Organic Final All – Flashcards

double bond is shorter and stronger than an alkene. 1) Partial Positive charge on the carbonyl carbon. Good electrophile. 2) Alpha hydrogens-hydrogens on the alpha carbon are surprisingly acidic-especially given that alkane hydrogens normally cannot be removed. 3) Electron donating/withdrawing groups. Donating groups decrease the reactivity of the carbonyl carbon and withdrawing groups increase its reactivity. 4) Steric hindrance-Bulky substituents attached to the carbonyl carbon decrease its reactivity 5) Planar Stereochemistry-sp² hybridized carbonyl carbon is planar and can therefore be attacked from either side. When the two substituents are NOT identical, an addition reaction could therefore create both R and S enantiomers in a racemic mixture.

Acidic due to resonance stabilization of the conjugate base. When there are two carbonyls separated by a single carbon, hydrogens on the middle carbon are even more acidic. The greater the partial positive charge on the carbonyl carbon , the more acidic its alpha hydrogens will be.

4. These structures differ in the substituents attached to the carbonyl carbon. An electron donating group on the carbonyl carbon will decrease its partial positive charge and thereby make it less able to stabilize the conjugate base. An electron withdrawing group will make the carbonyl better at stabilizing the conjugate base and therefore the strongest electron withdrawing group would indicate the strongest acid (because it produces the most stable conjugate base). Choice a) is the only electron withdrawing group, so it will have the most acidic alpha hydrogens. The weakest donating group of the other three is the methyl group on choice d). Amines (choice c) and hydroxyl groups (choice b) are both electron donating, but amines are better donating groups because nitrogen is less electronegative than oxygen. Therefore, in increasing order of acidity: c < b < d < a.

-al ending Aldehyde carbons are always considered carbon # 1 for numbering purposes. If a ketone must be named as a substituent it is called an oxo group, as in 4-oxopentanal.

HCOH

CH?OH

C?H?OH

CH?COH?

Alkanes are non-polar and therefore insoluble in water. Aldehydes and ketones can act as hydrogen-bond acceptors when dissolved in water, with water acting as the hydrogen bond donor. Therefore, aldehydes and ketones will be far more soluble than alkanes. Finally, alcohols can hydrogen bond as both a donor and acceptor with water, so they will be the most soluble. These trends assume comparable molecular weight, chain length, etc. This is important because a small alcohol such as methanol is miscible with water, but dodecanol is considered insoluble. The boiling point of alkanes will be the lowest because their only intermolecular attraction would be van der Waals forces. Aldehydes and ketones do NOT hydrogen bond with one another, but they are both polar and will therefore have much higher boiling points than alkanes. Finally, alcohols will have the highest boiling points due to intramolecular hydrogen bonding. Aldehydes and ketones can act as H-bond recipients, but NOT as H-bond donors.

In order for a substitution to occur there must be a leaving group. Acid derivatives all have leaving groups: Cl- in the case of acid chlorides, -OH in the case of carboxylic acids, a carboxylate ion in the case of anhydrides, etc. The stability of these groups after they leave varies widely, but in the case of an aldehyde or ketone there are no groups that would be reasonably stable and therefore no candidates to act as leaving groups. The aldehyde hydrogen will not leave as H:-, nor will an R group leave from a ketone as a carbanion R:- (excluding rare cases where substituents on the R group dramatically change its properties—as in -RX3). In both cases the leaving group would become a strong base. Recall that strong bases never make good leaving groups. Good leaving groups must be weak bases that are stable after they leave. For this reason, aldehydes and ketones only undergo addition reactions—lacking a suitable leaving group to undergo substitution.

Also, in white notebook for all of the acid and base catalyzed reactions with keto and enol forms

pg. 340 also in white notebook In reaction mechanism pocket-book pg. 63

Ketones or aldehydes can be prevented from reaction with a nucleophile or base by conversion to an acetal or ketal (which are unreactive in all but acidic conditions) Any terminal diol with at leas two carbons will work. 1) One end of the diol acts as the nucleophile described in steps 1-2 above 2) The other end of the diol acts as the second equivalent of alcohol described in steps 3-4 above. 3) Acidic conditions will return the acteal/ketal to the original aldehyde/ketone. Also, in white notebook

Substitution of a Br, Cl, or I for one of the alpha hydrogens on an aldehyde or ketone. Multiple halogenations often occur. 1) A base abstracts an alpha hydrogen, leaving a carbanion. 2) The carbanion attacks a diatomic halogen. Reaction type : Nucleophilic substitution Summary Under acidic reaction conditions, selective monohalogenation at the a-postion is observed. Chlorination, bromination and iodination all occur at the same rate. The reaction can be catalyzed by addition of acid, but it is actually autocatalytic as the by-product is the hydrogen halide. The reaction proceeds via the enol tautomer.

