Aldehydes, Ketones and Carboxylic Acids — JEE Mains Formulas and Traps

Preparation of Aldehydes and Ketones

Named routes to aldehydes and ketones

Name	Starting compound	Reagents	Product
Rosenmund reduction	Acyl chloride $\mathrm{RCOCl}$	$\mathrm{H_2}$ , Pd– $\mathrm{BaSO_4}$ (poisoned)	Aldehyde $\mathrm{RCHO}$
Stephen reduction	Nitrile $\mathrm{RC{\equiv}N}$	$\mathrm{SnCl_2}$ , HCl, then $\mathrm{H_3O^+}$	Aldehyde $\mathrm{RCHO}$ , through the imine $\mathrm{RCH{=}NH}$
Etard reaction	Toluene	$\mathrm{CrO_2Cl_2}$ in $\mathrm{CS_2}$ , then $\mathrm{H_3O^+}$	Benzaldehyde, through $\mathrm{C_6H_5CH(OCrOHCl_2)_2}$
Chromic oxide oxidation	Toluene	$\mathrm{CrO_3}$ in $\mathrm{(CH_3CO)_2O}$ , 273–283 K, then $\mathrm{H_3O^+}$	Benzaldehyde, through benzylidene diacetate
Gattermann–Koch reaction	Benzene	CO, HCl, anhydrous $\mathrm{AlCl_3}$ and CuCl	Benzaldehyde
Friedel–Crafts acylation	Benzene	$\mathrm{RCOCl}$ , anhydrous $\mathrm{AlCl_3}$	Aryl ketone $\mathrm{C_6H_5COR}$ ; with $\mathrm{C_6H_5COCl}$ , benzophenone
Dialkylcadmium route	Acyl chloride $\mathrm{RCOCl}$	$\mathrm{R'_2Cd}$	Ketone $\mathrm{RCOR'}$

Rosenmund starts from an acyl chloride, Stephen from a nitrile, Etard from toluene and Gattermann–Koch from benzene.

Reagents that stop at the aldehyde or ketone

Reagent	Acts on	Stops at
PCC in $\mathrm{CH_2Cl_2}$	1° alcohol $\mathrm{RCH_2OH}$	Aldehyde $\mathrm{RCHO}$
$\mathrm{CrO_3}$ – $\mathrm{H_2SO_4}$ (Jones) or $\mathrm{K_2Cr_2O_7}$ – $\mathrm{H_2SO_4}$	1° alcohol; 2° alcohol	Carboxylic acid $\mathrm{RCOOH}$ ; ketone
Hot $\mathrm{KMnO_4}$	1° alcohol or aldehyde	Carboxylic acid
Cu at 573 K	1° or 2° alcohol vapour	Aldehyde or ketone (dehydrogenation)
DIBAL-H at low temperature, then $\mathrm{H_2O}$	Ester $\mathrm{RCOOR'}$ or nitrile $\mathrm{RCN}$	Aldehyde $\mathrm{RCHO}$
$\mathrm{LiAlH_4}$ , then $\mathrm{H_3O^+}$	Ester $\mathrm{RCOOR'}$	Two alcohols, $\mathrm{RCH_2OH}$ and $\mathrm{R'OH}$
Dilute $\mathrm{H_2SO_4}$ , water	Ester $\mathrm{RCOOR'}$	Acid $\mathrm{RCOOH}$ and alcohol $\mathrm{R'OH}$ (hydrolysis)
$\mathrm{BH_3}$ ; $\mathrm{H_2O_2}$ , $\mathrm{OH^-}$ ; then PCC	Terminal alkene $\mathrm{RCH{=}CH_2}$	Aldehyde $\mathrm{RCH_2CHO}$
MnO at about 573 K	Benzoic acid vapour	Benzaldehyde, in one step

PCC and DIBAL-H are the two reagents built to stop at the aldehyde.

