Answers and Explanations - Analyzing Organic Reactions - MCAT Organic Chemistry Review

MCAT Organic Chemistry Review

Analyzing Organic Reactions

Answers and Explanations

1. C

NH3 and H2O are Lewis bases because nitrogen and oxygen can donate lone pairs. Ag+ is a Lewis acid because it can accept a lone pair into an unoccupied orbital.

2. C

Remember, good nucleophiles tend to have lone pairs or π bonds and are negatively charged or polarized. Alkoxide (RO) and hydroxide (OH) anions are strong nucleophiles. Alcohols (ROH) and carboxylic acids (RCOOH) are weak nucleophiles. The alkyl group of an alkoxide anion donates additional electron density, making it more reactive than the hydroxide ion. The carboxylic acid contains more electron-withdrawing oxygen atoms than the alcohol, making it less nucleophilic.

3. D

Good electrophiles are positively charged or polarized. CR+3 is a tertiary carbocation; it has a positive charge, which makes it very electrophilic. CH3Cl and CH3OH are both polarized; however, the leaving groups differ between these two. Cl is a weaker base than OH (HCl is a stronger acid than H2O). As such, Cl will be more stable in solution than OH, which increases the electrophilic reactivity of CH3Cl above CH3OH. CH3OCH3 has a much less stable leaving group, CH3O, and is therefore significantly less electrophilic.

4. A

Good leaving groups are weak bases, which are the conjugates of strong acids. Leaving groups must also be stable once they leave the molecule. H2O is, by far, the most stable leaving group and will be extremely unreactive once it leaves the molecule through heterolysis. Br is the conjugate base of HBr; HO is the conjugate base of water. HBr is a much stronger acid than water, so Br is a better leaving group than HO. Finally, hydride (H) is a very poor leaving group because it is extremely unstable in solution.

5. B

Carboxylic acids are the second most oxidized form of carbon (only carbon dioxide is more oxidized). In carboxylic acids, the carbon atom has three bonds to oxygen. In aldehydes, the carbon atom has two bonds to oxygen. In amines, the carbon atom has one bond to nitrogen. In an alkane, the carbon only has bonds to other carbons and hydrogens.

6. A

All that we need to know about cinnamaldehyde is that it is an aldehyde, and therefore will be reduced by a strong reducing agent like LiAlH4 to a primary alcohol.

7. C

Because 2-butanol is a secondary alcohol, oxidation by a strong oxidizing agent like dichromate will result in a ketone, 2-butanone.

8. C

Pyridinium chlorochromate is a weak oxidizing agent, and will oxidize an alcohol to an aldehyde. Stronger oxidizing agents are required to convert a primary alcohol to a carboxylic acid.

9. B

An SN1 reaction is a first-order nucleophilic substitution reaction. It is called first-order because the rate-limiting step involves only one molecule. Choice (A) is true, but does not explain why SN1 reactions have first-order kinetics; the rate-limiting step of an SN2 reaction is also the first (and only) step of that reaction, but SN2 reactions have second-order kinetics, not first-order. Choice (C) is a true statement as well, but again does not explain why the reaction is first-order. Finally, choice (D) is incorrect because it is the rate-limiting step, not the reaction overall, that involves only one molecule.


In a protic solvent, the protons in solution can attach to the nucleophile, decreasing its nucleophilicity. The larger the nucleophile, and the stronger its conjugate acid, the stronger the nucleophile will be. Of the options given, I will therefore be the strongest nucleophile because it is least likely to associate with the protons in solution.


Primary and secondary alcohols can undergo oxidation because the carbon can form additional bonds with oxygen while losing bonds to hydrogen. Choice (A), a tertiary alcohol; choice (B), a ketone; and choice (C), a carboxylic acid, cannot form additional bonds to oxygen because they have four bonds to other carbon or oxygen atoms already.


In this reaction, there has been an inversion of stereochemistry. The mostly likely explanation for this is that the reaction proceeded by an SN2 reaction mechanism. Inversion of stereochemistry is a hallmark of SN2 reactions, whereas racemization is a hallmark of SN1 reactions.


To carry out a nucleophile–electrophile reaction, the nucleophile must be able to dissolve in the solvent. Nucleophiles are nearly always polar, and often carry a charge. Polar solvents are therefore preferred for these reactions. Hexane is a nonpolar solvent and will not be useful for a nucleophile–electrophile reaction.


Aldehydes have one alkyl group connected to the carbonyl carbon, whereas ketones have two. This creates more steric hindrance in ketones, which lowers their reactivity to nucleophiles. Ketones are also less reactive because their carbonyl carbon has less positive charge character; the additional alkyl group can donate electron density—the opposite of choice (D)—which decreases the electrophilicity of the compound.


Remember, there is a hierarchy to the reactivity of carboxylic acid derivatives that dictates how reactive they are toward nucleophilic attack. In order from highest to lowest, this is anhydrides > carboxylic acids and esters > amides. In practical terms, this means that derivatives of higher reactivity can form derivatives of lower reactivity but not vice-versa. Nucleophilic attack of an ester cannot result in the corresponding anhydride because anhydrides are more reactive than esters.