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Aldehydes Ketones and Carboxylic Acids · TOPIC LESSON

Carboxylic Acids and Derivatives

Follow the acyl carbon through acid–base chemistry, addition–elimination, hydrolysis, reduction and carbon loss.

JEE Main · JEE Advanced · NEET UG51 min study guide10 indexed sections

BY THE END, YOU CAN

  • Identify acids, acyl halides, anhydrides, esters, amides and nitriles without confusing their connectivity.
  • Plan acid preparation with correct carbon accounting and reagent compatibility.
  • Explain nucleophilic acyl substitution and the usual derivative reactivity order.
  • Distinguish reversible acid-catalyzed ester chemistry from base-promoted saponification.
  • Predict reduction, organometallic addition, alpha bromination and selected decarboxylation products.

BEFORE YOU START

  • Carbonyl polarity, nucleophiles, leaving groups and curved electron-pair arrows.
  • Acid–base equilibria, resonance and basic organic nomenclature.
  • Alcohol, aldehyde, ketone and amine functional groups.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Identify the group attached to the acyl carbon

The carboxyl group is R–C(=O)–OH. The carbonyl carbon belongs to the acid’s parent carbon count: CH₃CH₂CO₂H is propanoic acid, not ethanoic acid with an extra unnamed carbon. A carboxylate RCO₂⁻ is the conjugate base, with its negative charge delocalized over two oxygens in the usual resonance description.

Common derivatives replace the acid OH by another group: an acyl chloride is RCOCl, an anhydride is RCO–O–COR′, an ester is RCOOR′, and an amide is RCONH₂, RCONHR′ or RCONR′R″. The atom immediately bonded to the carbonyl carbon controls the class. RCOCH₂R′ is a ketone, not an ester, because its second carbon substituent is not connected through oxygen.

An ester name separates an alcohol-derived alkyl group from an acid-derived carboxylate name. Butyl ethanoate is CH₃CO₂CH₂CH₂CH₂CH₃; ethyl butanoate is CH₃CH₂CH₂CO₂CH₂CH₃. They have the same molecular formula but different connectivity. Hydrolysis reveals different acid and alcohol fragments.

Nitriles R–C≡N are often discussed with acid derivatives because hydrolysis gives a carboxylic acid, but a nitrile has no carbonyl oxygen in its starting structure. Its nitrile carbon becomes the acid carbonyl carbon on hydrolysis. Lactones are cyclic esters and lactams are cyclic amides; hydrolysis opens their acyl–heteroatom connection rather than randomly breaking the carbon skeleton.

ORIGINAL WORKED EXAMPLE

Read an ester without swapping its fragments

Write the structure and complete alkaline-hydrolysis products of propyl butanoate.

  1. The butanoate acyl fragment is CH₃CH₂CH₂C(=O)–.
  2. The propyl alcohol-derived fragment is –OCH₂CH₂CH₃.
  3. Join them as CH₃CH₂CH₂CO₂CH₂CH₂CH₃.
  4. Under aqueous base, the acyl fragment becomes butanoate and the other fragment becomes propan-1-ol.

Answer CH₃CH₂CH₂CO₂CH₂CH₂CH₃ gives CH₃CH₂CH₂CO₂⁻ and CH₃CH₂CH₂OH under the stated base-promoted hydrolysis.

02 / UNDERSTAND & APPLY

2. Acidity changes both reactivity and physical behaviour

Carboxylic acids are much more acidic than typical alcohols because their conjugate bases distribute charge over two oxygens and benefit from the carbonyl-containing electronic structure. Electron-withdrawing groups usually strengthen an acid by stabilizing its conjugate base, with inductive effects commonly weakening as the substituent moves farther from the carboxyl group.

Reaction with hydroxide gives a carboxylate and water. A typical carboxylic acid also reacts with hydrogen carbonate to release CO₂ through formation and decomposition of carbonic acid. Ordinary alcohols and phenol are not strong enough to show that same simple bicarbonate behaviour under ordinary comparison conditions. Other acidic groups can react, so gas evolution alone is not a unique universal identification of CO₂H.

