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Organic Chemistry: Reactions & Mechanisms
SN1/SN2/E1/E2 mechanisms, addition reactions, oxidation/reduction, Grignard, EAS, and key reagent cards.
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Describe the SN2 mechanism: steps, stereochemistry, rate law, and energy diagram.
SN2 = Substitution, Nucleophilic, Bimolecular.
Mechanism: ONE concerted step — nucleophile attacks the electrophilic carbon from the BACKSIDE (180° from leaving group) while the leaving group departs simultaneously.
Stereochemistry: INVERSION of configuration (Walden inversion) — 100% inversion at the stereocenter.
Rate law: Rate = k[substrate][nucleophile] — second order (bimolecular).
Energy diagram: Single transition state (no intermediate), one energy maximum.
What conditions FAVOR SN2 reactions?
1. Substrate: Methyl > 1° > 2° (3° never — too sterically hindered).
2. Nucleophile: STRONG nucleophile required (e.g., CN⁻, I⁻, RS⁻, HO⁻).
3. Leaving group: Good leaving group (weak base after leaving: I⁻ > Br⁻ > Cl⁻ > F⁻).
4. Solvent: POLAR APROTIC solvent (DMSO, DMF, acetone, acetonitrile) — these don’t solvate/stabilize the nucleophile, keeping it reactive.
5. Concentration: High [Nu:] matters (it’s in the rate law).
Describe the SN1 mechanism: steps, stereochemistry, rate law, and energy diagram.
SN1 = Substitution, Nucleophilic, Unimolecular.
Step 1: Leaving group departs → forms CARBOCATION (rate-determining step).
Step 2: Nucleophile attacks carbocation from EITHER face.
Stereochemistry: RACEMIZATION — mixture of retention and inversion. Typically slight excess of inversion (~55/45) due to ion pair shielding.
Rate law: Rate = k[substrate] — first order (unimolecular). Nucleophile not in rate law.
Energy diagram: TWO transition states with a carbocation intermediate (energy minimum) between them.
What conditions FAVOR SN1 reactions?
1. Substrate: 3° > 2° >> 1° (never methyl). Stable carbocations required.
2. Nucleophile: WEAK nucleophile is fine (e.g., H₂O, ROH) — it’s not in the rate law.
3. Leaving group: Good leaving group needed (same trend as SN2).
4. Solvent: POLAR PROTIC solvent (water, methanol, ethanol, acetic acid) — stabilizes carbocation intermediate and leaving group through solvation.
5. Carbocation stability: 3° > 2° > 1°. Resonance-stabilized (allylic, benzylic) also favored.
Describe the E2 mechanism: steps, stereochemistry, and rate law.
E2 = Elimination, Bimolecular.
Mechanism: ONE concerted step — strong base abstracts a β-hydrogen while the leaving group departs simultaneously. Requires ANTI-PERIPLANAR geometry (H and LG are 180° dihedral).
Stereochemistry: Stereochemistry is determined by the anti-periplanar requirement; for many substrates the trans (E) product predominates, but this is substrate-dependent, not an inherent rule.
Rate law: Rate = k[substrate][base] — second order.
Product: Zaitsev product usually favored (most substituted alkene) unless bulky base used.
What conditions FAVOR E2 reactions?
1. Substrate: 3° > 2° > 1° (rate). Works on all substrates but competes with SN2 for 1° and 2°.
2. Base: STRONG, often BULKY base (e.g., t-BuOK, DBU, LDA). Bulky bases favor E2 over SN2.
3. Temperature: HIGH temperature favors elimination over substitution.
4. Anti-periplanar geometry required — in cyclohexanes, H and LG must both be axial (trans-diaxial).
5. Zaitsev rule: Most substituted alkene favored unless bulky base → Hofmann product (less substituted).
Describe the E1 mechanism: steps, stereochemistry, and rate law.
E1 = Elimination, Unimolecular.
Step 1: Leaving group departs → forms CARBOCATION (rate-determining step, same as SN1).
Step 2: A base (often solvent) abstracts a β-hydrogen → forms alkene.
Stereochemistry: Not stereospecific (carbocation is planar, β-H can be removed from any position). Mixture of E and Z possible.
Rate law: Rate = k[substrate] — first order.
Product: Zaitsev product favored (most substituted, most stable alkene). Often competes with SN1.
What conditions FAVOR E1 reactions?
