Alkanes, Cycloalkanes, and Elimination Reactions

Chemistry Flashcards: Alkanes, Cycloalkanes, and Elimination Reactions

What is the primary factor determining the orientation (position) of halogen attachment in the free radical halogenation of alkanes?

The orientation of halogen attachment is primarily determined by the stability of the intermediate alkyl radical formed during the propagation step. More stable radicals (tertiary > secondary > primary > methyl) form more readily, leading to the halogen attaching to the carbon that can best stabilize the radical.

Explain the relative stability of alkyl radicals (methyl, primary, secondary, tertiary).

Alkyl radical stability follows the order: tertiary (3°) > secondary (2°) > primary (1°) > methyl (CH₃•). This is due to hyperconjugation, where neighboring alkyl groups donate electron density to the electron-deficient radical center, stabilizing it. More alkyl groups mean more hyperconjugation and greater stability.

In the chlorination of propane, what are the expected major and minor products, and why?

Propane has primary and secondary hydrogens. Chlorination will yield both 1-chloropropane (from primary radical) and 2-chloropropane (from secondary radical). The 2-chloropropane is the major product (around 55%) because the secondary radical intermediate is more stable than the primary radical.

Compare the general reactivity order of halogens (F₂, Cl₂, Br₂, I₂) in alkane halogenation.

The reactivity of halogens decreases in the order: F₂ > Cl₂ > Br₂ > I₂.

• Fluorine (F₂): Extremely reactive, often violently, and difficult to control.

• Chlorine (Cl₂): Highly reactive, but controllable.

• Bromine (Br₂): Less reactive than chlorine, but more selective.

• Iodine (I₂): Generally unreactive with alkanes; iodination is often reversible and requires oxidizing agents.

Explain the concept of "selectivity" in halogenation reactions, particularly comparing chlorine and bromine.

Selectivity refers to the preference of a halogen radical to abstract a hydrogen from a specific type of carbon (e.g., tertiary over secondary, secondary over primary) when multiple options are available.

Chlorine (Cl•): More reactive and less selective. It reacts quickly with almost any available hydrogen, leading to a mixture of products, though it still shows a preference for more substituted hydrogens (e.g., 2° > 1° by a factor of ~3:1).

Bromine (Br•): Less reactive but significantly more selective. It preferentially abstracts hydrogen from the most stable radical intermediate (tertiary > secondary > primary), leading to a higher yield of the most substituted product (e.g., 2° > 1° by a factor of ~80:1).

Why is bromine more selective than chlorine in alkane halogenation?

Bromine is more selective because the hydrogen abstraction step by a bromine radical is more endothermic and has a higher activation energy compared to chlorine. This means the transition state for bromine abstraction is later (more product-like) and more closely resembles the stable alkyl radical, allowing the radical to "choose" the pathway that leads to the most stable intermediate. Chlorine's abstraction is more exothermic and has an earlier, more reactant-like transition state, making it less discriminating.

What is "ring strain" in cycloalkanes, and what are its main components?

Ring strain is the instability and high-energy state of cyclic organic compounds due to the distortion of bond angles and bond lengths from their ideal values. It results from a combination of three main factors:

Angle Strain (Baeyer Strain): Deviation of bond angles from the ideal tetrahedral angle of 109.5°.

Torsional Strain (Pitzer Strain): Caused by eclipsing interactions of bonds on adjacent atoms.

Steric Strain (Transannular Strain): Repulsion between non-bonded atoms that are forced too close to each other.

How is the stability of cycloalkanes experimentally measured or inferred?

The stability of cycloalkanes is often inferred by measuring their heat of combustion per CH₂ unit. More stable (less strained) cycloalkanes release less energy per CH₂ unit upon combustion, while less stable (highly strained) rings release more energy.

Rank the stability of common cycloalkanes (cyclopropane, cyclobutane, cyclopentane, cyclohexane) from least to most stable, and briefly explain why.

The general order of stability (least to most strained, or most to least stable) is: Cyclopropane > Cyclobutane > Cycloheptane (and larger rings initially) > Cyclopentane > Cyclohexane.

Cyclopropane: Most strained due to severe angle strain (60° vs. 109.5°) and high torsional strain (all C-H bonds eclipsed).

Cyclobutane: Highly strained, but less than cyclopropane, with angle strain (90° vs. 109.5°) and significant torsional strain, though it can pucker slightly to reduce some of it.

Cyclopentane: Has minimal angle strain (108° vs. 109.5°) but still experiences some torsional strain. It adopts an "envelope" conformation to reduce this.

Cyclohexane: The most stable cycloalkane. It can adopt a "chair" conformation where all bond angles are approximately 109.5° and all C-H bonds are staggered, effectively eliminating angle and torsional strain.

Larger Rings (C7+): As ring size increases beyond six, angle strain decreases, but conformational strain (torsional and steric) can increase due to more complex interactions, making them slightly less stable than cyclohexane but more stable than small rings.