Acid–base transfer is often the first event when a strong nucleophile is mixed with a free acid. A Grignard reagent consumes the acidic proton to form a hydrocarbon and a metal carboxylate rather than cleanly adding carbon to the free acid’s carbonyl. An amine likewise tends to form an ammonium carboxylate salt. Forming an amide generally requires activation or suitable dehydrating conditions, not merely mixing the acid and amine at room temperature.

Carboxylic acids can form strong hydrogen-bond associations, including dimers in suitable nonaqueous environments, affecting boiling points and apparent molecular masses. Water can disrupt or replace those associations. Small acids are often water-soluble; increasing a hydrocarbon portion tends to reduce neutral-acid water affinity. Ionization to a carboxylate often increases aqueous solubility, but the actual solvent, counterion and molecular size still matter.

RCO₂H+HCO₃⁻→RCO₂⁻+CO₂+H₂O

A typical carboxylic acid drives bicarbonate protonation followed by carbonic-acid decomposition.

03 / UNDERSTAND & APPLY

3. Build the acid and count where its carbon came from

Oxidation of a primary alcohol through an aldehyde can give a carboxylic acid under suitable aqueous oxidizing conditions. An aldehyde can also be oxidized directly. These transformations retain the carbon that originally carried OH or CHO as the carboxyl carbon; they do not add a carbon atom.

An alkyl side chain on a benzene ring can be oxidized to CO₂H under suitable strong benzylic-oxidation conditions if the benzylic carbon has an appropriate hydrogen. The carbon attached to the ring becomes the carboxyl carbon, while additional side-chain carbons may be lost in extensive oxidation. A tert-butyl substituent has no benzylic H and does not obey the ordinary side-chain oxidation rule under those standard conditions.

A nitrile can be hydrolyzed through an amide to an acid in aqueous acid, or to a carboxylate in aqueous base followed by an acidic workup for the free acid. If the nitrile was made by cyanide displacement on R–X, its carbon came from CN⁻, so the final RCO₂H has one more carbon than R–X. The substitution stage is suitable mainly for substrates that can undergo the required nucleophilic substitution; a tertiary halide is not a routine clean SN2 precursor.

Carboxylation of an organomagnesium reagent also adds one carbon: RMgX reacts with CO₂ to give a metal carboxylate, and a separate protonating workup gives RCO₂H. The organometallic stage must precede contact with water or acid. A substrate containing an unprotected OH, NH or CO₂H group can consume the organometallic reagent before the desired carbon–carbon bond formation.

ORIGINAL WORKED EXAMPLE

Distinguish carbon-retaining and carbon-adding routes

Compare oxidation of pentan-1-ol with hydrolysis of pentanenitrile and carboxylation of butylmagnesium bromide. What acid connectivity does each give?

  1. Pentan-1-ol has five carbons; its terminal alcohol carbon becomes CO₂H, giving pentanoic acid.
  2. Pentanenitrile has five carbons including the nitrile carbon; hydrolysis gives pentanoic acid.
  3. Butylmagnesium bromide has a four-carbon butyl group; CO₂ supplies one new carbon.
  4. After protonation, that route also gives pentanoic acid.

Answer All three give CH₃CH₂CH₂CH₂CO₂H by different carbon-accounting paths.

04 / UNDERSTAND & APPLY

4. Nucleophilic acyl substitution is addition followed by elimination

A nucleophile attacks the electrophilic carbonyl carbon of R–C(=O)–Y. The C=O π electron pair moves to oxygen, giving a tetrahedral intermediate. If an appropriate leaving group can depart, an oxygen lone pair reforms C=O while the C–Y bond electron pair moves to Y. The net result replaces Y by the incoming group.