Same conditions as SN1 (they always compete):
1. Substrate: 3° > 2°. Stable carbocations needed.
2. Base: WEAK base (or no base — solvent acts as base).
3. Solvent: Polar protic.
4. Temperature: HIGHER temperature shifts SN1/E1 competition toward E1 (elimination is entropically favored — ΔS is positive because 1 molecule → 2 molecules).
5. E1 always accompanied by SN1. Pure E1 is rare.
Given a 3° substrate with a strong base: predict SN1/SN2/E1/E2.
E2. Reasoning: 3° substrates are too sterically hindered for SN2 (backside attack blocked). Strong base drives E2 (bimolecular elimination). Even though 3° substrates can do SN1/E1, the STRONG BASE dominates the pathway → E2 wins.
Given a 1° substrate with a strong nucleophile/weak base in polar aprotic solvent: predict SN1/SN2/E1/E2.
SN2. Reasoning: 1° substrate is unhindered (good for backside attack). Strong nucleophile is in the rate law for SN2. Polar aprotic solvent doesn’t solvate the nucleophile, keeping it reactive. Weak base means elimination is not competitive.
Given a 3° substrate with a weak nucleophile in polar protic solvent at high temperature: predict.
E1 (major) + SN1 (minor). Reasoning: 3° substrate → cannot do SN2. Weak nucleophile → not E2. Polar protic solvent → favors carbocation formation (SN1/E1 conditions). High temperature → shifts equilibrium toward elimination (E1 > SN1).
Given a 2° substrate with a strong bulky base (like t-BuOK): predict.
E2. Reasoning: 2° substrate can do multiple pathways. Strong base → E2 favored over E1. Bulky base → cannot easily attack carbon for SN2, instead abstracts the more accessible β-hydrogen. E2 gives Hofmann product (less substituted alkene) with bulky bases.
How does the nucleophile/base distinction determine SN vs E?
NUCLEOPHILE = attacks CARBON (substitution). BASE = attacks HYDROGEN (elimination).
Strong nucleophile + weak base → SN2 (e.g., I⁻, RS⁻, CN⁻, N₃⁻)
Strong base + weak nucleophile → E2 (e.g., t-BuO⁻, LDA, DBN)
Strong nucleophile + strong base → SN2 or E2 depending on substrate and temperature (e.g., HO⁻, EtO⁻ — these are BOTH strong Nu and strong base; 1° → SN2, 3° → E2, 2° → mixture)
Weak nucleophile + weak base → SN1/E1 (e.g., H₂O, ROH)
What is Markovnikov’s Rule? Give the mechanism for HBr addition to propene.
Markovnikov’s Rule: In addition of HX to an unsymmetrical alkene, H adds to the LESS substituted carbon (the one with MORE H’s already), and X adds to the MORE substituted carbon.
Mechanism (propene + HBr):
1. Protonation: π electrons attack H⁺ of HBr → form the MORE STABLE carbocation (2° at C2, not 1° at C1).
2. Nucleophilic attack: Br⁻ attacks the carbocation at C2.
Product: 2-bromopropane (CH₃CHBrCH₃).
Rationale: The reaction goes through the most stable carbocation intermediate.
What is anti-Markovnikov addition? Give two examples with reagents.
Anti-Markovnikov: The nucleophilic group adds to the LESS substituted carbon of the alkene.
1. HBr + ROOR (peroxides): Radical chain mechanism. Br• adds first (anti-Mark), then H. Only works with HBr (not HCl or HI). Product: 1-bromopropane from propene.
2. Hydroboration-oxidation (BH₃·THF, then H₂O₂/NaOH): BH₃ adds B to less substituted carbon (syn addition). Oxidation replaces B with OH. Overall: anti-Markovnikov, SYN addition of water. Product: 1-propanol from propene.
Describe halogenation of alkenes (Br₂ addition). What is the stereochemistry?
Reagent: Br₂ (often in CH₂Cl₂ or CCl₄).
Mechanism:
1. Alkene π electrons attack Br₂ → form cyclic BROMONIUM ION intermediate + Br⁻.
2. Br⁻ attacks from the OPPOSITE face of the bromonium ion (backside attack, like SN2).
Stereochemistry: ANTI addition — the two Br atoms end up on opposite faces. This means trans-product from a cyclic alkene.
Product: vicinal dibromide. This reaction is also a TEST for unsaturation (Br₂/CCl₄ decolorizes).
What is catalytic hydrogenation? Reagents and stereochemistry?
