Organic Chemistry - Reaction Mechanisms & Carbonyl Chemistry Flashcards

Question 1

SN2 Reaction Mechanism & Stereochemistry

Accepted Answer

The SN2 (Substitution Nucleophilic Bimolecular) reaction is a concerted, one-step process where the nucleophile attacks the electrophilic carbon from the backside, simultaneously displacing the leaving group. This leads to **inversion of configuration** at the reacting stereocenter (Walden inversion). The rate law is $Rate = k[RX][Nu^-]$.

Question 2

SN2 Substrate & Solvent Preferences

Accepted Answer

What are the substrate and solvent preferences for SN2 reactions?
*   **Substrate:** Methyl > Primary ($1^\circ$) > Secondary ($2^\circ$). Tertiary ($3^\circ$) substrates are unreactive due to steric hindrance.
*   **Solvent:** Favored by **aprotic polar solvents** (e.g., DMSO, acetone, acetonitrile, DMF) which solvate cations but not anions, leaving the nucleophile "naked" and highly reactive.

Question 3

SN1 Reaction Mechanism & Stereochemistry

Accepted Answer

The SN1 (Substitution Nucleophilic Unimolecular) reaction is a two-step process.
1.  **Rate-determining step:** Ionization of the substrate to form a carbocation intermediate.
2.  Nucleophilic attack on the planar carbocation from either face.
This leads to **racemization** (a mixture of retention and inversion of configuration) if the reacting carbon is a stereocenter. The rate law is $Rate = k[RX]$.

Question 4

SN1 Substrate & Solvent Preferences

Accepted Answer

What are the substrate and solvent preferences for SN1 reactions?
*   **Substrate:** Tertiary ($3^\circ$) > Secondary ($2^\circ$). Primary ($1^\circ$) and methyl substrates are unreactive as they form unstable carbocations. Allylic and benzylic carbocations are also highly favored due to resonance stabilization.
*   **Solvent:** Favored by **protic polar solvents** (e.g., water, alcohols, carboxylic acids) which stabilize the carbocation intermediate and the leaving group through hydrogen bonding.

Question 5

Factors Favoring SN1 vs. SN2

Accepted Answer

Summarize the key factors that favor SN1 over SN2, and vice-versa.
*   **Substrate:** SN1 favors $3^\circ > 2^\circ$; SN2 favors $1^\circ > 2^\circ > \	ext{methyl}$.
*   **Nucleophile:** SN1 is independent of nucleophile strength; SN2 favors strong, unhindered nucleophiles.
*   **Leaving Group:** Both favor good leaving groups (e.g., halides, tosylates).
*   **Solvent:** SN1 favors protic polar; SN2 favors aprotic polar.
*   **Stereochemistry:** SN1 leads to racemization; SN2 leads to inversion.

Question 6

E2 Reaction Mechanism & Stereochemistry

Accepted Answer

The E2 (Elimination Bimolecular) reaction is a concerted, one-step process where a strong base removes a proton ($\beta$-hydrogen) simultaneously with the departure of the leaving group, forming a $\pi$-bond. It requires an **anti-periplanar** arrangement of the $\beta$-hydrogen and the leaving group. The major product typically follows **Zaitsev's Rule** (more substituted alkene), unless a bulky base is used (Hofmann product). The rate law is $Rate = k[RX][Base]$.

Question 7

E2 Substrate & Base Preferences

Accepted Answer

What are the substrate and base preferences for E2 reactions?
*   **Substrate:** Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$). More substituted carbons lead to more stable alkenes.
*   **Base:** Favored by **strong bases** (e.g., $OH^-$, $RO^-$, $DBN$, $DBU$). Bulky bases (e.g., $t-BuOK$) can favor the Hofmann product (less substituted alkene).

Question 8

E1 Reaction Mechanism & Stereochemistry

Accepted Answer

The E1 (Elimination Unimolecular) reaction is a two-step process:
1.  **Rate-determining step:** Ionization of the substrate to form a carbocation intermediate.
2.  Deprotonation of an adjacent $\beta$-hydrogen by a weak base (often the solvent) to form an alkene.
The major product typically follows **Zaitsev's Rule**. Stereochemistry is not as strictly defined as E2, as the carbocation is planar. The rate law is $Rate = k[RX]$.

Question 9

E1 Substrate & Solvent Preferences

Accepted Answer

What are the substrate and solvent preferences for E1 reactions?
*   **Substrate:** Tertiary ($3^\circ$) > Secondary ($2^\circ$). Primary ($1^\circ$) substrates are unreactive. Similar to SN1, stability of the carbocation is key.
*   **Solvent:** Favored by **protic polar solvents** (e.g., water, alcohols) which stabilize the carbocation intermediate.

