Chapter 6 School Material Flashcards
Kinds of Organic Reactions
Here, the chapter covers the main types of organic reactions:
Substitution reactions: One group is replaced by another.
Addition reactions: Atoms or groups add to a double or triple bond.
Elimination reactions: Atoms are removed, creating double or triple bonds.
Rearrangements: Structural reorganization of molecules.
Bond Breaking and Bond Making
Homolytic bond cleavage: Each atom takes one electron from a bond, forming radicals.
Heterolytic bond cleavage: One atom takes both electrons from a bond, forming ions. This section introduces reaction mechanisms, highlighting how bonds break and form.
Thermodynamics
Thermodynamics discusses how energy changes in reactions:
Exergonic reactions: Release energy (spontaneous).
Endergonic reactions: Require energy input (non-spontaneous). It also explains the relationship between Gibbs free energy, enthalpy, and entropy.
Enthalpy and Entropy
Enthalpy (ΔH)
: Heat released or absorbed during a reaction.
Entropy (ΔS): The measure of disorder in a system. These factors combine in the Gibbs free energy equation to determine spontaneity.
Energy Diagrams
Energy diagrams illustrate the energy changes during a reaction:
Reactants and products are shown at specific energy levels.
Transition states represent the highest energy point.
Activation energy (Ea) is the energy barrier that must be overcome for a reaction to proceed.
Kinetics explains the rate of a reaction, influenced by:
Concentration of reactants
Temperature
Catalyst presence
Rate-determining step in multi-step reactions
Analysis of Figure 6.1:
Ways to Write Organic Reactions
This figure demonstrates the notation used to represent organic reactions, including:
Reagent Placement:
The reagent (e.g., Br₂) can be written either:
On the left side of the reaction equation, combined with the reactant.
Above the arrow, indicating its addition to the reaction.
Reaction Parameters:
Additional information can be provided above or below the reaction arrow to specify:
Solvent: For instance, CCl₄ (carbon tetrachloride) is used here.
Conditions: Symbols are used to indicate specific requirements:
hν: Indicates light is needed for the reaction to proceed.
Δ: Indicates heat is required for the reaction.
Reaction Outcomes:
The products depend on the conditions and reactants involved. For example, bromination of a cyclic alkene results in the addition of bromine atoms to the double bond.
What does the symbol hν represent in an organic reaction?
Answer: hν represents the requirement of light (typically UV light) to initiate or drive the reaction.
If a reaction requires heat, how is that indicated in the reaction diagram?
Answer: Heat is indicated by the symbol Δ above or near the reaction arrow.
Why might a solvent like CCl₄ be used in an organic reaction?
Answer: CCl₄ is a non-polar solvent that stabilizes non-polar reagents like Br₂ and ensures the reaction occurs in a homogeneous phase, aiding proper contact between reactants.
In a bromination reaction, what kind of bond does Br₂ typically react with?
Answer: Br₂ typically reacts with double bonds (π bonds) in alkenes, undergoing an electrophilic addition reaction.
What would happen if no light or heat is provided in a reaction requiring hν or Δ?
Answer: The reaction would not proceed because the energy input (light or heat) is essential for breaking bonds or initiating intermediates. For instance, hν is needed to generate bromine radicals in radical reactions.
Rewrite the reaction using Br₂ as a reagent on the left side instead of above the arrow.
If a different solvent was used instead of CCl₄, how might this affect the reaction?
Answer: Using a different solvent could:
Change the solubility of the reactants and products.
Alter the reaction rate or mechanism. For example, polar solvents could stabilize ionic intermediates, leading to side reactions or a different pathway.
Explain why it’s important to specify reaction conditions (e.g., light, heat, solvent) in organic reactions.
Answer: Specifying reaction conditions ensures:
The reaction proceeds correctly under optimal conditions.
Reproducibility of results in different labs or experiments.
Prevention of side reactions by controlling energy input or stabilizing intermediates.
Analysis of Writing Equations for Sequential Reactions
This image illustrates the process of writing equations for sequential reactions, where two steps are carried out one after the other, without explicitly showing the intermediate compound in the reaction equation.
Key Points:
Sequential Reactions:
When two reactions happen in sequence, the reagents for each step are written above or below the reaction arrow and labeled in order.
Each step occurs separately and in sequence, not simultaneously.
Convention:
Step 1 occurs first, followed by Step 2. The reagents for each step are clearly identified.
This approach avoids drawing intermediates in the reaction, making the representation concise.
grignard reagent
Grignard reagents are extremely useful organometallic compounds in the field of organic chemistry. They exhibit strong nucleophilic qualities and also have the ability to form new carbon-carbon bonds. Therefore, they display qualities that are also exhibited by organolithium reagents and the two reagents are considered similar.
When the alkyl group attached to a Grignard reagent is replaced by an amido group, the resulting compound is called a Hauser base. These compounds are even more nucleophilic than their Grignard counterparts.
What is the purpose of numbering the reagents in a sequential reaction?
Answer: Numbering the reagents clearly indicates the order in which they are added. It helps to show that the first reagent reacts with the starting material, and the second reagent is introduced after the first step is completed.
Why is the intermediate compound often omitted in the reaction equation?
Answer:
The intermediate is often omitted for simplicity, as it is usually unstable and short-lived, and not the primary focus of the reaction scheme. This keeps the representation concise.
What type of reagent is
𝐶𝐻3𝑀𝑔𝐵𝑟, and what role does it play in the first step of the reaction?
Answer:
𝐶𝐻3𝑀𝑔𝐵𝑟 is a Grignard reagent, which acts as a nucleophile. In the first step, it adds to the carbonyl group of acetone (𝐶𝐻3𝐶𝑂𝐶𝐻3) to form a magnesium alkoxide intermediate.
Why is
𝐻2𝑂 added in the second step of this reaction?
Answer:
𝐻2𝑂 is added to hydrolyze the magnesium alkoxide intermediate, replacing the magnesium group with a hydrogen atom to produce the final alcohol product.
Write the full reaction mechanism for the conversion of acetone to the alcohol product, including the intermediate:
definition of substitution reaction
A substitution reaction occurs when one atom or group of atoms (𝑍) on a carbon atom is replaced by another atom or group (𝑌).
examples of substitution reaction
Examples:
Replacement of a halogen by a nucleophile in a nucleophilic substitution reaction.
Replacement of a hydrogen atom in an aromatic ring by an electrophile in an electrophilic substitution reaction.
general substitution reaction
What are the two main types of substitution reactions in organic chemistry?
Answer:
Nucleophilic Substitution (
𝑆𝑁1 and
𝑆n2
2): A nucleophile replaces a leaving group.
Electrophilic Substitution: An electrophile replaces a substituent, commonly in aromatic compounds.
How is
𝑍 characterized in a substitution reaction? What makes it leave the molecule?
Answer:
𝑍 is the leaving group, which must be stable after leaving the molecule. Stability is often due to resonance stabilization or being a weak base (e.g.,
𝐶𝑙−,𝐵𝑟−.
In electrophilic substitution, what type of group (electron-rich or electron-deficient) typically replaces
𝑍?
Answer: An electron-deficient group (electrophile), such as
𝑁𝑂2+
or
𝐵𝑟+
, typically replaces
𝑍 in electrophilic substitution reactions.
Why might a substitution reaction fail if the leaving group
𝑍 is not stable?
Answer: A substitution reaction might fail because an unstable leaving group does not detach easily, increasing the activation energy and potentially causing the reaction to favor other pathways.
Suggest a solvent that could favor a nucleophilic substitution reaction and explain why.
Answer:
For
𝑆𝑁1: A polar protic solvent (e.g., water, ethanol) stabilizes the carbocation intermediate and the leaving group.
For
𝑆𝑁2: A polar aprotic solvent (e.g., acetone, DMSO) increases the nucleophilicity of the attacking nucleophile by not solvating it strongly.
A polar aprotic solvent such as acetone, dimethyl sulfoxide (DMSO), or acetonitrile is ideal for favoring a nucleophilic substitution reaction, particularly SN2 reactions.
