CEM

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vibrational levels are subsets of the electronic levels

molecular energy levels excited electronic state rotational energy levels vibrational energy levels ground electric state rotational transition vibrational transition electronic transition microwave infrared uv visible there are sub divisions and within those are more subdivisions of quantized energy levels and so in the ground electronic state what we have is we have the possibility for a molecule to transition from one vibrational mode to another within each vibrational is a rotational energy level every time you go from one quantized energy state to another there's an absorption or emission of energy just so happens that vibrational transitions happen in the region of the infrared spectrum

the alkyl group migrates to the O and OH is expelled

repeat twice more attack by oh on the b oh picks up proton from h2o this reacts twice more to give 2 more alcohol molecules

the electromagnetic spectrum

shorter wavelength higher frequency higher energy gamma-ray nuclear spin transition ultraviolet infrared radio Xray microwave visible longer wavelength lower frequency lower energy

2-methylbutane

4 peaks there

.How many types of chemically equivalent H's (= signals in the 1H NMR)? A.1 B.2 C.3 D.4 E.5

1,1-dichloroethane If we move both chlorines to one carbon, the molecule is no longer symmetric, thus you should anticipate 2 hydrogen environments (B is the correct answer). there is a doublet and a quartet (small left) 2 and 4 signals

O-H (in alcohols or carboxylic acids)

3000-3500cm^-1

Note the integration #s and the downfield shift of the aldehyde H

9h 1h upfield 9h and downfield 1h 1h hydrogen is in a very electron poor area this carbonyl there's a big delta plus on it so its pulling electrons away from that hydrogen and so this signal at very low field is very indicative of an aldehyde hydrogen

Which has the lowest potential energy? Why? A. Eclipsed B. Gauche C. Eclipsed D. Anti

D The lowest potential energy would be the one where the methyl groups are as far apart as possible so there are fewer repulsions

energy diagrams for electron transitions

Each energy level has a quantum number The higher the number the higher the energy Energy levels are not orbits Electrons transition between energy levels by absorbing or emitting photons

Reaction Energy Diagram:

A reaction coordinate diagram, in which energy is on the y-axis and rxn progress is on the x is shown above. The reaction takes place in one continuous step with nucleophile addition and leaving group departure both captured in the transition state (which is why the rate is dependent on both). transition state products reactants gibbs free energy reaction coordinate

A specialized reagent that does not promote rearrangements is Hg(OAc)2/NaBH4

Here the Hg is the electrophile and adds first. It is later removed by reduction with sodium borohydride.

One way to calculate relative proportions of acid and base is to use the Henderson-Hasselbalch equation

Ka = ([H3O+][A-])/([HA]) + -log= PH = pka + log [A-]/[HA] = [A-]/[HA]=10(ph-pka)

IR Spectrum of ethanol fingerprint region

at 1500cm-1 below that is fingerprint region use for pattern matching but don't try to assign peaks

4-membered rings are reactive because of ring strain

Penicillin Cephalosporin Penicillin and other beta-lactam antibiotics contain four-membered rings as part of their structure (though these are amides and not cyclobutanes). The amides contained within these rings are fairly reactive due to the strained ring. Beta lactam antiobiotics interfere with the synthesis and replication of a major component of microbial cell walls (peptidoglycan). Most mammals have cell membranes rather than cell walls and so penicillin is (relatively) nontoxic to us.

Which is the major product? A.1 B.2 C. Both equally D. I don't know

The most substituted double bond will predominate (B)

What if we use an asymmetric alkene (which carbon does the nucleophile add to?)

The nucleophile adds to the most carbocation like (ie the one that has the lowest activation energy - the tertiary carbocation)

But if we change the conditions the order of reactivity reverses

The order of reactivity is completely reversed If put an alkyl halide in water, and heat the flask, we see the order of reactivity is reversed! Why might this be? Perhaps the reaction might not proceed in exactly the same manner as the SN2 substitution we have discussed previously ...

Predict the outcome CH3OH + H3O+

The pKa of hydronium is -1.8 and pKa of protonated methanol is -2.2, so you don't get very much buildup of "product" (in fact, equilibrium concentrations would slightly favor the products on the left). However, methanol now has a good leaving group and so chemistry may be done on the protonated "product". Doing such chemistry effectively removes the product, driving the production of more Le-Chatlier's principle.

So the product of this reaction is:

The product formed from the most stable carbocation: this is referred to as Markovnikov Addition

Here is a molecular model of CH4 methane If you try to draw it - it quickly becomes too complicated every bong angle and bond is the same

With a Lewis structure you have to INFER the 3D nature in plane out of plane behind plane Recall that Lewis structures are cartoons that do not capture the 3-dimensional shape of a given molecule (methane here looks flat but it is, of course, tetrahedral) Wedge dash representation

Now let's look at C2H6

With a Lewis structure, there is no implied arrangement of one carbon relative to another - but with the wedge-dash there is. c-c bonds allow rotation around them so it doesn't matter (single bonds)

H-1 NMR

You can see that, for H-NMR, the scale goes from 0-10 rather than from 0-220ppm. Delta is established relative to a standard peak at zero (not listed as most spectra we use are generated via software). Scale is now 0-10 ppm (now called δ)

Recall that at the atomic-molecular level energy levels are quantized They appear only in discrete levels you can't have an energy level in between an atomic energy level and the enxt

You have probably learned about atomic absorption and emission spectroscopy

Evidence from comparing rates of nucleophilic substitution reactions for different types of alkyl groups

You may have noticed that the electrophiles in our earlier examples were either methyl halides or some other type of primary halide. This is due to the fact that both the rate and mechanism of the reaction are very much substrate-dependent. As the number of methyl groups attached to the electrophilic carbon increases, the rate drops precipitously. In fact, virtually no Sn2 substitution happens on the tertiary alkyl halide. Why is this? Why does adding more methyl groups result in such a sluggish rate? 37 1.0 0.2 0.00008 rbr nacl dmf rcl nabr

IRS Spectrum of ethanol triple bonds

a little high frequency 2000 - 2800

The more alkyl groups on the carbocation the more stable (the lower in energy) that carbocation will be. The faster the carbocation will form.

alkyl groups are electron donating by conduction and hyper conjugation energy vs reaction progress

Predict the productt

an equal mixture

Cyclopentane

"Flat" bond angle closest to 109 But that would increase torsional (eclipsing) strain The molecule takes up an "envelope" conformation to relieve the strain Cyclopentane is significantly less strained than cyclobutane and cyclopropane. In fact, if the molecule were flat it would have approximately 109.5-degree bond angles between the ring carbons. However, all of the bonds in this orientation would be eclipsed, so torsional strain would be significant. To relieve this, cyclopentane bends into a series of envelope shapes (in dynamic and rapid motion)

Examples of Lewis Acid-Base reactions (which is the Lewis acid/base, draw the mechanism)!

(CH3)3B + N(CH3)3

There are two major factors that cause different chemical shifts

(a) deshielding due to reduced electron density (due electronegative atoms) and (b) anisotropy (due to π bonds). Coupling = Due to the proximity of "n" other equivalent H atoms, causes the signals to be split into (n+1) lines.

Cycloalkanes •What is the general formula of a (straight chain) alkane? •What is the general formula of a branched chain (non-cyclic) alkane? •What is the general formula of a cyclic alkane? A.CnH2n+2 B.CnH2n C.CnH2n-2 D.It depends

- A general alkane has the formula Cn H2n+s (A)- A branched alkane has the same formula (A)- In a ring, you lose two hydrogens by joining two carbons together (into the ring). The formula is thus CnH2n (or B)A ring, and a double bond, is called a "unit of unsaturation"

Carvone R or S? A.R B.S C.Don't know How many chiral centers in Carvone? A.0 B.1 C.2 D.3 E.4

- (R) carvone (spearmint) + (S) carvone (caraway seed)

Infra Red (Vibrational) Spectroscopy

-Molecular vibrations (and rotations) are quantized (just like everything else that has to do with the energies of atoms and molecules) - Molecules with more than 3 atoms can also bend - vibrations and bending are accompanied by a change in the potential energy of the molecule (why)? as the atom vibrates or bends it's requiring energy to stretch that bond (remember there is an attraction between nuclei there so if you're stretching it you have to put more energy into it) If you bend it's not quite as much energy and if you let it relax, the energy will be emitted. - The energy difference between vibrational states falls into the IR part of the system

Why does the equilibrium shift when we add base? (which species does the base react with?

-OH will react with the H3O+, and reduce its concentration. Therefore the rate of the back reaction will decrease (there are fewer molecules to collide) and the equilibrium position will shift to the right (since the forward reaction will initially still be proceeding at the same rate. Eventually the forward and backward rates will become the same and the system will reach a new equilibrium Le Chateliers Principle is a heuristic to remember this - if a change is exerted on a system at equilibrium then the system will shift to undo the change. This is not an explanation! It doesn't tell you why - but does allow you to predict.

Note: different numbers of peaks, different chemical shifts, different splitting patterns

1,2-dichloroethane 1,1-dichloroethane top spectrum all the hydrogens in the same environment we just see one peak and they all behave exactly the same but in the bottom spectrum what we see is that this hydrogen here which is on the carbon that's got two chlorine so it appears downfield but it's split into four signals called a quartet and this methyl group here is further upfield and it appears as a doublet

How many types of chemically equivalent H's (= signals in the 1H NMR)? A.1 B.2 C.3 D.4 E.5

1,2-dichloroethane This molecule is symmetric, so there is only 1 hydrogen environment (A is the answer) around 3.6

Methyl cyclohexane

1,3 diaxional interactions

Just like acid-base reactions nucleophilic substitution reactions are reversible. •For these reactions do you think the reaction would proceed as written or would it reverse? Why? •CH3Br + -OH •CH3OH + -Br A.Yes à B.No ß C.Don't know

1. 2.

Absorption of IR Radiation depends on:

1. Bond strength: The energy needed to stretch a bond (make it vibrate) depends on the strength of the bond. therefore multiple bonds appear at higher frequency (energy) than single bonds 2. Whether the vibrations involved/involve bond stretching or bending: it is easier to bend a molecule than to stretch a bond so bend appear at lower frequency 3. The masses of the atoms in a particular bond or group of atoms: bonds to very light atoms (particularly H) appear at higher frequency 4. The dipole moment of the molecule must also change

Assigning configurations: the Cahn Ingold Prelog Convention

1.Look at each atom connected to the chiral carbon and rank by atomic number, highest first. 2.If you can't make a decision here, go out to the next atom in each chain - and so on until you get to the first point of difference ... 3.Multiple bonds are equivalent to the same number of single-bonded atoms 4.Now place your eye so that you are looking down the bond from carbon to the lowest ranking group. •If the sequence high to low is clockwise: R (Rectus) •If the sequence high to low is counterclockwise: S (Sinister) 5.NOTE THIS HAS NOTHING TO DO WITH THE DIRECTION OF ROTATION OF PLANE POLARIZED LIGHT

For example

1.Rank the groups I>Br>Cl>H 2.Look down the C-H bond (luckily it is pointing back here) 3.The sequence ICl is clockwise, so the molecule is R.

Carbonyl (C=O)

1680-1800cm^-1

IR Spectrum of ethanol all

1700 carbonyl acetic oh is extremely hydrogen bonding so a broad peak very indicative of a carboxylic acid

There's the aldehyde peak! CH2 peaks can get a bit missy because, in fact, splitting does not totally end once you get beyond the adjacent hydrogens. For the exam, we will generate cleaner spectra for you.

1H-very downfield triplet/because its split by the ch2 2H-adjacent to another ch2 next to the carbonyl and 54 is furthest down quartet 2H- ch2 airfield at one is this one because its more complicated and because its further away from the carbonyl multiplets 3H- a 3h triplet

C-13 NMR

2 kinds of hydrogen c=o appears very low field c double o

Compare it to the low field peak in 2-propanol

2-propanol c at 60

Strain energy vs ring size

3 and 4 membered rings are highly strained no strain for a six-member ring relative to the open chained releases just as much or little energy as an open chain hydrocarbon

Which signal can be attributed to C-1? A B C

A chemical chift at 47ppm THE MOST DESHIELDED THE most downfield its easier to cause resonance because some electron density has been removed This is the first direct evidence we have for the inductive effect! The ppm at which the signal appears is called the chemical shift Downfield - nuclei are de-shielded, it is easier to cause resonance some electron density has been removed c2 is b cs is c evidence of inductive effects

Shifts and eliminationssssss

A hydride shift occurs here

The electronic environment of each nucleus affects the position of the peak pt2

A stronger magnetic field would need to be applied to make the nuclei flip

Evidence from studies on optically active substrates: There are three possibilities

A) R B) S C) R and S (a racemate) nal, dmf Sn2 reactions proceed with inversion of stereochemistry; the answer is B.

