Ochem 2 Flashcards

Question

* Aldehydes & Ketones * Keto-Enol Tautomerization

Answer 1

**_Keto-Enol Tautomerization:_** * 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 **carbonoxygen bond** to the bond between the **carbonyl carbon** and the **alpha carbon**

Answer 2

**_STEPS_**: * There are ***two possible ways*** to visualize this mechanism, based on which resonance form you start with: 1. With the 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

Answer 3

* The one on the ***left*** is clearly the more significant contributor to the actual structure * because it has ***no formal charges,*** compared to a charge separation in the structure on the right

Answer 4

**_Aldehyde_** * is any compound containing a carbonyl... * with one or more **hydrogen substituents** on the carbonyl carbon **_Ketone_** * is any compound containing a carbonyl... * with **two carbon substituents** on the carbonyl carbon

Answer 5

* Amides are the***MOST STABLE*** of all acid derivatives * Their carbonyl carbons are ***UN***reactive * This is because --NH₂ is ***NOT*** a good leaving group

Answer 6

* Primary and secondary amides can ***HYDROGEN BOND*** * ∴ amides are *water soluble* ***as long as:*** * _they lack long alkyl chains_ * Tertiary amides ***cannot*** H-bond **_Biochemistry Connection:_** * Amide hydrogen bonding is perfectly illustrated in the ***secondary structure of proteins*** * In an ***alpha helix*** every amine hydrogen forms an H-bond with the carbonyl four residues previous to it in the chain

Answer 7

* **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 ***SHORTENS (!!)*** the length of the carbon chain This reaction is important because it allows you to ***ADD AN AMIDE TO A TERTIARY CARBON***

Answer 8

* *Because the nitrogen in an amide donates its lone 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***

Answer 9

_Resonance (Double Bond Character) **LIMITS ROTATION:**_ * 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*

Answer 10

* Nitrogen has **less** electron density in an amide than it would in a normal amine * b/c of **donation of the lone pair** on the nitrogen into the conjugated system * Therefore, it will be ***LESS BASIC*** than comparable amines 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

Answer 11

**_Definition_** * An amide is any compound containing a carbonyl with an **amine substituent on the carbonyl carbon** **_Nomenclature_** * Named by replacing the **“-oic”** ending of the corresponding carboxylic acid with **“amide”** * i.e., benzoic acid⇒benzamide

Answer 12

Addition of Amines to Carbonyls.... aka Formation of ***ENAMINES*** and ***IMINES*** * Amines add to aldehydes and ketones to form imines and enamines **_STEPS:_** 1. The amine acts as a nucleophile * ...attacking the electrophilic carbonyl carbon 2. The oxygen is protonated twice * creating the ***good LG 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 pattern of the nitrogen)

Answer 13

* Primary (1°) amines: * yield ***IMINES*** * Secondary (2°) amines: * yield ***ENAMINES*** * Tertiary (3°) amines * ***DO NOT REACT***

Answer 14

_Definition_: * An amine is **any** organic compound that ***contains a basic nitrogen atom*** **_Properties_** * _Amines can act as either **bases** or **nucleophiles**_ * ****Primary or secondary amines usually act as *nucleophiles* * Tertiary amines ***ALWAYS*** act as *bases* * (because they are **too sterically hindered** to act as nucleophiles) * _Amine Basicity:_ * Basicity decreases from tertiary to secondary to primary to ammonia due to the electron donating effects of the R- groups * Amines attached to aromatic rings are significantly less basic than standard amines * _Amines with **four** substituents act as ***ELECTROPHILES***_ * as long as they have at least one hydrogen * ex: Ammonium, NH₄⁺ * _Amines are capable of ***hydrogen bonding***_

Answer 15

_STEPS:_ * The **phthalimide ion** (a reactive species with a *full negative charge on the nitrogen*) acts as a **nucleophile** * ...attacking the ***alkyl halide*** * via SN₂

Answer 16

Formation of a **1° amine** from a **1° alkyl halide** * Avoids the *side products of alkyl amine synthesis*

Answer 17

* ***ALCOHOLS*** will form stronger hydrogen bonds * because there is a **_greater difference in electronegativity_** between oxygen and hydrogen than there is between nitrogen and hydrogen * This ***greater dipole*** will create a ***stronger electrostatic attraction*** * ...and therefore a stronger hydrogen bond

