2310 Exam 3

Question 1

Cation Stabilities

Answer 1

Benzylic, allylic, tertiary, secondary, primary, methyl, vinylic

Question 2

Factors that influence substitution RXN’S

Answer 2

1. Structure of alkyl halide
2. Leaving Group
3. Nucleophile
4. Solvent

Question 3

Relative Basicities of Halide Ions

Answer 3

I < Br < Cl< F

I= weakest/most stable base; best LG

Question 4

Relative Reactivities of alkyl halides in Sn2 RXN

Answer 4

RI >RBr > RCl > RF

RI=most reactive, better LG.) (RF is too unreactive to do Sn2

Question 5

Good Nucleophiles

Answer 5

OH, RO, HS, RS, NH2, RNH, C=-N, RC+=-C, RCOO

Question 6

Basicity

Answer 6

Relative base strengths and relative nucleophilicites
NH2 > OH> F
Stronger base, better Nucleophile
NH2 is strongest base

Question 7

Predict
F
Cl
Br
I

Answer 7

More basic, better Nu
in aprotic solvent (sans H-bonds), F best, I worst
in protic (H-bonds) solvent, F worst, I best.

Question 8

Best to worst Nucleophile

Answer 8

RS- > I- > -C=-N, CH3O- > Br- > NH3 > Cl- > F- > CH3OH

Question 9

Aprotic Solvents

Answer 9

No H-bonds

DMSO, DMV, THF, Et2O

Question 10

Protic Solvents

Answer 10

H bonds

water, formic acid, methanl, ethanol, tert-butyl alcohol, acetic acid

Question 11

Sn2

Answer 11

one step mechanism
biomolecular rate determining step
rate is controlled by steric hinderance
product is always inverted
Leaving group: I>Br>Cl>F
better Nu= Faster rate

Question 12

Sn1

Answer 12

Two step mechanism with carbocation intermediate
unimolecular rate determining step
rate controlled by stability of carbocation
products are mixture of retained and inverted
leaving group: I > Br > Cl> F
Strength of nucleophile doesn’t affect rate of reaction

Question 13

Which mechanism does Primary Benzylic, or primary allylic use?

Answer 13

either

Question 14

Which mechanism does a secondary benzylic or allylic use?

Answer 14

either

Question 15

Which mechanism does a tertiary benzylic or allylic use?

Answer 15

Only Sn1

Question 16

When is there no rxn?

Answer 16

When X is attached at a double bond (think benzene)

Question 17

Zaitsev’s Rule

Answer 17

More stable and more substituted alkene will form

Question 18

Anti-Zaitsev Elimination

Answer 18

Bulky Base

Poor Leaving group (F) acts as carbanion (inverse stability of carbocations)

Question 19

What is requisite for E2

Answer 19

that an H be antiperiplanar

Question 20

E1 v E2

Answer 20

E2: Strong, Bulky bases
E1: weak, small bases

Question 21

Relative Reactivities of Alkyl Halides

Answer 21

Sn2 1>2>3
E2 3>2>1
Sn1 3>2>1
E1 3>2>1

Question 22

Primary Alkyl Halide

Answer 22

Primarily substitution, unless steric hindrance in R-X or Nu, in which case elimination is favored (always Sn2 or E2)

Question 23

Secondary Alkyl Halides

Answer 23

Both Sn2 and E2, stronger, bulkier the base and higher T, greater chance of E2. could be either E1 or Sn1

Question 24

Tertialy Alkyl Halide

Answer 24

Only E2, both E1 or Sn1

Question 25

Stereochemistry of Mechanisms

Answer 25

Sn1: both inverted and retained E1 Both E/Z formed, more with buliest groups on opposite sides of double bond) most stable alkene Sn2 Only inverted forms E2 Both E/Z form, more with bulkiest gropus on opposite sides of double bond. configuration depends on configuration of reactant (ANTIPERIPLANAR)

Question 26

Which is most stable resonance contributer?

Answer 26

1 All atoms have filled e- shells 2. Negative charges will be held by more electronegative atoms 3. more contributers= more stable.

Question 27

Factors of Nucleophilicty

Answer 27

Charge Electronegativity Nucleophilicity increases as EN decreases (C>N>O>F) however, F is the most electronegative and O is more EN than C

Question 28

E2 is....

Answer 28

Regioselective Stereoselective Stereospecific