Ch 18 Flashcards by James Berry

What do carbonyl resonance forms indicate regarding the dipole moments of ketones and aldehydes?

What hybridization is the carbonyl carbon?

Carbonyls have a large dipole moment because oxygen is more electronegative than carbon and bears a partial negative charge, while carbon bears a partial positive charge. Pi-bond electrons are not shared equally.

The carbonyl carbon is sp² hybridized. The three sigma-bonds are planar and oriented ~120 degrees apart.

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What is the IUPAC name?

2-hydroxycyclohexane-1,3-dione

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What is the IUPAC name?

cis-2,4-dimethylcyclopentanone

*Can also use R/S nomenclature for sterochem, but cis is allowed here*

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What is the IUPAC name?

3-oxocyclopentanecarbaldehyde

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What is the IUPAC and common name?

heptan-4-one

dipropyl ketone

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What is the IUPAC and common name?

propanone

(no need for propan-2-one because the ketone can only be in one spot)

acetone

(IUPAC also accepts “acetone”)

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What is the IUPAC name?

4-bromo-2-methylhexanal

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Rank the following carbonyl compounds in order of increasing equilibrium constant for hydration:

4 < 1 < 5 < 2 < 3

Steric hindrance* decreases hydrate formation.
Electron-withdrawing substituents* at the alpha-carbon increase hydrate formation.

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What is the priority of functional groups in naming organic compounds?

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What is the IUPAC name?

(1R,2R) 2-methoxycyclohexanecarbaldehyde

trans-2-methoxycyclohexanecarbaldehyde

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What is the IUPAC name?

6,6-dimethylcyclohexa-2,4-dienone

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What is the product of this Grignard reaction?

What does oxidation of the product create?

Grignard adds to the carbonyl carbon of the aldehyde to produce a secondary alcohol.

The alcohol intermediate can be oxidized, resulting in a ketone.

This reaction shows bleach as the oxidizing agent; “chromic acid” is the traditional lab reagent (H₂CrO₄).

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What reagents can be used to oxidize a primary alcohol to aldehyde AND prevent the over-oxidation to carboxylic acid?

PCC (Pyridinium chlorochromate) - traditional reagent ***know this one***

Sodium hypochlorite (NaOCl), i.e. bleach, with TEMPO

Swern oxidation (DMSO and COCl₂, hindered base like triethylamine in CH₂Cl₂)

Dess-Martin periodinane (DMP) reagent

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How does chromic acid oxidize a secondary alcohol into a ketone?

The mechanism involves the alcohol and chromic acid forming a chromate ester. Then, oxidation of the carbon atom and recution of the chromium atom results in a C=O bond and chromium leaving with an additional pair of electrons.

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What method can be used to turn an alkene into ketones and/or aldehydes?

Ozonolysis

BE CAREFUL: Oxidative Cleavage with Potassium Permanganate (KMnO₄) will give ketones and over-oxidize aldehydes into carboxylic acids

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What is(are) the product(s) of this reaction?

G = H, X, or activating group

Friedel-Crafts acylation can be used to make alkyl aryl ketones. Ring must not be deactivated.

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What is the product?

Friedel-Crafts acylation to produce a diaryl ketone.

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What is the product?

This reaction is a variant of what type of reaction?

Gatterman-Koch synthesis puts an aldehyde on a benzene ring and is a variant of Friedel-Crafts acylation.

Gatterman-Koch (like FC) requires benzene or activated benzene derivatives.

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What is the product?

Hydration of a terminal alkyne will give methyl ketones via an enol intermediate.

The enol is the Markovnikov product of terminal alkyne hydration (addition of water across the triple-bond). It quickly tautomerizes to its keto form.

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What is the product?

Hydroboration-oxidation of the terminal alkyne gives the anti-Markovnikov addition of water across the triple-bond. The unstable enol tautomerizes to the aldehyde form.

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What is one was to synthesize this product from starting materials containing no more than six carbons?

Add a terminal alkyne then do hydroboration.

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What is one way (that is not hydroboration) to synthesize this compound from starting materials containing no more than six carbon atoms?

Grignard addition to an epoxide results in a primary alcohol that can be oxidized to an aldehyde.

