MCAT Orgo 2

Overview of SN2

-1 step -polar aprotic solvents -1° > 2° > 3° -rate = k[Nu][RX] -optically active/inverted products

Overview of SN1

-2 steps -polar protic solvents -3° > 2° > 1° > methyl -rate = k[RX] -racemic products -favored w/ bulky Nu

More stable intermediate → more likely reaction

.CR₃ > .CR₂H > .CRH₂ > .CH₃ 3° > 2° > 1° > methyl

What are the products called when cold, dilute KMnO₄ adds -OH to each side of the double bond?

1,2 diols (vicinal diols), or glycols (syn orientation)

2 steps of SN1 reaction

1. Dissociation of molecule into a carbocation & a good leaving group 2. Combination of the carbocation with a nucleophile

reduction of alkynes to produce alkanes

1. can use lindlar's catalyst (produces cis isomer) 2. can use sodium in liquid ammonia below -33°C (produces trans isomer)

p-orbital

2 lobes symmetrically about nucleus, contains node in the nucleus (probability of finding electron is 0). dumbell, 3 orientations: along the x-, y-, and z-axis

how much s-character does sp³ have

25% s-character (and 75% p-character)

Termination (Step 3) of free radical halogenation

2X. → X₂ X. + R. → RX 2R. → R₂

d-orbital

4 symmetrical lobes & 2 nodes. 4 are clover shaped, 5th looks like a donut wrapped around the center of a p-orbital.

what is stronger - a pi bond or a sigma bond

A sigma bond is stronger in isolation, but a pi bond is always with a sigma bond, so double and triple bonds are stronger due to the additive strength

Rate of SN2 reaction

Affected by concentrations of both the substrate (which has the LG) and the nucleophile, because both these species react in the single step of an SN2 reaction. (2nd-order kinetics)

1 way to control E2 vs. SN2

Big bulky bases (Nu) - easy E2, hard SN2. Too big for backside attack. Easier to pluck of β-hydrogen.

pyrolysis of alkanes

C-C bonds are cleaved, producing smaller-chain alkyl radicals, which recombine to form a variety of alkanes: CH₃CH₂CH₃ -(heat)→ CH₃. + .CH₂CH₃ 2CH₃. → CH₃CH₃ 2 .CH₂CH₃ → CH₃CH₂CH₂CH₃

What should you do when a reaction involves heat?

Consider possibility of elimination reactions & free radicals

Controlling whether SN1 or E1 occurs

Control conditions: polarity, temp. SN2 depends more on substrate & base properties. E1 is less likely.

equation for combustion

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O + heat

Rate of SN1 reaction

Depends on the concentration of the original substrate molecule (1st-order reaction). Can be increased by anything that accelerates the formation of the carbocation.

Hydroboration

Diborane (B₂H₆ aka borane, BH₃) adds readily to double bonds. Boron attaches to less sterically hindered carbon, hydride transferred to adjacent carbon; oxidation-hydrolysis w/ peroxide & aqueous base transfers water to the bond with boron, produces alcohol with anti-mark, syn orientation.

Strength of nucleophile & electrophile in SN1 reactions

Do NOT require strong nucleophile. Need good LG. Carbocation is a strong electrophile.

Addition of HX

Double bond acts as Nu. H⁺ from HX yields carbocation intermediate (in asymmetrical alkenes, the proton adds to least substituted carbon atom (most protons) to make the adjacent carbon the most stable carbocation); the X⁻ from HX then combines with the carbocation to yield an alkyl halide.

Trick to determine if a reagent is an oxidizing agent

If it has a lot of oxygen in it

backside attack

In SN2 reactions, the nucleophile actively displaces that LG in an in-line attack

leaving group order

I⁻ > Br⁻ > Cl⁻ > F⁻

oxidizing agent - potassium permanganate

KMnO₄ cold, dilute: -OH added to each side of double bond hot, acidic: nonterminal alkenes cleaved to form 2 molar eqs of carboxylic acid; terminal alkenes cleaved to form carboxylic acid & CO₂

SN2 stereochemistry

Leads to a relative inversion of stereochemistry owing to backside attack (imagine umbrella turning out the wrong way). Be careful though, because the absolute configuration may remain the same if the LG and the nucleophile do not maintain the same priority.

SN1 stereochemistry

Leads to loss of stereochemistry; depending on where the nucleophile attacks, if the end product is chiral there can be 2 different products (R or S)

Requirements for SN2 reaction

Nucleophile must be strong, Substrate cannot be sterically hindered (1° substrates more likely to undergo SN2, which is the opposite trend of SN1)

What happens if you reduce the aldehyde or ketone products from ozonolysis?