Explanation:
• Polar aprotic solvents have a high polarity but lack hydrogen atoms bonded to electronegative atoms (like oxygen or nitrogen). This means they cannot form strong hydrogen bonds with nucleophiles.
• By not solvating the nucleophile, these solvents leave the nucleophile “naked” and free to attack the electrophilic carbon, enhancing the nucleophile’s reactivity.
• In SN2 reactions, the nucleophile directly displaces the leaving group in a single-step, bimolecular mechanism. A strong and unhindered nucleophile is essential for this.
Examples of Polar Aprotic Solvents:
1. DMSO – High polarity, stabilizes the transition state while not solvating the nucleophile.
2. Acetone – Good solvent for dissolving ionic nucleophiles without reducing their reactivity.
3. Acetonitrile (CH3CN) – Polar and aprotic, promotes SN2 mechanisms efficiently.
Why not Polar Protic Solvents?
Polar protic solvents like water or alcohols can form hydrogen bonds with nucleophiles, reducing their nucleophilicity and slowing down the substitution reaction. Protic solvents are better suited for SN1 mechanisms, where the nucleophile’s strength is less critical.
Thus, a polar aprotic solvent like DMSO is most suitable for favoring nucleophilic substitution reactions, particularly SN2.
Key Points
Bond Formation and Breaking:
Substitution reactions involve the breaking of one σ bond (single covalent bond) and the formation of another σ bond at the same carbon atom.
Leaving Group (
𝑍):
While 𝑍 can sometimes be a hydrogen atom, it is more commonly a heteroatom (e.g.,
𝐶𝑙, 𝐵𝑟, 𝐼
) that is more electronegative than carbon. This makes it a good leaving group.
Why must the leaving group (
𝑍) in a substitution reaction be more electronegative than carbon?
Answer: A leaving group must be able to stabilize the negative charge it acquires upon leaving. Electronegative atoms like
𝐶𝑙,
𝐵𝑟, and
𝐼 are good leaving groups because they can delocalize and stabilize the charge through their high electronegativity and size (in the case of
𝐼−).
What is the difference between
𝑆𝑁1 and
𝑆𝑁2 substitution mechanisms?
Answer:
𝑆𝑁1
: A two-step mechanism where the leaving group departs first, forming a carbocation intermediate, followed by nucleophilic attack. It occurs on tertiary or stabilized carbons.
𝑆𝑁2: A one-step mechanism where the nucleophile attacks the substrate from the opposite side as the leaving group, displacing it in a concerted reaction. It occurs on primary or less hindered carbons.
In the reaction
𝐶𝐻3𝐼 + 𝐶𝑙− → 𝐶𝐻3𝐶𝑙 + 𝐼−
, why is
𝐶𝑙− able to replace
𝐼−?
Answer:
𝐶𝑙− replaces
𝐼−
because iodine is a better leaving group than chlorine due to its larger size and ability to stabilize the negative charge more effectively after leaving.
Why do substitution reactions involving
𝑆𝑁2 favor less sterically hindered substrates like
𝐶𝐻3𝐼
Answer:
𝑆𝑁2 reactions require the nucleophile to attack the carbon from the opposite side of the leaving group. In sterically hindered substrates (e.g., tertiary carbons), the nucleophile cannot approach the carbon easily, reducing reaction efficiency. Less hindered substrates (e.g.,
𝐶𝐻3𝐼) provide better access for the nucleophile.
Suggest a reaction condition that would favor
𝑆𝑁1over
𝑆𝑁2
for a given substrate.
Answer:
𝑆𝑁1
is favored by:
Polar protic solvents (e.g., water, ethanol) that stabilize the carbocation intermediate and the leaving group.
Tertiary or stabilized carbons that can form stable carbocations.
Low nucleophile strength, as a strong nucleophile would favor
𝑆𝑁2
Analysis of Elimination Reactions
This slide explains elimination reactions, a key reaction type in organic chemistry, where elements are removed (lost) from a molecule, resulting in the formation of a double bond (
𝜋-bond).
Key Points
Definition:
An elimination reaction is a process in which two atoms or groups of atoms (typically
𝑋 and
𝑌) are removed from adjacent carbons of a molecule, resulting in the formation of a double bond (
𝐶=𝐶).
Bond Changes:
Two
𝜎-bonds (single bonds) are broken during the reaction.
A new
𝜋-bond (double bond) is formed between the two carbons.
General Reaction:
Starting material:
𝐶−𝐶 with attached
𝑋 and
𝑌.
Product: An alkene (
𝐶=𝐶) and a byproduct (
𝑋−𝑌), which may be a small molecule like
𝐻𝐶𝑙 or
𝐻2𝑂
What types of elimination reactions are most common in organic chemistry?
The two most common types of elimination reactions are:
E1 (Unimolecular elimination): A two-step process involving the formation of a carbocation intermediate before elimination.
E2 (Bimolecular elimination): A one-step, concerted process where the base abstracts a proton and the leaving group departs simultaneously.
What is the role of the reagent in an elimination reaction?
Answer: The reagent, typically a base, abstracts a proton (𝐻+) from a β-carbon (adjacent to the carbon with the leaving group), initiating the elimination process and facilitating the formation of the double bond.
What type of molecule is typically formed as the product in an elimination reaction?
Answer: The product is typically an alkene (
𝐶=𝐶), along with a small molecule byproduct like
𝐻𝐶𝑙,
𝐻𝐵𝑟, or
𝐻2𝑂
Write an example of an elimination reaction where
𝑋 = 𝐵𝑟 and
𝑌 = 𝐻
What are the two main elimination mechanisms, and how do they differ?
Answer:
E1 (Unimolecular Elimination):
The leaving group departs first, forming a carbocation intermediate.
A base then removes a proton (𝐻+) from the β-carbon, forming the double bond.
Favored by: Weak bases, polar protic solvents, and tertiary carbons.
E2 (Bimolecular Elimination):
The base abstracts a proton (𝐻+) while the leaving group departs simultaneously.
A concerted reaction with no intermediates.
Favored by: Strong bases, polar aprotic solvents, and less hindered substrates.
Explain why elimination reactions are favored under high-temperature conditions.
Answer: Elimination reactions are favored at high temperatures because the formation of a double bond (
𝐶 = 𝐶) and the loss of small molecules (e.g.,
𝐻𝐶𝑙,
𝐻2𝑂) increase entropy (Δ𝑆. High temperatures amplify the entropy term in the Gibbs free energy equation (
Δ𝐺 = Δ𝐻 − 𝑇Δ𝑆), making the reaction more thermodynamically favorable.
In a molecule with multiple β-hydrogens, how do elimination reactions decide which hydrogen is eliminated?
Answer: The elimination often follows Zaitsev’s Rule, which states that the most substituted (stable) alkene is the preferred product. However, with bulky bases (e.g., tert-butoxide), the less substituted alkene (Hofmann product) may be favored due to steric hindrance.
Suggest how the choice of base (strong vs. weak) influences the pathway of an elimination reaction.
Answer:
Strong bases (e.g., NaOH, NaOEt) favor the E2 mechanism because they can abstract protons quickly and efficiently.
Weak bases (e.g., water, alcohols) favor the E1 mechanism as they rely on the formation of a carbocation intermediate.
What are the roles of
𝑋 and
𝑌 in elimination reactions?
Answer:
𝑋 is typically a hydrogen atom that is removed from the β-carbon.
𝑌 is usually a leaving group (such as a halogen or
𝑂𝐻) that departs from the adjacent carbon. Together, their removal creates a double bond (
𝐶 = 𝐶) between the two carbons.
Why is a
𝜋-bond formed during an elimination reaction?
Answer:
The loss of two groups (
𝑋 and
𝑌) from adjacent carbons leaves two unshared electrons, which form the
𝜋-bond of the resulting double bond (
𝐶 = 𝐶).
