# Organic Chemistry I & II

Lead Author(s): Steven Forsey

Student Price: Contact us to learn more

Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

This content has been used by 15,690 students

## What is a Top Hat Textbook?

Top Hat has reimagined the textbook – one that is designed to improve student readership through interactivity, is updated by a community of collaborating professors with the newest information, and accessed online from anywhere, at anytime.

• Top Hat Textbooks are built full of embedded videos, interactive timelines, charts, graphs, and video lessons from the authors themselves
• High-quality and affordable, at a significant fraction in cost vs traditional publisher textbooks

## Comparison of Organic Chemistry Textbooks

Consider adding Top Hat’s Organic Chemistry textbook to your upcoming course. We’ve put together a textbook comparison to make it easy for you in your upcoming evaluation.

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

### Pricing

Average price of textbook across most common format

#### Up to40-60%more affordable

Lifetime access on any device

#### $219 Hardcover print text only ####$301

Hardcover print text only

#### $301 Hardcover print text only ### Always up-to-date content, constantly revised by community of professors Content meets standard for Introduction to Anatomy & Physiology course, and is updated with the latest content ### In-Book Interactivity Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost ### Customizable Ability to revise, adjust and adapt content to meet needs of course and instructor ### All-in-one Platform Access to additional questions, test banks, and slides available within one platform ## Pricing Average price of textbook across most common format ### Top Hat Steven Forsey, “Organic Chemistry”, Only one edition needed #### Up to40-60%more affordable Lifetime access on any device ### McGraw-Hill Carey & Giuliano, “Organic Chemistry”, 10th Edition ####$219

Hardcover print text only

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

#### $301 Hardcover print text only ### Wiley David R. Klein, “Organic Chemistry”, 3rd Edition ####$301

Hardcover print text only

## Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## About this textbook

### Lead Authors

#### Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

## Explore this textbook

Read the fully unlocked textbook below, and if you’re interested in learning more, get in touch to see how you can use this textbook in your course today.

# Chapter 14: Alcohols and Oxiranes

## Learning Objectives

• Recognize the reagents needed to convert carboxylic acids, ketones, aldehydes, esters and alkenes into 1°, 2° and 3° alcohols.
• Understand the mechanism and predict the products produced in the nucleophilic addition to carbonyl compounds (RCOR, RCHO, RCO2R) to produce alcohols. (Grignard reagents, alkyl lithium reagents, LiAlH4, NaBH4).
• Determine the relative base strength of Grignard and lithium reagents and predict the products produced in acid-base reactions.
• Recognize limitations of Grignard and alkyl lithium reagents.
• Understand and write the mechanism for the nucleophilic ring-opening of oxiranes to produce alcohols and predict the regiochemistry and stereochemistry of the products in both basic and acidic conditions.
• Recognize reagents needed to oxidize alcohols to ketones, aldehydes and carboxylic acids.
• Predict the products formed from the oxidation of 1° and 2° alcohols using different oxidizing reagents: K2Cr2O7/H2SO4, KMnO4/NaOH, PCC/CHCl3.
• Identify reagents used to convert alcohols into better leaving groups, such as Cl, Br, I, and sulfonates.
• Understand the limitation of using reagents SOCl2, PCl3, PIand PBr3 in converting alcohols to alkyl halides and predict the stereochemical changes when using these reagents.
• Predict the intermediates and the final product in multi-stepped synthesis questions.

## 14.1 Definitions

Hofmann Rule: This rule refers to a special β-elimination reaction in which the alkene having the smallest number of alkyl groups attached to the double bonded carbon atoms will be the predominant product (less stable alkene forms as the major product). The Hofmann rule is observed in elimination reactions with leaving groups like quaternary ammonium salts, tertiary sulfonium salts, and with bulky strong bases (NaOC(CH3)3) reacting with tertiary alkyl halides.

Markovnikov’s Rule: In the ionic addition of a polar reagent to an asymmetrical alkene, the positive portion of the reagent being added attaches itself to a carbon atom of the double bond forming the more stable carbocation as an intermediate. Older definition: the hydrogen becomes attached to the carbon atom of the double bond with the greater number of hydrogens (for addition of HX). It can also be stated; the hydrogen adds to the less substituted carbon.

Regiochemistry: A reaction that can occur with different regions on a molecule to produce different products. A region is defined as a site on a molecule where a reaction can occur. Regiochemistry is the difference in the reactivity of the various sites.