Answer 18

_Naming amines is a bit unique, so be sure you understand the system:_ 1. **Name the alkane** to which the N is attached e.g., propane) 2. Add **“amine”** in place of the **“e”** on the end of **“ane”** (e.g., propanamine) * It is also acceptable to separate the substituent name (e.g., 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-"** * e.g., 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 * e.g., N,N-diethylpropanamine, or N-ethyl-Nmethylpropanamine, or **N,N-dimethyl-N-ethylpropanamine (ATTACHED)**

Answer 19

_Nitro groups_ * can be reduced to the _associated primary amine_ * via **all** of the listed reducing agents * e.g., Li**AlH₄**, Na**BH₄**, and H₂/catalyst with pressure * Most O-Chem books focus on nitro groups being reduced by: * **metals in HCl M•HCl** * **M=Fe, Zn, Sn** _Nitrile groups_ * can be reduced to the _associated primary amine_ * via **all** of the above listed reducing agents * Most O-Chem books focus on nitriles being reduced by: * Li**AlH₄**

Answer 20

_Imines_ * can be reduced to the associated **1° amine** * via all of the above listed reducing agents * O-Chem books focus on Imines being reduced by: * **BH₃** * **^-:CN** or * **H₂ /** **catalyst** _Amides_ * reduce to the associated **1° amine** * via **LiAlH₄** ***ONLY***

Answer 21

***_Tautomerization_*** * An **enamine** and **imine** interchange * via a proton shift Analogous to the **keto-enol tautomerization**

Answer 22

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

Answer 23

* 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 LG than the water** formed by protonation of the carbonyl * ...and would therefore be kicked back off anyway

Answer 24

* Reduction of: 1. Am***IDES*** 2. Imines 3. Nitriles 4. Nitro groups * using common ***reducing agents*** such as: * LiAlH₄ * NaBH₄ * H₂/catalyst with pressure.

Answer 25

_Formation of an **alkylamine** from an **amine** and an **alkyl halide**_ ***NH₃ + CH₃Br ⇒ NH₂CH₃ + HBr*** **_STEPS_**: 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*** * This quenches 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**

Answer 26

***Protein Synthesis*** ## Footnote * the amine acts as a nucleophile * attacking the **carbonyl carbon** of **another amino acid** to form a peptide bond

Answer 27

***Anhydrides are excellent ELECTROPHILES!*** * The two carbonyl carbons are **highly** **reactive** to nucleophiles * because the leaving group is a ***resonance-stabilized carboxylate ion***

Answer 28

* **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

Answer 29

**_Decarboxylation_** * 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 retakes the hydrogen from the base, forming a keto-enol tautomer

Answer 30

**_Reaction of an alcohol with a carboxylic acid to form an ester_** (ROR) RCOOH + ROH ⇒ RCOOR + H₂O * The hydroxyl group will **never** leave **without** being *protonated first* * ....to form the “good leaving group water” * thus this reaction requires an _acid catalyst_ * Higher yields can be obtained by reacting an **anhydride** with an **alcohol** **_This is how triacylglycerols are formed:_** * A **glycerol** undergoes esterification with **three** fatty acids

Answer 31

RCOOH + H₂O ⇒ RCOOH₂⁺ + Nu:^-⇒ RCONu + H₂O

Answer 32

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

Answer 33

1. **_Resonance Stabilization_** * The carboxylate ion is uniquely stable due to resonance stabilization 2. **_Induction_** * Alpha substituents can either **donate** or **withdraw** from the carboxylate ion * **increasing** or **decreasing** acidity * REMEMBER: To predict acidity examine the stability of the conjugate base! * (A principle applicable to *any* acid) 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**

Answer 34

A carboxylic acid is any compound containing a carbonyl ***with a hydroxyl substituent on the carbonyl carbon*** ***_NOMENCLATURE_*** * Carboxylic acids are named with the ***“-oic acid”*** 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 * e.g., formic acid ⇒formate * If it is a salt formed between the carboxylate ion and a metal: * name the metal **first** * **then** the ion * e.g., benzoic acid⇒benzoate⇒sodium benzoate **_Common Names:_** * There are few common acids for which the MCAT will use non-IUPAC names * _These include:_ 1. Formic acid (HCOOH) 2. Acetic acid (CH₃COOH) 3. Benzoic acid (C₆H₅COOH)

