Carbenes, :CH

2 Preparation of simple carbenes 1.

2.

Mechanism of the a elimination.

carbene

Reactions of Carbenes, :CH

2

(not for synthesis)

Addition to double bond. Insertion into C-H bond Formation of ylide (later) liquid

Simmons Smith Reaction (

for synthesis

, addition to alkenes to yield cyclopropanes)

CH 2 I 2 + Zn(Cu)  ICH 2 ZnI Carbenoid, properties similar to carbenes.

Electronic Structure

Electrons paired, singlet

CH 2 N 2

Triplet and Singlet Methylene

Dominant form in solution Gas phase singlet carbene triplet carbene Rotation can occur around this bond.

pi electrons stereospecific addition CH 2 diradical + non-stereospecific

Aldehydes and Ketones Chapter 16

Structure

Aldehydes O R H Carbonyl group R Ketone O R pentanal O sp 2 2-pentanone O

Examples of Naming

OH O O O pentan-2-one O 3-oxopentanal O O (

E

)-3-(but-1-enyl)hexane-2,4-dione 2-(1-hydroxyethyl)pentanal H CH 3 CHO CHO H CO 2 H CH 3 2,3-dimethyl-4-oxobutanoic acid benzaldehyde

result

Resonance

O O   O

Extension of resonance O O

Boiling points

For compounds of comparable molecular weight… Alkanes, ethers < aldehydes, ketones < alcohols < carboxylic acids Dipole-dipole Hydrogen Bonding Dispersion Forces

Water Solubility

Ketones and Aldehydes, like ethers, can function as hydrogen bond acceptors and smaller compounds have significant water solubility.

Recall Preparation from Alcohols

OH OH Can also be done using KMnO 4 in base with heat or bleach in acid solution (HOCl).

HO CH 2 OH Na 2 Cr 2 O 7 acid, 35 deg.

O CO 2 H Be sure you can balance this kind of reaction.

Use PCC to limit oxidation of primary alcohol to the aldehyde. Secondary are oxidized to ketone.

Primary alcohol PCC RCH 2 OH RCH=O Secondary PCC R 2 CHOH R 2 C=O

Preparations, con’d • Reaction of acid chloride and Gilman

O R OH SOCl 2 Na 2 Cr 2 O 7 RCH 2 OH R O Cl O R R' 1. Li R'X 2. CuX R' 2 CuLi But where do we get these??

Note that we have two possible disconnects available O R' R HOCH 2 R HO O R + R'X R'OH R' O OH + RX ROH HOCH 2 R'

Example: Prepare 2-butanone from ethyl alcohol Requirement to start with ethanol suggests a disconnect into two carbon fragments.

O CH 3 CH 2 OH CH 3 CO 2 H + CH 3 CH 2 X via Gilman and acid chloride Done!

Aldehydes from carboxylic acids

Reduction R O SOCl 2 OH R O LiAlH(OC(CH 3 ) 3 ) 3 O Cl R H

And from alcohols, as before:

RCH 2 OH PCC RCHO Oxidation

A Common Sequence

RCH 2 OH PCC R'OH PX 3 RCHO RX Mg R'MgX Observe these parts at this moment.

OH R R' O Na 2 Cr 2 O 7 R R'

Reactions

Addition of a nucleophile: Nucleophilic Addition Good nucleophile, usually basic O Nu: O OH + O Nu tetrahedral intermediate Nu OH Attack of nucleophile occurs on both sides of carbonyl group.

Produces both configurations.

+ Nu Overall: H – Nu was added to carbonyl group double bond.

Nu Notice that the CO bond order was reduced from 2 to 1. The addition reduced the bond order. We will use this idea later.

Reaction can also be done in acid environment.

Nucleophiles not expected to be as strong ( why?) but the

oxygen may become protonated making the carbonyl a better electrophile ( why?

)

.

OH O acid OH OH + Nu: Nu Nu Very electronegative, protonated oxygen. Pulls the pi electrons into itself strongly.

Problem: If there is too much acid present the nucleophile may become protonated, deactivating it

Addition of Grignard (Trumpets Please)

Recall the formation of a Grignard and its addition to an oxirane O R-X Mg   R-Mg-X OMgX ether R O Carbonyls may be added to in same way… Mg   R-X R-Mg-X ether OMgX R + OMgX R If a new chiral center is created both configurations will be produced.

