organic - WordPress.com · Examples. cyclohexene 1,3 ... Since the −C ≡ C− group is very reactive, poly-ynes are not common. ... III. Functional Group Compounds. A. Organic

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ORGANIC CHEMISTRY

I. Introduction. A. General

1. Chemistry of the compounds of carbon except CO32-, SCN - OCN -, and CN - .

a. Carbon is the smallest element in group 14.

Tends to form strong homonuclear bonds

Bonds are slow to break ( high activation energies for C−C cleavage).

b. All organic compounds are unstable to combustion. They react with O2 to give

CO2, H2O, and oxides of other elements.

2. In compounds C can be bonded to 2, 3 or 4 other elements.

a. Bonding possibilities.

Lewis Diagram Bond Type Hybridization Geometry

− C ≡ triple+single sp linear

= C = 2 doubles sp linear

C = 2 singles+double sp2 trigonal planar

− C − 4 singles sp3 tetrahedral

B. Types of organic compounds.

1. Hydrocarbons − contain only C and H.

a. Alkanes −all C's form only single bonds, the C's are sp3 hybridized.

b. Alkenes − Have at least one C=C, at least two C's are sp2 hybridized.

c. Alkynes − Have at least one C≡ C, at least two C's are sp hybridized.

2. Functional group compounds.

a. Replace a H on a hydrocarbon with a group of atoms other than C and H.

b. Such groups are called functional groups. They impart the specific chemical

reactivity to the compound.

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II. Hydrocarbons.

A. Alkanes ( saturated hydrocarbons, paraffins).

1. CH4 Methane

a. All alkanes end in ane.

b. Lewis diagram: H

H − C − H

H

c. Methyl radical CH3−

Radical = fragment of a molecule.

To form a radical, remove a hydrogen .

Radicals end in yl. Drop the ane on the hydrocarbon name and add yl.

2. C2H6 Ethane

a. Structural Formula: CH3CH3

b. Lewis diagram

H C C H

H

H

H

H c. Ethyl radical: CH3CH2−

3. C3H8 Propane

a. Structural Formula: CH3CH2CH3

b. Lewis diagram:

H C

H

H

C

H

H

C

H

H

H

c. Propyl radical. Have two different kinds of H's on propane.

1) Remove one of the six end H's obtain CH3CH2CH2−

called the n - propyl radical ( n = normal ).

2) Remove one of the two interior H's obtain CH3CHCH3

called the isopropyl radical

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4. C4H10 Butane

a. Two different structures. CH3CH2CH2CH3 n - butane

CH3CHCH3 isobutane

CH3

b. Isomers = compounds with the same molecular formulas but with different forms.

All alkanes have the general formula CnH2n+2

This particular type of isomerism is called structural isomerism.

n - butane and isobutane are structural isomers.

c. n - butane = a straight chain hydrocarbon. C atoms are in a continuous chain.

isobutane = a branched chain hydrocarbon.

d. The shapes are shown using line diagrams

n - butane isobutane Note: each corner represents a C atom with the correct number of H's.

5. The number of structural isomers increases with the number of C's.

Formulas Name No. of Isomers

C5H12 pentane 3

C6H14 hexane 5

C7H16 heptane 9

C8H18 octane 18

C9H20 nonane 35

C10H22 decane 75

C20H42 eicosane 366, 319

C30H62 triacontane 4x109

B. IUPAC Rules of Nomenclature.

1. Rules.

a. The parent name of the compound is that of the longest continuous chain of

carbon atoms (l.c.c.). The names are those used above.

b. All other C's not in the l.c.c. are named as radicals.

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c. Number C's on the l.c.c to locate the positions of the radicals.

d. Indicate the number of times a radical appears by prefixes.

di = 2 tetra = 4 hexa = 6 octa = 8 deca = 10

tri = 3 penta = 5 hepta = 7 nona = 9

2. Examples.

a. CH3−CH2−CH2−CH3 butane

b. methylpropane

CH3 CH CH3

CH3

c. 3, 3, 5 - trimethlyhexane

CH3 CH

CH3

CH2 C

CH3

CH3

CH2 CH3

c. The isomers of C5H12.

CH3 CH2 CH2 CH2 CH3CH3 CH

CH3

CH2 CH3CH3 C

CH3

CH3

CH3

pentsne methylbutanedimethylpropane

Note: It was not necessary to number the l.c.c. in these isomers.

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d. Isomers of C6H14.

hexane 2-methylpentane

3-methylpentane

2,3-dimethylbutane2,2-dimethylbutane

C Cyclic Alkanes.