Which is a Lewis acid but not a Bronsted Acid? A.CH3CO2H B.BF3 C.Mg2+ D.Fe2+ E.B, C and D

A. Acetic acid is a Lewis Acid and Bronsted acid because of the H+ B. is a lewis acid no bronsted acid (no protons) has empty orbital sp2 empty p orbital that a base can donate electrons too C/D metal cations are also lewis acids not bronsted so E is the answer

Which bond do you think it would require more energy to stretch (make vibrate) A. C-H B. C-C C. Both the same

A. C-H we can use hookes law model bonds as springs and predict behavior v=1/2pi sqrtk/m frequency (v) is related to a constant which is the bond strength but is also inversely related to the mass of the atoms the lighter the atoms the more energy it takes to make it vibrate CH bonds C-H bonds take more energy to vibrate than Carbon carbon bonds

Which is more soluble in water? Why? A. Ethanol completely miscible B. Dimethyl ether 71g/L C. Both equally D. I don't know

A. Ethanol completely miscible Water is a polar solvent, usually the more polar the solute - the more soluble it is. There are more favorable interactions. "Like dissolves like" is a useful heuristic but an over-simplification! same molecular formula different structure ethanol has strong interactions that interact with the water

Which is the better nucleophile? A. -CH3 B. -NH2 C. -OH D. -F

A. Extremely reactive, wants to pick up a proton strongest nucleophile

How many C-13 signals do you predict for acetone A.1 B.2 C.3 D.I don't know

B.2

Which is a nucleophile? A.CH4 B.NH3 C.NH4+ D.B and C

A. Not a nucleophile because it doesn't have a lone pair B. ammonia there is a lone pair on N C ammonium ion no lone pair not nucleophilic not basic

Examples of Lewis Acid-Base reactions (which is the Lewis acid/base, draw the mechanism)

BF3 + H2O Example of a Lewis acid base complex boron gained electron so now it has a negative charge regularly reversible

Which compound has an electrophilic carbon? A.CH4 B.CH3OH C.CH3Li D.All of them E.I don't know

A. ch bonds are non polar so no electrophillic carbon B methanol polar bond oxygen has - c has + so c is electrophillic C lithium is an alkaline metal so L has + so c will have - so it is a carbanion exact opposite of electrophillic

which bond do you think it would require more energy A. stretching a bond B. bending/wagging etc C. both the same

A. stretching a bond pulling atoms apart takes more energy than it would take to bend a bond bend a molecule

Which has the highest potential energy? Why? A. Eclipsed B. Gauche C. Eclipsed D. Anti

A., the eclipsed conformation (by about 21 kJ/mol relative to the anti conformation Highest Potential Energy means the least stable A is the least stable because the methyl groups are really close and repelling each other

In C-13 NMR every chemically distinguishable type of carbon appears as a peak in the spectrum How many chemically distinguishable carbons in A.1 B.2 C.3 D.I don't know

A.1 There is only 1 type of carbon in this molecule (both carbons are in the exact same chemical environment

At pH 7 what is the ratio of CH3CO2- to CH3CO2H (pKa = 5)? A.100:1 B.1:100 C.2:1 D.1:2

A.100:1 At pH 7 for carboxylic acids the [conjugate base] > [conjugate acid] So [CH3CO2-] > [CH3CO2H] Carboxylic acids are mainly in their deprotonated form at pH 7

If we have a 0.1 M solution of the acid and measure the pH - we find it is ~ 3.0. What is the approximate concentration of H3O+? What is the approximate % of un-ionized acid? A.99% B.90% C.50% D.1% pH = -log[H3O+] [H3O+] = 10-pH [H3O+] = 0.001

A.99% 0.001X0.1=1% ionized

Which is the better leaving group? A.H2O B.NH3 C.Same D.I don't know

A.H2O because oxygen is more electronegative and can hold onto the electrons better

Which is the better leaving group? A.H2O B.-OH C.Same D.I don't know

A.H2O has positive charge kinda rare h2o goes away which is stabel and can except the electrons conjugate base to a stronger acid than OH conjugate acid OH is not a good leaving group

Consider butane again, as you rotate around the C2-3 bond notice what happens as the CH3 groups get closerWhat do you predict happens to the potential energy of the molecule?

A.Increase B.Decrease C.Same D.Don't know The potential energy increases as they repel!

Does a shielded nucleus require a stronger or a weaker magnetic field to make it flip (at the same frequency) A.Stronger B.Weaker C.Same Don't know

A.Stronger the more electron density their is around a nucleus shielded it is

Identifying chiral carbons and assigning configuration Does this molecule have a chiral carbon? A.Yes B.No I don't know R or S? A.R B.S C.Don't know

A.Yes A.R

Identifying chiral carbons and assigning configuration R or S? A.R B.S C.Don't know Does this molecule have a chiral carbon? A.Yes B.No C.I don't know

A.Yes A.R

Anti-Markovnikov addition

BH3 THF H2O2/OH H3C H2 O H C

Synthesis from alkynes .

Reactions that do not go via a carbocation

Addition of Br2 across a C=C. Typically Br2/CCl4 is used (a brown solution) when the reaction occurs the solution goes colorless.

Bonding

Alkenes Electron rich sp2 hybridized One pi bond Restricted rotation provides for a new type of stereoisomerism Cis/trans, or E/Z Alkynes Electron rich sp hybridized 2 pi bonds Cylindrical (so no cis/trans isomerism) Terminal H are acidic.

5-2: Reactions of Alkenes

Alkenes are electron rich: Which part of the reagent will react with an alkene first? 1. Electrophile (E+) adds first 2. Nucleophile (Nu-) 3. Either

The conformations taken up by molecules are a result of the attractions between polar parts of the molecule and repulsive interactions that occur when electron rich areas get too close

Although most C-C bonds allow full rotation, there are energy costs associated with rotating. As carbons rotate around the bond axis, the distance between groups attached to each carbon (even if they are hydrogen) changes. Recall that bonds are loci of negative charge and that negative charges repel one another. If two bonds get closer, they feel a greater repulsion than if they are relatively further apart. We can draw analogy here to our previous discussion of LDFs in CEM 141. Recall that as two atoms approach they initially attract due to the instantaneous dipole in one atom inducing a dipole in the other. Attractive interactions of this type lead to a decrease in potential energy of the system (repulsive interactions correspond to an increase).

Reaction with a stronger base (eg -NH2 (amide ion) will remove the H, pKa NH3 = 33)

H3C C H NANH2

Or by an alkyl shiftt

An alkyl shift

Potential energy changes

As centers of electron density get closer to each other, they repel - and the PE goes up. We typically draw longer chained alkanes in their "anti" conformation (hence the zig-zag looking chains of carbon that can be seen in textbook phospholipid depictions) conformations differ only in the rotation around the carbon

Review •Imagine you have a (big) molecule with parts that are partially charged. •What happens to the force of attraction as the oppositely charged parts get closer •What happens to the potential energy? A.Increases B.Decreases C.Stays the same D.Don't know

As oppositely charged pieces get closer together, the potential energy of the system decreases and the force of attraction increases (as it is distance-dependent).

4-5: Elimination Reactions

At the end of the last lecture, we began to think about all of the many things that could occur after a carbocation is generated. As it turns out, the propensity of these highly reactive electrophilic carbons to produce many products makes their generation problematic from a synthetic point of view. Add to this that most of the products would be very, very difficult to separate and you have a subset of reactions to most often be avoided.

4-6 Elimination Reactions (E1)

4-7 Rearrangements and eliminations of carbocation intermediates

Predict the outcome! CH3Br + NaN3

Azide anions are electron-rich and fairly nucleophilic (nitrogen does not have an overly high Zeff and can often donate LPs readily to form bonds). Azide adds while Br- leaves to give methyl azide azide ion methyl and sodium azide Carbon + Bromide- NA+ N3- -N=N=N1 (middle N is plus) Use the lone pair on the N to attack the Carbon and kick out the bromide everything that doesn't take part in the reaction stays bonded

Two chiral centers: 4 possible stereoisomers. How are they related? What is the relationship between I and II A.Identical B.Enantiomers C.Conformations Something else What is the relationship between I and III A.Identical B.Enantiomers C.Conformations D.Something else

B.Enantiomers D.Something else diasteromers 2/3 1/3 diastereomers

Fischer Projections: Mostly used for sugars Is this R or S? A.R B.S Is this R or S? A.R B.S

B.S A.R the horizontal bonds come towards you vertical bonds are going back in in this case, hydrogen is pointing out at us so the answer is the opposite

what is a property of a good nucleophile?

BASIC

Which is a nucleophile A. ch4 B. nh3 C.nh4+ D. B&C

B. NH3

Which is the better nucleophile? A.H2O B.-OH C.Same D.I don't know

B.-OH On the left, the conjugate base is substantially more nucleophilic than the corresponding acid (so the answer is -OH). On the right, carbon has the lowest effective nuclear charge of the bunch and is therefore the best base (it is very small however and has a relatively high effective nuclear charge compared to atoms in lower rows; CH3- is very, very basic and nucleophilic). The conjugate base is going to be more nucleophilic than the conjugate acid

At pH 7 what is the ratio of CH3NH2 (base) to CH3NH3+ (acid, pKa = 10) A.1000:1 B.1:1000 C.3:1 D.1:1

B.1:1000 At pH 7 for amines the [conjugate acid] > [conjugate base] So [CH3NH3+] > [CH3NH2] Amines are mainly in their protonated form at pH7 Base (CH3NH2) is much lower

The Effect of Polar Aprotic Solvents on the Nucleophile •Solvation lowers the energy of the solvated species, making it less reactive. An unsolvated nucleophile will be higher energy and more reactive (B) •If X is the energy of the solvated nucleophile, which is the most likely relative energy of the unsolvated nucleophile?

BR- IN DMF NOT SOLVATED MORE REACTIVE GOING TO REACT FASTER BR- ION CH3CH0H X(SOLVATED)

IR spectrum of acetic acid

BROAH OH PEAK VERY INDICITVE

The need for a reference standard

Because NMR spectra depend on the field strength of the magnet, we don't report absolute values for any peaks, but rather the change in field strength (or frequency of light) that would make the nuclei spin flip relative to TMS à. A peak at a chemical shift of, 20 ppm means that the carbon atoms which caused that peak need a magnetic field twenty millionths less than the field needed by TMS to produce resonance. We say that the peak at 20 ppm is downfield of TMS. Tetramethylsilane TMS

This is why we don't use strong bases for SN1 reactions (i.e with hindered substrates)

Because we get elimination reactions instead. The base can't get at the central carbon - and so it picks off a proton from one of the carbons on the outside. sodium hydroxide doesn't do nucleophilic attack its too hindered

In general organic compounds are soluble in organic solvents (similar interactions)

Biological systems are aqueous (water is the solvent). Typically polar and ionic substances are soluble in water Many organic molecules are mainly non-polar and insoluble in water - so how come we exist?

The D L naming convention:

Biomolecules are named for their "similarity" to the enantiomers of glyceraldehyde. The D or L nomenclature applies to the whole molecule. In general amino acids are L, and most sugars are D when there are n chiral centers there are a maximum of 2n stereoisomers

All nuclei in organic compounds are shielded to a greater of lesser extent (because they always have electrons around them). What might de-shield a nucleus?

Bonding to an electronegative element inductive effect

Effect of polar protic solvent

Both anion and cation are solvated, therefore they are more stable and less reactive

Fischer Projections What is the configuration at C2? A. R B. S C. I don't know What is the configuration at C3? A. RB. SC. I don't know

Both are R (A) start to number from the most oxidized carbon

SN1 mechanism explains why the rate depends only on the substrate (the first step is rate determining)

By reaction coordinate diagram, an SN1 mechanism satisfies the assembled evidence. The slowest step is the formation of the carbocation (which requires a fair bit of activation energy. This short-lived, reactive species then reacts with any available nucleophile (in a fast second step) to produce products. the reaction coordinate-free energy two-step sn1 reaction the substrate determines how fast its going

IRS Spectrum of ethanol double bonds

C double bonds carbon and carbonyl bonds around 1700 centimeters to the minus one carbonyls are really strong because they change the dipole moment a lot and they're good strong peaks 1900-1600 mostly double bonds carbonyl

Predict the outcome CH3CH2I + NaOH

C has a delta plus I has a delta minus The nucleophile is the hydroxide the sodium ion is just a counter ion so the nucleophile goes to electrophile carbon hydroxide attacks ethyl iodide and leaves us with ethanol Iodine is big, floppy, and easily able to accommodate excess electrons (it is a good leaving group). Hydroxide is negatively charged with ample electrons to donate (it is a strong base). Synchronous addition of OH- and departure of I- results in ethanol.