Answer 35

* Carboxylic acids 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)

Answer 36

"Electron lovers" * is ***attracted to e^-s / e^- rich centers*** * Usually ***POSITIVELY CHARGED***

Answer 37

* Is an **addition** reaction * Here, a compound with a **double** or **triple** bond (aka ***one that has 1 or 2 π bonds)*** * ...reacts with electron-rich reactant ("nucleophile") **_As a result:_** 1. Disappearance of the **double bond** 2. Creation of **two new single ("σ") bonds**

Answer 38

* Here, an electron **nucleophile** selectively bonds with (or attacks) the **positive or partially positive charge of an electrophile** or a group of atoms ## Footnote ***...to REPLACE** a so-called **"leaving group"***

Answer 39

_Saponification ***(HYDROLYSIS OF AN ESTER)***_ * Hydrolysis of an ester to yield: 1. an alcohol 2. 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”

Answer 40

**_Familiar Examples:_** * ATP, GTP, UTP, etc. are examples of inorganic ***triphosphate*** esters * FADH₂ and NADH are examples of inorganic ***diphosphate*** esters * FMN, DNA and RNA are examples of inorganic***monophosphate*** esters

Answer 41

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

Answer 42

* Reaction of an existing ester with an alcohol, creating a different ester * Also requires *_acid catalysis_* ## Footnote ***RCOOR_a + R_bOH⇒RCOOR_b + R_aOH***

Answer 43

**_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

Answer 44

**_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 * Be aware of ***both*** conventions

Answer 45

**_Definition:_** * An ester is any compound containing a carbonyl ***with an –OR group substituted on the carbonyl carbon*** **_Nomenclature:_** * Esters are named with the **"–oate"** ending, with the R portion of the –OR group named and placed **_in front of the name_** * ex: "methyl" pentan**oate** **_Common Names:_** * The common name applies to the ***second half of the name*** * The first portion of the name will be **different** for **different** R groups * We will use **methyl** for all three examples: 1. methyl formate 2. methyl acetate 3. methyl benzoate

Answer 46

* Ketones & Enols

Answer 47

the ***LEAST POLAR***

Answer 48

_Used to **isolate a specific molecule** or product based on:_ * a **VERY SPECIFIC affinity** or **binding interaction ** *To **elute,** the bound **target** molecules must **disrupt the binding interaction*** _This could be accomplished via:_ 1. a **SALT SOLUTION**, but *often* requires 2. a chemical **REVERSAL OF THE RXN** that ***BOUND*** the target **TO** the column

Answer 49

* The mixture to be separated is passed through a column packed with charged glass beads * or some other ***polar*** matrix * The solution is ***collected in fractions*** at the bottom of the column * i.e., the collecting tubes are changed at regular intervals In column chromatography, assuming a polar matrix, the ***MOST NON-POLAR*** substances will elute first * followed by increasingly more polar substances

Answer 50

**_Distillation_**: * To separate two substances by simple distillation they must have boiling points ***that are at least 25˚C apart*** 1. ******Simple Distillation 2. Fractional Distillation 3. Vacuum Distillation

Answer 51

**_How it Works:_** * The desired product (which still contains impurities) is **dissolved** in the minimum amount of hot solvent necessary to create a saturated solution * The solution is then **cooled** as slowly as possible * Because **pure** substances usually crystallize at a **higher** temperature than impure substances, * if the temperature is dropped just below the melting point of the product: * the crystals that form will be **product** * the impurities will **remain in solution** This is rarely a perfect process, but **repeated cycles** can produce an **increasingly pure** product

Answer 52

* The column or stationary phase is coated with cations or anions * The mixture is passed through and oppositely-charged ions adhere to the column * The target molecules can then be eluted * by washing with a ***salt*** solution

Answer 53

* Heat the mixture in a flask * the liquid with the ***lower boiling point evaporates first***, * enters a collecting arm, * cools, * and drops into a collecting flask _To improve separation process:_ 1. Cool the condensing arm or 2. Place the collecting flask on ice