1. Cyclopropane

CH2

H2C CH2

Ring is indicates by the prefix cyclo before the hydrocarbon name.

2. Other examples.

cyclobutane cyclohexane

methylcyclohexane 1,2-dimethylcyclohexane D. Alkenes (Olefins, Unsaturated Hydrocarbons).

1. C2H4 Ethene (ethylene)

a. Formulas CH2CH2 CH2=CH2

C C

H

H

H

H

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b. Alkenes end in ene. Drop ane and add ene.

C's are sp2 hybridized.

2. C3H6 Propene

a.CH3CH=CH2

b. Simple mono - olefins have the general formula CnH2n

3. C4H8 Butene

a. Two isomers

CH2=CHCH2CH3 1 - butene

CH3CH=CHCH3 2 - butene

b. Cis Trans Isomerism

1). Because of the π bonding, there is not free rotation about the double bond.

Therefore, the following isomers are possible:

C C

CH3

H

CH3

H cis - 2 - butene

C C

CH3

H

H

CH3 trans - 2 - butene

2). In the cis isomer the two CH3 's are on the same side of the double bond.

In the trans isomer the two CH3 's are on opposite sides of the double bond.

X

C C

XTRANS

X

C C

X

CIS

4. Branched Chain Alkenes.

CH3 CH C CH CH CH3

CH3

CH3 4, 5 - dimethyl - 2 - hexene

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CH3 CH C CH2 CH3

CH3

3 - methyl - 2 - pentene

b. In naming branched chain alkenes,

1) the l.c.c. must contain the double bond

2) the double bond is given the lowest number.

5. Polyenes − Have several double bonds.

a. CH2=CH−CH=CH2 1, 3 - butadiene

CH2=C=CHCH3 1, 2 - butadiene

b. Use prefixes to denote the number of double bonds and number the l.c.c. to

locate them.

6. Cyclic Olefins.

Use prefix cyclo and number to locate double bonds or radicals on the ring.

b. Examples.

cyclohexene1,3-cyclohexadiene

3,6-dimethylcyclohexadiene3-methylcyclohexene

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c. Benzene C6H6

1).The molecule has 6 π electrons in the ring. Because of the resonance

structures, the electrons are delocalized about the ring. This is

indicated by a circle in the hexagonal ring.

2). Due to the extensive delocalization, the π system is quite stable and benzene

does not act as a typical olefin. The π bonds are unreactive.

3). Compounds derived from benzene are called aromatic compounds.

Nonbenzenoid compounds are referred to as aliphatic compounds.

E. Alkynes (Unsaturated Hydrocarbon).

1. Simple alkynes.

a. C2H2 Ethyne ( acetylene )

1) Alkynes end in yne. Rules of nomenclature are the same as for the alkenes.

2) At least one triple bond ( 2 C's are sp hybridized)

H−C ≡ C−H linear molecule.

b. C3H4 CH3−C ≡ C−H propyne

C4H6

H−C ≡ C−CH2−CH3 1 - butyne

CH3C ≡ CCH3 2 - butyne

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2. Branched chain alkynes.

CH3 C C CH2 C CH3

CH3

CH3 5, 5 - dimethyl - 2 - hexyne

3. Cyclic Alkynes and Polyalkynes.

a. Because of the linear nature of the −C ≡ C− group, small ring alkynes are not

stable.

b. Since the −C ≡ C− group is very reactive, poly-ynes are not common.

CH3−C ≡ C−C ≡ C−CH2−CH3 2, 4 - heptadiyne

F. Reactions of the Hydrocarbons

1. Alkanes

a. Quite unreactive. Reactions involve the substitution of some other element for H.

b. Burned as fuel.

c. Used as nonpolar solvents.

2. Alkenes and Alkynes (Unsaturated Hydrocarbons)

a. Will add small molecules across the multiple bonds. The π bond breaks and two

σ bonds are formed ( addition reactions ).

b.Examples.

1) CH2=CH2 + H2 ---------> CH3CH3

C C H H H C C H

H

H

H

HH

H H

H

+

! bond breaks

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2) CH2=CH2 + X2 -------------> H2CX − CH2X

C C X X H C C H

X

H

X

HH

H H

H

+

! bond breaks

X = halogen

H C C H H C C H

X

X

X

X

X X+ 23)

( X = H, halogen ) c. If one adds small molecules of the form HX to an unsymmetric alkene or alkyne,

the addition is such that the H goes to the C having the greater number of H's. Examples: CH3CH=CH2 + HX ------> CH3CHXCH3 HC ≡ CH + 2HX -------> CH3CHX2

3. Aromatic Compounds. a. Aromatic compounds undergo substitution reactions rather than addition

reactions. b Example:

Cl

+ Cl2

AlCl3

+ HCl

III. Functional Group Compounds.

A. Organic Halides.

1. Halogen replaces a hydrogen on an alkane. Name halogen as a radical.

a. Drop the elemental ending on the halogen and add o.