Which is a shorter wavelength A. Radiowaves B. Xrays C. Gamma Rays D. UV Which is higher frequency? A. Radiowaves B. Xrays C. Gamma Rays D. UV Which is higher energy? A. Radiowaves B. Xrays C. Gamma Rays D. UV

C. Gamma Rays for all The shorter the wavelength the higher the frequency

Ethyl acetate How many signals in the 1H NMR? (Types of non-equivalent H?) A.1 B.2 C.3 D.4 E.8

C.3

How many chemically distinguishable carbons in CH3CH2CH2Cl (1-chloropropane) A.1 B.2 C.3 D.I don't know

C.3 the carbons are in a different environment bonded to different things

In C-13 NMR every chemically distinguishable carbon appears as a peak in the spectrum How many chemically distinguishable carbons in A.2 B.3 C.4 D.5 E.I don't know

C.4 There are 4 types of carbon in this molecule. The two methyl groups are chemically identical

IRS Spectrum of ethanol carbon hydrogren bonds

CH C-H 2800-3100

Generic Reaction- For example

CH3Br + -OH (+ Na+) à CH3OH + Br- (+ Na+) nucleophile electrophile leaving group sodium is counter iron

synthesis of an ether How would you make:

CH3CH2OCH3 nucleophile? O partial negative-CH3 Na+ Sodium Methoxide going under attack: Ch3CH2 (need a leaving group on there)Br Ch3Ch4

Generic Reaction

Carbon has delta plus with leading group L Nucleophile delta minus - and electrons nu comes in attracted to carbon and leading group leaves with a pair of electrons nucleophilic substiution

4-4: Carbocation Rearrangements

Carbocations are high energy intermediates: they will often undergo rearrangements of various kinds

Another mechanism for stabilization of carbocations involves rearrangements.

Carbocations can rearrange to give a more stable carbocation: by a hydride shift be ch3oh och3 1st. ionization of bromine with the nucleophile treatment

Predict the outcome!!! CH3CH2Cl + NH2CH3 A.CH3CH2NH2 + CH3Cl B.CH3CH2N+(CH3)H2 + Cl- C.CH3CH2CH3 + NH2Cl D.I don't know

Carbon delta plus Cl delta minus Ch3 w/e N is the nucleophile that attacks C and kicks out cl

We can change the direction of reactions by

Changing the reaction concentrations (Le Chatelier's) principle. Adding water will increase the rate of the back reaction and drive the formation of more alcohol Changing the temperature. Increasing the temperature will drive the formation of more alkene, because it will increase the contribution to ΔG from the T ΔS term in the reaction ΔG = ΔH - TΔS

SN1 mechanism explains why a chiral substrate produces a racemic mixture of products

Chiral centers racemize because the carbocations formed from them are sp2 hybridized the the nucleophile can attack from either the top or bottom

Partial reduction can produce either the cis or the trans alkene

Cis - via a "poisoned" catalyst The mechanism is the same as for alkenes and therefore the H's add from the same side of the molecule via adsorption on the catalyst. A poisoned catalyst will not reduce the less reactive alkene

Oxidation of alkenes:cis dihydroxylation

Cis because of cyclic intermediate that is not isolated permagnetic

formula of an alkene with n carbons

CnH2n

formula of an alkane with n carbons

CnH2n+2

formula of an alkyne with n carbons

CnH2n-2

Spin-spin splitting

Consider 2 H's that are non-equivalent, Ha and Hb Focus on Ha: It will experience effects of the magnetic field generated by Hb when nmr active nuclei are put in an external field then the spin states of any nmr active nuclei split and there'll be spin up spin state and spin down and each one of these squiggle across will generate its own magnetic field and so ha can experience also the magnetic fields generated by its neighboring hydrogens which has to be a hydrogen on the next carbon over so ha is going to experience two different magnetic fields because of its neighbor and so it will give two signals Hb can be spin up or spin down and so Ha will experience two different magnetic fields - and so will give two signals Hb can be against the field or with the field and we get a doublet ha appears as a doublet Two possible electronic environments produced by the hb proton Hb appears as a doublet

Make a model of cyclopropane: draw a Newman projection (down any C-C bond •Molecule must be planar (3 points define a plane) •All C-Hs is eclipsed •Bond angles highly strained - 60° rather than 109°

Cyclopropane is very strained indeed; it has a bond angle of 60 degrees instead of 109.5! everything is eclipsed

Naming alkenes

Count the longest chain with the double bond (both carbons) Start numbering so the double bond gets the lowest number Multiple double bonds - di, tri etc. Because of restricted rotation around double bonds - use cis and trans

I is a better nucleophile than F because it increases as we go down the group

Iodide is a weak base but a strong nucleophile

Which is the better nucleophile? A.F- B.Cl- C.Br- D.I- E.I don't know

D.I- Iodide is the most nucleophilic because it is the most polarizable

Which is the better leaving group? A.F- B.Cl- C.Br- D.I- E.I don't know

D.I- the least basic (even though its a nucleophile) therefore the best leaving group The ability of halides to serve as nucleophiles is inversely related to their basicity. Fluorine is the most basic and therefore the least likely to leave. Iodine, by contrast, is not very basic at all and able to easily accommodate extra negative charge. The orbital overlap between hydrogen and fluorine is fairly good as both are small atoms. By contrast, there is not much orbital overlap between hydrogen and iodine (which is much large). This results in a longer, weaker bond between iodine and hydrogen then exists between fluorine and hydrogen. Additionally, iodine has many, many electrons, another will not impact the stability of the species very much.

2-4 C-13 NMR

Detecting the different chemical environments for carbon in a molecule

Make a model of ethane C2H6 Look down the C-C bond what do you see? Newman Projection

Dihedral angle = 0° Newman Projections Dihedral angle = 60° Staggered Eclipsed In the Neuman projection, we look down the C-C bond acis. The large circle is the back carbon and the vertex at the middle of the three hydrogens is the front carbon. As it turns out, these two conformations are not equally stable. The staggered conformation is about 13.8 kJ/mol more stable than the eclipsed conformation (which makes sense as, in the eclipsed conformation, areas of electron density are in close proximity). This energy difference is easily overcome at room temperature but can be observed in low temp NMR experiments. furthest away hydrogens can get from one another is 60 degrees Front Carbon has 3 bonds that appear to be at 120 degrees (but its not it 109) the other f is the circle behind it

How to Draw Cyclohexane

Draw the chair conformation of cyclohexane and show a ring flip.The groups attached to the cyclohexane ring can assume either axial (straight up or down) or equatorial (closer to the plane of the ring). Axial Hydrogen straight up or down Equatorial hydrogen in the plane of the ring

Shifts and eliminationsss

Draw the mechanism that would lead to this product. Pay attention to carbons! An alkyl shift/ring expansion is required.

E or Z?

E Z neither

5-7: Alkynes For the most part alkynes behave similarly to alkenes

Electrophilic addition (via a vinylic carbocation)

E1 Elimination also obeys Zaitsev's rule (most substituted double bond is the major product)

Elimination reactions produce the most stable double bond; that is, the most substituted double bond. major minor h2so4

1,2-Dimethyl cyclohexane A. trans B. cis C. Both equally stab

1,2-Dimethyl cyclohexane

Equatorial substituents conformations are always favored over axial. Thus, a trans 2,3-dimethyl cyclohexane conformation in which both substituents are equatorial is the most stable than the cis conformation. Note that groups have not changed their cis or trans relationship when the ring flips. diaxial is unstable diequatorial is more stable cis-1,2 dimethyl cyclohexane trans-1,2 dimethyl cyclohexane 1 axial and 1 equatorial substituent are equally stable (but not as stable as diequatorial) Cis is the same

Good nucleophiles are generally negatively charged; they also have a counter-ion. e.g. NaBr, KCN, AgCl Which solvent will solvate the cation (counter-ion) better? A.Ethanol B.DMF C.Both will solvate it well Which solvent will solvate the anion (nucleophile) better? A.Ethanol B.DMF C.Both will solvate it well

Ethanol will solvate the nucleophile better on account of having a concentrated region of positive charge (A)Both will solvate the cation well as both DMF and ethanol have concentrated regions of negative charge (C) APROTIC SOLVENTS DONT SOLVATE NUCLEOPHILES VERY WELL AND ARE THEREFORE MORE REACTIVE SN2 ARE BETTER CARRIED OUT IN AN A POLAR PROTIC SOLVENT

What is the multiplicity of the signal indicated? A.Singlet B.Doublet C.Triplet D.Quartet E.Septet A B C

Ethyl acetate A3H B2H lower field because it's next to an oxygen C3H

In rings - for an H and the leaving group to be antiperiplanar, they must both be diaxial

For systems that can rotate freely, the antiperiplanar requirement of E2 eliminations is typically not a big deal. In rings, however, where rotation is constrained, it matters more. The proton must be trans-diaxial to the leaving group as pictured.

Oxidation and Reduction

Formal Definitions Oxidation - loss of electrons (Cu -> Cu2+ + 2e-) Reduction - gain of electrons (Cu2+ + 2e- -> Cu) but organic reactions rarely involve complete transfer of electrons More Useful Oxidation - forms more bonds to a more electronegative atom (C-H -> C-O) Reduction - forms more bonds to a less electronegative atom(C-C -> C-H, or C-O ->C-H)

The area under the curve (integrated) is proportional to the number of H's in that signal; This is not generally the case for C-13 NMR spectra

H-1 NMR In a proton NMR, the area under the peak is proportional to the number of protons that give rise to the signal. By integrating the area under the peak, we can generate an estimate of the relative numbers of hydrogens giving rise to each signal. The ratio of the peaks in this NMR is 1:3, and we know the smaller of the two peaks should have 2H, so the larger peak corresponds to 6H. C-13 signals are not reliably proportional to the number of equivalent carbons involved.

Reduction

H2/PD/C

Polar protic solvents

Have a concentrated d+ on the H (of the OH) and a concentrated d- on the O The highest concentration of negative charge is on the oxygen (A)The highest concentration of positive charge is the OH hydrogen (B) Both are fairly concentrated "point charges" POLAR PROTIC SOLVENTS CAN SOLVATE BOTH THE NUCLEOPHILE AND THE CATION (eg NaBR)

Many biological systems have Lewis acid-base complexes at their center

Heme (red) Chlorophyll- different ion in the middle and hanging off the side (green)

Experimental Set-up

Ho sample tube radio frequency output spectrum radio frequency input very strong magnetics

How many types of chemically distinct carbons? A.2 B.3 C.8 D.I don't know

Hopefully you are all amazing at this by now, from studying for the exam. Pay attention to those carbons attached to different things when discerning the number of carbon environments. In this case, there are 3 (B is correct).

Multiple chiral centers Generally when there are n chiral centers there are a maximum of 2n stereoisomers

How many chiral centers in D-glucopyranose? A.3 B.4 C.5 D.6 How many possible stereoisomers? A.6 B.16 C.32 2^5 D.64

If you have a mixture of benzoic acid C6H5CO2H toluene, C6H5CH3 and aniline. I.Which substance will be soluble in aqueous acidic solution II.which will be soluble in aqueous basic solution III.which will never be very soluble in water? IIII. all of them

I.Which substance will be soluble in aqueous acidic solution C. Aniline (basic substance) you'd get the nh3 with the benzene ring on that would be soluble in aqueous acid. II.which will be soluble in aqueous basic solution A. benzoic acid OH would kick off that proton and we'd get the corresponding carboxylate anion which is soluble in water III.which will never be very soluble in water? B. Toluene no acid or basic groups and therefore won't be affected when we add another stronger acid or base/ therefore it will always stay soluble in an organic solvent and never soluble in water

The coupling constant j separation here is called

Hydrogens that are coupled to each other have the same coupling constant no coupled hydrogens one coupled hydrogen two coupled hydrogen three coupled hydrogen any hydrogens that are split by each other have the same coupling constant

In methylcyclohexane I.If you have an axial methyl group, the H on that same carbon MUST be equatorial II.If you have a methyl group on the top side of the molecule, the H on that same carbon MUST be on the bottom side A.True B.False C.It depends D.I don't know

I. True (A)II. False (B)On a given cyclohexane carbon, a substituent is always either pointed up or down. For a substituent to change from "up" to "down", a C-C bond would need to be broken

What happens when you ring flip methyl cyclohexane? I.An axial methyl changes to an equatorial II.A methyl on the top side of the molecule changes to the bottom side A.True B.False C.It depends D.I don't know

I. TrueII. Also false (see before slide for the rationale)

Optical isomerism: detection of chirality in molecules Electromagnetic radiation can be considered as being composed of electrical and magnetic waves that oscillate perpendicular to each other. A given beam of light contains many such waves; the orientation of these waves is independent of the other waves in the beam. We can simplify an 2-m wave by passing the light beam through a polarizer - a transparent material composed of crystalline solid that blocks all e-m oscillations that do not align with the crystal structure. Thus light that passes through a polarizer is traveling in the same plane - it is called plane polarized lightIf you place a second polarizing filter, in the same orientation as the first, the path of the plane polarized light will be unaltered (the plane will pass through at the same angle). If you rotate the second filter by 90 degrees, the beam will be absorbed (blocked). Plane polarized sunglasses reduce glare by allowing light vibrating in only one direction to pass throughIf plane polarized light interacts with an asymmetric electric field (such as one generated by a given enantiomer), the plane of vibration will be rotated. We can measure the angle of rotation using another polarizing lens and thereby detect the presence of chiral molecules. Passing plane polarized light through a racemate or solution with non-chiral solute will not effect the direction of the polarization plane. A sample that rotates plane polarize light to the right is the + or dextrorotatory isomer and one that rotates the light to the left is the - or levorotatory isomer. The amount of rotation is dependent on the concentration of the substance in solution, the path length of the sample tube, the wavelength of incident light, and the nature of the substance itself (all of these variables are defined with rotates are reported in the literature).