When reaction is formed on a methyl ketone with sufficient halogen present to effect replacement of all three alpha hydrogens. 1) Complete the reaction above (a-halogenation of Aldehyde or Ketone) 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 transformed into a decent leaving group. When a strong hydroxide base, such as NaOH, is added, the OH attacks the carbonyl carbon, kicking the electrons from the C=O bond up into 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.

In white notebook, lots of examples. MCAT likes to give final product and ask what reactants are possible to form this product. 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

base catalyzed halogenation

The keto and enol forms are in an equilibrium with one another that strongly favors the keto form at room temperature. The keto form is more stable because the sum of its bond energies is greater than the sum of the bond energies in the enol form. The keto form has a C=O bond, a C-C bond, and a C-H bond that are replaced by a C-O bond, a C=C bond, and an O-H bond in the enol form. C-H and O-H bonds are quite close in bond energy, while a C=C has about 250 kJ/mol more bond energy than a C-C bond (not quite double). The real difference comes in the difference between a C-O bond and a C=O bond. A C=O bond has about 450 kJ/mole more bond energy! Of course, you are not expected to know these numbers—we just included them here to illustrate the point. What you should know is that carbonyl bonds are much shorter and stronger than alkene bonds. That is the most significant difference between the two forms and is the reason the keto form is favored

11. A 1,3-dicarbonyl can undergo an intramolecular hydrogen bond when one of the carbonyls is in the keto form and the other is in the enol form. This significantly stabilizes the enol compared to a stand-alone enol. In this condition the enol is acting as the hydrogen-bond donor and the carbonyl as the hydrogen bond acceptor. The MCAT loves alpha hydrogens so much, it wouldn't be right to mention 1,3-dicarbonyls without also pointing out that they have ultra-acidic alpha protons on the carbon between the two carbonyl carbons—double resonance stabilization!

12. If the halform ion (CX3:- ) were removed from the reaction mixture as it was formed, the carboxylic acid would remain intact. Normally, CX3:- acts as a base and abstracts the acidic proton from the acid as soon as it is formed.

Usually end product on Aldol condensation

In terms of the MCAT, you should think of an ?-?-unsaturated carbonyl as an electrophile. Nucleophiles either attack the beta carbon directly, or attack the carbocation that exists in the second resonance form. Nucleophiles that can't act as leaving gourps stay attached to the carbonyl carbon (hydrides and grignard reagents). Nucleophiles that can act as leavinggroups attack carbonyl carbon but get kicked back out and eventually attack the beta position and stay (amines, enolates, nitriles etc.). It might be tempting to think of the double bond between the alpha and beta carbons as a nucleophile that will undergo electrophilic addition. However, the withdrawing effect of the carbonyl decreases the electron density of the double bond deactivating it toward electrophilic addition. 1) With a 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 directly -starting with either resonance form the oxygen will get protonated to form an alcohol. Note that the protonated oxygen is really just the enol form of a keto-enol tautomer.

Transforms a carbonyl into an alkene The product will be an alkene, with the double bond formed between the carbonyl and the carbon to the polyphenyl group.

naming oic ending A carboxylate ion is the result of abstraction of a proton, leaving a negative charge on the oxygen. Carboxylate ions are named with an ate ending (formic acid?formate). If it is a salt formed between the carboxylate ion and a metal, name the metal first, then the ion (benzoic acid?sodium benzoate)

have very high boiling points due to their ability to form strong dimers involving two hydrogen bonds. Without long alkyl chains, they are soluble in water. Surprisingly, short-chain carboxylic acids are also soluble in many relatively non-polar solvents, such as chloroform (even though they are clearly polar).

15. Theoretically, the carboxylic acid dimer (pictured below) would have no net dipole moment. This explains its solubility in non-polar solvents.

