Chapter 6 School Material Flashcards

Question

definition of substitution reaction

Answer 1

A substitution reaction occurs when one atom or group of atoms (𝑍) on a carbon atom is replaced by another atom or group (𝑌).

Answer 2

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.

Answer 3

Nucleophilic Substitution ( 𝑆𝑁1 and 𝑆n2 2): A nucleophile replaces a leaving group. Electrophilic Substitution: An electrophile replaces a substituent, commonly in aromatic compounds.

Answer 4

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−.

Answer 5

Answer: An electron-deficient group (electrophile), such as 𝑁𝑂2+ or 𝐵𝑟+ , typically replaces 𝑍 in electrophilic substitution reactions.

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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−).

Answer 10

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.

Answer 11

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.

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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.

Answer 17

Answer: The product is typically an alkene ( 𝐶=𝐶), along with a small molecule byproduct like 𝐻𝐶𝑙, 𝐻𝐵𝑟, or 𝐻2𝑂

Answer 18

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.

Answer 19

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.

Answer 20

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.

Answer 21

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.

Answer 22

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.

Answer 23

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).

Answer 24

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.

Answer 25

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.

Answer 26

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).

Answer 27

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.

Answer 28

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.)

Answer 29

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.)

Answer 30

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

Answer 31

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.

Answer 32

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.

Answer 33

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.

Answer 34

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.

Answer 35

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

Answer 36

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.

Answer 37

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.

Answer 38

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.

Answer 39

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:).

Answer 40

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.

Answer 41

Answer: Reactive intermediates are high in energy because they have incomplete or unstable electronic configurations, making them highly reactive and short-lived.

Answer 42

𝐶 𝐻 3 𝐶 𝐻 2 𝐵 𝑟 + 𝑂 𝐻 − → 𝐶 𝐻 3 𝐶 𝐻 2 𝑂 𝐻 + 𝐵 𝑟 − CH 3 CH 2 Br+OH − →CH 3 CH 2 OH+Br − Answer: This is a concerted reaction because it follows an 𝑆 𝑁 2 S N 2 mechanism, where the nucleophile ( 𝑂 𝐻 − OH − ) attacks the carbon and the leaving group ( 𝐵 𝑟 − Br − ) departs simultaneously.

Answer 43

𝐶 𝐻 3 𝐶 𝐻 = 𝐶 𝐻 2 + 𝐻 𝐵 𝑟 → 𝐶 𝐻 3 𝐶 𝐻 + 𝐶 𝐻 3 → 𝐶 𝐻 3 𝐶 𝐻 𝐵 𝑟 𝐶 𝐻 3 CH 3 CH=CH 2 +HBr→CH 3 CH + CH 3 →CH 3 CHBrCH 3 Answer: The reactive intermediate is the carbocation ( 𝐶 𝐻 3 𝐶 𝐻 + 𝐶 𝐻 3 CH 3 CH + CH 3 ) formed after the addition of 𝐻 + H + to the double bond.

Answer 44

Answer: The stability of a reactive intermediate significantly affects the rate of a stepwise reaction. A more stable intermediate lowers the activation energy of its formation, increasing the overall reaction rate. For example, tertiary carbocations are more stable than primary carbocations, favoring faster reactions.

Answer 45

Answer: Diels-Alder Reaction follows a concerted mechanism. In this reaction, the bonds in the diene and dienophile rearrange simultaneously in a cyclic transition state, avoiding the formation of intermediates. This occurs because of the favorable overlap of 𝜋 π-orbitals in a single step.

Answer 46

Two Types of Bond Cleavage: Bonds can break by equal or unequal sharing of the electrons. Equal sharing leads to homolysis, while unequal sharing leads to heterolysis (covered in another section). Homolytic Cleavage (Homolysis): In homolysis, the bond breaks evenly, and each atom gets one electron from the bond. This generates two radicals, which are species with unpaired electrons. Example: 𝐴 − 𝐵 → 𝐴 ⋅ + 𝐵 ⋅ A−B→A ⋅ +B ⋅ (Each atom becomes a radical with one unpaired electron.) Radicals: Radicals are highly reactive due to the presence of an unpaired electron. Commonly formed in reactions involving heat ( Δ Δ) or light ( ℎ 𝑣 hv).