−F = fluoro −Cl = chloro −Br = bromo −I = iodo b. Name halogen as you would name any radical. Halogen is given the lowest possible number. 2. Examples. a. CH3Cl chloromethane ( methyl chloride )

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CH2Cl2 dichloromethane ( methylene chloride ) CHCl3 trichloromethane ( chloroform ) b. CH3CHCHCH3 | | 2 - chloro - 3 - methylbutane Cl CH3

c. Aromatic Halides.

Cl

chlorobenzene

3. Uses. a. Starting materials for other organic compounds. The halogen is fairly easy to remove. b. Solvents. The use of halogenated solvents is being phased out because of

environmental concerns. c. Coolants. CCl2F2 = Freon - 2. The widespread use of these chlorofluorocarbons is now thought to be one of the major causes for the decrease in the ozone layer. B. Alcohols. 1. Nomenclature and examples. a. Formed by replacing a H on an alkane by an OH group. General formula is ROH. Where R = hydrocarbon fragment. b. CH3OH Methanol ( methyl alcohol, wood alcohol ) CH3CH2OH Ethanol ( ethyl alcohol, grain alcohol ) c. Nomenclature 1). The parent name is taken from the l.c.c. having the OH. 2). Drop the e on the alkane name and add ol. 3). When necessary, number the l.c.c. to locate the OH. CH3CH2CH2OH 1 - propanol CH3CHCH3 | 2 - propanol (isopropanol) OH d. Branched chain alcohols. CH3CHCH2CHCH3 | | 4 - methyl - 2 - pentanol CH3 OH Note: the OH is given the lowest number. e. Poly alcohols ( Polyol's) CH2−CH2 | | 1, 2 - ethanediol ( ethylene glycol ) OH OH

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Use the prefixes di-, tri-, etc. to indicate the number of OH's in the molecule. CH2CH-CH2 | | | 1, 2, 3 - propanetriol ( glycerol or glycerine) OH OH OH 2. Classification of alcohols. a. Primary Alcohols = OH is on a C that is bonded to at least two H's. That is, the OH is on an end carbon. Examples: CH3OH, CH3CH2OH, CH3CH2CH2OH b. Secondary Alcohols = OH is on a C that is bonded to one H. That is, the C is bonded to two other carbons. Example: CH3CHCH3 | OH c. Tertiary Alcohols = OH is on a C that is not bonded to a H. That is, the C is bonded to three other carbons. CH3 | CH3CCH3 2 - methyl - 2 - propanol | ( t - butyl alcohol or tertiary butyl alcohol ) OH C. Oxidation of Primary Alcohols - Aldehydes and acids. 1. Oxidation Sequence. O O [O] || [O] || R−CH2−OH -------> R−C−H -------> R−C−O−H primary alcohol aldehyde acid O O [O] || [O] || CH3CH2OH -------> CH3−C−H −−−−> CH3−C−O−H Ethanol Ethanal Ethanoic Acid O || 2. Aldehyde R−C−H

a. Nomenclature.

1) Drop the ol on the alcohol name and add al.

2) O || H−C−H methanal ( formaldehyde.)

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O || CH3−C−H ethanal ( acetaldehyde ) O || 3) In numbering the l.c.c. the C–H carbon is C number 1. CH3 O | || CH3CH2CHCH2CHCH 2, 4 - dimethylhexanal | CH3 b.Uses 1) Used as solvents and preservatives. 2) Easy to oxidize. O O || || 3. Acids R –C-O-H (carboxylic acids, -C-O-H = carboxyl group ) a. Nomenclature

1) Drop e on the alkane and add oic acid.

2) The carboxyl carbon is carbon number 1.

