Ch 9 chart for organic chemistry PDF

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Summary

syllabus is good for this class. Read the syllabus....

Description

Ch 9 (Synthesis and Reaction of alkynes) Chapter 9 reactions deal with synthesis and reactions of alkynes. For the bare minimum, memorize what reagent is used, and what product is formed. For alkynes, you don’t have to memorize the mechanisms, but again it helps to at least understand it to make it easier to predict the products. I will try to emphasize which reactions are important, and what you should know for each reaction shown below. Synthesis of Alkynes Reaction

Example (Mechanism)

Synthesis of alkynes from terminal alkynes

-A terminal alkyne hydrogen is relatively acidic (pKa ~25), and can become deprotonated using a strong base (NH 2-, pKa ~35). The alkyne ion can then react as a good nucleophile in a SN2 reaction with a common electrophile (Must be a primary, unhindered, electrophile!). (You don’t need to specify stereochemistry)

-If reacting with an acetylide, specify 1 equivalent so you only extract 1 hydrogen

Addition of alkynes to carbonyl groups

-Similar to the previous reaction, NaNH2 can extract a hydrogen from a terminal alkyne and form a good nucleophile. The alkyne ion can also react with carbonyl groups (can be formaldehyde, aldehyde, or ketone) , become protonated to form an acetylenic alcohol

The degree of alcohol (1o, 2o, 3o), depends on what carbonyl group is used (formaldehyde = 1o, aldehyde = 2o, ketone = 3o

Reactants, Reagents, products Reactant – terminal alkyne (0 or 1 substituent attached) Reagent – NaNH2, then an electrophile Product – An substituted alkyne (1 more substituent than started off with) Reactant- Alkyne Reagent – NaNH2, then any carbonyl group (formaldehyde, aldehyde, or ketone) Product acetylenic alcohol (formaldehyde = 1o, aldehyde = 2o, ketone = 3o alcohol)

Stereochemistry/ Regiochemistry

Important things to note

Stereochemistry – No specific stereochemistry Regiochemistry – No specific regiochemistry

-Know this reaction! NaNH2 is an important reagent commonly used for formation of an alkyne ion -This is simply a SN 2 reaction, with the nucleophile being the alkyne ion -A strong base (NaNH 2) must be used to deprotonate the alkyne hydrogen -The electrophile must be primary, unhindered! -Not as important as other reactions, but still be able to recognize what happens if given the reagents. -Good for a synthesis question if given an alkyne with an alcohol on the same compound

Stereochemistry – No specific stereochemistry Regiochemistry – No specific regiochemistry

Synthesis of Alkynes by Elimination

A dihalide (alkane with 2 adjacent halogen substituents) can react with a strong base in very basic conditions to form an alkyne. NaNH 2 with heat is used for forming terminal alkynes (alkynes on the very end of the chain).

KOH under very basic and heated condition can be used to eliminate the dihalide to form internal alkynes (alkynes with 2 substituents). Reaction is non-stereospecfic

Reactant- Alkyl dihalide Reagent – NaNH2 for terminal alkyne, KOH for internal alkyne Product An alkyne (terminal or internal)

Stereochemistry – No specific stereochemistry Regiochemistry – No specific regiochemistry

-Not a common reaction you see, and most likely see as a test question, but recognize it and what happens just in case -However, this is a good reaction to know if you need to synthesize an alkyne from an alkene or alkane (alkene can react with Br2, then react with KOH or NaNH 2 with heat to form the alkyne) -A dihalide must be present to be able to do 2 eliminations

Reactions of Alkynes Reaction

Example (Mechanism)

Catalytic hydrogenat -ion to alkanes

-An alkyne can react with 2 equivalents of H2, with a metal catalyst (Pt, Pd or Ni), to hydrogenate the alkyne completely to an alkane. The catalyst is too strong: it will completely hydrogenate from alkyne to alkane, impossible to stop at the alkene

Catalytic hydrogenat ion of alkynes to cis alkenes

-An alkyne can react with a “poisoned” catalyst, called lindlar’s catalyst, which is much weaker catalyst than Pt. It can only add 1 addition of H2 in a syn addition, which forms a cis alkene. (Recognize what lindlar’s catalyst is, but for synthesis questions, just write H2 and Lindlar’s catalyst).