Answer 47

Answer: Homolytic cleavage: Electrons in the bond are divided equally between the two atoms, resulting in two radicals with unpaired electrons. Heterolytic cleavage: Electrons in the bond are divided unequally, with one atom receiving both electrons, forming a cation and an anion.

Answer 48

Answer: Radicals are highly reactive because they have an unpaired electron, which makes them unstable. They tend to react quickly to pair the unpaired electron and achieve a more stable configuration.

Answer 49

Answer: Homolytic cleavage occurs: 𝐶 𝑙 2 → ℎ 𝑣 𝐶 𝑙 ⋅ + 𝐶 𝑙 ⋅ Cl 2 hv Cl ⋅ +Cl ⋅ Ultraviolet light ( ℎ 𝑣 hv) provides the energy required to break the bond evenly, producing two chlorine radicals.

Answer 50

Answer: The initiation step of free-radical halogenation of methane involves homolysis: 𝐵 𝑟 2 → ℎ 𝑣 𝐵 𝑟 ⋅ + 𝐵 𝑟 ⋅ Br 2 hv Br ⋅ +Br ⋅ Bromine molecules absorb light energy and undergo homolysis to form two bromine radicals.

Answer 51

Answer: Heat ( Δ Δ) or light ( ℎ 𝑣 hv): Provide energy to break bonds evenly. Non-polar bonds: Homolysis is favored when the bond is between atoms with similar electronegativities (e.g., 𝐶 𝑙 − 𝐶 𝑙 Cl−Cl, 𝐶 − 𝐻 C−H). Gas phase or nonpolar solvents: These environments do not stabilize ions, favoring radical formation.

Answer 52

Answer: Homolysis is more likely when the resulting radicals are stable. Radical stability increases with: Delocalization: Resonance stabilization of the unpaired electron. Substitution: Tertiary radicals are more stable than secondary or primary radicals due to hyperconjugation. Electronegativity: Radicals on less electronegative atoms (e.g., carbon) are more stable than those on more electronegative atoms (e.g., oxygen).

Answer 53

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.

Answer 54

Answer: Fishhook arrows represent the movement of one electron, as opposed to full-headed arrows, which represent the movement of an electron pair. Homolysis involves the splitting of a bond where each atom takes one electron.

Answer 55

Answer: Radicals are formed, which are highly reactive because they have an unpaired electron, creating instability and driving them to react quickly to achieve a stable electronic configuration.

Answer 56

using fishhook arrows. Answer: 𝐻−𝐻→ ℎ𝑣 𝐻⋅+𝐻. Each hydrogen atom takes one electron from the bond, forming two hydrogen radicals.

Answer 57

Answer: UV light provides the energy needed to overcome the bond dissociation energy of the molecule, allowing the bond to break evenly and form radicals. This energy excites the electrons, making homolysis possible.

Answer 58

C(CH 3 (tertiary radical) is the most stable due to hyperconjugation and inductive effects from three methyl groups. 𝐶2𝐻5⋅ (primary radical) is less stable than the tertiary radical but more stable than 𝐶𝐻3⋅ 𝐶𝐻3⋅ (methyl radical) is the least stable as it lacks hyperconjugation and stabilizing groups.

Answer 59

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 − or B − ). The less electronegative atom becomes a cation ( 𝐴 + or 𝐵 + A + or B + ). Products of Heterolysis: The result of heterolysis is one positively charged ion (cation) and one negatively charged ion (anion). Example Reaction: 𝐻 − 𝐶 𝑙 → 𝐻 + + 𝐶 𝑙 − H−Cl→H + +Cl − : The more electronegative chlorine takes both electrons, forming a chloride anion ( 𝐶 𝑙 − Cl − ) and a proton ( 𝐻 + H + ).

Answer 60

Answer: Polar covalent bonds, where there is a significant difference in electronegativity between the two atoms, are more likely to undergo heterolysis. For example, bonds like 𝐻 − 𝐶 𝑙 H−Cl or 𝐶 − 𝐵 𝑟 C−Br.

Answer 61

Answer: The more electronegative atom has a greater affinity for electrons due to its higher electronegativity, making it more likely to attract and retain both electrons from the bond during heterolysis.

