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Part 2

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

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#1
Front (Question)

Nucleophilic Substitutions

Back (Answer)

Occurs in Alkanes. Nucleophiles are electron rich species that attracted to positively polarized atoms. Basicity: Stronger the base, stronger the nucleophile. Size/polarizability: dependson solvent. Protic = larger atoms. Aprotic = more basic atoms are better.

#2
Front (Question)

Best Leaving Groups

Back (Answer)

I > Br > Cl > F Weak bases make good leaving groups. Can accept electron pair and dissociate to form stable species.

#3
Front (Question)

Sn1 Reactions

Back (Answer)

Rate is dependent on ONE species. Rate determining step is the carbocation. Favored in POLAR protic solvents (water or acetone) rate = k[RX] Want stable carbocations: 3 > 2 > 1 > CH3 (favored with use of bulky nucleophiles) Produces Racemic products (lose optical activity) Leaving group = weak bases are best.

#4
Front (Question)

Sn2 Reactions

Back (Answer)

ONE step (no carbocation) rate = k[Nu][RX] (substrate and nucleophile) Usually attacks from the backside. Best Reactant: 1 > 2 > 3 Favored in polar APROTIC solvents. Optically active and inverted products. Need strong nucleophile and transition state.

#5
Front (Question)

Elimination Reactions

Back (Answer)

Used in synthesis of alkenes. Elimination reactions of either alcohols or alkyl halides. In these reactions, the carbon skeleton loses HX (X = halide) or a molecule of water, to form double bond. Two types E1 and E2

#6
Front (Question)

Unimolecular Elimination (E1)

Back (Answer)

TWO step process, proceeding through a carbocation intermediate. k = [RX] (substrate) Elimination of leaving group plus proton = double bond. • Leaving group departs, producing carbocation. • Proton is removed by a base. E1 is favored by same factors as SN1: highly polar solvents, Weak Nu, highly branched carbon chains (3), and good leaving groups. However, HIGH temps favor E1.

#7
Front (Question)

Bimolecular Elimination (E2)

Back (Answer)

ONE step process. k = [RX][Base] Strong base such as ethoxide ion (C2H5O-) removes a proton, while halide ion anti to proton leaves = double bond. Often two possible products, but more substituted preferred. Steric hindrance does NOT affect E2. Strong base favors E2 over SN2. SN2 favored over by E2 by weak Lewis bases. (Strong Nu)

#8
Front (Question)

Catalytic Hydrogenation

Back (Answer)

Reductive process of adding hydrogen to a double bond with aid of a metal catalyst. (platinum, palladium, and nickel) Product: Alkane with syn addition of H. Reaction takes place on metal surface, thus H atoms are added to same face (syn addition).

#9
Front (Question)

Addition of HX

Back (Answer)

An electrophilic addition that occurs in alkenes. Electron of double bond acts as Lewis base and reacts with electrophilic HX molecules. • Yields carbocation after double bond reacts with H+ • Halide ion combines with carbocation to give alkyl halide. Follows Markovnikov rule (add to the most substituted carbon) Product: Alkyl halide

#10
Front (Question)

Addition of X2

Back (Answer)

Addition of halogens to double bond. Rapid process. Nucleophile is the double bond, which attacks an X2 molecule, displacing X-. Forms intermediate cyclic halonium ion, which is then attacked by X- to make a dihalo compound. Anti-addition (attacks SN2). Product: dihalo alkane

#11
Front (Question)

Addition of H20

Back (Answer)

Water can be added to alkenes under acidic conditions. Protonated according to markovnikov. Performed at LOW temperatures. Product = alcohol

#12
Front (Question)

Free radical Additions

Back (Answer)

Alternate mechanism for addition of HX to double bond. Occurs when PEROXIDES, O2, or other impurities are present. Disobeys Markovnikov rule. Product: terminal alkyl halides.

#13
Front (Question)

Hydroboration

Back (Answer)

Diborane (B2H6) adds readily to double bonds. • Boron atoms is the Lewis Acd and attaches to less sterically hindered C atom. • Oxidation-hydrolysis with PEROXIDE/Aq. base produces alcohol with ANTI-markovnikov, syn orientation. Product: Alcohol with anti-Markovnikov.

#14
Front (Question)

Potassium Permangate (KMnO4)

Back (Answer)

Involved in Oxidation. • Cold, dilute KMnO4 produces 1, 2 diols (vicinal) with syn orientation. • Hot, basic KMnO4 plus acid produces: • Nonterminal alkenes = 2M COOH • Terminal - 1M COOH and 1 CO2 (if disubsituted it makes a ketone).

#15
Front (Question)

Ozonolysis

Back (Answer)

Involved in Oxidation. Treatment of alkenes with OZONE, followed by reduction with Zn/H20 results in CLEAVAGE of double bond. Product: Aldehyde. If reduced with NaBH4 instead of Zn/H2) will result in Alcohols.

#16
Front (Question)

Peroxycarboxylic Acids

Back (Answer)

Alkenes can be oxidized with peroxycarboxylic acids. e.g. Peroxyacetic acid (CH3CO3H) OR m-chloroperoxybenzoic acid (mcpba). Products = Epoxides or Oxiranes.

#17
Front (Question)

Polymerization

Back (Answer)

Creation of long, MW chains composed of repeating subunits in alkenes. Occurs through radical mechanism, requires HIGH TEMP and PRESSURE.

#18
Front (Question)

Synthesis of Alkynes

Back (Answer)

• Elimination of HX from geminal and vicinal dihalides with HEAT and BASE. • Add existing Triple bond to Nucleophile by removing acidic proton with STRONG BASE.

#19
Front (Question)

Reduction of Alkynes

Back (Answer)

Can be hydrogenated with a catalyst. Two Ways: 1. Lindlar’s catalyst (H2, Pd/BaSO4, Quinoline) Product: Alkane (cis) • Na, NH3 (liquid) -33 degrees C. Product: Trans alkene.

#20
Front (Question)

Electrophilic addition of alkynes.

Back (Answer)

Electrophilic addition to alkynes occurs in same manner as it does alkenes. (X2 added) Occurs according to Markovnikov. Can be stopped at alkene or proceed further. Product: Dihalo alkene or alkane

#21
Front (Question)

Free Radical Addition of Alkynes

Back (Answer)

Radicals add to triple bonds as with double bonds with ANTI-markovnikov (X-). Product: trans isomer. Because intermediate vinyl radical can isomerize to its more stable form.

#22
Front (Question)

Hydroboration Addition of Alkynes

Back (Answer)

Addition to triple bonds is same as double bonds. Addition is syn and BORON adds first. Boron can be replaced with proton from acetic acid = cis alkene. Product: syn alkene.

#23
Front (Question)

Oxidation Addition

Back (Answer)

Alkynes can be oxidatively cleaved with either basic KMnO4 followed by acidification or Ozone. KMnO4 Product: 2 M COOH Ozone Product: 2 M COOH

#24
Front (Question)

Huckel’s Rule

Back (Answer)

4n + 2 pi electrons Important indicator of aromaticity. n = any nonnegative integer; thus can be 2, 6, 10, 14, 18, etc.