Application Questions
for elimination reactions
Explain the mechanism of the elimination of
𝐻2𝑂 from cyclohexanol using sulfuric acid.
How does the strength of the base influence the elimination mechanism (
𝐸1
vs
𝐸2)?
Answer:
Strong Base: Favored in
𝐸2 reactions, where the base abstracts a proton and the leaving group departs in a single concerted step.
Weak Base: Favored in
𝐸1 reactions, where the leaving group departs first to form a carbocation, followed by deprotonation by the weak base.
Critical Thinking on elimination reactions
Analysis of Addition Reactions
This slide explains addition reactions, one of the fundamental types of organic reactions where new atoms or groups are added to a molecule, breaking a
𝜋-bond and forming two new
𝜎-bonds.
Key Points
Definition:
An addition reaction occurs when elements
𝑋 and
𝑌 are added to a molecule, usually across a double bond (
𝐶=𝐶) or triple bond (
𝐶≡𝐶).
Bond Changes:
The
𝜋-bond of the double bond is broken.
Two new
𝜎-bonds are formed—one for each of the added groups (
𝑋 and 𝑌).
General Reaction:
Starting Material:
𝐶=𝐶 (alkene or alkyne).
Reagent:
𝑋−𝑌
(e.g.,
𝐻𝐶𝑙,
𝐵𝑟2
, or
𝐻2o).
Product: A saturated compound with both
𝑋and 𝑌 attached to the carbons of the original double bond.
What type of bond is broken in an addition reaction, and what bonds are formed?
Answer:
A
𝜋
π-bond in a double bond (
𝐶
=
𝐶
C=C) is broken, and two new
𝜎
σ-bonds are formed between the carbons and the added groups (
𝑋
X and
𝑌
Y).
Why do addition reactions typically occur in molecules with
𝜋
π-bonds?
Answer:
𝜋
π-bonds are weaker and more reactive than
𝜎
σ-bonds, making them more susceptible to attack by electrophiles or nucleophiles. Breaking a
𝜋
π-bond allows the molecule to form new, stronger
𝜎
σ-bonds.
Write an example of an addition reaction where
𝑋
=
𝐻
X=H and
𝑌
=
𝐶
𝑙
Y=Cl.
Answer:
The reaction of ethene (
𝐶
2
𝐻
4
C
2
H
4
) with
𝐻
𝐶
𝑙
HCl:
𝐶
𝐻
2
=
𝐶
𝐻
2
+
𝐻
𝐶
𝑙
→
𝐶
𝐻
3
𝐶
𝐻
2
𝐶
𝑙
CH
2
=CH
2
+HCl→CH
3
CH
2
Cl
(Ethene reacts with hydrogen chloride to form chloroethane.)
Predict the product when ethene reacts with
𝐵
𝑟
2
Br
2
in a nonpolar solvent.
Answer:
The product is 1,2-dibromoethane:
𝐶
𝐻
2
=
𝐶
𝐻
2
+
𝐵
𝑟
2
→
𝐶
𝐻
2
𝐵
𝑟
−
𝐶
𝐻
2
𝐵
𝑟
CH
2
=CH
2
+Br
2
→CH
2
Br−CH
2
Br
(Bromine adds across the double bond, forming a vicinal dibromide.)
Describe the stepwise mechanism for the reaction of ethene with
𝐻
𝐶
𝑙
HCl.
Answer:
Step 1: The
𝜋
π-electrons of the ethene attack the hydrogen of
𝐻
𝐶
𝑙
HCl, forming a carbocation intermediate on one of the carbons.
𝐶
𝐻
2
=
𝐶
𝐻
2
+
𝐻
+
→
𝐶
𝐻
3
−
𝐶
𝐻
+
CH
2
=CH
2
+H
+
→CH
3
−CH
+
Step 2: The chloride ion (
𝐶
𝑙
−
Cl
−
) attacks the carbocation, forming chloroethane:
𝐶
𝐻
3
−
𝐶
𝐻
+
+
𝐶
𝑙
−
→
𝐶
𝐻
3
𝐶
𝐻
2
𝐶
𝑙
CH
3
−CH
+
+Cl
−
→CH
3
CH
2
Cl
What factors determine the regioselectivity of addition reactions (e.g., Markovnikov’s rule)?
Answer:
Regioselectivity is determined by the stability of the intermediate carbocation. According to Markovnikov’s rule, the hydrogen atom (from
𝑋
−
𝑌
X−Y) adds to the carbon with the most hydrogens already attached, forming the more stable carbocation. The nucleophile (
𝑌
Y) then adds to the more substituted carbon.
How would the outcome of an addition reaction differ if it occurred with an asymmetric alkene (e.g., propene)?
Answer:
For an asymmetric alkene like propene, the addition reaction follows Markovnikov’s rule:
Example: Reaction with
𝐻𝐶𝑙:
𝐶𝐻3−𝐶𝐻=𝐶𝐻2+𝐻𝐶𝑙→𝐶𝐻3−𝐶𝐻𝐶𝑙−𝐶𝐻3
The hydrogen adds to the less substituted carbon, and the chlorine adds to the more substituted carbon.
Suggest conditions that would promote an anti-addition product rather than a syn-addition product.
Answer:
Anti-addition can be promoted by:
Using halogens (
𝐵
𝑟
2
,
𝐶
𝑙
2
Br
2
,Cl
2
) in the presence of a nonpolar solvent, forming a cyclic halonium ion intermediate. The nucleophile attacks from the opposite side, leading to anti-addition.
Reactions involving hydroboration-oxidation or stereospecific reagents can also favor anti-addition.
Analysis of Addition Reactions (with Examples)
This slide expands on addition reactions by providing specific examples where a
𝜋
π-bond is broken and new
𝜎
σ-bonds are formed, demonstrating the addition of specific reagents to alkenes.
Key Points
Mechanism:
Addition reactions involve breaking a
𝜋
π-bond and forming two new
𝜎
σ-bonds between the carbon atoms and the added groups.
Examples in the Slide:
Example 1: Addition of HBr to ethene (
𝐶
2
𝐻
4
C
2
H
4
):
𝐶
𝐻
2
=
𝐶
𝐻
2
+
𝐻
𝐵
𝑟
→
𝐶
𝐻
3
𝐶
𝐻
2
𝐵
𝑟
CH
2
=CH
2
+HBr→CH
3
CH
2
Br
The double bond (
𝜋
π-bond) is broken.
Hydrogen adds to one carbon, and bromine adds to the other, forming bromoethane.
Example 2: Addition of water to cyclohexene under acidic conditions (
𝐻
2
𝑆
𝑂
4
H
2
SO
4
):
𝐶
𝑦
𝑐
𝑙
𝑜
ℎ
𝑒
𝑥
𝑒
𝑛
𝑒
+
𝐻
2
𝑂
→
𝐻
2
𝑆
𝑂
4
𝐶
𝑦
𝑐
𝑙
𝑜
ℎ
𝑒
𝑥
𝑎
𝑛
𝑜
𝑙
Cyclohexene+H
2
O
H
2
SO
4
Cyclohexanol
The double bond is broken.
A hydrogen adds to one carbon, and a hydroxyl group (
𝑂
𝐻
OH) adds to the other, forming cyclohexanol.
Describe how the reaction of ethene with HBr can be reversed.
Answer:
The addition of HBr to ethene:
𝐶
𝐻
2
=
𝐶
𝐻
2
+
𝐻
𝐵
𝑟
→
𝐶
𝐻
3
𝐶
𝐻
2
𝐵
𝑟
CH
2
=CH
2
+HBr→CH
3
CH
2
Br
can be reversed by providing conditions that favor elimination, such as:
Using a strong base (e.g.,
𝑁
𝑎
𝑂
𝐻
NaOH) to abstract a proton (
𝐻
+
H
+
) from the β-carbon.
