Chapter 8: Aldehydes, Ketones and Carboxylic Acids

Propionaldehyde (propanal) is $$\mathrm{CH_3-CH_2-CHO}$$. The carbon directly attached to the $$\mathrm{-CHO}$$ group is the $$\alpha$$-carbon. The prefix $$\alpha$$-methoxy places a methoxy group ($$\mathrm{-OCH_3}$$) on this $$\alpha$$-carbon.

Hence the structure is

$$\mathrm{CH_3-CH(OCH_3)-CHO}$$

(IUPAC name: 2-methoxypropanal)

Butanal is the four-carbon aldehyde $$\mathrm{\overset{4}{C}H_3-\overset{3}{C}H_2-\overset{2}{C}H_2-\overset{1}{C}HO}$$, the aldehyde carbon being C-1. The prefix 3-hydroxy places an $$\mathrm{-OH}$$ group on C-3.

Hence the structure is

$$\mathrm{CH_3-CH(OH)-CH_2-CHO}$$

In a cycloalkane carbaldehyde, the ring carbon bearing the $$\mathrm{-CHO}$$ group is numbered C-1. The prefix 2-hydroxy places an $$\mathrm{-OH}$$ group on the adjacent ring carbon, C-2.

To draw it: take a five-membered cyclopentane ring; attach a $$\mathrm{-CHO}$$ group to one ring carbon (C-1) and an $$\mathrm{-OH}$$ group to the neighbouring ring carbon (C-2).

Pentanal is the five-carbon aldehyde $$\mathrm{\overset{1}{C}HO-\overset{2}{C}H_2-\overset{3}{C}H_2-\overset{4}{C}H_2-\overset{5}{C}H_3}$$. The prefix 4-oxo indicates a keto group ($$\mathrm{>C=O}$$) at C-4.

Hence the structure is

$$\mathrm{CH_3-CO-CH_2-CH_2-CHO}$$

A secondary-butyl (sec-butyl) group is $$\mathrm{-CH(CH_3)CH_2CH_3}$$, in which the point of attachment is a secondary carbon. In di-sec-butyl ketone two such groups are joined to a carbonyl ($$\mathrm{>C=O}$$) carbon.

Hence the structure is

$$\mathrm{CH_3CH_2-CH(CH_3)-CO-CH(CH_3)-CH_2CH_3}$$

(IUPAC name: 3,5-dimethylheptan-4-one)

Acetophenone is methyl phenyl ketone, $$\mathrm{C_6H_5-CO-CH_3}$$. The ring carbon bearing the $$\mathrm{-COCH_3}$$ group is taken as C-1, so '4-fluoro' places a fluorine atom at the para (C-4) position.

Hence the structure is a benzene ring carrying a $$\mathrm{-COCH_3}$$ group at one position and an $$\mathrm{-F}$$ atom para to it:

$$\mathrm{p\text{-}F\text{-}C_6H_4\text{-}COCH_3}$$

This is a Friedel-Crafts acylation. In the presence of anhydrous $$\mathrm{AlCl_3}$$ (a Lewis acid catalyst, with $$\mathrm{CS_2}$$ as the solvent), the acyl chloride $$\mathrm{C_2H_5COCl}$$ (propanoyl chloride) introduces an acyl group ($$\mathrm{-COC_2H_5}$$) onto the benzene ring.

$$\mathrm{C_6H_6 + C_2H_5COCl \xrightarrow{Anhyd.\ AlCl_3,\ CS_2} C_6H_5-CO-C_2H_5 + HCl}$$

The organic product is ethyl phenyl ketone (propiophenone, 1-phenylpropan-1-one).

Dialkylcadmium reagents react with acyl chlorides to give ketones. Here dibenzylcadmium supplies two benzyl groups, each of which couples with the acetyl group of acetyl chloride.

$$\mathrm{(C_6H_5CH_2)_2Cd + 2\,CH_3COCl \rightarrow 2\,C_6H_5CH_2-CO-CH_3 + CdCl_2}$$

The organic product is benzyl methyl ketone (1-phenylpropan-2-one).

