Organic Chemistry 1 – Comprehensive Study Guide
2025 | Detailed Notes, Exam Questions & Verified
Answers | Reaction Mechanisms, Structures &
Concepts Explained
Which is stronger, a sigma bond or a pi bond? Why? - ANSW✔-Sigma bonds are significantly stronger
than pi bonds. This is because sigma bonds allow for electron density to be concentrated to a much
larger degree between the two nuclei. The lowest energy state for electrons—being electrostatically
attracted to both nuclei simultaneously, is between those two nuclei and as close to each nuclei as
possible. In a pi bond the p orbitals overlap above and below the atom, localizing the electrons above
and below the plane of the bond—a higher energy state compared to the head-on overlap of a sigma
bond. You can also conceptualize that pi bonds are weaker simply because we know those electrons are
in a higher-energy state. It is universally true that when a bond is HIGHER in energy it will require LESS
energy to break it.
The first bond to form between two elements... - ANSW✔-is ALWAYS a sigma bond (ex. a single bond,
or the first bond or a double or triple bond) and involves head to head overlap of two atomic orbitals.
The second bond or third bond to be formed between two elements... - ANSW✔-are ALWAYS pi bonds
and involves side to side overlap of two p-orbitals. Because pi bonds require side-to-side overlap, the
atoms must be fairly close to one another.
Which is the weaker bond c=o or c=n? - ANSW✔-c=n, because the radii are further apart.
As the radius of either atom increases, the p orbitals are spread apart, resulting in less overlap and a
weaker pi bond.
Remember that atoms with smaller radius ( like Fluorine) will be able to hug other things more easily
because of its small radius.
,Important Note on Pi Bonds - ANSW✔-Pi bonds themselves are weaker associations than are sigma
bonds. However, a double bond (one sigma bond plus one pi bond ) is stronger together than a single
bond (one sigma bond only). Also, triple bonds are much shorter, stronger bonds than are single bonds,
but are also more reactive. Reactivity has to do with the tendency of the third bond (a pi bond) to react.
Bond strength is at measure of the energy needed to completely break the two atoms apart (ex. break
all three bonds)
Relative characteristics of sigma and pi bonds. - ANSW✔-
How does rotational limitation of pi bonds relate to our previous study of proteins? - ANSW✔-The
amide functional group in a protein (created when the amino group of one amino acid attacks the
carboxylic acid group of another amino acid) exhibits resonance and therefore both the C-O and C-N
bonds have double-bond character. This prevents rotation—a key characteristic of peptide bonds.
Hybridization definition - ANSW✔-Atoms, when bonded, hybridize (mix) their higher and lower energy
valence orbitals to form "hybrid" orbitals with intermediate energy. Carbon has 2 e's in s and p orbitals.
However, Carbon will form 4 orbitals with sp3 energy.
How do you determine Hybridization? - ANSW✔-Count the number of sigma bonds and add the
number of lone PAIRS (of unbonded electrons). This number will equal the sum of the superscripts on
one of the following hybridizations:
sp, sp2, sp3, sp3d, sp3d2
(Note: no superscript indicates a superscript of one).
What is the percent "s" character of the hybrid oxygen orbital in water? - ANSW✔-When orbital
hybridize they do so in a weighted manner such that the character of the hybrid orbital is an average of
its contributors. Therefore, an sp hybrid orbital has 50% s character and 50% p character. By this same
logic, the oxygen in water—which is sp3 hybridized, exhibits 25% s character and 75% p character.
Mathematically, 1s orbital and 3p orbital's equal 4 orbitals (1s + 3p = 4 orbitals). 1 out of 4, 1⁄4 , or 25%
is from the s orbital.
,What is VESPR? - ANSW✔-Valence Shell Electron Pair Repulsion Theory. Predicts the shape molecules
will take due to the repulsion of lone pairs of electrons. Hybridization state determines the possible
shapes for the molecule.
Shape and bond angle are determined by? - ANSW✔-1). Hybridization
2). Lone pairs of electrons
What is the Hybridization of BeCl2? - ANSW✔-Linear
BeCl2 is going to have: Be has a sigma bond with each of the chlorines. Which gives it a linear
Hybridization.
Be --> 2 valence e-
Cl --> 7 valence e- (2x)
Total = 16 valence e-
Each chlorine has 6 valence electrons (nonbonding) and 2 valence electrons (bonding with the Be).
