ochem 3
Draw the mechanism for enamine/imine tautomerization.
.
Provide a reactant that will form each of the following when reacted with an acid chloride: a) an ester b) an amide c) an anhydride d) a carboxylic acid
.
Provide a structure for each of the following carboxylic acids: 1) Formic acid 2) Acetic acid 3) Benzoic acid
.
Describe what an acetal/Hemiacetal and ketal/hemiketal is. How does the MCAT sometimes refer to Hemiacetals? Describe the steps to the Formation of Acetals/Hemiacetals and Ketals/Hemiketals reaction.
. Acetals/Ketals have TWO -OR substituents and Hemiacetals/Hemiketals have ONE -OR substituent plus one alcohol substituent (-OH group). The MCAT will sometimes refer to both Hemiacetals and Acetals as simply "Acetals." On other occasions they have differentiated the two. Steps: 1) An alcohol acts as the nucleophile, attacking the electrophilic carbonyl carbon and pushing electrons from the C=O bond up onto the oxygen. 2) The negatively charged oxygen is protonated to form an alcohol and the original alcohol is deprotonated to form an ether. This yields a hemiacetal if it was originally an aldehyde, or a hemiketal if it was a ketone. 3) The alcohol is protonated again to form the good leaving group water, and a second equivalent of alcohol attacks the central carbon via SN2, kicking off water as the leaving group. 4) Deprotonation of the second alcohol results in another ether, yielding an acetal if it was originally an aldehyde or a ketal if it was a ketone.
Describe the product with the addition of amines to carbonyls. Describe the steps of the reaction and specifically what product will form when a primary, secondary, and tertiary is used as the reactant.
. Amines add to aldehydes and ketones to form imines and enamines. Steps: 1) The amine acts as a nucleophile, attacking the electrophilic carbony carbon. 2) The oxygen is protonated twice, creating the good leaving group water. 3) A base abstracts a hydrogen from the nitrogen and kicks off water in an E2 mechanism. This forms either an imine or an enamine (depending on the substitution of the nitrogen). Primary amines: yield imines Secondary amines: yield enamines Tertiary amines: DO NOT REACT.
Describe the Wolf-Kishner Reduction reaction. What are the products/reactants and what are the steps?
. Complete reduction of an aldehyde or ketone to an alkane (replace the carbonyl with C-H₂) via an imine intermediate. The reaction begins the same as the addition of amine to carbonyl reaction but starting with hydrazine (H₂N-NH₂) as the amine, the product is an amine-substituted imine. Subsequent addition of a hot, strong base (i.e., KOH/heat) replaces the imine with two hydrogens, yielding an alkane. Steps: 1) CH₃COCH₃ + H₂N-NH₂ → imine 2) Imine + KOH/∆ → CH₃CH₂CH₃
Describe the Gabriel Synthesis. List the products and reactants, and the steps.
. Formation of a primary amine from a primary alkyl halide; AVOIDS SIDE PRODUCTS of alkyl amine synthesis. Steps: 1) The pthalimide ion, a reactive species with a full negative charge on the nitrogen, acts as a nucleophile, attacking the alkyl halide via SN2. 2) The resulting intermediate is then hydrolyzed with aqueous base, releasing a primary amine.
Describe the Hofmann Elimination reaction. What are the products/reactants? Why is this reaction important?
. Formation of an alkene via elimination of an amine. Important because you get the LEAST substituted alkene.
How do you protect Ketones/Aldehydes from reaction?
. Ketones or aldehydes can be prevented from RXN with a nucleophile or base by conversion to an acetal or ketal (which are unreactive in all but acidic conditions). Any terminal diol (HO-CH₂-CH₂-OH) with at least two carbons will work. Steps: 1) One end of the diol acts as the nucleophile (attacking the electrophilic carbonyl carbon and pushing electrons from the C=O bond up onto the oxygen. The negatively charged oxygen is protonated to form an alcohol and the original alcohol is deprotonated to form an ether.) 2) The other end of the diol acts as the "second equivalent of alcohol" (The alcohol is protonated again to form the good leaving group water, and a second equivalent of alcohol attacks the central carbon via SN2, kicking off water as the leaving group. Deprotonation of the second alcohol results in another ether, yielding an acetal if it was originally an aldehyde or a ketal if it was a ketone.)