Obtain alcohols

Second step of E1

Proton on adjacent (β) carbon is removed by a weak base; Double bond formed with the electrons from broken C-H bond

order of nucleophilic strength if attacking atom is oxygen (how much will it attract a positively charged proton?)

RO⁻ > HO⁻ > RCO₂⁻ > ROH > H₂O

Addition of X₂

Rapid process. Double bond acts as Nu, attacks 1/2 of X₂, displacing X⁻ from other side. Results in cyclic halonium ion. X⁻ attacks ion on opposite face, where the already attached X isn't blocking (anti-addition), results in dihalo compound.

What is the hallmark of the SN1 reaction? What is the hallmark of the SN2 reaction?

SN1 - carbocation SN2 - transition state

What factors favor E1

Same as SN1 - polar protic solvents; ability to form stable carbocation; highly branched carbon chains; good LGs; no good Nu; higher temps

First step of E1

Same as SN2 - Departure of LG; Formation of carbocation intermediate

Ozonolysis

Strong oxidative process; cleaves double bond; oxidizes the carbon to an aldehyde (or ketone if substrate is disubstituted) under reducing conditions. If it is under oxidizing conditions, products will be same as hot, acidic KMnO₄.

Peroxycarboxylic acids

Strong oxidizing agents; (CH₃CO₃H & mcpba commonly used) products = epoxides

Another way to control E2 vs SN2

Stronger base favors E2 over SN2. (strong base pulls off β-hydrogen before reaching α-carbon) Weak Lewis base (strong Nu) favors SN2 over E2.

What happens in elimination reactions?

The carbon backbone kicks off ("eliminates") a hydrogen and a halide (dehydrohalogenation) or a molecule of water (dehydration), and forms a double bond.

Rate limiting step in SN1 reaction

The dissociation of the molecule to form the carbocation. (i.e. the formation of the carbocation)

Why do E2 reactions often have 2 possible products?

The double bond can form on either side of the departing halide. More substituted double bond = larger % of products

When is pyrolysis most commonly used?

To reduce the average molecular weight of heavy oils and increase the production of the more desirable volatile compounds.

Nature of leading group affects rate of SN1 reaction

Weak bases dissociate more easily from the alkyl chain and thus make better LGs , increasing the rate of carbocation formation.

Propagation (Step 2) of free radical halogenation

X. + RH → HX + R. R. + X₂ → RX + X. free radical + alkane makes HX and an alkyl radical (R.); Alkyl radical reacts with X₂ to make alkyl halide (RX) which also generates another free radical (X.) This step can occur multiple times before final step occurs.

Adding HX to alkenes using free-radical intermediate

X. adds first to double bond, producing the most stable free radical; halogen ends up on least substituted carbon; useful for HBr, but not HCl or HI

Initiation (Step 1) of free radical halogenation

X₂ → 2X. where X. is a free radical (neutral species with an unpaired electron, very reactive) and the halogens are cleaved using either heat or UV light

disproportionation

a radical transfers a hydrogen atom to another radical, producing an alkane and an alkene: .CH₃ + .CH₂CH₃ → CH₄ + H₂C=CH₂

triple bond

a sigma bond and two pi bonds

In a free radical halogenation reaction that produces 2 different products, which halogen is involved? a. Cl₂ b. Br₂ c. I₂ d. F₂

a. Cl₂ (free-radical halogenation reactions are only practical for bromine & chlorine. bromine radicals react slow compared to chlorine, and so only form most stable alkyl radical - only one bromination product. chlorine radicals react fast and product multiple products)

which hybridization does the Be atom in BeH₂ assume? a. sp b. sp² c. sp³ d. none of the above

a. sp

X

abbreviation used to represent any halogen

In orgo, think of reduction as...

adding bonds to hydrogen

In orgo, think of oxidation as...

adding bonds to oxygen

addition reaction

adding compounds to double bonds while leaving carbon skeleton intact (electrophiles can easily attack the pi bond because they want the electron pair and pi bonds are weak compared to sigma bonds)

syn addition

addition to same side of double bond (so that when looking down the double bond, they are eclipsed)

alkenes

aka olefins. C=C stereochemistry described by cis, trans, E, Z. common names used: ethylene, propylene, isobutylene.

Markovnikov's rule

alkyl substituents stabilize carbocations; formation of most stable carbocation - addition of H to carbon atom on either side of the double bond that has more protons (H) attached creates a positive charge on the other carbon, which becomes the carbocation intermediate

hydrogens attached to carbon atoms

always carry the same name as their parent carbon atom (e.g. an H on 1° carbon is a 1° H)

free radical addition to alkynes

anti-mark orientation. reaction product is trans isomer.