Answer 62

Answer: 𝐶𝐻3𝐼→𝐶𝐻3+ + 𝐼− The iodine atom, being more electronegative, takes both electrons, forming an iodide anion ( 𝐼−) and a methyl cation ( 𝐶𝐻3+ ).

Answer 63

Answer: 𝐻 2 𝑆 𝑂 4 H 2 SO 4 undergoes heterolysis to produce: 𝐻 2 𝑆 𝑂 4 → 𝐻 + + 𝐻 𝑆 𝑂 4 − H 2 SO 4 →H + +HSO 4 − The 𝐻 + H + is a proton (cation), and the 𝐻 𝑆 𝑂 4 − HSO 4 − is the hydrogen sulfate anion.

Answer 64

Answer: Heterolytic cleavage is more likely in polar solvents because they can stabilize the charged ions (cation and anion) that result from the cleavage. Nonpolar solvents lack this stabilizing effect, making heterolysis less favorable.

Answer 65

Answer: Heterolytic cleavage can generate a nucleophile and a leaving group, which are essential for substitution reactions. Example: In 𝐶 𝐻 3 𝐶 𝑙 CH 3 Cl: 𝐶 𝐻 3 𝐶 𝑙 → 𝐶 𝐻 3 + + 𝐶 𝑙 − CH 3 Cl→CH 3 + +Cl − The chloride ion ( 𝐶 𝑙 − Cl − ) acts as a leaving group, and the positively charged methyl group can undergo substitution with a nucleophile.

Answer 66

Factors Favoring Heterolytic Cleavage: Electronegativity Difference: A bond between two atoms with a large difference in electronegativity (e.g., 𝐻 − 𝐶 𝑙 H−Cl) is more likely to undergo heterolysis. Polar Solvents: Solvents like water ( 𝐻 2 𝑂 H 2 O) or ethanol ( 𝐶 2 𝐻 5 𝑂 𝐻 C 2 H 5 OH) stabilize the resulting ions, making heterolytic cleavage more favorable. Bond Polarity: The more polarized a bond, the more likely it is to break heterolytically (e.g., 𝐶 − 𝐵 𝑟 C−Br in alkyl halides). Role in Substitution Reactions: Heterolytic cleavage generates electrophiles and nucleophiles, both of which drive substitution reactions. Example: The reaction of 𝐶 𝐻 3 𝐵 𝑟 CH 3 Br with 𝑂 𝐻 − OH − : 𝐶 𝐻 3 𝐵 𝑟 + 𝑂 𝐻 − → 𝐶 𝐻 3 𝑂 𝐻 + 𝐵 𝑟 − CH 3 Br+OH − →CH 3 OH+Br − Here, 𝐵 𝑟 − Br − is the leaving group, and 𝑂 𝐻 − OH − is the nucleophile. Role in Acid-Base Chemistry: Heterolytic cleavage explains the dissociation of acids and bases. For example, in water: 𝐻 𝐶 𝑙 → 𝐻 + + 𝐶 𝑙 − HCl→H + +Cl − The 𝐻 + H + acts as a proton donor, and 𝐶 𝑙 − Cl − is the conjugate base.

Answer 67

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.

Answer 68

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.

Answer 69

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.

Answer 70

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

Answer 71

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.

Answer 72

Answer: Electronegativity difference: If one atom is significantly more electronegative, heterolysis is favored. Solvent polarity: Polar solvents stabilize ions, promoting heterolysis, while nonpolar solvents favor homolysis. Bond strength: Stronger bonds are less likely to break homolytically. Reaction conditions: Heat or UV light promotes homolysis; acidic or polar conditions favor heterolysis.

Answer 73

Answer: Nonpolar solvents do not stabilize ions formed during heterolysis but allow for the generation and stabilization of neutral radicals.

Answer 74

Answer: Carbocations are the most reactive because they are highly electron-deficient, seeking electrons to stabilize their positive charge. Radicals are moderately reactive due to their unpaired electron. Carbanions are reactive but less so compared to carbocations because they have excess electron density that can be stabilized by resonance or electronegative groups.

Answer 75

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.

Answer 76

Answer: Radicals are electron-deficient because they lack a full octet of electrons. This deficiency makes them behave like electrophiles, as they seek electrons to achieve stability.

