ORG CHEM 4 Flashcards
Alcohols react using both acid and base catalysis: please elaborate.
Acids catalyze hydrolysis and ether formation from alcohols. Bases polarize the OH bond, even to the point of forming an alkoxide salt.
Acids are used as catalysts in the Williamson ether synthesis and for hydrolysis of ethers. Bases dissolve in ethers for use in other reactions.
Acids and bases will react in alcohols to form salts that are useful for ether synthesis.
Acids catalyze alcohol dimerization to ethers while bases hydrolyse them.
Acids catalyze hydrolysis and ether formation from alcohols. Bases polarize the OH bond, even to the point of forming an alkoxide salt.
The Williamson ether synthesis involves which of the following?
The two Williamson reagents normally used in combination with heat and light.
An alkyl alcohol heated up with acid to distill off water.
An alkoxide salt of an alcohol reacted with a primary or secondary halide.
An alkoxide salt of an alcohol reacted with the alcohol it was made from.
An alkoxide salt of an alcohol reacted with a primary or secondary halide.
Epoxides are a special kind of ether that:
are reactive because of ring string (3 membered ring).
react faster with acid or base catalysis.
can react with amines, alcohols and acids.
all of the above.
react faster with acid or base catalysis.
Ethers can react with which of the following?
With base and water to give two alcohols.
With acid and another alcohol to give transetherification- another ether and a different alcohol from the starting one.
With ethyl bromide to give an ethyl ether and a new alkyl bromide.
With a carboxylic acid and heat to give an ester and an alcohol.
With acid and another alcohol to give transetherification- another ether and a different alcohol from the starting one.
Alcohols are unique in forming ethers because:
they can react with themselves in the presence of heat and acid to give ethers.
they have an exchangeable hydrogen (the OH one).
they have higher boiling points than hydrocarbons or ethers with the same number of carbon atoms.
they are all of the above.
they are all of the above.
Ether alcohols (molecules with both an ether and a beta alcohol) can best be made by:
acid catalyzed reaction of an epoxide and an alcohol with maybe a little heat.
addition of an alcohol to an alkene with base catalysis and heat.
acid catalyzed addition of water to a vinyl ether (ether with a double bond next to the oxygen).
addition of an alcohol to a ketone in the presence of acid and heat.
acid catalyzed reaction of an epoxide and an alcohol with maybe a little heat.
There are many ways to make an ether including:
reaction of two alcohols with base and heat catalysis.
reaction of an epoxide with an alkene using acid catalysis.
reaction of an alkoxide and an alkyl bromide.
addition of water to an epoxide with base catalysis.
reaction of an alkoxide and an alkyl bromide.
Pick the synthetic conditions that will make an ether.
Reaction of an epoxide with a carboxylic acid.
Conversion of an alcohol with base catalysis and heat, distilling off the ether as it forms.
Reaction of an alkyl halide with NaOH in water.
the conditions for a Williamson synthesis (alkoxide plus halide).
the conditions for a Williamson synthesis (alkoxide plus halide).
An epoxide reacts with an alkoxide to give an ether plus something else:
and the something else is another alkoxide.
and the something else is NaOH.
and the something else is another epoxide.
and there is nothing else formed
and the something else is another alkoxide.
Ethers are pretty good solvents for:
alkyl halides.
lithium alkyl and sodium alkoxide salts.
alcohols.
all of the above to some extent.
lithium alkyl and sodium alkoxide salts.
Why are thiols and their anionic conjugate bases more reactive than their oxygen analogs?
They aren’t really, since alcohols and acids are more reactive.
Because the sulfur is big, and the bigger an atom is with unpaired outer shell electrons, the more reactive it is.
Because sulfur has a bigger, less tightly packed nucleus.
Because it’s smaller than oxygen and has more localized outer shell electrons which are more nucleophilic.
Because it’s smaller than oxygen and has more localized outer shell electrons which are more nucleophilic.
Why does sulfur oxidize so easily to sulfoxides, sulfones, sulfonic and sulfuric acid?
No one knows, it’s one of the mysterious of the universe.
Oxygen is more electron rich, and wants to share electrons with any atom it can bind to.
Oxidation is just “burning,” and we know that sulfur burns well (high sulfur coal?).
Sulfur just loves to share its outer shell electrons, and the more sharing it does, the happier it is.
Oxygen is more electron rich, and wants to share electrons with any atom it can bind to.