OH mild acid OH R R

Common Reactions of Grignards

R 2 R'COH a tertiary alcohol Examine reaction with ester further.

R'CO 2 Et an ester R-H ROH  RX  RH(D) acid, weak acid ROH + R’CH 2 OH  RX + R’CO 2 H  R 2 C(OH)R’ O R-Mg-X O H R O OH primary alcohol H Both of these reactions extend carbon chain & keep -OH functionality at end of chain. Can extend further.

R' R'' O RCH 2 OH primary alcohol CO 2 R' H RR'R''C-OH tertiary alcohol ROH + R’R”CHOH  RX + R’R”CO  RR’C (OH)R” RCHR' OH secondary alcohol RCO 2 H carboxylic acid ROH  RX  RCO 2 H ROH + R’CH 2 OH  RX + R’CHO  RCH(OH)R’

Grignard Reacting with an Ester.

Look for two kinds of reactions.

O O R-Mg-X Any alcohol will do here.

R' OEt R' R OEt R' Substitution EtOH R'COCl Acid chloride But where does an ester come from?

SOCl 2 R'CO 2 H Perhaps this carboxylic acid comes from the oxidation of a primary alcohol or reaction of a Grignard with CO 2 .

O R R-Mg-X R' R OMgX OH R R' R Addition R

Synthetic Planning… Use of epoxides and carbonyls offer different disconnect sites.

1. RMgX O 2. H 2 O, HCl epoxide HO New bond. Disconnect site.

R Pattern HO-C-C-R Nucleophile O R-Mg-X OH carbonyl R New bond. Disconnect site Pattern HO-C R Want this to be the nucleophile (Grignard).

Patterns to recognize: carbonyl vs oxirane We can create the following fragments of target molecules by using an organometallic (carbon nucleophile)

reaction with a carbonyl dil. acid O RMgX OH R The difference is the extra CH 2 an oxirane.

when using reaction with a oxirane dil. acid O RMgX OH R

Synthetic Planning… Three different disconnects possible Give synthetic routes to OH R 1 R 2 C R 3 R 1 O R 2 + R 3 -Mg-X R 2 O R 3 + R 1 -Mg-X R 1 O R 3 + R 2 -Mg-X If none of the Rs are H then these three synthetic routes may be available.

O Example: Synthesize from ethanol CH 3 CH 2 OH CH 3 CH 2 X OH Done OH MgX CH 3 CO 2 Et 2 O CH 3 CH 2 MgX CH 3 CO 2 H CH 3 CH 2 OH CH 3 CH 2 OH

Preliminary Analysis

•Hmmm, even number of carbons, at least that is good; ethanol is a two carbon molecule. •Now the problem is to divide it up into smaller fragments. •Ether linkage is easily constructed. Williamson.

•Two butyl groups attached to the central 2 carbon fragment. Grignard + ester.

Bisulfite Addition O HO S O O O HO S O O OH O S O O Addition product.

Practical importance: liquid carbonyl compounds can be difficult to purify. The bisulfite addition products will be crystalline and may be recrystallized.

Addition of Organolithium Compounds to Carbonyls

Generally the reactions are the same as for Grignards but the lithium compounds are more reactive (and more difficult to handle).

O O OH Br Li Li mild acid bromocyclohexane Decreased reactivity of electrophile due to steric hindrance to attack. So we used the alkyl lithium instead of a Grignard.

Nucleophiles derived from terminal alkynes

For example, once formed, the new alkynyl alcohol can be hydrated in two ways, Markovnikov and

anti

Markovnikov.

Carefully observe the structure of the products, the relationship of the OH and the carbonyl.

Can do all the reactions of an alkyne and an alcohol. But remember that we have two acidic groups: the more acidic OH and the less acidic terminal alkyne. We discussed this problem earlier.

Note that the regioselectivity used here is only effective if this alkyne is terminal. Otherwise get a mixture.

Addition of hydrogen cyanide

basic Think of what the mechanism should be….

Followed by protonation of the alkoxide ion (perhaps by unionized HCN).

pH issue.

Slightly basic media so that HCN has partially ionized to cyanide ion, the actual nucleophile.

Follow-up reactions on the cyanohydrins…

H 2 O R R O NaCN CH 2 R N RHC dil acid OH R CH 2 R acid catalysis dehydration N We saw this hydrogenation before.

