Specific Reactions and Named Reactions - Organic Chemistry

First step: Friedel-Crafts acylation of benzene

Second step: Formation of enolate

Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)

Fourth step: neutralization of anion and dehydration forming alkene

The reaction uses a Grignard reagent's nucleophilic carbon to attack the carbon in carbon dioxide. Following treatment with water the resulting molecule (a carboxylate anion, of the form ) the carboxylate oxygen will be protonated, and the result is a carboxylic acid. Thus, the resulting molecule is a benzene ring with a carboxyl group attached (this is also known as benzoic acid).

This is a Diels-Alder reaction; these reactions happen between a nucleophilic diene, shown in blue below, and an electrophilic dienophile, in green. Diels-Alder reactions install a set of bonds that connect each external carbon of the diene system to an alkene carbon in the dienophile system to create a new six-membered ring. All remaining structure of the two reactants are retained, including the six- and five-membered rings below. The red bonds are the newly installed bonds.

The electrons from one of the double bonds on the 1,3-dibutene create a new single bond. The other new single bond is created from the electrons in the double bond of the other reactant. These two new single bonds join the reactants to create a cyclic product.

The electrons from the other double bond in the 1,3-dibutene move between the carbon 2 and 3. Thus, the final product is a 6-carbon cycloalkene with a halogen substituent.

This is a standard Diels-Alder reaction. Diels-Alder reactions are driven solely by adding heat to the reagents. By looking at the reagents and the product, we can tell that this is a Diels-Alder reaction. For Diels-Alder, we need a cis-diene and an alkene as reactants. When these reactants are stimulated by heat, they form a cyclohexene product.

Diels-Alder reactions create cyclohexene rings (eliminate III, IV, and V), and starting dienophile is trans (E conformation), so product is E (Eliminate I).

This is a classic Diels-Alder reaction and it consists of a diene (cyclopentadiene) and a dienophile (ethene). The bicyclic structure forms if the electrons are moved in a circular fashion.

The Diels-Alder reaction converts a conjugated diene and a substituted alkene into a six-membered ring containing cyclohexene (a substituted cyclohexene system). In Hoffmann elimination, tetra-alkyl ammonium salts undergo elimination to form the least substituted alkene. The Wittig reaction uses phosphorus ylides, aldehydes, or ketones to form an alkene and a triphenylphosphine oxide. Lastly, Gabriel synthesis forms primary amines via the reaction of a phthalimide with an alkyl halide, followed by cleavage with hydrazine.

This is a classic esterification reaction. Esterfication occurs when a carboxylic acid and an alcohol are reacted together. Only one answer choice is an ester.

This is an example of a Grignard reagent reaction. Because we are adding three carbons to our chain, the Grignard reagent we need must have three carbons on it. We can therefore rule out methyl grignard and ethyl grignard.

N-propyl is the straight-chained 3-carbon alkane, while isopropyl is branched. Looking at our final product, we can see the carbon chain we have added is straight-chained, and thus N-propyl Grignard is the best option. Because Grignard reagents are relatively basic, we must add an hydronium ion workup to protonate our alcohol.

The reaction of a Grignard reagent with oxirane (a type of epoxide) in addition to the work-up with dilute acid will yield a primary alcohol solely because there is a work-up with dilute acid. This shows that there is an excess of hydrogen, which will yield to a primary alcohol versus a secondary or tertiary alcohol.

Grignard reagents (often formed in-situ due to their highly non-specific reactivity) act as reducing agents by forming carbanions, which are strong bases and nucleophiles. Nitriles may react with Grignard reagents to form imines. The reaction proceeds in an analogous fashion to a standard nucleophilic carbonyl addition, converting the triple bond to a double bond by forming an intermediate in which nitrogen carries a negative charge. A protic solvent, such as ethanol, is then added to neutralize the intermediate. Thus, the correct answer is the only compound in which an imine is formed, which is found in compound I.

Step 1 converts the carboxylic acid into an ester.

Step 2 adds two equivalents of Grignard reagent: one to turn the molecule into acetone, and a second one to turn it into tert-butoxide. There is no way to stop the reaction after the first addition of Grignard reagent.

Step 3 will neutralize the base, leaving only t-butyl alcohol (I).

The haloform reaction requires a carbonyl with a terminal alpha carbon. A hydrogen gets abstracted, and the enolate is formed. A halogen attacks the alpha carbon, and the ketone is reformed. This occurs three more times until the carbon has bonds to three halogens. Then the carbon leaves, forming a carbanion, and the base attacks the carbonyl carbon. An ester is formed.

Only bromocyclohexane is created because there is only one bromine group that bonds with one of the carbons, while the other carbon is bonded with hydrogen group.

The most acidic hydrogen of the ester gets abstracted and the enolate form of the compound is attained. The electrons from the carbon-carbon double bond attack the carbonyl carbon of the other ester. Deesterfication occurs in this same step. The final product has one ketone and one ester.

During acid catalyzed hydration, a hydroxy group replaces one of the bonds in the triple bond and a double bond is formed. This is called an enol. The enol naturally turns into a ketone in a process called tautomerization. The hydroxy group can attach to either carbon across the double bond, and naming is done so that substituents have the lowest numbers. Only on 5-decyne will result in a single product, as no matter which carbon the hydroxy group bonds to, it is still on carbon 5. Thus, the only final product is 5-decone.

The other answer options will still react, but will form multiple products due to lack of symmetry.

First step: bromination across the double bond

Second step: double dehydrohalogenation and removal of terminal alkyne hydrogen

Third step: neutralization of the molecule

Fourth/fifth step: hydroboration/oxidation, followed by keto/enol tautomerization

The Ag2O oxidizes the alcohol to form the alkoxide ion. The Ag2O is a reducing agent, and it oxidizes the alcohol because the alcohol loses a proton. Ag2O is a strong base, meaning it will accept protons readily. The given reaction is an example of Williamson ether synthesis. This reaction occurs via an SN2 pathway.