Be sure to use a mild oxidizing agent in the final step like bleach or PCC. Chromic acid will give a carboxylic acid.

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What is the product of this reaction?

Why are 2 equivalents of the organolitium reagent required?

The first equivalent generates the carboxylate salt (it reacts with H⁺ of the COOH).

The second equivalent attacks the carbonyl group; subsequent protonation and dehydration gives the ketone.

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What type of reagent can be used to reduce nitriles to aldehydes?

Give a specific example of a reagent that is commonly used.

Aluminum hydrides can reduce nitriles to aldehydes.

Diisobutylaluminum hydride, abbr. (i-Bu)₂AlH (DIBAL-H) is commonly used.

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Why must a milder reducing agent be used when reducing a carboxylic acid to an aldehyde?

Strong reducing agents will continue the reduction of carboxylic acid all the way to a primary alcohol (not stop at the aldehyde).

When going from a carboxylic acid to an aldehyde, why is SOCl₂ then a bulky lithium hydride used, instead of LiAlH₄?

LiAlH₄is too strong and will reduce the -COOH all the way to a primary alcohol. A milder bulky reducing agents react _faster_ with acid chlorides than with aldehydes, thus giving good yields of aldehydes.

What is the product?

DIBAL-H reduces esters directly to aldehydes at dry ice temperature.

How many times does a Gilman reagent (lithium dialkylcuprate) react with an acid chloride? How many times does a Grignard reagent (and organolithium reagents) react with acid chlorides? Which is better for synthesizing ketones?

* Gliman reagents* are weaker organometallic reagents and react faster with acid chlorides than with ketones, so they give good yields of ketones. They react ***once*** with an acid chloride to give a ketone. * Grignard* and organolithium reagents are stronger and react fast with acid chlorides to give ketones, _but also react fast with ketones_ to give tertiary alcohols. They react ***twice*** with an acid chloride to give tertiary alcohols. **Gliman reagents are better for synthesizing ketones from acid halides.**

Which reducing agent is stronger, LiAlH₄ or NaBH₄?

LiAlH₄. Aluminum is *less electronegative* than boron (remember periodic trends), so more of the negative charge in AlH₄^- is taken on by the hydrogen atoms, making them ready to go and react as a hydride (**hydride reduction**). Alternatively, Al and B are both in Group 3A, but Al is one row down and is larger than B; that means it is more *polarizable* and a better nucleophile (better at delivering a hydride). NaBH₄ is useful because it is a more selective reducing agent (weaker) and will reduce aldehydes and ketones selectively.

Is the oxygen on the carbonyl group a weak acid or a weak base?

It is a **weak base** because it can become *protonated in acidic conditions*. The addition of a proton (or some other electrophile) makes the carbonyl strongly electrophilic (resonance forms help show this). This makes the protonated carbonyl susceptible to attack by even weak nucleophiles like water and alcohols under acidic conditions.

What's the difference between base-catalyzed addition to a carbonyl and acid-catalyzed addition to a carbonyl?

BASE-CATALYZED **Step 1:** _Strong Nuc attacks_ carbonyl carbon. **Step 2:** A _weak acid_ (if the strong Nuc is also a strong base, its conjugate acid is weak), _protonates_. ACID-CATALYZED **Step 1:** _Protonation_. Carbonyl oxygen is a weak base and accepts a proton under acidic conditions. This makes the carbonyl carbon a strong electrophile. **Step 2:** _Weak Nuc attacks_ carbonyl carbon. **Takeaway:** Think *"is there something strong?"* and *make* *it act first*! If the Nuc is strong, it ain't waiting - it's gonna attack the carbonyl right from the start. If the Nuc is weak (like water or alcohol), then it needs some help. We gotta make the carbonyl a strong electrophile first. Therefore, protonate first.

Which carbonyl is more electrophilic, an aldehyde carbonyl or ketone carbonyl? Why? How does the electrophilic strength of the carbonyl relate to the equilibrium constants for hydration of ketones and aldehydes?