Reagent: H₂ gas with a metal catalyst (Pd/C, Pt/C, or Ni).
Reaction: Adds H₂ across a double bond (or triple bond) — reduces alkenes to alkanes.
Stereochemistry: SYN addition — both H atoms add to the same face of the alkene (because both H atoms are delivered from the catalyst surface).
Note: Pd/C with H₂ reduces C=C but usually NOT C=O. Lindlar’s catalyst (Pd/CaCO₃, poisoned) reduces alkynes to CIS alkenes only (partial reduction).
What does mCPBA do? Mechanism and stereochemistry of epoxidation.
mCPBA (meta-chloroperoxybenzoic acid) is a peroxyacid that converts alkenes to EPOXIDES.
Mechanism: Concerted — the peroxyacid delivers an oxygen atom to the alkene in a single step (butterfly mechanism).
Stereochemistry: SYN addition — the oxygen adds to one face. A cis-alkene gives a cis-epoxide; a trans-alkene gives a trans-epoxide. The stereochemistry of the alkene is RETAINED in the product.
What is syn-dihydroxylation? Reagent and product?
Reagent: OsO₄ (osmium tetroxide), usually with NMO (N-methylmorpholine N-oxide) as a co-oxidant to make OsO₄ catalytic.
Reaction: Adds two –OH groups across a double bond.
Stereochemistry: SYN addition — both OH groups add to the SAME face through a cyclic osmate ester intermediate.
Product: cis-1,2-diol (vicinal diol).
Alternative: KMnO₄ (cold, dilute, basic) also gives syn-dihydroxylation but is harder to control.
What is ozonolysis? Reagents and products?
Reagents: 1. O₃ (ozone), 2. Reductive workup: Zn/AcOH or (CH₃)₂S (dimethyl sulfide). Oxidative workup: H₂O₂.
Reaction: CLEAVES the C=C double bond completely.
Products (reductive workup):
- Each C of the double bond becomes a C=O.
- R₂C=CR₂ → two ketones
- RHC=CHR → two aldehydes
- R₂C=CHR → one ketone + one aldehyde
- RCH=CH₂ → aldehyde + formaldehyde (HCHO)
Oxidative workup: Aldehydes → carboxylic acids.
Useful for determining alkene structure from fragments.
What does PCC (pyridinium chlorochromate) do?
PCC (Cr-based oxidant, mild, used in CH₂Cl₂) selectively oxidizes:
1° alcohol → ALDEHYDE (stops here — does NOT overoxidize to carboxylic acid)
2° alcohol → KETONE
3° alcohol → NO REACTION
Key: PCC is the reagent of choice when you need to stop oxidation at the aldehyde stage. It works in anhydrous conditions (CH₂Cl₂), which prevents overoxidation. Dess-Martin periodinane (DMP) is a modern alternative.
What does Jones reagent (CrO₃/H₂SO₄/acetone) do?
Jones reagent is a STRONG oxidant:
1° alcohol → CARBOXYLIC ACID (goes all the way — oxidizes through aldehyde to acid)
2° alcohol → KETONE
3° alcohol → NO REACTION (under normal conditions; harsh conditions can cleave C–C bonds)
Jones reagent works in aqueous acetone. The water present allows overoxidation of the aldehyde intermediate to the carboxylic acid via the hydrate.
What does KMnO₄ do to alkenes under different conditions?
Cold, dilute, basic KMnO₄: SYN dihydroxylation → cis-1,2-diol (with brown MnO₂ precipitate). Baeyer test for unsaturation.
Hot, concentrated, acidic KMnO₄: OXIDATIVE CLEAVAGE of C=C bond → carboxylic acids and/or ketones.
- RHC=CHR → 2 carboxylic acids
- R₂C=CR₂ → 2 ketones
- RCH=CH₂ → RCOOH + CO₂ (terminal CH₂ fully oxidized)
Similar cleavage results to ozonolysis but harsher.
Compare NaBH₄ and LiAlH₄ as reducing agents: selectivity and conditions.
NaBH₄ (sodium borohydride): MILD, SELECTIVE reducing agent.
- Reduces: aldehydes → 1° alcohols, ketones → 2° alcohols
- Does NOT reduce: esters, carboxylic acids, amides, epoxides (usually)
- Solvent: MeOH, EtOH, or water (tolerates protic solvents)
LiAlH₄ (lithium aluminum hydride): POWERFUL, NON-SELECTIVE reducing agent.