1) Resonance stabilization: The carboxylate ion is uniquely stable due to resonance stabilization. 2) Induction: Pay careful attention to alpha substituents; they can either donate or withdraw from the carboxylate ion, increasing or decreasing acidity. Remember: To predict acidity examine the stability of the conjugate base! 3) Hydrogen bonding: It is worth mentioning twice; don't forget that carboxylic acids can not only hydrogen bond, but do it twice-to form dimers.

RCOOH + H?O?RCOOH?? + Nu:??RCONu +H?O

the loss of CO? molecule from a beta-keto carboxylic acid, leaving behind a resonance-stabilized carbanion. The process usually requires catalysis by a base.

Without a base catalyst, this mechanism can also be visualized as a 6-member, concerted, "ring-like" intramolecular reaction that does not require protonation of the enolate ion. Help your students visualize how his might happen. The carboxylate ion usually takes the hydrogen back from the base, forming a keto-enol tautomer.

RCOOH + ROH?RCOOR +H?O Reaction of an alcohol with a carboxylic acid to form an ester. The reaction requires an acid catalyst to form a good leaving group (the water). Higher yields can be obtained by reacting an anhydride with an alcohol.

any compound containing a carbonyl with a chlorine substituent on the carbonyl carbon. -oyl chloride, as in propanoyl chloride.

not an actual acid chloride. It is a chlorine attached to a benzyl group.

RCOOH +PCl??RCOCl +H?O Three reagents readily produce acid chlorides when added to carboxylic acids: PCl?, PCl?, and SOCl?

A chloride ion is capable of attacking the carbonyl carbon of a carboxylic acid. However, when the electrons in the carbonyl bond are kicked up onto the oxygen, and then collapse back down, the substituent that is the best leaving group will leave, regardless of the original structure of the molecule. Chloride ion is more stable (i.e., it is a weaker base) than hydroxide ion, so the chlorine will be kicked off to reform the acid.

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, and 2) the fact that chloride ion is a superb leaving group.

a) ROH ; b) RNH2 ; c) RCOOH ; d) H2O

compound with two acyl groups connected to one another by a single oxygen. Viewed another way is an ester where the R group is a carbonyl. replacing the oic ending of corresponding carboxylic acid with oic anhydride (benzoic acid?benzoic anhydride) Mixed acid anhydrides are named alphabetically (ethanoic methanoic anhydride)

mixed anhydride made from ethanoic acid and methanoic acid.

are excellent electrophiles! The two carbonyl carbons are highly reactive to nucleophiles because the leaving group is a resonance stabilized carboxylate ion.

replacing oic ending of corresponding carboxylic acid with amide (benzoic acid?benzamide).

most stable of all acid derivatives. You can consider the carbonyl carbons unreactive for MCAT. This is because NH? is not a good leaving group.

can hydrogen bond are thus water soluble as long as they lack long alkyl chains.

cannot H-bond

the lone pair on the amide nitrogen resonates with the carbonyl double bond, giving both the C-O and the C-N bonds double bond character. This prevents rotation.

Because the nitrogen in an amide donates its electron pair via resonance, both the C-O and the C-N bond have double-bond character. Therefore the hydbridization of the nitrogen will be closer to sp2 than to sp3.

For the same reasons given in the previous question—namely the donation of the lone pair on the nitrogen into the conjugated system—nitrogen has less electron density in an amide than it would in a normal amine. Therefore, it will be less basic. One could also consider the effect of induction. The carbonyl carbon has a strong partial positive charge, which will withdraw electron density from the amine through the sigma bond.

Primary amides (amides with only hydrogens on the nitrogen) react in strong, basic solutions of Cl? or Br? to form primary amines. The mechanism includes decarboxylation and thus shortens the length of the carbon chain. This reaction is important because it allows you to add an amine to a tertiary carbon!

a carbonyl with an -OR group substituted on the carbonyl carbon. named with -oate ending, with the R portion of the -OR group named and placed in front of the name, as is methyl pentanoate.

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.

Reaction of an existing ester with an alcohol, creating a different ester. Also requires acid catalysis. RCOOR? + R?OH?RCOOR? +R?OH

Hydrolysis of an ester to yield an alcohol and the salt of a carboxylic acid 1) The hydroxide ion (NaOH or KOH) attacks the carbonyl 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. Pg. 344

Formation of a ketone from a ?-keto ester. 1) A base abstracts the acidic alpha hydrogen, leaving a carbanion. 2) The carbanion attack 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.

specifically referring to oxo acids such as phosphoric acid, sulfuric acid, and nitric acid, plus their associated esters. I need to be able to draw each of these esters. Other Examples: ATP, GTP, UTP are examples of inorganic triphosphate esters. FADH? and NADH are examples of diphosphate esters. FMN, DNA, and RNA are examples of monophosphate esters.