This mechanism differs from concerted SN2 substitution at a saturated carbon. The acyl carbon changes from approximately trigonal planar to tetrahedral and then returns to trigonal planar. A tetrahedral intermediate is part of the pathway; a simultaneous backside displacement picture at the original carbonyl carbon is not the usual instructional mechanism.

Proton transfers depend on the medium and the nucleophile. Neutral alcohol or amine attack can give a positively charged incoming heteroatom, which must lose a proton at some stage. In acid catalysis, protonation can activate the carbonyl and convert a poor departing OH or OR group into water or alcohol. Never draw free OH⁻ as the departing species from a strongly acidic protonated intermediate when water is the chemically appropriate leaving molecule.

Aldehydes and ketones usually undergo addition instead because their substituents are H or carbon groups, which are not ordinary leaving groups in these conditions. Acid derivatives possess a potentially replaceable heteroatom group. The presence of a carbonyl alone therefore does not determine whether the net reaction is addition or substitution.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Electron-pair flow in nucleophilic acyl substitutionA generic anionic nucleophile attacks R-C(=O)-Y. One curved arrow begins at the Nu lone pair and ends at the acyl carbon; another starts at the carbonyl pi bond and ends at oxygen. The tetrahedral intermediate has R,Nu,Y and O-minus bonded to carbon. In collapse, an oxygen lone pair reforms C=O while the C-Y bond pair moves onto Y, producing R-C(=O)-Nu and Y-minus. Formal net charge stays minus one. Only relevant electron pairs are shown.Acyl substitution: add, then eliminate1. Nu⁻ adds; the C=O π pair moves to OCORYNu⁻2. Tetrahedral intermediate collapsesCO⁻RNuYO lone pair → C=O; C–Y bond pair → YCORNu+ Y⁻Net: RCOY + Nu⁻ → RCONu + Y⁻Suitable Nu and leaving group Y are assumed.Relevant pairs shown; proton transfers depend on medium.
Generic addition–elimination connectivity scheme, not a literal planar geometry for the tetrahedral intermediate. A suitable leaving group and anionic nucleophile are assumed; neutral nucleophiles and acid catalysis need additional proton transfers.

05 / UNDERSTAND & APPLY

5. Derivative interconversion follows leaving-group and resonance effects

For common neutral derivatives under comparable nucleophilic acyl-substitution conditions, a useful general reactivity order is acyl chloride > anhydride > ester > amide. The trend reflects both leaving-group ability and how strongly the attached heteroatom donates electron density into the carbonyl. It is not a universal rate law independent of steric hindrance, solvent or catalysis.

Chloride is a much weaker base and better leaving group than an ordinary alkoxide or amide anion. Nitrogen in an amide also donates strongly into the carbonyl, reducing carbonyl electrophilicity and giving the C–N bond partial double-bond character. Amides consequently often need stronger hydrolysis conditions than esters, while acyl chlorides hydrolyze readily.

Carboxylates are comparatively unreactive toward ordinary nucleophilic addition because of their negative charge and resonance stabilization. A free carboxylic acid can first neutralize a basic reagent. For synthesis, activation can convert the acid OH into a better leaving group or use an acid catalyst to facilitate a reversible substitution.

Thionyl chloride is a standard conceptual reagent for converting RCO₂H to RCOCl, with SO₂ and HCl in the overall balance. An acyl chloride can then react with an alcohol to form an ester or with ammonia/primary/secondary amines to form amides. A base can capture the acid by-product. When the amine alone also serves as that base, two equivalents appear in the overall acyl-chloride aminolysis balance.

Anhydrides contain two acyl groups joined by oxygen. Hydrolysis gives the corresponding acids; alcoholysis gives an ester plus an acid-derived by-product; aminolysis gives an amide plus an acid-derived by-product. An unsymmetrical anhydride can present competing acyl sites, so a unique product cannot always be inferred without selectivity information.

ORIGINAL WORKED EXAMPLE

Account for the second amine equivalent

Write the ideal overall reaction of ethanoyl chloride with ethylamine when ethylamine also neutralizes the HCl by-product.