Formal Charge = (valence e-) - (nonbonding e-) - (bonding e- / 2)
Each Cl and Be has a formal charge of 0, which lets me know that this is the correct structure.
Hybridization - ANSW✔-sp= linear (180)
sp2 = Trigonal planar or bent (120)
sp3 = Tetrahedral, trigonal pyramidal, or bent (109.5)
sp3d= Trigonal bipyramidal, seesaw, T-shaped, or linear (90, 120, or 180)
, sp3d2= Octahedral, square pyramidal, or square planar (90)
Lone pairs of electrons - ANSW✔-The presence or absence of unpaired electrons determines the exact
shape from among those choices. For example, an sp2 hybridized atom can take on the trigonal planar
shape (no lone pairs) or the bend shape (one lone pair). Sp hybridized atoms are always linear.
1) The H-O-H bond angle in water is measured experimentally to be 104.4°. The related angle H-N-H in
ammonia is expected to be:
A) larger due to the greater electron repulsion in ammonia
B) larger due to the greater electron repulsion in water
C) smaller due to the greater electron repulsion in water
D) smaller due to the greater electron repulsion in ammonia - ANSW✔-Solution: Both molecules are
approximately tetrahedral in shape, but experience some distortion due to the repulsion of lone pairs of
electrons. In water, two lone pairs push both hydrogen substituents away from the plane shared by the
two lone pairs, decreasing the H-O-H bond angle. Ammonia has only one lone pair, so the same effect is
observed, but to a lesser degree. Thus, the bond angle is larger in ammonia due to the greater repulsion
in water than in ammonia, or answer B.
Rank the following according to decreasing bond length: a) triple bonds
b)double bonds
c)single bonds - ANSW✔-Single > Double > Triple [bond length]
Rank the bonds listed above according to increasing stability. - ANSW✔-When you see the term
"stability" on the MCAT they mean "thermodynamic stability"—which in
essentially every case will be reflected by the strength of the bond. Remember that strong bonds are a
"lower-energy" state and weaker bonds represent a "higher-energy" state. Therefore, TRIPLE BONDS are
the MOST thermodynamically STABLE, meaning it will require the most energy to break them apart.
In increasing order of stability they would be ranked: single < double < triple. (Be careful not to confuse
stability with reactivity.)
2025 | Detailed Notes, Exam Questions & Verified
Answers | Reaction Mechanisms, Structures &
Concepts Explained
Which is stronger, a sigma bond or a pi bond? Why? - ANSW✔-Sigma bonds are significantly stronger
than pi bonds. This is because sigma bonds allow for electron density to be concentrated to a much
larger degree between the two nuclei. The lowest energy state for electrons—being electrostatically
attracted to both nuclei simultaneously, is between those two nuclei and as close to each nuclei as
possible. In a pi bond the p orbitals overlap above and below the atom, localizing the electrons above
and below the plane of the bond—a higher energy state compared to the head-on overlap of a sigma
bond. You can also conceptualize that pi bonds are weaker simply because we know those electrons are
in a higher-energy state. It is universally true that when a bond is HIGHER in energy it will require LESS
energy to break it.
The first bond to form between two elements... - ANSW✔-is ALWAYS a sigma bond (ex. a single bond,
or the first bond or a double or triple bond) and involves head to head overlap of two atomic orbitals.
The second bond or third bond to be formed between two elements... - ANSW✔-are ALWAYS pi bonds
and involves side to side overlap of two p-orbitals. Because pi bonds require side-to-side overlap, the
atoms must be fairly close to one another.
Which is the weaker bond c=o or c=n? - ANSW✔-c=n, because the radii are further apart.
As the radius of either atom increases, the p orbitals are spread apart, resulting in less overlap and a
weaker pi bond.
Remember that atoms with smaller radius ( like Fluorine) will be able to hug other things more easily
because of its small radius.
,Important Note on Pi Bonds - ANSW✔-Pi bonds themselves are weaker associations than are sigma
bonds. However, a double bond (one sigma bond plus one pi bond ) is stronger together than a single
bond (one sigma bond only). Also, triple bonds are much shorter, stronger bonds than are single bonds,
but are also more reactive. Reactivity has to do with the tendency of the third bond (a pi bond) to react.