Reactants, Reagents, products Reactant – Any Alkyne Reagent – 2 H2, and a metal catalyst (Pt, Pd, Ni) Product – An alkane

Stereochemistry/ Regiochemistry

Important things to note

Stereochemistry – No specific stereochemistry Regiochemistry – No specific regiochemistry

-Don’t need to know the mechanism, but how it reacts and what forms! -Can’t stop at the alkene, will go straight to the alkane

Reactant – Any Alkyne Reagent – Lindlar’s catalyst, or nickel boride (Ni2B) Product – A “cis” alkene

Stereochemistry – Syn addition – The hydrogens will add in a syn addition forming a cis alkene (same side) Regiochemistry – No specific regiochemistry

-Don’t need to know the mechanism, but know what reacts and what forms! -Will only get the cis alkene! -Be able to recognize what Lindlar’s catalyst is. Another possibility for a reagent they can use is Nickel boride (Ni2B). It does the same thing.

Metalammonia reduction of alkynes to trans alkenes

-The alkyne will react with an electron in the solution, forming an anion and a radical. Due to sterics, the charges will place the substituents trans to each other. For a terminal alkyne, the radical will form on the most substituted carbon, and the anion will form on the least to help stabilize the intermediate (for a terminal alkyne, it doesn’t matter). Note that where the anion and radical form does not change the final product.

-The carboanion will extract a hydrogen from NH 3, forming a neutral charge with the radical still on the alkene

-Similar to the previous steps, another electron will be donated to the radical, forming a carboanion, and another NH 3 will donate a hydrogen, forming a neutral trans-alkene

-The overall reaction is shown below

Reactant – Any Alkyne Reagent – Na/liquid NH 3 Product – A trans alkene

Stereochemistry – -Anti addition – The hydrogens that are attached will attach on the opposite side of the double bond (substituents will be going trans to each other) Regiochemistry – No specific regiochemistry

-Don’t need to know the entire mechanism, but know what reacts and what forms! -Must react with Na/NH3! - Forms a trans alkene - the substituents will be trans to each other

Addition of hydrogen halides

An alkyne can react with 1 equivalent with HBr (can be HCl, HBr, HI), forming a carbocation. Similar to alkenes, the hydrogen will bond to the least substituted end of the alkyne to form the most stable carbocation. Bromine will then attach onto the alkene. The reaction can stop after the formation of the alkene with the halide

Reactant – Any Alkyne Reagent – Acid halide (HCl, HBr, or HI) Product – An alkyl dihalide, with the halides on the most substituted carbon

Stereochemistry – No specific stereochemistry Regiochemistry – Markovnikov Addition. The halides (any addition, 1 or 2) will attach to the most substituted carbon due to the formation of the most stable carbocation.

-Don’t need to know this mechanism. but I recommend you do, because it is really similar to the alkene equivalent reaction! -Markovnikov addition: will form the most stable carbocation, with any halide forming on the most substituted carbon. -Can stop after 1 addition, or go straight to 2

Reactant – Any Alkyne Reagent – Halogen (Cl2, Br2, I2) Product – A tetrahalide alkane (not alkene!)

StereochemistryNo specific stereochemistry Regiochemistry – No specific stereochemistry

-Don’t need to know the entire mechanism, but know what reacts and what forms! -Will not stop at the alkene, will go straight to the alkane!

-A second addition of HBr can react on the alkene, forming a carbocation intermediate on the more substituted carbon (still most stable carbocation). Bromine will react to form 2 halides on the most substituted carbon. There is no specific stereochemistry (will get a mixture of cis and trans).

-To go straight from an alkyne to an alkyl dihalide, just write 2 equivalents of HBr, or write excess HBr. To go from an alkyne to an alkene, specify 1 equivalent of HBr!!

Addition of halogens

-Unlike HBr, Br2 will completely halogenate the alkyne straight to the alkane, forming a tetrahalide alkane. A mixture of syn or anti addition will result.

Mercuric ioncatalyzed hydration of alkynes

-An alkyne can bond with Hg, with Hg binding on the least substituted carbon, forming the more stable carbocation. Water will then attack the carbocation forming an alcohol attached to the alkene.

-The alkene with the Hg is in equilibrium inside the acidic solution, and H+ will displace the Hg, forming an enol (an enol is an alcohol adjacent to an alkene)

-The enol can now undergo a keto-enol tautomerism, which converts the enol form of the product to the keto form (think ketone for keto). Notice that the equilibrium favors the keto form as opposed to the enol. The keto is the more stable form! If given a question that asks for the most stable product, draw the keto form!