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 attacking nucleophile is the better LG, so the original acid derivative is simply reformed.

-Cl>-OCOR>-OH>-OR>-NH?

Chlorine is the best leaving group because it is a weak base that is remarkably stable while holding a formal negative charge. This ability to hold a charge is due to the size of chlorine's electron cloud. The negative charge is spread out across a much larger distance than it would be for a smaller atom. For this reason, bromine would be an even better leaving group, and iodine better yet. The -OCOR group is the leaving group portion of an anhydride—the portion kicked off when a nucleophile attacks one of the carbonyl carbons. It is a carboxylate ion stabilized by resonance. That stabilization makes it much more stable than other leaving groups that place a negative charge on an oxygen. Hydroxide and alkoxide are best compared together. It should be noted that under acidic conditions, hydroxide gets the nod as the better leaving group because -R groups are weak donating groups. Hydrogen, by comparison is considered neither donating, nor withdrawing. Therefore, the -R group on an alkoxide donates more electron density to an oxygen that already bears a full negative charge. Under basic conditions the acidic hydroxide gets deprotonated and does not act as a leaving group, therefore only the alkoxide RO- can act as a leaving group. Finally, an amine makes a relatively awful leaving group because when it leaves it forms the strong base NH2-. This is a stronger base than hydroxide or alkoxide and therefore is the most unstable bearing a full negative charge. This is the reason that amides are the least reactive of the acid derivatives.

pattern is exact opposite of reactivty-the better the leaving group the more unstable the acid derivative. Amide ; Ester ; Carboxylic Acid ; Anhydride ; Acid Chloride

can act as a 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)

Basicity decreases from tertiary to secondary to primary to ammonia due to the electron donating effect of the R- groups. Amines attached to aromatic rings are significantly less basic than standard amines.

Aromatic amines are less basic because they donate their electron pair into the ring forming a conjugated system with the ring.

(ammonium) act as electrophiles (as long as they have at least one hydrogen)

of hydrogen bonding. This has the expected impact on boiling point solubility.

Protein synthesis

1) Name the alkane to which the N is attached (propane) 2) Add amine in place of the e on the end of ane (propananime). It is also acceptable to separate the substituent name (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 (N-ethylpropanamine) 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-methlypropanamine, or N,N-dimethyl-N-ethylpropanamine)

An enamine and imine interchange via a proton shift analogous to the keto-enol tautomerization: Pg. 346

NH?+ CH??NH?CH? + HBr Formation of an alkylamine from an amine and an alkyl halide 1) Ammonia acts as a nucleophile attacking the alkyl halide via SN2 and kicking off the halide ion. 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.

Formation of a primary amine from a primary alkyl halide; avoids the side products of alkyl amine synthesis. The pthalimide ion, a reactive species with a full negative charge on the nitrogen, acts as a nucleophile, attacking the alkyl halide via SN2. A single hydrazine molecule will react at both carbonyl carbons to release a primary amine and create the closed-ring diamide product shown. See the diagram below for electron flow: Pg. 346

Nitro Groups-can be reduced by LiAlH?, NaBH?, and H?/catalyst with pressure. Most O-chem boos focus on nitro groups being reduced by metals in HCl (M*HCl M=Fe, Zn, Sn)

Nitrile Groups: LiAlH?, NaBH?, and H?/catalyst with pressure. However, focus on nitriles being reduced by LiAlH?.

Imines: LiAlH?, NaBH?, and H?/catalyst with pressure. Most books focus on imines being reduced by NaBH?CN or H? and a catalyst.

Amides: Only LiAlH?

Amines add to aldehydes and ketones to form imines and enamines. 1) The amine acts as a nucleophile, attacking the electrophilic carbonyl 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 enamine (depending on the substitution pattern of the nitrogen)

imines

yield enamines

do not react

27. Tertiary amines are not good nucleophiles because they are too sterically hindered. They are more likely to act as a base. Even if they were to attack the carbonyl carbon, this would form an unstable quaternary amine with a full positive formal charge. This would be a better leaving group than the water formed by protonation of the carbonyl and would therefore be kicked back off anyway

Complete reduction of an aldehyde or ketone to an alkane via imine intermediate. Starting wit hydrazine as the amine, the product is an amine-substituted imine. Subsequent addition of a hot, strong base (KOH/heat) replaces the imine with two hydrogens, yielding an alkane. 1) CH?COCH? + H?N-NH??imine 2) Imine + KOH/?? CH?CH?CH?