OH H 2 NH 2 catalyst R CH 2 R Hydrolysis. Acid, heat or base, heat OH R CH 2 R CO 2 H

Let’s see what we can do with the mechanism of the

hydrolysis of the nitrile group to a carboxylic acid.

Overall aq. acid, heat + NH 3 R R CO 2 H N The action is at the nitrile group, CN --> CO 2 H.

But how does a nitrile group behave? What could be happening?

We are breaking the CN bond; bond order goes from 3 to 0. Probably stepwise. Chemically speaking: the nitrogen of the nitrile is basic (lone pair) and can be protonated. This makes it a better electrophile (cf. carbonyl). Multiple bond can undergo addition (cf. carbonyl) reducing bond order.

Goal: Break the C to N bonding and create C-O bonds.

Considerations: neither the electrophile (RCN) nor the nucleophile (water) is very reactive. Since we are in acid protonate the CN group to make it a better electrophile. Then attack it with the water nucleophile to add water. This results in reduction of C-N bond order and creation of C to O bonds .

Again, we are in acid environment. Let’s protonate something…. Protonate the multiple bonded N atom to make better electrophile and attack with the nucleophile, water. Note the bonding pattern here. We have seen it before.

R acid R N NH HO similar to NH HO CH 2 which tauermerizes in acid or base, keto-enol HO CH 2 O R R R R NH H 2 O NH HO NH + H + O NH 2 + H + HOH What have done so far? Reduced the CN bond order from 3 to 2 and added one O to the C.

Moving in the right direction! Want to reduce the CN bond order to zero and introduce more O on the C. Keep going! To induce the water to attack again (adds another O) need to increase the reactivity of the electrophile. Protonate again!! On the O.

HOH R acid NH 2 O O Initial equilibrium with acid H R NH 2 R HO OH 2 NH 2 Now want to get rid of the NH 2 . We have all the O’s we need. We know what we have to do. Have to get the N protonated to make it a good leaving group.

OH 2 H R HO NH 2 reposition the H + R O R + NH 3 O NH 3 HO HO Done.

Wittig Reaction

O H I PPh 3 Bu-Li or NaH Substitution Elimination R Example, synthesize CHCH 3 R R R

or combine them the other way…

Wittig Reaction Mechanism

H I H Acidic hydrogen Ph 3 P Ph 3 P: + Nucleophilic substitution H Ph 3 P strong base, BuLi Ph 3 P Nucleophilic center Phosphonium ylide R O R O O R R + Ph 3 P R R Ph 3 P Ph 3 P oxaphosphetane betaine R R R O R Ph 3 PO Ph 3 P oxaphosphetane

Friedel Crafts Acylation

O R AlCl 3 Cl R O And then all the reactions of ketones…

Formation of Hydrates, carbonyls and water.

Carbonyl side of equilibrium is usually favored.

Hemiacetals and Acetals, carbonyls and alcohols

Addition reaction.

(Unstable in Acid; Unstable in base) (Unstable in Acid;

Stable in base

) Substitution reaction

Formation of Hemiacetals, catalyzed by either acid or base. Let’s do it in Base first.

But first let’s take stock.

We have an addition reaction.

Just mixing a carbonyl and an Use Base to reaction.

set-up good better reactant. Good nucleophile better electrophile by protonating in acid.

Alcohol can become a better nucleophile in base by ionization.

hemiacetal An addition of the alcohol to the carbonyl has taken place. Same mechanism as discussed earlier.

Alternatively, hemiacetal formation in Acid

Protonation of carbonyl (making the oxygen more electronegative) Attack of the (poor) nucleophile on (good) electrophile.

Deprotonation Overall, we have added the alcohol to the carbonyl.

Hemiacetal to Acetal, Acid Only

Protonate the hemiacetal, setting up leaving group.

Departure of leaving group.

Attack of nucleophile Deprotonation Substitution reaction, cf S N 1.

Equilibria

Generally, the hemiacetals and acetals are only a minor component of an equilibrium mixture. In order to favor formation of acetals the carbonyl compound and alcohol is reacted with acid in the absence of water. Dry HCl) The acetals or hemiacetals maybe converted back to the carbonyl compound by treatment with water and acid.

An exception is when a cyclic hemiacetal can be formed (5 or 6 membered rings).

