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Thursday, May 25, 2023 by Guilford Techno Consultants, Inc. | Name Reactions

An interesting reaction that is sometimes covered in undergraduate Organic Chemistry is the Cannizzaro reaction, discovered by Stanislaus Cannizzaro and dating back to 1853. The reaction involves the disproportionation of a non-enolizable aldehyde under basic conditions to yield a carboxylic acid and an alcohol. Disproportionation refers to a redox reaction in which a molecule is simultaneously oxidized to one product (in this case, a carboxylic acid) and reduced to another (an alcohol). A non-enolizable aldehyde is an aldehyde that doesn’t contain an alpha carbon with acidic protons.

In terms of mechanism (found below), the reaction is straight forward. Nucleophilic acyl attack by the hydroxide anion on the carbonyl carbon of the aldehyde produces a tetrahedral intermediate. Deprotonation of the OH group gives a dianion, which reforms the carbonyl and causes a hydride to shift to the carbonyl carbon of a second mole of the aldehyde. The alkoxide formed from the hydride shift is protonated by water to give the alcohol. In a second acid step, the carboxylic acid is produced through protonation of the carboxylate. The Crossed Cannizzaro Reaction involves the reaction of an aldehyde with formaldehyde as reducing agent. Practice problems can be found under the OChem II tab. For each reaction, predict the product and propose a mechanism. 

Tuesday, May 16, 2023 by Guilford Techno Consultants, Inc. | Mechanism

Acid catalyzed dehydration of alcohols yields alkenes, while acid catalyzed dehydration of vicinal diols (glycols) yields ketones or aldehydes. The mechanism, known as the pinacol rearrangement, involves the protonation of one of the hydroxyl groups, followed by loss of water to produce a carbocation. Rearrangement of the carbocation gives rise to a resonance stabilized cation. Deprotonation of this cation forms the final product. Below is shown the acid catalyzed dehydration of 2,3-dimethyl-2,3-butanediol (pinacol) to 3,3-dimethyl-2-butanone (pinacolone). The mechanism involves the rearrangement of a tertiary carbocation to a resonance stabilized cation through migration of a methyl group. In the case of a symmetrical diol, the same carbocation is generated after loss of water on either hydroxyl bearing carbon. When the diol is not symmetrical, the most stable carbocation is initially formed prior to rearrangement. Rearrangements can involve migration of alkyl groups and hydrides, as well as ring expansion or compression. Practice problems can be found under the OChem I tab. For each reaction, predict the product. For extra practice, propose a mechanism for each reaction.  

Sunday, April 30, 2023 by Guilford Techno Consultants, Inc. | Substitution and Elimination Reactions

Most first semester students find substitution and elimination reactions to be overwhelming, in part because of the sheer volume of information needed to understand these reactions. In terms of mechanism, they are relatively simple in comparison to many of the other reactions of the course. Some of  the difficulty stems from the fact that substitution and elimination reactions compete with each other, making product prediction a challenge. Many students have trouble distinguishing between a nucleophile and a base, and additionally determining what constitutes a strong or weak nucleophile or base. In E2 elimination, also known as beta elimination, a strong base (i.e., hydroxide, methoxide, t-butoxide) deprotonates a carbon that is beta (next to) a carbon with a leaving group (alpha carbon). The electrons from the C-H bond push towards the carbon with the leaving group, thus forming a C-C double bond with loss of the leaving group. Below is the mechanism for an E2 reaction.

One of the problem spots of the E2 reaction is the requirement that the proton that is removed from the beta carbon is antiperiplanar to the leaving group, meaning that the proton and the leaving group must be roughly 180° apart. If this is not the case, the bond between the alpha and beta carbons must be rotated to meet this requirement. This requirement determines the stereochemical outcome (cis or trans) of the product. When I have to do a bond rotation to get the leaving group and the proton antiperiplanar to each other, I like to use a Newman projection. You can also do the bond rotation without a Newman projection if you prefer. Below, I illustrate both methods. Practice problems can be found under the OChemI tab. For each reaction, predict all of the possible products from E2 elimination.