Do eletrophilic compounds always react at the sulfur, no matter what?
Yes, except when oxygens are attached to the sulfur, then the oxygens react first.
Yes, except when carbons are attached to the sulfur, then substitution occurs at the carbon.
Yes, except when bromine is involved.
Pretty much, yes, even when oxygens are attached to the sulfur.
Pretty much, yes, even when oxygens are attached to the sulfur.
The Swern oxidation involves using DMSO (remember this great solvent?) to do what?
React with oxygen to form dimethylsulfone, the di-oxygen derivative of sulfur.
React with primary or secondary alcohols to give aldehydes or ketones.
Convert a primary alcohol to a carboxylic acid using oxygen as the co-reactant.
Convert amines to oxides such as nitric oxide using peroxide as co-reactant.
React with primary or secondary alcohols to give aldehydes or ketones
What is the difference between a thiol and a mercaptan?
There is none; they’re the same and they both have SH bonds.
A thiol is the sulfur analog of an ether; a mercaptan is like an alcohol (SH bond).
A mercaptan is like an alcohol (SH bond) while a thiol is like an ether.
There is none; they’re both ether analogs (C-S-C bonds).
There is none; they’re the same and they both have SH bonds.
It’s possible to make a polymer from a conjugated double bond diene. Pick the right diene below and the polymer that it forms.
1, 5-hexadiene (gives a polymer with four carbon backbone and two-carbon pendent group).
2-methyl-1, 3-butadiene (also called isoprene), gives polyisoprene like the natural rubber used in tires.
2, 4-hexadiene (gives a polymer with four carbon backbone and two pendent methyl groups).
1, 2-propadiene (gives a polymer with three carbon backbone).
2-methyl-1, 3-butadiene (also called isoprene), gives polyisoprene like the natural rubber used in tires.
Diels-Alder reactions usually involve:
some kind of diene with pendent ester groups reacting with some kind of alkene to give a substituted cyclohexene.
1, 3-butadiene reacting with ethylene to give cyclohexene.
1, 3-cyclohexadiene reacting with carbon monoxide to give a cyclic ether-alkene.
some kind of alkene with pendent ester groups (e.g., maleic acid diethyl ester) reacting with some kind of diene (like 1, 3-hexadiene) to give a substituted cyclohexene.
some kind of alkene with pendent ester groups (e.g., maleic acid diethyl ester) reacting with some kind of diene (like 1, 3-hexadiene) to give a substituted cyclohexene.
Dienes can react with bromine (the diatomic molecule, that is) to give a product; which combination below is correct?
1, 3-butadiene plus 1 equivalent of bromine gas to give mostly 1, 3-dibromo-3-butene.
1, 3-butadiene plus 1 equivalent of bromine gas to give mostly 1, 4-dibromo-2-butene.
1, 5-hexadiene plus excess bromine to give mainly 1, 2-dibromo-5-hexene.
1, 2-propadiene plus one equivalent of bromine to give 1, 3-dibromopropene
1, 3-butadiene plus 1 equivalent of bromine gas to give mostly 1, 4-dibromo-2-butene.
Dienes will readily react with chlorine, especially under the right conditions: which is the best answer below?
2, 4-hexadiene with chlorine gas under a strong uv light to give the dichlorohexene.
1, 3-butadiene added dropwise to liquid chlorine to give 1, 2-dichloro-3-butene.
1equivalent of frozen chlorine (solid) added to cold 1, 4-cyclohexadiene to give 2, 5-dichloro-1, 4-cyclohexadiene..
chlorine gas dissolved in cold water with 1, 3-butadiene added dropwise to give dichlorobutanediols.
chlorine gas dissolved in cold water with 1, 3-butadiene added dropwise to give dichlorobutanediols.
Conjugated double bonds in dienes often lead to an especially stable intermediate during reactions; pick a good explanation below for why.
Any reactive species whether radical, cationic or anionic will give a resonance stabilized “allylic” intermediate.
The first step in the reaction of conjugated double bonds is the same as for isolated double bonds, and the radical, cation or anion formed will be stabilized by the alkyl substituent.
The intermediate will have both the 1- and the 4-carbons of the starting conjugated double bonds attached to the attacking electrophile.
Actually, the reactions are so fast that no intermediate is expected no matter what electrophile is involved.
Any reactive species whether radical, cationic or anionic will give a resonance stabilized “allylic” intermediate.
