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CHEM 212 Exam 3 Part 1

Professor Kelly Boebinger 6.7 Stability

 Cis are less stable than trans alkenes  Compare heat given off on : Ho  Less stable isomer is higher in energy, and gives off more heat  Comparing heat given off when C=C is converted to C-C  More stable alkene gives off less heat  Trans generates 5 kJ less heat than cis- butene

2 Alkene Stability

 Alkenes become more stable with increasing substitution  Stability Order: Most to Least stable

This order is due to two factors: 1) Hyperconjugation 2) Bond strengths

3 Hyperconjugation

 An interaction that results from the overlap of a vacant p orbital with a neighboring bond. The more the substituents present, the more opportunities for hyperconjugation.

4 Bond Strengths

 sp2 to sp3 is stronger than sp3-sp3

5 Problem: Which is more stable? Why?

More substitution mono- vs di- & sp2-sp3 bonding

Trans more stable than cis

More substitution tri- vs di- 6 Writing Organic Reactions

 No established convention – shorthand  Can be formal kinetic expression  Not necessarily balanced  Reactants can be before or on arrow  , temperature, details, on arrow 7 6.8 Electrophilic Addition of HX to Alkenes ( X = Br, Cl, I)

 For unsubstituted alkenes or for symmetrically substituted alkenes, the addition of HX across a double bond appears to proceed in an identical fashion as the addition of halogens across a double bond. The reaction is successful with HCl and with HBr. H H H H ether C C + Br H H C C Br

H H H H

Complete the reaction:

CHCHCHCH H3CH3C CHC3 H3 3 3 3 3 etheretherether C C C C ++ Br BrH H H3CH3C C C C C Br Br

H CH C CHCH 3 3 3 3 H H CHC3 H3 8 6.8 Electrophilic Addition of HX to Alkenes ( X = Br, Cl, I)

 Note that HI is generated from KI and phosphoric acid

9 Electrophilic Addition of HX to Alkenes ( X = Br, Cl, I)

 For unsymmetrical alkenes, the addition of HX across a double bond occurs with regiochemistry (Regiospecific – one product forms where two are possible)

H C H CH3 H CH3 H 3 ether C C + HCl H3C C C H and H3C C C H

H3C H Cl H H Cl

100 % 0 %

H C CH CH3 CH3 CH3 CH3 3 3 ether C C + HCl H3C C C H and H3C C C H

H3C H Cl H H Cl

mostly this very little Let’s look at background on why this happens 10 6.9, 6.10 Orientation of Electrophilic Addition: Markovnikov’s Rule

 In an unsymmetrical alkene, HX reagents can add in two different ways, but one way may be preferred over the other  If one orientation predominates, the reaction is regiospecific  Markovnikov observed in the 19th century that in the addition of HX to alkene, the H attaches to the carbon with the most H’s and X attaches to the other end (to the one with the most alkyl substituents)  Markovnikov’s rule: Hydrogen will add to the side that will produce the more stable carbocation. 11 Carbocation Structure and Stability

 Hydrogen adds to the double bond first. It will add in a way that leaves the most stable carbocation, ready for the halogen to add and complete the reaction.

12 Carbocation Structure and Stability

+  Stability of carbocations: 3º > 2º > 1º > CH3

13 Energy of Carbocations and Markovnikov’s Rule  More stable carbocation forms faster  Tertiary cations and associated transition states are more stable than primary cations

14 Electrophilic Addition: Markovnikov’s Rule

15 Back to the reactions, now the mechanisms!

H C CH CH3 CH3 CH3 CH3 3 3 ether C C + HCl H3C C C H and H3C C C H

H3C H Cl H H Cl mostly this little of this

- H-X X X CH CH 3 CH 3 H3C 3 H3C C C + C C C C H C H C H H C 3 3 3 H H C H H 3 H

- H-X X H X H H C CH3 3 + CH3 C C C C C C H C H C H C H 3 3 CH 3 H H C 3 3 H H3C 16 6.11 The Hammond Postulate: Transition State Structures

 A transition state is the highest energy species in a reaction step  By definition, its structure is not stable enough to exist for one vibration  But the structure controls the rate of reaction  So we need to be able to guess about its properties in an informed way  We classify them in general ways and look for trends in reactivity – the conclusions are in the Hammond Postulate  The Hammond Postulate helps explain Markovnikov’s rule by explaining the relationship between reaction rate and intermediate stability.  “The structure of a transition state resembles the structure of the nearest stable species…” 17 The Hammond Postulate

 (a) The transition state of an endergonic reaction step structurally resembles the product  (b) The transition state of an exergonic reaction step structurally resembles the reactant.

