Alkynes

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Structure of Alkynes

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Alkynes

• Functional group - Carbon-carbon triple bond• Simplest alkyne is ethyne (C2H2)

• Commonly called acetylene

• Each C—H bond is formed by the overlap of a 1s orbital of hydrogen with an sp orbital of carbon

• Unusually strong because of its 50% s-character

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Nomenclature Alkynes

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Nomenclature of Alkynes: IUPAC Names

• Infix -yn- is used to show the presence of C≡C bonds• Location of the triple bond is shown by the number of its first carbon

• Longest carbon chain with the triple bond is numbered from the end that gives lower numbers to the triply bonded carbons

• Exceptions - Ethyne and propyne

• If a hydrocarbon chain contains more than one triple bond, the infixes -adiyn-, -atriyn-, and so forth are used

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• Draw skeletal structures for:Acetylene Dimethylacetylene

1-butyne 2-butyne

Propargyl chloride

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IUPAC Nomenclature for Alkynes

• Write the IUPAC name of each compound

• Solutiona. 2-Pentyneb. 3-Chloropropynec. 2,6-Octadiyne

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Physical Properties Alkynes

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Physical Properties of Alkynes

• Similar to those of alkanes and alkenes of comparable molecular weight and carbon skeletons

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Cycloalkynes

• Cyclooctyne• Smallest isolated cycloalkyne• Quite unstable and polymerizes at room temperature• C—C≡C bond angle is approximately 155° (rather than the optimal angle of 180°),

indicating a high angle strain

• Cyclononyne• Stable at room temperature• C—C≡C bond angles are

approximately 160°

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Acidity of 1-Alkynes

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Acidity of Terminal Alkynes • Do you recall the pKa’s of the following species?

• Can you explain why the hydrocarbons have such different pKa’s?

• Calculate the equilibrium constant of the following reaction:

• The resulting species, often called an acetylide anion, is an excellent nucleophile for SN2 reactions. This might be a good way to make carbon-carbon bonds. . . do you recall the other ways of making carbon-carbon bonds?

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Acidity of 1-Alkynes

• The pKa of acetylene and terminal alkynes is approximately 25, which makes them stronger acids than ammonia but weaker acids than alcohols (Section 4.1)

• H bonded to a triply bonded carbon atom of a terminal alkyne is sufficiently acidic that it can be removed by a strong base (sodium amide NaNH2) to give an acetylide anion

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• Sodium hydride and lithium diisopropylamide (LDA) are strong bases used to form acetylide anions

• Because water is a stronger acid than acetylene, hydroxide ion is not a strong enough base to convert a terminal alkyne to an alkyne anion

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• Which of the following is NOT a reason for alkyne hydrogens to be more acidic than alkene hydrogens?

1. The lone pair on alkynyl anion has more s character2. The alkynyl anion is more stable 3. The lone-pair electrons on alkynyl anion are closer to the nucleus4. Hybridization differences make the vinyl anion less stable5. The alkynyl anion is resonance stabilized

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Preparation of Alkynes

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Alkylation Reaction of Acetylide Anions• Nucleophilic substitution reaction

• Practical only with methyl and 1° halides

• Convenient method to synthesize terminal alkynes and internal alkyne

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HC CHHC C- Na+

H

Br

Sodium acetylide

Bromo-cyclohexane

Acetylene Cyclohexene

Na+Br -++ +elimin-

ation(Ch 9)

Preparation of Alkynes from Alkenes• Treating a dihaloalkane with two moles of a strong base such as NaNH2 in

liquid NH3 results in two successive dehydrohalogenations• Dehydrohalogenation: Removal of HX from a haloalkane

• For the conversion of 1-hexene to 1-hexyne, 3 moles of sodium amide are required