Describe the Hofmann Degradation reaction. What are the reactants/products? Why is this reaction important?
. Primary amides (amides with only hydrogens on the nitrogen) react in strong, basic solutions of Cl₂ or Br₂ to form primary amines. The mechanism includes decarboxylation, and thus reduces the carbon chain. R-C(=O)-NH₂ → R-NH₂ This reaction is IMPORTANT because it allows you to add an amine to a TERTIARY CARBON. This is impossible via "Synthesis of an Alkyl Amine" reaction. (The Synthesis or an Alkyl Amine reaction occurs via an SN2 reaction; therefore, it cannot proceed using a tertiary carbon due to steric hindrance.)
Describe what an esterification is. What are the reactants/products? How can you obtain higher yields with this reaction?
. Reaction of an alcohol with a carboxylic acid to form an ester. You've probably realized by now that the hydroxyl group will never leave without being protonated first to form the "good leaving group water;" thus this RXN requires an acid catalyst. Higher yields can be obtained by reacting an anhydride with an alcohol.
Describe the transesterification reaction. What are the reactants/products?
. Reaction of an existing ester with an alcohol, creating a different ester. Requires an acid catalyst.
What is the Aldol Condensation reaction? What are the reactants/products? What are the steps to the reaction? What is important to remember with this reaction for the MCAT?
. The condensation of one aldehyde or ketone with another aldehyde or ketone. Steps: 1) A base abstracts an alpha hydrogen, creating a carbanion 2) The carbanion will attack any carbonyl carbon in the solution 3) The oxygen is protonated to form an alcohol Be extremely careful not to lose track of one carbon when combining the two species. The MCAT is very likely to ask you to predict the product. The MCAT loves the Aldol Condensation! It is addressed in one way or another on almost every exam.
Describe what tautomerization is. Show the mechanism.
. This is the process by which an alpha hydrogen adjacent to an aldehyde or ketone becomes bonded to the carbonyl oxygen, while the double bond is switched from the carbon-oxygen bond to the carbonyl carbon-alpha carbon bond.
Describe the Formation of Acid Chlorides reaction. What reagents can be used? Why does addition of chloride to a carboxylic acid not produce an acid chloride? Why are acid chlorides important?
. Three reagents readily produce acid chlorides when added to carboxylic acids: PCl₃, PCl₅, and SOCl₂. Addition of chloride to a carboxylic acid DOES NOT produce an acid chloride. The reason involves forming good leaving groups. Acid chlorides are important because they are the most reactive of the carboxylic acid derivatives. Their reactivity is due to: 1) the withdrawing power of the chlorine, which makes the partial positive charge on the carbonyl larger than normal 2) the fact that chloride ion is a superb leaving group
What is the Wittig Reaction? What is the reactant/product? (Don't need to know the mechanism)
. Transforms a carbonyl into an alkene. The product will always be an alkene, with the double bond formed between the carbonyl carbon and the carbon attached to the polyphenyl group.
Describe the Haloform reaction. What are the reactants/products? What are the steps?
. When the Halogenation reaction is done to a methyl ketone with sufficient halogen present to effect replacement of all three alpha hydrogens the trihalo product reacts further to produce a carboxylic acid and a haloform (chloroform, CHCl₃; bromoform, CHBr₃; or iodoform, CHI₃). Steps: 1) Complete the Halogenation reaction using a methyl ketone and enough halogen to replace all three alpha hydrogens. 2) The tri-substituted alpha carbon has a large partial positive charge and is thus a good leaving group. When a strong hydroxide base, sucah as NaOH, is added, the -OH attacks the carbonyl carbon, kicking the electrons from the C=O bond up onto the oxygen. 3) The electrons from the oxygen collapse down, reforming the double bond and kicking off the haloform as a leaving group. The result is a carboxylic acid. Note: To get a good yield of carboxylic acid, the haloform (-CBr₃) must be constantly removed as it forms. Otherwise, you get only enolate ions.