18 Statement of the Hammond Postulate

 A transition state should be similar to an intermediate that is close in energy

In a reaction involving a carbocation, the transition states look G like the intermediate carbocation

Reaction

19 6.12 Evidence of Mechanism of Electrophilic Addition: Rearrangements of Carbocations

 The most likely evidence of these two step mechanisms are that some products are formed from rearrangement (shifts), these products cannot be formed if it was a single step reaction.  There is evidence of a hydride shift in some reactions, and a methyl shift in others.

20 CHEM 212 Exam 3 Part 2

Professor Kelly Boebinger 7. Alkenes: Reactions and Synthesis

X CH2Cl2 C C + X2 C C

X

X2 = Cl2, Br2 anti addition Diverse Reactions of Alkenes

 Alkenes react with many to give useful products by addition (often through special reagents)  alcohols (add H-OH)  alkanes (add H-H)  halohydrins (add HO-X)  1,2-dihalides (add X-X)  halides (add H-X)  1,2-diols (add HO-OH)

 cyclopropane (add :CH2)  carbonyl

23 Diverse Reactions of Alkenes

24 7.1 Preparation of Alkenes: A Preview of Elimination Reactions

 Alkenes are commonly made by  elimination of HX from alkyl halide (dehydrohalogenation)  Uses KOH and ethanol

H Br KOH C C K Br HO H ethanol C C + +

25 7.1 Preparation of Alkenes: A Preview of Elimination Reactions

 Alkenes are commonly made by  elimination of H-OH from an alcohol (dehydration)

 require strong acids (sulfuric acid, 50 ºC)

H OH + H3O ,  C C C C + HO H

26 7.2 Addition of Halogens to Alkenes

 Bromine and chlorine add to alkenes to give 1,2- dihaldes, an industrially important process

 F2 is too reactive and I2 does not add + -  Cl2 reacts as Cl Cl , Br2 is similar X CH2Cl2 C C + X2 C C

X

X2 = Cl2, Br2 anti addition

Antistereochemistry: Anti which two ends of a double bond are attacked from different sides. 27 Addition of Br2 to Cyclopentene proves anti addition

 Addition is exclusively trans

Mechanism

28 Problems: Complete the following reactions:

A. HH OOHH HHOO++, ,  33 HH33CC HH HH33CC CC CHCH22 CC CC ++ HH OOHH

CCHH33 HH33CC HH B. KOH CH CH CH CH CH Br CH CH CH CH=CH K Br 3 2 2 2 2 3 2 2 2 HO H ethanol + +

C. Write the bromination reaction of 2-methylpropene CH CH3 Br CH CH Cl Br CH3 CH2Cl2 H2C C CH3 Br2 H C C CH 2 3 + 2 H C C CH3 H Br 29 7.3 Halohydrin Formation

 This is formally the addition of HO-X to an alkene (with +Br as the ) to give a 1,2-halo alcohol, called a halohydrin

 The actual reagent is the dihalogen (Br2 or Cl2 in water in an organic solvent)

Anti addition

Markonivnikov regiochemistry observed, Br adds first to less substituted carbon 30 Mechanism of Formation of a Bromohydrin

 Br2 forms bromonium ion, then water adds (acts as a ) and opens the ring  Aromatic rings do not react

31 An Alternative to Bromine

 Bromine is a difficult reagent to use for this reaction since few alkenes are soluble in water  N-Bromosuccinimide (NBS) produces bromine in organic and is a safer source

32 7.4 Addition of Water to Alkenes: Oxymercuration

 Hydration of an alkene is the addition of H-OH to to give an alcohol  Acid catalysts are used in high temperature industrial processes: is converted to ethanol

Regiochemistry corresponds to Markovnikov addition

33 Oxymercuration Intermediates

 For laboratory-scale hydration of an alkene  Use mercuric acetate in THF followed by sodium borohydride  Markovnikov orientation  via mercurinium ion

34 Problems: Complete the following reactions:

A. Br OOHH CH Br22 CH33 CH ++ CH33 HBr H O ++ HBr H22O BrBr HH CH B. CH 3 CH 3 CH 3 1. Hg(OAc)2, H2O 3 H C 1. Hg(OAc) , H O H3C H C3 CH 2 2 H C 3 3 3 CH CH3 3 2. NaBH4 CH3 2. NaBH4 OH OH