-The overall reaction is summarized below

Reactant – Any Alkyne, terminal or internal Reagent – Mercury and water in acidic solution. (Hg+, H2O and H+) Product – A ketone (keto form)

StereochemistryNo specific stereochemistry Regiochemistry – Markovnikov addition -The initial alcohol will form on the most substituted carbon due to the more stable. The carbonyl will form on the substituted carbon

- Don’t need to know the entire mechanism, but know what reacts and what forms! -However, know the enolketo tautomerism (the product is in equilibrium in the two forms). -The carbonyl will form on the most substituted carbon (forming a ketone) -Remember the keto form is the most stable form! -You need mercury, water and an acidic solution for this reaction to occur -Similar reagents used compared to the mecuration reaction of the alkene in ch 8

Hydroborat ionoxidation of alkynes

-An alkyne can react with disiamylborane (Sia2BH, don’t need to know what it looks like), initially forming an alkene with Sia2BH attaching on the least substituted carbon of the alkyne. Any boron compound with 2 bulky substituents can be used. Notice that the Sia2BH and Hydrogen add in a syn addition (this will not affect the final product, but realize this is similar to the hydroboration reaction of alkenes)

Reactant – Any Alkyne Reagent – Sia2BH (any bulky boron compound), and then H2O2 and NaOH Product – The keto form of the alcohol (on a terminal alkyne, this is always an aldehyde)

StereochemistrySyn-Addition – Sia2BH and a hydrogen will add to the same side of the alkyne. This will not change the final product though Regiochemistry – Anti-Markovnikov addition –The initial alcohol formed is attached on the least substituted carbon of the alkyne, with the keto form, or carbonyl forming on the least substituted carbon

- Don’t need to know the entire mechanism, but know what reacts and what forms! -Also remember the product is also in equilibrium in the enol-keto form, with the keto form favored -Need a bulky boron compound (Sia2BH is the most common) -Similar to the hydroboration reaction in ch 8 of alkenes -Sia2BH and a hydrogen will add to the same side of the alkyne. This will not change the final product though

Reactant – Any Alkyne (terminal and internal) Reagent – KMnO4 (either neutral conditions, or basic and heated conditions)

StereochemistryNo specific stereochemistry Regiochemistry – No specific stereochemistry

- Don’t need to know the entire mechanism, but know what reacts and what forms! Specifically, know what happens given KMnO 4 under different conditions, and how it will react with the alkyne (internal or terminal)

- Sia2BH can then be cleaved off by H 2O2, and NaOH will attach an OH onto the double bond, forming the enol form of the product. The equilibrium will favor the keto form, so the enol will switch to the keto form. In a terminal alkyne, an aldehyde is produced

The overall reaction can be written below

Oxidation of alkynes using permangan te

-In neutral conditions, KMnO4 can react with an internal alkyne to form a diketone shown below.

-In neutral conditions, KMnO4 can react with an terminal alkyne to form a ketoaldehyde shown below. The aldehyde will continue to oxidize to a carboxylic acid

-In heated, basic conditions, KMnO4 can react with an internal alkyne, initially forming a diketone. The diketone will then be cleaved, and form 2 carboxylic acids shown below. (It doesn’t cleave like that, I just show it like that to make it easier to see where the products will separate)

- In heated, basic conditions, KMnO4 can react with an terminal alkyne, initially forming a keto-aldehyde. The aldehyde is oxidized to a carboxylic acid. The keto-acid will then be cleaved, and attach OH to each compound that was cleaved.

- The carboxylic acid with 2 OH’s is carbonic acid. This will released as H2O and CO2

-The overall reaction of a terminal alkyne reacting in basic heated KMnO4 is shown below

Product – Dependent on the conditions of KMnO4 and the type of alkyne Internal, neutral: diketone Terminal, neutral: keto-acid (ketone with adjacent carboxylic acid) Internal, basic, heated: 2 carboxylic acids Terminal, basic, heated: A carboxylic acid, CO2 and H2O

-Remember, it is basic and heated conditions that cause KMnO4 to cleave the alkyne to carboxylic acids. This is different from alkene reactions, which requires heat and acidic conditions to cleave the alkene! -These are kinda tricky, go through these reactions carefully, remember that KMnO4 oxidizes aldehydes to carboxylic acids, similar to alkene reactions!

Ozonolysis of Alkynes

-An alkyne can react with ozone and water, cleaving the alkyne and forming 2 carboxylic acids

-In a terminal alkyne, ozone will cleave the alkyne and form a carboxylic acid and a formic acid (aldehyde with OH). This will remain formic acid (The hydrogen will not oxidize to OH)

Reactant – Any Alkyne Reagent – Ozone, then water Product – Carboxylic acids

StereochemistryNo specific stereochemistry Regiochemistry – No specific stereochemistry

- Don’t need to know the entire mechanism, but know what reacts and what forms! -Similar to alkene reactions, but forms carboxylic acids instead of ketones. Any formic acid (HCO2H) will not continue to oxidize!...