3) Examples O || H-C-O-H ethanioc acid ( formic acid ) O || CH3-C-O-H ethanoic acid ( acetic acid ) O || CH3CH2C-O-H propanoic acid O || CH3CHCH2CH2COH 4 - methylpentanoic acid | CH3 b. Weak acids Acid Ka Methanoic HCHO2 1.8x10 - 4 Ethanoic HC2H3O2 1.8x10 - 5 Propanoic HC3H5O2 1.4x10 - 5

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D. Oxidation of Secondary Alcohols 1. General reaction OH O | [O] || R−C−R' ----------> R−C−R'

| H Secondary Ketone Alcohol 2. Nomenclature a. Drop the ol on the alcohol and add one. When necessary, number the l.c.c. to locate the = O group. b. Exanples. O || CH3CCH3 propanone ( acetone ) O || CH3CH2CCH2CH3 3 - pentanone O CH3 || | CH3CCH2CHCH3 4 - methyl - 2 - pentanone 3. Difficult to oxidize. Used extensively as solvents 4. Tertiary alcohols are not oxidized like primary and secondary alcohols. E. Other Functional Groups. 1. Amines R−NH2 a. Can think of amines as arising by substituting alkyl groups for H's on ammonia. H H | | H−N| ammonia R−N| primary amine | | H H H R | | R−N| secondary amine R−N| tertiary amine

| | R R b. NH2 group = amino group. Name as a radical. CH3NH2 methyl amine ( aminomethane )

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(CH3)2NH dimethyl amine CH3CHCH2CHCH3 | | 4 - methyl - 2 - aminopentane CH3 NH2 c. Weak ammonia - bases H H + | | CH3N| + H2O CH3N−H + OH - Kb = 4.4x10 - 4 | | H H d. Nitrogen can be incorporated into an aromatic ring

N

Pyridine (an example of a heterocyclic compound) Pyridine is a weak base, Kb = 1.7x10 - 9 2. Ethers R−O−R'

a. An alkyl group replaces the H on an alcohol. Can be prepared from alcohols. H2SO4 2ROH ----------> R−O−R + H2O (Conc. H2SO4 is a catalyst) b. CH3O - = methoxide ion ( very strong base ) CH3OCH3 = methyl methoxide ( dimethyl ether ) c. Used as solvents. Highly flammable and toxic. d. Cyclic Ethers.

CH2

CH2

O

H2C

H2C

O

Tetrahhdrofuran (THF) isan important solvent.

IV. Condensation Reactions. A. General. 1. A reaction that joins two organic molecules together by splitting out a water molecule between them. H2SO4 The reaction 2ROH ----------> R−O−R + H2O is an example of a condensation reaction 2. This is one of the most important general reactions in organic synthesis. Will consider only a few example. B. Condensation Reaction Between an Acid and an Alcohol. Esters. 1. Reaction.

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O –H2O O || H+ || R−C−O−H + HOR' ------------> R−C−O−R' + H2O Acid Alcohol Ester a. Example. O O || H+ || CH3CH2−C−O−H + HOCH3 ----------> CH3CH2−C−O−CH3 + H2O Propanoic acid methanol methyl propanoate b. Nomenclature. O || R − C − O − R' RC(O)OR’ acid part alcohol part O || 1) Name alcohol part ( R' ) first, then the acid part ( R−C−O− ) . 2) Name the alcohol part as a radical. 3) In naming the acid part, drop the ic ending of the acid and add ate. c. Examples. O || H−C−O−CH2CH3 ethyl methanoate ( ethyl formate ) O || CH3−C−O−CH3 methyl ethanoate ( methyl acetate ) d. Esters generally have pleasant odors. The odors of fruits is due to esters.

Ester Fruit Butyl ethanoate CH3C(O)OCH2CH2CH2CH3 Bananas Ethyl butanoate CH3CH2CH2C(O)OCH2CH3 Pineapples Octyl ethanoate CH3C(O)OCH2(CH2)6CH3 Oranges Pentyl butanoate CH3CH2CH2C(O)O(CH2)4CH3 Apricots 2. Mechanism.

In forming the H2O molecule, the OH comes from the acid and the H comes from the alcohol.

O – H2O O || || R−C−O−H H−O−R' −−−−−> R−C−O−R

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3. Hydrolysis of Esters ( Saponification ). a. Can hydrolyze an ester by reaction with a base. O O || || R−C−O−R' + NaOH ----------> R−C−O - Na+ + R'−O−H

b. In the hydrolysis the O on the carboxyl group comes from the OH - ion. 4. Fats ( Triglycerides ). Biologically Important Compounds. a. Esters of the tri - alcohol glycerol and long chain(fatty) acids. glycerol = CH2−CH−CH2 | | | OH OH OH O || H2C−O−C−R R = long chain alkyl group O R = CH3(CH2 )16 . Acid = Stearic Acid. || Fat = stearin ( animal fat ) HC−O−C−R R = CH3(CH2)14 . Acid = Palmitic Acid. O Fat = palmitin (animal fat) || R = CH3(CH2)7CH=CH(CH2)7 . Acid = Oleic Acid. H2C−O−C−R Fat = Olein ( vegetable fat or oil )

b. Characteristics of biologically important fatty acids. 1) Contain an even number of carbons 2) Are straight chain acids. 3) In animal fats the acids are saturated, while in vegetable fats the acids contain at least one double bond. 4) Saturated fatty acids and their corresponding fats have higher melting points than the unsaturated ones. Therefore animal fats are solids at room temperature while vegetable fats are liquids at room temperature. c. Saponification of fats. O || H2C−O−C−R O Η2C−O−H O || || HC−O−C−R + 3NaOH ---------> HC−O−H + 3 R−C−O - Na+ Ο | || H2C−O−H H2C−O−C−R Τhe resulting sodium salts of the long chain acids are soaps.

