CHEM 211

Guidelines for “Arrow Pushing”

"Arrow pushing" is a scheme for rationalizing and predicting the products of organic reactions. Real molecules may not react in simple stages as we imply with arrows, but the method is founded on fundamental principles of chemistry and is justified by its success. Since electrons do not vanish during the course of a chemical reaction, arrow pushing serves as a bookkeeping device. When properly carried out, it not only distinguishes plausible reactions from impossible ones, but it provides a means of creating new chemistry. In the past, the student of organic chemistry was required to learn an enormous number of seemingly unrelated name reactions. Although there is still a substantial amount of material to remember, we concern ourselves not so much with the name of a reaction as with the class of mechanism to which it belongs. Thus, arrow pushing provides a means of systematizing organic reactions.

1. An arrow represents movement of an electron pair. A chemical bond is an electron pair. The movement of atoms is not represented by arrows, except incidentally.

2. Single barbed arrows are used for movement of single electrons and will be discussed further at the end of this document.

3. The result of a particular arrow-push will be to create a new chemical bond between two atoms or between one atom and a lone electron pair on another atom. Simultaneously, one bond or electron pair will vanish from the drawing. The total number of electrons does not change.

4. The atom at the origin of the arrow, or series of arrows, will become more positive by one charge and that at the head more negative by one. The reason is that an arrow represents movement of an electron pair.

A B A B H NR3 H+ NR3++ - +

5. Unless you can write correct Lewis structures, you will never be able to write reasonable arrow

pushing mechanisms. No intermediate may possess more than one bond to hydrogen or more than four bonds to carbon. For all second row elements, the total of lone pairs and bonds shall not exceed four, because such elements may have no more than eight outer shell electrons. Third row elements, however, may sometimes have more than eight outer shell electrons.

Group III IV V VI VII

Second Row B C N O F Third Row A1 Si P S Cl

Br I

6. To deduce how many electrons surround a particular atom, add the number of lone electrons

plus two electrons for each bond to the atom.

2

7. To calculate the charge on an atom, first add the number of lone electrons plus one electron for each bond to the atom. The result is given a negative sign because you are counting negative charges. Add this number to the group number of the element in the periodic table, which is given a positive sign.

Common Organic Structures

Structure Name Total Outer Shell Electrons Charge

B

borane 6 0

C carbene 6 0

C

carbonium ion 6 +1

C

carbanion 8 –1

C

carbon radical 7 0

N nitrenium ion 6 +1

N nitrene 6 0

N

nitrogen radical cation 7 +1

N

trivalent nitrogen 8

0

N

nitrogen anion (amide) 8 –1

O divalent oxygen 8 0

O

oxonium ion 8 +1

O oxygen anion (alkoxide) 8 –1

O oxy radical 7 0

3

Example 1. How many electrons surround oxygen and nitrogen in the molecule below? What is the formal charge on each atom?

H3CN

H3CO

Solution. Number of electrons surrounding nitrogen = 2 lone electrons + 6 electrons in bonds (3 bonds with 3 electrons each) = 8 electrons. Number of electrons surrounding oxygen = 5 lone electrons + 2 electrons in the N - O bond = 7 electrons. Charge on nitrogen = 2 lone electrons + 3 electrons in bonds = negative charge of 5. Group number of nitrogen is V. 5 + 5 = 0 net charge. Charge on oxygen = 5 lone electrons + 1 electron in N – O bond = 6. Oxygen is in group VI so net charge is zero. We conclude that this species is a neutral free radical because it has no net charge, but it has one unpaired electron.

Example 2. Answer the above questions for the following molecule. (To be worked on your own.)

NO

8. In writing reaction intermediates, remember that electronegative atoms are best able to bear negative charges. Likewise, electropositive atoms are best able to bear positive charges. As an example,

is vastly more probable than

9. Atoms with lone electron pairs can stabilize adjacent positive charges by resonance, but more electronegative atoms do this less effectively.

C F CC F O C N C NC O+ + + + ++

least stable most stable

10. In writing organic reaction mechanisms, we must take four things into consideration:

a) Analogy with mechanisms previously learned.

b) The attraction of opposite charges. Thus, electron pairs tend to move toward positively charged centers (see below).

R O H R O + H+

R O H R O + H–X

4

CO

N

H+δ+δ-

c) The question of which atoms need to become bonded to which in order to make the product.

d) Whether acidic or basic conditions prevail during the reaction. There cannot be an appreciable concentration of a strong acid in a basic solution. Likewise, there cannot be an appreciable concentration of a strong base in an acidic solution.

Example 3: Acid catalyzed hydration of a ketone.

–OH cannot exist in acidic solution because it is immediately converted to water

H2O can exist in acidic solution and it provides the needed OH group indirectly 11. Free radical reactions and the fishhook notation

A single barbed arrow represents the movement of single electrons. Every electron is accounted for. If the overall reaction loses or gains an electron, you have made an error. Atom transfer

Cl• H CH3 Cl H •CH3+

Even though an atom moves, we focus on electron movement.

C

O H3O+

C

OH –OH

C

HO OHbad

mechanism

C

O H3O+

C

OH H2O

C

HO OH2H2O

C

HO OH

C

OH–OH2

C

OH OH2

C

HO O

C

HO OHH

H

OH2

– H3O+

goodmechanism

proper arrow pushing

5

Homolytic dissociation of a bond: one electron to each side

O OO•

O•Δ

Radical recombination reaction

H3C•CH3 •CH3 CH3 Radical addition

•CH3 H2C CH2 H3C

H2C

CH2• Most of the mechanisms we encounter in CHEM 211 will be heterolytic, involving movement of electron pairs, not single electrons.