Describe what a carbonyl is. Describe two key features of carbonyls Describe the bond strength of the carbonyl double bond. Describe whether aldehydes, ketones, carboxylic acids, amides, esters, and anhydrides undergo substitution or addition. Discuss the importance of alpha hydrogens.
. A carbonyl is any compound containing the C=O functional group. Key features: 1) Planar (can be attacked from both sides) 2) Partial positive charge on the carbonyl carbon (very electrophilic) Bond Strength: The carbonyl double bond is shorter and stronger than an alkene. Aldehydes and Ketones: always undergo nucleophilic ADDITION Carboxylic Acids/Amides/Esters/Anhydrides: always undergo nucleophilic SUBSTITUTION Alpha Hydrogens: This is a very important feature of all carbonyls. Hydrogens on a carbon one away from the carbonyl carbon are acidic due to resonance stabilization of the conjugate base; if there are two carbonyls separated by one carbon, hydrogens on this middle carbon are VERY acidic.
What are the four steps to determining if a group is electron donating or withdrawing? Label the following as electron donating or electron withdrawing: 1) alkyl groups 2) nitro groups 3) cyano groups 4) sulphones 5) amines 6) carboxylic acids 7) esters 8) alcohols 9) quaternary amines
1) Look at the first atom from the point of attachment. Compare its electronegativity to the atoms bound to it. If it is more electronegative, it will bear a partial negative charge and if it is less electronegative, it will bear a partial positive charge. 2) Atoms will full or partial positive charges WITHDRAW from whatever they are attached to. Atoms with full or partial negative charges DONATE to whatever they are attached to. 3) Hydrogen is considered NEITHER electron donating nor withdrawing. 4) Alkenes are weakly electron withdrawing (memorize this). 1) alkyl groups: donating 2) nitro groups: withdrawing 3) cyano groups: withdrawing 4) sulphones: withdrawing 5) amines: donating 6) carboxylic acids: withdrawing 7) esters: withdrawing 8) alcohols: donating 9) quaternary amines: withdrawing
List the 4 steps in naming an amine.
1) Name the alkane to which the N is attached (i.e., propane) 2) Add "amine" in place of the "e" on the end of "ane" (i.e., propanamine). It is also acceptable to separate the substituent name (i.e., propyl amine). 3) If the amine is secondary, the longest chain is included in the name as indicated above. The other chain is added at the beginning, proceeded by the letter "N-" (i.e., N-ethylpropananmine) 4) If the amine is tertiary or quaternary, add additional substituents to the front of the name in alphabetical order, all with the prefix N- included (N,N-diethylpropanamine, or N-ethyl-N-methylpropanamine, or N,N-dimethyl-N-ethylpropanamine)
How do you determine whether a molecule will act as a base or a nucleophile (3 ways)?
1) Steric hindrance favors basicity over nucleophilicity; Nucleophiles MUST have very little hindrance. Primary nucleophiles are most common. Secondary atoms often act as bases or nucleophiles, depending on conditions. However, tertiary atoms will ONLY act as bases. 2) Reactivity (low stability) favors basicity over nucleophilicity (i.e., NH₂⁻ and RO⁻). If an atom has a full negative charge it will almost always act as a base (unless it's a halide). 3) Highly electronegative atoms make good bases, but not good nucleophiles. A nucleophile must SHARE its electrons in a new bond. Bases abstract PROTONS and create highly polar bonds with them (i.e., very little, or unfair sharing).
13. Certain diols with a specific bond-to-bond connectivity convert to a single ketone or aldehyde when heated. Such a reaction must involve which of the following processes? A. oxidation B. hydride shift C. methyl shift D. substitution
13) C; The process described in the question stem is the pinacol rearrangement. It requires a vic-diol and involves the spontaneous disassociation of one of the protonated alcohols followed by a methyl shift. Following the methyl shift, the other alcohol's hydrogen is abstracted and the electrons condense to form a carbonyl and quench the carbocation.