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C. Amides 1. Condensation of an acid and an amine. O - H2O O || || R−C−O−H + H−N−R ---------> R−C−N−R + H2O | | R' R' Acid Amine Amide 2.The amine can be either a primary or a secondary amine. 3. The amide linkage is biologically important. IV. Polymers and Polymerization reactions. A. General. 1. Polymer = large molecule made by joining a number of smaller (monomer) molecules together. 2. Can classify polymers in terms of the type of reaction used to couple the monomer molecules together. B. Addition Polymers. 1. Couple alkenes and alkynes by breaking the π bonds and adding one unsaturated molecule to another. a.Polymerization of ethene ( ethylene ). H H H H H H H H H H H H | | | | | | | | | | | | C = C + C = C + C = C + … -[−C − C − C − C − C − C−]n- | | | | | | | | | | | | H H H H H H H H H H H H ethylene Polyethylene b. Polymerization occurs through a free radical mechanism. R−O−O−R -------> 2R−O• initiation ( R−O−O−R = an initiator)

RO• + H2C = CH2 -------> ROCH2CH2 • ROCH2CH2• + H2C = CH2 -----> ROCH2CH2CH2CH2• propagation steps etc. -CH2CH2• + •CH2CH 2 ----> -CH2CH2CH2CH2 termination c. Most of the common plastics are addition polymers of alkenes ( see attached ).

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d. Conducting Polymers. Polyacetylene. H−C≡C−H + H−C≡C−H + H−C≡C−H + H−C≡C−H −CH=CH−CH=CH−CH=CH−CH=CH− Polyacetylene In polyacetylene the π electrons are delocalized along the carbon chain. Therefore, polyacetylene is a good conductor of electricity ( nonmetallic conductor ). 2. Elastomers = Polymers possessing the ability to stretch reversibly to several times their lengths. a. Most elastomers are addition polymers of the diene H2C CH2 C − C cis - 1, 3 - butadiene H H b. Polymerization reaction. CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2

C − C + C − C + C − C + C − C H H H H H H H H −CH2 CH2 − CH2 CH2 − CH2 CH2 − CH2 CH2−

C = C C = C C = C C = C H H H H H H H H c. All natural and synthetic rubbers are polymers of derivatives of cis - 1, 3 - butadiene. 1). Natural rubber is the polymer of CH2 CH2

C − C ----------------> polyisoprene H CH3 2 - methyl - 1, 3 - butadiene ( isoprene )

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2). Neoprene Rubber is a polymer of CH2 CH2 C − C -----------------> neoprene H Cl 2 - chloro - 1, 3 - butadiene ( chloroprene ) 3). SBR Rubber = copolymer of cis - 1, 3 - butadiene and styrene in a 3 : 1 molar ratio. This rubber comprises about 45% of the U.S. market. 4). Vulcanization = treatment of rubber by heating with sulfur. Crosslinks the polymer chains with - S - S - bonds. 3. Polyurethanes. a. Addition polymerization does not have to take place only between alkenes and alkynes. H O | || 1) R−N = C = O + R'−O−Η -------------> R − N − C − O − R' isocyanate alcohol urethane 2) The urethane linkage is formed by the addition of an alcohol across the N = C double bond. b. Polyurethanes = polymers of di - isocyanates and dialcohols. O H H O || | | || n O = C = N −R−N = C = O + n H−Ο−R'−O−H -----> −(−C−N−R−N−C−O−R'−O−)n− a polyurethane c. Depending on the nature of R and R' and the method of production, polyurethanes can be used as high impact and abrasion resistant coatings, bedding and upholstery

material, and fabrics for clothing that must stretch ( swim suits,ski wear, etc.). C. Condensation Polymers. 1. Condensation polymerization = polymerization based on a condensation reaction. a. Important both industrially and biologically. b. Includes polyesters, polyethers, polyamides and polypeptides. 2. Polyesters. a. Formed by the reaction of a diacid and a dialcohol. The general equation is: O O O O || || || || n H-O-C-R-C-O-H + n H - O- R'- O - H -----> -[-C-R-C-O-R'-O-]n- b. A specific example is Dacron which is the polymer formed between the dialcohol,

ethylene glycol ( HOCH2CH2OH ), and the diacid, terethphalic acid.