Formation of an alkene via elimination of an amine. Important because it yields the LEAST substituted alkene.

The greater the difference in electronegativity of two atoms in a bond, the more polar the bond. This shows us that oxygen forms stronger hydrogen bonds than does nitrogen.

Comparing electronegativity, we see that the partial positive charge on the carbon in a C=O bond is greater than the partial positive charge on the carbon in a C=N bond. This indicates that carbonyls are more reactive than comparable imines.

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 with 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 or withdrawing 4) Alkenes are weakly electron withdrawing

Alkyl groups are weakly electron donating; nitro groups are strong withdrawing groups; cyano groups are electron withdrawing; sulphones are electron withdrawing (assuming connection to the sulfur); amines are electron donating; carboxylic acids are electron withdrawing, esters are electron withdrawing (assuming connection to the carbonyl; if connected to the oxygen of the -OR group, an ester is electron donating); alcohols are electron donating; quaternary amines are strong electron donators.

Looking at the stability of the conjugate base, we see that a carbanion on the alpha carbon of a methyl would experience electron withdrawal due to the carbonyl, but no additional inductive effects. However, if we examine ethyl side we see that the same carbanion on that alpha carbon would experience withdrawal due to the carbonyl, but also induction by a weakly donating methyl group (i.e., the beta carbon). This would destabilize the carbanion.

Always look at the stability of the conjugate base! This stability of lack thereof, will often be affected by electron donating or withdrawing groups (alcohols are weaker acids than water due to the donating effect of the -R group). Resonance Stabilization: This explains why the alcohol group on a carboxylic acid is more acidic than other hydroxyl groups and why alpha hydrogens are acidic.

Electron donating groups increase basicity, while electron withdrawing groups decrease basicity.

Steric hindrance favors basicity over nucleophilicity; nucleophiles must have very little hindrance. Primary nucleophiles are most common. Secondary atoms can often act as bases or nucleophiles, depending on conditions. However, tertiary atoms will ONLY act as BASES. Reactivity (low stability) favors basicity over nucleophilicity (NH?? vs RO?). If an atom has a full negative charge it will almost always act as a base (halide are notable exception).

are always electron poor. They will always have a full or partial positive charge. Electrophiles always get attacked by electron rich species, NOT the other way around!

1) Carbocations: The mechanism will always proceed through the most stable carbocation (unless peroxide is present). Carbocation stability =tertiary>secondary>primary 2) Steric Hindrance: If more than one mechanism, intermediate, electrophile, or nucleophile is possible, the one that involves the LEAST steric hindrance will be favored. 3) Count your Carbons: (eg grignard synthesis, Hofmann degradaton, aldol condensation, acetoacetic ester synthesis)

The most powerful influence in determining boiling point is hydrogen bonding, which is an intermolecular force. Branching certainly has an influence, but more so on melting point. You may be tempted to think "but if molecular weight differences are HUGE, then hydrogen bonding won't matter." This may be true, but to consider which is the greatest influence, we must ask ourselves: If there is a large difference in molecular weight, could a small difference in hydrogen bonding have an affect? The answer is yes. However, if there is a large difference in hydrogen bonding, a small difference in molecular weight will NOT have an effect. You may be more familiar with this than you realize. Consider water, which has a very small mw of 18g/mol and a Bp of 100 degrees Celsisus. Then consider how many much, much larger mw hydrocarbons have lower boiling points.

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

First, draw out the structure described. Cold, dilute acid will turn the alkene into an alcohol, but will allow for rearrangement if it produces a more stable carbocation. After the alkene is protonated, a carbocation will exist on carbon 3. A methyl shift will turn this secondary carbocation into a tertiary carbocation and addition of water followed by removal of the acidic hydrogen results in answer D. You may be tempted to do a hydrogen shift instead, but note that there isn't actually a hydrogen on carbon 4.

You may be tempted to think that two equivalents of oxidizing agent added to an alcohol always yields the associated carboxylic acid. However, this is only true for primary alcohols. Thus, the secondary alcohol is simply oxidized to a ketone, Answer D.