Carbonyl compounds from alkynes, gem-dihalides and alkenes

Starting compound	Reagents	Product
Ethyne $\mathrm{HC{\equiv}CH}$	$\mathrm{H_2O}$ , $\mathrm{HgSO_4}$ , dilute $\mathrm{H_2SO_4}$	Ethanal $\mathrm{CH_3CHO}$
Terminal alkyne $\mathrm{RC{\equiv}CH}$	$\mathrm{H_2O}$ , $\mathrm{HgSO_4}$ , dilute $\mathrm{H_2SO_4}$	Methyl ketone $\mathrm{RCOCH_3}$
Terminal gem-dihalide $\mathrm{RCHCl_2}$	Aqueous KOH (hydrolysis)	Aldehyde $\mathrm{RCHO}$
Internal gem-dihalide $\mathrm{RCCl_2R'}$	Aqueous KOH (hydrolysis)	Ketone $\mathrm{RCOR'}$
Toluene	$\mathrm{Cl_2}$ and light, then water at 373 K	Benzaldehyde, through $\mathrm{C_6H_5CHCl_2}$
Alkene	$\mathrm{O_3}$ , then Zn and water	Aldehydes or ketones, one from each end of the C=C
Alkene $\mathrm{RCH{=}CH_2}$	CO and $\mathrm{H_2}$ , cobalt or rhodium catalyst	Aldehyde $\mathrm{RCH_2CH_2CHO}$ , one carbon longer
Methane	$\mathrm{O_2}$ over a molybdenum oxide catalyst, heat	Methanal $\mathrm{HCHO}$

Only ethyne gives an aldehyde on hydration; an end-carbon gem-dihalide gives an aldehyde on hydrolysis.

Common traps

The Stephen reduction needs the water step

\mathrm{SnCl_2}

and HCl stop at the imine salt. Only hydrolysis with

\mathrm{H_3O^+}

turns it into the aldehyde, so a scheme without that step does not give RCHO yet.

Etard and Gattermann–Koch start from different rings

Etard oxidises a methyl group already on the ring (toluene). Gattermann–Koch adds a new CHO group to benzene itself. A match list often swaps these two.

PCC stops at the aldehyde; the Jones reagent does not

Both are chromium(VI) reagents. PCC is used without water and stops at RCHO. The Jones reagent is aqueous and takes a 1° alcohol on to RCOOH.

Hydroboration puts the oxygen on the end carbon

Acid-catalysed hydration or

\mathrm{HgSO_4}

hydration follows Markovnikov's rule and leads to a ketone. Only hydroboration–oxidation puts OH on the terminal carbon, so only that route leads to the aldehyde.

Alkyne hydration gives an aldehyde only from ethyne

Every other alkyne gives a ketone, because the OH goes to the more substituted carbon. Propanal, for example, cannot be made this way.

The position of the two halogens decides the product

Both halogens must sit on one carbon. On an end carbon they give an aldehyde, on a middle carbon a ketone. A 1,2-dihalide is not a gem-dihalide and does not give a carbonyl compound this way.

Nucleophilic Addition and Carbonyl Derivatives

Learn this subtopic in the notes

Reactivity towards nucleophilic addition

Order of reactivity towards nucleophiles

\mathrm{HCHO > RCHO > RCOR'} \qquad \mathrm{RCHO > ArCHO} \qquad \text{ring: EWG} > \text{H} > \text{EDG}

Cyanohydrins and what they turn into

Cyanohydrin, then hydrolysis or reduction

\mathrm{R_2C{=}O \xrightarrow{HCN,\ OH^-} R_2C(OH)CN} \qquad \xrightarrow{H_3O^+} \mathrm{R_2C(OH)COOH} \qquad \xrightarrow{LiAlH_4} \mathrm{R_2C(OH)CH_2NH_2}