Question 10

Competition: SN1/E1 vs. SN2/E2

Accepted Answer

How do substrate, nucleophile/base strength, and solvent influence the competition between substitution and elimination reactions?
*   **Substrate:** $1^\circ$ favors SN2 (strong Nu) or E2 (strong, bulky base). $3^\circ$ favors SN1/E1 (weak Nu/base, protic solvent) or E2 (strong base). $2^\circ$ is complex and depends on other factors.
*   **Nucleophile/Base:** Strong, unhindered nucleophiles favor SN2. Strong, hindered bases favor E2. Weak bases/nucleophiles favor SN1/E1.
*   **Solvent:** Protic polar favors SN1/E1. Aprotic polar favors SN2.

Question 11

Electrophilic Addition to Alkenes: HBr (Markovnikov)

Accepted Answer

Describe the mechanism and regioselectivity of HBr addition to an unsymmetrical alkene.
This is an electrophilic addition. The $\pi$-electrons of the alkene attack the proton of HBr, forming the more stable carbocation intermediate (Markovnikov's Rule). The bromide ion then attacks the carbocation. The overall addition is **Markovnikov regioselective**, meaning the hydrogen adds to the carbon with more hydrogens, and the bromine adds to the more substituted carbon.

Question 12

Electrophilic Addition to Alkenes: H2O/H+ (Markovnikov)

Accepted Answer

Describe the mechanism and regioselectivity of acid-catalyzed hydration of an alkene.
This is an electrophilic addition. The alkene $\pi$-electrons attack a proton from the acid, forming a carbocation (Markovnikov's Rule). Water then acts as a nucleophile, attacking the carbocation. Finally, a proton is removed from the protonated alcohol by water, yielding an alcohol. The addition is **Markovnikov regioselective**.

Question 13

Electrophilic Addition to Alkenes: Hydroboration-Oxidation (Anti-Markovnikov)

Accepted Answer

Describe the mechanism and regioselectivity of hydroboration-oxidation of an alkene.
This two-step process (1. $BH_3 \cdot THF$, 2. $H_2O_2, NaOH$) results in the **anti-Markovnikov** addition of water across the double bond and **syn stereoselectivity**. Borane adds to the less hindered carbon (anti-Markovnikov), and the subsequent oxidation replaces the boron with a hydroxyl group, retaining the stereochemistry.

Question 14

General Mechanism: Nucleophilic Addition to Carbonyls

Accepted Answer

Outline the general mechanism for nucleophilic addition to aldehydes and ketones.
The carbonyl carbon is electrophilic due to resonance stabilization and the polarity of the C=O bond. A nucleophile attacks the carbonyl carbon, pushing the $\pi$-electrons onto the oxygen, forming a tetrahedral alkoxide intermediate. This intermediate is then protonated (often by an acid workup) to yield the final product (e.g., an alcohol).

Question 15

Grignard Reaction with Aldehydes/Ketones

Accepted Answer

Describe the Grignard reaction with aldehydes or ketones, including the type of product formed.
Grignard reagents ($RMgX$) are strong nucleophiles and strong bases. They attack the electrophilic carbonyl carbon of aldehydes or ketones, forming an alkoxide intermediate. Subsequent acidic workup (e.g., $H_3O^+$) protonates the alkoxide to yield an alcohol.
*   Formaldehyde $\ightarrow$ Primary alcohol
*   Other aldehydes $\ightarrow$ Secondary alcohol
*   Ketones $\ightarrow$ Tertiary alcohol

Question 16

Wittig Reaction Mechanism & Product

Accepted Answer

Describe the Wittig reaction, its key reagent, and the type of product formed.
The Wittig reaction converts an aldehyde or ketone into an alkene using a phosphorus ylide (a triphenylphosphonium ylide, $Ph_3P=CR_2$). The ylide attacks the carbonyl carbon, forming a betaine intermediate, which then cyclizes to an oxaphosphetane. This intermediate collapses to yield an alkene and triphenylphosphine oxide ($Ph_3P=O$). It's useful for forming new C=C bonds without carbocation rearrangements.

Question 17

Aldol Condensation: Enolate Formation

Accepted Answer

Explain how enolates are formed from aldehydes or ketones, and why they are important for Aldol reactions.
An enolate is formed when a strong base (e.g., $LDA$, $NaOH$, $NaOEt$) deprotonates an $\alpha$-hydrogen of an aldehyde or ketone. The resulting carbanion is resonance-stabilized by the adjacent carbonyl group. Enolates are powerful nucleophiles, making them crucial for carbon-carbon bond formation reactions like the Aldol condensation.