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Allenes• In dehydrohalogenation of a haloalkene with at least one hydrogen on each

adjacent carbon, a side reaction occurs, namely the formation of an allene

• Compounds containing a C=C=C functional group• 1,2-propadiene is the simplest allene and is commonly called allene• Most are less stable than their isomeric alkynes

02 2 3

02 3 3 3

CH = C = CH CH C CH H –6.7 kJ (–1.6 kcal) / mol

CH = C = CHCH CH C CCH H = –16.7 kJ (–4.0 kcal) / mol

¾¾® º D =

¾¾® º D 19

Example - Synthesis of Alkynes

• Show how you might convert 1-pentene to 1-pentyne• Solution

• Can be done in two steps• Treating 1-pentene with one mole of bromine gives 1,2-dibromopentane• Treating this dibromoalkane with three moles of sodium amide followed by H2O

gives 1-pentyne

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Test Yourself

• Which compound can most easily undergo elimination to give an alkyne?

1. 2. 3.

4. 5.

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• Consider the carbocations shown below• Which carbocation is more stabilized and why?

1. The vinyl cation is more stabilized because of resonance2. The vinyl cation is more stabilized because it is a stronger Lewis acid3. The vinyl cation is more stabilized because of hyperconjugation4. The ethyl cation is more stabilized because of a σ*C-H/p orbital interaction5. The ethyl cation is more stabilized because of a σC-H/p orbital interaction

H

HH

H

H

H

HH

Test Yourself

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Electrophilic Addition to Alkynes

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Addition of HX to Alkynes • Draw a complete curved-arrow mechanism for the following reaction:

• Predict the product and draw a complete curved-arrow mechanism for the following reaction:

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Addition of HX• Alkynes undergo regioselective addition of either 1 or 2 moles of HX,

depending on the ratios in which the alkyne and halogen acid are mixed

• Step 1: Make a new bond between a nucleophile (p bond) and an electrophile—add a proton

• Step 2: Make a new bond between a nucleophile and an electrophile

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• In the addition of the second mole of HX, Step 1 is the reaction of the electron pair of the remaining π bond with HBr to form a carbocation

• Of the two possible carbocations, the resonance-stabilized 2° carbocation is favored

• Addition of one mole of HCl to acetylene gives chloroethene (vinyl chloride)• Vinyl chloride is the monomer in the production of the polymer poly(vinyl chloride),

abbreviated PVC

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Addition of Halogens (X2) to Alkynes • Draw a complete curved-arrow mechanism for the following reaction:

• Predict the product and draw a complete curved-arrow mechanism for the following reaction. Be sure to show the stereochemistry of the product!

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Addition of Br2 to an Alkyne• Adding one mole of Br2 to an alkyne gives a dibromoalkene

• Addition shows anti stereoselectivity• Carrying out the bromination in acetic acid with added bromide ion increases the

preference for anti addition

• Bridged bromonium ion intermediate is formed, which is then attacked by bromide ion from the face opposite that occupied by the positively charged bromine atom

• Addition of a second mole of Br2 gives a trtrabromoalkane

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• What is the product of the following reaction?

1. 2.

3.

4. 5.

HBr

excess

Test Yourself

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• Which reaction sequence would be best to carry out the following transformation?

1. HBr (1 equivalent) followed by Br2/CCl42. HBr (2 equivalents) followed by Br2/hν3. Br2/CH2Cl2 (2 equivalents) followed by H2/Pd/C4. Br2/CH2Cl2 (2 equivalents) followed by NaOH5. Br2/H2O followed by HBr (2 equivalents)

Br

Br Br

Test Yourself

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• Which of the following would be the expected product if propyne reacts with Br2/H2O following the patterns of electrophilic-addition mechanisms?

1. 2. 3.

4. 5.

Test Yourself

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Hydration of Alkynes to Aldehyde and Ketones

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Hydroboration/Oxidation of Alkynes: Enol Formation • The following reaction produces an unstable intermediate known as an enol (alkene +

alcohol = enol). Draw a complete curved-arrow mechanism for this reaction up to the formation of the enol. You should also explain why the alkene geometry is trans (that is, you should explain the stereochemistry!)