Acetals, oximes, hydrazones, semicarbazones and enamines

Reagent	Product with a carbonyl compound	What to remember
One $\mathrm{R'OH}$ , dry HCl	Hemiacetal $\mathrm{R_2C(OH)OR'}$	Usually reverts; cyclic hemiacetals (sugars, lactols) are stable
Two $\mathrm{R'OH}$ , dry HCl	Acetal (from a ketone, a ketal) $\mathrm{R_2C(OR')_2}$	Stable to base; dilute acid gives the carbonyl back
Ethane-1,2-diol, dry HCl	Cyclic acetal (ethylene ketal)	Protects a C=O while another group reacts
Hydroxylamine $\mathrm{NH_2OH}$	Oxime $\mathrm{R_2C{=}NOH}$	An aldoxime loses water with $\mathrm{P_2O_5}$ to give a nitrile
Hydrazine $\mathrm{NH_2NH_2}$	Hydrazone $\mathrm{R_2C{=}NNH_2}$	First step of the Wolff–Kishner reduction
Phenylhydrazine $\mathrm{C_6H_5NHNH_2}$	Phenylhydrazone $\mathrm{R_2C{=}NNHC_6H_5}$	Crystalline; used to identify the carbonyl compound
2,4-Dinitrophenylhydrazine (2,4-DNP)	2,4-Dinitrophenylhydrazone	Yellow to orange precipitate: the test for any aldehyde or ketone
Semicarbazide $\mathrm{NH_2NHCONH_2}$	Semicarbazone $\mathrm{R_2C{=}NNHCONH_2}$	Bonds through the NH₂ of the NH–NH₂ end; the product keeps all three N
Primary amine $\mathrm{R'NH_2}$	Imine (Schiff base) $\mathrm{R_2C{=}NR'}$	The C=N carries the amine's R group
Secondary amine $\mathrm{R'_2NH}$	Enamine, C=C–NR′₂	Needs an α-hydrogen on the carbonyl compound

Every entry is addition to C=O followed by loss of water.

Common traps

A small donor group still slows addition

A para-methyl group donates electrons by hyperconjugation, so 4-methylbenzaldehyde is less reactive than benzaldehyde. Do not rank by size of the molecule; rank by the charge on the carbonyl carbon.

Aryl ketones are the slowest

An aryl ketone has both handicaps: two groups on the carbon and a ring in conjugation. It sits below every aldehyde and below simple dialkyl ketones.

HCN addition does not give an amine

The product of HCN with a carbonyl compound is the cyanohydrin. An amine appears only after a separate reduction of the CN group, for example with

\mathrm{LiAlH_4}

.

Racemic, not optically active

A cyanohydrin with a new stereocentre is formed as a 50:50 mixture of enantiomers. Unless a chiral reagent is used, the product shows no optical rotation.

Acetals survive base because alkoxide leaves badly

An assertion–reason item may say acetals are stable in base because alkoxide leaves easily. The reason is reversed: they are stable because alkoxide is a poor leaving group.

Which end of semicarbazide bonds

The product is

\mathrm{R_2C{=}N{-}NH{-}CO{-}NH_2}

. A structure written as

\mathrm{R_2C{=}N{-}CO{-}NH{-}NH_2}

has bonded through the amide nitrogen, which is not nucleophilic.

Grignard Reagents with Carbonyls and Nitriles

Learn this subtopic in the notes

Grignard addition to aldehydes, ketones and esters

Grignard addition, then hydrolysis

\mathrm{R'COR'' \xrightarrow{(i)\ RMgX,\ (ii)\ H_3O^+} R'R''C(OH)R} \qquad \mathrm{HCHO} \to 1^\circ,\ \mathrm{R'CHO} \to 2^\circ,\ \text{ketone or ester} \to 3^\circ

Grignard reagents with nitriles, carbon dioxide and water

One R group only: nitrile to ketone, CO₂ to acid

\mathrm{R'C{\equiv}N \xrightarrow{(i)\ RMgX,\ (ii)\ H_3O^+} R'COR} \qquad \mathrm{RMgX \xrightarrow{(i)\ CO_2,\ (ii)\ H_3O^+} RCOOH}

Common traps

Esters take two equivalents, plus one for each acidic H

Count the ester first (two), then add one for every OH, NH, COOH or terminal alkyne C–H in the molecule. The acidic H reacts first, before any addition.

A terminal alkyne is an acid to a Grignard

\mathrm{RC{\equiv}CH + CH_3MgBr \to RC{\equiv}CMgBr + CH_4}

. The reagent is consumed and methane is given off, even though no C=O has reacted.