14. Which of the following statements is(are) true regarding epoxides and their reactivity? I. The C-O-C bond angle is approximately 60 degrees II. They are highly reactive due to ring strain III. Reaction with a nucleophile can proceed via an SN2 mechanism IV. Reaction with a nucleophile can proceed via an SN1 mechanism A. I and II B. I, II, and III C. I, II, and IV D. I, II, III and IV
14) D; If you decided that you had to choose between Statements III and IV because they appear to be opposites and therefore only one must be true, you need to STOP using tricks. There is a difference between strategies and tricks. Strategies are ways of approaching questions and the test in general that will ALWAYS help you utilize the knowledge you do have in a more effective and productive way. Tricks are non-academic rules that aren't based at all on your knowledge (although they are often called 'strategies' by other prep companies). Tricks rely on trends or idiosyncrasies of the test writers. Kaplan and others use tricks a lot. In my experience, tricks work less than half the time and will drive you insane when they cause you to second-guess yourself and get a question wrong that you would have otherwise gotten right. The point? Ignore trends and evaluate every option equally, always keeping your mind open. All four of these statements are true of epoxides. They can either be protonated first and then proceed via SN1, or undergo back side attack from a nucleophile in a single-step SN2.
17. Which of the following structures will be the first intermediate formed during the decarboxylation of a beta-keto acid? A. a carbanion; formed by the abstraction of an acidic alpha hydrogen B. a resonance-stabilized carbanion C. an enolate ion; formed via removal of the -R group from the ester D. a carbon dioxide substituted ketone
17) B; The best way to answer this question is to draw out the structures and try to push electrons to make the described reaction occur. A base has to abstract the hydrogen on the acid and in one step the electrons collapse down forming another double bond to the carbon and breaking the bond between the carbonyl carbon and the alpha carbon. This leaves a carbanion on the alpha carbon that is stabilized by resonance onto the neighboring ketone. Answer A sounds logical because it could certainly happen, but it would just produce the original structure minus a hydrogen and would not be a step toward decarboxylation. Answer C makes no sense because the R group is not an electrophile and can't be abstracted like a hydrogen. Answer D is impossible because carbon dioxide cannot form another bond to attach as a substituent to anything. This leaves answer B, which is exactly what happens.
21. The first step in the aldol condensation of 2-butanone is the abstraction of a hydrogen from: A. carbon 1 B. carbon 2 C. carbon 3 D. carbon 4
21) A; Draw out 2-butanone and label the carbons as you would when naming the compound. Carbons 1 and 3 are the only alpha carbons, so answer D is obviously false. There isn't a hydrogen on carbon 2, so that's just silly. The most acidic of the two alpha hydrogen positions will be abstracted. To decide which is more acidic, look at the conjuagate bases. In the case of carbon 3, an R-group is attached that will weakly donate electrons making the carbanion even more unstable. In the case of carbon 1 no such R-gorup exists. Thus C1 will be the alpha that is abstracted. Resonance is unimportant because they both benefit from resonance to the carbonyl equally.
27. If an imine-substituted benzene is reacted with LiAlH₄, a primary benzyl amine can be produced. Which of the following will be an intermediate formed during this reaction? A. a carbanion B. a carbocation C. a nitrogen anion D. a nitrogen cation
27) C; As always, draw out the structure and do the proscribed electron pushing. Recall that the strong reducing agent is basically a hydride ion. The hydride ion will attack the partial-positively charged carbon and kick the double bond electrons up onto the nitrogen. This is best reflected by answer C, a nitrogen anion. The nitrogen is certainly not positively charged as in Answer D, and the carbon never bears a charge as indicated by Answers A and B.
28.
28) A; You may have never seen this reaction before, but don't sweat it. The MCAT will definitely ask you to do this. All you need to do is focus on correctly identifying the nucleophile and electrophile. Carbonyls are almost exclusively electrophiles when you see them. The sulfur in hydrogen sulfide is the nucleophile. It will attack the carbonyl carbon. The oxyanion will be protonated to create the good leaving group water. The first hydrogen on the sulfide is removed to quench the charge, and the collapse of the second one kicks off the water, forming a thioketone (i.e., R2C=S).