  1. One ethylamine molecule supplies nitrogen to the amide.
  2. The product connectivity is CH₃CONHCH₂CH₃.
  3. A second ethylamine accepts the proton and becomes ethylammonium chloride.
  4. The acyl carbon and both carbons of the incorporated ethyl group remain in the amide.

Answer CH₃COCl+2CH₃CH₂NH₂→CH₃CONHCH₂CH₃+[CH₃CH₂NH₃]Cl.

06 / UNDERSTAND & APPLY

6. Fischer esterification: activate, add, transfer protons and lose water

In Fischer esterification, a carboxylic acid and an alcohol reversibly form an ester and water under acid catalysis. Protonation of the carbonyl oxygen increases electrophilicity. The alcohol oxygen then attacks the carbonyl carbon while the π pair moves to oxygen, generating a tetrahedral intermediate.

Proton transfers neutralize the incoming alcohol oxygen and convert an appropriate OH group into OH₂⁺. Collapse of the tetrahedral intermediate expels water and reforms C=O; deprotonation restores the catalyst. The mechanism conserves atoms and charge through each step, rather than combining unrelated elimination and addition arrows into one unexplained jump.

Increasing alcohol activity or removing water can favour ester formation; excess water favours the reverse hydrolysis. The acid catalyst accelerates approach to equilibrium in both directions and does not change the equilibrium constant at a fixed temperature. A numerical equilibrium problem must specify a model for activities or concentrations rather than treating every real concentrated liquid mixture as ideal.

In the usual acyl-substitution pathway, the alcohol oxygen becomes the ester’s single-bonded OR oxygen. An isotope label on that alcohol oxygen therefore tracks into that position in the product. Prolonged acid-catalyzed oxygen exchange, alternative mechanisms of special substrates and scrambling can complicate real experiments; state the usual pathway assumption.

Intramolecular esterification of a hydroxy acid can form a lactone when the ring size and conditions are favourable. Ring counting includes the carbonyl carbon and the ester oxygen. A five-membered lactone is not a four-membered carbon ring merely because one of its ring atoms is oxygen.

RCO₂H+R′OH⇌RCO₂R′+H₂O

Acid-catalyzed reversible esterification; acid–alcohol equilibrium differs from an irreversible conversion claim.

ORIGINAL WORKED EXAMPLE

An idealized esterification equilibrium

Initially 1.00 mol each of acid and alcohol, with no ester or water, react in an idealized constant-volume single-phase model. If Kc=4.00 for acid+alcohol⇌ester+water, find the equilibrium ester amount.

  1. Let x mol each of ester and water form; acid and alcohol each become 1−x mol.
  2. The common volume cancels from Kc=x²/(1−x)².
  3. Take the physically positive root: x/(1−x)=2.
  4. Solve x=2−2x, giving x=2/3 mol.

Answer 0.667 mol ester, with 0.333 mol each reactant remaining, within the stated idealized model.

07 / UNDERSTAND & APPLY

7. Hydrolysis products depend on pH and workup

Acid-catalyzed ester hydrolysis usually follows the reverse of Fischer esterification. Water attacks an activated carbonyl, proton transfers prepare the departing alcohol, and C=O is restored. The process is reversible, with excess water helping favour acid and alcohol.

In base-promoted hydrolysis, OH⁻ adds to the ester carbonyl and a tetrahedral oxyanion forms. Collapse expels an alkoxide in the standard acyl-cleavage pathway. The resulting carboxylic acid is deprotonated to carboxylate, and the alkoxide becomes alcohol. Formation of the resonance-stabilized carboxylate helps drive the net reaction under the usual aqueous conditions.

The net saponification balance is RCO₂R′+OH⁻→RCO₂⁻+R′OH. Hydroxide is consumed stoichiometrically in this net conversion; calling it merely a regenerated catalyst hides the acid–base step. A later acidic workup converts carboxylate to RCO₂H. Before that workup, the carboxylate is the appropriate principal acid-derived product.

