Lab Report 8

I H H3C NOC XT :0: MET H₂C base H3C The aldol reaction is a carbonyl condensation reaction between two carbonyl partners and involves a combination of nucleophilic addition and a-substitution steps. One partner is converted into an enolate ion nucleophile and adds to the electrophilic carbonyl group of the second partner. In the classic aldol reaction, the carbonyl partners are aldehydes or ketones, although aldehydes are more reactive. The product is a ß-hydroxy carbonyl compound. H OH q H3C H₂C H Under reaction conditions slightly more vigorous than those employed for the aldol reaction, the ß-hydroxyl group is eliminated in an E1cB dehydration to give an a,ß-unsaturated carbonyl compound. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions heat H H + H₂O

2 Moin H3C H H₂C C⇒x= base :0: OH Hori H3C H. The aldol reaction is a carbonyl condensation reaction between two carbonyl partners and involves a combination of nucleophilic addition and a-substitution steps. One partner is converted into an enolate ion nucleophile and adds to the electrophilic carbonyl group of the second partner. In the classic aldol reaction, the carbonyl partners are aldehydes or ketones, although aldehydes are more reactive. The product is a B-hydroxy carbonyl compound. Under reaction conditions slightly more vigorous than those employed for the aldol reaction, the ß-hydroxyl group is eliminated in an E1cB dehydration to give an a,ß-unsaturated carbonyl compound. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions H₂C Ἡ :0: heat H Home H3C + H₂O

base H3C H OH heat H + H₂O H The aldol reaction is a carbonyl condensation reaction between two carbonyl partners and involves a combination of nucleophilic addition and a-substitution steps. One partner is converted into an enolate ion nucleophile and adds to the electrophilic carbonyl group of the second partner. In the classic aldol reaction, the carbonyl partners are aldehydes or ketones, although aldehydes are more reactive. The product is a ẞ-hydroxy carbonyl compound. In the mixed aldol reaction, the best results are obtained when one of the partners does not have an a-hydrogen. It cannot form an enolate anion and can therefore not undergo self-reaction. The enolizable aldehyde is added slowly to a basic solution of the first aldehyde so that the mixed product is preferentially formed. Under reaction conditions slightly more vigorous than those employed for the aldol reaction, the ẞ-hydroxyl group is eliminated in an E1cB dehydration to give an a,ß-unsaturated carbonyl compound.…

H3C. 00 H3C. CH3 XII HO: OCH3 Aldehydes and ketones react reversibly with two equivalents of alcohol in the presence of an acid catalyst to give acetals. Alcohols are poor nucleophiles, and so protonation of the carbonyl oxygen is used to make the carbonyl carbon a stronger electrophile. Addition of the first equivalent of alcohol gives a hemiacetal, a hydroxyether. Addition of the second equivalent of alcohol is accompanied by loss of water to yield the product acetal. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions 2 CH₂OH HCI catalyst CH3 H3C H-OH₂ H3CO OCH3 CH3 H3C. H₂O: OCH3 CH3 H₂O Ча

3 Treatment of 1-aminoadamantane, C„H„N, with methyl 2,4-dibromobutanoate in the presence of a nonnucleophilic base, R,N, involves two successive S,2 reactions and gives compound A. Propose a structural formula for compound A. R&N NH, + Br. C15H93NO, + 2 R,NH Br OCH Br 1-Aminoadamantane Methyl 2,4-dibromobutanoate A

H3C OH 1. Br2, PBг3 2. H₂O H3C OH Br The a-bromination of carbonyl compounds by Br₂ in acetic acid is limited to aldehydes and ketones because acids, esters, and amides don't enolize to a sufficient extent. Carboxylic acids, however, can be a-brominated by first converting the carboxylic acid to an acid bromide by treatment with PBr3. Following enolization of the acid bromide, Br2 reacts in an a-substitution reaction. Hydrolysis of the acid bromide completes the reaction. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions Br Br зь P-Br Br H₂C H₂C SH Br Br

In nucleophilic acyl substitution what is true? V addition to the carbonyl by a nucleophile is followed by loss of the leaving group. O Tetrahedral intermediate is formed. O loss of the leaving group is followed by rearrangement of the carbocation. O an SN2 reaction occurs. O protonation of the carbonyl is followed immediately by loss of the leaving group. 4>A 144 & 7 8 R T Y

H3C- H3C. OCH 3 H3CO CH3 H Aldehydes and ketones react reversibly with two equivalents of alcohol in the presence of an acid catalyst to give acetals. Alcohols are poor nucleophiles, and so protonation of the carbonyl oxygen is used to make the carbonyl carbon a stronger electrophile. Addition of the first equivalent of alcohol gives a hemiacetal, a hydroxyether. Addition of the second equivalent of alcohol is accompanied by loss of water to yield the product acetal. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions NA 2 CH₂OH HCI catalyst CH3 HỘ H3CQ H3C- H3CO CH3 H3C. OCH3 CH3 H₂O*: 45

What is the major product of the following reaction?

Which of the following molecules is most nucleophilic? OA. NaOCH₂CH3 OB. CH3CH₂OH OC. NaOC6H5 OD. C₂H5OH 1 E.CH3COONa

H₂C=CHCCH3 then H3O+ The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an a,ß-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the ẞ-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions N 9a

Ph- OC XT :OH₂ Primary amines add to aldehydes and ketones to give imines. Imines are formed in a reversible, acid-catalyzed process that begins with nucleophilic addition of the primary amine to the carbonyl group, followed by transfer of the proton to yield a neutral carbinolamine. Protonation of the hydroxyl group converts it into a good leaving group and an E1-like loss of water yields an iminium ion. Deprotonation yields the product imine and regenerates the acid catalyst. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions NHOH CH3 CH3 NH₂OH OH HN I Ph CH3 OH H₂O CH3 за

1. NaOEt CO₂Et 2. H3O+ 3. heat Br H3C CO₂H CO2 H3C CO₂Et The malonic ester synthesis is a carbonyl alkylation reaction. It is used to prepare carboxylic acids from primary alkyl halides, lengthening the carbon chain by two atoms. Thus, the product can be visualized as being a "substituted acetic acid." The reaction consists of three steps: generation of the enolate anion followed by SN2 reaction with a primary alkyl halide, ester hydrolysis under acid conditions, and decarboxylation. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions 22 CX EtO₂C EtO₂C HOCH2CH3 HOCH2CH3 EtO₂C EtO₂C

Identify A, B, and C, intermediates in the synthesis of the five-membered ring called an α- methylene-γ-butyrolactone. This heterocyclic ring system is present in some antitumor agents.

Williamson Ether Synthesis of Phenacetin Background: Synthesize phenacetin from acetaminophen (Tylenol) via a Williamson ether synthesis.The Williamson ether synthesis is one of the simplest methods used to make ethers. The reaction is named after Alexander Williamson, who discovered the reaction in 1850. In this reaction, an alkoxide anion reacts with an alkyl halide via a substitution reaction.                         RO-Na+   +   R` – X   →   R-O-R`   +   NaX Alkoxides are strong nucleophiles that tend to react via an SN2 mechanism. As a result, the reaction works best with primary alkyl halides. In a traditional Williamson ether synthesis, the alkoxide is generated by reacting sodium hydride (NaH) with an alcohol. ROH   +   NaH    →    RO-Na+   +   H2 While this reaction is very efficient, it does pose a fire hazard, as it is possible for hydrogen gas to ignite. In this lab, the alkoxide will be created using sodium hydroxide (NaOH). This is a much safer option, eliminating a…