Is DMF a protic or aprotic solvent?

aprotic

In a double-bonded carbon atom: a. hybridization occurs between the s-orbital and one p-orbital b. hybridization occurs between the s-orbital and two p-orbitals c. hybridization occurs between the s-orbital and three p-orbitals d. no hybridization occurs between the s- and p-orbitals

b. hybridization occurs between the s-orbital and two p-orbitals

SN1 reactions show 1st order kinetics because: a. the rate-limiting step is the 1st step to occur in the reaction b. the rate-limiting step involves only 1 molecule c. there is only 1 rate-limiting step d. the reaction involves only 1 molecule

b. the rate-limiting step involves only 1 molecule

Molecular orbitals can contain a maximum of: a. one electron b. two electrons c. four electrons d. 2n² electrons, where n is the principal quantum number of the combining atomic orbitals

b. two electrons (like atomic orbitals, MOs each can contain a max of 2 electrons with opposite spins)

Pi bonds are formed by which of the following orbitals? a. two s-orbitals b. two p-orbitals c. one s- and one p-orbital d. all of the above

b. two p-orbitals (pi bonds are formed by the parallel overlap of unhybridized p-orbitals)

E2

bimolecular elimination reaction; 1 step; rate depends on substrate and base (Nu); often has 2 possible products

SN2 reactions

bimolecular nucleophilic substitution; involves a strong nucleophile pushing its way into a compound while simultaneously displacing the LG, in one concerted step.

quaternary (4°) carbon

bonded to four carbons

primary (1°) carbon

bonded to only one other carbon

tertiary (3°) carbon

bonded to three carbons

secondary (2°) carbon

bonded to two carbons

pi bond

bonding molecular orbital formed when 2 p-orbitals line up parallel and their electron clouds overlap

branched vs. straight chain

branched = less surface area for interactions = slightly lower BP than straight chain isomers

Which would be the best solvent for an SN2 reaction? a. H₂O b. CH₃CH₂OH c. CH₃SOCH₃ d. CH₃CH₂CH₂CH₂CH₂CH₃

c. CH₃SOCH₃ (polar aprotic solvent)

Which of the following conditions, listed as LG & Nu, respectively, would most favor nucleophilic substitution? a. I⁻, Cl⁻ b. EtO⁻, tosylate c. tosylate, CN⁻ d. OH⁻, H₂O

c. tosylate, CN⁻ (Reaction is SN1. Tosylate is an excellent LG, and cyanide is an excellent Nu. Do not want LG to also be a good Nu, so iodide is wrong. Tosylate is a weak Nu. Ethoxide is a poor LG. OH⁻ is a poor LG, water is a poor Nu.)

azimuthal quantum number l

can equal 0, 1, 2, 3, etc. corresponds to type of orbital (s, p, d, f, etc.) orbital shape.

quantum number n

can equal 1, 2, 3, etc. corresponds to energy levels. essentially measure of size.

polymerization

creation of polymers; usually occurs thru radical mechanism; e.g. polyethylene from ethylene (ethene), requires high temps and pressures

2 atomic orbitals combine to form: I. a bonding molecular orbital II. an antibonding molecular orbital III. new atomic orbitals a. I only b. I, II, & III c. III only d. I & II only

d. I & II only (2 atomic orbitals combine to form one low-energy bonding MO and one high-energy anti bonding MO - new orbitals do not form)

The hybridization of the carbon atom and the nitrogen atom in the ion CN- are respectively: a. sp³ & sp³ b. sp³ & sp c. sp & sp³ d. sp & sp

d. sp & sp

Addition of H₂O

double bond protonated (mark's rule) → most stable carbocation; carbocation reacts with water, yielding protonated alcohol, which loses proton to become alcohol; performed at low temps

nucleophiles

electron-rich, attracted to positively charged or positively polarized atoms (which are called electrophiles)

1 way to make triple bond

elimination of germinal (same C) or vicinal (adjacent C) dihalide; requires high temp & strong base

What does the most common method of synthesizing alkenes involve?

elimination reactions of either alcohols or alkyl halides

IUPAC name for acetylene

ethyne

terminal alkynes

fairly acidic (for a carbon atom); can stabilize neg. charge fairly well.

molecular orbital

formed by 2 atomic orbitals combining. obtained by adding or subtracting the wave functions.

Reduced species

gets more hydrogen

Do weak bases make good or bad LGs?

good

Leaving group

good leaving group makes nucleophilic substitution easier. good leaving groups are weak bases, or stable ions or neutral species - easily can accommodate electron pair.

shorter bond length =

greater strength of the bond

antibonding orbital

higher energy (less stable) - produced if the signs of the wave function are different

Solvent effects affect rate in SN1 reaction

highly polar solvents better at surrounding & isolating ions than less polar solvents. polar protic solvents (e.g. water) work best because solvation stabilizes the intermediate state.