Transesterification exchanges the OR group using another alcohol under suitable acid or alkoxide catalysis. It is an equilibrium process; excess incoming alcohol or removal of the outgoing alcohol can shift it. Basic conditions with substantial water can instead promote hydrolysis, so replacing one reagent word without considering the medium is not a complete synthetic plan.

Amide hydrolysis is usually more demanding. Acidic conditions give a carboxylic acid and an ammonium form of the nitrogen fragment; basic conditions give carboxylate and ammonia or an amine. In either case the carbonyl carbon remains with the acid-derived fragment. Hydrolysis of a lactam opens the ring into an amino-acid-related connectivity, with protonation states set by the medium.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Ester acyl-cleavage and reduction fragment accountingThe upper panel tracks an oxygen-18 label originally in R-C(=O)-18O-CH2CH3 through the usual aqueous-base acyl-cleavage pathway: the carboxylate RCO2-minus and labeled ethanol CH3CH2-18OH form. Subsequent acid workup protonates carboxylate. The lower independent panel shows complete ordinary-ester hydride reduction followed by workup giving R-CH2OH plus R-prime-OH. The original acyl carbon remains in the first alcohol. The two starting ester fragments do not become one joined alcohol.Follow the ester oxygen and carbonR–C(=O)–¹⁸O–CH₂CH₃The label is on the single-bonded ester oxygen.Aqueous OH⁻: usual acyl-bond cleavageRCO₂⁻ + CH₃CH₂¹⁸OHThe label remains with the alcohol fragment.Acid workup protonates RCO₂⁻ to RCO₂H.Different reaction: complete hydride reductionR–C(=O)–O–R′R–CH₂OH + R′OHOriginal acyl carbon remains in R–CH₂OH.Do not join both ester fragments into one alcohol.Separate reduction and workup; ordinary ester scope.Label tracking assumes no subsequent isotope exchange.
Separate reaction panels. Hydrolysis uses the usual acyl-cleavage mechanism and excludes isotope exchange; reduction assumes suitable complete hydride reduction and separate workup.

ORIGINAL WORKED EXAMPLE

Track an oxygen label through saponification

An ordinary alkyl ester R–C(=O)–¹⁸O–CH₂CH₃ undergoes the usual aqueous-base acyl-cleavage hydrolysis. Where does the labeled oxygen go?

  1. OH⁻ attacks the acyl carbon, not the ethyl carbon in the assumed pathway.
  2. Collapse breaks the acyl C–¹⁸O bond.
  3. The labeled ethoxide fragment is protonated to labeled ethanol.
  4. The label stays attached to the ethyl group throughout this mechanism.

Answer The label appears in CH₃CH₂¹⁸OH, assuming the stated pathway and no subsequent isotope exchange.

08 / UNDERSTAND & APPLY

8. Reduction and organometallic addition must be matched to the derivative

LiAlH₄ followed by a suitable separate workup reduces ordinary carboxylic acids and esters to primary alcohols at the original acyl carbon. An ester also releases its alcohol-derived OR′ fragment as R′OH after workup. For RCO₂R′, the products are RCH₂OH and R′OH; do not join both fragments into a single longer alcohol.

Amides behave differently under strong hydride reduction: the carbonyl oxygen is removed from the organic framework while the C–N connection is retained, giving an amine. RCONHR′ becomes RCH₂NHR′. This keeps the original acyl carbon, unlike the one-carbon-shorter Hofmann rearrangement of a primary amide.

Ordinary NaBH₄ conditions commonly reduce aldehydes and ketones but leave many simple esters and carboxylic acids unreduced; free acids can also consume hydride through acid–base chemistry. Borane can selectively reduce carboxylic acids in suitable substrates, but other functional groups and conditions still require checking. No reagent label guarantees complete chemoselectivity in every molecule.