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CC OH

O

HO

O

Terethphalic Acid 3. Polyamides. a. Nylon. O O H H O O || || | | || || n H2N(CH2)6NH2 + n HOC(CH2)6COH ------> -[-N(CH2)6N-C(CH2)6C-]n- 1, 6 - diaminohexane adipic acid 6 : 6 Nylon 4. Polypeptides - Proteins. a. Alpha amino acids. 1) Structure. general formula of an alpha amino acid is H O | || R − C − C − O − H | NH2

The term alpha is derived from a old system of nomenclature that designates the carbon next to the carboxyl group as the alpha ( α ) carbon.] 2) Nomenclature. H O | || H − C− C − O − H aminoethanoic acid (glycine) | ΝΗ2

H O | || CH3− C − C − O − H 2 - aminopropanoic acid (alanine) | NH2

In naming the amino acids the common names are usually used instead of the IUPAC ones. 3) For all alpha amino acids other than glycine the alpha carbon is surrounded by four different groups. Whenever a tetrahedral center is surrounded by four different groups a type of isomerism called optical isomerism exists. Consider such a center and its mirror image shown below.

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`

A

B

C

D

C

B

D

A

B

C

D

ARotate 180°

Note that the two mirror images are nonsuperimposible. The two are optical isomers or enantiomers. The compound is said to be optically active or chiral. 4) For simple molecules, the only difference between the two compounds is their ability to rotate the plane of polarized light. One isomer will rotate the plane to the right ( is dextrorotatory) and is called the D isomer, while the other isomer, the L isomer, rotates the plane a like amount to the left( is levorotatory). A 50:50 mixture of the two isomers, called a racemic mixture, will not exhibit optical activity. 5) In more complex reactions, such as enzyme catalysed reactions, the stereoisomers can be distinguished. 6) Whenever a carbon is bonded to four different groups the particular molecule is chiral. Some examples (the chiral carbon is in bold): H H Cl | | | CH3CH2CCH2CH2CH2CH3 CH3CCH2CH3 CH3C–H | | | CH3 OH Br

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b. Proteins are polymers of L amino acids. H H O H H O H H O | | || | | || | | || H - N - C - C - O - H H - N - C - C - O - H H - N - C - C - O - H | | | R R R

H H O H H O H H O | | || | | || | | || -[- N - C - C - N - C - C - N - C - C -]n- | | | R R R O H || | c. The - C - N - linkage is referred to as the peptide linkage and simple proteins are called polypeptides.

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Organic Problems

1. Draw and name the isomers of C7H16 . 2. Name the following. a. CH3 d. CH3 CH3 | | | CH3CHCH2CCH3 CH2=CCH2CCH3 | | | CH3 CH3 CH3 b. CH3 CH3 e. CH3 | | | CH3CH2CCH2CCH3 HC≡CCCH2CH3 | | | CH3 CH3 CH3 c. CH3 CH3 f. CH3 CH3 | | | | CH3CCH2CHCHCH3 CH2 CH2 | | | | CH2 CH3 CH3CHCH2CHCHCH3 | | CH3 CH3 3. Name the following

a. b.

c. d.e.

4. Draw structural formulas for the following. a. 2,3,5,6-tetramethyloctane b. cis-3,4-dimethyl-3-hexene c. 1-ethyl-3-methylcycloheptane d. 4,4-dimethyl-2-pentyne e. 6-methyl-2,4-nonadiene f. ethylcyclopentane g. trans-2,3-dimethyl-1,3-butadiene 5. Draw and name 12 isomers that have the formula C6H12. 6. Complete and balance the following.

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a. CH3CH=CH2 + HCl ----------> b. CH3CH=CHCH3 + Br2 --------> c. CH3C≡CH + HBr ----------> d. CH3C≡CCH3 + F2 ---------> e. 1,3-cyclohexadiene + Cholrine ----------> AlCl3 f. 1,4-dimethylbenzene + Chlorine ------------> 7. Name the following. O ||

a. CH2CH2 b. CH3CCH2CHCH3 | | | Br Br CH3 CH3 CH3 O | | || c. CH3CH2CHCH2CHCHCH3 d.CH3CH=CHCH2CCH3 e.CH3CH2CHC–H | | | | CH2 CH3 CH3 CH3 | CH3 OH CH3 CH3 | | | f. CH3CHCH2CH g. CH3CHCH2CH3 h.HC≡CCH2CHCH2CH3 | | CH3 NH2