Heating the reaction mixture (
Δ
Δ) to increase entropy and drive the reaction towards the formation of ethene:
𝐶
𝐻
3
𝐶
𝐻
2
𝐵
𝑟
→
Δ
,
base
𝐶
𝐻
2
=
𝐶
𝐻
2
+
𝐻
𝐵
𝑟
CH
3
CH
2
Br
Δ,base
CH
2
=CH
2
+HBr
Predict the product of the elimination of HBr from bromoethane (
𝐶
𝐻
3
𝐶
𝐻
2
𝐵
𝑟
CH
3
CH
2
Br).
Answer:
The elimination of HBr from bromoethane results in the formation of ethene (
𝐶
𝐻
2
=
𝐶
𝐻
2
CH
2
=CH
2
):
𝐶
𝐻
3
𝐶
𝐻
2
𝐵
𝑟
+
𝑁
𝑎
𝑂
𝐻
→
Δ
𝐶
𝐻
2
=
𝐶
𝐻
2
+
𝑁
𝑎
𝐵
𝑟
+
𝐻
2
𝑂
CH
3
CH
2
Br+NaOH
Δ
CH
2
=CH
2
+NaBr+H
2
O
This reaction occurs via an
𝐸
2
E2 mechanism when a strong base is present, and high temperatures favor the elimination pathway.
Top Row
a. Cyclohexanol (
𝑂
𝐻
OH) to Bromocyclohexane (
𝐵
𝑟
Br):
Reaction Type: Substitution.
The
𝑂
𝐻
OH group is replaced by
𝐵
𝑟
Br.
b. Cyclohexanone (
𝐶
=
𝑂
C=O) to Cyclohexanol (
𝑂
𝐻
OH):
Reaction Type: Addition.
A hydrogen atom and hydroxyl group (
𝐻
+
𝑂
𝐻
H+OH) are added to the carbonyl group.
c. Acetone (
𝐶
𝐻
3
𝐶
𝑂
𝐶
𝐻
3
CH
3
COCH
3
) to 2-Chloropropane (
𝐶
𝐻
3
𝐶
𝐻
𝐶
𝑙
𝐶
𝐻
3
CH
3
CHClCH
3
):
Reaction Type: Substitution.
The chlorine atom replaces the carbonyl oxygen.
d. 2-Butanol (
𝐶
𝐻
3
𝐶
𝐻
2
𝐶
𝐻
(
𝑂
𝐻
)
𝐶
𝐻
3
CH
3
CH
2
CH(OH)CH
3
) to Butene (
𝐶
𝐻
3
𝐶
𝐻
=
𝐶
𝐻
𝐶
𝐻
3
CH
3
CH=CHCH
3
):
Reaction Type: Elimination.
Water (
𝐻
2
𝑂
H
2
O) is eliminated, forming a double bond.
Bottom Row
a. Cyclohexane-1,2-diol (
𝐻
𝑂
𝐶
𝐻
−
𝐶
𝐻
𝑂
𝐻
HOCH−CHOH) to Cyclohexanone (
𝐶
=
𝑂
C=O):
Reaction Type: Elimination.
A water molecule (
𝐻
2
𝑂
H
2
O) is eliminated, forming a ketone (
𝐶
=
𝑂
C=O).
b. Cyclohexane (
𝐻
H) to Chlorocyclohexane (
𝐶
𝑙
Cl):
Reaction Type: Substitution.
A hydrogen atom is replaced by a chlorine atom.
c. Acetaldehyde (
𝐶
𝐻
3
𝐶
𝐻
𝑂
CH
3
CHO) to Ethanol (
𝐶
𝐻
3
𝐶
𝐻
2
𝑂
𝐻
CH
3
CH
2
OH):
Reaction Type: Addition.
A hydrogen atom (
𝐻
2
H
2
) adds to the carbonyl group.
d. Cyclohexanoyl chloride (
𝐶
𝑂
𝐶
𝑙
COCl) to Cyclohexanal (
𝐶
𝐻
𝑂
CHO):
Reaction Type: Substitution.
The chlorine atom is replaced by a hydrogen atom, forming an aldehyde.
types of reaction
One-step Reaction (Concerted Reaction):
The reaction occurs in a single step, with all bond-breaking and bond-forming happening simultaneously.
No intermediates are formed.
Example:
𝐴
→
𝐵
A→B.
Stepwise Reaction:
The reaction occurs in multiple steps.
Involves the formation of an unstable intermediate, known as a reactive intermediate, before the product is formed.
Example:
𝐴
→
Intermediate
→
𝐵
A→Intermediate→B.
Reactive Intermediates:
Reactive intermediates are short-lived, high-energy species that exist only during the course of the reaction. Examples include:
Carbocations (
𝐶
+
C
+
).
Free radicals (
⋅
⋅).
Carbanions (
𝐶
−
C
−
).
Carbenes (
𝐶
:
C:).
What distinguishes a concerted reaction from a stepwise reaction?
Answer:
A concerted reaction occurs in a single step, where all bond-breaking and bond-forming happen simultaneously, with no intermediate formed.
A stepwise reaction occurs in multiple steps and involves the formation of one or more reactive intermediates before the product is formed.
Bond Breaking – Homolytic Cleavage (Advanced)
This slide expands on the concept of homolytic cleavage, detailing the mechanism using curved arrows and emphasizing the role of radicals.
Key Points
Mechanism Representation:
The movement of a single electron is represented using a half-headed curved arrow, also called a fishhook.
In homolysis, two fishhook arrows are needed:
One for each electron in the bond that is split evenly.
Products of Homolysis:
Homolysis generates two uncharged radicals, each with an unpaired electron.
Example:
𝐴−𝐵→Δ,ℎ𝑣
𝐴⋅ + 𝐵⋅
Radicals:
A radical is a reactive intermediate with one unpaired electron.
Radicals are highly unstable because they do not have a full octet of electrons, making them highly reactive and short-lived.
Suggest how homolysis can be used in polymerization reactions.
Bond Breaking – Heterolytic Cleavage
This slide explains heterolysis or heterolytic cleavage, where electrons are unequally divided when a bond is broken.
Key Points
Heterolysis Definition:
Heterolysis is the breaking of a bond where the bonded electrons are transferred unequally to one of the atoms.
This typically happens when the two atoms in the bond have different electronegativities.
Electron Distribution:
The more electronegative atom takes both bonding electrons, forming an anion (
𝐴
−
or
𝐵
−
A
−
orB
−
).
The less electronegative atom becomes a cation (
𝐴
+
or
𝐵
+
A
+
orB
+
).
Products of Heterolysis:
The result of heterolysis is one positively charged ion (cation) and one negatively charged ion (anion).
Example Reaction:
𝐻−𝐶𝑙→𝐻++𝐶𝑙−H−Cl→H + +Cl −
: The more electronegative chlorine takes both electrons, forming a chloride anion (
𝐶
𝑙
−
Cl
−
) and a proton (
𝐻
+
H
+
).
Key Points
Arrow Representation:
A full-headed curved arrow is used to show the movement of an electron pair during heterolysis.
This arrow points to the more electronegative atom, which receives the electron pair.
Products of Heterolysis:
Cations (
𝐴
+
A
+
): Formed when an atom loses the electron pair.
Anions (
𝐵
−
B
−
): Formed when an atom gains the electron pair.
Special Intermediates:
Carbocations:
Positively charged carbon species.
Highly unstable because carbon is left with only six electrons in its valence shell.
Carbanions:
Negatively charged carbon species.
Unstable because carbon, not being very electronegative, poorly stabilizes the extra electron density.
Why is a carbocation more unstable than a neutral carbon atom?
Answer:
A carbocation is more unstable because it has only six valence electrons, leaving it electron-deficient and unable to achieve an octet. This deficiency creates a high-energy, unstable intermediate.
How can a carbanion become stabilized?
Answer:
A carbanion can be stabilized by:
Electron-withdrawing groups (e.g., fluorine, carbonyl groups) that reduce electron density on the negatively charged carbon.
Resonance: Delocalization of the negative charge across multiple atoms.
Hybridization: Carbanions in sp-hybridized states are more stable than those in sp
2
2
or sp
3
3
states.