Bond strength is at measure of the energy needed to completely break the two atoms apart (ex. break
all three bonds)
Relative characteristics of sigma and pi bonds. - ANSW✔-
How does rotational limitation of pi bonds relate to our previous study of proteins? - ANSW✔-The
amide functional group in a protein (created when the amino group of one amino acid attacks the
carboxylic acid group of another amino acid) exhibits resonance and therefore both the C-O and C-N
bonds have double-bond character. This prevents rotation—a key characteristic of peptide bonds.
Hybridization definition - ANSW✔-Atoms, when bonded, hybridize (mix) their higher and lower energy
valence orbitals to form "hybrid" orbitals with intermediate energy. Carbon has 2 e's in s and p orbitals.
However, Carbon will form 4 orbitals with sp3 energy.
How do you determine Hybridization? - ANSW✔-Count the number of sigma bonds and add the
number of lone PAIRS (of unbonded electrons). This number will equal the sum of the superscripts on
one of the following hybridizations:
sp, sp2, sp3, sp3d, sp3d2
(Note: no superscript indicates a superscript of one).
What is the percent "s" character of the hybrid oxygen orbital in water? - ANSW✔-When orbital
hybridize they do so in a weighted manner such that the character of the hybrid orbital is an average of
its contributors. Therefore, an sp hybrid orbital has 50% s character and 50% p character. By this same
logic, the oxygen in water—which is sp3 hybridized, exhibits 25% s character and 75% p character.
Mathematically, 1s orbital and 3p orbital's equal 4 orbitals (1s + 3p = 4 orbitals). 1 out of 4, 1⁄4 , or 25%
is from the s orbital.
,What is VESPR? - ANSW✔-Valence Shell Electron Pair Repulsion Theory. Predicts the shape molecules
will take due to the repulsion of lone pairs of electrons. Hybridization state determines the possible
shapes for the molecule.
Shape and bond angle are determined by? - ANSW✔-1). Hybridization
2). Lone pairs of electrons
What is the Hybridization of BeCl2? - ANSW✔-Linear
BeCl2 is going to have: Be has a sigma bond with each of the chlorines. Which gives it a linear
Hybridization.
Be --> 2 valence e-
Cl --> 7 valence e- (2x)
Total = 16 valence e-
Each chlorine has 6 valence electrons (nonbonding) and 2 valence electrons (bonding with the Be).
Formal Charge = (valence e-) - (nonbonding e-) - (bonding e- / 2)
Each Cl and Be has a formal charge of 0, which lets me know that this is the correct structure.
Hybridization - ANSW✔-sp= linear (180)
sp2 = Trigonal planar or bent (120)
sp3 = Tetrahedral, trigonal pyramidal, or bent (109.5)
sp3d= Trigonal bipyramidal, seesaw, T-shaped, or linear (90, 120, or 180)
, sp3d2= Octahedral, square pyramidal, or square planar (90)
Lone pairs of electrons - ANSW✔-The presence or absence of unpaired electrons determines the exact
shape from among those choices. For example, an sp2 hybridized atom can take on the trigonal planar
shape (no lone pairs) or the bend shape (one lone pair). Sp hybridized atoms are always linear.
1) The H-O-H bond angle in water is measured experimentally to be 104.4°. The related angle H-N-H in
ammonia is expected to be:
A) larger due to the greater electron repulsion in ammonia
B) larger due to the greater electron repulsion in water
C) smaller due to the greater electron repulsion in water
D) smaller due to the greater electron repulsion in ammonia - ANSW✔-Solution: Both molecules are
approximately tetrahedral in shape, but experience some distortion due to the repulsion of lone pairs of
electrons. In water, two lone pairs push both hydrogen substituents away from the plane shared by the
two lone pairs, decreasing the H-O-H bond angle. Ammonia has only one lone pair, so the same effect is
observed, but to a lesser degree. Thus, the bond angle is larger in ammonia due to the greater repulsion
in water than in ammonia, or answer B.
Rank the following according to decreasing bond length: a) triple bonds
b)double bonds
c)single bonds - ANSW✔-Single > Double > Triple [bond length]
Rank the bonds listed above according to increasing stability. - ANSW✔-When you see the term
"stability" on the MCAT they mean "thermodynamic stability"—which in
essentially every case will be reflected by the strength of the bond. Remember that strong bonds are a
"lower-energy" state and weaker bonds represent a "higher-energy" state. Therefore, TRIPLE BONDS are
the MOST thermodynamically STABLE, meaning it will require the most energy to break them apart.
In increasing order of stability they would be ranked: single < double < triple. (Be careful not to confuse
stability with reactivity.)