The ketone appears only after water

Before hydrolysis the product is the imine salt, so the ketone never meets the Grignard reagent. That is why a nitrile, unlike an ester, gives a ketone and not a 3° alcohol.

Carbon dioxide adds a carbon

The acid from

\mathrm{RMgX}

and

\mathrm{CO_2}

has one more carbon than R. Propylmagnesium bromide gives butanoic acid, not propanoic acid.

Reductions: Clemmensen, Wolff-Kishner and Hydrides

Learn this subtopic in the notes

Clemmensen and Wolff–Kishner reductions

Feature of the substrate	Clemmensen: Zn-Hg, conc. HCl	Wolff–Kishner: NH₂NH₂, KOH, glycol, heat
Aldehyde or ketone C=O	Reduced to $\mathrm{CH_2}$	Reduced to $\mathrm{CH_2}$ , with loss of $\mathrm{N_2}$
Medium	Strongly acidic, aqueous	Strongly basic, about 470 K
Isolated C=C	Unchanged	Unchanged
COOH group	Unchanged	Unchanged (present as the carboxylate until acidified)
3° or benzylic OH	Dehydrated; avoid this method	Unchanged; use this method
C–Cl bond in the chain	Survives the acid; use this method	Substituted or eliminated by hot base; avoid this method
Amide $\mathrm{CONH_2}$	Hydrolysed to COOH by the hot acid	Hydrolysed to the carboxylate by the hot base

Same result on the C=O; the rest of the molecule decides the method.

How far LiAlH₄, NaBH₄ and DIBAL-H reduce each group

Group	LiAlH₄, then H₃O⁺	NaBH₄	DIBAL-H at low temperature, then H₂O
Aldehyde $\mathrm{RCHO}$	$\mathrm{RCH_2OH}$	$\mathrm{RCH_2OH}$	$\mathrm{RCH_2OH}$
Ketone $\mathrm{RCOR'}$	$\mathrm{RCH(OH)R'}$	$\mathrm{RCH(OH)R'}$	$\mathrm{RCH(OH)R'}$
Ester $\mathrm{RCOOR'}$	$\mathrm{RCH_2OH + R'OH}$	No reaction	$\mathrm{RCHO + R'OH}$
Lactone (cyclic ester)	Diol	No reaction	Hydroxy aldehyde (or its lactol)
Nitrile $\mathrm{RC{\equiv}N}$	$\mathrm{RCH_2NH_2}$	No reaction	$\mathrm{RCHO}$
Isolated C=C	Unchanged	Unchanged	Unchanged

NaBH₄ is the selective one; DIBAL-H is the one that stops at the aldehyde.

Common traps

Neither method stops at the alcohol

Clemmensen and Wolff–Kishner give

\mathrm{CH_2}

, not CHOH. An option showing the alcohol is the product of a hydride such as

\mathrm{NaBH_4}

, not of these reagents.

Choose the method by what else is in the molecule

A statement that a molecule 'can be reduced' by one of these methods is false if the medium destroys another group: acid dehydrates a 3° alcohol, hot base removes a chlorine.

NaBH₄ leaves esters, acids and amides alone

In a molecule with a ketone and an ester,

\mathrm{NaBH_4}

reduces the ketone only.

\mathrm{LiAlH_4}

would reduce both.

DIBAL-H must be cold

DIBAL-H stops at the aldehyde only at low temperature with one equivalent. Warm, or in excess, it reduces the ester on to the alcohol.