C. Write the hydration reaction of 2-methylpropene using the industrial method.

CH 3 H CH H3PO4, catalyst 3 H2C C CH3 + H2O H C C CH 250°C 3 H OH 35 7.5 Addition of Water to Alkenes:

 Borane (BH3) is electron deficient is a Lewis acid  Borane adds to an alkene to give an organoborane  Boron attaches to the less highly substituted carbon (steric reasons)

General Overall Reaction H OH 1. BH3, THF C C C C - R 2. H2O2, OH R 36 Syn addition, anti Markovnikov Orientation in Hydration via Hydroboration

 Regiochemistry is opposite to Markovnikov orientation  OH is added to carbon with most H’s  H and OH add with syn , to the same face of the alkene (opposite of anti addition) -  H2O2, OH inserts OH in place of B  Retains syn orientation

37 7.6 Addition of Carbenes to Alkenes

KOH

 The carbene functional group is “half of an alkene”  Carbenes are electrically neutral with six electrons in the outer shell  They symmetrically add across double bonds to form cyclopropanes  This reaction happens in a single step without

intermediates. 38 7.6 Addition of Carbenes to Alkenes

Overall reaction Cl Cl KOH C C C + CHCl3 + KCl + H2O C C

 One of the simplest methods for generating a substituted carbene is with treatment of chloroform, with a strong base.  The addition is stereospecific. Starting from a cis- alkene, a cis- disubstituted cyclopropane is produced, trans- gives a trans- product.

+ HOH

39 Simmons-Smith Reaction

 Best method for preparing nonhalogenated cyclopropanes. Equivalent of addition of CH2:  Reaction of diiodomethane with zinc-copper alloy produces a carbenoid species

CH Zn(Cu) 2 + ZnI H C CH CH CH + CH I H C HC CH CH 2 3 3 2 2 ether 3 3

 Forms cyclopropanes by

Zn(Cu) + CH I + ZnI 2 2 ether 2

40 7.7 Reduction of Alkenes: Hydrogenation

H H Pd or Ni or Pt C C + H2 C C

 Addition of H-H across C=C

 Reduction in general is addition of H2 or its equivalent  Requires Pt or Pd or Ni as powders on carbon and H2  Hydrogen is first adsorbed on catalyst  Reaction is heterogeneous (process is not in solution)

41 Mechanism of Catalytic Hydrogenation  Heterogeneous – reaction between phases  Addition of H-H is syn

42 Problems: Complete the following reactions:

A. 1. BH , THF 1. BH33, THF H3C CH2 OHOH 2. H O , OH-- 2. H22O2, OH

ClClCl Cl B. KOHKOH H CH CH CH3 3 3 CHCl 3 KCl + HOH H3C CHCl+ 3 + KCl H3C + + HH3C H CH3 Trans reactant  trans product 3

C. Write the hydrogenation reaction of 2-methylpropene CH 3 CH CH3 3 Pd Pdor or Ni Ni or or Pt H CH3 H C C HCCH C CH H HH2 2 2 3 + 3 + H HC3CC CCHH3 CH3 H H

43 7.8 Oxidation of Alkenes: Hydroxylation and Cleavage

 Hydroxylation adds OH to each end of C=C  Catalyzed by osmium tetroxide  Stereochemistry of addition is syn  Product is a 1,2-dialcohol or diol (also called a glycol)

44 Alkene Cleavage: Ozone

 Ozone, O3, adds to alkenes to form molozonide  Reduce molozonide to obtain and/or aldehydes

45 Structure Elucidation With Ozone

 Used in determination of structure of an unknown alkene  Cleavage products reveal an alkene’s structure

 What is the structure of the alkene reactant?

1. O3

+ 2. Zn, H3O

46 Permangante Oxidation of Alkenes

 Oxidizing reagents other than ozone also cleave alkenes

 Potassium permanganate (KMnO4) can produce carboxylic acids and carbon dioxide if H’s are present on terminal C=C

H H O KMnO4 C C + CO2 + C H H3O OH

What if alkene is internal not terminal?

KMnO4 O O

+ + H3O OH HO 47 Bonus Slide!