Regioselective reaction: When there is more than one reactive site a regioselective reaction can occur. A regioselective reaction is one that produces one major product when many possible products are possible. For example, the molecule 2-bromo-2-methylbutane has eight hydrogens on β carbons that can undergo elimination reaction in the presence of a base such as sodium ethoxide. Removal of a Ha hydrogen produces the minor product 2-methylbut-1-ene while removal of a Hb hydrogen produces the major product.

Stereoselective reaction: A reaction in which a single reactant can produce two or more stereoisomeric products and one of these products is preferred over another. For example, the reaction of either enantiomer of 2-bromobutane will stereoselectively produce (E)-but-2-ene as the major product. Notice that it does not matter which enantiomer the starting material is; the product ratio is the same.

Stereospecific reaction: A reaction in which a single reactant can produce two or more stereoisomeric products and one of these products is exclusively formed over the other(s). For example; the elimination reaction of (1R,2R)-1-bromo-1,2-diphenylpropane with a strong base produces the (Z) stereoisomer whereas the elimination of the (1R,2S) diastereomer produces the (E) stereoisomer.

Similarly, a SN2 reaction with 2-bromobutane and sodium methanethiolate (CH3SNa) will produce either (S)-sec-butyl(methyl)sulfane or (R)-sec-butyl(methyl)sulfane depending on which enantiomer you start with. This is another example of a stereospecific reaction.

Zaitsev’s Rule: An empirical rule that states; when two or more alkenes can be produced in an elimination reaction, the thermodynamically most stable alkene will predominate. The most thermodynamically stable alkene will be the alkene that has the most alkyl groups attached to the alkene carbons.

## 14.2 Carbocation Stability

Before examining reactions with alkenes, we need to review carbocation stability because carbocation intermediates play an important role in determining the regiochemistry of the products.

Neighboring functional groups can stabilize carbocations through hyperconjugation, adjacent lone pairs and adjacent π bonds. Carbocations can also be destabilized by nearby partial positive charges.

### 14.2.1 Hyperconjugation

The stability of a carbocation increases as the number of alkyl groups attached to the electron deficient carbon increases. Review: Substitution-elimination chapters: Effect of structure on the rate of reaction.

This can be explained by the electron donation of the adjacent C-C or C-H sigma bonds into the empty carbocation’s p orbital, which stabilizes the electron deficient carbon. Alkyl groups donate electron density inductively through σ bond conjugation or hyperconjugation.

### 14.2.2 Adjacent Lone Pairs

Atoms adjacent to a carbocation that have a lone pair (N,O, S, and halogen) will stabilize the carbocation through resonance.

You may be thinking that electronegative atoms such as oxygen and chlorine would destabilize the carbocation, but in most cases, resonance with the heteroatom has a greater influence than inductive effects.

### 14.2.3 Adjacent π Bonds

Carbocations that are adjacent to π bonds (allyl, benzyl and cyano) are stabilized through resonance, which delocalizes the charge onto different atoms.

### 14.2.4 Inductive Effects

Carbocations are destabilized by electronegative atoms that inductively create a partial positive charge adjacent to the carbocation.

## 14.3 Oxidation States of Carbon - Review from Chapter 12.10

In the next sections, we will be looking at specific reactions involving the oxidation and reduction of carbon atoms using various oxidizing and reducing reagents.

When a carbon is oxidized, it 'loses electrons' by gaining bonds to atoms that are more electronegative than itself, such as oxygen or other electronegative atoms like nitrogen. When a carbon is reduced it 'gains electrons' by being bonded to an atom that is less electronegative than itself such as hydrogen or silicon.

However, generally speaking, the reduction of a molecule increases its hydrogen content, and oxidation increases its oxygen content. Or, if you think about the oxidation or reduction of a specific carbon, the oxidation state changes with the number of times it is bonded to an oxygen or a hydrogen as shown below.

As mentioned above, the oxidation state of carbon does not only change with the number of times it is bonded to oxygen or hydrogen. The oxidation state changes when carbon is bonded to any element that is more or less electronegative than it. For example, the chlorination of methylcyclohexane is an oxidation of methylcyclohexane.

### 14.3.1 Calculation of the Oxidation States

The oxidation state of an atom is the number of electrons that an atom gains or loses, or appears to gain or lose, in bonding with other atoms in compounds. This is not a formal charge; it is another way of electron bookkeeping.

To calculate the oxidation state of a single carbon atom, proceed as follows:

• Assign -1 to a carbon bonded to each hydrogen or anything less electronegative than carbon
• Assign +1 to a carbon bonded to each nitrogen, oxygen or anything more electronegative than carbon
• Assign 0 to a carbon bonded to each carbon.