If the light turns to the right: + To the left: - An equal mixture of enantiomers ± (no rotation) racemate light is vibrating pass it through a polarizing filterwindow and you get planed polarized light that can interact with a chiral molecule differently

Cyclohexane

If the molecule were flat it would have torsional (angle) strain (120) And all the C-Hs are eclipsed (torsional strain) Does not exist like this. Make a model of cyclohexane

SN2 reactions proceed with inversion of configuration

Imagine an umbrella turning inside-out in the wind walden inversion

Fischer Projections For a chain - put the backbone carbons vertical and the other bonds horizontal, the most oxidized carbon at the top (carbonyl or carboxylic acid)

In general: a molecule with n chiral centers has 2n possible stereoisomers How many possible stereoisomers in this sugar?A. 1 B. 2 C. 4 D. 8 There are 2^2 or four possible stereoisomers steroisomer2^chiral center

Shifts and eliminations

In this reaction, three products are produced. Draw the mechanism that would lead to each. protonate the alcohol turn into a good leaving group

UV-Visible spectroscopy tells us something about electron transitions in atoms and molecules

Infra-Red spectroscopy tells us transitions between vibration NMR spectroscopy Longer wavelength nuclear spin

The hydrochloride is more soluble in water

It is ionic and there are strong favorable ion-dipole interactions. Add HCL to it NH is basic so it will pick up a proton and you now form a very ionic group which now makes the hydrochloride more soluble Ionic compounds tend to be soluble in water because they are highly favorable interactions We can determine the solubility of various organic molecules by changing the pH if they've got acidic or basic groups on them

What does this evidence tell us?

It tells us that both the substrate and the nucleophile are present in the rate determining step for the reaction

Zaitsev's Rule: Elimination produces the most substituted double bond

Just as with E1 reactions, E2 reactions will produce the most substituted product preferentially. This is due to stabilizing interactions between the LUMO of the double bond and HOMO of the alkyl group

H-1 NMR Which is the peak that corresponds to the CH2? A. B. C. I don't know

Knowing the trend for deshielding in carbon, which proton NMR peak do you think corresponds to the CH2 group? Note that the CH2 is closer to an electronegative atom than the CH3. The answer is A.

Mass Spectrometry

Mass spectrometry is the technique that enables us to ascertain the molecular mass of a given molecule. A sample is injected into the instrument, ionized (by bombarding it with electrons), accelerated through a magnetic field and deflected by another magnetic field. Lighter ions are deflected more than heavier ions (sometimes to such an extent that they do not hit the detector). Ions are separated by charge and mass (and displayed as a mass/charge ratio). The highest peak is typically the molecular weight of the injected sample (called the molecular ion). Observing what fragments are generated from electron bombardment can also be useful in figuring out what compound has been injected.

2-3 Nuclear Magnetic Resonance (NMR)

NMR tells us about the environment of atoms in molecules and how the atoms are connected.

Nucleophilic Attack at an sp2 hybridized carbon

NOT at an alkene - too electron-rich At a carbonyl - because the C has a large delta + We can draw a resonance form when there's a delta plus on the c and delta minus on the o that means that a carbonyl is going to be susceptible to a nucleophilic attack h3c h c o has got a pi bond and a sigma bond so it won't be attacked by something with a negative charge or extra electrons

Anti-Markovnikov addition of H2O:Hydroboration/oxidation also does not go via a carbocation

NOTE: The H and the OH end up on the same side of the molecule. This reaction is regiospecific and stereospecific. In this case B is the electrophile that adds first and is later removed and replaced by OH boron electrophile

BUT it is important to know when resonance comes into play (and when it doesn't)

NR vinyl position

Trans via a stepwise reduction of e- and H+

Na metal in liquid ammonia is a source of electrons Because the reaction is stepwise, the more stable trans product is formed. NA/NH3

Reduction of alkenes (addition of hydrogen across the double bond)

Need a catalyst, addition is syn (both H's from the same side)

Make a model of cyclopropane

Note from your model that six of the hydrogens on cyclopropane are eclipsed; that is, their loci of negative charge are quite close (a very high energy state). To relieve some of the bond angle strain of a 60-degree bond angle, cyclopropane has bent bonds. The electron density of cyclopropane bulges out rather than being located between two carbons (making the band a bit larger). Both torsional strain (from eclipsing H's) and angle strain contribute to cyclopropane's t instability everything is eclipsed

Effect of polar aprotic solvent

Only the cation is solvated, therefore the anion (nucleophile) is less stable and more reactive. The "naked" nucleophile is more reactive. VERY DIFFUSE WHICH DOES NOT SOLVATE THE ANION WELL BR

Confusing isn't it?

Nucleophile/base strength Methyl Primary Secondary Tertiary Strong/strong e.g. -OCH3 SN2 SN2 E2 E2 Strong/weak e.g. RSH, halide ions SN2 SN2 SN2 NR Weak/strong e.g. -OtBu, NaH NR E2 E2 E2 Weak/weak e.g. H2O, CH3OH NR NR SN1/E1 SN1/E1

n+1 rule: n equivalent H's will split an adjacent signal into an n+1 multiplet

Number of neighbors multiplicity relative intensities of individual peaks doublet triplet quartet quintet sextet septet table

If we add Br2 to cyclohexane?

Only A (the trans product is formed) Addition of Br2 is stereospecific

Note: NaOCH3 sodium methoxide is made by adding solid sodium (or potassium) metal to CH3OH CH3OH + Na(s) à CH3O- Na+ + 1/2H2(g)

Or by using the non-nucleophilic base, sodium hydride (NaH) CH3OH + NaH à CH3O- Na+ + H2(g)

Note: NaOCH3 sodium methoxide is made by adding solid sodium (or potassium) metal to CH3OH CH3OH + Na(s) à CH3O- Na+ + 1/2H2(g) What kind of reaction is this? A.Acid base B.Nucleophilic substitution C.Redox D.Don't know

Or by using the non-nucleophilic base, sodium hydride (NaH) CH3OH + NaH à CH3O- Na+ + H2(g) The first reaction is a redox reaction: sodium is oxidized (goes from zero to +1) and hydrogen is reduced (goes from +1 to 0); the answer is CReaction of sodium hydride with methanol is an acid-base reaction (with H- as the base)

Recall: C-C sigma bonds allow free rotation

Or do they???? Does it matter how you draw C2H6?

Optical Isomers: have none-superimposable mirror images:they are chiral. Two compounds that are mirror images of each other are called enantiomers

Other objects beyond molecules can have chirality. The example everyone likes to use is your hands (As they too are nonsuperimposable mirror images of each other ... probably). Two compounds that are nonsuperimposable mirror images of each other are called enantiomers

Solvent diagram Polar aprotic (has no protons attached to electronegative atoms) Polar protic (does have protons attached to electronegative atoms)

Polar aprotic (has no protons attached to electronegative atoms) •Acetone •Dimethyl Formamide (DMF) •Dimethyl sulfoxide (DMSO) Polar protic (does have protons attached to electronegative atoms) •Ethanol, methanol •Water

5-5 Polymerization of Alkenes

Polymers are formed from smaller units (monomers). There are huge numbers of biologically and commercially important polymers

Reactions that go via a carbocation result in racemic products if a chiral center is formed. Addition of HBr is regiospecific(we know which end h+ adds to) but not stereospecfic(both possible)

Predict the product

There are two general mechanisms for elimination reactions: E1

Rate = k[(CH3)3CBr) - this is an E1 reaction: only the alkyl halide is present in the rate-determining step e1 depends only on the concentration of the substrate

There are two general mechanisms for elimination reactions: E2

Rate = k[(CH3)3CBr] [-OH](base) This is an E2 reaction: both the alkyl halide and the OH are present in the rate-determining step

SN2 reactions are accelerated by polar aprotic solvents

Reactions are almost all carried out in a solvent of one sort or another and the character of that solvent has significant impacts on the rate and likelihood of a given process occuring.

Reaction of a ketone with a Grignard

Reagent 1: ch3mgbr, dry ether(because carbanion CH3- is highly reactive and if there was water it would just simply protonate but if no water it will attack the carbonyl ) (methyl magnesium bromide) Reagent 2: h3o+ So the electrons go off onto the oxygen and we form a new carbon-carbon bond from the attack the magnesium bromide counter ion off to the side Adding an aqueous acid to protonate the oxygen and end up with a tertiary alcohol took a carbonyl with three carbon units and made it into a four carbon unit on a different functional group

Recall (from right before the exam) that SN2 reactions are second order; that is, both the nucleophile and the substrate take part in the rate determining (i.e. slowest) step of the reaction. Thus, SN2 reactions proceed in a single step in which the nucleophile forms a bond to the electrophilic carbon as the leaving group leaves. The transition state for this process is flat and sp2 hybridized. Conversion inverts as the nucleophile forms a bond opposite the leaving group. The last point we have not yet discussed; SN2 reactions are accelerated by polar solvents without acidic protons (like DMF).

5-3 More Reactions of Alkenes

Recall all the reactions we have discussed so far occur via a carbocation intermediate Therefore these kinds of reactions may suffer from the problems we saw with carbocation rearrangements.

SN2 Transition State

Recall that all the reactants are dissolved in a solvent and that reactant molecules are both colliding with each other and the solvent. For a nucleophile and electrophile to react, they first have to collide with each other. Further, that collision must transfer enough energy so that the complex (nucleophile + electrophile) can form the transition state. To complicate things still more, formation of the transition state requires molecules collide in a particular orientation. Remember that, from our prior discussions, the transition state has the central carbon connected to five groups: the incoming nucleophile, the leaving group, and three other substituents that do not change during the reaction. As the bond forms between the nucleophile and electrophile, and the bond between the carbon and leaving group breaks, the carbon changes hybridization ... In other words, in the transition state, the carbon is attached to non-reacting substituents by sp2 orbitals. The nucleophile and leaving group are thus interacting with different sides of a p orbital.

Protic solvents aid in pulling off the leaving group, and stabilizing it sn1 accelerated by a protic solvent

Recall that polar protic solvents (like water and ethanol) contain a dipole; a partially positive and partially negative region. The dipole on these solvents serves two functions 1) it can solvate the leaving group, in effect helping to remove it from the carbocation through interactions with the positive end of the solvent dipole. 2) It can solvate the carbocation through interactions with the solvent dipole's negative region (lowering the energy of the intermediate) protic solvent assists ionization of leaving group negative end of solvent dipole stabilizes the carboncation intermediate

2.5 Intro to1H NMR Spectroscopy

Recall that several of the examples we examined when discussing C-13 NMR had one or more possible structures (i.e. an aldehyde or ketone with the same number of chemically distinguishable carbons). As you might imagine, this will not do if we want to know exactly what compound has emerged from our reaction mixture "witch's brew". Enter H1-NMR, which will enable us to differentiate between different kinds of hydrogen!

2.7 1H NMR Spectroscopy practice and mass spectrometry

Terminal alkynes are acidic

Recall that the C is sp hybridized. The negative charge on the resulting anion is held closer to the nucleus than in sp3 carbon - and so more stable. (s orbitals have a higher probability of electron density closer to the nucleus) PKA>55 PKA 45 PKA 26

Conformations: the ways that molecules arrange themselves by rotating around single bonds

Recall that two parts of a molecule can rotate around a sigma bond. In fact, at room temperature, the carbons in most sigma bonds are rotating freely and rapidly with respect to each other (though, of course, we have to portray them as static in our drawings). different conformations have different energies

Mechanism of exchange

Recall what we discussed regarding protonations and deprotonations being very, very rapid? You might protonate the alcohol with deuterium from D2O and deprotonate the hydrogen to give ROD, which would not be NMR active (thus, the disappearance of the alcohol HNMR peak). our oh can exchange by picking up deuterium of d2o heavy water giving this protonated species and this very small equilibrium amount of that species but there is some immediately will carry on every exchange a proton and those more or less than equal chance that the deuterium and hydrogen will go back and if the hydrogen exchanges and what we end up with its the deuterated product which now you don't see that oh signal nmr and finally

Remember:

Repulsive interactions (same charges) raise the potential energy - making it less stable. Attractive interactions (of opposite charges) lower the potential energy of a molecule - making it more stable

Carbocations are high energy intermediates: any mechanism that can stabilize the carbocation may allow SN1 reactions.

