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Organic Chemistry

Mario Lintz

Mario.Lintz@ucdenver.edu

303-946-5838

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Organic Chemistry I

Functional Groups Molecular Structure Hydrocarbons Substitution and Elimination Oxygen Containing Compounds Amines

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Functional GroupsList #1- Critical for the MCAT Alkane Alkene Alkyne Alcohol Ether Amine Aldehyde Ketone Carboxylic Acid Ester Amide

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Functional GroupsList #2- Memorize as well Alkyl Halogen Gem-dihalide Vic dihalide Hydroxyl Alkoxy Hemiacetal Hemiaketal Mesyl group Tosyl group Carbonyl Acetal Acyl

Anhydride Aryl Benzyl Hydrazine Hydrazone Vinyl Vinylic Allyl Nitrile Epoxide Enamine Imine Nitro Nitroso

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Bonds

Types: Ionic: complete transfer of electrons

Covalent: shared electrons Coordinate covalent bonds- One atom provides both

electrons in a shared pair. Polar covalent: unequal sharing of electrons

Hydrogen Bonds: bonds between polar molecules containing H and O, N, or F

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Bonds

In the pi bond of an alkene, the electron pair have:

A. 33% p character and are at a lower energy level than the electron pair in the o bond.

B. 33% p character and are at a higher energy level than the electron pair in the o bond.

C. 100% p character and are at a lower energy level than the electron pair in the o bond.

D. 100% p character and are at a higher energy level than the electron pair in the o bond.

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Covalent Bonds

Sigma

Pi

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Covalent Bonds

Sigma Between s orbitals Small, strong, lots of rotation

Pi

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Covalent Bonds

Sigma Between s orbitals Small, strong, lots of

rotation Pi

Between p orbitals Discreet structure,

weaker than sigma, no rotation

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Covalent Bonds

Sigma Between s orbitals Small, strong, lots of

rotation Pi

Between p orbitals Discreet structure,

weaker than sigma, no rotation

Always add to sigma bonds creating a stronger bond

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When albuterol I dissolved in water, which of the following hydrogen-bonded structures does NOT contribute to its water solubility?

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Dipole Moments(Solely responsible for Intermolecular Attractions)

Charge distribution of bond is unequal Molecule with dipole moment = polar Molecule without dipole moment = nonpolar Possible to have nonpolar molecules with polar bonds

Induced Dipoles Spontaneous formation of dipole moment in nonpolar molecule Occurs via: polar molecule, ion, or electric field

Instantaneous Dipole Due to random e- movement

Hydrogen Bonds Strongest dipole-dipole interaction Responsible for high BP of water

London Dispersion Forces Between 2 instantaneous dipoles Responsible for phase change of nonpolar molecules

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Lewis Dot Structures Rules for writing

Find total # valence e- 1 e- pair = 1 bond Arrange remaining e- to satisfy duet and octet rules

Exceptions Atoms containing more than an octet must come from the

3rd period, (vacant d orbital required for hybridization) Not very popular on the MCAT

Formal Charge # valence e- (isolated atom) - # valence e- (lewis structure) Sum of formal charge for each atom is the total charge on

the molecule (actual charge distribution depends on electronegativity)

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Structural Formulas

Dash Formula Condensed Formula Bond-line Formula Fischer projection Newman projection Dash-line-wedge Ball and stick

All Images courtesy of Exam Krackers

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Hybridization

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Hybrid Bonds

Suffix C bonds Hybridization

Percent

S:P

Bond Angle

Bond Length

Bond Strength

-ane

-ene

-yne

-yl

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Hybrid Bonds

Suffix C bonds Hybridization

Percent

S:P

Bond Angleo

Bond Length

(pm)

Bond Strength

(kJ/mol)

-ane C-C sp3 25:75 109.5 154 346

-ene C=C sp2 33:66 120 134 612

-yne C=C sp 50:50 180 120 835

-yl Side chain

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Hybrid Bonding in Oxygen and Nitrogen

Nitrogen- Lone pair occupies more space than N-H Causes compression of the bond angle. Bond

angles are 107.3 as opposed to 109.5 Oxygen-

2 sets of lone pair electrons Causes greater compression than in Nitrogen.