Controlled partial reduction of an ester with an appropriate DIBAL-H regime can give an aldehyde after workup; excess or unsuitable conditions can lead to further reduction. Acid chlorides can also be converted to aldehydes under suitably moderated reduction conditions, such as the Rosenmund reaction. These selective cases should not be confused with ordinary excess LiAlH₄, which usually continues to the alcohol.

An ordinary ester or acid chloride often reacts twice with a Grignard reagent: addition–elimination first gives a ketone, then a second addition gives an alkoxide and, after workup, a tertiary alcohol with two reagent-derived groups. Formate esters are a special carbon-count case and can give secondary alcohols because the retained substituent is H. An appropriate organocuprate can instead convert an acid chloride to a ketone under suitable conditions; using an ordinary Grignard reagent is not a reliable way to stop at that ketone.

ORIGINAL WORKED EXAMPLE

Identify both products of ester reduction

Predict the organic products after complete LiAlH₄ reduction and workup of ethyl 2-methylpropanoate.

  1. The ester is (CH₃)₂CHCO₂CH₂CH₃.
  2. The acyl carbon becomes CH₂OH, giving (CH₃)₂CHCH₂OH.
  3. The O-ethyl fragment becomes ethanol.
  4. Carbon accounting gives four carbons in the first product and two in the second, matching six in the ester.

Answer 2-Methylpropan-1-ol and ethanol.

09 / UNDERSTAND & APPLY

9. Reactions beside the carbonyl and loss of the carboxyl carbon

The Hell–Volhard–Zelinskii reaction alpha-brominates a carboxylic acid with an alpha hydrogen through an acid-bromide intermediate. In the conventional Br₂/PBr₃ description, acid-bromide formation enables enolization; the enol reacts with bromine, and hydrolysis restores the alpha-bromo acid. A carboxylic acid with no alpha H cannot follow that ordinary enol-based alpha substitution.

The alpha carbon is the carbon immediately next to CO₂H, not the carboxyl carbon itself. Starting from butanoic acid, the ordinary alpha-brominated product is 2-bromobutanoic acid, not 4-bromobutanoic acid. If a new stereocenter forms through an achiral planar intermediate and no chiral influence is present, a racemic product is commonly expected.

Decarboxylation removes the carboxyl carbon as CO₂. Beta-keto acids readily fit a useful mechanistic pattern: a cyclic proton-transfer/C–C-cleavage arrangement gives an enol, which tautomerizes to a ketone. For CH₃COCH₂CO₂H, the product skeleton is CH₃COCH₃ plus CO₂. The carboxyl carbon leaves, while the ketone carbonyl carbon remains.

The presence of a CO₂H group alone does not make every acid decarboxylate easily on warming. Stabilization of the developing intermediate and a favourable pathway matter. Malonic-acid derivatives provide another familiar activated case. Strong soda-lime decarboxylation of appropriate carboxylate salts is a different set of conditions and should not be drawn as the same mild beta-keto-acid mechanism.

In the elementary soda-lime balance RCO₂Na+NaOH→RH+Na₂CO₃, the organic product has one fewer carbon. For a sodium benzoate example, benzene retains the six ring carbons and the seventh, carboxyl carbon enters carbonate. Do not call ordinary ester hydrolysis decarboxylation: hydrolysis retains the acyl carbon in a carboxylate or acid.

ORIGINAL WORKED EXAMPLE

Which carbon disappears?

A beta-keto acid is CH₃CH₂COCH₂CO₂H. Predict its ordinary thermal decarboxylation product after tautomerization.

  1. Identify the carboxyl carbon at the CO₂H end.
  2. Loss of that carbon as CO₂ converts the adjacent CH₂ position into the terminal carbon of the enol/ketone product.
  3. The original internal ketone carbonyl remains.
  4. The product is CH₃CH₂COCH₃, which has four carbons rather than five.

Answer Butan-2-one plus CO₂.