Describe the specific types and extent of ring strain in cyclopropane.

Cyclopropane is highly strained due to:

Angle Strain: Its internal bond angles are 60°, a significant deviation from the ideal 109.5° tetrahedral angle. This forces the C-C bonds to be "bent" (banana bonds), leading to poor orbital overlap and weaker bonds.

Torsional Strain: All adjacent C-H bonds are in an eclipsed conformation, which cannot be relieved by rotation due to the rigid ring structure. This contributes significantly to its instability.

Steric Strain: Minimal, as the ring is too small for significant non-bonded interactions. The total ring strain in cyclopropane is estimated to be around 27.6 kcal/mol (115 kJ/mol).

Describe the specific types and extent of ring strain in cyclobutane.

Cyclobutane is also highly strained, but less so than cyclopropane:

Angle Strain: Its internal bond angles are approximately 90°, still a significant deviation from 109.5°, but less severe than cyclopropane.

Torsional Strain: While still present, cyclobutane can adopt a slightly "puckered" (or "bent") conformation, which helps to reduce some of the eclipsing interactions between adjacent C-H bonds, thereby reducing torsional strain compared to a planar cyclobutane.

Steric Strain: Very low, similar to cyclopropane, as it's generally not large enough for significant steric hindrance between substituents. The total ring strain in cyclobutane is estimated to be around 26.3 kcal/mol (110 kJ/mol), slightly less than cyclopropane.

What is the central postulate of Baeyer's Strain Theory?

Proposed by Adolf von Baeyer in 1885, Baeyer's Strain Theory postulates that all cycloalkane rings are planar (flat). Based on this assumption, he suggested that any deviation of the internal bond angles from the ideal tetrahedral angle of 109.5° (or 109°28') causes "angle strain" in the molecule, which reduces its stability. The greater the deviation, the greater the strain and the lower the stability.

According to Baeyer's Strain Theory, which cycloalkane should be the most stable, and why?

According to Baeyer's original theory, cyclopentane should be the most stable cycloalkane because its internal bond angles (108° for a regular pentagon) are closest to the ideal tetrahedral angle of 109.5°. He predicted that larger rings would become increasingly strained due to angles deviating further from 109.5° (becoming larger than 109.5°).

What are the major limitations of Baeyer's Strain Theory?

The major limitations of Baeyer's Strain Theory are:

Planarity Assumption: The theory incorrectly assumes that all cycloalkanes are planar.

Larger Rings: It fails to explain the observed high stability of cyclohexane and larger rings. According to Baeyer, rings larger than cyclopentane should become increasingly strained and unstable due to increasing angle deviation, but in reality, cyclohexane is very stable, and larger rings are also stable.

Observed vs. Predicted Stability: It predicts cyclopentane to be more stable than cyclohexane, which is contrary to experimental observations (cyclohexane is more stable).

What is the core idea of the Sachse-Mohr Theory of Strainless Rings, and how did it address the limitations of Baeyer's theory?

The Sachse-Mohr Theory (proposed by Sachse in 1890 and confirmed by Mohr in 1918) states that cycloalkanes with six or more carbon atoms are not planar but exist in puckered, three-dimensional conformations. This allows the carbon atoms to maintain their ideal tetrahedral bond angles of 109.5° and avoid angle strain. This theory successfully explained the observed high stability of cyclohexane (e.g., chair and boat conformations) and larger rings, which Baeyer's theory could not.

Give examples of "strainless" conformations predicted by the Sachse-Mohr theory for cyclohexane.

For cyclohexane, the Sachse-Mohr theory introduced the concept of non-planar, strain-free conformations, most notably the chair conformation and the boat conformation. The chair form is the most stable, completely eliminating angle and torsional strain.

What is the general reaction for the dehydration of alcohols to alkenes, and what reagents are typically used?

Dehydration of alcohols involves the elimination of a water molecule (–OH and an adjacent –H) to form an alkene.

General Reaction: R–CH(OH)–CH₂–R' → R–CH=CH–R' + H₂O

Reagents: Typically requires heating the alcohol in the presence of a strong acid catalyst, such as concentrated sulfuric acid (H₂SO₄) or phosphoric acid (H₃PO₄).

Outline the general mechanism for the acid-catalyzed dehydration of secondary and tertiary alcohols.

Secondary and tertiary alcohols typically dehydrate via an E1 mechanism (Elimination, unimolecular), which involves three steps:

Protonation of the Alcohol: The oxygen atom of the –OH group is protonated by the acid, forming a good leaving group (alkyloxonium ion, –OH₂⁺).

R–OH + H⁺ → R–OH₂⁺

Loss of Water to Form a Carbocation: The protonated alcohol loses a molecule of water, forming a carbocation intermediate. This is the slowest, rate-determining step.