H2O bond angles are 104.5 vs 109.5.

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A) sp2, sp2 B) sp2, sp3 C) sp3, sp3 D) sp3, sp2

For the molecule 1,4 pentadiene, what type of hybridization is present in carbons # 1 and # 3 respectively?

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VSEPR

valance shell electron pair repulsion Prediction of shape

Minimize electron repulsion

1. Draw the Lewis dot structure for the molecule or ion

2. Place electron pairs as far apart as possible, then large atoms, then small atoms

3. Name the molecular structure based on the position of the atoms (ignore electron pairs)

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molecule Lewis structure Shape molecule Lewis structure Shape

BeCl2 Linear, sp

SF4 Seesaw SO3

Trigonal planar,

sp2

ICl3 T shaped NO2- Bent

CH4Tetrahedral,

sp3NH3

Trigonal Pyramid

al

PCl5

Trigonal bipyramidal,

dsp3

SF6Octahedral, d2sp3

IF5Square

PyramidalICl4

- Square Planar

VSEPR1. Draw the Lewis dot structure for the molecule or ion

2. Place electron pairs as far apart as possible, then large atoms, then small atoms

3. Name the molecular structure based on the position of the atoms (ignore electron pairs)

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Delocalized e- and Resonance passage 25

Resonance forms differ only in the placement of pi bond and nonbonding e-

Does not suggest that the bonds alternate between positions

Neither represent the actual molecule, rather the real e assignment is the intermediate of the resonant structures. The real structure is called a resonance hybrid (cannot be seen on paper)

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Organic Acids and Bases

Organic Acids- Presence of positively charged H+ Two kinds

present on a OH such as methyl alcohol present on a C next to a C=O such as acetone

Organic Bases- Presence of lone pair e to bond to H Nitrogen containing molecules are most common Oxygen containing molecules act as bases in presence of

strong acids

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Stereochemistry Isomers: same elements, same proportions.

Different spatial arrangements => different properties. Structural (constitutional): Different connectivity.

Isobutane vs n-butane Both C4H10

Conformational (rotational): Different spatial arrangement of same molecule

Chair vs. boat Gauche vs Eclispsed vs Antistaggered vs Fully Eclipsed

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Stereochemistry-isomers

Stereoisomers: different 3D arrangement Enantiomers: mirror images, non-

superimposable. Same physical properties (MP, BP, density, solubility,

etc.) except rotation of light and reactions with other chiral compounds

May function differently; e.g. thalidomide, sugars, AA Have chiral centers

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Stereochemistry-isomers

Stereoisomers: different 3D arrangement Diastereomers: not mirror images (cis/trans)

Different physical properties (usually), Can be separated Chiral diastereomers have opposite configurations at

one or more chiral centers, but have the same configuration at others.

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Stereochemistry-isomers What kind of isomers are the two compounds below?

A. Configurational diastereomers B. Enantiomers C. Constitutional isomers D. Cis -trans diastereomers

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Stereochemistry-polarization of light Excess of one enantiomer causes rotation of plane-polarized light.

Right, clockwise, dextrarotary (d), or + Left, counterclockwise, levarotary (l), or –

Racemic: 50:50 mixture of 2 enantiomers, no net rotation of light

RELATIVE Configuration: configuration of one molecule relative to another. Two molecules have the same relative configuration about a carbon if they differ by only one substituent and the other substituents are oriented identically about the carbon.

Specific rotation [: normalization for path length (l) and sample density (d). ocm3/g

[] = / (l*d)

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Stereochemistry-Chiral moleculespassage 27

Achiral=plane or center of symmetry ABSOLUTE Configuration: physical orientation of atoms

around a chiral center R and S:

1. Assign priority, 1 highest, 4 lowest H < C < O < F higher atomic #, higher priority If attachments are the same, look at the atoms (ethyl beats methyl)

2. Orient 4 away from the observer3. Draw a circular arrow from 1 to 2 to 3

R = clockwise S = counterclockwise

This has nothing to do with the rotation of light! E and Z: Different than cis and trans

Z= same side of high priority groups E=opposite side of high priority groups

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IUPAC Naming Conventions IUPAC Rules for Alkane Nomenclature1.   Find and name the longest continuous carbon

chain. 2.   Identify and name groups attached to this chain.3.   Number the chain consecutively, starting at the

end nearest a substituent group. 4.   Designate the location of each substituent group

by an appropriate number and name.5.   Assemble the name, listing groups in alphabetical

order. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.