Write the heterolytic cleavage for
𝐶𝐻3𝐶𝑙, identifying the carbocation and anion formed.
Heterolytic cleavage of
𝐶𝐻3𝐶𝑙:
𝐶𝐻3𝐶𝑙 →𝐶𝐻3+ + 𝐶𝑙−
𝐶𝐻3+
: Carbocation (methyl cation), where carbon has a positive charge.
𝐶𝑙−
: Chloride anion, which takes both electrons from
Reactive Intermediates from Breaking a C-Z Bond
This slide explains the types of reactive intermediates formed when a bond between carbon (
𝐶
C) and another atom (
𝑍
Z) breaks, emphasizing homolysis and heterolysis.
Key Points
Homolysis (Homolytic Cleavage):
The bond breaks evenly, with each atom receiving one electron.
Represented using half-headed arrows.
Products: Radicals (
𝐶
⋅
C
⋅
and
⋅
𝑍
⋅Z).
Radicals are highly reactive intermediates often involved in radical chain reactions.
Heterolysis (Heterolytic Cleavage):
The bond breaks unevenly, with one atom receiving both electrons.
Represented using full-headed arrows.
Two scenarios:
Formation of a carbocation (
𝐶
+
C
+
) and an anion (
:
𝑍
−
:Z
−
).
Formation of a carbanion (
𝐶
−
C
−
) and a cation (
𝑍
+
Z
+
).
Reaction Context:
Radicals are intermediates in radical reactions, often initiated by heat or light.
Carbocations and carbanions are intermediates in polar reactions, often stabilized by solvents or surrounding groups.
Bond Breaking – Reactive Intermediates
This slide provides insight into the behavior and roles of radicals, carbocations, and carbanions, which are common intermediates in organic reactions.
Key Points
Radicals:
Structure:
𝐶
⋅
C
⋅
A carbon atom with a single unpaired electron.
Electrophilic behavior: Radicals are electron-deficient because they lack a full octet of electrons. This makes them reactive toward electron-rich species (nucleophiles).
Carbocations:
Structure:
𝐶
+
C
+
A carbon atom with a positive charge and only six valence electrons.
Electrophilic behavior: Carbocations are highly reactive as they seek electrons to complete their octet. They are stabilized by electron-donating groups or resonance.
Carbanions:
Structure:
𝐶
−
C
−
A carbon atom with a negative charge, possessing a full octet and a lone pair of electrons.
Nucleophilic behavior: Carbanions are electron-rich and seek electrophilic species to donate their electrons. They are stabilized by electron-withdrawing groups or resonance.
Suggest a reaction mechanism involving a radical intermediate.
Bond Formation
This slide discusses how bonds are formed between atoms or ions and highlights two primary mechanisms of bond formation.
Key Points
Two Mechanisms of Bond Formation:
Formation from Two Radicals:
Each atom contributes one electron to form a shared two-electron bond.
Represented with half-headed curved arrows showing the movement of single electrons.
Formation from Two Ions:
The negatively charged ion donates both electrons to the bond.
Represented with a full-headed curved arrow indicating the transfer of an electron pair.
Bond Formation Always Releases Energy:
The process of bond formation is exothermic, as energy is released when a stable bond is formed.
Suggest a reaction mechanism where bond formation from radicals is the final step.
Analysis and Answers
This problem involves heterolysis of the carbon-heteroatom bond, considering the electronegativity differences between the atoms. We need to draw the products and classify the organic reactive intermediates as carbocation or carbanion.
a.
(
𝐶
𝐻
3
)
3
𝐶
−
𝑂
𝐻
(CH
3
)
3
C−OH
Heteroatom: Oxygen (O).
Electronegativity Difference: Oxygen is more electronegative than carbon.
Heterolysis:
(
𝐶
𝐻
3
)
3
𝐶
−
𝑂
𝐻
→
(
𝐶
𝐻
3
)
3
𝐶
+
+
𝑂
𝐻
−
(CH
3
)
3
C−OH→(CH
3
)
3
C
+
+OH
−
Classification:(𝐶𝐻3)3𝐶+(CH 3 ) 3 C +
: Carbocation (electron-deficient carbon).
𝑂
𝐻
−
OH
−
: Hydroxide ion.
b.
𝐶
6
𝐻
11
−
𝐵
𝑟
C
6
H
11
−Br
Heteroatom: Bromine (Br).
Electronegativity Difference: Bromine is more electronegative than carbon.
Heterolysis:
𝐶
6
𝐻
11
−
𝐵
𝑟
→
𝐶
6
𝐻
11
+
+
𝐵
𝑟
−
C
6
H
11
−Br→C
6
H
11
+
+Br
−Classification:𝐶6𝐻11+C 6 H 11+
: Carbocation (electron-deficient carbon).
𝐵
𝑟
−
Br
−
: Bromide ion.
c.
𝐶
𝐻
3
𝐶
𝐻
2
−
𝐿
𝑖
CH
3
CH
2
−Li
Heteroatom: Lithium (Li).
Electronegativity Difference: Carbon is more electronegative than lithium.
Heterolysis:
𝐶
𝐻
3
𝐶
𝐻
2
−
𝐿
𝑖
→
𝐶
𝐻
3
𝐶
𝐻
2
−
+
𝐿
𝑖
+
CH
3
CH
2
−Li→CH
3
CH
2
−
+Li
+Classification:𝐶𝐻3𝐶𝐻2−CH 3 CH 2−
: Carbanion (carbon with a lone pair).
𝐿
𝑖
+
Li
+
: Lithium cation.
This problem asks for the products of homolysis or heterolysis of each indicated bond, considering the electronegativity differences. Each carbon reactive intermediate needs to be classified as a radical, carbocation, or carbanion.
a. Homolysis of
𝐶
𝐻
3
−
𝐶
−
𝐻
CH
3
−C−H
Bond:
𝐶
−
𝐻
C−H
Process: Homolysis splits the bond equally, with each atom retaining one electron.
Products:
𝐶
𝐻
3
−
𝐶
⋅
+
𝐻
⋅
CH
3
−C
⋅
+H
⋅
Classification:𝐶𝐻3−𝐶⋅CH 3 −C ⋅
: Radical (contains one unpaired electron).
𝐻
⋅
H
⋅
: Hydrogen radical (neutral, unpaired electron).
b. Heterolysis of
𝐶
𝐻
3
−
𝑂
−
𝐻
CH
3
−O−H
Bond:
𝑂
−
𝐻
O−H
Electronegativity Difference: Oxygen is more electronegative than hydrogen.
Process: Heterolysis assigns both electrons to the more electronegative oxygen atom.
Products:
𝐶
𝐻
3
−
𝑂
−
+
𝐻
+
CH
3
−O
−
+H
+Classification:𝐶𝐻3−𝑂−CH 3 −O −
: Alkoxide ion (negatively charged oxygen).
𝐻
+
H
+
: Proton (positively charged hydrogen).
c. Heterolysis of
𝐶
𝐻
3
−
𝑀
𝑔
𝐵
𝑟
CH
3
−MgBr
Bond:
𝐶
−
𝑀
𝑔
C−Mg
Electronegativity Difference: Carbon is more electronegative than magnesium.
Process: Heterolysis assigns both electrons to the carbon atom.
Products:
𝐶
𝐻
3
−
+
𝑀
𝑔
𝐵
𝑟
+
CH
3
−
+MgBr
+Classification:𝐶𝐻3−CH 3−
: Carbanion (negatively charged carbon with a lone pair).
𝑀
𝑔
𝐵
𝑟
+
MgBr
+
: Magnesium bromide cation (positively charged complex).
Analysis and Steps for Each Reaction
Here are the step-by-step explanations and details for the provided reactions. We’ll analyze each one and use either full-headed or half-headed curved arrows to represent electron movement.
a.
Reaction:
Cyclohexylbromide
→
Cyclohexylcarbocation
+
Bromideion
Cyclohexylbromide→Cyclohexylcarbocation+Bromideion
Mechanism: Heterolytic cleavage.
