4.5 Steps to Problem Solving

Learning Objectives

After Chapter 4.5, you will be able to:

With all of these rules in hand, we can now apply our knowledge in a systematic way to simplify organic chemistry reactions that appear on the MCAT. These steps are described below.

MCAT Expertise

While studying organic chemistry for the MCAT, don’t permit yourself to simply nod along with the reaction mechanisms—get involved! During each step of a mechanism, ask yourself how the trends and problem-solving steps in this chapter play out.

Step 1: Know Your Nomenclature

Before you can even start to understand what reactions will occur and what products will form, it is vital to know which compounds IUPAC and common names refer to! If you’re still having trouble with nomenclature, be sure to review Chapter 1 of MCAT Organic Chemistry Review.

Step 2: Identify the Functional Groups

Look at the organic molecules in the reaction. What functional groups are in the molecules? Do these functional groups act as acids or bases? How oxidized is the carbon? Are there functional groups that act as good nucleophiles, electrophiles, or leaving groups? This step will help define a category of reactions that can occur with the given functional groups.

Step 3: Identify the Other Reagents

In this step, determine the properties of the other reagents in the reaction. Are they acidic or basic? Are they suggestive of a particular reaction? Are they good nucleophiles or a specific solvent? Are they good oxidizing or reducing agents?

Step 4: Identify the Most Reactive Functional Group(s)

Once you’ve identified the functional groups in the compound and the other reagents present, this step should be relatively quick. Remember that more oxidized carbons tend to be more reactive to both nucleophile–electrophile reactions and oxidation–reduction reactions. Note the presence of protecting groups that exist to prevent a particular functional group from reacting.

Step 5: Identify the First Step of the Reaction

If the reaction involves an acid or a base, the first step will usually be protonation or deprotonation. If the reaction involves a nucleophile, the first step is generally for the nucleophile to attack the electrophile, forming a bond with it. If the reaction involves an oxidizing or reducing agent, the most oxidized functional group will be oxidized or reduced, accordingly.

Once you know what will react, think through how the reaction will go. Did the protonation or deprotonation of a functional group increase its reactivity? When the nucleophile attacks, how does the carbon respond to avoid having five bonds? Does a leaving group leave, or does a double bond get reduced to a single bond (like the opening of a carbonyl)?

Step 6: Consider Stereospecificity/ Stereoselectivity

Though not all reactions are stereospecific or stereoselective, these possibilities should be considered when predicting products. For stereospecificity, consider whether the configuration of the reactant necessarily leads to a specific configuration in the product, as seen in SN2 reactions. Stereoselectivity, on the other hand, occurs in reactions where one configuration of product is more readily formed due to product characteristics. Stereoselectivity is seen in many reactions, as different products often possess different traits which affect their relative stability. If there is more than one product, the major product will generally be determined by differences in strain or stability between the two molecules. More strained molecules (with significant angle, torsional, or nonbonded strain) are less likely to form than molecules without significant sources of strain. Products with conjugation (alternating single and multiple bonds) are significantly more stable than those without.

Example Reactions

Now, we’ll apply these rules to three novel reactions. Focus on the decision-making element of this process so that you will be able to apply the same logic to reactions that appear on Test Day.

Reaction 1

We’ll start with a series of reactions involving ethyl 5-oxohexanoate. First, it is reacted with 1,2-ethanediol and p-toluenesulfonic acid in benzene; second, with lithium aluminum hydride in tetrahydrofuran, followed by a heated acidic workup. What are the intermediates and final product?

Let’s go through the steps:

  1. First, let’s draw out the reactants and reaction conditions.
    ethyl 5-oxohexanoate with three sequential reaction arrows listing the conditions mentioned in reaction 1
  2. This molecule has an alkane backbone, a ketone, and an ester. Both of the carbonyl carbons are electrophilic targets for nucleophiles. The carbonyl oxygens can also be reduced. Acidic α-hydrogens are also present.
  3. For the first part of the reaction, we have a diol, which is commonly used as a protecting group for aldehydes or ketones. Diols are nucleophiles because of lone pairs on the oxygens in the hydroxyl groups. For the second reaction, we have a reducing agent in an organic solvent. Finally, we have an acidic workup—which is often used to remove protecting groups. We’re starting to get hints as to what is happening here.
  4. In the first reaction, both the ketone and ester carbonyls are highly reactive. One or both of these functional groups will react in the first step.
  5. The diol is a good nucleophile because it contains lone pairs on the oxygen atoms in the hydroxyl groups. Further, the presence of a diol hints at protecting the ketone carbonyl because diols are commonly used for this function. This gives our first intermediate—the ketone carbonyl will be replaced by a protected diether. The second reaction, then, will only be able to proceed on the ester. LiAlH4 is a strong reducing agent, so the next reaction will be reduction of the carbonyl all the way to an alcohol. In the final reaction, the protecting group will be removed by acidic workup, leaving us with our original ketone group.
  6. The product and intermediate have no stereoselectivity, so this won’t be a consideration.