HO NBSNBS HO NBS HO H2O/CH3SOCH3 (DMSO) H22O/CH33SOCH33 (DMSO) H O/CH SOCH (DMSO) 2 3 3 BrBr Br

1. O3 O 1. O33 O O 1. O + OO + O 2. Zn,3 H O++ + 2.2. Zn, Zn, H H3OO 3 + O + 2. Zn, H3O OH OOHH OO + OH ++ O KMnO4, H3O KMnO44, H33O KMnO , H O+ ++ 4 3 OH OO + OOHH O OH

48 Cleavage of 1,2-diols

 Reaction of a 1,2-diol with periodic (per-iodic) acid, HIO4 , cleaves the diol into two carbonyl compounds

 Sequence of diol formation with OsO4 followed by diol cleavage is a good alternative to ozonolysis OH OH 1. OsO4 HIO4 C C C C C O + O C 2. NaHSO3 H2O, THF

49 Problems: Complete the following reactions:

OH A. 1. OsO4 OH 1. OsO4 H3C CH2 OH H3C CH2 OH 2. NaHSO3 H3C 2. NaHSO3 H3C

H C H3C H C H3C 3 1. O 1. O 3 CH 3 3 O O CH B. CH2 2 O + O+ CH2 2 2. Zn,+ H O+ 2. Zn, H3O 3 CH CH CH3 3 CH3 3

C. Write the oxidation reaction of 1-butene and KMnO4 O KMnO O KMnO4 4 + CO + HO + CO2 2 H O + HO H3 3O D. OHOH O O H C CH HIO4 H3C3 CH3 3 HIO4 H C O H3C3 O + CH + CH3 3 H O, THF 50 OHOH H2O,2 THF Leah4Sci Video: Alkene Reactions Products and Shortcuts

 https://youtu.be/avF559OtpPw  http://leah4sci.com/alkene-reactions presents: 18 alkene reactions in 15 minutes - a thorough overview of alkene reaction products including regioselectivity and stereoselectivity of products - Syn/Anti additions as well as Markovnikov and Anti-Markovnikov Products.

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51 7.9 Biological Alkene Addition Reactions  Biological O-Chem takes place in aqueous medium inside cells, and involves complex catalysts, called enzymes. Example: One step of citric acid cycle.

52 7.10 Addition of Radicals to Alkenes: Polymers

 A polymer is a very large molecule consisting of repeating units of simpler molecules monomers, formed by polymerization  Alkenes react with radical catalysts to undergo radical polymerization  Ethylene is polymerized to polyethylene, for example

53 Free Radical Polymerization of Alkenes

 Alkenes combine many times to give polymer  Reactivity induced by formation of free radicals

54 Common polymers and their uses

55 56 CHEM 212 Exam 3 Part 3

Professor Kelly Boebinger 8. Alkynes: An Introduction to Organic Synthesis Alkynes

 Hydrocarbons that contain one or more carbon-carbon triple bonds  Acetylene, the simplest alkyne is produced industrially from methane and steam at high temperature

 Simple alkynes have the general formula

CnH2n-2

59 8.1 Electronic Structure of Alkynes  Carbon-carbon triple bond result from sp orbital on each C forming a sigma bond and unhybridized pX and py orbitals forming a  bond  The remaining sp orbitals form bonds to other atoms at 180º to C-C triple bond.

 The bond is shorter and stronger than single or double  Breaking a  bond in acetylene (HCCH) requires 318 kJ/mole (in ethylene it is 268 kJ/mole)

60 8.2 Naming Alkynes

 Alkynes are named in the same way you named alkenes except they have the –yne ending.

61 8.2 Naming Alkynes: Provide the name for the following:

ethyne (acetylene) propyne

HC CH HC C CH3

3-methylbutyne cyclodecyne

HC

C CH3 CH

CH3

4-methyl-2-pentyne H3C

CH C C CH3

H3C

62 Diynes, Enynes, and Triynes

 A compound with two triple bonds is a diyne  An enyne has a double bond and triple bond  A triyne has three triple bonds  Number from chain that ends nearest a double or triple bond  Double bonds gets lower number if both are present in the same relative position, the last name is -yne

Alkynes as substituents H3C CH2 C C are called “alkynyl” 1-butynyl 63 Problems: Give the IUPAC name for each of the following:

2,2-dimethyl-3-hexyne 2,5-octadiyne

CH3CH2C C C(CH3)3 H3C C C CH2C C CH2CH3

3,6-dimethyl-2-hepten-4-yne 3,3-dimethyl-1,5-hexadiyne

CH HC

1,3-hexadien-5-yne 3,6-diethyl-2-methyl-4-octyne

CH

64 8.3 Preparation of Alkynes: Elimination Reactions of Dihalides

 Treatment of a 1,2 dihaloalkane with KOH or NaOH produces a two-fold elimination of HX. Vinyl halides yield alkynes when treated with a strong base. (Compare to Sec 7.1)

H H 2 KOH, EtOH 2 H O + 2 KBr R C C R R C C R + 2 or 2 NaNH2, NH3 Br Br

H Br NaNH2, NH3 R C C R + NaBr R C C R or KOH, EtOH

 Synthesis from alkene: (Recall reaction 7.2) Vicinal dihalides are formed from addition of bromine or chlorine to an alkene.