i

!!!!!!!!!!!! ! CH3!!!!O

| ||

j. CH3CH2CHCH2C-O-H 8. Draw structural formulas for the following: a.3-hexanone b. trans-3-hexene c. 2,3-pentadiol d.propylbutanoate e.2,3,4-trimethylhexane f.3-methyloctanal g.2-pentyne h. 1,2,3-triethylcyclohexane i.butylpropanoate j. The oxidation product of 2-pentanol k. The fat formed by hexanoic acid 9.Write formulas for the products and name all organic compounds in the following: a. CH3C=CH2 + HI ----> | CH3

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O || H+ b. CH3CH2CH-C-OH + CH3OH ----> | CH3

[O] c. CH3CH2CHCH2CH3 ----> | OH

[O] [O] d. CH3CH2CH2CH2OH ----> ----> f. CH3C≡CH + HCl ----> O || g. CH3CH2C-O-CH3 + NaOH ---->

10. Draw at least three repeating units of the polymers formed by each of the following monomers.

a. 2,3-dimethyl-2-pentene O O || || b. H-O-C-CH2-C-O-H + HOCH2CH2CH2OH O || c. CH3CH2CHCOH | NH2 d. cis-2,5-dimethyl-2,4-hexadiene

e. CH3C≡CCH3

11. Which compounds in questions 7 and 8 would be optically active (chiral)?

e, + Cl2 ---->

27

Organic Problems

1. Draw and name the isomers of C7H16 . (only C’s shown) Heptane 2-methylhexane 3-methylhexane C-C-C-C-C-C-C-C C-C-C-C-C-C C-C-C-C-C-C | | C C 2,2-dimethylpenane 2,3-dimethylpentane 2,4-dimethylpentane C C C | | | C-C-C-C-C C-C-C-C-C C-C-C-C-C | | | C C C 3,3-dimethylpentane 3-ethylpentane 2,2,3-trimethylbutane C C-C C C | | | | C-C-C-C-C C-C-C-C-C C-C-C-C | | C C 2. Name the following. a. CH3 d. CH3 CH3 | | | CH3CHCH2CCH3 CH2=CCH2CCH3 | | | CH3 CH3 CH3 2,2,4-trimethylpentane 2,4,4-trimethyl-1-pentene b. CH3 CH3 e. CH3 | | | CH3CH2CCH2CCH3 HC≡CCCH2CH3 | | | CH3 CH3 CH3 2,2,4,4-tetramethylhexane 3,3-dimethyl-1-pentyne c. CH3 CH3 f. CH3 CH3 | | | | CH3CCH2CHCHCH3 CH2 CH2 | | | | CH2 CH3 CH3CHCH2CHCHCH3 | | CH3 CH3 3,3,5,6-tetramethylheptane 3,4,6-trimethyloctane

28

3. Name the following

a. b.

c. d. e.

4-methyl-2-pentene 2,3,4-trimethylhexane

3,6-dimethyl-1,4-cyclohexadiene

2,4-heptadiene

1,2-dimethylcyclopentane 4. Draw structural formulas for the following. a. 2,3,5,6-tetramethyloctane C C C C | | | | C-C-C-C-C-C-C-C b. cis-3,4-dimethyl-3-hexene

C C

C

C

C

C

C C

c. 1-ethyl-3-methylcycloheptane

H2C

H3C

CH3

d. 4,4-Dimethyl-2-pentyne C | C-C≡C-C-C | C

29

e. 6-methyl-2,4-nonadiene C | C-C=C-C=C-C-C-C-C f. ethylcyclopentane g. trans-2,3-dimethyl-1,3-butadiene

CH2-CH3

C C

C

C

C

C

5. Draw and name 12 isomers that have the formula C6H12.

C=C-C-C-C-C C-C=C-C-C-C C-C-C=C-C-C cyclohexane 1-hexene 2-hexene 3-hexene C C C C | | | | C=C-C-C-C C=C-C-C-C C=C-C-C-C C-C=C-C-C 2-methyl-1pentene 3-methyl-1-pentene 4-methyl-1-pentene 2-methyl-2-pentene C C C C C | | | | | C-C=C-C-C C-C=C-C-C C=C-C-C C=C-C-C | C 3-methyl-2-pentene 4-methyl-2-pentene 2,3-dimethyl-1-butene 3,3-dimethyl-1-butene C C | | C-C=C-C 2,3-dimethyl-2-butene (cis & trans isomers possible) 6. Complete and balance the following. Cl | a. CH3CH=CH2 + HCl ----------> CH3CHCH3 Br Br | |

b. CH3CH=CHCH3 + Br2 -------->CH3CH-CHCH3 c. CH3C≡CH + HBr ----------> CH3CBr2CH3 d. CH3C≡CCH3 + F2 --------->CH3CF2CF2CH3