You should recall that strong reducing agents such as Lithium Aluminum Hydride essentially produce basic hydride ions that can attack electrophiles. The carboxylic acid will be attacked by the hydride pushing the double bond up onto the oxygen as a lone pair. In any protic solvent this will be protonated to form a hydroxyl group. A gem diol is an exact description of the intermediate, two hydroxyl groups attached to the same carbon. Recall that two hydroxyl groups on neighboring carbons is called a vic diol.

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.

Carboxylic acids, due to resonance stabilization of the conjugate base, are the most acidic on the list. This narrows it to answers B or D because we are listing them in INCREASING order. Aldehydes are less acidic than either water or alcohols and that may be something you'll just need to know. Water and alcohols are similar in acidity, but you can compare them by looking at the conjugate bases. The alcohol's conjugate has an electron donating group destabilizing the negatively charged oxygen, making it less stable and therefore less acidic.

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. Some students have pointed out that you can also draw the mechanism without forming the carbanion, by pushing the electrons into the bond between the carbonyl carbon and the alpha carbon (forming a double bond) and pushing the carbonyl double bond up onto the oxygen. True. Your best clue that you shouldn't worry about that valid alternative is . . . it isn't one of the answer choices! Work with the exam and not against it. Students who try to understand how the author of the question likely intended it to be answered tend to significantly outperform students who pride themselves on finding exceptions.

; Answer B is false because tosyl chloride is an electrophile and cannot react with a carbonyl. Answer C is false because a grignard reagent actually reduces a carbonyl to an alcohol and is not easily reversed. Answer D is false because mesyl chloride, like tosyl chloride, is an electrophile. Answer A is the correct answer. The diol shown will form a cyclic structure with a carbonyl and then can easily be reversed by acid. Tosyl chloride is traditionally used to protect alcohols or other nucleophiles. The alcohol attacks the tosyl chloride and becomes tosylated.

This is an excellent exercise in the ever-important MCAT skill of being CAREFUL and thinking thru things CAREFULLY. If you are good at working thru things carefully, this should be a fairly easy question. First decide where the new bond could be formed to combine another ketone with 2-butanone to get the structure shown. The clue is to look at where the alcohol is located. The carbon to which the alcohol is attached USED TO BE the carbonyl. The bond that connects that carbon with the section of the structure that bears the other carbonyl is the new bond formed during the reaction. Separate the structure at this point, change the alcohol back to a carbonyl, and name it. You should get Answer C, 3-methyl-2-butanone.

You should instantly recall that if you want to turn a ketone or aldehyde into an alkene, you'll want to use the Wittig reaction. Recall that the carbonyl double bond and whatever is attached to the Ph3P-? fuse together in a new alkene bond and the oxygen is removed (it ends up back on the tri-phenol phosphorous species). An elimination reaction would make no sense because we have no good leaving groups. The Halofrom reaction would NOT produce an alkene and neither would Wolf-Kishner.

; 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.

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).

Draw out the structure and you'll see it's a simple gem-halide. It can react via SN2 with two alcohols to create an acetal. There are no acidic hydrogens on the product, so A is clearly false. Answer B may be tempting, but an alcohol addition results in an -OR group, not an -OH group. Finally, Answer C, a dienol has no logical connection to the two species reacting.

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.

Nitro groups, Nitriles, and Imines can be reduced to primary amines by all of the common reducing agents. Amides however, can only be reduced to a primary amine by so A is the correct answer. Just memorization here, but you don't want to miss such an easy question!

The reaction shown is a decarboxylation. The pericyclic transition state described in the stem is drawn below for reference. A basic catalyst is not required because the carbonyl double bond acts as the nucleophile, abstracting the hydrogen from the neighboring carboxylic acid to form an enol. The enol then converts to the keto form to yield the product shown. Carbon dioxide is the only other product and no additional reagents are required, so answer D is the best choice. Many students tend to think that additional hydrogen is necessary. However, the hydrogen for the enol comes from one of the carboxylic acids, and that same hydrogen then moves onto the alpha carbon when the enol converts to its keto form.

Draw out everything! You are answering one of the very last questions from one of the final chapters of this course. If you didn't draw out all four structures you may be in big trouble. Drawing out condensed structures, IUPAC names, even creating your own tables, flowcharts, etc., is an absolute must for the MCAT. Doing so helps your brain process the information and make the necessary connections. Once you draw out all four structures, you see that compound I, II, and III will not decarboxylate. Compound II cannot decarboxylate because there is no resonance to stabilize the carbanion (where the carbonyl should be we have a hydroxyl group instead). Compound III will not decarboxylate because the ketone is in the alpha position instead of the beta position. Compound IV will decarboxylate and the product will be a ketone plus carbon dioxide. The answer is therefore A: IV only.