32. The compound (COCH₃)₃CBr undergoes substitution reactions that other bromides do not primarily because of: A. increased electron density around the carbon due to induction B. activation of the central carbon due to electron withdrawal C. increased acidity due to strong withdrawing groups D. steric hindrance around the bromine-bearing carbon
32) B; The carbon is a very strong electrophile because its electrons are being withdrawn thru all four bonds. This makes it highly reactive. Answer B best describes this fact. Answer A is false because electron density around the carbon would only increase if DONATING groups were attached. Answer C is false because there are no acids present. Answer D is false because steric hinderance interferes with reactivity; we're looking for a reason this bromide is more reactive than others.
36. A chemist is attempting to identify an unknown. He discovers that the unknown is soluble in dilute acid, but not dilute base. The unknown most likely contains which of the following functional groups? A. carbonyl B. carboxylic acid C. amine D. ether
36) C; Remember this question! It is a point that isn't usually emphasized in O-Chem class. When trying to identify an unknown, something that is soluble in acid, but not is base is SUSPECTED to be an amine. This is not going to be true all the time, but it is a good way to get started in the right direction. Meanwhile, something that is soluble in base, but not in acid, is SUSPECTED to be a carboxylic acid. Thus, C is the best answer.
37. All of the following functional groups can be reduced to a primary amine when reacted with NaBH₄, EXCEPT: I. RNO₂ II. R₂C=NH III. RCONH₂ IV. RC≡N A. III only B. I and III C. I and IV D. II and III
37) D; Nitros and nitriles can be reduced by all reducing agents. Imines require catalytic hydrogenation ONLY and amides require LiAlH4 ONLY. Just memorization here, but you don't want to miss such an easy question!
6. A student is examining an unknown. She is told by her lab instructor the the unknown contains both alkene and carbonyl functional groups. Which of the following will not produce at least one alcohol functional group when reacted with the unknown? A. HgOAc₂/NaBH₄ B. BH₃/H₂O₂ C. H⁺/H₂O (hot, concentrated) D. CH₃MgBr
6) C; Answer A is true because oxymercuration/demercuration will turn an alkene into an alcohol. Answer B is true because hydroboration will also produce an alcohol (although a different one) from an alkene. Answer D is true because Grignard reagents react with carbonyls to produce alcohols. Answer C is false because although acid can turn an alkene into an alcohol, it must be cold, dilute acid; hot, concentrated acid favors the alkene.
8. All of the following statements are true regarding a molecule of benzene, EXCEPT: A. It is an example of a conjugated pi system B. It has a plane of symmetry C. During reactions, electron donating groups direct new substituents to the ortho and para positions D. It will undergo addition reactions with several electrophiles
8) D; Answer A is true; the p orbitals above and below the ring structure combine to form a completely conjugated system that looks much like two donuts with the ring structure sandwiched between them. Answers B and C are also both true. You should know that electron withdrawing groups direct new substituents to the meta position. Answer D, however, is false. You will never see an addition to a conjugated system because it will disrupt the resonance. Benzene will, however, undergo substitution reactions.
Describe the reaction with α-β Unsaturated Carbonyls. What are the products/reactants? What are the steps of the reaction?
A carbonyl with a double bond between the alpha and beta carbons. A nucleophile will bond to the beta carbon and an alcohol will form from the carbonyl oxygen. Two possible resonance structures exist. Steps: There are two possible ways to visualize this mechanism, based on which resonance form you start with: 1) WIth teh double bond between the alpha and beta carbons, the nucleophile attacks the beta carbon, pushing the double bond over one carbon and forcing the C=O electrons up onto the oxygen. 2) With a carbocation on the beta carbon, the nucleophile simply attacks the beta carbon, quenching the charge (least contributor since there is a charge in the resonance structure). 3) Starting with either resonance form, the oxygen will get protonated to form an alcohol.
With substitution reactions of acid derivatives how do you determine which substituent will leave (-NH₂, -Cl, -OH, -OCOR, -OR)? Why? List the stability of carboxylic acid derivatives in order (Acid Chloride, Carboxylic Acid, Amide, Ester, Anhydride). How do you determine which is more stable?