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Hydrocarbons

# of C Root Name # of C Root Name

1 meth 6 hex

2 eth 7 hept

3 prop 8 oct

4 but 9 non

5 pent 10 dec

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Hydrocarbons Saturated: CnH(2n+2)\

Unsaturated: CnH[2(n-u+1)] ; u is the # of sites of unsaturation

Primary, secondary, tertiary, and quaternary carbons

Know and be able to recognize the following structures

n-butyl sec-butyl

iso-butyl tert-butyl

n-propyl

Iso-propyl

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Alkanes

Physical Properties: Straight chains: MP and BP increase with length

(increased van Der Waals interactions) C1-4: gas C5-17: liquid C18+: solid

Branched chains: BP decreases (less surface area, fewer vDW) When compared to the straight chain analog, the straight

chain will have a higher MP than the branched molecule. BUT, amongst branched molecules, the greater the branching, the higher the MP.

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Alkanes-Important ReactionsVery Unreactive

Combustion: Alkane + Oxygen + High energy input (fire) Products: H2O, CO2, Heat

Halogenation Initiation with UV light

Homolytic cleavage of diatomic halogen Yields a free radical

Propagation (chain reaction mechanisms) Halogen radical removes H from alkyl Yields an alkyl radical

Termination Radical bonds to wall of container or another radical

Reactivity of halogens: F > Cl > Br >>> I Selectivity of halogens (How selective is the halogen in choosing a position on an alkane):

I > Br > Cl > F more electronegative (Cl) means less selective (Br)

Stability of free radicals: more highly substituted = more stable aryl>>>alkene> 3o > 2o > 1o >methyl

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Halogenation

In the halogenation of an alkane, which of the following halogens will give the greatest percent yield of a tertiary alkyl halide when reacted with 2-methylpentane in the presence of UV light.

A. F2

B. Cl2

C. Br2

D. 2-methylpentane will not yield a tertiary product

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Cycloalkanes

General formula: (CH2)n or CnH2n As MW increases BP increases though MP

fluctuates irregularly because different shapes of cycloalkanes effects the efficiency in which molecules pack together in crystals.

Ring strain in cyclic compounds:

Bicyclic Molecules:

http://www.chem.uh.edu/Courses/Thummel/Chem3331/Notes/Chap3/

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CycloalkanesNaming

Find parent Count C’s in ring vs longest chain. If # in ring is equal to or

greater than chain, then name as a cycloalkane. Number the substituents and write the name Start at point of attachment and number so that subsequent

substituents have the lowest # assignment If two or more different alkyl groups are present, number them

by alphabetic priority If halogens are present, treat them like alkyl groups Cis vs Trans

Think of a ring as having a top and bottom If two substituents both on top: cis It two substituents and 1 top, 1 bottom: trans

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Cycloalkanes Ring Strain

Zero for cyclohexane (All C-C-C bond angles: 111.5°) Increases as rings become smaller or larger (up to cyclononane)

Cyclohexane Exist as chair and boat conformations Chair conformation preferred because it is at the lowest energy. Hydrogens occupy axial and equatorial positions.

Axia (6)l- perpendicular to the ring Equatorial (6)- roughly in the plane of the ring

Neither energetically favored When the ring reverses its conformation, substituents reverse their

conformation Substituents favor equatorial positions because crowding occurs most

often in the axial position.