Br2 R CH CH R a vicinal dibromide R CH CH R CH2Cl2 Br Br 65 8.3 Preparation of Alkynes: Elimination Reactions of Dihalides

 Complete the following reactions

H H BrBr22 H3C CHH3CCHCHCHC2 CHHCH3 2 CH3 HH33CC CCHCCHCHCH2 C2 HC3H3 CH Cl CH22Cl22 BrBrBrBr

Cl H Cl H NaNH , NH 2 NaNH3 2, NH3 H C C C CH CH H C C C CH CH + NaCl 3 H3C C2 C CH3 2 CH3 3 H3C C 2 C CH3 2 CH3

66 8.4 Reactions of Alkynes: Addition

of HX and X2  Addition reactions of alkynes are similar to those of alkenes  Intermediate alkene reacts further with excess reagent Will state excess if alkane product  Regiospecificity according to Markovnikov is wanted X = Br or Cl X H HX X H HX X C C H R C C H C C ether ether R H R H

Stereospecific addition of X2 , Initial addition gives trans intermediate (anti addition). Tetrahalide product with excess reagent (or 2 mole X2).

X2 = Br2 or Cl2 X X X 2 X R X2 R C C R R C C R C C CH Cl 2 2 R X CH2Cl2 X X 67 anti addition Mechanism for Addition of HX to Alkynes Involves Vinylic Carbocations  Comparing addition of HX to alkenes vs. alkynes.  Addition of H-X to alkyne should produce a vinylic carbocation intermediate  Secondary vinyl carbocations form less readily than primary alkyl carbocations

68 Complete the following reactions:

ClClClCl CH2Cl2 HHCCCCCHCH2C2CHH3 3 ++2 2Cl Cl2 2 HHCCCCCHCH2C2CHH3 3 ClClClCl

ether CHCH22 CHCH22 CC CCHH ++ HHBrBr CC CCHH22 BrBr

BrBr HH ether ++ HHBrBr ++ Same as HH BrBr EE and and Z Z products products 69 8.5 Hydration of Alkynes

 Two methods of addition of H-OH  First: Mercury (II) catalyzes Markovinikov oriented addition, forms a . O H2SO4, H2O OH R C CH O H2SO4, H2O OH C R C CH2 R C CH HgSO R C CH3 4 R C CH2 HgSO R CH3 terminal alkyne 4 yields a methyl ketone terminal alkyne yields a methyl ketone

O O H O, H SO O O 1 2 2H O, 2H SO4 R 1C C R 2 2 2 4 2 1 R C C R R 2 + R 1 HgSO R + R 4 1 2 HgSO4 R 1 R R 2 internal alkyne R internal alkyne yields a mixture yields a mixture 70 Keto-enol Tautomerism

 Isomeric compounds that can rapidily interconvert by the movement of a proton are called tautomers and the phenomenon is called tautomerism  Enols rearrange to the isomeric ketone by the rapid transfer of a proton from the hydroxyl to the alkene carbon  The keto form is usually so stable compared to the enol that only the keto form can be observed 71 Leah4Sci Video: Keto Enol Tautomerism Acid and Base Reaction and Mechanism

 https://youtu.be/TJrTytxucus  Keto Enol Tautomerism - KET reaction and mechanism in acidic and basic conditions by http://leah4sci.com

72 Hydration of Alkynes

Second method:  Hydroboration-oxidation gives the non-Markovnikov product, forms an aldehyde

1. BH3 O R CH R 2. H2O2

73 What alkyne would you start with to prepare the following?

A. O O

1. BH3, THF1. BH3, THF C C H C C H CH2 C CHH2 C H 2. H2O2, NaO2H. H, H2O2O2, NaOH, H2O O B O HgSO , H SO 4 2HgS4O , H SO CH3CH2C CHCHCH C CH 4 2CH34CH2CCCHH3 CH CCH 3 2 H2O 3 2 3 H2O