R–CH(OH₂⁺)–R' → R–C⁺H–R' + H₂O

Deprotonation to Form Alkene: A base (often the conjugate base of the acid or another water molecule) abstracts a proton from a carbon adjacent to the carbocation (a β-hydrogen), forming the alkene and regenerating the acid catalyst.

R–C⁺H–CH₂–R' + B⁻ → R–CH=CH–R' + BH

How does the reactivity of primary, secondary, and tertiary alcohols differ in dehydration, and why?

The ease of dehydration follows the order: Tertiary (3°) > Secondary (2°) > Primary (1°).

This order is due to the stability of the carbocation intermediate formed in the E1 mechanism (tertiary carbocations are the most stable, primary are the least stable).

Primary alcohols typically undergo an E2 mechanism (concerted, one-step) due to the instability of primary carbocations, requiring higher temperatures.

What is the Hofmann Elimination reaction, and what are its key starting materials and products?

The Hofmann Elimination (also known as Hofmann Degradation or Exhaustive Methylation) is a chemical reaction that converts quaternary ammonium salts into tertiary amines and alkenes through a two-step process.

Exhaustive Methylation: An amine is treated with excess methyl iodide (CH₃I) to form a quaternary ammonium iodide salt.

Elimination: The quaternary ammonium iodide salt is treated with silver oxide (Ag₂O) and water, followed by heating, which results in a β-elimination reaction to form the alkene and a tertiary amine.

What is the characteristic regioselectivity of the Hofmann Elimination, and what rule governs it?

The Hofmann Elimination characteristically follows Hofmann's Rule, which states that the major alkene product formed is the least substituted (and typically less stable) alkene. This is in contrast to Zaitsev's Rule, which favors the more substituted alkene.

Why does the Hofmann Elimination favor the least substituted alkene (Hofmann product)?

The Hofmann Elimination favors the least substituted alkene primarily due to steric hindrance caused by the bulky quaternary ammonium leaving group (–NR₃⁺).

The bulky leaving group makes it sterically more difficult for the base to abstract a proton from a more hindered (more substituted) β-carbon.

Instead, the base preferentially abstracts a proton from the most accessible (least hindered) β-carbon, which typically leads to the formation of the least substituted alkene.

The reaction proceeds via an E2 mechanism.

What is a "vicinal dihalide," and what is the dehalogenation reaction of vicinal dihalides?

A vicinal dihalide is an organic compound where two halogen atoms are attached to adjacent carbon atoms. Dehalogenation of vicinal dihalides is a reaction that converts these dihalides into an alkene by removing the two halogen atoms and forming a carbon-carbon double bond between the carbons that previously held the halogens.

What common reagent is used for the dehalogenation of vicinal dihalides, and what is the general reaction equation?

The dehalogenation of vicinal dihalides typically uses zinc metal (Zn) as a reducing agent, often in an alcohol solvent like ethanol.

General Reaction: R–CHX–CHX–R' + Zn → R–CH=CH–R' + ZnX₂

(Where X is a halogen, commonly Br or Cl).

Briefly describe the mechanism of dehalogenation of vicinal dihalides with zinc.

The dehalogenation of vicinal dihalides with zinc is believed to proceed through an E2-like mechanism or involve an organozinc intermediate.

The zinc metal effectively removes the two halogen atoms, leading to the formation of a double bond.

One proposed pathway involves the oxidative insertion of Zn(0) into a C-X bond, forming an organozinc species, which then undergoes elimination.

The reaction often shows anti-stereoselectivity, supporting an E2-like concerted removal of the two halogens.

State Hofmann's Rule in the context of elimination reactions.

Hofmann's Rule states that in certain elimination reactions, particularly the Hofmann Elimination of quaternary ammonium salts, the major alkene product formed is the least substituted alkene (i.e., the one with the fewest alkyl groups attached to the double bond carbons).

How does Hofmann's Rule contrast with Zaitsev's Rule?

Feature	Zaitsev's Rule	Hofmann's Rule
Product	Favors the more substituted (more stable) alkene.	Favors the least substituted (less stable) alkene.
Conditions	Typically occurs with smaller, less hindered bases and good leaving groups (e.g., halogens).	Occurs with bulky bases or bulky leaving groups (e.g., quaternary ammonium salts, alkyl fluorides).
Driving Force	Thermodynamic stability of the alkene.	Steric hindrance in the transition state.

Provide an example of a reaction where Hofmann's Rule would be observed.

A classic example is the Hofmann Elimination of a quaternary ammonium hydroxide, such as (CH₃)₃N⁺CH₂CH₂CH₃ OH⁻. When heated, the major product will be propene (CH₂=CHCH₃), the least substituted alkene, rather than any more substituted isomer if available. Another instance is E2 elimination with a bulky base (e.g., potassium tert-butoxide, t-BuOK), which also favors the Hofmann product.