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Cyclohexanes

In a sample of cis-1,2-dimethylcyclohexane at room temperature, the methyl groups will:

A. Both be equatorial whenever the molecule is in the chair conformation.

B. Both be axial whenever the molecule is in the chair conformation.

C. Alternate between both equatorial and both axial whenever the molecule is in the chair conformation

D. Both alternate between equatorial and axial but will never exist both axial or both equatorial at the same time

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Substitutions

Substitution: one functional group replaces another Electrophile: wants electrons, has partial + charge

Nucleophile: donates electrons, has partial – charge

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Substitution SN1: substitution, nucleophilic, unimolecular

Rate depends only on the substrate (i.e. leaving group) R=k[reactant] Occurs when Nu has bulky side groups, stable carbocation

(3o), weak Nu (good leaving group) Carbocation rearrangement Two step reaction

1. spontaneous formation of carbocation (SLOW)

2. Nucleophile attacks carbocation (chiral reactants yield racemic product mixtures)

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Substitution

SN2: substitution, nucleophilic, bimolecular Rate depends on the substrate and the nucleophile R=k[Nu][E] Inversion of configuration Occurs with poor leaving groups (1o or 2o) One step reaction

1. Nu attacks the C with a

partial + charge

http://www.mhhe.com/physsci/chemistry/carey5e/Ch08/ch8-4.html43

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Which of the following carbocations is the most stable?

A.CH3CH2CH2CH2

B.CH3CH2CH2CHCH3

C. (CH3)3C

D.CH3

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Benzene Undergoes substitution not addition Flat molecule Stabilized by resonance Electron donating groups activate the ring and are ortho-para

directors Electron withdrawing groups deactivate the ring and are meta

directors Halogens are electron withdrawing, however, are ortho-para

directors

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BenzeneSubstituent Effects

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Oxygen Containing Compounds

Alcohols Aldehydes and Ketones Carboxylic Acids Acid Derivatives

Acid Chlorides Anhydrides Amides

Keto Acids and Esters

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Alcohols

One of the most common reactions of alcohols is nucleophilic substitution. Which of the following are TRUE in regards to SN2 reactions:

I. Inversion of configuration occursII. Racemic mixture of products resultsIII. Reaction rate = k [S][nucleophile]

A. I onlyB. II onlyC. I and III onlyD. I, II, and III

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Alcohols Physical Properties:

Polar High MP and BP (H bonding) More substituted = more basic

(CH3)3COH: pKa = 18.00 CH3CH2OH: pKa = 16.00 CH3OH: pKa = 15.54

Electron withdrawing substituents stabilize alkoxide ion and lower pKa. Tert-butyl alcohol: pKa = 18.00 Nonafluoro-tert-butyl alcohol: pKa = 5.4

IR absorption of OH at ~3400 cm-

General principles H bonding Acidity: weak relative to other O containing compounds

(CH groups are e- donating = destabilize deprotonated species) Branching: lowers BP and MP

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AlcoholsNaming

Select longest C chain containing the hydroxyl group and derive the parent name by replacing –e ending of the corresponding alkane with –ol.

Number the chain beginning at the end nearest the –OH group.

Number the substituents according to their position on the chain, and write the name listing the substituents in alphabetical order.

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Alcohols-Oxidation & Reduction

Oxidation

Reduction

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Alcohols-Oxidation & Reduction

Common oxidizing and reducing agents Generally for the MCAT

Oxidizing agents have lots of oxygens Reducing agents have lots of hydrogens

Oxidizing Agents Reducing Agents

K2Cr2O7 LiAlH4

KMnO4 NaBH4

H2CrO4 H2 + Pressure

O2

Br2

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Reduction Synthesis of Alcohols

Reduction of aldehydes, ketones, esters, and acetates to alcohols.

Accomplished using strong reducing agents such as NaBH4 and LiAlH4

Electron donating groups increase the negative charge on the carbon and make it less susceptible to nucleophilic attack. Reactivity:

Aldehydes>Ketones>Esters>Acetates Only LiAlH4 is strong enough to reduce esters and

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Alcohols: Pinacol rearrangement

Starting with Vicinal Diol Generate ketones and aldehydes Formation of most stable carbocation Can get ring expansion or contraction

http://www.cem.msu.edu/~reusch/VirtualText/rearrang.htm54

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Alcohols-Protection

Alcohol behaves as the nucleophile. (As is often the case)

OH easily transfer H to a basic reagent, a problem in some reactions.