C. O

H3C CH2 C CH2 CH3 H2O, H2SO4 H C C C CH CH 3 2 3 + O HgSO4

H3C C CH2 CH2 CH3

74 8.6 Reduction of Alkynes

 Addition of H2 over a metal catalyst (such as palladium on carbon, Pd/C) converts alkynes to alkanes (complete reduction)

 The addition of the first equivalent of H2 produces an alkene, which is more reactive than the alkyne so the alkene is not observed

H H Pt,or Ni, or Pd R C C R + 2 H2 R C C R H H

75 Conversion of Alkynes to cis- Alkenes

 Addition of H2 using chemically deactivated palladium on calcium carbonate as a catalyst (the Lindlar catalyst) stops at a cis alkene  The two hydrogens add syn (from the same side of the triple bond)

Lindlar R R R C C R C C + H2 catalyst H H

76 Conversion of Alkynes to trans- Alkenes

 Anhydrous ammonia (NH3) is a liquid below -33 ºC  Alkali metals dissolve in liquid ammonia and function as reducing agents  Alkynes are reduced to trans alkenes with sodium or lithium in liquid ammonia

Li or Na R H R C C R C C + H2 NH3 H R

77 8.7 Oxidative Cleavage of Alkynes

 Strong oxidizing reagents (O3 or KMnO4) cleave internal alkynes, producing two carboxylic acids

O O KMnO4 1 2 1 2 R C C R R C + R C or O 3 OH OH  Terminal alkynes are oxidized to a carboxylic acid and carbon dioxide O KMnO4 R C C H R C + CO or O 2 3 OH  Neither process is useful in modern synthesis

78 Leah4Sci Video: Alkyne Reactions Products and Shortcuts

 https://youtu.be/LUjUOc69w_0  http://leah4sci.com/AlkyneReactions presents: 11 Alkyne Reactions in 11 Minutes. Overview of Organic Chemistry 1 Alkyne reactions and product shortcuts to help you work through them quickly!

79 8.8 Alkyne Acidity: Formation of Acetylide Anions  Terminal alkynes are weak Brønsted acids (alkenes and alkanes are much less acidic).  Reaction of strong anhydrous bases with a terminal acetylene produces an acetylide ion

+ Na+ + NH R C C H + NH2Na R C C 3 acetylide anion

 The sp-hydbridization at carbon holds negative charge relatively close to the positive nucleus.

80 8.9 Alkylation of Acetylide Anions

 Alkylation: a new alkyl group is attached, any terminal alkyne can be converted. Makes compounds with more carbons.  Effective only with 1o alkyl bromides or iodides  The formation of terminal alkyne must use acetylene

 Formation of internal alkyne

81 Mechanism of Alkylation of Acetylide Anions

 Negative charge and unshared electrons make this anion strongly nucleophilic.  Reaction with a primary alkyl halide produces a hydrocarbon that contains carbons from both partners, it is a way to make larger carbon chains.

82 8.10 An Introduction to Organic Synthesis  Synthesis as a Tool for Learning Organic Chemistry. A synthesis combines a series of proposed steps to go from a defined set of reactants to a specified product

 In order to propose a synthesis you must be familiar with reactions  What they begin with  What they lead to  How they are accomplished  What the limitations are

83 Synthesis Example: Prepare hexane from 1-butyne  1st step: Strategy

1-butyne hexane

HC≡CCH2CH3 CH3CH2CH2CH2CH2CH3 (starting material) (product)

 Need to add 2 more carbons  Need to remove triple bond and add hydrogens

84 Synthesis Example: Prepare hexane from 1-butyne  2nd step: Add two carbons (alkylate rxn)

CH3CH2Br NaNH2 +Θ HC≡CCH2CH3 Na C ≡CCH2CH3 NH3 CH3CH2C≡CCH2CH3

 3rd step: Remove triple bond and add hydrogens

Pt CH3CH2C≡CCH2CH3 + 2H2 CH3(CH2)4CH3

85 Common Format for Synthesis: Prepare Pentane from 1-butyne

NaNH2, NH3 +Θ HC≡CCH2CH3 Na C ≡CCH2CH3

+ 2 H2, Pt

CH3C≡CCH2CH3 CH3(CH2)3CH3

86 Build these models ahead of time and bring these models to the next lecture after this exam.

You will need two like these

8787 CHEM 212 Exam 3 Part 4

Professor Kelly Boebinger