Answer 54

* A fractionating column is placed between the heating flask and the condensing arm * The mixture is heated to **slightly** **above** *the boiling point of the more volatile liquid* * The gas rises through a column of glass beads or metal shards * This causes any impurities in the vapor (i.e., molecules from the liquid with the higher boiling point) to **condense** and **fall back into the flask** * resulting in a better separation * The more volatile component will be above its boiling point * and therefore will ***NOT*** condense * This approach allows separation of compounds with boiling points less than 25˚C apart

Answer 55

**_How it Works:_** * The two solvents used will ideally have widely different **polarities** and **densities**, and will therefore create a distinct _line of separation_ * If the target product is already dissolved in a solvent that fits these requirements, it can serve as one of the two immiscible liquids * If not, the mixture containing the target will be added to the separatory funnel along with two other immiscible solvents 1. The contents of the funnel are gently mixed or swirled 2. The mixture is then allowed to sit until the two solvents are fully separated * If the target is a polar compound that was synthesized in a non-polar solvent then the target should easily move into the polar layer during mixing (like dissolves like) * If the target is a non-polar product it should readily move into the non-polar solvent 3. After the two layers are fully separated, the bottom layer (usually aqueous) is drained out through the stopcock 4. The solvent layer containing the product is then evaporated to obtain the target

Answer 56

**_The separation can be aided by one or more of the following:_** 1. Repetition: * Rarely is a full separation obtained **after only one cycle** 2. Fractional Extraction: * Extracting 5mL ten ***separate*** times will produce a much better separation than extracting 50mL all ***at once*** 3. Addition of an **acid** * to *protonate* the product 4. Addition of a **base** * to *deprotonate* the product

Answer 57

* An acid would **protonate bases** in the mixture, * potentially adding a formal charge and dramatically increasing solubility in the aqueous layer * i.e., protonated amines * Bases could similarly abstract protons to create a negatively charged species—with the identical impact upon solubility * i.e. deprotonated carboxylic acids

Answer 58

1. Mixing ***TOO*** vigorously * This can result in the formation of emulsions that are difficult to separate 2. Reactive solvents * The chosen solvents should not be reactive with: 1. Each other 2. The targeted product 3. High boiling point * If a solvent with a high boiling point is used, it will be difficult to evaporate off the solvent to obtain the product

Answer 59

**_Purpose:_** * **__**Used to: 1. Separate two compounds 2. To remove a desired product from a reaction mixture Two **immiscible** ("un-mixable") **liquids**, a **polar** (aqueous) and a **non-polar** (organic) are used * The separation depends on the target molecule having ***differential solubility in the two solvents***

Answer 60

**_Gas Chromatography:_** * A liquid is used as the stationary phase * The mixture is dissolved into a heated gas and then passed through the liquid * Various components reach the exit port at different rates based on 1. Boiling point, and 2. Polarity * *Only* consider polarity if the two substances have ***almost identical boiling points*** On a gas chromatograph, there will be ***one peak for each unique compound*** in the mixture * The **height** of the peak in a gas chromatograph is relative to the ***ABUNDANCE*** of that component

Answer 61

* *Physical* separation of a **_solid_** (either crystallized product or solid impurities) **_from a liquid_** *by passing it **through filtration paper*** * Fluted filter paper is recommended Fluted filter paper ***increases the surface area*** for filtration

Answer 62

**_Whether you would want to filter hot or cold would be determined by the conditions_** * If you are filtering out a solid impurity, filtering hot could prevent your product from crystallizing on the filter paper * However, if you are close to the freezing point of the solid then filtering hot could bring impurities back into solution * In other cases you may be trying to capture product that crystallized upon cooling of your reaction mixture * In that case you would certainly want to fitler cold or else you would redissolve your product