Structural factors affecting rate in SN1 reaction

highly substituted alkyl halides allow for distribution of the positive charge over a greater number of carbon atoms, forming the most stable carbocations.

First thing to do when you see a substitution reaction

identify the nucleophile and the electrophile

nucleophile strength in protic solvent

larger atom = better nucleophile CN⁻ > I⁻ > RO⁻ > HO⁻ > Br⁻ > Cl⁻ > F⁻ > H₂O

Oxidized species

loses hydrogen, thus gaining double bonds and often, oxygen

bonding orbital

lower energy (more stable) - produced if the signs of the wave functions are the same

hybridization

mixing different types of orbitals. way of making all the bonds to a central atom equivalent to each other.

symmetry of molecule and MP

more symmetrical = higher MP

aprotic

no protons

pyrolysis (aka cracking)

occurs when a molecule is broken down by heat.

halogenation

one or more hydrogen atoms are replaced w/ a halogen atom via a free-radical substitution mechanism.

double bond

one pi bond on top of an existing single bond

Lindlar's catalyst

palladium on barium sulfate (BaSO₄) with quinoline (a heterocyclic aromatic poison that stops the reaction at the alkene stage)

Typical catalysts

platinum (Pt); palladium (Pd); Nickel (Ni)

free-radical chlorination reactions

rapid process, so is more likely to replace 1° hydrogens if they are the most abundant type present, despite their relative instability

combustion

reaction of alkanes with molecular oxygen to form carbon dioxide, water, and heat

stereospecific reaction

reaction where only one stereoisomer is formed

catalytic hydrogenation

reducing an alkene by adding molecular hydrogen to the double bond with the aid of a metal catalyst. results in syn addition of the 2 hydrogens.

sigma (σ) bond

resulting bond when a molecular orbital is formed by head-to-head or tail-to-tail overlap. all single bonds are sigma bonds.

order of nucleophilic strength in aprotic solvent

same as base strength, opposite of in protic solvent F⁻ > Cl⁻ > Br⁻ > I⁻

electrophilic addition to alkynes

same manner as alkenes. products form according to mark's rule. can be stopped at intermediate alkene stage or can go all the way the alkane stage w/ 2 equivalents of reactants.

sp²

seen in alkenes 3 sp² orbitals are oriented 120° apart (3 things)

sp

seen with triple bonds 2 sp orbitals oriented 180° apart (2 things)

s-orbital

spherical & symmetrical, centered on nucleus

Unequal electron distribution in alkenes

sp³ carbons donate electrons to sp² carbons (dipole moment points from electropositive alkyl groups toward the electronegative alkene)

Is CH₃O⁻ (methoxide) a strong or weak base?

strong

Only step of E2

strong base removes proton; a halide ion anti to the proton leaves; double bond forms

More practical method of making triple bond

terminal triple bond converted to Nu by removing its acidic proton with a strong base, produces acetylide ion. Then ion performs nucleophilic displacements on primary alkyl halides.

sp³

tetrahedral - ex: CH₄ (4 things)

rate-determining step of SN1 reaction

the dissociation of the substrate (original molecule) to form a stable, positively charged ion called a carbocation.

What can you expect from free radical reactions if peroxides or UV light are present?

they won't follow mark's rule

MP of trans vs. cis

trans = higher MP (more symmetric)

BP of trans vs. cis

trans = lower BP (less polar)

transition state in SN2 reaction

trigonal bipyramidal (sp²)

E1

unimolecular elimination reaction; 2 steps; rate depends on substrate only

SN1 reaction

unimolecular nucleophilic substitution; rate of reaction depends only on substrate.

How can the carbon in the substrate molecule be coaxed into becoming the carbocation?

using polar protic solvents w/ lone electron pairs. more highly substituted cation = more stable.

Is I⁻ a a strong or weak Nu?

very strong (but poor base = good for SN2)

Is ⁻CN a strong or weak Nu?

very strong (but poor base = good for SN2)

MP & BP of alkenes

↑ w/ molecular weight (size)

How do physical properties change as molecular weight increases?

↑BP ↑MP ↑density

overview: physical properties of alkanes

↑chain length: ↑BP, ↑MP, ↑density ↑branching: ↓BP, ↓density

How can you determine the best leaving group if you are given the pKa of the LG's conjugate acid?

↓ pKa of conjugate acid = better LG (A conjugate acid is the LG w/ an added proton (H))