30

AlCl3 e. 1,3-cyclohexadiene + 2 Cholrine ---------->

f. 1,4-dimethylbenzene + Chlorine ------------>

7. Name the following. O ||

a. CH2CH2 b. CH3CCH2CHCH3 | | | Br Br CH3 1,2-dibromoethane 4-methyl-2-pentanone CH3 CH3 O | | || c. CH3CH2CHCH2CHCHCH3 d.CH3CH=CHCH2CCH3 e.CH3CH2CHCH | | | | CH2 CH3 CH3 CH3 | CH3 5,5-dimethyl-2-hexene 2-methylbutanal 5-ethyl-2,3-dimethyl-heptane

OH CH3 CH3 | | | f. CH3CHCH2CH g. CH3CHCH2CH3 h.HC≡CCH2CHCH2CH3 | | CH3 NH2 4-methyl-2-pentanol 2-aminobutane 4-methyl-1-hexyne

CH3

+ HCl

CH3

Cl

Cl

Cl

Cl

Cl

31

i.

!!!!!!!!!!!! ! CH3!!!!!!O

| ||

h. CH3CH2CHCH2C-O-H 1,3-cyclopentadiene 3-methylpentanoic acid 8. Draw structural formulas for the following: a.3-hexanone b. trans-3-hexene c. 2,3-pentadiol O || C-C-C-C-C-C

CC

C C

C

C

OH

HO d.propylbutanoate e.2,3,4-trimethylhexane f.3-methyloctanal O C C C C O || | | | | || C-C-C-C-O-C-C-C C-C-C-C-C-C C-C-C-C-C-C-C-C-H g.2-pentyne h. 1,2,3-triethylcyclohexane

C C C C C

CH2CH3

CH2CH3

CH2CH3 i.butylpropanoate j. The oxidation product of 2-pentanol

O

||

CH3CH2COCH2CH2CH2CH3

O

||

CH3CCH2CH2CH3

32

k. The fat formed by hexanoic acid O || H2C-O-C-CH2CH2CH2CH2CH3 O || H-C-O-C-CH2CH2CH2CH2CH3 O || H2C-O-C-CH2CH2CH2CH2CH3

9.Write formulas for the products and name all organic compounds in the following: I |

a. CH3C=CH2 + HI ---> CH3CCH3 | | CH3 CH3 Methyl propene 2-iodo-2-methylpropane

O O || H+ || b. CH3CH2CH-C-OH + CH3OH ----> CH3CH2CH-C-OCH3 | | CH3 CH3 2-methylbutanoic acid methanol methyl 2-methylbutanoate [O] c. CH3CH2CHCH2CH3 ----> CH3CH2CCH2CH3 | || OH O 3-pentanol 3-pentanone O O [O] || [O] || d. CH3CH2CH2CH2OH ----> CH3CH2CH2CH ----> CH3CH2CH2COH 1-butanol butanal butanoic acid

33

Cl

Cl

e. + Cl2

cyclohexene 1,2-dichlorocyclohexane f. CH3C≡CH + 2 HCl ----> CH3CCl2CH3 Butyne 2,2-dichloropentane O O || ||

g. CH3CH2C-O-CH3 + NaOH ----> CH3CH2C-O Na+ + CH3OH

Methylpropnaoate Sodium propanoate Methanol

10. Draw at least three repeating units of the polymers

formed by each of the following monomers.

a. 2,3-dimethyl-2-pentene

C C C C C C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

O O || || b. H-O-C-CH2-C-O-H + HOCH2CH2CH2OH O O O O O O || || || || || || -O-C-CH2-C-O-CH2-CH2-CH2-O-C-CH2-C-O-CH2-CH2-CH2-O-C-CH2-C-O-

O || c. CH3CH2CHCOH | NH2

34

O O O || || || -NH-CH-C-NH-CH-C-NH-CH-C- | | | CH2CH3 CH2CH3 CH2CH3

d. cis-2,5-dimethyl-2,4-hexadiene

e. CH3C≡CCH3 C C C C C C | | | | | | -C=C-C=C-C=C- 11. 7 b,c,e,f,g,h,j 8 c,e,f,j

C

C C

C C

C C

C C

C C

CCC C C C

C C C CC

C C