Question 18

Aldol Condensation: Base-Catalyzed Mechanism

Accepted Answer

Outline the mechanism of the base-catalyzed Aldol condensation.
1.  **Enolate formation:** A strong base deprotonates an $\alpha$-hydrogen, forming an enolate.
2.  **Nucleophilic attack:** The enolate (nucleophile) attacks the carbonyl carbon of another aldehyde or ketone molecule (electrophile).
3.  **Protonation:** The resulting alkoxide is protonated by water to form a $\beta$-hydroxy carbonyl compound (the Aldol product).
*   Under heating, the $\beta$-hydroxy carbonyl can undergo dehydration to form an $\alpha,\beta$-unsaturated carbonyl (Aldol condensation product).

Question 19

Aldol Condensation: Acid-Catalyzed Mechanism

Accepted Answer

Outline the mechanism of the acid-catalyzed Aldol condensation.
1.  **Enol formation:** Acid protonates the carbonyl oxygen, followed by deprotonation of an $\alpha$-hydrogen, forming an enol (the nucleophile).
2.  **Nucleophilic attack:** The enol attacks the protonated carbonyl of another aldehyde or ketone molecule.
3.  **Deprotonation/Dehydration:** Subsequent deprotonation and dehydration (loss of water) lead to the $\alpha,\beta$-unsaturated carbonyl compound.

Question 20

Claisen Condensation Mechanism

Accepted Answer

Describe the Claisen condensation, including the type of reactants and products.
The Claisen condensation is a carbon-carbon bond forming reaction between two esters (or an ester and a ketone) in the presence of a strong base (typically an alkoxide, $RO^-$). It involves nucleophilic acyl substitution.
1.  Enolate formation from one ester.
2.  Nucleophilic attack of the enolate on the carbonyl carbon of a second ester molecule.
3.  Elimination of an alkoxide leaving group.
The product is a $\beta$-keto ester.

Question 21

General Mechanism: Nucleophilic Acyl Substitution

Accepted Answer

Outline the general mechanism for nucleophilic acyl substitution reactions.
This is a two-step process for carboxylic acid derivatives:
1.  **Nucleophilic attack:** A nucleophile attacks the electrophilic carbonyl carbon, forming a tetrahedral intermediate. The $\pi$-electrons move to the oxygen.
2.  **Leaving group departure:** The carbonyl reforms as the leaving group (L) departs from the tetrahedral intermediate.
This differs from aldehyde/ketone nucleophilic addition because a leaving group is present.

Question 22

Reactivity Order of Carboxylic Acid Derivatives

Accepted Answer

Rank the common carboxylic acid derivatives by their reactivity towards nucleophilic acyl substitution and explain why.
The order of reactivity is: **Acid Chlorides > Acid Anhydrides > Esters > Amides**.
*   **Explanation:** Reactivity is determined by the electron-withdrawing ability and the leaving group ability of the substituent attached to the carbonyl. Good leaving groups (e.g., $Cl^-$) and strong electron-withdrawing groups make the carbonyl more electrophilic and facilitate the departure of the leaving group. Amides are the least reactive due to the poor leaving group ($NH_2^-$) and resonance donation from nitrogen.

Question 23

Imine and Enamine Formation

Accepted Answer

Describe the formation of imines and enamines from aldehydes/ketones.
*   **Imine formation:** Aldehydes or ketones react with primary amines ($RNH_2$) under acid catalysis to form imines ($R_2C=NR'$), via a carbinolamine intermediate followed by dehydration.
*   **Enamine formation:** Aldehydes or ketones react with secondary amines ($R_2NH$) under acid catalysis to form enamines ($R_2C=CR-NR'_2$), via a carbinolamine intermediate followed by dehydration and deprotonation of an $\alpha$-carbon.

Question 24

Cyanohydrin Formation

Accepted Answer

Describe the formation of cyanohydrins from aldehydes or ketones.
Aldehydes and ketones react with hydrogen cyanide ($HCN$) in the presence of a base (e.g., $KCN$) to form cyanohydrins. The cyanide ion ($CN^-$) acts as a nucleophile, attacking the carbonyl carbon. The resulting alkoxide is then protonated by HCN or water. Cyanohydrins are useful synthetic intermediates as the nitrile group can be hydrolyzed to a carboxylic acid or reduced to an amine.

Question 25

Reduction of Carbonyls (Aldehydes/Ketones)

Accepted Answer

What reagents are commonly used to reduce aldehydes and ketones, and what products are formed?
Aldehydes and ketones can be reduced to alcohols using hydride reducing agents.
*   **Lithium aluminum hydride ($LiAlH_4$)**: A very strong reducing agent, reduces aldehydes to primary alcohols and ketones to secondary alcohols. It also reduces carboxylic acids and their derivatives.
*   **Sodium borohydride ($NaBH_4$)**: A milder reducing agent, selectively reduces aldehydes to primary alcohols and ketones to secondary alcohols, without affecting esters, carboxylic acids, or amides.

Organic Chemistry - Reaction Mechanisms & Carbonyl Chemistry

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