Conversion of the OH to a removable functional group without an acidic proton protects the alcohol

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Protection-Deprotection

An example of TMS-alcohol protection in a synthesis

http://www.chem.umd.edu/courses/spring05/chem241fribush/chsum/354,60,Protection-Deprotection 56

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Alcohols to Alkylhalidesvia a strong acid catalyst R-OH + HCl RCl + H20

Alcohol is protonated by strong acid, (it takes a strong acid to protonate an alcohol).

-OH is converted to the much better leaving group, H2O

Occurs readily with tertiary alcohols via treatment with HCl or HBr.

Primary and secondary alcohols are more resistant to acid and are best converted via treatment with SOCl2 or PBr3

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Alcohols to Alkylhalidesreactions with SOCl2 and PBr3

Halogenation of alcohols via SN1 or SN2 OH is the Nu, attacking the halogenating agent It is not OH that leaves, but a much better leaving

group -OSOCl or –OPBr2, which is readily expelled by backside nucleophilic substitution.

Does not require strong acids (HCl, HBr)

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Alcohols-preparation of mesylates and tosylates OH is a poor leaving

group, unless protonated, but most Nu are strong bases and remove such a proton

Conversion to mesylates or tosylates allow for reactions with strong Nu

Preparation SN1: no change of stereogenic center. Reaction SN2: inversion of configuration

http://www.oglethorpe.edu/faculty/~m_wolf/PowerPoint/CareyOrgPP/sections1st/380,36,Tosylates allow control of stereochemistry and http://www.chem.uh.edu/Courses/Thummel/Chem3331/Wade/wade11.pdf

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Alcohols: Esterification

Fischer Esterification Reaction: Alcohol + Carboxylic Acid Ester + Water

Acid Catalyzed- protonates –OH to H2O (excellent leaving group)

Alcohol performs nucleophilic attack on carbonyl carbon

These bonds arebroken

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Alcohols: Inorganic Esters passage 30

Esters with another atom in place of the carbon

1. Sulfate esters: alcohol + sulfuric acid

2. Nitrate esters: alcohol + HNO3 (e.g. nitroglycerine)

3. Phosphate esters: DNA

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Upon heating 2,3-Dimethyl-2,3-butanediol with aqueous acid, which of the following products would be obtained in the greatest amount?

a) 3,3-Dimethyl-2-butanone

b) 2,2-Dimethyl-3-butanone

c) 2,3-Dimethyl-3-butanone

d) 2,3-Dimethyl-2-butanone

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In the reaction above, what is the purpose of using the 1,2-ethanediol in the first step?

a) Heterogeneous catalyst

b) Homogeneous catalyst

c) Alcohol protection

d) Oxidizing agent

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In the reaction above, if the reagents in the first step were replaced with LiAlH4, what product would result?

a) c)

b) d)

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OH

OH

OH

O

OH

OH

OHHO

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CarbonylsCarbon double bonded to Oxygen

Planar stereochemistry Partial positive charge on Carbon

(susceptibility to nucleophilic attack) Aldehydes & Ketones (nucleophilic addition) Carboxylic Acids (nucleophilic substitution) Amides

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Aldehydes and Ketones Physical properties:

Carbonyl group is polar Higher BP and MP than alkanes because of dipole-dipole interactions More water soluble than alkanes Trigonal planar geometry, chemistry yields racemic mixtures

IR absorption of C=O at ~1600

General principles: Effects of substituents on reactivity of C=O: e- withdrawing increase

the carbocation nature and make the C=O more reactive Steric hindrance: ketones are less reactive than aldehydes Acidity of alpha hydrogen: carbanions , unsaturated carbonyls-resonance structures

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Aldehydes and KetonesNaming

Naming Aldehydes Replace terminal –e of corresponding alkane with –al. Parent chain must contain the –CHO group The –CHO carbon is C1 When –CHO is attached to a ring, the suffix carbaldehyde

is used. Naming Ketones

Replace terminal –e of corresponding alkane with –one. Parent chain is longest chain containing ketone Numbering begins at the end nearest the carbonyl C.