Resonance stabilization can help primary carbons undergo SN1 reactions allyl position be h2o what happens when the water solvent pulls off the bromide

Conformations of rings If these molecules were flat, what would the bond angles be? Cyclopropane Cyclobutane Cyclopentane Cyclohexane A.60 B.90 C.~109 D.120

Rings are very interesting with respect to conformational analysis. First, they do not allow free rotation about their C-C bonds (as their bonds are locked into a ring). For 360 degree rotation, you would have to break a covalent bond, which requires more energy than is available through thermal collisions. There are, however, a range of rotations that are accessible Cyclopropane 120, cyclobutane 90, cyclopentane, 72, cyclohexane 60. But are they flat? You should know from the bond line drawings above that each of the carbons in the above molecules are sp3 hydbridized; that is, their most stable bond angle is 109.5! Cycloalkanes adopt a range of different conformations to minimize angle strain and repulsions between functionalgroups. Ring strain can be experimentally determined by measuring the heats of combustion for cycloalkanes: the energy released by this reaction is a proxy for the stability of the particular cycloalkane.

Another view - now the H is not pointing back

STRATEGIES TO ASSIGN CHIRALITY WHEN THE LOWEST RANKING GROUP IS NOT POINTING BACK ● 1.Make a model (and be sure to make the correct enantiomer), then rotate the model so the H is pointing back ● 2.Switch two groups so that the H is pointing back - this will give you the opposite enantiomer - and you must switch back from the configuration you assign. ● 3.Imagine yourself moving around the back of the molecule so you are looking down the C-H bond ● 4.Rotate the molecule in your head so you are looking down the C-H bond ● Always use at least 2 methods to check on yourself.

Protonation in the absence of a poor nucleophile (SO42-) CH3OH + H2SO4 à CH3OCH3

Sulfate is a poor nucleophile because it's not reactive not polarizable and a weak base protonated dimethyl ether than can be detonated easily by water or alcohol

Sn2 reactions occur more rapidly when carried out in a polar aprotic solvent - i.e. a solvent that is polar, but has no acidic protons. Acetone, DMF, and DMSO are all polar aprotic solvents, while water and ethanol, which have acidic hydrogens off their OH groups, are polar protic. Polar aprotic solvents solvate cations quite well (as they have a concentration of negative charge, usually on their carbonyl oxygen) but are less good at solvating anions (due to the diffuse positive charge in the molecules). Thus, they act to solvate the counterions of nucleophiles like NaCN. This leaves the anion unsolvated, increasing its reactivity and energy.

C-13 spectra usually recorded between 0 and 200 ppm

Spectra are recorded in ppm, relative to a standard at 0 ppm (this is so that we can compare spectra from different spectrometers) Downfield (low field) Upfield (high field)

Four possible electronic environments produced by the hb proton Hb appears as a quartet

Splitting from 3 equivalent hydrogens all 3 hb against field one with two against two with one against al three hb with field hb appears as a quartet

Alkenes may also react via radicals (species with unpaired electrons)

Step one: initiation of a radical - usually via breaking a weak O-O bond Step two: addition of the radical to the double bond (to give the most stable radical) Step three: chain propagation Step four: chain termination - when the chain meets another radical R. hydrogen peroxide a highly reactive radical (unpaired electron species)

Recall that under SN2 conditions the order of reactivity DECREASES AS YOU INCREASE THE NUMBER OF ALKYL GROUPS AROUND THAT CENTRAL CARBON because you can't get the nucleophile through the clouds of electron density but if you change the solvent to water and heat it up slightly the order of reactivity is completely reversed

Steric hinderance! The nucleophile cannot easily access the the electrophilic carbon opposite the leaving group.

Examples of Lewis Acid-Base reactions (which is the Lewis acid/base?)

Tetrahydrofuran (THF)

Thalidomide Which is the R isomer? A. B.

Thalidomide was originally prescribed for morning sickness. Now we know the R isomer is the sedative, but the S isomer is teratogenic. The drug was administered as a racemate with terrible consequences in the 1960's. Now we know that even the pure R enantiomer racemizes at physiological pH. Thalidomide is now used to treat leprosy. The reason carvone enantiomers smell differently is living systems are asymmetric on the molecular level. That is, biomolecules have active sites with particular shapes that might interact with one enantiomer of a compound and not the other. The most famous, or perhaps infamous, example of this is Thalidomide. The R enantiomer of thalidomide is an effective sedative and was used in Europe to combat morning sickness. Unfortunately, the S enantiomer causes birth defects associated with the development of limbs. A large number of children were born with limb deformities while thalidomide was still on the market. A is the R isomer

The E2 reaction has a stereroelectronic requirement:

The H and the leaving group should be antiperiplanar

At pH 5 what is the ratio of CH3CO2- (pKa = 5)to CH3CO2H? A.10:1 B.1:10 C.1:1 D.1:2

The Henderson-Hasselbalch equation also tells you when there are equal amounts of conjugate acid and conjugate base (When the pH = = pKa the [acid]=[base])

Polar protic solvents and solvation If we dissolve NaBr in solution which part of the molecule will interact with the Na+ A.the O B.the H C.the CH2 D.the CH3 Which part of the molecule will interact with the Br-? A.the O B.the H C.the CH2 D.the CH3

The Na+ will interact with the partial negative charge on the oxygen (A.)The Br- will interact with the partial positive charge on the hydrogen (B) Polar protic solvents can solvate both positive and negative charges well due to having concentrated regions of both positive and negative charge. Recall that solvation is an interaction that lowers the energy of a species, making it less reactive (thus polar protic solvents can blunt the reactivity of nucleophiles).

Polar protic solvents can solvate both the nucleophile and the cation (eg NaBr) The solvent can interact strongly with both the cation (Na+) and the nucleophile (Br-)

Which proton will be removed? A B C-both

The answer is still Hc; hopefully re-drawing this as a chair will help matters. Elimination of Hc will produce product A

Newman projections of cyclohexane

The bending of cyclohexane can completely relieve all angle and torsional strain. See, all of the bonds are staggered in the chair conformation! In fact, cyclohexane is as stable as hexane; that is, there is no ring strain associated within a six membered ring

Identification of acidic H's (bonded to O, or N - usually)

The deuterated solvent can be used to identify particularly acidic hydrogens in a molecule. In this case, we have an OH (pKa around 15), which we might like to get evidence for. Upon treatment with D2O, we can see the OH peak vanish! How might this be done? any hydrogen that is bonded to an electronegative element is constantly being exchanged with other hydrogens in that same system PArtial 1h nmr spectrum of menthol in cdcl2 if you add a drop of d2o heavy water the oh group exchanges with deuterium and that peak goes away nice way of figuring out which peak is the oh, nh peak

As 1H is the most abundant isotope of hydrogen, it is a certainty that a given molecule will have more than 1 1H within it (thus the spectra can be obtained in high resolution from less sample in less time). Also, as we will see, NMR active nuclei interact with one another and can provide rich information about specific connectivity in a molecule (i.e. what is next to what).

The effect of shielding and de-shielding is exactly the same in 1H NMR as it is in 13C NMR. That is, more deshielded (i.e. electron poor) hydrogens appear downfield and more electron-rich hydrogens appear upfield.

In any spectroscopic technique we are measuring: The energy of a molecule The energy of an atom The energy change from one quantum state to another The energy to break a bond

The energy change from one quantum state to another

When you dissolve acetic acid in water there is a proton transfer The equilibrium lies to A.The right (as drawn) B.In the middle C.To the left D.Impossible to predict If we add acetic acid to water what is the pH? A.Below 7 B.7 C.Above 7 D.Impossible to predict

The equilibrium lies to the left because the hydronium ion is a much stronger acid Hydronium cation is much more acidic than acetic acid (only about 3% of which is deprotonated at neutral pH). The equilibrium thus lies on the left to a large degree. If we add acetic acid to water what is the pH? If we add acetic acid to water would it make the solution more acidic than pure water? yes but not much so below 7 which is acidic.

Optical Isomers Here are two models - they look the same. - but are they?

The have 4 different groups attached and are mirror images. They are chiral Undoubtedly, some of you made compounds that are not exactly the same but are mirror images

The electronic environment of each nucleus affects the position of the peak

The external magnetic field nucleus an isolated proton fields only the external field exposed to this external magnetic field from the magnet magnetic field induced by the electron a proton surrounded by electron density the increased field decreases the strength of the magnetic field felt by the nucleus the nucleus is shielded if external field goes up where the electron is actually reducing the amount of external field by a particular amount so the nucleus doesn't feel the strong external field we say that the nucleus is shielded

Polar aprotic solvents

The highest concentration of negative charge Is on the O But the positive charge is distributed among the N and the two CH3 - it is delocalized The highest region of negative charge is on the oxygen (A). This is a fairly concentrated area of high electron density The positive charge on DMF, by contrast, is diffuse and localized on the N and two CH3s (D).

Polar protic solvents Where is the highest concentration of negative charge? A.On the O B.On the H C.On the CH2 D.On the CH3 Where is the highest concentration of positive charge? A.On the O B.On the H C.On the CH2 D.On the CH3

The highest concentration of negative charge is on the oxygen (A)The highest concentration of positive charge is the OH hydrogen (B) Both are fairly concentrated "point charges"

Polar aprotic solvents Where is the highest concentration of negative charge? A.On the O B.On the N C.On the CH3 D.On the H next to the C=O Where is the highest concentration of positive charge? A.On the O B.On the N C.On the two CH3 D.On the N and the two CH3

The highest region of negative charge is on the oxygen (A). This is a fairly concentrated area of high electron density The positive charge on DMF, by contrast, is diffuse and localized on the N and two CH3s (D).

Which mechanism predominates?A. SN2 B. E2

The less hindered base will favor an SN2 substitution (A) The more hindered base will favor an E2 elimination (B)

E2 eliminations: Rate - k[RX][B]

The loss of the H+ and Br- occur at the same time. We know this because the rate of this reaction = k[tBuBr][OH] This is an E2 reaction (Elimination, second order) The H+ is removed from the beta carbon beta elimination second order

Monomers MUST have more than one reactive site (so that a chain can form)

The monomers form a repeating unit a+b b

Substitution, Nucleophilic, 1st order: Proposed Mechanism

The rate-determining step (slowest step) is breaking of the carbon bond from the carbon to the leaving group It then gives you a carbocation which is very reactive and only has three centers of electron density, its sp2 hybridized, very reactive So ita going to react to what's there, in this case, is the solvent (H2O) which attacks the carbocations center to give this protonated alcohol that then can be deprotonated by some more water and end up switching the bromine with an oh group Given that the observed reaction rate is dependent only on substrate concentration, the slowest step of the process must involve only substrate. Bond breaking must be involved as that is all that can happen with only one molecule involved in the RDS. Thus, perhaps the bond to the leaving group breaks leaving a high energy carbocation. Carbocations are intermediates, not transition states. They can be detected by spectroscopic methods whereas transition states exist only for one molecular vibration. Carbocations will react readily with any potential nucleophile (water, in this case). Although water is not a great nucleophile, it will react with a carbocation to give the protonated form of the product (which can be deprotonated by another molecule of water).

Steric Hindrance Productive collisions are very difficult once the electrophilic carbon is surrounded by methyl groups. This phenomenon is called steric hinderence and explains the correlation between rate and the bulk of the electrophile

The reactants may collide - but the collisions do not lead to the product because the nucleophile can't "get to" the electrophilic carbon accessibility of electrophilic carbon atom to nucleophilic attack reaction rate in sn2 reactions

What configuration would the product be? A. R B. S C. I don't know

The reaction must occur at the chiral center to invert the stereochemistry! It would be R (see next slide)

Now make a model of butane (C4H10) and look down the C2-3 bond. Draw a picture

The repulsion we observed in ethane is more extreme in the case of butane as methyl groups are much more massive than hydrogens. Eclipsed two methyl groups too close together equal a high energy conformation we can rotate through 60 degrees (Gauche) where the two methyl groups are right next to each other but they are not eclipsed if we rotate again we get another eclipsed conformation where the hydrogens and methyl group are now eclipsed rather than the two methyl groups if we rotate again the methyl groups are as far apart as they can possibly be therefore the electron density is not impinging on one another (anti conformation)

This energy gap also depends on the electronic environment of the nucleus

The stronger the magnetic field the larger the energy gap between the spin states Increasing mag field strength widens energy gap increasing strength of the field

How many types of chemically distinct hydrogens? (Don't worry about rotations around C-C bonds for now) A.2 B.3 C.8 I don't know

There are two different kinds of hydrogen here (the methyl groups are equivalent, as are the Hs on the CH2). The answer is A. A.2 scale from 0-10 hydrogen nmr carbon nmr scale o-200

Make a model of ethane C2H6 Look at the molecule from a side angle, what do you see? (draw a picture)

These are each different ways to draw the same molecule to highlight particular structural features

atomic absorption and emission spectra

These spectra are the evidence for the existence of quantized energy levels Caused by electron transitions between quantized energy levels in atoms The energies recorded are the energy difference between energy levels if the electron is promoted to a higher level of energy (n=1 to n=2) it has to absorb energy therefore that energy is missing from the spectrum but if the electron falls back down again it will emit exactly the same wavelength frequency of light and you'll see a bright light instead absorption electrons are promoted up emission spectra electrons are moving down

Elimination reactions often compete with nucleophilic substitutions

These two types of reactions can occur on the same substrate - that is a compound that has a good leaving group on it eliminate h and br to get an alkene

Predict the outcome CH3CH2I + NaSCN

Thiocyanate can act as a nucleophile either at sulfur or nitrogen (draw a resonance structure to convince yourself of this). However, given that it is substantially larger (i.e. has more core electrons, valence electrons feel less effective nuclear charge), sulfur is typically more nucleophilic than nitrogen and thus we can expect sulfur addition to predominate

The transition state for the rate determining step is more like the intermediate carbocation than the starting material

This is called Hammond's postulate. As the transition state for the RDS is closer in energy to the intermediate than the starting materials, it more closely resembles the structure of the intermediate. As the formation of the intermediate carbocation is the RDS, and the transition state for this formation resembles the carbocation itself, the structure of the carbocation (and relative stability of the carbocation) directly impacts the activation energy for the RDS.