Answer 77

Carbocations are stabilized by electron-donating groups such as alkyl groups (via hyperconjugation and inductive effects) or resonance structures. Carbanions are stabilized by electron-withdrawing groups such as halogens, carbonyl groups, or nitriles, which reduce the negative charge density on the carbon.

Answer 78

Answer: The stability order is: ( 𝐶 𝐻 3 ) 3 𝐶 + > ( 𝐶 𝐻 3 ) 2 𝐶 𝐻 + > 𝐶 𝐻 3 + (CH 3 ) 3 C + >(CH 3 ) 2 CH + >CH 3 + Reason: Tertiary carbocations ( ( 𝐶 𝐻 3 ) 3 𝐶 + (CH 3 ) 3 C + ) are more stable due to greater hyperconjugation and inductive effects from three alkyl groups, while methyl carbocations ( 𝐶 𝐻 3 + CH 3 + ) are the least stable as they lack stabilizing alkyl groups.

Answer 79

Answer: The carbanion in 𝐶 𝐻 3 − + 𝐶 𝐻 3 𝐼 → 𝐶 𝐻 4 + 𝐶 𝐻 2 𝐼 − CH 3 − +CH 3 I→CH 4 +CH 2 I − acts as a nucleophile, attacking the electrophilic carbon in 𝐶 𝐻 3 𝐼 CH 3 I.

Answer 80

Radicals: Moderate reactivity due to their neutral nature but electron deficiency; they participate in radical reactions. Carbocations: Highly reactive electrophiles due to extreme electron deficiency, making them highly unstable. Carbanions: Reactive nucleophiles due to their excess electron density; they are less reactive than carbocations because the extra electrons provide some stability.

Answer 81

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.

Answer 82

Answer: Bond formation releases energy because the atoms or ions involved achieve a more stable state. The system moves to a lower energy configuration, and the excess energy is released as heat or light.

Answer 83

Answer: In radical bond formation, each atom contributes one electron to the bond. In ionic bond formation, the negatively charged ion donates both electrons to form the bond.

Answer 84

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: ( 𝐶 𝐻 3 ) 3 𝐶 + (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: 𝐶 6 𝐻 11 + 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: 𝐶 𝐻 3 𝐶 𝐻 2 − CH 3 CH 2 − : Carbanion (carbon with a lone pair). 𝐿 𝑖 + Li + : Lithium cation.

Answer 85

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: 𝐶 𝐻 3 − 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: 𝐶 𝐻 3 − 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: 𝐶 𝐻 3 − CH 3 − : Carbanion (negatively charged carbon with a lone pair). 𝑀 𝑔 𝐵 𝑟 + MgBr + : Magnesium bromide cation (positively charged complex).

Answer 86

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: Cyclohexyl bromide → Cyclohexyl carbocation + Bromide ion Cyclohexyl bromide→Cyclohexyl carbocation+Bromide ion 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: Acetone chloride → Acetone + Chloride ion Acetone chloride→Acetone+Chloride ion 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 → Ethyl bromide + :Br − Ethane+Br 2 →Ethyl bromide+: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).

Answer 87

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 ∘ = Bond Dissociation Energy ΔH ∘ =Bond Dissociation Energy. Summary: Bond Breaking: Requires energy (endothermic, positive BDE). Bond Formation: Releases energy (exothermic).

Answer 88

A: Bond dissociation energy (BDE) is the energy required to homolytically cleave a covalent bond, resulting in the formation of two radicals.

Answer 89

A: Bond breaking is always endothermic because it requires energy to break the bond.

Answer 90

A: Bond formation always releases energy, making it an exothermic process.

Answer 91

A: Homolysis produces two radicals, each carrying one unpaired electron.

Answer 92

A: It is represented as Δ 𝐻 ∘ ΔH ∘ , indicating the energy required to break a bond.

Answer 93

A: The bond dissociation energy for the H-H bond is + 435 kJ/mol +435kJ/mol (or + 104 kcal/mol +104kcal/mol).

Answer 94

A: Breaking the H-H bond is an endothermic process because energy is required to cleave the bond.

Answer 95

A: − 435 kJ/mol −435kJ/mol (or − 104 kcal/mol −104kcal/mol) of energy is released when the H-H bond is formed.