The alcohol

Hemiacetal of D-Glucose

The carbonyl Try following the stereochemistry here for yourself The hemiacetal can form with two different configurations at the carbon of the carbonyl group. The carbon is called the anomeric carbon and the two configurations are called the two anomers. The two anomers are interconverted via the open chain form.

Stabilities of the Anomers…

Here note the alternating up-down relationships.

More stable b form, with the OH of the anomeric carbon is equatorial Less stable a form.

Here see the cis relationship of these two OH groups, one must be axial.

Acetals as Protecting Groups

Synthetic Problem

, do a retrosynthetic analysis Target molecule

E N

The nucleophile could take the form of an organolithium or a Grignard reagent. The electrophile would be a carbonyl.

Form this bond by reacting a nucleophile with an electrophile. Choose Nucleophile and Electrophile centers.

Grignard would react with this carbonyl.

Br-Mg Do

you

see the problem with the approach??

Use

Protecting Group for the carbonyl

… Acetals are stable (unreactive) in neutral and

basic

solutions.

Create acetal as protecting group.

protect Now create Grignard and then react Grignard with the aldehyde to create desired bond.

Remove protecting group.

Same overall steps as when we used silyl ethers: protect, react, deprotect.

react deprotect

Tetrahydropyranyl ethers (acetals) as protecting groups for alcohols.

Recall that the key step in forming the acetal was creating the carbocation as shown… There are other ways to create carbocations……

Recall that we can create carbocations in several ways

: 1. As shown above by a group leaving.

An acid 2. By addition of H + to a C=C double bond as shown next.

This cation can now react with an alcohol to yield an acetal. The alcohol becomes part of an acetal and is protected.

This resonance stabilized carbocation then reacts with an alcohol molecule to yield the acetal.

Sample Problem Provide a mechanism for the following conversion O HO OH HCl/H 2 O O O OH First examination: have acid present and will probably protonate Forming an acetal. Keep those mechanistic steps in mind.

Ok, what to protonate? Several oxygens and the double bond. Protonation of an alcohol can set-up a better leaving group. Protonation of a carbonyl can create a better electrophile.

We do not have a carbonyl but can get a similar species as before.

The protonation of the C=C O H + O Strongly electrophilic center, now can do addition to the C=O O H Now do addition, join the molecules HO HO OH O O O Product Now must open 5 membered ring here. Need to set-up leaving group.

HO O O H + HO H O O

Leaving group leaves….

HO H O O HO H O O HO H O O Followed by new ring closure.

Done. Wow!

Consider formation of acetal Sulfur Analogs O OH dry HCl OH acetaldehyde ethanal O O Sulfur Analog O SH dry HCl SH acetaldehyde ethanal S S dithiane

The aldehyde hydrogen has been made acidic S S H Bu Li S S + BuH Why acidic?

Sulfur, like phosphorus, has 3d orbitals capable of accepting electrons: violating octet rule.

S S S S

Recall early steps from the Wittig reaction discussed earlier H H I Ph 3 P Ph 3 P: + H strong base, BuLi Ph 3 P Ph 3 P This hydrogen is acidic.

Why acidic? The P is positive and can accept charge from the negative carbon into the 3d orbitals PH

Some Synthetic Applications

Umpolung – reversed polarity What we have done in these synthetic schemes is to reverse the polarity of the carbonyl group; change it from an electrophile into a nucleophile.

O CN O CN electrophile O O OH O S S nucleophilic Can you think of two other examples of Umpolung we have seen?

Nitrogen Nucleophiles

Mechanism of Schiff Base formation

Attack of nucleophile on the carbonyl Followed by transfer of proton from weak acid to strong base.

Protonation of –OH to establish leaving group.

Leaving group departs, double bond forms.

Hydrazine derivatives

Note which nitrogen is nucleophilic

H 2 N N H O NH 2 H 2 N N H O NH 2 Nucleophilic nitrogen Favored by resonance Less steric hinderance

Reductive Amination

Pattern: R 2 C =O + H 2 N-R ’   R 2 CH-NH R’

Enamines

Recall

primary amines

react with carbonyl compounds to give Schiff bases (imines), RN=CR 2 .

Primary amine But

secondary amines

react to give enamines See if you can write the mechanism for the reaction.

Secondary Amine

Acidity of

a

Hydrogens

a hydrogens are weakly acidic Weaker acid than alcohols but stronger than terminal alkynes.