The transition state for the rate-determining step is more like the intermediate carbocation than the starting material

We can use the principle of steric hindrance to stop nucleophilic substitutions on unhindered substrates

Using the strongly hindered base potassium tertiary butoxide (KOtBu)

A B C nacn dmf

This requires some flipping about of molecules, but the answer is B (which is S to the original compounds R

Hindered bases also include:

Triethylamine Di-isopropylamine

How many possible different alkenes? A.1 B.2 C.3 D.4

Two (elimination of one of the methyl groups, or elimination of one of the adjacent CHs) (B)

SN1 mechanism explains why the rate depends on the type of substrate:

Two mechanisms can stabilize carbocations and both predict that the more alkyl groups attached to the positive C, the more stable the carbocation will be. One is induction; the alkyl groups attached to the central carbon are polarizable and their electron density is attracted toward the positive charge. The more alkyl groups attached to a positive carbon, the more pronounced this stabilizing influence will be. Hyperconjugation is another mechanism of positive charge stabilization. The electron density from any adjacent C-H or C-C bond can overlap with the empty p orbital on the sp2 hybridized carbocation, forming a sort of pi bond, and again delocalizing the positive charge over the rest of the molecule. Hyperconjugation stabilizes the carbocation by spreading out the charge. It is delocalized by overlap of the C-H sigma bond with the empty p orbital on the central carbon.

As we introduce new functional groups, we will discuss their spectra

Typical C-13 NMR chemical shifts table

The presence of the pi bond restricts rotation - it would break the bond

Unless energy is added to break the bond cis-2-butene trans-2-butene

aspartic acid

carboxylic acid is deprotonated and amino group is protonated -1 at ph7

Let's Review

We already discussed a significant amount of the chemistry behind nucleophilic substitutions. Recall that a nucleophile with a concentration of negative charge and accessible lone pair can form a bond with an electrophile at the same time a bond between the electrophile and a leaving group breaks. Electrostatic attraction is the driving force behind these processes. The negative or partial negative charge on the nucleophile is attracted to the positive or partial positive charge on the electrophile. We spent a good deal of time discussing what characteristics made a good nucleophile, electrophile and leaving group back in chapter 1.

These attractive forces can include •Hydrogen bonding •Dipole-dipole •London dispersion forces •Collectively -Van der Waals forces -Non-covalent interactions

We have discussed IMFs extensively in 141 and 142 and mentioned H-bonding at the beginning of this course. Recall that they are all driven by electrostatic interaction and that attractive interactions lead to a decrease in system potential energy. in picture molecules are attracted to one another through hydrogen bonding

But didn't we just see the reverse (elimination) reaction in chapter 4?ΔG = ΔH - TΔS

We know that this reaction is endothermic ΔH is positive (because it is the reverse of the addition reaction). We also know, because the reaction happens, that ΔG is negative. This means that ΔS must be positive. This makes sense because we are increasing the number of possible arrangements - by increasing the number of molecules

5-6 Reversible Reactions

We know that water adds across the double bond ΔG = ΔH - TΔS We also know (from experiment) that this reaction is exothermic ΔH is negative. We also know, because the reaction happens, that ΔG is negative.

Often nucleophilic substitutions at secondary and tertiary substrates will be accompanied by competing elimination reactions

We spent most of last lecture discussing the substitution products of carbocation generation and rearrangement. This is just part of the story, you can also generate a product by elimination! After a tertiary carbocation is generated, the hydrogens on the carbons adjacent to it become MUCH more acidic (recall that alkyl CHs have a pKa of around 50). This is due to the electron density in the C-H bond being drawn toward the adjacent positive charge weakening the bond and increasing the positive charge on the hydrogen. As a base (ethanol in this case) forms a bond to the hydrogen on a beta carbon, the electrons from this bond go to the empty p-orbital on the adjacent carbocation. At the same time the beta carbon rehybridizes from sp3 to sp2.

The most common spin ½ nuclei in organic chemistry are C-13 and H-1

We will start with C-13 NMR which is only ~ 1% of the natural abundance of C (which is mostly C-12)

Cyclohexane Conformations

We're changing the nomenclature from Lewis acid-base to nucleophile and electrophile. As you will see, nuleophiles are Lewis bases and electrophiles are Lewis acids ...

Nucleophilic Substitution: Part IIEvidence for the Mechanism

We're changing the nomenclature from Lewis acid-base to nucleophile and electrophile. As you will see, nuleophiles are Lewis bases and electrophiles are Lewis acids ...

SN2 Reactions and the effect of solvent 4-2

We're changing the nomenclature from Lewis acid-base to nucleophile and electrophile. As you will see, nuleophiles are Lewis bases and electrophiles are Lewis acids ...

Up to now we have not offered any evidence about this reaction, except for showing the reactants and products

What evidence do we have about the mechanism?

Amino acids - have an amino group and an acid group! A never this one B acididc solution C pH 7 D basic solutiom

What form does glycine appear in at pH 7? Amino group h2n carboxilic acid co2h at ph 7 what form does glycine take C. Amino group is protonated carboxylic acid is deprotonated if d, the amino group would deprotonate first if it was b, both groups are protonated

Compounds with a plane of symmetry

What is the relationship between I and III? A.Identical B.Enantiomers C.Conformations D.Diastereomers What is the relationship between I and II A.Identical meso compunds B.Enantiomers C.Conformations D.Diastereomers

Anti-Markovnikov addition of H2O

What reagents would be used for Anti-Markovnikov addition? A. H3O+ B. Hg(OAc)2: NaBH4 C. BH3/THF: H2O2/NaOH D. O3 What is the product of AM addition of H2O to

Repulsive forces come into play:

When parts of the molecule get too close and their electron density starts to overlap When same charge (or partial charged) species get too close structure of ethane can make different structures

E/Z nomenclature

When there are more than 2 groups we can't use cis/trans. Instead we rank the groups on each sp2 carbon as high or low, using the Cahn Ingold prelog convention. Then if the high groups are on the same side Z (Zusammen, together) on opposite sides E (Entgegen, away)

the nuclei are electrically charged and many have spin that cause them to behave like a magnet

a nucleus with spin at higher energy generates a magnetic field in the opposite direction to the external magnetic fields Energy gap corresponds to radio frequency A nucleus with spin at lower energy generates a magnetic field in the direction of the external magnetic fields External magnetic field the case of the spin-1/ nucleus energy separates

For now we will consider only very reactive nucleophiles - for example a "carbanion"

a stabilized carbanion ch3mgbr Grignard reagent - in which we take an alkyl halide R-BR or CH3-BR and if you treat that with magnesium what happens is that you get magnesium kind of inserting into the carbon bromine bond and because magnesium is a metal it's got a huge delta plus on it positive charge so carbon has a big delta minus carbanions are extremely reactive you can't make them by deprotonating an alkane but you can by kind of activating the carbon with a metal and pulling the electrons off of it c mg br mg 2+ Alkyl halide treat it with lithium and you get methyl lithium CH3-Br arrow (2Lithium) to CH3 Lithium & Lithium Bromide

Order of stabilization of carbocations is

a tertiary carbocation is the Most stable(stabilized by induction plus hyperconjugation) Least stable methyl carbocation are the least stable

good leaving group

able to leave with an electron should be stable not reactive example is a weak base

What does the very low field peak tell us?

acetone c at 200 very de-shielded a carbon double bonded to oxygen is pulling electrons very very strongly away from that carbon one is way low field very de-shielded a carbon double bond to oxygen is pulling electrons very very strongly away from that carbon this is going to allow us to identify some species

But what if we change the pH - how do we predict the position of equilibrium? For example if we add some base? (change the pH to >7). A.The equilibrium position will shift to the right B.The equilibrium position will shift to the left C.Nothing D.I don't know

add o h oh A.The equilibrium position will shift to the right -OH will react with the H3O+, and reduce its concentration. Therefore the rate of the back reaction will decrease (there are fewer molecules to collide) and the equilibrium position will shift to the right (since the forward reaction will initially still be proceeding at the same rate. Eventually the forward and backward rates will become the same and the system will reach a new equilibrium Le Chateliers Principle is a heuristic to remember this - if a change is exerted on a system at equilibrium then the system will shift to undo the change. This is not an explanation! It doesn't tell you why - but does allow you to predict.

good electrophile

attracted to positive negative charge has some kind of positve or delta positive on the reacting atom c must be bonded to an electronegative atom

Energy diagram how much energy is required to move an electron from n=1 to n=3 a. 13.6 eV b. 12.1 eV c. 10.2eV d. 1.51 eV (1 electron volt (eV) =1.6X10^-19J)

b. 12.1 eV energy levels for the hydrogen atom ground state ionization level energy (eV) hydrogen

what functional group is most likely present here? A. Alkyne B. Carbonyl C.Alkyl Halide D. Alcohol

big peak around 1700 that tells us this is a carbonyl and sure enough this is acetone B. Carbonyl c=o 1700 carbonyl absorbs the most light at this peak

what functional group is most likely present here? A. Alkyne B. Carbonyl C.Alkyl Halide D. Alcohol .

big peak around 2100 the alkyne region triple bond A. Alkyne big peak around 2100 triple bond alkyne region

Three possible electronic environments produced by the hb proton Hb appears as a triplet

both of these hb atoms can be both against the field one with and one against both hb protons with the field hb appears as a triplet

Competing Elimination Reactions

bromide ionizes

carboxylic acid

c double bond o really acidic oh group

aldehyde

c double bond o h/c simplest formaldehyde

amide

c double bond o nh2 not basic because lone pair is bonded to carbonyl lone pairs on and o we don't see

ester

c double bond o o c

ketones

c double bond o two bonds to carbon carbonyl carbonyl bonded to two other carbons simplest ketone is acetone

Absorption of IR radiation depends on several factors: Which bond do you think it would require more energy to stretch (make vibrate) a. C-C b. C=C c. triple bond c d. all the same e. i don't know

c. triple bond c is stronger, harder to break, and therefore it takes more energy to make it vibrate as well

How would you make

can't do with nucleophilic attack right now instead, an attack on the carbonyl if we add together then add aqueous acid

alkyne

carbon carbon triple bonds. Sp 180 linear yne triple bonds

CO2 does not have a permanent dipole (as shown in a) but the asymmetric stretch (vibration) does change the dipole moment (in b and c) and is therefore "IR" active

carbon dioxide is linear doesn't have a dipole moment so they cancel each other our and so if this was a symmetrical stretch and both ends stretched out at the same time we wouldn't see that but what you would see it in infrared is the asymmetric stretch

alkane doesnt have a functional group

carbon hydrogen and carbon carbon single bonds doesnt react much at all gives off a lot of energy sp3

Which compound gives rise to this spectrum? What would the other one look like? A B C D 2H 3H 3H

don't see an H peak around 10, so this can't be an aldehyde a 3h singlet that tells us there's a methyl group isolated not next to any other compound only one of the possibilities has a ch3 group all by itself next to the carbonyl so we can predict that a is the answer

alkene

does have some functionality sp2

But didn't we just see the reverse (elimination) reaction?

e1 elimination electrophillic addition

Draw how the potential energy changes with dihedral angle

eclipse gauche eclipse most stable anti

in the ultraviolet and the physical

electron transitions and molecules infared spectroscopy nuclear

Diastereoisomers are stereoisomers that are not superimposable on their mirror image. For chiral species the are the same configuration at (at least) one carbon and different at (at least) one carbon

enatiomers

spectrum

energy source frequency type of radiation wavelength longer higher frequency and more energy inner electron transitions is atoms outer electron transitions in atoms Light going from x ray to visible

Epoxides are strained and will undergo trans ring opening via SN2 reactions

epoxide trans ring opening

mass of molecular ion: 108

ethyl bromide c25hbr you can see two molecular ions here because of the bromine bromine exists as two isotopes in about equal amounts \atomic weight of 79 and 81 the results of that there are two molecular ions because bromine has about 50/50 of the two isotopes