You can also artificially treat each sigma bond as an ionic bond by heterolytically breaking each bond, giving the more electronegative atom the electrons from the sigma bond.

For example, methane is bonded to four hydrogens. Since carbon is more electronegative than hydrogen, assign -1 to the carbon for each hydrogen. Thus, this carbon would have an oxidation state of -4. Go through the examples below.

To determine if a molecule is being oxidized or reduced, you calculate the oxidation states of the products and reactants. If there is a decrease in a molecule's oxidation state in going from reactant to product, the molecule is gaining electrons or being reduced. If the oxidation state becomes more positive, the molecule has been oxidized. For example, the oxidation state for each carbon in ethene is -2. When ethene is treated with hydrogen gas and a catalyst, ethane is produced. The oxidation state of each carbon has decreased from -2 to -3. Thus each carbon has gained one electron and the molecule has been reduced.

If there is an increase in a molecule's oxidation state, the molecule is losing electrons or being oxidized.

## 14.4 Introduction to Alcohols

Alcohols are weak bases and weak acids. Molecules that are both acidic and basic are called amphoteric.

Alcohols are weak bases, and strong acids such as sulfuric acid (H2SO4) are needed to protonate the alcohol. Once formed, alkoxonium ions are strong acids as seen by their small pKa values.

To deprotonate alcohols, strong bases such as sodium hydride (NaH) or sodium amide (NaNH2) must be used because alkoxide ions are themselves strong bases.

Alkoxides can also be produced through an oxidation-reduction reaction with a metal such as Na, K or Li.

We have looked at the acid/base properties of alcohols in Chapter 4: Acids and Bases. Can you answer the following question?

Q14.1 - Level 1

Which of the following equilibria would be shifted to the left?

A

a

B

b

C

c

### Alcohol Chemistry & Breaking Bad

Although organic synthesis is used to prepare the majority of medicines used to benefit humanity, regrettably, illegal drugs can also be made.

One example of illegal drug synthesis from pop culture is the synthesis of methamphetamine (aka “meth”), as shown in the hit TV show “Breaking Bad”. Methamphetamine is a very dangerous, highly addictive drug that systematically destroys the body and causes memory loss and psychotic behavior.

It is estimated that nearly 500 metric tons of amphetamine-type stimulants are produced each year with over 24 million abusers worldwide.

Have you ever been “carded” when purchasing the popular over-the-counter decongestant Sudafed? Or, if you watched “Breaking Bad”, you’ll recall several references to Sudafed in the earlier episodes.

Sudafed, and several other major brand-named medicines, all contain pseudoephedrine, which is structurally very similar to methamphetamine. In fact, pseudoephedrine can be converted to methamphetamine, which is why sales of Sudafed are now monitored.

Keeping it Real Q14.1

Would you classify this transformation involving the removal of an alcohol as an oxidation process, a reduction process, or neither?

A

Oxidation

B

Reduction

C

Neither

## 14.5 Synthesis of Alcohols

You have learned a number of ways to synthesize alcohols in previous chapters. How much do you remember?

### 14.5.1: Review of SN1, SN2, Acid-Catalyzed Hydration, Hydroboronation-Oxidation, Oxymercuration-Demercuration and Halohydrin Reactions

SN2 and SN1 Reactions: Chapter 7, Chapter 9 and Chapter 11

Q14.2 - Level 1

Match the appropriate reagent and mechanism to the following substitution reactions.

Premise
Response
1

1)

A

Reagent: NaOH; Reaction mechanism: S$_N$2

2

2)

B

Reagent: NaOH; Reaction mechanism: S$_N$1

C

Reagent: H$_2$O; Reaction mechanism: S$_N$1

D

Reagent: H$_2$O; Reaction mechanism: S$_N$2

Hydroboronation-Oxidation: Chapter 12.5

Q14.3 - Level 2

What are the major products produced in the following reaction?

A

1) and 2)

B

3) and 4)

C

1) and 3)

D

2) and 4)

E

All are produced

Oxymercuration-demercuration: Chapter 12.6

Q14.4 - Level 2

What is the major product formed in the following reaction?

A

1)

B

2)

C

3)

D

4)

Halohydrins: Chapter 12.7

Q14.5 - Level 2

What are the major products formed in the following reaction?

A

All products are produced equally

B

1) and 2) are the major products

C

3) and 4) are the major products

D

1) and 3) are the major products

E

2) and 4) are the major products

Diols: Chapter 12.12

Q14.6 - Level 2

Which of the given compounds are produced in the oxidation reaction?