Aldehydes possess the more electrophilic carbonyl carbon. The ketone's carbonyl group is stabilized by its **two** adjacent alkyl groups, but an aldehyde only has **one** stabilizing alkyl group. Since aldehydes have less EDG's, they are more electron-poor, thus more electrophilic. Ketones have smaller hydration K_eq values than aldehydes because the carbonyl group for ketones is more stable and less reactive. It is happier with its two adjacent EDG's. An aldehyde, with its more electrophilic carbonyl group, is more likely to undergo hydration, so it has a larger K_eq. K_eq = [products]/[reactants]

What is the mechanism for ketone hydration in acidic conditions?

This is just acid-catalyzed addition. Remember the two steps: **Step 1:** Protonation. Carbonyl oxygen is a weak base and accepts a proton under acidic conditions. This makes the carbonyl carbon a strong electrophile. **Step 2:** Weak Nuc attacks carbonyl carbon. In this case water. *After step two, it should be clear to deprotonate and get the + charge off oxygen. Get the first TWO steps down, and you're golden.*

What is the mechanism for ketone hydration in basic conditions?

This is just base-catalyzed addition. Remember the two steps: **Step 1:** _Strong Nuc attacks_ carbonyl carbon. *Hydroxide in this case.* **Step 2:** _A weak acid_ (if the strong Nuc is also a strong base, its conjugate acid is weak), _protonates._ *Water in this case is the weak acid.*

Cyanohydrin formation is an example of what type of reaction?

Cyanohydrin formation is just _base-catalyzed addition_ to the carbonyl group. The cyanide ion is a *strong base* and *strong nucleophile*. Remember the two steps. Think *"is there something strong?"* and make it strike first. BASE-CATALYZED **Step 1:** _Strong Nuc_ attacks carbonyl carbon. **Step 2:** A _weak acid_ (if the strong Nuc is also a strong base, its conjugate acid is weak), _protonates_.

Imine formation is optimal at pH = 4.5. What type of mechanism is likely at work at some point during the reaction?

pH of 4.5 should be a clue that acid-catalyzed addition is likely at work. ## Footnote **Imine formation** is an important mechanism, but it is _nothing new_ and can be broken down into *two smaller, familiar mechanisms*: **_1: ACID-CATALYZED ADDITION_** **Step 1:** _Protonation._ Carbonyl oxygen is a weak base and accepts a proton under acidic conditions. This makes the carbonyl carbon a strong electrophile. **Step 2:** _Nuc attacks_ carbonyl carbon. *Before moving on, recognize the + charge and deprotonate real quick.* **_2: ACID-CATALYZED DEHYDRATION_** (from Ch 11) **Step 1:** _Protonation._ Adding a H⁺ to the hydroxy group makes it a good LG. **Step 2:** _Loss of H₂O_ (dehydration). The resulting cation is resonance-stabilized. *The resonance form with the + charge on nitrogen is the major form (octets), so deprotonate and be done!*

What are the _six_ steps of imine formation?

1. Proton _on_ (resonance-stabilized cation) 2. Nucleophile _attacks_ 3. Proton _off_ 4. Proton _on_ 5. LG _leaves_ (resonance-stabilized cation) 6. Proton _off_ See how **4 of 6** steps are _proton transfers_. Break it down into managable chucks. * First three steps are acid-catalyzed addition. * Last three steps are acid-catalyzed dehydration.

What is the product? What type of reaction is this?

This is just _imine formation_ where the R-group on the primary amine is another amine.

What is the product? What type of reaction is this?

This is just _imine formation_ where the R-group on the primary amine is a hydroxy group.

What is the product? What type of reaction is this?

This is just _imine formation_ where the R-group on the primary amine is a secondary amine attached to a phenyl group.

Formation of acetals can be divided into two easy processes. What are they?

Acetal formation can be broken down into **acid-catalyzed addition** and **S_N1 substitution**.

What is the mechanism to turn a ketone into a hemiacetal?

**_ACID-CATALYZED ADDITION_** - turns a ketone into a hemiacetal. Step 1: Protonation. Step 2: Nuc attacks carbonyl carbon. Step 3: Deprotonation.

What is the mechanism for turning a hemiacetal into an acetal?

**_S_N1_** **Step 1:** Protonation - adding a proton to the hydroxy groups makes a good LG. **Step 2:** Loss of water - forms a resonance-stabilized cation. **Step 3:** Second alcohol adds - attacks the site of the carbocation. **Step 4:** Deprotonation - forms an acetal.

