Arrow Pushing Review

8/10/2019 Arrow Pushing Review

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Arrow-Pushing and Solving Mechanism Problems

A.

General Points:

1. It is important that you understand exactly what the various arrows represent:

flow of two electrons flow of one electron reaction arrow

resonance arrowequilibrium arrow retrosynthetic arrow

2. Why curved arrows?

(a)One curved arrow = two electrons = one bond (when a bond is being formed or broken) or one

lone pair of electrons.(b)The arrow starts where the two electrons originate, and points toward the atom with which the

electrons are reacting or end up on (this means that an arrow can never start at H+).

Et3N H Cl Et3N H Cl

O

H3C H OH

O

H3CH2O

Example 1 Example 2

B.

Specific Tactics and Hints:

OHH3C

H3CO OHO

H3C OHcat OH

Example 3

1. Draw out the starting materials, reagents, and products as clear structures, including formal charges

and lone pairs.

2. Compare the starting material(s) and product(s).

(a)

Count the atoms in the starting material(s) and product(s). Determine what is lost or gained. Thishelps you keep track of all the atoms in the reaction. Moreover, it may give you a hint as to whattype of reaction occurs. For Example 3, the starting material has 7 carbons, 3 oxygens, and 16

hydrogens. The product has 6 carbons, 2 oxygens, and 12 hydrogens. This tells you that you are

losing 3 carbons, 1 oxygen, and 4 hydrogens you might recognize that this is H3COH

(methanol).

OHH3C

H3CO OHO

H3C OHcat OH

+ H3COH


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(b)Determine which bonds are formed and which are broken. Doing so gives you an idea of whatreactions might occur. For Example 3, in order to form a cyclic structure a new C-O bond must be

formed as indicated in the product. We also have identified that the starting material loses

methanol, thus the C-OMe bond probably breaks.

OH3C

H3CO OHO

H3C OHcat OH

+ H3COHH

(c)Determine which pieces of the product(s) correspond to which part of the starting materials. InExample 3, it is most likely that the methanol comes from the OCH3group in the starting material

as we have highlighted in blue.

OH3C

H3CO OHO

H3C OHcat OH

+ H3COHH

(d)Determine which pieces of your starting material(s) are lost and what remains unchanged. We

have already identified that we lose methanol. We can also see that the carbon indicated by anasterisk (*) forms a new bond to the alcohol (*).

* OH*

H3C

H3CO OHO

**

H3C OHcat OH

+ H3COH

(e)Number the atoms. This exercise may help you if the starting materials and products look

drastically different. First, assign each atom in the starting material a number, especially the

carbon atoms. Next, try to match up one or two key atoms in the starting material with theircounterparts in the product(s), and give them the same numbers. For example, there is only one

methyl group in our starting material and product so we can guess that this is the same carbon #1

in both product and starting material as assigned below.

OHH3C

H3CO OHO

H3C OHcat OH

+ H3COH

12

34

56

1

2

3

4

5

67

7

3. Consider the reaction conditions and how the starting materials are affected by them. Ask what do

these kinds of molecules do? For example, what are the nucleophilic and electrophilic sites of mystarting materials? Or under acidic conditions, what is the most basic site of my starting material? InExample 3, you might recognize that the starting material is a hemiacetal. Under basic conditions,

hemiacetals are in equilibrium with a carbonyl and an alcohol. Also, since the reaction is under basicconditions, it is likely that something will get deprotonated. The two alcohol protons are the most

acidic protons in the starting material so these are likely candidates.

4. Start pushing arrows. In our analysis of Example 3, we suggested that deprotonation of one of thealcohols would be a reasonable first step under basic conditions. We also recognized that our starting

material contains a hemiacetal. It therefore might make sense to begin by deprotonating the hydroxyl


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group of the hemiacetal. This generates water and an alkoxide. Since we need to lose CH3OH, we can

now do that by having the lone pair on the alkoxide add to the hemiacetal carbon, breaking the C-OCH3bond and placing the electrons from the breaking bond on oxygen. We have produced a ketone

intermediate that has the exact same number of carbons, hydrogens, and oxygens as the desired

product. From here, the mechanism for product formation should be recognizable. Intramolecular(within the same molecule) hemiacetal formation leads to product as shown below.

OHH3C

H3CO O

OHH

OHH3C

H3CO O

+ H2O

OH3C

O

+ OCH3H

OH3C

O

O

H3C O

H OHO

H3C OH

OHH3COH+

+ +

(a)REMEMBER: write out every step of the mechanism and every intermediate.

(b)It can help to draw your starting material in a similar orientation to your product.

(c)Reevaluate your mechanism at intermediate stages to assess whether you are making productive

changes (i.e. getting closer to the structure of the product).

(d)If you get stuck, analyze your intermediates in the mechanism as you did for the starting material

in part 2

5. Double check your mechanism:

(a)Do not invoke strongly basic intermediates under acidic conditions. Do not invoke strongly acidicintermediates under basic conditions. For Example 3, you should not generate + charges on any of

the intermediates.

(b)The net charge in the overall system must be conserved from starting material(s) to product(s).

(c)Check the numbering of carbon atoms in the starting material and product and determine if it isconsistent with your arrow-pushing.

(d)Carbon should never have more than 8 valence electrons. In general watch out for the octet rule(it helps to draw out all implicit hydrogens at the reaction site(s)).


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Arrow Pushing ExercisesO

OHH3C

OHO

OH3CH2O

Br

Br

CH3

H3C ClCH3

CH3

H3C CH3Cl

Br

H

H

H HH

OH H2C CH2

Br

CH3H

HH CH3

CH3

CH2H

HH CH3

CH3Br

OH

H2O Br

H3C

OO

H

H3C

O O

Ph

O

CH3Li CH3

O

CH3

Li

CH3

Ph

O

CH3 Al

H

H HH

Li

Ph

O

CH3

LiAlH3

O

H3C CH3OCH3

H3C CH3

O OCH3

H3C CH3

HO OH O

CH3H3C

H

OH

O

H3C CH3

H2O

H3C CH3

HO OPh

H OH2H3C CH3

HO OPh

HO

CH3H3C

H

PhOH

O

CH3

H Me2CuLi

O

H

CH3H3C

O

H

CH3H3C

H2O

Li

1

2

3

4

5

6

7

8

9

10

11

12


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Mechanism Problems

O

OHH+, CH3OH OH

H3CO OH

OHO

OH

OHOOH

pH 7

O

O

OH

OH

O

OH+

O OH

HO

O

OH

OHOH (cat)

O

OH

OH

NaBH4, MeOH

1.

2.

3.

4.

5.

OH

HO

OH

Arrow Pushing Review

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