Oxidation and Identification Tests

Learn this subtopic in the notes

The iodoform test and the haloform reaction

Haloform reaction of a methyl ketone

\mathrm{RCOCH_3 + 3I_2 + 4NaOH \to RCOONa + CHI_3\downarrow + 3NaI + 3H_2O}

Tollens', Fehling's and the 2,4-DNP test

Test and reagent	Positive sign	Positive for	Negative for
Tollens': $\mathrm{[Ag(NH_3)_2]^+}$ , $\mathrm{OH^-}$	Silver mirror	All aldehydes, aliphatic and aromatic; methanoic acid; α-hydroxy ketones; reducing sugars	Simple ketones; carboxylic acids other than methanoic acid
Fehling's: $\mathrm{Cu^{2+}}$ , tartrate, NaOH	Red-brown precipitate of $\mathrm{Cu_2O}$	Aliphatic aldehydes; α-hydroxy ketones such as fructose	Aromatic aldehydes; simple ketones
Benedict's: $\mathrm{Cu^{2+}}$ , citrate, $\mathrm{Na_2CO_3}$	Red-brown precipitate of $\mathrm{Cu_2O}$	Aliphatic aldehydes; α-hydroxy ketones such as fructose	Aromatic aldehydes; simple ketones
2,4-DNP	Yellow, orange or red precipitate	Any aldehyde or ketone	Carboxylic acids, esters, amides, alcohols, ethers
Iodoform: $\mathrm{I_2}$ , NaOH	Yellow precipitate of $\mathrm{CHI_3}$	$\mathrm{CH_3CO{-}}$ on C or H; $\mathrm{CH_3CH(OH){-}}$	Ketones and alcohols without these groups; acetic acid and its esters
$\mathrm{NaHCO_3}$ solution	Effervescence of $\mathrm{CO_2}$	Carboxylic acids; picric acid	Aldehydes, ketones, alcohols, most phenols

Tollens' catches every aldehyde; Fehling's catches only aliphatic ones.

Common traps

Aromatic aldehydes fail Fehling's but pass Tollens'

Benzaldehyde and its ring-substituted relatives give a silver mirror but no red precipitate. Counting 'aldehydes' is not enough for a Fehling's count.

2,4-DNP does not separate aldehydes from ketones

Both give the orange precipitate. The test only shows that a C=O of an aldehyde or ketone is present.

Acetic acid and its esters are negative

They contain

\mathrm{CH_3CO}

, but it is joined to oxygen. The test needs

\mathrm{CH_3CO}

joined to carbon or hydrogen.

Ethanol and ethanal are the only positives in their classes

Ethanol is the only 1° alcohol that gives the test and ethanal the only aldehyde. Methanol and methanal are both negative.

Enols and Aldol Condensation

Learn this subtopic in the notes

Acidity of α-hydrogens and enol content

Acidity of the α-hydrogen

\mathrm{RCOCH_2COR > RCOCH_2COOR' > R'OOCCH_2COOR' > RCOCH_3}

Self-aldol condensation: predicting the product

Aldol addition, then dehydration

\mathrm{2\,RCH_2CHO \xrightarrow{dil.\ OH^-} RCH_2CH(OH)CH(R)CHO \xrightarrow{\Delta,\ -H_2O} RCH_2CH{=}C(R)CHO}

Intramolecular aldol: which ring closes

Ring size

\text{ring atoms} = \text{enolate carbon to attacked carbonyl carbon, both counted}; \quad 5 \text{ or } 6 \text{ wins}

Common traps

The most acidic H sits between two C=O groups

In a 1,3-dicarbonyl compound the

\mathrm{CH_2}

between the carbonyls is far more acidic than a terminal

\mathrm{CH_3}

next to only one. In a 1,4- or 1,5-diketone no carbon has two C=O neighbours.

An ester group helps less than a ketone group

Replacing one ketone of a β-diketone by an ester lowers the acidity by about two pKa units, and replacing both lowers it further.

Number the product from the new chain

Join the α-carbon of one molecule to the carbonyl carbon of the other, draw the whole chain, then number from the C=O that survives. Naming each half separately gives wrong locants.

No α-hydrogen left, no dehydration

Water leaves from the OH and an α-hydrogen. If the α-carbon of the aldol carries two alkyl groups and the C=O, the aldol cannot condense further.

Count atoms in the ring, not bonds

Include both the enolate carbon and the carbonyl carbon. Miscounting by one turns a five-membered ring into a 'four' or a 'six' and sends you to the wrong option.

Pick the enolate that makes a five- or six-membered ring

A diketone usually has two or more α-carbons. Try each, count the ring, and keep only the one that gives five or six atoms.