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Hydroboration/Oxidation of Alkynes: Enol Hydrolysis • The enol is not the final product of hydroboration/oxidation of an alkyne. The enol

reacts rapidly with the aqueous base (in the oxidation step) to form a carbonyl compound: in this case, an aldehyde. Provide a complete curved-arrow mechanism for the following reaction:

• Show the complete synthetic transformation from an alkyne to an aldehyde:

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Hydroboration-Oxidation• Addition of borane to an internal or a terminal alkyne gives a trialkenylborane

• Addition is syn stereoselective

• In reaction with terminal alkynes, the alkenyl group reacts further with borane to undergo a second hydroboration

• Reaction is difficult to stop at this stage• To prevent second hydroboration, a sterically hindered disubstituted borane, such

as (sia)2BH, is used• Treatment of a terminal alkyne with (sia)2BH results in stereoselective and

regioselective hydroborations

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Hydroboration-Oxidation (continued)

• Treating an alkenylborane with H2O2 in aqueous NaOH gives an enol

• Enol: A compound containing an —OH group bonded to a doubly bonded carbon atom• Is in equilibrium with a keto form by migration of a hydrogen from oxygen to carbon and

rearrangement of the C=C to form a C=O• Less stable than the keto forms

• Keto-enol tautomerism: Isomerism involving keto (from ketone) and enol tautomers• Tautomers: Constitutional isomers that are in equilibrium with each other

• Differ only in the location of a hydrogen atom or another atom and a double bond relative to a heteroatom (most commonly O, N, or S)

• Hydroboration-oxidation of an internal alkyne gives a ketone• Hydroboration-oxidation of a terminal alkyne gives an aldehyde

3-Hexanone3-Hexyne

1. BH32. H2O2, NaOH

O

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Example - Hydrobo ration-Oxidation of Alkynes• Hydroboration-oxidation of 2-pentyne gives a mixture of two ketones, each with

the molecular formula C5H10O• Propose structural formulas for these two ketones and for the enol from which

each is derived• Because each carbon of the triple bond in 2-pentyne has the same degree of

substitution, very little regioselectivity occurs during hydroboration• Two enols are formed, and the isomeric ketones are formed from them

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Addition of Water to Alkynes: Enol Formation • The following reaction produces an unstable intermediate known as an enol

(alkene + alcohol = enol). Draw a complete curved-arrow mechanism for this reaction up to the formation of the enol.

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Addition of Water to Alkynes: Enol Hydrolysis • The enol is not the final product of the reaction of an alkyne with water. The enol

reacts rapidly with the aqueous acid to form a carbonyl compound: in this case, a ketone. Provide a complete curved-arrow mechanism for the following reaction:

• Show the complete synthetic transformation from an alkyne to a ketone:

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Acid-Catalyzed Hydration• In the presence of concentrated sulfuric acid and Hg(II) salts, alkynes undergo

addition of water in a reaction analogous to the oxymercuration of alkenes

• Step 1: Make a new bond between a nucleophile (p bond) and an electrophile• Step 2: Make a new bond between a nucleophile and an electrophile

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• Step 3: Take a proton away• Proton transfer to solvent gives an organomercury enol

• Step 4: Keto-enol tautomerism of the enol gives the keto form

• Step 5: Add a proton• Proton transfer to the carbonyl oxygen gives an oxonium ion

• Steps 6 and 7: Break a bond to give stable molecules or ions followed by keto-enol tautomerism

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Examples - Reactions of Alkynes• Show reagents to bring about the following conversions

a. Hydration of this monosubstituted alkyne using a mercuric ion catalyst gives an enol that is in equilibrium with the more stable keto form

b. Hydroboration using di-sec-isoamylborane followed by treatment with alkaline hydrogen peroxide gives an enol that is in equilibrium with the more stable aldehyde

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• A reaction of propyne with Hg2+ is run in methanol (CH3OH) in the presence of a strong acid (H+), and in the absence of water, with a non-nucleophilic counterion

• If this reaction follows a typical mechanism of electrophilic addition, what would be the expected product?