Answer 63

* Notice how the **O-H** and **C=O** absorbances ***OVERLAP***

Answer 64

* **The bond roughly follows Hooke’s Law** * and either atom can be compared to the *mass* * A ***_STRONGER_*** bond * is analogous to a spring with a **larger spring constant (k)** * (∴ a weaker bond would be analogous to a spring with a smaller spring constant) * The ***MW*** of the atom * is analogous to the size of the mass on the spring ***_PER HOOKE'S LAW:_*** Stronger spring (larger k) **increases oscillation** frequency according to: * f = ½π √k/m* * A **weaker** bond would therefore vibrate at a **lower** frequency A **smaller** atom *(**lower MW)* has a**higher** vibrational frequency * ∴ a **larger** atom has a **lower** vibrational frequency

Answer 65

_The Vibrational Frequency is determined by:_ * 1) **Strength** of the bond * 2) **MW** of the bonded atoms IR absorbances do ***NOT*** represent the entire molecule * They only represent a single polar bond in the molecule

Answer 66

* Notice how shallow the amine absorption is

Answer 67

* Notice how **sharp and deep** the absorbance for the C-N triple bond (nitrile) is

Answer 68

1. **Carbonyl, C=O** * 1700 cm^-1 * sharp, deep 2. **Alcohol, OH** * 3300 cm^-1 * broad, separate from CH 3. **Saturated Alkane, CH** * 2800 cm^-1 * sharp, deep 4. **Carboxylic Acid, OH** * 3000 cm^-1 * broad, overlaps CH 5. **Amine, NH** * 3300 cm^-1 * broad, shallow 6. **Amide, NH** * 3300 cm-1 * broad, deep 7. **Nitriles,** **C≡N** * 2250 cm^-1 * sharp, deep

Answer 69

notice the OH stretch

Answer 70

**_IR (Infrared) Spectroscopy_** How it Works: * If a bond has a ***DIPOLE***, exposing that bond to an external electric field will cause the atoms to move within that field * much like a charged particle attached to a spring * Infrared radiation is used to create an oscillating electric field * which in turn causes dipolar bonds to oscillate at a specific "**vibrational frequency"** * To produce an IR spectrum, we slowly vary the frequency of the IR radiation. * When the IR radiation exactly matches the frequency of vibration of a particular bond, that bond is said to be in “resonance” * and **it will absorb some of the IR energy** * This absorbance is what is picked up by the detector * A bond with **no dipole** will **NOT** be detected by IR

Answer 71

* The sample is usually a mixture of components which is spotted onto the paper or TLC plate near its bottom edge * The top of the solvent should always be **below** this line where the sample is spotted * so that the mixture does not simply dissolve into the solvent

Answer 72

_R_f is a ratio of the distance traveled by the component over the distance traveled by the solvent_ * So a number **close to one** means that the polarity of the component is **similar to that of the solvent** * Because **non-polar solvents are normally used**, the component is therefore relatively _non-polar_

Answer 73

* Smears are usually caused when the solvent is **unable to dissolve** and carry the components as it moves up the paper via capillary action * This can be because too much of the sample was spotted onto the paper and the solvent could not dissolve ***ALL*** of the sample * It can also occur if there is **too big of a difference in polarity** between the sample and the solvent * If the polarities are too distant the solvent cannot dissolve the sample well enough to carry it * By a similar token, if they are too similar, the components may never come out of solution to form a spot and be carried all the way to the top of the paper

Answer 74

* Higher and sharper melting points indicate **better purity** * Impurities **lower** melting point * They also broaden the range across which the crystals melt

Answer 75

**_How it Works:_** * The energy difference (in Joules) between two adjacent molecular orbitals, on average, just happens to be about the **same amount of energy** created by electromagnetic radiation in the UV spectrum * Thus, when UV radiation is shone on a molecule, electrons within that molecule will often absorb that energy and “excite” to the next highest energy level * This absorbance is recorded on a UV spectrum

Answer 76

* Molecules containing only single bonds show **low or no** UV absorbance * Double and triple bonds absorb UV **strongly** * ****double * Conjugated systems absorb UV **even more strongly** than do isolated double or triple bonds. * The greater the degree of conjugation, **the farther to the right** the species will absorb * i.e., at a higher wavelength * A UV spectrum is a graph of **absorbance** vs. **wavelength** (nm)

Answer 77

* Vacuum filtration is performed with a Hirsch or Buchner funnel * A vacuum is created inside of the flask which creates suction to *pull the filtrate through the filter paper* * The filter usually has holes in it which are covered by the filter paper The primary advantage is that it is ***FASTER*** than gravity filtration