Crossed Aldol and Cannizzaro Reactions

Learn this subtopic in the notes

Crossed aldol: counting and naming the products

Number of aldol products (ignoring stereoisomers)

N = (\text{partners with an }\alpha\text{-H}) \times (\text{all partners})

Cannizzaro reaction

Cannizzaro and crossed Cannizzaro

\mathrm{2\,ArCHO \xrightarrow{conc.\ OH^-} ArCOO^- + ArCH_2OH} \qquad \mathrm{HCHO + ArCHO \xrightarrow{conc.\ OH^-} HCOO^- + ArCH_2OH}

Common traps

Name each crossed product by which partner is the enolate

The enolate supplies the α-carbon, and the acceptor supplies the carbon that ends up doubly bonded to it. Swapping roles gives a different product, so write both crossed pairs separately.

A partner with no α-hydrogen is never the enolate

Aromatic aldehydes and methanal can only be attacked. Any option that needs them as the enolate is not formed.

Concentrated alkali, not dilute

Dilute base with an aldehyde that has α-hydrogens gives an aldol. The Cannizzaro reaction needs concentrated alkali and an aldehyde with no α-hydrogen.

The transferred hydrogen comes from carbon

Run in

\mathrm{D_2O}

with NaOD, the alcohol is

\mathrm{ArCH_2OD}

: the

\mathrm{CH_2}

keeps both hydrogens, because the hydride came from the other aldehyde, not from the solvent.

Carboxylic Acids: Acidity and Reactions

Learn this subtopic in the notes

Ranking the strength of carboxylic acids

pKa of some acids (lower pKa, stronger acid)

\mathrm{CF_3COOH}\ (0.23) < \mathrm{CCl_3COOH}\ (0.65) < \mathrm{ClCH_2COOH}\ (2.86) < \mathrm{HCOOH}\ (3.75) < \mathrm{C_6H_5COOH}\ (4.19) < \mathrm{CH_3COOH}\ (4.76)

Routes that end at a carboxylic acid

Starting compound	Reagents	Product
1° alcohol $\mathrm{RCH_2OH}$	Alkaline $\mathrm{KMnO_4}$ , then $\mathrm{H_3O^+}$ ; or Jones reagent	$\mathrm{RCOOH}$ , same carbons
Aldehyde $\mathrm{RCHO}$	Tollens' reagent, $\mathrm{K_2Cr_2O_7/H^+}$ or bromine water	$\mathrm{RCOOH}$ , same carbons
Alkylbenzene with a benzylic H	Hot alkaline $\mathrm{KMnO_4}$ , then $\mathrm{H_3O^+}$	Benzoic acid, whatever the chain length
Nitrile $\mathrm{RCN}$	$\mathrm{H_3O^+}$ and heat (or $\mathrm{OH^-}$ , then acid)	$\mathrm{RCOOH}$ , through the amide $\mathrm{RCONH_2}$
Grignard reagent $\mathrm{RMgX}$	Dry ice $\mathrm{CO_2}$ , then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , one carbon more
Methyl ketone $\mathrm{RCOCH_3}$	$\mathrm{I_2}$ and NaOH, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , one carbon fewer, and $\mathrm{CHI_3}$
1,1,1-Trihalide $\mathrm{RCCl_3}$	Aqueous KOH, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ , same carbons
Ester, acid chloride or anhydride	Water with acid or alkali, then $\mathrm{H_3O^+}$	$\mathrm{RCOOH}$ (an ester also gives the alcohol)

Grignard plus CO₂ adds a carbon; the haloform reaction removes one; the rest keep the count.