Let this question be a prominent, all-encompassing example of MCAT O-Chem. This is the most common MCAT O-Chem question type. It illustrates exactly what we try to emphasize at the close of the O-Chem 3 lesson: namely, that the MCAT will present you with various reactions and require you to make accurate predictions. Predictions, predictions, predictions—you cannot overemphasize the need to develop a strong, confident ability to predict things such as: nucleophile, electrophile, place of attack, most reactive carbon, most acidic hydrogen, products, intermediates, alternate products given changes to the reactants, mechanism or conditions, and so forth. In this case, the nucleophile will be the amine nitrogen. The only logical electrophile candidates are carbon 2 or carbon 4. Based on the Altius rules for electron donating and electron withdrawing groups, we can see that the carbonyl carbon of a ketone is more reactive than the carbonyl carbon of an ester, telling us Carbon 2 will be the target of the nucleophilic attack. Whenever the product is given (which it often, if not usually, is on the MCAT) use it to your advantage. If you look carefully at the carbon backbone of the product you can see that an attack at Carbon 3 or 4 will not produce the product shown. A is therefore the correct answer. You do not need to be able to visualize or predict exactly how this ring-closing reaction proceeds. You DO need to draw out everything as we have been teaching you from the beginning. A very good trick on these complex molecules is to count carbons. Is there any chance that length is added to or taken away from the carbon chain in this case? In some cases the # of carbons could definitely change, but in this case it does not. Two 6-carbon molecules combine to form one 5-membered ring with a total of 12-carbons (plus nitrogen as a heteroatom) in the product. If you draw the nitrogen bound to the carbonyl carbon you can see that the methyl ketone is the 2nd carbon from the nitrogen. You can also see that the carbonyl carbon of the ester is three carbons away from the nitrogen. Both of these are also true in the final product. If you attempt to draw the nitrogen attacking the ester you see that these relative placements do NOT match. It can also be helpful to notice that the product still has TWO esters, but no ketones. Hmmm . . . that might be a good indication that the ketones are reacting and the esters are not. Did this seem like a hard problem? Did you take way longer than a minute to complete it? Don't get frustrated. Get with your tutor and work on similar problems. Do practice until you can feel confident doing this kind of work because you will see these problems on the MCAT and you will only have one minute to process them. You can do it!

Answering this question requires the ability to accurately predict the most likely mechanism for this reaction. You are greatly aided in this task by having the product before you. In the product we see two ester moieties. Before confirming it with additional info, it would be logical to assume those probably come from the one ester in the reactant and the other ester in the compound drawn above the reaction arrow. To test this assumption, label the carbons in both molecules. Then count the carbons carefully in the product. Take careful note to see if the nitrogen atom and other substituent are in the expected places with respect to the carbonyl carbons of each ester. These checks confirm that the two methyl groups are the "tails" left out of the ring structure as the two compounds combine to create the heterocycle. They do NOT both come from Carbon 1 because we do not combine two of the compounds above the reaction arrow to create Compound A. Rather, the second methyl group comes from the amine reactant shown. This makes D the only correct answer.

The ring-forming reaction involves two steps that both feature nucleophilic addition to a carbonyl with resultant loss of the carbonyl oxygen. That oxygen is lost after it is protonated twice to become the good leaving group water. Therefore, a complete list of products would include two equivalents of H2O.

For the ring-closing reaction to proceed, Electrons are donated from the nitrogen making carbon 3 a carbanion. The resulting carbanion then attacks carbon 5 to form a 5-membered hetero-ring that includes nitrogen. The base aids in the condensation of the Aldol like reaction. The main clues, once again, are given by the reactants and products. We suggest you tackle a question such as this by looking at the product and counting and/or labeling each atom. By doing so, almost without understanding the nucleophile or electrophile, you can predict what two atoms must be connected to create the product out of the given intermediate. Once again, those unreacted esters are an excellent clue. They remain in the product, so we know that the carbanion must attack carbon 5, not carbon 6. If it did attack carbon 6, the ring would be too large, and we would be missing an ester functional group.