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Organic 2 > Exam 1 Material > Flashcards

Exam 1 Material Flashcards

1
Q

SN2 reactions

A
  • one step
  • leaving group leaves as nucleophile attacks (opposite side of LG).
  • If LG is on a chiral carbon, you will see inversion of stereochemistry (R vs. S)
  • Rate= [substrate][nucleophile]
  • VERY sensitive to nature of starting materials (LG).
  • methyl>primary>secondary>tertiary
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2
Q

R vs. S

A 
R= clockwise
S= counterclockwise
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3
Q

The elimination reaction

A

The beta proton is removed by the base, the electrons from that bond form a double bond, and the leaving group gets kicked off.
*** concerned with strong bases not strong nucleophiles!

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4
Q

E2 Reaction

A
  • second order (one step)
  • Rate = [substrate][base]
  • can proceed quickly with tertiary substrates because the steric hindrance is not as important since a proton is being targeted.
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5
Q

Cis vs. trans

A

Trans is always more stable because there is les steric hindrance.

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6
Q

E vs. Z

A
  • E means that the priority groups are on different sides of the double bond.
  • Z means that they are on the same side.
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7
Q

Stability of alkenes

A
  • Trans is more stable than cis
  • E is more stable than Z
  • substitution= how many groups are attached to the double bond.
    • substitution stability= tetra> tri> di> mono
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8
Q

Regioselectivity of E2 reactions

A
  • when small bases (NaOEt, NaOMe) are used, the more substituted product ZAITSEV will be favored.
  • when bulky bases (t-Bu- Ok, diisopropyl amine, triethyl amine) are used, the less substituted HOFMANN product will be favored.
  • this is because hydrogens on terminal carbons are easier for a sterically hindered base to attack.
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9
Q

Stereoselectivity of E2 Reactions

A

Need to consider that rotation can occur at double bonds, therefore the trans product will be the major product over the cis.

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10
Q

Stereospecificity of E2 reaction

A
  • In Newman projections and chair conformations (carbon rings), the beta proton must be anti- periplanar to the leaving group ( in the same plane)
  • ** if the leaving group is on a wedge then the beta proton must be on a dash and vice versa.
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11
Q

SN1 Reaction

A
  • stepwise process
  • Rate= k[substrate]
  • the leaving group must first leave, resulting in a carbocation. The nucleophile then attacks.
  • if a chirality center is involved, can result in a race ic mix because the nucleophile can attack either side of the molecule.
  • tertiary action> secondary> primary
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12
Q

Carbocation Rearrangement

A
  • common for SN1
  • goal is to make a more stable carbocation ( tertiary over secondary)
    1) hydride shift- movement of a hydrogen and the electrons from its bond from one carbon to another.
    2) methyl shift- migration of a methyl group and its bonded electrons from one carbon to another.
    3) alkyl shift- migration of an alkyl group ( something other than a methyl) often happens to alleviate ring strain.
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13
Q

Nucleophile for SN1

A

Rate is not dependent on the nucleophile, therefore it DOES NOT need to be strong.
Ex) H2O, MeOH, EtOH.
*** if water attacks, always assume that another water molecule is in solution and takes the extra H to form an alcohol.

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14
Q

E1 Reaction

A
  • Stepwise Process
  • Leaving group leaves and a carbocation is formed. THEN the base attacks the beta proton and a double bond is formed.
  • Rate=k[substrate]
  • substrate trend is the same as with SN1… tertiary>secondary> primary
  • ** because most nucleophiles can also be good bases, a mixture of SN1 and E1 can be obtained.
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15
Q

Regioselectivity for E1 reactions

A
  • Due to carbocation formation, the more substituted alkene (zaitsev) is generally favored.
  • **If conc. H2SO4 and heat are reactants, you automatically know that the elimination product is formed!
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16
Q

Solvent effects for ALL reactions

A

SN2 + E2: polar aprotic ( no H connected to electronegative atom)
** ex) DMSO, DMF, THF
- this is because these mechanisms use strong nucleophiles and strong bases, so you don’t want them to become surrounded by a solvent shell if they need to attack an electrophilic center.
SN1 + E1: polar protic ( has an H attached to an electronegative atom)
*** ex) H2O, MeOH, EtOH, acetic acid, NH3
- this is because when the leaving group leaves, you want it to be surrounded by a solvent shell so that it doesnt re-attach to substrate.