Types of vibration

every time these bonds stretch or compress the energies change in the molecule and those energies are quantized and therefore we can measure the energy that it takes to do this

Newman Methyl cyclohexane

everything is staggered ,ethyl is as far away as it could anti conformation

Radical addition to alkenes: the Br adds first

formation of most stable radical (tertiary) Formation of anti-Markovnikov product and chain propagation

Organic Bases

generic amino acid adenine morphine cocaine

Consider epinephrine: It is not very soluble in blood. Why is it administered as the hydrochloride?

got some polar regions but a lot more non polar regions administered as a hydroclhoride

Which part of each reagent will add first to the alkene?

h br

Hydration (acid catalyzed addition of water)

h3o h3c c o keto tetromerization

In organic chemistry we deal with molecules not atoms

in molecules, not only are the energies of the electrons quantized (as they occupy molecular orbitals), but also the vibrations and rotations of the molecules

explanation for ph pka

henderson hassleback equation calculated it or carboxylic acid at ph blank is going to be deprotonated any biological species with ph blank with a carboxylic acid on is going to be negatively charged

IR spectra are good for identifying functional groups

how do we identify what is attached to the group? infrared absorption data table if table

Chapters 2-1 : Spectroscopy

how we know what we know about chemical structure

region that corresponds to vibrational energy transitions

infrared rays lower energy light can promote and move molecules from one vibrational state to another rather than having the electron move

Radical addition to alkenes: initiation generates a bromine radical

initiation: the o-o bond is broken by adding energy in the form of light a highly reactive radical (unpaired electron species) Formation of the bromine radical

O H OH

is a nucleophile Is attracted to a positive charge likes positive charge

alcohols

lone pairs oh bond

the two spectra together

look very different ethanol is on top (OH Peak) and acetic acid is on the bottom (carbonyl and broad OH peak part of the carboxylic acid) tells you that these OH peaks don't behave the same it is doing something different acetic acid has broad oh peak compared to ethanols

Metals are very electronegative elements

make carbons

mass of molecular ion: 58

molecular weight of acetone theres a bit more after 58 due to c 13 isotopes maybe some deuterium its not very much but just a reminder that although all these atoms have also minor isotopes that are going along with them

Chapter 3 Conformations and Configurations

molecules are 3-dimensional! As it turns out, the three-dimensional structure of a molecule can significantly affect its stability, reactivity, and potential to engage in intermolecular interactions hydrogens behind each other

acetic acid oh broader than co

more hydrogen bonding

ether more sulble in aqueous acid

more polar

h2o is a better leaving group that oh

more stable

Glucose exists as six membered rings: note the preponderance of equatorial OH groups!

most of the substituents (oh groups) are in the equatorial position

Substitution, Nucleophilic, 1st order: Proposed Mechanism answer

nicely drawn out the first step is the rate determining step

1-7 Electrophiles and Nucleophiles

nucleophiles are Lewis bases and electrophiles are Lewis acids ...

ether

o 2 c dimethyl ether solvent but not soluble in water very much

IR Spectrum of OH NH O-H N-H bonds

oxygen nitrogen hydrogen bonds here 3200- 4000

Alkenes are electron rich

pi orbitals overlap to form pi bond

If we draw an energy diagram (G vs reaction progress) for this reaction, knowing that the equilibrium lies to the left...

pka -1.8 pka 14 pka 4.8 DG is positive, the equilibrium position lies to the left If we know that the equilibrium lies to the left and that 99% of it is unionized then we can say delta G is positive

SN1 mechanism explains why a chiral substrate produces a racemic mixture of productss

planar carbocation attacks top and bottom equally enantiomers racemic mixture Chiral centers racemize because the carbocations formed from them are sp2 hybridized the the nucleophile can attack from either the top or bottom

Alkenes potentially have two reactive sites (each C in C=C).They react with both electrophiles and nucleophiles sites.

polypropylene (monomer propene)

region that corresponds to nuclear spin transitions

radio tv long radio waves very low energy

nuclear spin

radio tv long rw low energy

Oxidation of alkenes: epoxidation

rco3h epoxide rco3h is a per-acid

Where does the nucleophile attack? Why?

rco3h epoxide oh sh nash

amine

secondary amine to bons to carbon alcolate oxygen

Which is the better nucleophile? A.-CH3 B.-NH2 C.-OH D.-F

series of basis that are across the periodic table so c,n,o,f increasing electronegativity so F is more stable and least reactive so ch3 is the most reactive A. -CH3

Similarly organic acids are more soluble as their conjugate base salt

sodium benzoate solubility 60 g/100mL benzoic acid solubility 0.3g/100 mL NaOH carboxylic acid group very low solubility but if we react with sodium hydroxide which will pick off that hydrogen to the left it raises the solubility to 60 grams per 100 millitliters

We can alter the solubility of acidic and basic compounds by changing the pH to form the more soluble salt (acid or base form).The original substance can be regenerated by restoring the pH

sodium benzoate solubility 60 g/100mL benzoic acid solubility 0.3g/100 mL NaOH HCL by adding sodium hydroxide we can make the water-soluble but if we then add hydrochloric acid back we can protonate and reverse the reaction the benzoic acid itself is soluble in organic solvents so this procedure allows us to separate out species into different solvents

Specialized reagents deliver hydride ion (H-)(very reactive/ need specialized reagents to do this) as a nucleophile

sodium borohydride lithium aluminum hydride na+ bh4- li+ alh4- nabh4/ch3oh h3c

What forces/interactions and energy changes occur when something goes into solution? (CLUE CH6)

solvent is less ordered than solute. when solvent comes into contact with the solute interactions occur. if the interactions are strong enough they can overcome the interactions that are in the solute forming solvent solute interactions and breaking solute solute interactions end up with interaction where solute is dispersed

in which directions does the energy of the vibrational transition increase in an IR spectrum (of acetone)

starts at right arrow points to left

Alkynes can also be prepared by elimination of HBr (twice)

strong base (nanh2) h3c c ch3 nh2 hbr

Common addition polymers

styrene polystyrene acrylonitrile free radical vinyl polymerization polyacrylonitrile vinyl chloride polyvinyl chloride

Vision depends upon light catalyzed cis-trans isomerism

sunlight arrives on the rose rhodopsin dounle bond sigma plus reformed pi bond

Oxidation of alkenes:ozonolysis

the alkene is cleaved at the double bond

Typical C-13 NMR chemical Shifts

table

How would you make: CH3SCH3

take a substrate: ch3br and a nucleophile :hsch3 and react them you'd get ch3sch3

Look again at the NMR spectrum of 1,1-dichloroethane

the 3h signal is split by the 1h so it appears as a doublet in contrast the 1h split by the 3h methyl group and so it appears as a quartet

Ring Flips

the axial hydrogen becomes equatorial and vice versa flagpole interaction is bad

in any spectroscopic technique we are meausring

the energy change from one quantum state to another

Syn addition

the heterogenous catalyst is a solid double bond gets broken

Spectroscopy:

the interaction of light (energy) and matter Spectroscopy provides about atomic/molecular structure- but how?

table of infrared absorption data

the ir absorption peaks of substances that can H-bond tend to be broad because many of the O-H groups are involved to different extents in H-Bonding Bond Wavelength

The case of the spin-1/2 nucleus

the nuclei are electrically charged and many have a spin that causes them to behave like a magnet No field

spectrum of two butyne

the triple bond peak is really small here because its a symmetrical molecule very small peak ir absorptions occur when there is a change in the dipole- this alkyne is symmetrical ch peaks around 3000

sp3 hybridized carbon is 3-Dimensional

there are many ways to represent that

a good nucleophile has an available lone pair

they use the one pair to attack the carbon lewis base is almost the smae thing specificially talking about reactivity carbon nucleophile not too electronegative we dont want to attract the lone pairs to itself enc should be not be very high each of the electrons in the valence shell sees or is attracted by more protons in the nucleus than there are for oxygen or nitrogen or carbon positive and negative charges attract force of attraction and repulsion

Ethyl acetate What is the multiplicity of the signal indicated? A.Singlet B.Doublet C.Triplet D.Quartet E.Septet

this methyl has group has no adjacent hydrogens/ there are no hydrogens on the carbon next to it so it is going to appear as a singlet A.Singlet for H3c ch3 bonded to oxygen ch3 is adjacent to carbon with three hydrogens on so we expect a quartet 2h quartet for methyl group adjacent to ch2 so a triplet 3h triplet

1,2-Dimethyl cyclohexane Both groups on the same "side" (above or below) = cis Groups on opposite sides = trans

trans cis Once we have two (or more) substituents, we need to consider whether these substituents are on the same or opposite sides of the ring. This is easiest to depict if we draw the ring as flat (which of course it is not), and depict substituents as wedges or dashes. We do not know from these projections whether a given substituent is axial or equatorial (though we can of course figure it out). To interconvert between cis and trans isomers we would need to break a C-C bond. These are called geometric isomers (that is, they have the same formula and connectivity but different permanent relationships in space)

Stability of Alkenes

trans 2-butene is more stable than cis-2-butene - this is a steric effect fewer repulsions

1H NMR allows us to identify the connectivity of molecules. With these spectra we can't distinguish between

two molecules/same number of carbons both have carbonyl groups 4 carbons in c13 nmr can't tell the difference between them

Absorption (or emission) of a quantum electromagnetic radiation in the radio-wave frequency will cause the nuclei to transition to the other spin state.

very very low energy

region that corresponds to electronic transitions

visible light

IRS Spectrum of ethanol

y axis= transmittance the amount of light that is transmitted so if it goes down to 0 that means all light has been absorbed the stronger the signal the bigger the peak the x-axis is labeled in wave numbers centimeters to the minus one or actually one over distance essentially they are directly related to frequency so the higher frequency means higher energy transmittance at 0 means all light has been absorbed

ΔG tells us whether the reaction will occur - ΔG depends on the actual conditions and can be + or -

ΔGo is the standard Gibbs energy change - and it refers to specific conditions (1M concentrations of reactants and products, at 298K)

The more substituted the alkene the more stable it is:

ΔH -26.9 kcal/mol most stable ΔH -28.5 kcal/mol ΔH -30.3 kcal/mol less stable more erergy released Potential energy

Lewis Acid-Base Reactions

•(CH3)3B + N(CH3)3

Chirality Optical isomers have a property called chirality - they are non-superimposable mirror images of one another. You just constructed a model of the simplest chiral organic molecule

•A chiral object is distinguishable from its mirror image: it cannot be superimposed onto it. •The simplest example in organic chemistry is a carbon compound that has 4 different groups attached to it.

Cyclic compounds have two types of strain

•Angle strain (the energy required to distort the "natural" bond angle (usually ~ 109) •Torsional strain (from eclipsed groups - with overlapping electron density) •We can calculate the ring strain by comparing the measured combustion energy (i.e. reaction with O2) with that calculated by using average bond energies.

Determining chemically distinguishable carbons

•Chemically distinguishable carbons have different chemical environments (different bonds, electron density) •Look at the connections to each carbon •Look for symmetry elements (e.g. mirror planes)

Signals in 1H NMR

•Depend on the electronic environment of the H (just like C-13) •Also depend on the number of chemically equivalent H's -the area under the signal is proportional to the number of H's -Most 1H NMR spectra also provide this data (by integrating the area under the peak) -NOTE: in 13C spectra the peak areas are sometimes not proportional to the number of equivalent carbons - and you can't use the peak area to figure out how many carbons there are

Diastereomers

•Diastereoisomers are stereoisomers that are not superimposable on their mirror image. •For chiral species the are the same configuration at (at least) one carbon and different at (at least) one carbon •But what we have been calling geometric (cis/trans) isomers are also diasteromers. •Diasteromers have different properties - because the arrangement of the atoms in space is different - and therefore the distribution of electron density is different

Cyclohexanee

•Exists in the "chair" conformation (all substituents are staggered) •Can interconvert to the other chair by a ring flip which goes through the high energy boat conformation •There are two types of substituents - axial or equatorial •Axial is higher energy because of 1,3 diaxial interactions. •Equatorial is lower energy more energy not such strong reactions •Since the ring is constrained there are now two sides that cannot be converted by ring flipping If the molecule were flat it would have torsional (angle) strain (120) And all the C-Hs are eclipsed (torsional strain) Does not exist like this.

Kinetics Review

•Experiments vary the concentrations of the reactants to determine how the rate varies with concentration - to give: •A rate law: Rate = k[A][B] •The concentrations in the rate law equation are those present in the rate determining step •The order of the reaction is the number of reactants present in the rate law -Rate = k[A] first order -Rate = k[A][B] second order

Evidence from kinetic studies In the Sn2 reactions we have looked at so far, the rate is dependent on the concentration of both the nucleophile and the electrophile. This means both must be present in the rate determining step. The simplest explanation for this is the mechanism we have already proposed: nucleophilic attack and leaving group departure happen at the same time.