Let’s see what we came up with. The first intermediate will have a protective diether at the ketone carbonyl. The second will show the reduction of the ester to an alcohol, with the protecting group still present. The third will be our final product.

full reaction schema described in reaction 1


Reaction 2

If ethanol is reacted in acidic solution with potassium dichromate, what will the end product be?

Let’s go through the steps again.

  1. First, let’s draw out our molecules.
    ethanol reacted with potassium dichromate in sulfuric acid


  2. This molecule has an alkane backbone and a primary alcohol. Alcohols make good nucleophiles and can also be oxidized. The hydroxyl group can also act as a leaving group, especially if it gets protonated.
  3. Next, the reagents. Dichromate is a good oxidizing agent.
  4. The alcohol carbon is most likely to react because it is the most oxidized.
  5. The primary product of a primary alcohol with a strong oxidant like dichromate will be a carboxylic acid. One other possible product could have been an aldehyde, so this could trip us up! Remember, however, that primary alcohols can only be oxidized to aldehydes by reagents specifically designated for this purpose, like pyridinium chlorochromate (PCC). If we start with ethanol, we’ll obtain ethanoic acid (acetic acid) after reaction with dichromate.
  6. Stereospecificity again isn’t a consideration and won’t change the outcome in this reaction.

Therefore, the primary product of this reaction will be ethanoic acid.

Reaction 3

Determine the product of a reaction between 2-amino-3-hydroxypropanoic acid and 2,6-diaminohexanoic acid in aqueous solution.

Let’s go through the steps one last time.

  1. First, let’s draw out both molecules.
    2-amino-3-hydroxypropanoic acid and 2,6-diaminohexanoic acid
  2. Both of these molecules have a carboxylic acid (which has an acidic hydrogen and an electrophilic carbonyl carbon) and an amino group (which is nucleophilic). The first molecule also has a hydroxyl group; the second has an additional amino group and a long alkane chain.
  3. There are no additional reagents listed. Therefore, it will be the properties of the two reactants alone that determine how the reaction will proceed.
  4. Either of the two molecules could act as the nucleophile in this reaction, and either could be the electrophile. The most reactive species are likely the nucleophilic amino groups attacking the electrophilic carbonyl carbon.
  5. The first step of this reaction will be nucleophilic attack by the amino group on the electrophilic carbonyl carbon. Carbon cannot have five bonds, so the carbonyl group will have to open up. The hydroxyl group on the carboxylic acid is a poor leaving group, but proton rearrangements in the molecule turn the hydroxyl group into water, improving its leaving group ability. Then, the carbonyl will reform, kicking off the water molecule as a leaving group.
  6. We might ask why the hydroxyl group on 2-amino-3-hydroxypropanoic acid doesn’t react. Remember that more oxidized groups tend to be more reactive, and the carboxylic acid is significantly more oxidized than the hydroxyl group. Another question to consider is which of the amino groups of 2,6-diaminohexanoic acid will react. This is a question that is perhaps best answered retrospectively; in this case, the amino group closer to the carbonyl will react because the resulting product will be stabilized by resonance.

Does this reaction look familiar? It should! 2-amino-3-hydroxypropanoic acid and 2,6-diaminohexanoic acid are serine and lysine, respectively—in this reaction, we are forming a peptide bond. If we treat them as generic amino acids, this is the reaction:

two amino acids forming a peptide bond by nucleophilic acyl substitution

We’ve worked through a few problems here to get a handle on how to use this method. Once you have read further chapters and learned specific mechanisms that we did not touch on here, be sure to come back and see how these rules apply to novel reactions.