10 / UNDERSTAND & APPLY

10. Solved practice and a mechanism audit

For each product prediction, mark the original acyl carbon, decide whether it survives, identify the leaving group and state the protonation condition. All examples are conceptual chemistry questions rather than laboratory procedures.

  • The carboxyl or nitrile carbon belongs in the parent carbon count.
  • Base hydrolysis gives carboxylate before acidic workup.
  • An ester has two fragments; track both on cleavage or reduction.
  • Do not draw acyl substitution as ordinary saturated-carbon SN2.
  • A catalyst changes rate, not the equilibrium constant at fixed temperature.
  • An amide reduction retains C–N; ester reduction releases the OR fragment.
  • HVZ needs an alpha H; decarboxylation needs an appropriate substrate and conditions.

ORIGINAL PRACTICE · 1

What are the principal organic products of complete aqueous NaOH hydrolysis of methyl pentanoate before acid workup?

Reveal answer & reasoning

Pentanoate and methanol.

The five-carbon acyl fragment becomes CH₃CH₂CH₂CH₂CO₂⁻; the methyl group bonded through oxygen becomes CH₃OH.

ORIGINAL PRACTICE · 2

Does mixing propanoic acid with ethylamine at room temperature automatically give N-ethylpropanamide?

Reveal answer & reasoning

No; acid–base salt formation is the immediate expectation.

The amine accepts a proton to form an ethylammonium propanoate pair. Amide synthesis requires appropriate activation or dehydrating conditions.

ORIGINAL PRACTICE · 3

Which is generally more reactive toward comparable nucleophilic acyl substitution: an acid chloride or an amide?

Reveal answer & reasoning

The acid chloride.

It has a better departing chloride group and less stabilizing nitrogen-to-carbonyl resonance than the amide.

ORIGINAL PRACTICE · 4

Hydrolysis of CH₃CH₂CH₂C≡N gives which acid after acidic workup?

Reveal answer & reasoning

Butanoic acid.

Count the nitrile carbon as the fourth carbon; it becomes the carboxyl carbon. Hydrolysis does not remove it.

ORIGINAL PRACTICE · 5

Under the usual Fischer mechanism, where does the oxygen of CH₃¹⁸OH appear in a methyl ester?

Reveal answer & reasoning

In the single-bonded –¹⁸OCH₃ group.

The alcohol oxygen attacks the acyl carbon and remains as the ester OR oxygen, assuming no later isotope scrambling.

ORIGINAL PRACTICE · 6

What amine results from complete strong-hydride reduction of CH₃CH₂CONHCH₃?

Reveal answer & reasoning

CH₃CH₂CH₂NHCH₃.

The amide C=O becomes CH₂ while the C–N bond and N-methyl group remain. The product retains all four carbons.

ORIGINAL PRACTICE · 7

Can 2,2-dimethylpropanoic acid undergo the ordinary alpha-H-dependent HVZ reaction?

Reveal answer & reasoning

No.

Its alpha carbon is bonded to three methyl groups and the carboxyl carbon and has no hydrogen to remove in the required enolization step.

ORIGINAL PRACTICE · 8

A 0.150 mol sample of a pure beta-keto monocarboxylic acid completely loses one CO₂ per molecule. How much CO₂ forms?

Reveal answer & reasoning

0.150 mol, or 6.60 g using M(CO₂)=44.0 g mol⁻¹.

The stated decarboxylation stoichiometry is one-to-one. Multiply 0.150 by 44.0 for the mass.

KEEP THE BIG IDEAS

Quick revision

  • Acyl derivatives differ by the group bonded directly to the carbonyl carbon.
  • Nucleophilic acyl substitution proceeds through addition and elimination with medium-dependent proton transfers.
  • Carboxylic acid–base chemistry can dominate over the intended carbonyl reaction.
  • Hydrolysis and reduction products must be assigned with carbon accounting and workup conditions.
  • Esterification equilibria, selective reductions and decarboxylations have explicit limits.
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