Given the overall reaction for acetal and ketal formation, what does removal (by distillation) of the the water produced during the reaction do to the equilibrium? What would adding excess water do to the equilibrium?

Removal of water as it is produced as a reaction by-product _forces the equilibrium to the right_ and promotes acetal or ketal formation. Excess water in dilute acid promotes the reverse reaction, _shifts the equilibrium to the left_, and promotes aldehyde and ketone formation.

Compare the steps of acetal formation to imine formation.

Imine formation is _1) acid-catalyzed addition_ (of amine) and _2) dehydration_. Acetal formation is _1) acid-catalyzed addition_ (of alcohol) and _2) S_N1_ (add second alcohol). Acetal formation has **seven** steps, and the **first five** are _similar to imine formation_. 1. Proton _on_ (resonance-stabilized cation) 2. Nucleophile _attacks_ 3. Proton _off_ 4. Proton _on_ 5. LG _leaves_ (resonance-stabilized cation) 6. Second nucleophile _attacks_ (only step different from imine formation) 7. Proton _off_

What alcohol(s) and carbonyl give the follow:

Why can we NOT use Grignard reagents with an aldehyde or ketone functional group in synthesis? What can be done to fix the situation?

A Grignard reagent with an aldehyde and/or ketone functional group will **_attack itself_**. Its own carbonyl group will react with its own nucleophilic organometallic group. To prevent self-attack, the carbonyl can be _protected as an acetal_, which is unreactive toward a Grignard reagent. After the Grignard reaction is complete, _dilute aqueous acid hydrolyzes the acetal back_ to the deprotected carbonyl.

What is *selective acetal formation* (what is the reactivity concept behind it) and how can it be employed when dealing with a compound that contains both an aldehyde and a ketone?

Remember aldehydes are *less stabilized* than ketones (one EDG vs two EDGs) and are, therefore, *more reactive*. Aldehydes form acetals more readily than ketones, so we can preferentially turn the aldehyde into an acetal while leaving the ketone alone. Then we can modify the ketone, and finally deprotect the aldehyde. ## Footnote **Step 1:** Preferentially _protect aldehyde_ (acetal formation) **Step 2:** _Modify ketone_ in neutral or basic conditions (acidic conditions will reverse acetal too soon). **Step 3:** _Deprotect the aldehyde_ by hydrolyzing the acetal in dilute aqueous acid.

Show how to accomplish the following synthesis: (you may use whatever additional reagents you need)

What is the "net effect" of the Wittig Reaction?

The Wittig reaction converts the carbonyl group of a ketone or aldehyde into a C==C alkene. C==O → C==C (there can be an R-group on the alkene depending on the ylide used)

How is the phosphorus ylide synthesized for Wittig reactions? If a new R-group is desired on the alkene product, where is that R-group coming from?

Phosphorus ylides are produced by: 1. **S_N2 attack** of an unhindered (usually primary) alkyl halide. 2. Phosphonium salt **treated with strong base** (usually Bu-Li) to nab hydrogen from carbon bonded to phosphorus. The new R-group on the alkene product is going to come from the alkyl halide portion of the ylide.

Phosphorus ylide has two resonance forms. Which is the major form, and what does it enable the ylide to do?

The ylide resonance form with the _negative charge on the carbon and the positive charge on the phosorus is the major form_. The double-bond between P and C requires 10 electrons in the phosorus valence shell (requires help from *d* orbital to overcome octet limitation). The orbital overlap in the pi-bond between C and P is poor because of significant size differences, so the **pi-bond is weak**. _The resonance form with the charge separation satisfies the octet rule - it is the major form_. The negative charge major resonance form puts a _huge partial negative charge on the ylide carbon_ (carbanion character) making it a _great nucleophile to attack the electron-poor carbonyl carbon_.

What are the ***three*** steps of the Wittig Reaction?

**Step 1:** _Y__lide carbon attacks carbonyl carbon_. Strong partial negatively charged carbon (Nuc) attacks partial positively charged carbon (E⁺). **Step 2:** _Opposite charges attract_. Positively charge phosphorus and negatively charged oxygen bond to form four-membered ring (large ring-strain). **Step 3:** _Ring collapses_. Triphenylphosphine oxide product is *very stable* and is the driving force for collapsing the unstable 4-membered ring into more stable porducts.