- Reduces: ALL of the above PLUS esters → 1° alcohols, carboxylic acids → 1° alcohols, amides → amines, epoxides → alcohols
- Solvent: Anhydrous ether or THF ONLY (reacts violently with water/protic solvents)
- Workup: Aqueous acid quench after reaction
What is the selectivity difference between NaBH₄ and LiAlH₄ for reducing an ester vs a ketone in the same molecule?
NaBH₄ will SELECTIVELY reduce the ketone to an alcohol while LEAVING the ester untouched. LiAlH₄ will reduce BOTH — the ketone to a 2° alcohol AND the ester to two 1° alcohols. This selectivity makes NaBH₄ invaluable for chemoselective reductions.
What are Grignard reagents and what do they react with?
Grignard reagent: RMgX (made from R–X + Mg in anhydrous ether or THF). The carbon is a strong NUCLEOPHILE and BASE.
Reactions with carbonyls (add R group, then acid workup gives alcohol):
- HCHO (formaldehyde) + RMgX → 1° alcohol
- RCHO (aldehyde) + R’MgX → 2° alcohol
- R₂CO (ketone) + R’MgX → 3° alcohol
- Ester + 2 eq RMgX → 3° alcohol
- CO₂ + RMgX → carboxylic acid (after acid workup)
- Epoxide + RMgX → alcohol (adds 2 carbons, opens at less substituted end)
CRITICAL: Grignards are destroyed by any acidic proton (H₂O, ROH, NH, COOH). Reaction must be ANHYDROUS.
What is the Grignard reagent incompatibility rule?
Grignard reagents (RMgX) react with ANY acidic hydrogen (pKa < ~45). The molecule being reacted with a Grignard CANNOT contain:
- –OH (alcohol, phenol)
- –NH (amine, amide)
- –SH (thiol)
- –COOH (carboxylic acid)
- terminal alkyne C≡C–H
If these groups are present, they will protonate the Grignard (RMgX + HA → RH + MgXA) before it can do the desired nucleophilic addition. Protect these groups first or use a different strategy.
What is the Aldol Reaction / Aldol Condensation?
Aldol reaction: An enolizable aldehyde (or ketone) reacts with base (NaOH) to form an ENOLATE, which then attacks a second aldehyde/ketone carbonyl → gives a β-hydroxy aldehyde (or ketone) = aldol product.
Aldol condensation: Upon HEATING, the aldol product undergoes DEHYDRATION (–H₂O) → gives an α,β-unsaturated carbonyl (conjugated enone).
Crossed aldol: Between two different carbonyl compounds — gives mixtures unless one partner has no α-hydrogens (e.g., benzaldehyde, formaldehyde) or directed conditions (LDA, kinetic enolate).
What is Fischer Esterification? Reagents, mechanism, and equilibrium.
Reaction: Carboxylic acid + alcohol ⇌ ester + water (acid-catalyzed, REVERSIBLE).
Catalyst: H₂SO₄ or HCl (acid catalyst).
Mechanism:
1. Protonation of carbonyl oxygen
2. Nucleophilic attack of alcohol on protonated carbonyl
3. Proton transfer
4. Loss of water (leaving group)
5. Deprotonation → ester product
Key: REVERSIBLE (equilibrium). Drive forward by: excess alcohol, removing water (Dean-Stark trap), or excess acid. Uses acid catalyst — no strong nucleophiles or bases involved.
Compare acid-catalyzed vs base-catalyzed hydrolysis of esters.
ACID-CATALYZED (reverse of Fischer esterification):
Ester + H₂O + H⁺ catalyst ⇌ carboxylic acid + alcohol. REVERSIBLE.
BASE-CATALYZED (saponification):
Ester + NaOH → carboxylate salt (RCOO⁻Na⁺) + alcohol. IRREVERSIBLE.
Why irreversible? The carboxylate anion (resonance-stabilized) is a very poor electrophile — the reverse reaction doesn’t occur. This is why saponification goes to completion.
Saponification of fats (triesters of glycerol) produces SOAP (carboxylate salts) + glycerol.
What is the general mechanism for Electrophilic Aromatic Substitution (EAS)?
Step 1: Generation of the electrophile (E⁺).
Step 2: Electrophilic attack — π electrons of benzene attack E⁺ → form ARENIUM ION (sigma complex/Wheland intermediate). This is a resonance-stabilized carbocation (3 resonance structures). This is the RATE-DETERMINING STEP.
