysis
① conformation equilibrium
I. particularly stable conformations are called conformers
II. Boltzmann distribution of conformations is governed by the energy of that state
(i.e. the most stable conformer is the most populated)
② ethane
I. ethane can adopt infinite conformations by differing degree of rotation around C-C
bond
II. there are two extreme conformations: staggered & eclipsed
III.
IV. torsional angles (τ)
,V.
VI. Not to be confused with average bond angles, which are still 109°
VII. Staggered is lower in energy than eclipsed, because:
a. Minimise the steric interactions of the hydrogen atoms (not the main reason,
as the hydrogen atoms are too small to get in each other’s way)
b. there are some stabilising σ to σ* interaction between the C–H σ bonding
orbital on one carbon and the C–H σ*antibonding orbital on the other carbon
(maximised when exactly parallel, in staggered conformation)
c. the repulsion between the electrons in the bonds is at a minimum in the
staggered conformation
VIII. potential energy diagram
, ③ propane
I.
II. Methyl is electron donating, so has more steric effect. But steric factor is still not
important
III. The repulsion (van der Waals repulsion) is greater in propane than ethane as methyl
group is larger than H.
IV. Van der Waals repulsion is caused by the additional internal energy in the molecule
caused by the valance electron shells of two atoms competing for the same space
V. The larger the groups the greater this interaction (a large group requires more
space)
VI. The energy graph for bond rotation in propane would look exactly the same as that
for ethane except that the barrier is now 14 kJ mol–1.
① n-butane
I. the two methyl groups are large enough to get in the way of each other so steric
factor is important