Answer 78

**_Paper or Thin-Layer Chromatography (TLC):_** * Paper chromatography uses paper as the stationary phase. * Thin-Layer Chromatography (TLC) is nearly identical * *_...but uses manufactured glass or plastic sheets coated with silica, alumina, etc._* ***R_f*** = * distance traveled by *component*, OR * distance traveled by *solvent*

Answer 79

**_Vacuum Distillation_** The air inside the apparatus is evacuated to create a vacuum * Vacuum filtration is used because it dramatically **lowers** the boiling point, * allowing you to work at more manageable temperatures with substances that have much higher boiling points at atmospheric pressure * This observation is supported by the fact that liquids can be said to boil at the point where vapor pressure equals atmospheric pressure * By creating the vacuum, we essentially drive ***down*** the atmospheric pressure to ***meet*** the vapor pressure

Answer 80

Separation of one or more compounds by dissolving them in a “mobile phase” and then passing that phase through or across a “stationary phase” * The substances in the mobile phase interact to varying degrees with the stationary phase ***based on their polarity*** * The more interaction that occurs between the two phases * the ***SLOWER*** the substance will move * Substances that do not interact with the stationary phase (or do so to the least degree), will move the ***FASTEST***

Answer 81

* A **linear** increase in the magnetic field will create a **linear** increase in the force exerted on the particles according to: ***F = qvBsinθ*** * For a curved flight tube with a detector at the far end, in order for any particle to strike the detector: * It must experience **exactly the right centripetal acceleration** to trace the precise curvature that results in striking the detector * As the centripetal force **increases**, this will occur for increasingly **massive** particles The graph below shows the linear relationship between magnitude of the B field and mass of particles at the detector

Answer 82

* **_The Height of each peak_** * gives the **relative abundance** of that fragment * **_The Parent peak_** * represents the original molecule minus one electron * a.k.a. "Molecular Ion Peak" * **_The Base peak_** * is the most common fragment * is usually **the most stable fragment generated** * It is defined as existing at 100% relative abundance * In other words, it will be the highest peak on the spectra and the height of all *other* peaks will be a function of how abundant ***THAT*** fragment is **compared to the base peak**

Answer 83

**_C¹³-NMR Absorbances you should know:_** * **C – C** * 0-50ppm * **C – O** * 50-100ppm * **C = C** * 100-150ppm * **C = O** * 150-200ppm **_Somewhat Analogous to IR:_** * Whereas H-NMR peaks represent unique hydrogen environments, C¹³-NMR peaks represent carbon ***functional groups*** * This is more similar to IR, where the number associated with the absorbance helps you predict the functional group * In H-NMR we do not memorize numbers because although they ***DO*** tell us the relative degree of shielding, they ***DON'T*** directly correlate with *specific functional groups*

Answer 84

***MAGNETIC FIELD, B***

Answer 85

* All nuclei with an odd atomic or mass number have what is called **"****nuclear spin"** * ****—a concept analogous to electron spin * When an external magnetic field is applied to a group of nuclei *they will align their **own** magnetic fields* with the direction of the external field * Then, if photons are shone onto the nuclei, some of the nuclei will be able to **absorb** this photon energy and **flip their orientation** * **** so that they are lined up ***AGAINST*** the external field * It is ***THIS*** absorbance of energy that is picked up by NMR * Different nuclei will require a different **frequency** photon (i.e., different *amount of energy*) to cause this **flip in orientation** * These differences are caused by the degree to which each nucleus is shielded by neighboring hydrogens

Answer 86

* Absorbance Range: 0 – 12 ppm * 12 ppm is ***D***ownfield = ***D***eshielded * 0 ppm is *upfield* = *shielded* * The reference compound tetramethyl silane * Si(CH₃)⁴, or "TMS" * is defined as 0 ppm

Answer 87

**_Important Differences:_** 1. ***NO*** spin-spin splitting * i.e., all peaks are singlets 2. ***NO*** integration * i.e., area under the curve does ***NOT*** indicate the relative number of carbons * Absorbance Range: 0 – 220 ppm * 220 ppm is ***D***ownfield = ***D***eshielded * 0 ppm is upfield = shielded Reference compound= Tetramethyl silane * "TMS," or ***Si(CH₃)⁴*** * is defined as 0 ppm