A nucleophile can be added to any carboxylic acid or acid derivative and it will attack the carbonyl carbon. Only SOMETIMES however, will the result be substitution of that nucleophile for the existing substituent. Recall that the intermediate is an oxygen anion that collapses down to re-form the carbonyl. Which substituent will leave as a result of this collapse depends solely on its quality as a leaving group. Often, the nucleophile is the better LG, so the original acid derivative is simply reformed. Ranking of LGs (best to worst): -Cl > -OCOR > -OH > -OR > -NH₂ A strong base makes a poor leaving group. The equilibrium will favor the formation of the compound whose leaving group is a strong base. Stability of Carboxylic Acid Derivatives: This pattern is the exact opposite of the previous one-the better the leaving group the more unstable the acid derivative. Amide > Ester > Carboxylic Acid > Anhydride > Acid Chloride
How do you determine the strength of an acid?
ALWAYS look at the stability of the conjugate base! The stability, or lack thereof, will often be affected by electron donating or withdrawing groups (i.e., alcohols are weaker acids that water due to the donating effect of the -R group). Resonance Stabilization: This explains why the OH group on a carboxylic acid is more acidic than than other hydroxyl groups and why alpha hydrogens are acidic.
What do aldehydes and ketones function as (nucleophile, electrophile, base, acid)? Which is more acidic, the alpha hydrogen of a ketone, or the alpha hydrogen of an aldehyde? Why?
Aldehydes and Ketones function mostly as electrophiles, when their carbonyl carbon is attacked by a nucleophile. They also function as Lewis Acids, accepting electrons when a base abstracts an alpha hydrogen. Because alkyl groups are electron donating and a ketone has two alkyl groups attached to the carbonyl, the carbonyl carbon of the conjugate base of the ketone is less able to distribute negative charge and is slightly less stable than that of an aldehyde. Thus aldehydes are slightly more acidic than ketones. Both aldehydes and ketones are less acidic than alcohols. Any electron withdrawing groups attached to the α-carbond or the carbonyl tend to stabilize the conjugate base and thus increase acidity.
Describe amine basicity, which amines (primary, secondary, etc.) are basic, nucleophilic, electrophilic? How does basicity decrease (primary, secondary, etc.), and why?
All primary amines and some secondary amines are nucleophilic; all tertiary amines are basic. Basicity decreases from tertiary to secondary to primary to ammonia due to the electron donating effects of the R-groups. Quaternary amines act as electrophiles (as long as they have at least one hydrogen).
Describe the stability of amides, are their carbonyl carbons reactive or unreactive?
Amides are the MOST stable of all acid derivatives. For the MCAT, you can consider their carbonyl carbons as unreactive.
What is an amine? Do they act as bases, nucleophiles, or electrophiles?
Amines are derivatives of ammonia, wherein one or more hydrogen atoms have been replaced by a substituent such as an alkyl or aryl group. Important amines include amino acids, biogenic amines, trimethylamine, and aniline. Amines can act as either bases or nucleophiles. Primary or secondary amines usually act as nucleophiles and tertiary amines always act as bases (because they are too sterically hindered to act as nucleophiles).
What is an amide?
An amide is any compound containing a carbonyl with an amine substituent on the carbonyl carbon.
When predicting Reactions and Products, what are the three things to remember based on carbocations, steric hindrance, and counting carbons (which reactions add to or take away from the carbon chain)?
Carbocations: The mechanism will always proceed thru the most stable carbocation (unless peroxide is present). Carbocation stability = tertiary > secondary > primary 2) Steric Hindrance: If more than one mechanism or electrophile or nucleophile is possible, the one that involves the LEAST steric hindrance will be favored. 3) Count Your Carbons: Be sure that the product has the correct number of carbons. Remember that many nucleophiles and other reactants ADD TO, or TAKE AWAY from the carbon chain (i.e., Grignard Synthesis, Hofmann Degradation, Aldol Condensation, Acetoacetic Ester Synthesis).
What type of a reaction do carboxylic acids and their derivatives (acid chlorides, esters, amide, anhydrides) prefer? What type of a reaction do aldehydes and ketones prefer? (nucleophilic substitution/addition)
Carboxylic acids and their derivatives undergo nucleophilic substitution; aldehydes and ketones prefer nucleophilic addition.