Addition of water to an alkyne in the presence of $$\mathrm{Hg^{2+}/H_2SO_4}$$ follows Markovnikov's rule, giving an unstable enol that tautomerises to a carbonyl compound. For propyne the $$\mathrm{-OH}$$ adds to the more substituted (internal) carbon.

$$\mathrm{H_3C-C\equiv C-H + H_2O \xrightarrow{Hg^{2+},\ H_2SO_4} \left[H_3C-\underset{\underset{\textstyle OH}{|}}{C}=CH_2\right] \rightarrow CH_3-CO-CH_3}$$

Only ethyne gives an aldehyde (acetaldehyde) on hydration; all other alkynes give ketones. Hence the product is propanone (acetone).

Chromyl chloride ($$\mathrm{CrO_2Cl_2}$$) oxidises a ring methyl group to a chromium complex which, on hydrolysis with $$\mathrm{H_3O^+}$$, gives the corresponding aldehyde. This is the Étard reaction; the oxidation stops cleanly at the aldehyde stage.

$$\mathrm{p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3 \xrightarrow[2.\ H_3O^+]{1.\ CrO_2Cl_2} p\text{-}O_2N\text{-}C_6H_4\text{-}CHO}$$

The product is p-nitrobenzaldehyde.

In a nucleophilic addition reaction the nucleophile attacks the electrophilic carbonyl carbon. The reactivity is lowered by two factors: (a) the +I (electron-releasing) effect of alkyl groups, which reduces the positive charge on the carbonyl carbon, and (b) the steric crowding of bulky groups, which hinders the approach of the nucleophile.

Aldehydes carry only one alkyl group on the carbonyl carbon while ketones carry two; hence aldehydes are more reactive than ketones. Among the aldehydes, ethanal ($$\mathrm{CH_3CHO}$$) is more reactive than propanal ($$\mathrm{CH_3CH_2CHO}$$), since the larger ethyl group exerts a greater +I effect and more steric hindrance than methyl. Among the ketones, propanone ($$\mathrm{CH_3COCH_3}$$) is more reactive than butanone ($$\mathrm{CH_3COCH_2CH_3}$$), which carries an additional electron-releasing, bulkier ethyl group.

Hence the increasing order of reactivity in nucleophilic addition is:

$$\mathrm{Butanone < Propanone < Propanal < Ethanal}$$

Acetophenone ($$\mathrm{C_6H_5COCH_3}$$) is a ketone — it bears two groups on the carbonyl carbon, so it is the least reactive of the four.

The other three are aromatic aldehydes; their reactivity is decided by the ring substituent. An electron-withdrawing group raises the positive charge on the carbonyl carbon (more reactive), while an electron-releasing group lowers it (less reactive).

• In $$p$$-nitrobenzaldehyde the $$\mathrm{-NO_2}$$ group is strongly electron-withdrawing ($$-I$$ and $$-R$$), making it the most reactive.
• In $$p$$-tolualdehyde the $$\mathrm{-CH_3}$$ group is electron-releasing ($$+I$$ and hyperconjugation), so it is less reactive than benzaldehyde.
• Benzaldehyde, having no ring substituent, lies between the two.

Hence the increasing order of reactivity is:

$$\mathrm{Acetophenone < p\text{-}Tolualdehyde < Benzaldehyde < p\text{-}Nitrobenzaldehyde}$$

Hydroxylamine ($$\mathrm{H_2N-OH}$$) is an ammonia derivative. It adds to the carbonyl group of the ketone and then eliminates a molecule of water (a nucleophilic addition-elimination reaction), so that the $$\mathrm{>C=O}$$ group is replaced by $$\mathrm{>C=N-OH}$$. The product is an oxime, and a trace of acid catalyses the reaction.

$$\mathrm{Cyclopentanone + H_2N-OH \xrightarrow{H^+} Cyclopentanone\ oxime + H_2O}$$

The product is cyclopentanone oxime — the cyclopentane ring in which the ring carbon now carries a $$\mathrm{=N-OH}$$ group in place of $$\mathrm{=O}$$.

2,4-Dinitrophenylhydrazine ($$\mathrm{H_2N-NH-C_6H_3(NO_2)_2}$$) reacts with the carbonyl group of the ketone by nucleophilic addition followed by elimination of water. The $$\mathrm{>C=O}$$ group is converted into $$\mathrm{>C=N-NH-C_6H_3(NO_2)_2}$$.

$$\mathrm{Cyclohexanone + H_2N-NH-C_6H_3(NO_2)_2 \rightarrow Cyclohexanone\ 2,4\text{-}dinitrophenylhydrazone + H_2O}$$

The product, cyclohexanone 2,4-dinitrophenylhydrazone, separates as an orange-red solid (this is the basis of the 2,4-DNP test for carbonyl compounds).