Reaction of 1, 3-butadiene with maleic anhydride gives a single product that is:
3-vinylcyclobutane-1, 2-dicarboxylic acid anhydride.
3-(3-butenyl)succinic anhydride.
4-cyclohexene-cis-dicarboxylic acid anhydride.
a 1:1 copolymer of the two reactants.
4-cyclohexene-cis-dicarboxylic acid anhydride.
You unexpectedly are given hundreds of tons of 1, 3-cyclohexadiene. You decide to convert it to benzene by:
first reacting with chlorine gas and UV light followed by KOH in alcohol.
treating with NaOH in water to make the alcohol followed by distillation from concentrated sulfuric acid.
carefully adding excess chlorine followed by excess KOH in alcohol.
heating it in the presence of a hydrogen and a catalyst such as palladium.
first reacting with chlorine gas and UV light followed by KOH in alcohol.
You can make butene alcohol by:
simply heating 1, 3-butadiene with NaOH in water.
carefully adding one equivalent of HCl followed by NaOH in water.
carefully adding one equivalent of bromine followed by NaOH in water.
reacting with hydrogen peroxide and acetic acid.
carefully adding one equivalent of HCl followed by NaOH in water.
How would you best make a conjugated diene?
Starting with a linear alkane with at least four carbons, first halogenate once and dehydrohalogenate to an alkene; do radical allylic halogenation to give an allyl halide; then dehydrohalogenate again with KOH in alcohol to the diene.
Take any 4-carbon alkene, add bromine across the double bond and dehydrohalogenate twice with KOH in alcohol.
React a four carbon (at least) linear alkane with two equivalents of chlorine under UV light, then dehydrohalogenate twice with KOH in alcohol.
First chlorinate and dehydrochlorinate butane, then deprotonate with butyl lithium or concentrated KOH in alcohol, and pour the mixture into dilute acid to neutralize the base.
Starting with a linear alkane with at least four carbons, first halogenate once and dehydrohalogenate to an alkene; do radical allylic halogenation to give an allyl halide; then dehydrohalogenate again with KOH in alcohol to the diene.
You decide to make 1, 3-cyclopentadiene from cyclopentene you happen to have by:
first reacting with one equivalent of chlorine gas and uv light followed by treating with KOH in alcohol.
treating with concentrated NaOH in water followed by dehydration with concentrated sulfuric acid.
adding one equivalent of bromine and treating the product with two equivalents of KOH in alcohol.
reacting it with HOCl (bleach) to give the halohydrin, and dehydrating with concentrated sulfuric acid.
first reacting with one equivalent of chlorine gas and uv light followed by treating with KOH in alcohol.
Please name the diene that is a cyclic six-carbon ring with two double bonds (conjugated).
1, 2-cyclohexadiene
1, 4-cyclohexadiene
1, 5-cyclohexadiene
1, 3-cyclohexadiene
1, 3-cyclohexadiene
Huckel was one of the first scientists to describe aromaticity, and he used which simple rule?
involving 4n + 2 carbon atoms in a cyclic array.
involving 4n + 2 electrons in a continuous cyclic array of pi-bonds and/or p-orbitals.
requiring conjugated carbon atoms in a cyclic array like benzene or naphthalene.
with a continuous cyclic array of pi-bonds.
involving 4n + 2 electrons in a continuous cyclic array of pi-bonds and/or p-orbitals.
Aromatic molecules include which of the following?
Benzene, naphthalene and higher analogs with 4n + 2 electrons in the conjugated pi-bonds.
Furan which has four carbons and one sp2 hybridized oxygen totalling five atoms and 6 pi or non-bonded electrons.
Cycloheptatriene possessing 7 carbon atoms with three conjugated pi-bonds containing 6 pi electrons
Cyclopentadiene cation with 5 carbons, two pi-bonds and one sp2 carbon containing 0 electrons in a p-orbital.
Benzene, naphthalene and higher analogs with 4n + 2 electrons in the conjugated pi-bonds
Naphthalene (10 carbons with two fused benzene rings) is aromatic according to Huckel because:
there is a continuous array of p-orbitals from pi-bonds around the outside of the fused rings AND there are 4n + 2 electrons in them (10).
fused benzene rings are always aromatic.
10 carbons are an even number and there are 5 double bonds with 8 pairs of electrons in them.
none of the above.
there is a continuous array of p-orbitals from pi-bonds around the outside of the fused rings AND there are 4n + 2 electrons in them (10).