Answer 96

Is bond formation endothermic or exothermic? A: Bond formation is exothermic because it releases energy.

Answer 97

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).

Answer 98

A: Because homolysis occurs, the shared pair of electrons in the bond is split equally between the two hydrogen atoms, forming two radicals, each with one unpaired electron.

Answer 99

A: The H-F bond has the highest bond dissociation energy at 569 kJ/mol 569kJ/mol (or 136 kcal/mol 136kcal/mol) because fluorine is highly electronegative, resulting in a very strong polar covalent bond with hydrogen.

Answer 100

A: The I-I bond has the lowest bond dissociation energy at 151 kJ/mol 151kJ/mol (or 36 kcal/mol 36kcal/mol) because iodine atoms are large and have weak overlapping of orbitals, resulting in a weaker bond.

Answer 101

A: The bond dissociation energy for H-Cl is 431 kJ/mol 431kJ/mol (or 103 kcal/mol 103kcal/mol). The bond dissociation energy for H-I is 297 kJ/mol 297kJ/mol (or 71 kcal/mol 71kcal/mol). Reason: The H-Cl bond is stronger because chlorine is smaller and forms a stronger bond due to better orbital overlap compared to the much larger iodine atom.

Answer 102

A: Bond dissociation energy decreases because the atomic size of the halogens increases down the group, leading to weaker overlap of orbitals and consequently weaker bonds.

Answer 103

A: Primary (e.g., CH 3 -H CH 3 -H): 435 kJ/mol 435kJ/mol (or 104 kcal/mol 104kcal/mol). Secondary (e.g., ( CH 3 ) 2 CH-H (CH 3 ) 2 CH-H): 397 kJ/mol 397kJ/mol (or 95 kcal/mol 95kcal/mol). Tertiary (e.g., ( CH 3 ) 3 C-H (CH 3 ) 3 C-H): 381 kJ/mol 381kJ/mol (or 91 kcal/mol 91kcal/mol). Explanation: The bond strength decreases from primary to tertiary because the C-H bond in tertiary carbons is more shielded and destabilized by adjacent alkyl groups.

Answer 104

A: The bond dissociation energy for CH ≡ CH CH≡CH is 523 kJ/mol 523kJ/mol (or 125 kcal/mol 125kcal/mol) because triple bonds consist of one sigma bond and two pi bonds, making the bond stronger and requiring more energy to break.

Answer 105

(e.g., CH 3 CH 2 -H CH 3 CH 2 -H to CH 3 CH 2 CH 2 -H CH 3 CH 2 CH 2 -H)? A: The bond dissociation energy remains nearly constant at 410 kJ/mol 410kJ/mol (or 98 kcal/mol 98kcal/mol), indicating that the chain length does not significantly affect the C-H bond strength in saturated hydrocarbons.

Answer 106

Compare the bond dissociation energy of O-H bonds in water ( HO-OH HO-OH) and in alcohols ( CH 3 -OH CH 3 -OH). A: HO-OH HO-OH: 213 kJ/mol 213kJ/mol (or 51 kcal/mol 51kcal/mol). CH 3 -OH CH 3 -OH: 389 kJ/mol 389kJ/mol (or 93 kcal/mol 93kcal/mol). Explanation: The bond in water is weaker due to the repulsion between lone pairs on oxygen atoms in HO-OH HO-OH, whereas the single O-H bond in alcohols is stronger due to reduced repulsion and stabilization from the alkyl group.

Answer 107

: R-F bonds (e.g., CH 3 -F CH 3 -F) have a higher dissociation energy ( 456 kJ/mol 456kJ/mol) compared to R-I bonds ( 234 kJ/mol 234kJ/mol) because fluorine is smaller and more electronegative, leading to a stronger bond compared to the larger and less electronegative iodine atom.

Answer 108

A: Bond dissociation energy is directly proportional to bond strength. A higher bond dissociation energy indicates a stronger bond because more energy is required to break it.

Answer 109

A: CH 3 -F CH 3 -F: 456 kJ/mol 456kJ/mol CH 3 -Cl CH 3 -Cl: 351 kJ/mol 351kJ/mol CH 3 -Br CH 3 -Br: 293 kJ/mol 293kJ/mol CH 3 -I CH 3 -I: 234 kJ/mol 234kJ/mol Explanation: The trend shows decreasing bond dissociation energy as the size of the halogen increases. Larger halogens have weaker bonds due to poorer orbital overlap with carbon.