What are the two combinations of possible ylide and carbonyl reactants in the compound formed via Wittig reaction? Which is the preferred set of reactants and why?

Ylides come from an alkyl halide and triphenylphosphine. Triphenylphosphine is a *super bulky nucleophile* (3 benzene rings), so it _needs an unhindered alkyl halide to attack_ (primary or methyl). Secondary halides react super slow and give poor yields of ylides. Look to use a ylide that has a **negatively charged carbon with one alkyl and one hydrogen _instead_ of two alkyl groups** because it comes from a less hindered alkyl halide. Note: a methyl halide will give a ylide where the negatively charged carbon has two hydrogens.

The Tollens reagent tests for what functional group?

Tollens reagent tests for the presence of an aldehyde functional group on an unknown compound. If an aldehyde is present, the _silver-ammonia complex oxidizes the aldehyde into a carboxylate_ (carboxylic acid anion). Tollens reagent does not react with simple hydrocarbons, ethers, ketones, and alcohols.

What is the product? Why is NaBH₄ used instead of LiAlH₄?

Sodium borohydride reduces aldehydes to *primary alcohols* and ketones to *secondary alcohols*. Sodium borohydride is a milder reducing agent. LiAlH₄ also accomplishes these reductions, but it is more powerful and will reduce other functional groups as well, like carboxylic acids and esters.

What type of reaction is this? What is the product?

This reaction is reduction using Raney nickel. Note: Alkene bonds (C==C) are reduced easily by Raney nickel, as well, and will reduce first. So Raney nickel will reduce _carbonyl bonds to alcohols_ and _alkene bonds to alkane bonds_.

What is **deoxygenation**, and what are **two** reduction methods we can use to "deoxygenate" a carbonyl?

Deoxygenation (as the name implies) **removes the oxygen from the carbonyl bond** of _ketones_ or _aldehydes_ and replaces it with two hydrogens. _Clemmensen Reduction_ and _Wolff-Kishner Reduction_ both deoxygenate ketones and aldehydes. Clemmensen Reduction under acidic conditions. Wolff-Kishner Reduction under basic conditions.

What type of reaction is this? What is the product? What type of Electrophilic Aromatic Substitution reaction is it often paired with to make longer alkyl substituents on benzene rings without allowing rearrangement?

Clemmensen Reduction It is _often paired with Friedel-Crafts acylation_ because it converts acylbenzenes to alkylbenzenes. Friedel-Crafts alkylation is prone to rearrangement and over alkylation, but Friedel-Crafts acylation cannot rearrange and deactivates the ring after the first acylation. Then Clemmensen Reduction can be used to reduce the carbonyl and give the alkane.

The Wolff-Kishner Reduction can be broken down into two separate mechanisms. What are they, and what are the reaction conditions?

**First Mechanism: Imine formation** (specifically an imine with another amine as an R-group = hydrazone) **Second Mechanism: Reduction** (specifically two tautomeric proton transfers from N to C) The first mechanism is acid-catalyzed. The second mechanism takes place under heated basic conditions. Strong bases like **KOH** or ***tert*-butoxide** often used. Solvents with high bp used to facilitate temp conditions (ethylene glycol or diethylene glycol often used).

What type of reaction is this? What is the product?

Wolff-Kishner Reduciton.

Wolff-Kishner Reduction involves turning a carbonyl group into an imine (specifically hydrazone), then reduction of the hydrazone all the way to an alkane. What is the mechanism for this second step?

The reduction step involves two tautomeric proton transfers. Just think of _swaping the hydrogens from the N to the C and kicking out N₂_. 1. **Proton off** (remove from N) 2. **Proton on** (place on C) 3. **Proton off** (remove from N) 4. **Lose N₂**. Electron shift to form N₂ breaks the sigma-bond between N and C. N₂ is extremely stable (think about what is in the air all around us) and wants to form and leave. 5. **Proton on** (place on C) The mechanism is _FOUR proton transfers_ and _ONE loss of a LG_.

Ch 18 Flashcards

(69 cards)