Answer 88

FALSE. ## Footnote * C¹³-NMR on a large steroid would **NOT** detect **every single carbon** * *because it **only** detects the carbon-13 isotope* * which has a relative abundance of only about 1%

Answer 89

**_Peaks:_** * Each peak represents all of the hydrogens in a molecule **that share an indistinguishable chemical environment** * These are called "equivalent hydrogens" **_Spin-spin splitting:_** * The presence of “**neighbors**” (*non*-equivalent hydrogens attached to a neighboring carbon) **causes splitting** of the peak * The peak for a set of equivalent hydrogens will be split into exactly ***n + 1*** sub-peaks * where n is the number of non-equivalent hydrogen neighbors * The number of these sub-peaks **tells you how many hydrogens are represented by that peak** * (# of sub-peaks – 1) * This will assist you in deducing which parts of a molecule are represented by each peak

Answer 90

* This technique is used to differentiate molecules based on the differing chemical environments of: * their hydrogen nuclei (H-NMR), or * their carbon nuclei (C 13 -NMR) * On the MCAT, however, ***H-NMR*** will be tested about ***ninety percent of the time*** * An atom must have either: * an odd ***atomic number*** or * an odd ***mass number*** ***...*** to register on an NMR * For the MCAT, ***think of MRIs as an NMR _of the human body_***

Answer 91

* Ketones or aldehydes can be prevented from reaction with a nucleophile or base ***by conversion to an acetal or ketal*** (which are ***UN***reactive in ***all but acidic*** conditions) * Any terminal diol with ***at least two carbons*** will work ## Footnote **_STEPS:_** 1. One end of the diol acts as the nucleophile 1. An alcohol acts as the nucleophile, attacking the electrophilic carbonyl carbon and pushing the pi 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. 2. The other end of the diol acts as the “second equivalent of alcohol” 1. The alcohol is protonated again to form the good leaving group water, and a second equivalent of alcohol attacks the central carbon as shown below 2. 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. 3. **Acidic** conditions will return the acetal/ketal to the **original** aldehyde/ketone

Answer 92

* 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**

Answer 93

* 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***

Answer 94

(Best LG) ***Cl^- \> ^-OCOR \> ^-OH \> ^-OR \> ^-NH₂*** (Worst LG) _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 butt * 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

Answer 95

***ALKENE*** * An alkene involves a sigma and a pi bond between **two carbons** ***CARBONYL*** * A carbonyl involves a sigma and a pi bond between a **carbon** and an **oxygen** Whenever you think about pi bonds, one of the first things you should consider is the **amount of overlap of the p orbitals** * Compared to two carbons, a carbon and an oxygen **can get closer to one another** * because oxygen has a *smaller atomic radius* * This allows for **more pi overlap** and therefore a stronger bond

Answer 96

**_Area Under the Peak:_** * This is a relative representation of the **number of hydrogens** *accounted for by that peak* * An ***"Integral trace"*** (a step-wise line superimposed across the spectrum) makes this easier to determine * The relative area under each curve **is given by the height** of each step. * By “relative” we mean that if one peak is 2 units high and the other is 6 units high, this tells you only that the latter is accounted for by ***3 times the number of hydrogens***—***NOT*** that the first peak accounts for 2 hydrogens and the second for 6 hydrogens

Answer 97

* There are two resonance structures for the conjugate base formed by removal of an alpha hydrogen * Remember that ***neither form actually exists*** * ******—the actual structure is a permanent hybrid of the two * In this case, the hybrid will look ***slightly more like the second structure*** * because the negative charge is on the more electronegative oxygen

Answer 98

***3*** ## Footnote * 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 * **#1** * ****is the ***only electron withdrawing group***, so it will have the ***most acidic*** alpha hydrogens * Bromines are BIG! * "Suck away" electrons The weakest **donating** group of the other three is the methyl group on **#4** * Amines (**#3**) and hydroxyl groups (**#2**) are both ***electron donating*** * but amines are **better** donating groups * because nitrogen is **less electronegative** than oxygen