•For the reactions we have been considering: i.e. methyl or primary alkyl groups with good leaving groups and good nucleophiles, in polar aprotic solvents (more on that later). •Rate = k[RL][Nuc] •eg for RBr + NaCl à RCl + NaBr •Rate = k[RBr][NaCl] •We call this an SN2 (Substitution, Nucleophilic, 2nd order) reaction

Improving the leaving group

•Halides ( I-, Br- and Cl-) are good leaving groups •But -OH is not (it is too strong a base - and is too reactive) •We have improved OH as a leaving group by protonating, but not everything is stable in acidic solution •If we use a strong base with OH leaving group it will just deprotonate the strong base

Chiral molecules Two molecules that are mirror images of each other have identical physical and chemical properties with the exception of the ability to rotate plane polarized light. Needless to say, enantiomers are very difficult to separate and it is often preferable to make only one. An entire field of synthetic organic chemistry, called asymmetric synthesis, is dedicated to this extremely difficult task.

•Have identical intermolecular distances •Have identical physical and chemical properties •UNLESS - they are interacting with another chiral species -E.g. in biological systems •An equal mixture of two optical isomers is called a racemate, or racemic mixture.

Extension to carbon

•When reactions involve a carbon center •The Lewis acid is called an electrophile (loves electrons or - charge) •The Lewis base is called a nucleophile (loves nucleus or + charge) •The group that leaves is called the "leaving group"

What mechanism can we propose that makes sense of this?

•It must contain only the substrate (carbon electrophile) in the rate determining step •It must explain why tertiary substrates are much faster than secondary •It must explain why chiral centers are racemized. •It must explain why the reaction is accelerated in protic solvents

Lewis acid-base reactions

•Lewis acid - electron pair acceptor example mg2+ B with empty orbital or h-0 with a plus on h -empty orbital (or a delta +) N with lone pair •Lewis base - electron-pair donor

Make a model of cyclobutane

•Molecule puckers to remove both angle and torsional strain Cyclobutane is more stable than cyclopropane because the ring is bigger and better able to bend (to alleviate both torsional and angle strain). At room temperature, the cyclobutane ring is constantly bending so that each carbon can be relieved of a little of the strain. puckered planar

Mass Spectrometryy

•Molecules are ionized by bombarding with electrons •Resulting ions are accelerated through a magnetic field •Separated by their charge to mass ratio e/m •Identified by fragmentation patterns and •Molecular ion (the highest peak - usually at the molecular weight) detector magnet sample enters here electron beam source tons accelerated heater vaporizes sample sample enters here magnetic field deflects lightest ions most

1H also has a spin 1/2

•More abundant (most H is 1H, whereas 13C is only 1% abundance) •Spectra are recorded easier and faster •Similar effects are seen on the position of the chemical shift due to electron withdrawing effects. •The spectra look more complicated (which is why we introduced C-13 first)

Solvent considerations

•Most NMR spectra are recorded in a solvent •This solvent cannot have any protons in it (otherwise it will be difficult to see the peaks from the actual substance) •Common solvents have any H's replaced by D ( isotope of H) which is not NMR active •Eg CDCl3, DMSO d6 (dimethylsulfoxide) Deuterium has one proton and one neutron/ same chemical species but its called heavy hydrogen its not nmr active cdcl3 The vast majority of NMR are recorded in a solution. This presents a problem as, in typical organic solvents, there is quite a lot of H1 and solvent concentration is almost always much higher than sample concentration. We can minimize solvent peaks by using solvent enriched with hydrogens that are not NMR active, that is deuterium (1 proton and 1 neutron). Recall that nuclei with even numbers of "nucleons" are not NMR active. Deuterated chloroform (CDCl3) is the most common such solvent used. More exotic deuterated solvents (like deuterated toluene) can be VERY expensive. cant run an nmr in proton in solvents that have hydrogens in them

What makes a good nucleophile?? Recall that good nucleophiles need a lone pair to donate (that is, they are Lewis bases). Recall also that strong bases are good nucleophiles. Decreasing electronegativity enables greater lone pair accessibility and deprotonation of a conjugate base increases negative charge density (both increase nucleophilicity). Nucleophile strength increases down a group as electron clouds get larger and more "floppy". Larger electron cloud = more accessible lone pair and can form bonds earlier as the large cloud distorts toward loci of positive charge.

•Need an available lone pair •Strong bases are good nucleophiles - so all the trends in base strength follow for nucleophile strength e.g. •N>O>F (decreases across a row as electronegativity increases) •-OH >H2O (conjugate base is always better than the corresponding acid •Nucleophile strength (but not base strength) increases down a group S>O, Br>Cl

What makes a good nucleophile?

•Need an available lone pair •Strong bases are good nucleophiles - so all the trends in base strength follow for nucleophile strength e.g. •N>O>F (decreases across a row as electronegativity increases) N better nucleophile and base lone pair more available bc electronegativity increases as you go across and the atoms hold on more tightly to their electrons •-OH >H2O (conjugate base is always better than the corresponding acid) conjugate base is always better than the conjugate acid it is more reactive and therefore more basic and nucleophilic •This brings in a new potential nucleophile -CH3 •Nucleophile strength (but not base strength) increases down a group some nucleophiles don't align with the base strength and this is typically when comparing nucleophile strength as you go down a group in the periodic table

Leaving Groups A good leaving group needs to be able to support an additional lone pair. Weak bases are thus better leaving groups (as they are less likely to donate their lone pair to an acid and thereby re-add to the electrophile or deprotonate something). Thus conjugate acids are better leaving groups then their corresponding base. Also, more electronegative elements are better able to support additional negative charge and so will tend to be better leaving groups. This DOES NOT hold true for the halides as fluorine is a crappy leaving group (on account of being small and forming short, relatively strong bonds with carbon). The other halides are pretty decent as leaving groups.

•Need to be able to support the electron pair from the broken bond •Weak bases are better leaving groups •H2O > -OH (the conjugate acid is a better leaving group) •H2O > NH3 (the most electronegative is a better leaving group) •Halides (except for F- are good leaving groups) •I->Br->Cl->>F-

What makes a good leaving group?

•Need to be able to support the electron pair from the broken bond example nu cl the leading group goes away with a sometimes negative charge and electron pairs and stable •Weak bases are better-leaving groups •H2O > -OH (the conjugate acid is a better leaving group) •H2O > NH3 (the most electronegative is a better leaving group) •All Halides (except for F- are good leaving groups) •I->Br->Cl->>F- Leaving groups need to be able to support excess electrons (high effective nuclear charge, can stabilize anion via induction and/or resonance). Weak bases tend to be better leaving groups (strong bases tend to be good nucleophiles and just re-add). Conjugate acids are better leaving groups than their conjugate bases More electronegative elements (higher effective nuclear charge) are better leaving groups than less electronegative elements Going down a column, trends are dominated by polarizeability and bond strength (remember, the H-I bond is much weaker than the H-F bond)

Splitting occurs

•When non-equivalent H-atoms are on adjacent (or the same) carbon (they have to be close to "feel" the induced magnetic field) •Usually (but not always) if Hs are further away the effect is too weak to be measured. this hydrogen will feel the effect of this hydrogen but not that one this magnetic interaction for hydrogens that are on the carbons that are next to each other

If we change the conditions we find the following evidence:

•RBr + H2O à ROH •A nucleophilic substitution occurs without needing a good nucleophile! •The rates of reaction are now tert > sec > primary •Rate = k[RBr] (First order) •(no nucleophile in the rate law) •The strength of the nucleophile does not affect the rate. •If R is chiral the product is racemic •Protic solvents accelerate the reaction In this scenario, water is both the nucleophile and the solvent. Water, as you may recall, is both a fairly weak nucleophile and aprotic solvent. Under these conditions, substitution takes place, but it differs from an SN2 substitution in several observable ways. More substituted alkyl halides react faster, not slower The rate of the reaction is dependent only on the concentration of the substrate The strength of the nucleophile does not affect the rate (represented by the nucleophile not being in the rate law) If substitution occurs at a stereocenter, the product of the reaction is a racemic mixture (i.e. a 1:1 mixture of both enantiomers)

Recall SN2 reactions

•Second order -TS for the rate determining step includes both electrophile and nucleophile •Proceed with inversion if the reaction takes place at the chiral center. •Accelerated by polar aprotic solvents.

Why are the peaks split??

•Since most hydrogens are 1H, in a given molecule there will be almost all 1H nuclei •Therefore each 1H nucleus can detect the magnetic fields generated by all the near-by 1H nuclei. •(NOTE: For 13C, the probability is that there will be no more than 1 13C per molecule - so none of this type of splitting is seen)

The Effect of Protic Solvents on the Nucleophile

•Solvation lowers the energy of the solvated species, making it less reactive •If X is the energy of the unsolvated nucleophile, which is the most likely relative energy of the solvated nucleophile? A; solvation lowers the energy of the system through attraction between positive and negative charges (between the Nu- and a partial positive region of a solvent molecule, for instance)

Non-basic nucleophiles

•Some weak bases are strong nucleophiles -Typically they are more polarizable - that is the electron cloud is "floppy" and can become polarized towards an electrophilic carbon. example ch3L I- I- is a big nucleophile with a floppy cloud and what can happen is that as it gets near to the carbon which has a positive charge it can get polarized it starts to distort and can more easily attach at the carbon so as we go down the periodic table to things like iodine and bromide we see that they are more nucleophilic even though they are less basic and sulfur and phosphorous are also analogous •I-, Br- are examples, as are S and P containing compounds. •Down a group the trend is for increasing nucleophilicity Nonbasic nucleophiles are typically large, floppy (i.e. polarazable) atoms that can have a strong transient dipole attract them to an electrophilic carbon.

Types of Isomers We have previously looked and conformational (i.e. rotational) and geometric (different permanent orientations in space but same connectivity) isomers. We will now introduce another type of isomer that is of especial interest to those who develop pharmaceuticals. We are referring to optical isomers.

•Structural Isomers -Same formula different connectivity • •Stereoisomers -Same formula, same connectivity, a different arrangement in space. •Conformational isomers- can be interconverted by rotations •Cis and trans isomers- -cannot be interconverted by rotations And ...

•Isomers: Same molecular formula diagram

•Structural isomers: Different connectivity •Stereoisomers: same connectivity, a different arrangement in space •Optical Isomers: non-superimposable mirror images •Conformational Isomers: interconverted by C-C rotations •Diasteromers: geometric isomers (cis/trans)

SN1 or SN2? SN2 versus

•Substrate -CH3>primary>sec>tertiary •Leaving group - Increases rate by lowering activation energy •Nucleophile -Strong nucleophiles •Solvent -Polar aprotic solvates the cation leaving the anion more reactive •Stereochemistry -Inversion •Typically no rearrangements

SN1 or SN2? SN1 versus

•Substrate -Tert>sec>primary •Leaving group - Increases rate by lowering activation energy •Nucleophile -Non-(Bronsted) basic nucleophiles, weak nucleophiles •Solvent -Polar solvates C+, protic helps ionize the leaving group (either is OK) •Stereochemistry -Racemization •Often accompanied by rearrangements

Now lets talk about the substrate (the electrophile) Electrophilic carbons must be bonded to something electronegative in order that their electron density be reduced and they have a partial positive charge. A good leaving group must be attached. Remember the 3rd point about clutter around the electrophilic carbon? We're coming to that now.

•The C-X bond must be polarized with the δ+ on the carbon. •The group that leaves must be able to support a negative charge •There must not be too much clutter around the reactive center (we will get to that now).

What are the factors that make a carbon electrophilic? (susceptible to reaction with a nucleophile?)

•The C-X bond must be polarized with the δ+ on the carbon. to make it interact with the nucleophile •The group that leaves must be able to support a negative charge (or the pair of electrons from the broken bond) leading group must be stable and comfortable •There must not be too much clutter around the reactive center (we will get to that later). •sp3 hybridization (no pi bonds - too much electron density) bonded to 4 things not expect nucleophile to attack c c double bonds

Diastereomers have different properties

•They can be used to resolve enantiomers •In general the racemic mixture is reacted (in a reversible reaction - often an acid-base reaction) with an optically pure resolving agent. •This produces a mixture of diastereomers that can be physically separated •The original compounds are then regenerated have different properties

Why don't 13C NMR Spectra Exhibit Coupling?

•They do! •But to simplify the spectrum the sample is irradiated with a radiofrequency that saturates the high spin state •Since all the H's are in the same state now, the Carbons only "feel" one magnetic field, and so no signal splitting occurs •"Broadband decoupling"

Some nuclei are "NMR active"

•We can consider nuclei as spinning (positively) charged particles. •This generates a magnetic field •Some nuclei have different spin quantum states (in particular C-13 and H-1) •When these nuclei are placed in an external magnetic field the spin states split into two •The energy required to flip from one state to another lies in the radio-wave region of the electromagnetic spectrum. This is called "resonance" (hence Nuclear Magnetic Resonance NMR).

CEM

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