1. 2. 3.

4. 5.

Test Yourself

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• Which of the following molecules cannot be synthesized from propyne using only one reaction?

1. 2.

3.

4. 5.

Test Yourself

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• Which is the best starting material to prepare 2-pentanone?1. Pentane2. 1-Pentene3. 2-Pentene4. 1-Pentyne5. 2-Pentyne

Test Yourself

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• Double bonds can be made by elimination of HX with the help of a base or by elimination of H—OH with the help of an acid

• Similarly, triple bonds can be made by elimination of two HX molecules with a base, but they cannot be produced by elimination of two molecules of water with the help of an acid

• What is the main reason for this failure?

Test Yourself

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Which statement is correct?

1. 1,2-Diols cannot be protonated by acids2. In diols, water can be eliminated only with the help of a base3. Formation of a triple bond requires elimination of three molecules of

water4. The intermediate carbocation is too unstable5. The intermediate enol tautomerizes to a carbonyl compound

Test Yourself

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• What is the structure of intermediate I?

1. 2. 3.

4. 5.

BD3

THF

H2O2

NaOHI product

Test Yourself

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Reduction of Alkynes

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Hydrogenation of Alkynes: Catalytic Hydrogenation • Ordinarily, the reaction of an alkyne with H2 / Pd adds two equivalents of

hydrogen and yields an alkane. Show the intermediate and final product in the following transformation. Be sure to consider the stereochemistry!

• However, we can use a poisoned catalyst (for example, a Lindlar catalyst) to halt the reaction at the alkene stage. This is one of the best ways of making a cis-alkene:

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Reduction of Alkynes: Catalytic Reduction• Treatment of an alkyne with H2 in the presence of a transition metal catalyst,

most commonly Pd, Pt, or Ni, converts the alkyne to an alkane

• With the Lindlar catalyst, the reduction stops at syn stereoselective addition of one mole of H2

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Reduction of Alkynes: Hydroboration-Protonolysis• Syn stereoselective addition of borane to an internal alkyne gives a

trialkenylborane

• Treatment of a trialkenylborane with acetic acid results in stereoselective replacement of B by H

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Hydrogenation of Alkynes: Reduction with Na in NH3• We can convert an alkyne into a trans-alkene by using a unique reducing agent:

sodium metal in liquid ammonia (!). Ammonia is a gas at room temperature; it boils at –33°C. When it is condensed into a liquid it is a tremendously interesting and useful solvent. It dissolves sodium metal to produce a deep blue solution that contains solvated electrons:

• Write a balanced equation showing how Na dissolves in NH3 to produce solvated electrons e–

• These solvated electrons react with the alkyne via an interesting free-radical mechanism that yields a trans-alkene as the final product. (You can remember the steps of this mechanism if you note that it involves only 2 kinds of reactions: adding an electron to form an anion, and protonating the anion using NH3.)

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Reduction of Alkynes: Dissolving-Metal Reduction• Reduction of an alkyne to an alkene by Na or Li in liquid NH3 is

stereoselective• Involves anti addition of two hydrogen atoms to the triple bond

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Reduction of an Alkyne by Sodium in Liquid Ammonia• Step 1: A one-electron reduction of the alkyne gives an alkenyl radical

anion

• Step 2: Add a proton• Alkenyl radical anion (a very strong base) abstracts a proton from ammonia (a very

weak acid) to give an alkenyl radical

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• Step 3: A one-electron reduction gives an alkenyl anion• Trans-alkenyl anion is more stable than its cis isomer• This step establishes the stereochemistry of the final product

• Step 4: Add a proton• A second proton-transfer reaction gives the trans-alkene

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• 2-Hexyne is allowed to react with Li in NH3(liq)• The obtained product is treated with Cl2 in CCl4• Which of the following isomer is the main product?