Cyclopropene has three carbons in a ring with one double bond: is it aromatic?
No because there’s only 2 electrons in the pi-bond (2 p-orbitals).
No, because there’s no continuous array of p-orbitals around the ring.
No, because there needs to be a positive charge somewhere in the ring.
No, because there’s no continuous array of p-orbitals around the ring.
Substituents on aromatic molecules can have long-range effects: explain why.
Sigma bonds transmit electronic effects by electronic polarization effects.
Resonance causes spreading of the electronic effects of various substituents throughout the ring through sigma bonds.
Resonance of the electron donating or withdrawing affects of a substituent can stabilize or destabilize charge at ortho and para positions.
Through space interactions of a substituent with sigma bonds in the ring affect individual carbon electron densities.
Resonance of the electron donating or withdrawing affects of a substituent can stabilize or destabilize charge at ortho and para positions.
A good explanation of aromatic stabilization involves which of the following?
A more even distribution of pi electrons occurs in aromatic molecules.
Rapid resonance of pi electrons spreads the electron density over more atoms.
The molecular orbitals that form from a conjugated cyclic array of p-orbitalswith just the right number of electrons in them are substantially lower in energy than for non-conjugated systems.
There is no good explanation that we know of, it’s just a fact of nature.
The molecular orbitals that form from a conjugated cyclic array of p-orbitalswith just the right number of electrons in them are substantially lower in energy than for non-conjugated systems.
Because of their unique resonance stabilization and molecular structures, aromatic molecules possess unique properties: select the best description below.
They are more stable and less reactive then non-aromatic molecules.
They react with unique mechanisms such as electrophilic aromatic substitution and nucleophilic aromatic substitution.
They do not undergo simple addition reactions that occur with simple alkanes.
All of the above.
All of the above.
Benzene possess a unique structure: select from those below.
It has six carbons like cyclohexane and therefore has both chair and boat confirmations.
It consists of two fused cyclobutadiene rings that are basically flat.
The six carbons in the ring all occupy the same plane with pi bond orbitals above and below that plane.
Benzene has carbons connected by six sigma and three pi bonds, with the pi bonds being significantly shorter than the ordinary sigma bonds in the benzene ring.
The six carbons in the ring all occupy the same plane with pi bond orbitals above and below that plan
Heteroatom’s and charges on atoms can affect aromaticity in what way?
Their orbitals can be hybridized in such a way as to participate in the cyclic array of p-orbitals needed for 4n+2 electrons.
They can certainly change the stability of an aromatic ring.
They modify the behavior of the ring in strange and mysterious ways.
They decrease the reactivity of the rings in which they are located.
Their orbitals can be hybridized in such a way as to participate in the cyclic array of p-orbitals needed for 4n+2 electrons.
Substituents on aromatic ring can affect each other: Pick the answer that best illustrates this.
Meta-nitrobenzoic acid is more acidic than the para-nitro isomer because the nitro is closer to the acid group.
3, 4-Dinitrobenzoic acid is readily made by nitration of benzoic acid with excess fuming nitric acid.
The order of acidity of substituted benzoic acid increases for para-substituents nitro, chloro, hydroxy.
The more electron withdrawing groups on the ring, the more acidic is benzoic acid; i.e., trichloro more than dichloro more than monochlorobenzoic acid.
The more electron withdrawing groups on the ring, the more acidic is benzoic acid; i.e., trichloro more than dichloro more than monochlorobenzoic acid.
Phenol is an aromatic alcohol and differs from aliphatic alcohols in what ways?
The OH group is less acidic and less easily exchanged with deuterium because of resonance.
The OH Group has an enormous effect on reactivity and aliphatic alcohols but not in phenols.
Reactions on carbons alpha or beta to the OH in phenols is strictly controlled by electronegativity affects.
Because of resonance stabilization of intermediates, phenols are more easily brominated and nitrated.
Because of resonance stabilization of intermediates, phenols are more easily brominated and nitrated.
The reactivities of substituents on aromatic rings are affected by the aromatic ring as illustrated by which of the following answers?
The methyl group of toluene readily undergoes free radical reaction and oxidation.
Benzoic acid is more acidic than acetic acid.
Styrene or vinylbenzene is a better free radical monomer then propene.
All of the above.
All of the above.