Reactions of carboxylic acids and their derivatives

Reagent	Product from RCOOH	Remember
$\mathrm{NaHCO_3}$ solution	$\mathrm{RCOONa + CO_2 + H_2O}$	Effervescence separates acids from phenols
$\mathrm{R'OH}$ , conc. $\mathrm{H_2SO_4}$ , heat	Ester $\mathrm{RCOOR'}$	Reversible; nucleophilic acyl substitution
$\mathrm{SOCl_2}$ (or $\mathrm{PCl_5}$ , $\mathrm{PCl_3}$ )	Acid chloride $\mathrm{RCOCl}$	With $\mathrm{SOCl_2}$ the by-products $\mathrm{SO_2}$ and HCl are gases
$\mathrm{P_2O_5}$ , heat; or heat alone for a suitable diacid	Anhydride $\mathrm{(RCO)_2O}$	cis-Butenedioic (maleic) acid gives a cyclic anhydride on heating; the trans acid cannot
$\mathrm{NH_3}$ , then heat	Amide $\mathrm{RCONH_2}$	Through the ammonium salt $\mathrm{RCOONH_4}$
$\mathrm{LiAlH_4}$ or $\mathrm{B_2H_6}$ , then $\mathrm{H_3O^+}$	1° alcohol $\mathrm{RCH_2OH}$	$\mathrm{NaBH_4}$ does not reduce COOH
Sodium salt with NaOH and CaO (soda lime), heat	Alkane $\mathrm{RH}$	Decarboxylation: one carbon fewer
Electrolysis of the sodium salt (Kolbe)	Alkane $\mathrm{R{-}R}$	Two R groups join
$\mathrm{X_2}$ and red phosphorus, then water (Hell–Volhard–Zelinsky)	α-Halo acid $\mathrm{RCH(X)COOH}$	Only the α-carbon is halogenated; it needs an α-hydrogen
Conc. $\mathrm{HNO_3}$ and conc. $\mathrm{H_2SO_4}$ (on benzoic acid)	3-Nitrobenzoic acid	COOH is meta-directing and deactivating

The first six change only the COOH group; soda lime and Kolbe remove it; HVZ acts at the α-carbon and nitration on the ring.

Common traps

Distance weakens the inductive effect

A halogen two or three carbons away from COOH has a small effect. Rank by position before counting halogens along a chain.

Picric acid behaves like a carboxylic acid with NaHCO₃

Three nitro groups make 2,4,6-trinitrophenol a strong acid, so it releases

\mathrm{CO_2}

from

\mathrm{NaHCO_3}

. Other phenols do not.

Mild hydrolysis of a nitrile stops at the amide

A nitrile needs vigorous hydrolysis to reach the acid. Under mild conditions the product is

\mathrm{RCONH_2}

, not RCOOH.

Count the carbons

Grignard carboxylation adds one carbon and the haloform reaction removes one. A route that gives the right functional group but the wrong chain length is the wrong answer.

HVZ halogenates only the α-carbon

The halogen goes to the carbon next to COOH, never further along the chain. An acid with no α-hydrogen, such as 2,2-dimethylpropanoic acid, does not react.

Soda lime removes a carbon

Decarboxylation of

\mathrm{RCOONa}

gives

\mathrm{RH}

, with one carbon fewer than the acid. Sodium propanoate gives ethane, not propane.

Aldehydes, Ketones and Carboxylic Acids formulas

Preparation of Aldehydes and Ketones

Named routes to aldehydes and ketones

Reagents that stop at the aldehyde or ketone

Carbonyl compounds from alkynes, gem-dihalides and alkenes

Common traps

Nucleophilic Addition and Carbonyl Derivatives

Reactivity towards nucleophilic addition

Cyanohydrins and what they turn into

Acetals, oximes, hydrazones, semicarbazones and enamines

Common traps

Grignard Reagents with Carbonyls and Nitriles

Grignard addition to aldehydes, ketones and esters

Grignard reagents with nitriles, carbon dioxide and water

Common traps

Reductions: Clemmensen, Wolff-Kishner and Hydrides

Clemmensen and Wolff–Kishner reductions

How far LiAlH₄, NaBH₄ and DIBAL-H reduce each group

Common traps

Oxidation and Identification Tests

The iodoform test and the haloform reaction

Tollens', Fehling's and the 2,4-DNP test

Common traps

Enols and Aldol Condensation

Acidity of α-hydrogens and enol content

Self-aldol condensation: predicting the product

Intramolecular aldol: which ring closes

Common traps

Crossed Aldol and Cannizzaro Reactions

Crossed aldol: counting and naming the products

Cannizzaro reaction

Common traps

Carboxylic Acids: Acidity and Reactions

Ranking the strength of carboxylic acids

Routes that end at a carboxylic acid

Reactions of carboxylic acids and their derivatives

Common traps

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