Answer 110

A: Fluorine is the smallest and most electronegative halogen, resulting in strong orbital overlap and a very strong bond with carbon. This requires more energy to break.

Answer 111

A: Halomethanes with lower bond dissociation energy (e.g., CH 3 -I CH 3 -I) are more reactive because their bonds are weaker and easier to break. Halomethanes with higher bond dissociation energy (e.g., CH 3 -F CH 3 -F) are less reactive due to stronger bonds.

Answer 112

A: Bond breaking is endothermic and requires energy equal to the bond dissociation energy. Bond formation is exothermic and releases energy. The overall reaction's enthalpy ( Δ 𝐻 ΔH) depends on the balance between energy absorbed (for breaking bonds) and energy released (for forming bonds).

Answer 113

A: The enthalpy change is calculated as: \Delta H^\circ = \text{sum of } \Delta H^\circ \text{ of bonds broken} - \text{sum of } \Delta H^\circ \text{ of bonds formed}

Answer 114

A: If \Delta H^\circ is positive: • The reaction is endothermic. • More energy is required to break bonds in the reactants than is released when new bonds are formed in the products. • The bonds in the starting materials are stronger than the bonds in the products.

Answer 115

A: If \Delta H^\circ is negative: • The reaction is exothermic. • More energy is released in forming bonds in the products than is required to break bonds in the reactants. • The bonds formed in the products are stronger than the bonds broken in the starting materials.

Answer 116

A: Bond dissociation energy determines the energy required to break bonds in the reactants and the energy released in forming bonds in the products. These values directly influence whether a reaction is endothermic or exothermic.

Answer 117

Calculate the overall enthalpy change ( \Delta H^\circ ). A: \Delta H^\circ = (+435) - (-498) = -63 \, \text{kJ/mol} The reaction is exothermic.

Answer 118

• If reactants have stronger bonds than products: \Delta H^\circ is positive (endothermic). • If products have stronger bonds than reactants: \Delta H^\circ is negative (exothermic).

Answer 119

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} .

Answer 120

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.

Answer 121

A: Bond dissociation energies present overall energy changes for bond-breaking processes. However, they do not provide information about the reaction mechanism or the rate at which a reaction occurs.

Answer 122

A: Bond dissociation energies are measured in the gas phase. Most organic reactions occur in a liquid solvent, where solvation energy also contributes to the overall enthalpy change of the reaction.

Answer 123

A: Using BDEs to calculate \Delta H^\circ is considered imperfect because: 1. BDEs do not account for solvation effects present in liquid-phase reactions. 2. They provide only an approximation of energy changes and ignore other factors like molecular interactions or intermediate states.

Answer 124

A: BDEs are useful because they provide a straightforward and practical method to approximate the energy changes in a reaction, helping to predict whether a reaction is endothermic or exothermic.

Answer 125

A: 1. The equilibrium must favor the products. 2. The reaction rate must be fast enough to form the products within a reasonable time.

Answer 126

A: Thermodynamics describes: • How the energies of reactants and products compare. • The relative amounts of reactants and products at equilibrium.

Answer 127

A: Kinetics describes the rate of the reaction, i.e., how fast the reaction occurs.

Answer 128

Thermodynamics determines whether the reaction is energetically favorable (equilibrium shifts toward products), while kinetics determines how quickly the reaction proceeds. Both are essential for a reaction to be feasible in real-world conditions.

Answer 129

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!

Answer 130

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!

Answer 131

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!

Answer 132

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. Let me know if you need more questions and answers based on this content!

Answer 133

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. Let me know if you need further clarification!

Answer 134

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. Let me know if you have additional questions!

Answer 135

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.

Answer 136

A: The equatorial position is favored because it provides more space for the substituent, minimizing steric hindrance. This stability leads to a lower energy conformation.

Answer 137

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.

Answer 138

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.

Answer 139

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 . Let me know if further clarification is needed!

Answer 140

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 . Let me know if further clarification is needed!

Answer 141

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 . Let me know if further clarification is needed!

Answer 142

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. Let me know if you need further clarification!