Step 3: Deprotonation — a base removes H⁺ from the sp³ carbon → restores aromaticity.
Key: SUBSTITUTION not addition — aromaticity is preserved. The aromatic ring acts as a nucleophile.
What are the reagents for bromination of benzene (EAS)?
Reagents: Br₂ + FeBr₃ (Lewis acid catalyst).
Mechanism: FeBr₃ polarizes Br₂ → generates electrophilic Br⁺ (or Br–Br–FeBr₃ complex). Benzene attacks Br⁺ → arenium ion → deprotonation → bromobenzene + HBr.
FeBr₃ is regenerated. AlBr₃ also works as the Lewis acid.
Note: Without Lewis acid catalyst, Br₂ does NOT react with benzene (not electrophilic enough). Exception: activated rings like phenol or aniline react with Br₂/H₂O without catalyst.
What are the reagents for nitration of benzene (EAS)?
Reagents: HNO₃ + H₂SO₄ (mixed acid).
Electrophile generated: NO₂⁺ (nitronium ion), formed by protonation and dehydration of HNO₃ by H₂SO₄.
Product: Nitrobenzene.
The nitro group is a strong deactivating, meta-directing group — so further nitration is slower and gives m-dinitrobenzene.
What are Friedel-Crafts Alkylation and Acylation?
ALKYLATION: ArH + RCl + AlCl₃ → ArR + HCl
Electrophile: R⁺ carbocation (generated from RCl + AlCl₃). Problem: Carbocation can rearrange (1° → 2° → 3°). Also, product is MORE reactive than starting material → polyalkylation.
ACYLATION: ArH + RCOCl + AlCl₃ → ArCOR + HCl
Electrophile: Acylium ion (RC≡O⁺, resonance-stabilized — NO rearrangement). Product ketone is deactivated → NO polyacylation.
Limitation: Neither reaction works on strongly deactivated rings (–NO₂, –CF₃, –COR, –SO₃H, –NR₃⁺).
What are activating and deactivating groups in EAS? What are the directing effects?
ACTIVATING (make ring MORE reactive than benzene, donate electron density):
- Strong activating (ortho/para directors): –NH₂, –NHR, –NR₂, –OH, –OR
- Moderate activating (ortho/para): –NHCOR, –OCOR
- Weak activating (ortho/para): –CH₃, –R (alkyl groups — hyperconjugation/induction)
DEACTIVATING (make ring LESS reactive):
- Weak deactivating (ortho/para director!): –F, –Cl, –Br, –I (halogens — special case!)
- Moderate/strong deactivating (META directors): –NO₂, –CN, –SO₃H, –COR, –COOR, –COOH, –CF₃, –NR₃⁺
Key exception: Halogens are deactivating but ortho/para directing (inductive withdrawal but lone pair donation by resonance).
Why are halogens ortho/para directors but deactivating?
Two competing effects:
1. INDUCTIVE EFFECT: Halogens are electronegative → withdraw electron density through σ bonds → deactivating (makes ring less electron-rich overall → slower reaction than benzene).
2. RESONANCE EFFECT: Lone pairs on halogen can donate into the ring by resonance → stabilizes the arenium ion intermediate ONLY for ortho/para attack (not meta).
The inductive effect dominates for RATE (deactivating), but resonance dominates for REGIOCHEMISTRY (ortho/para directing). Net result: slower than benzene, but substitution occurs at ortho/para positions.
Describe the mechanism of radical halogenation of alkanes.
Three stages:
INITIATION: Homolytic cleavage of X₂ by heat (Δ) or light (hν) → 2 X• radicals.
PROPAGATION (chain-carrying):
Step 1: X• + R–H → HX + R• (H abstraction). Rate-determining step.
Step 2: R• + X₂ → R–X + X• (halogen transfer). Regenerates X•.
TERMINATION (any two radicals combine):
R• + R• → R–R
R• + X• → R–X
X• + X• → X₂
Selectivity: Br• is MORE selective than Cl• (Br• preferentially abstracts 3° > 2° > 1° H). Hammond postulate: Br• abstraction is endothermic with late TS → more selective.
What is the selectivity of radical bromination vs chlorination?
CHLORINATION: Relatively NON-selective.
Relative reactivity: 3°:2°:1° ≈ 5:4:1
Gives mixtures of products. Useful mainly for methane → CH₃Cl.