Bromine is more electronegative than carbon, so it takes both bonding electrons.
Arrow Type: Full-headed curved arrow from the bond to bromine.
Products:
Cyclohexyl carbocation (
C
+
C
+
).
Bromide anion (
:Br
−
:Br
−
).
b.
Reaction:
Acetonechloride
→
Acetone
+
Chlorideion
Acetonechloride→Acetone+Chlorideion
Mechanism: Heterolytic cleavage.
Chlorine, being more electronegative, captures the bonding electrons.
Arrow Type: Full-headed curved arrow from the bond to chlorine.
Products:
Acetone (
C=O
C=O).
Chloride ion (
:Cl
−
:Cl
−
).
c.
Reaction:
⋅
CH
3
+
⋅
Cl
→
CH
3
−
Cl
⋅CH
3
+⋅Cl→CH
3
−Cl
Mechanism: Radical combination.
Both radicals combine by sharing one electron each to form a single covalent bond.
Arrow Type: Half-headed curved arrows (fishhooks) showing each radical contributing one electron.
Products:
Methyl chloride (
CH
3
−
Cl
CH
3
−Cl).
d.
Reaction:
Ethane
+
Br
2
→
Ethylbromide
+
:Br
−
Ethane+Br
2
→Ethylbromide+:Br
−
Mechanism: Substitution.
Bromine substitutes for a hydrogen atom in ethane.
Arrow Type:
Full-headed arrow from the
Br
2
Br
2
bond to one bromine atom.
Full-headed arrow from the carbon-hydrogen bond to hydrogen.
Products:
Ethyl bromide (
C
2
H
5
Br
C
2
H
5
Br).
Bromide anion (
:Br
−
:Br
−
).
e.
Reaction:
Bromoethane
+
OH
−
→
Ethanol
+
:Br
−
Bromoethane+OH
−
→Ethanol+:Br
−
Mechanism: Nucleophilic substitution (SN2).
Hydroxide ion attacks the carbon atom attached to bromine, displacing bromide.
Arrow Type:
Full-headed arrow from
:OH
−
:OH
−
to carbon.
Full-headed arrow from the carbon-bromine bond to bromine.
Products:
Ethanol (
CH
3
CH
2
OH
CH
3
CH
2
OH).
Bromide ion (
:Br
−
:Br
−
).
f.
Reaction:
Isobutane
+
:OH
−
→
Isobutene
+
H
2
O
Isobutane+:OH
−
→Isobutene+H
2
O
Mechanism: Elimination.
Hydroxide ion abstracts a proton, and a double bond forms between adjacent carbons.
Arrow Type:
Full-headed arrow from the hydroxide ion to hydrogen.
Full-headed arrow from the carbon-hydrogen bond to the adjacent carbon-carbon bond.
Products:
Isobutene (
CH
2
=
C
(
CH
3
)
2
CH
2
=C(CH
3
)
2
).
Water (
H
2
O
H
2
O).
The slide explains Bond Dissociation Energy (BDE) and highlights key points about bond breaking and formation:
Key Points:
Definition:
The bond dissociation energy is the energy required to homolytically cleave a covalent bond.
This process involves splitting the bond evenly, resulting in two radicals (unpaired electrons).
Energy Requirement:
Homolysis requires energy, making the bond dissociation energy always a positive value.
Since breaking bonds absorbs energy, homolysis is always endothermic.
Bond Formation:
The reverse of bond breaking is bond formation.
Bond formation releases energy, making it an exothermic process.
Energy Representation:
The energy required for homolysis is represented as
Δ
𝐻
∘
=
BondDissociationEnergy
ΔH
∘
=BondDissociationEnergy.
Summary:
Bond Breaking: Requires energy (endothermic, positive BDE).
Bond Formation: Releases energy (exothermic).
What is the bond dissociation energy for the H-H bond?
A: The bond dissociation energy for the H-H bond is
+
435
kJ/mol
+435kJ/mol (or
+
104
kcal/mol
+104kcal/mol).
Is breaking the H-H bond an endothermic or exothermic process? Why?
A: Breaking the H-H bond is an endothermic process because energy is required to cleave the bond.
How much energy is released when the H-H bond is formed?
A:
−
435
kJ/mol
−435kJ/mol (or
−
104
kcal/mol
−104kcal/mol) of energy is released when the H-H bond is formed.
Is bond formation endothermic or exothermic?
A: Bond formation is exothermic because it releases energy.
What is the relationship between the energy required to break a bond and the energy released when the same bond is formed?
A: The energy required to break a bond (endothermic process) is equal in magnitude but opposite in sign to the energy released when the bond is formed (exothermic process).
Analysis: Determining \Delta H^\circ for a Reaction
Q1: What is the enthalpy change ( \Delta H^\circ ) for the given reaction?
A: \Delta H^\circ = -3 \, \text{kJ/mol} .
This value is calculated by summing the bond dissociation energies for bonds broken and subtracting the sum of bond dissociation energies for bonds formed:
\Delta H^\circ = (+829) + (-832) = -3 \, \text{kJ/mol}.
Q2: Is this reaction endothermic or exothermic? Why?
A: The reaction is exothermic because \Delta H^\circ is negative ( -3 \, \text{kJ/mol} ). This indicates that more energy is released during bond formation than is required for bond breaking.
Q3: Which bonds are broken in this reaction?
A:
1. (CH_3)_3C{-}Cl bond ( +331 \, \text{kJ/mol} ).
2. H{-}OH bond ( +498 \, \text{kJ/mol} ).
The total energy needed to break these bonds is +829 \, \text{kJ/mol} .
Q4: Which bonds are formed in this reaction?
A:
1. (CH_3)_3C{-}OH bond ( -401 \, \text{kJ/mol} ).
2. H{-}Cl bond ( -431 \, \text{kJ/mol} ).
The total energy released in forming these bonds is -832 \, \text{kJ/mol} .
Q5: Why are the bonds formed in the products stronger than the bonds broken in the reactants?
A: The total bond dissociation energy for bonds formed ( -832 \, \text{kJ/mol} ) is greater in magnitude than the energy required to break the bonds ( +829 \, \text{kJ/mol} ). This means the bonds in the products are more stable (stronger) than those in the reactants.
Q6: Why is the energy change only -3 \, \text{kJ/mol} despite being exothermic?
A: The energy released during bond formation ( -832 \, \text{kJ/mol} ) is almost equal to the energy required for bond breaking ( +829 \, \text{kJ/mol} ), leading to a small net energy release of -3 \, \text{kJ/mol} .
Analysis: Enthalpy Changes in Oxidation Reactions
Q1: Why is \Delta H^\circ negative for both oxidation reactions?
A: \Delta H^\circ is negative for both reactions because they are exothermic, meaning they release energy. The bonds formed in the products (e.g., CO_2 and H_2O ) are stronger than the bonds in the reactants, leading to a net release of energy.
Q2: How much energy is released during the oxidation of isooctane?
A: The oxidation of isooctane releases \Delta H^\circ = -5447 \, \text{kJ/mol} .
Q3: How much energy is released during the oxidation of glucose?
A: The oxidation of glucose releases \Delta H^\circ = -2872 \, \text{kJ/mol} .
Q4: Why do both isooctane and glucose release energy on oxidation?
A: Both isooctane and glucose release energy on oxidation because the products formed, CO_2 and H_2O , have stronger bonds compared to the bonds in the reactants. Breaking the weaker bonds in isooctane or glucose requires less energy than is released when the stronger bonds in the products are formed.
Q5: Why is the energy released from isooctane oxidation greater than that from glucose oxidation?
A: Isooctane has a higher energy content per molecule because it contains more high-energy C{-}H and C{-}C bonds than glucose. As a result, its complete oxidation yields more energy compared to glucose.
Q6: What do these negative \Delta H^\circ values indicate about the stability of the products compared to the reactants?