1. 2. 3.

4. 5.

Test Yourself

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• 2-Butyne is allowed to react with H2 over the Lindlar catalyst• Obtained product is treated with OsO4/pyridine followed by

NaHSO3/H2O• Which of the following compounds is the end result of this sequence of

reactions?

Test Yourself

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Summary: Reactions of Alkynes

• Show how each of the following products could be synthesized from an alkyne:

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Organic Synthesis

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Organic Synthesis

• Series of reactions by which a set of organic starting materials is converted to a more complicated structure

• A successful synthesis:• Provides the desired product in maximum yield with a maximum control of

stereochemistry at all stages• Should not produce or release byproducts harmful to the environment (“green”

synthesis)

• Best strategy is to work backward from the target product

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Retrosynthetic Analysis: Analysis of Target Molecule1. Count the carbon atoms of the carbon skeleton of the target molecule

• Challenge - Determining how to build the carbon skeleton from available starting materials

• Only method to date for forming a new C—C bond is the alkylation of acetylide anions with methyl or 1° halides

2. Analyze the functional groups• What are they, how can they be changed to facilitate formation of carbon skeleton,

and how can they then be used in forming the final set of functional groups in the target molecule?

3. Work backward• Use retrosynthesis

• Retrosynthesis: A process of reasoning backward from a target molecule to a set of suitable starting materials• An open arrow to symbolize a step in a retrosynthesis

targetmolecule

startingmaterials 62

Organic Synthesis of 2-Heptanone

• Analysis

• Synthesis

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Example - Synthesis Using Alkynes

• How might the scheme for the synthesis of 2-heptanone be modified so that the product is heptanal?

• Solution• Steps 1 and 2 are the same and give 1-heptyne• Instead of acid-catalyzed hydration of 1-heptyne, treat the alkyne with (sia)2BH

followed by alkaline hydrogen peroxide (Section 7.7)

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Reactions of Functional Groups in Complex Molecules• Synthesis of fexofenadine illustrates two

important points about how functional group analysis is used to develop systematic syntheses of complex molecules

• Alkyne was the only functional group to react under the conditions used, and it reacted as expected

• Other functional groups often have an influence over the regiochemistry or stereochemistry of a reaction

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Sample Synthesis Problem (from old exam) • You should be able to find at least three completely different synthetic routes for the following problem!

Show all reagents and intermediates for the multi-step synthesis below. You may use any organic or inorganic reagents in your synthesis, but you must begin with the indicated starting material. You do not need to show any mechanisms. The correct answer will require fewer than 5 steps. • Starting material

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Organic Synthesis Using Alkynes • Provide a synthetic route to the following product from starting materials that contain no

more than 2 carbons:

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Test Yourself

• Which reagents are best to carry out the following reaction?

1. BH3/THF followed by NaOH/H2O2

2. HgSO4/H2O/H2SO4

3. NaNH2 followed by CH3I, followed by HgSO4/H2O/H2SO4

4. OsO4 followed by NaHSO3/H2O

5. KMnO4/H+ followed by CH3I

O

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Test Yourself

• Which compound cannot serve as an immediate precursor to the synthetic target shown below?

H

OOH

D

target molecule

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Problem 7.28

• An acid-catalyzed domino reaction that ultimately leads to the synthesis of progesterone is depicted in the image

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• We focus on the acid-catalyzed step that leads to cyclization of the four-ring system, which is characteristic of steroids

• Assume that this reaction is initiated by protonation of the 3° alcohol followed by loss of water to give a carbocation

• The series of reactions initiated by this carbocation gives compound B

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Tandem reactions, cascade sequencesJohnson’s elegant cation-π cyclization