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Aldehydes and Ketones-Acetal and Ketal Formationnucleophilic addition at C=O bond

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Aldehydes and Ketones-Imine Formationnucleophilic addition at C=O bond

Imine R2C=NR Primary amines (RNH2) + aldehyde or ketone R2C=NR Acid Catalyzed protonation of –OH H2O

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Aldehydes and Ketones-Enamine Formationnucleophilic addition at C=O bond

Enamine (ene + amine) R2N-CR=CR2 Secondary amine (R2N) + aldehyde or ketone R2N-

CR=CR2 Acid catalyzed protonation of –OH H2O

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Aldehydes and Ketones-reactions at adjacent positions Haloform: trihalomethane

Halogens add to ketones at the alpha position in the presence of a base or acid.

Used in qualitative analysis to indicate the presence of a methyl ketone. The product, iodoform, is yellow and has a characteristic odor.

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Aldehydes and Ketones-reactions at adjacent positions

Aldol (aldehyde + alcohol) condensation: Occurs at the alpha carbon Base catalyzed condensation Alkoxide ion formation (stronger than –OH, extracts H from H2O to

complete aldol formation) Can use mixtures of different aldehydes and ketones

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Aldehydes are easy to oxidize because of the adjacent hydrogen. In other words, they are good reducing agents. Potassium dichromate (VI): orange to green Tollens’ reagent (silver mirror test): grey ppt.

Prevents reactions at C=C and other acid sensitive funtional groups in acidic conditions.

Fehlings or benedicts solution (copper solution): blue to red

Ketones, lacking such an oxygen, are resistant to oxidation.

Aldehydes and Ketones-Oxidation (Aldehydes Carboxylic acids)

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Aldehydes and Ketones

Keto-enol Tautomerism: Keto tautomer is preferred (alcohols are more

acidic than aldehydes and ketones).

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Aldehydes and Ketones Internal H bonding: 1,3-dicarbonlys

Enol tautomer is preferred (stabilized by resonance and internal H-bonding)

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Guanine, the base portion of guanosine, exists as an equilibrium mixture of the keto and enol forms. Which of the following structures represents the enol form of guanine?

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Aldehydes and Ketones Organometallic reagents:

Nucleophilic addition of a carbanion to an aldehyde or ketone to yield an alcohol

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Acetoacetic Ester Synthesis Alkyl Halide + Acetoacetic Ester Methyl Ketone

Acetoacetic ester synthesis: Use acetoacetic ester (ethyl acetoacetate) to

generate substituted methyl ketones Base catalyzed extraction of α H

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Aldehydes and Ketones

Wolff-Kishner reduction: Nucleophilic addition of hydrazine (H2N-NH2) Replace =O with 2 H atoms

+ H2O

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In which of the following reactions would

the formation of an imine occur?

a) Methylamine + propanol

b) Methylamine + propanal

c) Dimethylamine+ propanal

d) Trimethylamine + propanal

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In which of the following reactions would

the formation of an enamine occur?

a) Methylamine + propanol

b) Methylamine + propanal

c) Dimethylamine+ propanal

d) Trimethylamine + propanal

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In an organic chemistry class a group of students are trying to determine the identity of an unknown compound. In the haloform reaction the reaction mixture turned yellow indicating a positive result. Which of the following is true of the unknown compound?

a) It contains an aldehyde

b) It contains an alcohol

c) It contains a methyl ketone

d) It contains a carboxylic acid

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Oxygen Containing Compounds-Carboxylic Acids Physical Properties:

Acidic Trigonal planar geometry Higher BP and MP than alcohols

Polarity, dimer formation in hydrogen bonding increases size and VDW interactions

Solubility: small (n<5) CA are soluble, larger are less soluble because long hydrocarbon tails break up H bonding

IR absorption of C=O at ~1600, OH at ~3400

General Principles: Acidity Increases with EWG (stabilize carboxylate) Acidity decreases with EDG (destabilize carboxylate) Relative reactivity Steric effects Electronic effects Strain (e.g. -lactams: 3C, 1N ring; inhibits bacterial cell wall formation) 83

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Carboxylic AcidsNaming

Carboxylic acids derived from open chain alkanes are systematically named by replacing the terminal –e of the corresponding alkane name with –oic acid.