The reagent $$\mathrm{H_2N{-}CO{-}NH{-}NH_2}$$ is semicarbazide. It reacts with the carbonyl ($$\mathrm{C{=}O}$$) group only — the carbon-carbon double bond ($$\mathrm{C{=}C}$$) of the $$\alpha,\beta$$-unsaturated aldehyde is not attacked. By nucleophilic addition-elimination the $$\mathrm{-CHO}$$ group is converted into $$\mathrm{-CH=N-NH-CO-NH_2}$$ with loss of water.

$$\mathrm{R-CH=CH-CHO + H_2N{-}CO{-}NH{-}NH_2 \xrightarrow{H^+} R-CH=CH-CH=N-NH-CO-NH_2 + H_2O}$$

The product is the semicarbazone of the unsaturated aldehyde; the carbon-carbon double bond remains intact.

A primary amine ($$\mathrm{CH_3CH_2NH_2}$$, ethanamine) adds to the carbonyl group of the ketone acetophenone and then loses a molecule of water. The $$\mathrm{>C=O}$$ group is replaced by $$\mathrm{>C=N-R}$$, giving a substituted imine (a Schiff's base). The reaction is catalysed by a trace of acid.

$$\mathrm{C_6H_5COCH_3 + CH_3CH_2NH_2 \xrightarrow{H^+} C_6H_5-C(CH_3)=N-CH_2CH_3 + H_2O}$$

The product is $$N$$-ethyl-1-phenylethanimine (a Schiff's base).

The compound is $$\mathrm{C_6H_5-CH_2-CH_2-COOH}$$. The $$\mathrm{-COOH}$$ group is the principal characteristic group, and the carboxyl carbon is C-1. The longest chain bearing $$\mathrm{-COOH}$$ has three carbons: $$\mathrm{\overset{1}{C}OOH-\overset{2}{C}H_2-\overset{3}{C}H_2-}$$, so the parent acid is propanoic acid.

A phenyl group is attached to C-3. Hence the name is 3-phenylpropanoic acid.

The compound is $$\mathrm{(CH_3)_2C=CH-COOH}$$. Numbering from the carboxyl carbon: $$\mathrm{\overset{1}{C}OOH-\overset{2}{C}H=\overset{3}{C}(CH_3)-CH_3}$$. The longest chain through $$\mathrm{-COOH}$$ contains four carbons (parent: butenoic acid), with the double bond between C-2 and C-3 and a methyl branch on C-3.

Hence the name is 3-methylbut-2-enoic acid.

When a $$\mathrm{-COOH}$$ group is attached directly to a ring, the compound is named as a cycloalkanecarboxylic acid, and the ring carbon bearing $$\mathrm{-COOH}$$ is numbered C-1.

The methyl group sits on the adjacent ring carbon, C-2. Hence the name is 2-methylcyclopentane-1-carboxylic acid (commonly written 2-methylcyclopentanecarboxylic acid).

A $$\mathrm{-COOH}$$ group on a benzene ring makes the compound a benzoic acid, and the ring carbon bearing $$\mathrm{-COOH}$$ is C-1. The three nitro groups occupy positions 2, 4 and 6.

Hence the name is 2,4,6-trinitrobenzoic acid.

An alkyl side-chain on a benzene ring is oxidised to a $$\mathrm{-COOH}$$ group (irrespective of the length of the chain) by a hot, strong oxidising agent such as alkaline $$\mathrm{KMnO_4}$$. The whole ethyl group is degraded to a single carboxyl carbon.

$$\mathrm{C_6H_5CH_2CH_3 \xrightarrow[\Delta]{KMnO_4,\ KOH} C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH}$$

Acidification of the potassium benzoate formed gives benzoic acid.