17
Q

Alkene

A
  • incorporates a Carbon- carbon double bond
  • 3 separate bonds ( 2 sigma and one pi)
  • each carbon is sp2 hybridized
18
Q

Alkyne

A
  • Has a carbon- carbon triple bond
  • 3 separate bonds: one sigma and two pi
  • each carbon is sp hybridized
19
Q

Geminal dihalides

A

When there are two halide atoms on the same carbon

20
Q

Vicinal halides

A

When There are two halide atoms in a molecule but they are on different carbons.

21
Q

Addition reactions

A

The addition of two groups/ atoms across a pi bond.

22
Q

Hydrohalogenation

A
  • the addition of a hydrogen and a halogen across ( Br, CL, etc.) a pi bond
  • the halogen will attach to the more substituted side of the double bond (markovnikov product)
  • ** if a chirality center is produced, then we will get a racemic mixture of products.
  • reactants: an alkene or alkyne and H-Cl, H-Br, etc.
  • with an alkyne, if the reactant is present in excess, then it will react twice to form a single bond with two halogen atoms attached to the markovnikov position.
23
Q

Acid- catalyzed hydration

A
  • the addition of H and OH across a pi bond.
  • reactants: an alkene or alkyne and H3O+
  • **must have an additional water molecule take the extra H off of water to form OH
24
Q

Acid- catalyzed hydration of alkyne

A
  • Reactants: H2SO4, H2O… HgSO4 increases the rate of the reaction (catalyst)
  • the transition state of this reaction is an enol and then tautomerization occurs to form a ketone.
25
Q

Acid- catalyzed tautomerization

A

Start with an enol and form a ketone. The double bond in the enol attacks a hydrogen on H3O+ , this hydrogen adds to the end and there is no longer a carbon- carbon double bond. Next, a lone pair from the O-H forms a double bond between the C and the O and the electrons shared between the O-H move to the O, removing the H and forming a ketone.

26
Q

Base- catalyzed tautomerization

A

Enol to ketone using OH-… The O in OH- attacks the H on the OH group in the enol, those electrons are left on the O as the H is taken. Next, these electrons form a double bond between the O and the C, this causes the c-c double bond to break forming an anion on the terminal carbon. An H attached here and forms a ketone.

27
Q

Hydroboration Oxidation

A
  • anti-markovnikov addition of H2O… places OH group at less substituted carbon
  • reactants: 1)BH3 and THF AND 2) H2O2 and NaOH
  • OH and H add syn (same side).
28
Q

Hydroboration- oxidation of alkyne

A
  • enol- aldehyde
  • use 9-BBN instead of BH3, THF
  • this is because you want to prevent 2 additions on BH3 across an alkyne, so you use a sterically hindered borane reagent.
29
Q

Catalytic hydrogenation

A
  • addition of molecular hydrogen (H2) across a double bond
  • use metal catalyst (Pd, Ni, Pt)
  • **reduce alkene to alkane
  • both hydrogens add to the same face of pi bond (syn)
  • ** if you have an alkyne and you see Pd, Ni, Pt… you reduce alkyne to alkane
30
Q

Reduction of alkynes

A

1) catalytic hydrogenation: reduces all the way to alkane (no stereochemistry since you are adding two sets of H’s.
2) poisoned catalyst: H2 and Lindlar catalyst…forms a cis alkene (like a dog playing dead)
3) dissolving metal reduction: Na and NH3(l) forms a trans alkene.

31
Q

Halogenation

A
  • Addition of X2 across pi bond
  • happens with Cl2 and Br2 with CCl4 as solvent
  • results in anti- addition (opposite faces of double bond… one on wedge and one on dash)
  • Br2 and Cl2 are not polar but they are polarizable when placed by a nucleophile(double bond) making it an electrophile for the double bond to attack.
  • transition state is bromonium cation (VERY UNSTABLE )
32
Q

Halogenation of alkynes

A
  • Can add one or two rounds of X2
  • if one round, result is an alkene ( no stereochemistry because only three things attached)
  • if two rounds (xs),result is an alkane
33
Q

Halohydrin formation

A

-Adding X and OH across a double bond
Reactants: X2 and H2O
- OH and X2 add anti ( opposite face of double bond)
- transition state is bromonium cation
- Br anion that is left gets surrounded by solvent shell so that it cannot continue to react
- after water adds to the molecule, another water molecule deprotonates it so that it forms OH
- result is a racemic mix.

Organic 2 (4 decks)

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