Compounds that have a –CO2H group bonded to a ring are named using the suffix –carboxylic acid.

The –CO2H group is attached to C #1 and is not itself numbered in the system.

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Carboxylic Acids-important reactions

Carboxyl group reactions: Nucleophilic attack:

Carboxyl groups and their derivatives undergo nucleophilic substitution. Aldehydes and Ketones undergo addition because they lack a

good leaving group.

Must contain a good leaving group or a substituent that can be converted to a good leaving group.

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Carboxylic Acids-important reactions

Reduction: Form a primary alcohol LiAlH4 is the reducing agent

Unlike oxidation, cannot isolate the aldehyde

CH3(CH2)6COOH CH3(CH2)6CH2OHLiAlH4

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Carboxylic Acids-important reactions

Carboxyl group reactions: Decarboxylation:

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Carboxylic Acids-important reactions

Fischer Esterification Reaction: Alcohol + Carboxylic Acid Ester + Water

Acid Catalyzed- protonates –OH to H2O (excellent leaving group)

Alcohol performs nucleophilic attack on carbonyl carbon

These bonds arebroken

H+

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Carboxylic Acids-reactions at two positions

Substitution reactions: keto reactions shown, consider enol reactions

To make ->

SOCl2

   or PCl3

Heat, -H2OR'OH, heat,

H+-

R2NH

heatHO-

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Carboxylic Acids-reactions at two positions passage 26

Halogenation: enol tautomer undergoes halogenation

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Acid Derivatives- Acid Chlorides, Anhydrides, Amides, Esters Physical Properties:

Acid chlorides: acyl chlorides React violently with water Polar Dipole attractions (no H bonds) Higher BP and MP than alkanes, lower than alcohols

Anhydrides Large, polar molecules Dipole attractions (no H bonds) Higher BP than alkanes,

lower than alcohols

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Acid Derivatives

Physical Properties: Amides:

Highest BP and MP Soluble in water (H bonds)

Esters: Poor to fair H bond acceptors Sparingly soluble in water Weakly basic H on alpha C weakly acidic

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Acid DerivativesNaming

Acid Halides (RCOX) Identify the acyl group and then the halide Replace –ic acd with –yl, or –carboxylic acid with –carbonyl

Acid Anhydrides (RCO2COR’) Symmetrical anhydrides or unsubstituted monocarboxylic acids and cyclic anhydrides of dicarboxylic acids are

named by replacing the word acid with anhydride. 2 acetic acid acetic anhydride

Anhydrides derived from substituted monocarboxylic acids are named by adding the prefix –bis to the acid name. 2 chloroacetic acid bis(chloroacetic) anhydride

Unsymmetrical anhydrides- those produced from two different carboxylic acids- are named by citing the two acids alphabetically. Acetic acid + benzoic acid acetic benzoic anhydride

Amides (RCONH2) Amides with an unsubstituted –NH2 group are named by replacing the –oic acid or ic acid ending with amide, or by

replacing the –carboxylic acid ending with carboxamide. Acetic acid acetamide

If the nitrogen atom is further substituted, the compound is named by first identifying the substituent groups and then the parent amide. The substituents are preceded by the letter N to identify them as being directly attached to nitrogen. Propanoic acid + methyl amine N-Methylpropanamide

Esters (RCO2R) Identify the alkyl group attached to oxygen and then the carboxylic acid. Replace the –ic acid ending with -ate

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Acid Derivatives-Relative Reactivity and Reactions of Derivatives

A more reactive acid derivative can be converted to a less reactive one, but not vice versa

Only esters and amides commonly found in nature.