The acidity of phenol is greater than cyclohexanol because:
an aromatic ring is inherently electron withdrawing, pulling electrons away from the O-H bond.
resonance stabilization of the ring bond to oxygen makes the bond between O and H weaker.
resonance stabilization of the conjugate base (phenolate anion) is greater than is stabilization of the cyclohexanol alkoxide by the attached ring.
alkyl carbons are electron donating compared to an aromatic ring, baking the cyclohexanol OH bond stronger than that of phenol.
resonance stabilization of the conjugate base (phenolate anion) is greater than is stabilization of the cyclohexanol alkoxide by the attached ring.
Compare the reactivity of the oxygen to acetylation with acetic anhydride for phenol versus cyclohexanol by picking the best answer below.
Electron donation by the cyclohexane ring increases nucleophilic attack by the oxygen of cyclohexanol versus phenol.
Resonance is greater for the phenol acetyl oxonium intermediate than for the cyclohexanol analog.
Base catalyzed formation of the phenolate is faster than for cyclohexanol due to resonance stabilization, and the anion formed is more reactive.
In fact, they both react at the same rate.
Electron donation by the cyclohexane ring increases nucleophilic attack by the oxygen of cyclohexanol versus phenol.
You find to your surprise that the reaction of phenol with acetic anhydride and aluminum trichloride gives a mixture of two disubstituted products. Pick the best reason for why.
The OH oxygen reacts twice to give the bisester derivative.
Ring acylation occurs twice in the ortho-ortho’ and ortho-para positions.
The first acylation product is in the ortho and para positions which deactivates the ring and causes meta-substitution with the second acetyl group to give ortho-meta and meta-para diacetylphenols.
The first reaction is at the oxygen to give phenyl acetate followed by acetylation in the ortho and para positions to give a mixture of products.
The first reaction is at the oxygen to give phenyl acetate followed by acetylation in the ortho and para positions to give a mixture of products.
EAS reactions involve attack by something seeking:
all of the below.
electrons, and those are almost always relatively “available” ones in pi-orbitals.
a nucleus to bond to, making the reaction also a “nucleophilic” attack.
a one-step reaction involving loss of leaving group at the same time the attacking group makes a new bond.
electrons, and those are almost always relatively “available” ones in pi-orbitals.
All EAS reactions involve a “tetrahedral” or Td intermediate. Which of the following is true of this statement?
Can’t happen with an aromatic ring carbon which by definition is sp2 or trigonal planar.
There is no intermediate in EAS which simple involves attack at electrons.
No, the leaving group leaves at the same time the attacking group forms a bond.
Almost all do: an sp2 carbon becomes sp3 in the intermediate.
Almost all do: an sp2 carbon becomes sp3 in the intermediate.
The best way to describe the “arrow” showing the mechanism of EAS is:
it goes from the attacking atom (electrophile) to the carbon being attacked.
it goes from the attacking atom to the new bond being formed.
it goes from the electron pair in the pi-bond on the ring to the new sigma bond being formed.
it involves a half-head arrow from the attacking species and a half-head arrow from the pi-bond to the new sigma bond to be formed.
it goes from the electron pair in the pi-bond on the ring to the new sigma bond being formed.
general, electrophilic aromatic substitution (EAS) happens:
between any aromatic molecule and an appropriate electrophile, assuming aromaticity follows the 2(n + 2) rule.
with any pi-bond containing molecule and an electrophile.
with chlorine or bromine and iron catalysis on any pi-bond species.
when an electrophile bonds to a nucleophile of some kind.
between any aromatic molecule and an appropriate electrophile, assuming aromaticity follows the 2(n + 2) rule
EAS with chlorine gas bubbled into chlorobenzene containing aluminum trichloride would:
not happen because chlorine is deactivating.
gives mostly meta-dichlorobenzene with mild heating.
gives mostly ortho-dichlorobenzene with mild heating.
generates HCl and mostly para-dichlorobenzene.
generates HCl and mostly para-dichlorobenzene.
Fuming nitric acid converts phenol into:
pure trinitrophenol (also called “picric acid” which has a really low pKa for a phenolic compound).
2, 4-dinitrophenol because of deactivation with each nitro group added.
para-nitrophenol since the OH group on an aromatic ring is activating and ortho-para directing.
a complex mixture of ortho- and para-substituted phenols with di- and trinitro derivatives predominating.
a complex mixture of ortho- and para-substituted phenols with di- and trinitro derivatives predominating.