Describe the Bp and the Mp of carboxylic acids. Describe the solubility of carboxylic acids.
Carboxylic acids are able to make strong double hydrogen bonds to form a dimer. The dimer significantly increases the boiling point of carboxylic acids by effectively doubling the molecular weight of the molecules leaving the liquid phase. Without long alkyl chains, they are soluble in water. Surprisingly they are ALSO soluble in non-polar solvents (although they are clearly polar) because the dimer form allows the carboxylic acid to solvate without disrupting the hydrogen bonds of the dimer.
Would you expect amides to be more or less basic than comparable amines? Why?
Compared to amines, amides are very weak bases. While the conjugate acid of an amine has a pKa of about 9.5, the conjugate acid of an amide has a pKa around -0.5. Therefore amides don't have as clearly noticeable acid-base properties in water. This lack of basicity is explained by the electron-withdrawing nature of the carbonyl group where the lone pair of electrons on the nitrogen is delocalized by resonance. On the other hand, amides are much stronger bases than carboxylic acids, esters, aldehydes, and ketones.
How do you determine the strength of a base?
Electron donating groups increase basicity, while electron withdrawing groups decrease basicity.
Do esters H-bond? Are they soluble in water? What is their water solubility compared to acids or alcohols?
Esters act as H-bond recipients, but not donors. Without long alkyl chains, they are slightly soluble in water; but less soluble than acids or alcohols.
Describe the Acetoacetic Ester Synthesis reaction. What are the reactants/products? What are the steps of the reaction?
Formation of a ketone from a β-keto ester. Steps: 1) A base abstracts the acidic alpha hydrogen, leaving a carbanion. 2) The carbanion attacks an alkyl halide (R-X), resulting in addition of the -R group to the alpha carbon. 3) Hot acid during workup causes loss of the entire -COOR group.
What is a Saponification (hydrolysis) of an Ester? What are the reactants/products? What are the steps?
Hydrolysis of an ester to yield an alcohol and the salt of a carboxylic acid. Steps: 1) The hydroxide ion (NaOH or KOH) attacks the carbonyl carbon and pushes the C=O electrons up onto the oxygen. 2) The electrons collapse back down and kick off the -OR group. 3) Either the -OR group, or hydroxide ion, abstracts the carboxylic acid hydrogen, yielding a carboxylate ion. This associates with the Na⁺ or K⁺ in the solution to form "soap."
Describe the nomenclature is the hydrogen is missing from a carboxylic acid, leaving a negative charge on the oxygen. What is the nomenclature if a salt is formed by removing the hydrogen and replacing with a metal?
If the hydrogen is missing, leaving a negative charge on the oxygen, it is named with an "-ate" ending (i.e., formic acid→formate). If there is a salt formed between the carboxylate ion and a metal, name the metal first, then the ion (i.e., benzoic acid → benzoate → sodium benzoate).
Describe the common reducing agents and which are used to reduce amides, imines, nitriles and nitro groups to amines?
LiAlH₄, NaBH₄, and H₂/pressure Nitro groups: reduce to the associated primary amine via all of the normal reducing agents Nitrile groups: reduce to the associated primary amine via all of the normal reducing agents Imines: reduce to the associated primary amine via catalytic hydrogenation ONLY (Requires a heterogenous catalyst such as H₂-Nickel). Amides: reduce to the associated primary amine via LiAlH₄ ONLY
Describe how the melting point and boiling point changes with amines due to molecular weight and branching. Describe the solubility of amines.
MP and BP increase with MW and decrease with branching. Amines without long alkyl chains are very soluble in water due to hydrogen bonding.
Describe the polarity, BP, and solubility properties of aldehydes and ketones.
More polar and thus higher Bp than equivalent alkanes; Act as H-bond recipients, but not donors. Lower Bp than water. Soluble in water unless they have a long alkyl chain.
Describe the mechanism for the synthesis or alkyl amines.