Premise
Response
1

1)

A

Not produced

2

2)

B

Produced

3

3)

C

Not produced

4

4)

D

Produced

5

5)

E

Not produced

6

6)

F

Not produced

Here is a question to make sure you know the differences between the above reactions.

Q14.7 - Level 2

Match the reagent needed to accomplish the following transformations.

Premise
Response
1

1)

A

Reagent b

2

2)

B

Reagent d

3

3)

C

Reagent a

4

4)

D

Reagent c

### 14.5.2 Synthesis of Alcohols with Grignard and Organolithium Reagents: Nucleophilic Additions to Carbonyl Compounds

When an alkyl halide is added to magnesium or lithium metal in an ether solvent, an organometallic reagent is formed. If magnesium is used, the reagent is called a Grignard reagent. The reagent was named after Victor Grignard (1871-1935) who extensively determined the chemistry of the reagent. Grignard and organolithium reagents can be made from alkyl halides, aryl halides, and vinyl halides.

Grignard and organolithium reagents are not simple structures. In the case of Grignard reagents, the ether solvent is essential and incorporates two solvent molecules into its structure.

It is generally accepted that the mechanism follows a series of single electron transfers.

Grignard reagents can be made with chloro, bromo, and iodoalkyl halides but not fluroalkyl halides.

Organolithium reagents are more complex than Grignard reagents. Organolithium reagents form oligomeric structures, which consist of a few monomer units. Depending on the size of the alkyl group, aggregates are formed.

However, when thinking about the chemical reactivity of these reagents, the structures can be simplified to RMgX and RLi. The most important feature about these reagents is that the bond between carbon and the metals is a highly polarized bond that is slightly ionic in character. This is because carbon is far more electronegative than Mg or Li and carbon withdraws electron density from the metals. The carbon-lithium bond has about 40% ionic character and the magnesium-carbon bond 35%. We can characterize this strongly polarized bond by the resonance structures given below.

Thus Grignard and organolithium reagents are essentially carbanions and are very strong bases.

For comparison, the structure of a carbocation, carbon radical, and a carbanion are given below. Because of charge repulsion, carbanions are tetrahedral in structure.

They are also strong nucleophiles that are capable of attacking electrophiles, including the carbonyl group of aldehydes, ketones, and esters.

### 14.5.2.1 Reaction with Carbonyl Groups to Form Alcohols

The carbonyl group contains a short, strong and very polar bond. The hybridization of the C and O atoms of the carbonyl group is sp2 hybridized. The C, O and the two atoms attached to the carbon all lie in the same plane. The bond angles about the carbonyl are about 120°. The π bond of the carbonyl is created by the sideways overlap of the p orbitals on the carbon and oxygen.

Carbonyl groups are susceptible to attack by both electrophiles and nucleophiles.

### Question 14.8

Q14.8 - Level 1

Click on the atom that would be attacked by a nucleophile.

Click here to see the answer to Question 14.8.

What is the mechanism of a Grignard Reagent attacking a carbonyl group?

Q14.9 - Level 1

Grignard reagents react as carbanions (negatively charged carbons) and are strong nucleophiles. Click on the arrows that represent the first step in the mechanism of a Grignard reagent reacting with a ketone.

As seen from the question above, the first step in the mechanism is a nucleophilic attack by the negatively charged carbon of the Grignard (or organolithium) reagent on the partially positive carbonyl carbon (negative goes to positive) to form a halomagnesium alkoxide. To complete the synthesis and produce the alcohol, diluted acid or water is added to the halomagnesium alkoxide.

General Reactions to Produce 1°, 2° and 3° Alcohols and Carboxylic Acids

When a Grignard or organolithium reagent is reacted with formaldehyde, a primary alcohol is produced.

When a Grignard or organolithium reagent is reacted with an aldehyde a secondary alcohol is produced.

When a Grignard or organolithium reagent is reacted with a ketone, a tertiary alcohol is produced.

When a Grignard or organolithium reagent is reacted with carbon dioxide, a carboxylic acid is produced. This is not an alcohol but is included here to show the full scope of the organometallic reagent.

When a Grignard or organolithium reagent is reacted with an ester, a tertiary alcohol is formed and interestingly the tertiary alcohol produced has the alkyl group of the reagent added twice to it. The reason for this is that a ketone is generated as an intermediate that then reacts with a second equivalence of the Grignard reagent, as shown in the mechanism below.

The first step is the nucleophilic addition of the reagent to form a tetrahedral intermediate.

In the second step, an alkoxide leaves to form a ketone. Typically, alkoxides are not good leaving groups, however the tetrahedral intermediate itself is an alkoxide and loss of RO- produces a more thermodynamically stable carbonyl compound.