A: The negative \Delta H^\circ values indicate that the products ( CO_2 and H_2O ) are more stable than the reactants (isooctane and glucose). This increased stability arises from the stronger bonds in the products.
Q7: What is the significance of these reactions being exothermic in biological or industrial processes?
A:
1. Biological significance: The exothermic oxidation of glucose is the primary energy source for living organisms, powering cellular processes.
2. Industrial significance: The oxidation of hydrocarbons like isooctane releases significant energy, making them valuable fuels in engines and power generation.
Analysis: Equilibrium Constant and Thermodynamics
Q1: What is the equilibrium constant (K_{eq})?
A:
The equilibrium constant (K_{eq}) is a mathematical expression that relates the concentrations of the products to the concentrations of the starting materials at equilibrium for a chemical reaction.
Q2: How is the equilibrium constant (K_{eq}) expressed for a reaction?
A:
For a general reaction:
[ A + B \rightleftharpoons C + D ]
The equilibrium constant is expressed as:
K_{eq} = \frac{[C][D]}{[A][B]}
Where:
• [C] and [D] are the concentrations of the products.
• [A] and [B] are the concentrations of the starting materials.
Q3: What does a high value of K_{eq} indicate about the reaction?
A:
A high K_{eq} value indicates that the reaction favors the formation of products at equilibrium, meaning the concentration of products is significantly greater than that of the reactants.
Q4: What does a low value of K_{eq} indicate about the reaction?
A:
A low K_{eq} value suggests that the reaction favors the reactants, meaning the concentrations of starting materials are higher than those of the products at equilibrium.
Let me know if further clarification is needed!
Analysis: Equilibrium Constant and Thermodynamics
Q1: What is the equilibrium constant (K_{eq})?
A:
The equilibrium constant (K_{eq}) is a mathematical expression that relates the concentrations of the products to the concentrations of the starting materials at equilibrium for a chemical reaction.
Q2: How is the equilibrium constant (K_{eq}) expressed for a reaction?
A:
For a general reaction:
[ A + B \rightleftharpoons C + D ]
The equilibrium constant is expressed as:
K_{eq} = \frac{[C][D]}{[A][B]}
Where:
• [C] and [D] are the concentrations of the products.
• [A] and [B] are the concentrations of the starting materials.
Q3: What does a high value of K_{eq} indicate about the reaction?
A:
A high K_{eq} value indicates that the reaction favors the formation of products at equilibrium, meaning the concentration of products is significantly greater than that of the reactants.
Q4: What does a low value of K_{eq} indicate about the reaction?
A:
A low K_{eq} value suggests that the reaction favors the reactants, meaning the concentrations of starting materials are higher than those of the products at equilibrium.
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Analysis: Equilibrium Constant and Thermodynamics
Q1: What is the equilibrium constant (K_{eq})?
A:
The equilibrium constant (K_{eq}) is a mathematical expression that relates the concentrations of the products to the concentrations of the starting materials at equilibrium for a chemical reaction.
Q2: How is the equilibrium constant (K_{eq}) expressed for a reaction?
A:
For a general reaction:
[ A + B \rightleftharpoons C + D ]
The equilibrium constant is expressed as:
K_{eq} = \frac{[C][D]}{[A][B]}
Where:
• [C] and [D] are the concentrations of the products.
• [A] and [B] are the concentrations of the starting materials.
Q3: What does a high value of K_{eq} indicate about the reaction?
A:
A high K_{eq} value indicates that the reaction favors the formation of products at equilibrium, meaning the concentration of products is significantly greater than that of the reactants.
Q4: What does a low value of K_{eq} indicate about the reaction?
A:
A low K_{eq} value suggests that the reaction favors the reactants, meaning the concentrations of starting materials are higher than those of the products at equilibrium.
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Analysis: The Equilibrium Constant
Q1: What does the size of  indicate?
A:
• The size of  expresses whether the starting materials or products predominate once equilibrium is reached:
• : Equilibrium favors the products and lies to the right as the equation is written.
• : Equilibrium favors the starting materials and lies to the left as the equation is written.
Q2: What is required for a reaction to be useful?
A:
For a reaction to be useful, the equilibrium must favor the products, and  must be greater than 1 ().
Q3: What determines the position of equilibrium?
A:
The position of the equilibrium is determined by the relative energies of the reactants and products.
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Analysis: Relationship Between Equilibrium Constant and Free Energy
Q1: What is the relationship between \Delta G^\circ and K_{eq}?
A:
The relationship is expressed by the equation:
\Delta G^\circ = -2.303RT \log K_{eq}
where:
• R = 8.314 J/(K·mol), the gas constant.
• T = temperature in Kelvin.
Q2: What happens when K_{eq} > 1?
A:
• \log K_{eq} is positive.
• \Delta G^\circ becomes negative.
• Energy is released (exergonic process).
• Equilibrium favors the products.
Q3: What happens when K_{eq} < 1?
A:
• \log K_{eq} is negative.
• \Delta G^\circ becomes positive.
• Energy is absorbed (endergonic process).
• Equilibrium favors the reactants.
Q4: Why does K_{eq} depend on the energy difference between reactants and products?
A:
The equilibrium constant K_{eq} is influenced by the relative stability of the reactants and products. Lower energy (more stable) species are favored, shifting the equilibrium position.
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Analysis: Energy Difference and Equilibrium
Q1: Why does equilibrium favor lower-energy compounds?
A:
Compounds that are lower in energy are more stable. When products are more stable (lower in energy) than reactants, the equilibrium shifts toward the products, favoring their formation.
Q2: How does \Delta G^\circ influence the equilibrium constant (K_{eq})?
A:
• A small change in \Delta G^\circ corresponds to a large difference in K_{eq} due to the logarithmic relationship:
\Delta G^\circ = -2.303RT \log K_{eq}
This explains why even slight variations in free energy can lead to significant differences in equilibrium composition.
Q3: What is the relationship between \Delta G^\circ, K_{eq}, and the relative amounts of A and B?
A:
From the table:
• When \Delta G^\circ > 0, K_{eq} < 1, and reactants (A) dominate.
• When \Delta G^\circ = 0, K_{eq} = 1, and A and B are present in equal amounts.
• When \Delta G^\circ < 0, K_{eq} > 1, and products (B) dominate.
Q4: What does the data in Table 6.3 imply?
A:
The table illustrates how:
• A \Delta G^\circ of +18 kJ/mol results in almost entirely reactants (K_{eq} = 10^{-3}).
• A \Delta G^\circ of -18 kJ/mol results in almost entirely products (K_{eq} = 10^{3}).
• Small changes in free energy (\Delta G^\circ) significantly affect the equilibrium composition.
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What are the two conformations of monosubstituted cyclohexanes?
A:
Monosubstituted cyclohexanes exist as two chair conformations:
1. Axial substituent position (A)
2. Equatorial substituent position (B)
These conformations rapidly interconvert at room temperature.
How is the equilibrium constant (K_{eq}) related to the energy difference (\Delta G^\circ)?
A:
The equilibrium constant is calculated using the equation:
\Delta G^\circ = -2.303RT \log K_{eq}
In this case, \Delta G^\circ = -12.1 \, \text{kJ/mol}, leading to K_{eq} \approx 100. This indicates that the equatorial conformation (B) is about 100 times more abundant than the axial conformation (A) at equilibrium.
What is the significance of \Delta G^\circ = -12.1 \, \text{kJ/mol}?
A:
This value indicates the energy difference between the two conformations, favoring the equatorial form. The negative sign shows that moving toward the equatorial conformation releases energy, making it thermodynamically preferred.
Analysis and Answers:
Q1: Which K_{eq} corresponds to a negative value of \Delta G^\circ , K_{eq} = 1000 or K_{eq} = 0.001 ?
• Answer: K_{eq} = 1000
A negative \Delta G^\circ corresponds to an equilibrium that favors the products, meaning K_{eq} > 1 . Since K_{eq} = 1000 is much greater than 1, it aligns with a negative \Delta G^\circ .