Learn this table….

Keto-Enol Tautomerism

(Note: we saw tautomerism before in the hydration of alkynes.) Fundamental process O CH 3 acid or base catalysis HO CH 2 keto form enol form usually small component Mechanism in base: O :OH O CH 3 O Negative carbon, a carbanion,

basic, nucleophilic carbon

.

CH 2 CH 2 basicity and nucleophilicity.

H-O-H HO Additional resonance form, stabilizing anion, reducing CH 2 Protonation to yield enol form.

Details…

Base strength Alkoxides will not cause appreciable ionization of simple carbonyl compounds to enolate.

Strong bases (KH or NaNH 2 ) will cause complete ionization to enolate.

O Double activation (1,3 dicarbonyl compounds) will be much more acidic.

For some 1,3 dicarbonyl compounds the enol form may be more stable than the keto form.

H H O

nucleophilicity O

More details…

Nucleophilic carbon CH 2 Some examples: O O base O O R-X O O O CH 3 :OH O CH 2 Br-Br O CH 2 Br R

Some reactions related to acidity of

a

hydrogens

Racemization Exchange

Oxidation: Aldehyde



Carboxylic

Recall from the discussion of alcohols.

Milder oxidizing reagents can also be used RCHO Ag(NH 3 ) 2 + RCO 2 - + Ag Tollens Reagent test for aldehydes

“Drastic Oxidation” of Ketones

O dichromate, etc at high temperature Obtain four different products in this case.

CO 2 H HO 2 C CO 2 H HO 2 C

Reductions: two electron

O NaBH 4 Or LiAlH 4 OH H O H 2 /Pt OH H

Reductions: Four Electron

acid O Clemmenson Zn(Hg), HCl H H base O Wolf-Kishner H 2 N-NH 2 KOH, heat H H

Mechanism of Wolf-Kishner, C=O  CH 2

Recall

reaction of primary amine and carbonyl to give Schiff base.

Here is the formation of the Schiff base

. We expect this to happen.

Weakly acidic hydrogen removed. Resonance occurs. Same as keto/enol tautomerism.

N H N O H N N H 2 N-NH 2 OH H H-O-H Protonation (like forming the enol) N N H N N H H These hydrogens are

weakly acidic

, just as the hydrogens a to a carbonyl are acidic.

H N H C N C O Here is the resonance for the anion from the keto-enol system N H N C H C C C H H O O Perform an elimination reaction to form N 2 . N N H H OH N N H N N H H-O-H H H

O

Haloform Reaction, overall

CH 3 X 2 NaOH O CX 3 NaOH CO 2 + HCX 3 O CH 3 a methyl The last step which produces the haloform, HCX 3 an a methyl only occurs if there is group, a methyl directly attached to the carbonyl.

If done with iodine then the formation of iodoform, HCI 3 , a bright yellow precipitate, is a test for an a methyl group (iodoform test).

Steps of Haloform Reaction

O CX 3 O CH 3 The first reaction: X 2 NaOH O X 2 NaOH CH 2 X All three H’s replaced by X.

This must happen stepwise, like this: O CH 3 Pause for a sec: We have had three mechanistic discussions of how

elemental halogen

, X 2 , reacts with a hydrocarbon to yield a new C-X bond. Do you recall them? Radical Reaction: R .

+ X-X  R-X + X .

(initiation required) Br Addition to double bond: C=C + X-X  O Nucleophilic enolate anion: CH 3 :OH O + Br (alkene acts as nucleophile, ions) O Br-Br CH 2 CH 2 Br

Mechanism of Haloform Reaction-1

Using the last of the three possibilities :OH Br O O O One H has been replaced by halogen.

C H 3 C H 2 Br-Br CH 2 Br R R R O Repeat twice again to yield C Br 3 R Where are we? The halogens have been introduced. First reaction completed. But now we need a substitution reaction. We have to replace the CBr 3 OH. group with

Mechanism of Haloform - 2

O O CH 3 X 2 NaOH R Here’s how: Attack of hydroxide nucleophile. Formation of tetrahedral intermediate. Anticipate the attack… O R R Reform the carbonyl double bond. CX 3 is ejected. The halogens stabilize the negative carbon. O R CX 3 OH Neutralization.