Acid halides and anhydrides react rapidly with water and do not exist in living organisms

•Hydrolysis- +water carboxylic acid•Alcoholysis- +alcohol ester•Aminolysis- +ammonia or amine amide•Reduction- + H- aldehyde or alcohol•Grignard- + Organometallic ketone or alcohol

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Acid Derivatives-important reactions

Preparation: replace OH

Nucleophilic Substitution:

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Acid DerivativesHoffman Degredation Hoffman degradation (rearrangement) of amides; migration of an aryl

group 1° Amides + Strong basic Br or Cl soln 1° Amines + CO2

http://users.ox.ac.uk/~mwalter/web_04/resources/name_reactions/hofmann.shtml

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Acid DerivativesTransesterification

Transesterification: exchange alkoxyl group with ester of another alcohol

Alcohol + Ester Different Alcohol + Different Ester

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Acid Derivatives-Saponification

Saponification- ester hydrolysis in basic solutions

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Acid Derivatives-Hydrolysis of Amides passage 33

Hydrolysis of amides: Acid or base catalyzed

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Acid DerivativesStrain (e.g., β-lactams) Lactams- cyclic amides Although amides are most stable acid derivative, β-lactams are highly

reactive due to ring strain. Subject to nuclephilic attack.

Found in several types of antibiotics Inhibits bacterial cell wall formation.

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Keto-Acids and Esters Keto acids contain a ketone and a carboxyl group (alpha

and beta) Amino acids degraded to alpha keto acids and then go

into the TCA

Esters have distinctive odors and are used as artificial flavors and fragrances

Beta-keto esters have an acidic alpha hydrogen

Consider keto-enol tautomerism

Naming Esters Esters are named by first determining the alkyl

group attached to the oxygen and then the carboxylic acid from which the ester is derived.

EX: Methyl Propanoate is derived from propanoic acid and a methyl group 101

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Keto Acids and Esters- important reactions

Decarboxylation

Acetoacetic ester synthesis: see aldehydes and ketones

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Amines Important functions in amino acids, nucleotides, neurotransmitters 1o, 2o, 3o, 4o based on how many carbons bonded to Can be chiral, rarely have 4 side groups

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Amines Important functions in amino acids, nucleotides, neurotransmitters 1o, 2o, 3o, 4o based on how many carbons bonded to Can be chiral, rarely have 4 side groups Physical properties:

Polar Similar reactivity to alcohols Can H bond, but weaker H bond than alcohols MP and BP higher than alkanes, lower than alcohols

IR absorption: 2800-3000

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Amines Important functions in amino acids, nucleotides, neurotransmitters 1o, 2o, 3o, 4o based on how many carbons bonded to Can be chiral, rarely have 4 side groups Physical properties:

Polar Similar reactivity to alcohols Can H bond, but weaker H bond than alcohols MP and BP higher than alkanes, lower than alcohols

IR absorption: 2800-3000 General principles:

Lewis bases when they have a lone electron pair NR3 > NR2 > NR > NH3 (least basic)

Stabilize adjacent carbocations and carbanions Effect of substituents on basicity of aromatic amines:

Electron withdrawing are less basic Electron donating are more basic

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Amines-major reactions

Amines are basic and fairly nucleophilic

Amide formation: proteins

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Amines-major reactions Reactions with nitrous acid (HONO):

Distinguishes primary, secondary, and tertiary Primary: burst of colorless, odorless N2 gas

Secondary: yellow oil, nitrosamine-powerful carcinogen

Tertiary: colorless solution, amine forms an ion, e.g. (CH3)3NH+

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Amines- Alkylation Alkylation: SN2 with amine as the nucleophile and alkyl halides as

the electrophile

Reaction with 1° alkyl halide

Alkylation of 1° and 2° are difficult to control and often lead to mixtures of products

Alkylation of 3° amines yield quaternary ammonium salts

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Amines-Hoffman Elimination

Elimination of amine as a quaternary ammonium salt to yield an alkene.

Does not follow Zaitsev’s rule. Less highly substituted alkene predominates

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For Next Time…Last Slide (Hooray!!!!)

Functional Group Quiz Spectra Separations and Purifications Biological Molecules

Carbohydrates Amino acids and proteins Lipids Phophorous containing compounds

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