Q2: Which K_{eq} corresponds to a lower value of \Delta G^\circ , K_{eq} = 10^{-2} or K_{eq} = 10^{-5} ?
• Answer: K_{eq} = 10^{-5}
A lower K_{eq} value (closer to 0) indicates a higher positive \Delta G^\circ , meaning the starting materials are more favored. Thus, K_{eq} = 10^{-5} corresponds to a lower \Delta G^\circ .
Q3: Given the following values, is the starting material or product favored at equilibrium?
a. K_{eq} = 5.5
b. \Delta G^\circ = 40 \, \text{kJ/mol}
• Answer:
• a. K_{eq} = 5.5 : The product is favored because K_{eq} > 1 .
• b. ( \Delta G^\circ = 40 \, \text{kJ/mol}: The starting material is favored because a positive ( \Delta G^\circ ) indicates reactants are thermodynamically more stable.
Q4: Given the following values, is the starting material or product lower in energy?
a. \Delta G^\circ = 8.0 \, \text{kJ/mol}
b. K_{eq} = 10
c. \Delta G^\circ = -12 \, \text{kJ/mol}
d. K_{eq} = 10^{-3}
• Answer:
• a. ( \Delta G^\circ = 8.0 \, \text{kJ/mol}: The starting material is lower in energy because ( \Delta G^\circ > 0 ).
• b. K_{eq} = 10 : The product is lower in energy because K_{eq} > 1 .
• c. ( \Delta G^\circ = -12 \, \text{kJ/mol}: The product is lower in energy because ( \Delta G^\circ < 0 ).
• d. K_{eq} = 10^{-3} : The starting material is lower in energy because K_{eq} < 1 .
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Analysis and Answers:
Q1: Which K_{eq} corresponds to a negative value of \Delta G^\circ , K_{eq} = 1000 or K_{eq} = 0.001 ?
• Answer: K_{eq} = 1000
A negative \Delta G^\circ corresponds to an equilibrium that favors the products, meaning K_{eq} > 1 . Since K_{eq} = 1000 is much greater than 1, it aligns with a negative \Delta G^\circ .
Q2: Which K_{eq} corresponds to a lower value of \Delta G^\circ , K_{eq} = 10^{-2} or K_{eq} = 10^{-5} ?
• Answer: K_{eq} = 10^{-5}
A lower K_{eq} value (closer to 0) indicates a higher positive \Delta G^\circ , meaning the starting materials are more favored. Thus, K_{eq} = 10^{-5} corresponds to a lower \Delta G^\circ .
Q3: Given the following values, is the starting material or product favored at equilibrium?
a. K_{eq} = 5.5
b. \Delta G^\circ = 40 \, \text{kJ/mol}
• Answer:
• a. K_{eq} = 5.5 : The product is favored because K_{eq} > 1 .
• b. ( \Delta G^\circ = 40 \, \text{kJ/mol}: The starting material is favored because a positive ( \Delta G^\circ ) indicates reactants are thermodynamically more stable.
Q4: Given the following values, is the starting material or product lower in energy?
a. \Delta G^\circ = 8.0 \, \text{kJ/mol}
b. K_{eq} = 10
c. \Delta G^\circ = -12 \, \text{kJ/mol}
d. K_{eq} = 10^{-3}
• Answer:
• a. ( \Delta G^\circ = 8.0 \, \text{kJ/mol}: The starting material is lower in energy because ( \Delta G^\circ > 0 ).
• b. K_{eq} = 10 : The product is lower in energy because K_{eq} > 1 .
• c. ( \Delta G^\circ = -12 \, \text{kJ/mol}: The product is lower in energy because ( \Delta G^\circ < 0 ).
• d. K_{eq} = 10^{-3} : The starting material is lower in energy because K_{eq} < 1 .
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Analysis and Answers:
Q1: Which K_{eq} corresponds to a negative value of \Delta G^\circ , K_{eq} = 1000 or K_{eq} = 0.001 ?
• Answer: K_{eq} = 1000
A negative \Delta G^\circ corresponds to an equilibrium that favors the products, meaning K_{eq} > 1 . Since K_{eq} = 1000 is much greater than 1, it aligns with a negative \Delta G^\circ .
Q2: Which K_{eq} corresponds to a lower value of \Delta G^\circ , K_{eq} = 10^{-2} or K_{eq} = 10^{-5} ?
• Answer: K_{eq} = 10^{-5}
A lower K_{eq} value (closer to 0) indicates a higher positive \Delta G^\circ , meaning the starting materials are more favored. Thus, K_{eq} = 10^{-5} corresponds to a lower \Delta G^\circ .
Q3: Given the following values, is the starting material or product favored at equilibrium?
a. K_{eq} = 5.5
b. \Delta G^\circ = 40 \, \text{kJ/mol}
• Answer:
• a. K_{eq} = 5.5 : The product is favored because K_{eq} > 1 .
• b. ( \Delta G^\circ = 40 \, \text{kJ/mol}: The starting material is favored because a positive ( \Delta G^\circ ) indicates reactants are thermodynamically more stable.
Q4: Given the following values, is the starting material or product lower in energy?
a. \Delta G^\circ = 8.0 \, \text{kJ/mol}
b. K_{eq} = 10
c. \Delta G^\circ = -12 \, \text{kJ/mol}
d. K_{eq} = 10^{-3}
• Answer:
• a. ( \Delta G^\circ = 8.0 \, \text{kJ/mol}: The starting material is lower in energy because ( \Delta G^\circ > 0 ).
• b. K_{eq} = 10 : The product is lower in energy because K_{eq} > 1 .
• c. ( \Delta G^\circ = -12 \, \text{kJ/mol}: The product is lower in energy because ( \Delta G^\circ < 0 ).
• d. K_{eq} = 10^{-3} : The starting material is lower in energy because K_{eq} < 1 .
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Analysis and Answers
Q8: Given each value, determine whether the starting material or product is favored at equilibrium.
• a. K_{eq} = 0.5 : Starting material is favored because K_{eq} < 1 .
• b. \Delta G^\circ = -100 \, \text{kJ/mol} : Product is favored because \Delta G^\circ < 0 .
• c. \Delta H^\circ = 8.0 \, \text{kJ/mol} : Not directly related to equilibrium; this indicates an endothermic process. No conclusion about K_{eq} .
• d. K_{eq} = 16 : Product is favored because K_{eq} > 1 .
• e. \Delta G^\circ = 2.0 \, \text{kJ/mol} : Starting material is favored because \Delta G^\circ > 0 .
• f. \Delta H^\circ = 200 \, \text{kJ/mol} : Not directly related to equilibrium; this indicates a highly endothermic process. No conclusion about K_{eq} .
• g. \Delta S^\circ = 8 \, \text{J/(K·mol)} : Positive entropy favors products but does not indicate the equilibrium directly without temperature or \Delta G^\circ .
• h. \Delta S^\circ = -8 \, \text{J/(K·mol)} : Negative entropy favors the starting material but does not indicate the equilibrium directly without temperature or \Delta G^\circ .
Q9:
• a. Which value corresponds to a negative value of \Delta G^\circ : K_{eq} = 10^{-2} or K_{eq} = 10^2 ?
Answer: K_{eq} = 10^2
A negative \Delta G^\circ corresponds to K_{eq} > 1 , which indicates the products are favored.
• b. In a unimolecular reaction with five times as much starting material as product at equilibrium, what is the value of K_{eq} ? Is \Delta G^\circ positive or negative?
Answer: K_{eq} = 0.2 , \Delta G^\circ > 0 .
If the starting material is favored, K_{eq} < 1 , which corresponds to a positive \Delta G^\circ .
• c. Which value corresponds to a larger K_{eq} : \Delta G^\circ = -8 \, \text{kJ/mol} or \Delta G^\circ = 20 \, \text{kJ/mol} ?
Answer: \Delta G^\circ = -8 \, \text{kJ/mol} .
A more negative \Delta G^\circ corresponds to a larger K_{eq} , favoring the products more.
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