CX 3 NaOH CX 3 OH O R O + HCX 3 O R CX 3 OH O OH R + :CX 3 This is a

substitution step

; OH anion.

replaces the CX to become the carboxylate O O R + HCX 3 3 and then ionizes

Cannizaro Reaction

Overall: conc. KOH 2 RCHO heat RCO 2 + RCH 2 OH Restriction: no a hydrogens in the aldehydes.

O CHO H CHO H 3 C a hydrogens No a hydrogens Why the restriction? The a hydrogens are acidic leading to ionization.

Mechanism

What can happen? Reactants are the aldehyde and concentrated hydroxide.

Hydroxide ion can act both as Base, but remember we have no acidic hydrogens (no a hydrogens).

Nucleophile, attacking carbonyl group.

O O O R H HO : Attack of nucleophilic HO R R H R OH OH H O H Re-establish C=O and H eject H which is immediately received by second RCHO + R O R O O + H R H OH Acid-base

Experimental Evidence

KOH, H 2 O 2 RCDO RD 2 OH + RCO 2 These are the hydrogens introduced by the reaction. They originate in the aldeyde and do not come from the aqueous hydroxide solution.

Kinetic vs Thermodynamic Contol of a Reaction

Examine Addition of HBr to 1,3 butadiene H H HBr + Br Br 1,2 product 1,4 product

Mechanism of reaction.

Allylic resonance H H H-Br Br H Br Br 1,2 product Br 1,4 product H But which is the dominant product?

Nature of the product mixture depends on the temperature.

H HBr + Br Br 1,2 product 1,4 product Product mixture at -80 deg 80% 20% Product mixture at + 40 deg 20% 80% H Goal of discussion: how can temperature control the product mixture?

When two or more products may be formed in a reaction A  X or A  B Thermodynamic Control: Most stable product dominates Kinetic Control: Product formed fastest dominates Thermodynamic control assumes the establishing of equilibrium conditions and the most stable product dominates.

Kinetic Control assumes that equilibrium is not established. Once product is made it no longer changes.

Equilibrium is more rapidly established at high temperature. Thermodynamic control should prevail at high temperature where equilibrium is established.

Kinetic Control may prevail at low temperature where reverse reactions are very slow.

Nature of the product mixture depends on the temperature.

H HBr + Br Br 1,2 product 1,4 product Product mixture at -80 deg 80% 20% Product mixture at + 40 deg 20% 80% H Thermodynamic Control Kinetic Control Product formed most quickly, lowest E a More stable product

Formation of the allylic carbocation. Can react to yield 1,2 product or 1,4 product.

Most of the carbocation reacts to give the 1,2 product because of the smaller E a leading to the 1,2 product. This is true at all temperatures.

At low temperatures the reverse reactions do not occur and the product mixture is determined by the rates of forward reactions. No equilibrium.

Most of the carbocation reacts to give the 1,2 product because of the smaller E a leading to the 1,2 product. This is true at all temperatures.

At higher temperatures the reverse reactions occur leading from the 1,2 or 1,4 product to the carbocation. Note that the 1,2 product is more easily converted back to the carbocation than is the 1,4. Now the 1,4 product is dominant.

Diels Alder Reaction/Symmetry Controlled Reactions Quick Review of formation of chemical bond.

Electro n donor Electron acceptor Note the overlap of the hybrid (donor) and the s orbital which allows bond formation.

For this arrangement there is no overlap. No donation of electrons; no bond formation.

Diels Alder Reaction of butadiene and ethylene to yield cyclohexene.

We will analyze in terms of the pi electrons of the two systems interacting. The pi electrons from the highest occupied pi orbital of one molecule will donate into an lowest energy pi empty of the other. Works in both directions: A donates into B, B donates into A.

B HOMO donates into A LUMO LUMO acceptor HOMO donor A B LUMO acceptor Note the overlap leading to bond formation HOMO A HOMO donates into B LUMO Note the donor overlap leading to bond formation

Try it in another reaction: ethylene + ethylene  cyclobutane LUM O HOMO LUMO HOMO Equal bonding and antibonding interaction, no overlap, no bond formation, no reaction

Reaction Problem Br Br excess sodium methoxide

Synthesis problem HO OEt using only compounds having two carbons as the source of all carbons in the target molecule

Mechanism Problem Give the mechanism for the following reaction. Show all important resonance structures.

Use curved arrow notation

.

O OH aq. acid + EtOH heat OEt