NH₃ + CH₃Br → NH₂CH₃ Formation of an alkylamine from an amine and an alkyl halide. Steps: 1) Ammonia acts as a nucleophile, attacking the alkyl halide via SN2 and kicking off the halide ion. (Since the reaction is an SN2 reaction, the carbon cannot be a tertiary carbon.) 2) The halide ion acts as a base, abstracting a hydrogen to quench the charge on the nitrogen. Note: This reaction results in many side products because the resultant amine is still a good nucleophile and can react again.
How do you name amides?
Named by adding the "-amide" ending to the "-ic" ending of the corresponding carboxylic acid (i.e., benzoic acid → benzoamide).
Can amides hydrogen bond?
Primary and secondary amides can hydrogen bond and are thus water soluble as long as they lack long alkyl chains. Tertiary amides cannot H-bond and are thus considered to have low water solubility.
What competes with SN1? What competes with SN2? What are the major factors that determine which mechanism will be used? A 3⁰ carbon is required to undergo what reactions?
SN1 competes with E1 and SN2 competes with E2. The major factors that determine which mechanism will be used are: the solvent, steric hindrance and the strength of the nucleophile or base. 1) E1: favored by weak bases; 3⁰ carbons only; polar protic solvents (stabilize the intermediate) 2) SN1: favored by poor nucleophiles; 3⁰ carbons only; polar protic solvents (stabilize the intermediate) 3) E2: favored by strong bases 4) SN2: favored by good nucleophiles; methyl, 1⁰, or 2⁰ carbons
Describe the Halogenation of an Aldehyde or Ketone reaction. What are the steps of the reaction?
Substitution of a Br, Cl or I for one of the alpha hydrogens on an aldehyde or ketone. Multiple halogenations often occur. Steps: 1) A base abstracts an alpha hydrogen, leaving a carbanion 2) The carbanion attacks a diatomic halogen
Who makes stronger hydrogen bonds, alcohols or amines?
The N-H bonds in amines are somewhat polar. As we might guess from considering electronegativities (estimated from positions in the periodic table), the N-H bond is more polar than the C-H bond and less polar than the O-H bond. This polarity shows up in a comparision of physical properties of amines and alcohols. The simplest examples are water and ammonia. Water boils at 100oC, while ammonia boils at -33oC. This is interpreted to mean that it takes a good deal more energy to boil water than ammonia. Correspondingly, the forces between molecules of water (which resist separating them in the boiling process) are much stronger than those between molecules of ammonia. The most important of these forces is called the hydrogen bond.
Why are alpha hydrogens on the methyl side of a methyl-ethyl ketone more acidic than those on the ethyl side?
The ethyl group is electron donating, thus decreasing the acidity and increasing the basicity. Therefore, the alpha hydrogens on the methyl side will be more acidic because there is not an electron donating group attached to the carbon.
How does polarity increase in a bond? Which atom forms stronger hydrogen bonds, oxygen or nitrogen? How does polarity help us tell the relative reactivity of the carbons in a C=O vs. a C=N bond? How does polarity help identify the nucleophile and electrophile in a reaction?
The greater the difference in the electronegativity of two atoms in a bond, the more polar the bond. This shows us that oxygen forms stronger hydrogen bonds than does nitrogen. The relative reactivity of a C=O bond is greater than a C=N bond since there is a greater difference in electronegativity. Polarity results in partial positive and partial negative charges, which are clues when trying to identify the nucleophile and electrophile in a reaction. To be a base or nucleophile, it MUST have a full or partial negative charge. To act as an electrophile, it MUST have a full or partial positive charge.
Describe the decarboxylation reaction. What are the reactants/products? Describe the steps for the reaction and the energy of activation and enthalpy.
The loss of a CO₂ molecule from a beta-keto carboxylic acid, leaving behind a resonance-stabilized carbanion. The process usually requires catalysis by a base. The carboxylate ion usually takes the hydrogen back from the base, forming a keto-enol tautomer. Decarboxylation has a very high Ea and is highly exothermic.
How do you determine if a molecule/atom will act as an electrophile?
This is the easy one. Electrophiles are ALWAYS electron poor. They will always have a full or partial positive charge. DO NOT start a reaction arrow at an electrophile. Electrophiles always get attacked by electron rich species, NOT the other way around!