BROMINATION: Highly SELECTIVE.
Relative reactivity: 3°:2°:1° ≈ 1600:80:1
Almost exclusively reacts at the most substituted position (most stable radical).
Reason (Hammond Postulate): H-abstraction by Br• is endothermic → late, product-like transition state → TS stability reflects radical stability → high selectivity.
What does Br₂ / FeBr₃ do?
Electrophilic aromatic substitution (EAS): Bromination of benzene ring. Adds –Br to aromatic ring. Requires Lewis acid catalyst (FeBr₃ or AlBr₃) for unactivated rings. Activated rings (phenol, aniline) react with Br₂/H₂O without catalyst.
What does Br₂ / CCl₄ (or CH₂Cl₂) do?
Halogenation of alkenes: ANTI addition of Br₂ across C=C double bond via bromonium ion intermediate → vicinal dibromide (1,2-dibromide). Also a qualitative test for unsaturation — orange Br₂ solution decolorizes.
What does HBr alone vs HBr/ROOR do?
HBr alone: Markovnikov addition to alkene. H adds to less substituted C, Br to more substituted C. Mechanism: electrophilic addition via carbocation.
HBr/ROOR (peroxides): ANTI-Markovnikov addition. Br adds to less substituted C, H to more substituted C. Mechanism: radical chain. Note: This only works with HBr — HCl and HI do not undergo radical addition (thermodynamic reasons).
What does BH₃·THF followed by H₂O₂/NaOH do?
Hydroboration-oxidation: Anti-Markovnikov, SYN addition of H₂O across alkene.
Step 1: BH₃ adds to less substituted carbon (concerted, syn addition — B and H add to same face).
Step 2: H₂O₂/NaOH oxidizes C–B bond → C–OH with retention of configuration.
Overall: Produces the anti-Markovnikov alcohol with syn stereochemistry. Complementary to acid-catalyzed hydration (which gives Markovnikov product).
What does H₂/Pd(C) do? What about H₂/Lindlar’s catalyst?
H₂/Pd(C): Catalytic hydrogenation. Reduces C=C to C–C (syn addition). Reduces C≡C to C–C (fully). Also removes Bn (benzyl) protecting groups. Does NOT reduce C=O in most cases.
H₂/Lindlar’s catalyst (Pd/CaCO₃ + quinoline poison): PARTIAL reduction of alkyne → CIS alkene (stops at alkene stage because poisoned catalyst). Syn addition gives Z-alkene.
For trans alkene from alkyne: Use Na/NH₃(l) (dissolving metal reduction) → TRANS (E) alkene via radical anion mechanism.
What does OsO₄ (catalytic) / NMO do?
Syn-dihydroxylation of alkenes → cis-1,2-diol (vicinal diol). OsO₄ forms a cyclic osmate ester (syn addition), then NMO (co-oxidant) reoxidizes Os(VI) back to Os(VIII) making it catalytic. Hydrolysis gives the cis-diol. Stereochemistry: SYN — both OH groups on same face.
What does O₃ followed by Zn/AcOH (or DMS) do?
Ozonolysis with reductive workup: Cleaves C=C double bond completely → produces aldehydes and/or ketones.
- Disubstituted end → ketone
- Monosubstituted end → aldehyde
- Unsubstituted end (=CH₂) → formaldehyde (HCHO)
The Zn/AcOH or (CH₃)₂S prevents overoxidation of aldehydes to carboxylic acids. If oxidative workup (H₂O₂) used instead, aldehydes → carboxylic acids.
What does PCC (pyridinium chlorochromate) in CH₂Cl₂ do?
MILD oxidation:
1° alcohol → aldehyde (STOPS here — key feature)
2° alcohol → ketone
3° alcohol → no reaction
Must use anhydrous conditions (CH₂Cl₂ solvent). The absence of water prevents further oxidation of the aldehyde to carboxylic acid. Alternatives with same selectivity: Dess-Martin periodinane (DMP), Swern oxidation.
What does NaBH₄ in MeOH do?
Mild, selective REDUCTION:
- Aldehyde → 1° alcohol
- Ketone → 2° alcohol
Does NOT reduce: esters, carboxylic acids, amides, alkenes.
Safe to use in protic solvents (MeOH, EtOH). Delivers H⁻ (hydride) to electrophilic carbonyl carbon. Workup: aqueous acid.
What does LiAlH₄ in THF do?