The above reaction is in equilibrium and the alkoxide can act as a nucleophile and reverse the process. However, once the ketone is formed, the stronger nucleophile, which is the Grignard reagent, will preferentially react with the ketone to form a new C-C bond and convert the ketone into a tertiary alcohol.

The final step is not in equilibrium since a leaving group is not present.

Simplifying the Mechanism (Nucleophilic Addition to a Carbonyl Carbon)

Notice that all of the reactions involve a nucleophilic attack on the carbonyl carbon to form an alkoxide.

Thus for all Grignard and organolithium reactions follow the simplified mechanism.

• Identify the carbanion of the reagent. It is the carbon bonded to the metal.
• Draw an arrow from the carbanion to the carbonyl’s electrophilic carbon. This forms the new carbon–carbon bond.
• Draw another arrow from the π bond to the oxygen. This forms a negatively charged alkoxide ion.
• Step 2 protonates the alkoxide to form the desired alcohol.

If the carbonyl is an ester, the nucleophile adds twice.

• Identify the carbanion of the reagent. It is the carbon bonded to the metal.
• Draw an arrow from carbanion to the carbonyl’s electrophilic carbon.This forms the new carbon-carbon bond,
• Draw another arrow from the π bond to the oxygen.This forms a negatively charged alkoxide ion.
• The alkoxide leaving group leaves to form a ketone.
• Draw an arrow from carbanion to the carbonyl’s electrophilic carbon.This adds the alkyl group for the second time.
• Draw another arrow from the π bond to the oxygen.This forms a negatively charged alkoxide ion.
• Step 2 protonates the alkoxide to form a tertiary alcohol with the alkyl group from the reagent added twice.
Q14.10 - Level 2

What is the major product produced in the given reactions?

A

1)

B

2)

C

3)

D

4)

Q14.11 - Level 1

What type of alcohol is produced in the given reactions?

Premise
Response
1

1)

A

Secondary alcohol

2

2)

B

Primary alcohol

3

3)

C

Tertiary alcohol

D

Primary alcohol

E

Tertiary alcohol

F

Secondary alcohol

Q14.12 - Level 2

What is the major product produced in the given reaction?

A

1)

B

2)

C

3)

D

4)

### 14.5.2.2 Grignard and Organolithium Reagents as Strong Bases

Grignard and organolithium reagents behave like carbanions and are much more basic than amides and alkoxides, because carbon is less electronegative than either nitrogen or oxygen. Because alkylmetals are strong bases, organometallic reagents are sensitive to moisture and react rapidly with water. It is very important that all reagents are dry when using organometallic reagents. If water is present in the reaction vessel, the yield of the desired alcohol will be dramatically decreased because the reagent will react as a Brønsted-Lowry base and abstract a proton from water to generate an undesired alkane. The example below depicts the undesirable side reaction with water that would decrease the yield of 1-phenylethanol.

This can be turned into a useful reaction if the desired product is a deuterated compound.

### Question 14.13

Q14.13 - Level 2

What is the major product produced in the given reactions?

A

1)

B

2)

C

3)

D

4)

Click here to see the answer to Question 14.13.

### 14.5.2.3 Limitations of Grignard and Organolithium Reagents

Grignard and organolithium reagents are strong bases. Because of this, they cannot be prepared from compounds that contain acidic groups (-OH, -NH2, -NHR, -SH, -C≡CH, RCO2H) or compounds that have carbonyl groups.

### Question 14.14

Q14.14 - Level 1

Which of the following compounds could be used successfully to prepare a Grignard reagent for alcohol synthesis by subsequent reaction with an aldehyde or ketone?

A

A)

B

B)

C

C)

D

D)

Click here to see the answer to Question 14.14.

Q14.15 - Level 1

Which of the given syntheses of a Grignard reagent would fail to form as written?

A

1)

B

2)

C

3)

D

4) all would succeed

### 14.5.3 Synthesis of Acetylenic Alcohols

Magnesium acetylides can be readily made by using an alkyl Grignard reagent because the sp hybridized ≡C-H bond is more acidic than the sp2 hybridized C-H bond, Chapter 4.2.7.

Lithium acetylides can be made by the addition of lithium metal or using strong bases such as butyllithium (LiCH2(CH2)2CH3).

Acetylides can also be made using sodium amide.

Acetylides are nucleophilic and will react with carbonyl groups. Again, the attack of an acetylide follows the simplified general mechanism where the nucleophile is RC≡C.

Q14.16 - Level 2

What is the final product produced in the following reactions?

A

1)

B