- Friedel-Crafts acylation of ortho-nitrotoluene with acetyl chloride and aluminum trichloride gives:
a mixture of acetophenones with the nitro group always ortho or para to the carbonyl.
almost exclusively a single product, 3-nitro-4-methyl-acetophenone.
2-methyl-acetophenone by nitro group displacement.
no reaction since the nitro group is so deactivating.
almost exclusively a single product, 3-nitro-4-methyl-acetophenone.
Common explosives developed by Alfred Nobel use trinitro-aromatics such as:
trinitrophenol made by low-temperature nitration of phenol.
trinitrobenzene made by heating benzene in fuming nitric acid.
2, 5-dinitro-1, 4-dichlorobenzene made by chloronating the dinitrobenzene with ferric chloride as catalyst.
TNT or trinitrotoulene made by careful exhaustive nitration of toluene.
TNT or trinitrotoulene made by careful exhaustive nitration of toluene.
Friedel-Crafts alkylation of benzene with 1-chloro-2-methylpropane and aluminum trichloride gives:
isobutylbenzene (or 1-phenyl-2-methylpropane).
t-butylbenzene (or 1-phenyl-1, 1-dimethylethane).
dimethylstyrene (or phenyl-dimethylethene).
paramethylpropylbenzene.
t-butylbenzene (or 1-phenyl-1, 1-dimethylethane).
You decide to make meta-nitrophenol by doing which of the following steps?
Reacting phenol with fuming nitric acid in an ice bath.
Reacting benzene with fuming nitric acid, then reacting the product with chlorine/iron, and last heating with NaOH in water.
Reacting benzene with chlorine/iron, then reacting the product with nitric acid followed by NaOH in water.
Reacting benzene with fuming nitric acid to make trinitrobenzene and then with NaOH in water.
Reacting benzene with fuming nitric acid, then reacting the product with chlorine/iron, and last heating with NaOH in water.
You acquire a mixture of meta- and para-fluoronitrobenzene and want to react it with a nucleophile. What should you do first?
You pick sodium methoxide because it’s more reactive than a sulfur anion, sodium hydrosulfide (NaSH).
You choose NaCl in ethanol since that’s a pretty good nucleophile system.
You decide on sodium methoxide dissolved in water to make the para-nitrophenyl methyl ether.
You mix NaSH with ethanol and then add the aromatic molecule to make the aryl sulfide.
You mix NaSH with ethanol and then add the aromatic molecule to make the aryl sulfide.
You want to make para-hydroxybenzoic acid, so you decide to do both EAS and NAS as follows:
You react para-chlorotoulene with hot KMnO4, isolate the product and then react it with hot methanol and HCl catalyst. The product of this you react with neat NaOH in the melt.
You react toluene with concentrated NaOH in water at reflux to do NAS formation of the hydroxide, and then oxidize the methyl group with hot permanganate to give the acid.
You first react toluene with hot permanganate to make benzoic acid, then react that with chlorine/iron to make the para-chlorobenzoic acid which you heat with NaOH in water.
You react benzoic acid with chlorine and iron catalyst, then treat with concentrated NaOH in water and heat.
You react para-chlorotoulene with hot KMnO4, isolate the product and then react it with hot methanol and HCl catalyst. The product of this you react with neat NaOH in the melt.
There are two components to a nucleophilic aromatic substitution reaction with the order of reactivity of:
halide leaving groups on the aromatic ring being F>Cl>Br, I not reactive.
the nature of the attacking nucleophile being unimportant.
other activating groups that are NOT displaced but stay on the ring such as ester, nitro, sulfone and carbonyl.
all of the above.
halide leaving groups on the aromatic ring being F>Cl>Br, I not reactive
The Kolbe and Riemer-Tieman reactions are special types of EAS reaction in which:
chloroform is the electrophile and it reacts with the sodium salt of phenol (sodium phenolate).
bubbling carbon dioxide through a solution of phenol in water generates the para-hydroxybenzoic acid.
sodium phenolate (sodium salt of phenol) reacts directly with carbon dioxide (high temperature and pressure) to give ortho- and para-hydroxybenzoic acids as their sodium salts.
phenol dissolved in ethanol with iron chloride reacts rapidly with carbonic acid to form the ortho- and para-hydroxybenzoic acids.
sodium phenolate (sodium salt of phenol) reacts directly with carbon dioxide (high temperature and pressure) to give ortho- and para-hydroxybenzoic acids as their sodium salts