Answer 99

* This scenario would occur **if the carbonyl was of the form R₁COR₂** * ...and the nucleophile was something such as a Grignard Reagent **(R-MgBr)** * **...**that added **either** R₁ ***OR*** R₂ to the carbonyl carbon * Because **two** of the substituents would be identical, **_it could not be chiral_** * This could also happen if water or hydroxide were added to the bond, **creating two –OH substituents**

Answer 100

1. **_Partial positive charge on the carbonyl carbon_** * This makes the carbon a ***good electrophile*** 2. **_α hydrogens_** * Hydrogens on the α carbon are **_surprisingly acidic_** * --*especially* given that alkane hydrogens normally *CANNOT* be removed 3. **_Electron donating/withdrawing groups_** * The reactivity of a carbonyl with a nucleophile is dramatically affected by the presence of electron donating or electron withdrawing groups on the carbonyl carbon * Donating groups: * **decrease** the reactivity of the carbonyl carbon * Withdrawing groups: * **increase** its reactivity 4. **_Steric hindrance_** * Bulky substituents attached to the carbonyl carbon **decrease** its reactivity 5. **_Planar stereochemistry_** * The sp² hybridized carbonyl carbon is **planar** * and can therefore be attacked from **either** side * When the two substituents on the carbonyl carbon are ***NOT*** identical, an addition reaction could therefore create both R and S enantiomers **in a racemic mixture**

Answer 101

* Hydrogens on a carbon adjacent to a carbonyl carbon are ***ACIDIC*** * ******due to *resonance stabilization of the conjugate base* * When there are two carbonyls separated by a single carbon (i.e., 1,3-dicarbonyls), 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 * For this reason, be on the lookout for *electron donating* and *withdrawing* groups on the carbonyl carbon

Answer 102

_A carbonyl is a ***carbon double bonded to an oxygen***_ * For the MCAT, you can treat most carbonyl analogues, such as **S=O** or **N=O**, as you would carbonyls However, it is quite possible that the MCAT would require you to use your basic knowledge of electronegativity, bond polarity, atomic radius, etc., to predict difference in reactivity between an analogue and a carbonyl

Answer 103

**_How it works:_** * The molecules of the sample are bombarded with electrons * causing them to both **break apart** into smaller pieces, and **ionize** * This will happen in a random way * producing fragments with different **masses** and **charges** * These fragments are accelerated through a narrow curved magnet called a “flight tube”. * Only particles with a certain mass-to-charge ratio (m/z) will follow the exact curved path necessary to **NOT hit the walls** and exit onto the detector at the end of the flight tube. * The strength of the magnetic field is varied from low to high * ...changing the curvature of each fragment until **all** of the fragments have struck the detector

Answer 104

* It will turn it into an **ALCOHOL**

Answer 105

Most **SUBSTITUTED** (Think of alkenes)

Answer 106

* The formation of an **additional** C-C bond Grignard Reagent: **RMgX**

Answer 107

Ketones is **NOT** oxidized further

Answer 108

First gets oxidized to **ALDEHYDES** With 2+ equiv of oxidizing agent, gets oxidized again into **CARBOXYLIC ACIDS**

Answer 109

* 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

Answer 110

**_Pinacol rearrangement_** * Requires a **VIC**-diol * OH's on *neighboring C's* Involves: 1. the spontaneous **disassociation** of one of the protonated alcohols 2. followed by a **methyl shift** 3. Following the methyl shift, the other alcohol’s hydrogen is **abstracted** 4. and the electrons condense to **form a carbonyl** * ...and quench the carbocation

Answer 111

Aldehydes are **LESS** acidic than either water **OR** alcohols! * Water and alcohols are **similar** in acidity, but you can compare them by looking at the conjugate bases * The alcohol’s conjugate base has an electron donating group ("R") destabilizing the negatively charged oxygen * making it **less** stable * therefore **less** acidic * Water doesn't have an e' donating group, therefore the negatively charged Oxygen is **more** stable (remember: Oxygen **LOVES** having a (-) charge!) * thus **more** acidic

Answer 112

Tosyl chloride is traditionally used to protect **alcohols** or other **nucleophiles** The alcohol attacks the tosyl chloride and becomes "**tosylated"**

Ochem 2 Flashcards

(143 cards)