Powerful, non-selective REDUCTION:
- Aldehyde → 1° alcohol
- Ketone → 2° alcohol
- Ester → 1° alcohol (+ alcohol from OR’ group)
- Carboxylic acid → 1° alcohol
- Amide → amine
- Epoxide → alcohol (at less hindered carbon)
MUST use anhydrous solvent (THF or ether). Reacts violently with water and protic solvents. Workup: careful aqueous quench (e.g., 1. H₂O, 2. NaOH, 3. H₂O — Fieser workup).
What does RMgBr (Grignard) + carbonyl, then H₃O⁺ workup give?
Nucleophilic addition of R group to carbonyl carbon → new C–C bond:
- RMgBr + HCHO → 1° alcohol (one more carbon)
- RMgBr + R’CHO → 2° alcohol
- RMgBr + R’₂CO → 3° alcohol
- RMgBr + ester → 3° alcohol (2 equivalents of RMgBr add)
- RMgBr + CO₂ → carboxylic acid (after H₃O⁺ workup)
- RMgBr + ethylene oxide → 1° alcohol with 2 extra carbons
REQUIRES absolutely anhydrous conditions.
What does SOCl₂ (thionyl chloride) do to alcohols and carboxylic acids?
With alcohol: R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑
Converts alcohol to alkyl chloride. Good leaving groups (SO₂ and HCl are gases — drive reaction forward). Retention or inversion depending on mechanism (SNi for some, SN2 with added base).
With carboxylic acid: RCOOH + SOCl₂ → RCOCl + SO₂ + HCl
Converts carboxylic acid to acid chloride (acyl chloride). Essential for making reactive acyl derivatives.
What does H₂SO₄ (concentrated) at high temperature do to alcohols?
DEHYDRATION of alcohols → alkenes (elimination).
Conditions: Concentrated H₂SO₄, high temperature (typically 180°C for 1° alcohols, lower for 2° and 3°).
Mechanism: E1 for 3° and 2° alcohols (via carbocation). E2 for 1° alcohols.
Regiochemistry: Zaitsev product (most substituted alkene) is major product.
Carbocation rearrangements possible (1,2-hydride or methyl shifts for 2° and 3°).
What does NaOH/H₂O (aqueous base) do to acid chlorides, anhydrides, and esters?
Nucleophilic acyl substitution — hydrolysis:
- Acid chloride + NaOH → carboxylate salt + Cl⁻ (fast, irreversible)
- Anhydride + NaOH → 2 carboxylate salts (fast)
- Ester + NaOH → carboxylate salt + alcohol (saponification, irreversible because carboxylate is stabilized)
Reactivity order: acid chloride > anhydride > ester > amide. The better the leaving group, the more reactive the acyl derivative.
What does Na/NH₃(l) — dissolving metal reduction — do to alkynes?
Reduces internal alkynes to TRANS (E) alkenes. Mechanism: Radical anion pathway — Na donates electrons to the alkyne → vinyl radical anion → protonation by NH₃ → vinyl radical → second electron from Na → vinyl anion → second protonation → trans-alkene. The trans selectivity arises because the more stable trans-vinyl anion is formed preferentially. Complementary to Lindlar (which gives cis).
What does mCPBA do to alkenes?
Epoxidation: Converts alkene to an epoxide (oxirane). Concerted mechanism — stereochemistry of the alkene is preserved (cis-alkene → cis-epoxide, trans-alkene → trans-epoxide). The oxygen is delivered from the peroxyacid to ONE face of the alkene (syn). Alternative epoxidation reagents: DMDO, Sharpless asymmetric epoxidation (with chiral Ti catalyst for enantioselective).
Rank the reactivity of carboxylic acid derivatives toward nucleophilic acyl substitution.
Most reactive → least reactive:
Acid chloride (RCOCl) > Acid anhydride ((RCO)₂O) > Ester (RCOOR’) > Amide (RCONR₂) > Carboxylate (RCOO⁻)
Reasoning: Reactivity depends on leaving group ability AND resonance donation to carbonyl:
- Cl⁻ is a good LG, minimal resonance donation → most reactive
- Carboxylate (RCOO⁻) is a moderate LG
- OR is a moderate LG, moderate resonance
- NR₂ is a poor LG, strong resonance donation → least reactive
- O⁻ is a terrible LG → carboxylates are unreactive
You can go DOWN the series (more reactive → less reactive) but NOT up without special activation.
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