Acetophenone, $$\mathrm{C_6H_5COCH_3}$$, is a methyl ketone. On vigorous oxidation with hot alkaline $$\mathrm{KMnO_4}$$ the $$\mathrm{-COCH_3}$$ group is degraded to a $$\mathrm{-COOH}$$ group:

$$\mathrm{C_6H_5COCH_3 \xrightarrow[\Delta]{KMnO_4,\ KOH} C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH}$$

Alternatively, being a methyl ketone, acetophenone undergoes the haloform (iodoform) reaction with $$\mathrm{NaOI}$$ (or $$\mathrm{NaOX}$$): $$\mathrm{C_6H_5COCH_3 + 3NaOI \rightarrow C_6H_5COONa + CHI_3 + 2NaOH}$$, and acidification of sodium benzoate again gives benzoic acid.

Bromobenzene is first converted into its Grignard reagent, which is then treated with carbon dioxide (solid $$\mathrm{CO_2}$$, dry ice). The Grignard reagent adds to $$\mathrm{CO_2}$$ to give a salt; acidic hydrolysis then liberates the carboxylic acid.

$$\mathrm{C_6H_5Br \xrightarrow{Mg,\ dry\ ether} C_6H_5MgBr}$$

$$\mathrm{C_6H_5MgBr \xrightarrow{CO_2} C_6H_5COOMgBr \xrightarrow{H_3O^+} C_6H_5COOH}$$

Styrene, $$\mathrm{C_6H_5-CH=CH_2}$$, has an unsaturated side chain. Hot acidic (or alkaline) $$\mathrm{KMnO_4}$$ oxidatively cleaves the side-chain double bond; the terminal $$\mathrm{=CH_2}$$ carbon is lost as $$\mathrm{CO_2}$$ and the benzylic carbon is oxidised to $$\mathrm{-COOH}$$.

$$\mathrm{C_6H_5-CH=CH_2 \xrightarrow[\Delta]{KMnO_4,\ H^+} C_6H_5COOH + CO_2 + H_2O}$$

The strength of a carboxylic acid is measured by the stability of its conjugate base (the carboxylate ion). An electron-withdrawing group disperses the negative charge of the carboxylate, stabilises it, and so increases acidity.

In $$\mathrm{CH_3CO_2H}$$ the methyl group is weakly electron-releasing ($$+I$$). In $$\mathrm{CH_2FCO_2H}$$ (fluoroacetic acid) the fluorine atom is strongly electron-withdrawing ($$-I$$), which stabilises the carboxylate ion.

Hence $$\mathrm{CH_2FCO_2H}$$ is the stronger acid.

Both molecules have one halogen on the $$\alpha$$-carbon, so the comparison rests on the magnitude of the $$-I$$ (inductive) effect of the halogen.

Fluorine is more electronegative than chlorine, so it exerts a stronger $$-I$$ effect. It therefore withdraws electron density more effectively and stabilises the carboxylate ion to a greater extent.

Hence $$\mathrm{CH_2FCO_2H}$$ is the stronger acid.

Both acids contain one fluorine atom, so what matters is how far the fluorine is from the $$\mathrm{-COOH}$$ group. The inductive ($$-I$$) effect falls off rapidly with distance, so a halogen nearer the carboxyl group stabilises the carboxylate ion more strongly.

Numbering from the carboxyl carbon (C-1):

• $$\mathrm{CH_2FCH_2CH_2CO_2H}$$: $$\mathrm{\overset{4}{C}H_2F-\overset{3}{C}H_2-\overset{2}{C}H_2-\overset{1}{C}O_2H}$$ — fluorine is on C-4 ($$\gamma$$-carbon).
• $$\mathrm{CH_3CHFCH_2CO_2H}$$: $$\mathrm{\overset{4}{C}H_3-\overset{3}{C}HF-\overset{2}{C}H_2-\overset{1}{C}O_2H}$$ — fluorine is on C-3 ($$\beta$$-carbon).

Fluorine is closer to $$\mathrm{-COOH}$$ in $$\mathrm{CH_3CHFCH_2CO_2H}$$, so its $$-I$$ effect is felt more strongly.

Hence $$\mathrm{CH_3CHFCH_2CO_2H}$$ is the stronger acid.

Both are para-substituted benzoic acids; the substituent decides the acid strength.

In $$p$$-$$\mathrm{F_3C-C_6H_4-COOH}$$ the $$\mathrm{-CF_3}$$ group is strongly electron-withdrawing ($$-I$$). It pulls electron density away from the carboxylate ion, stabilises it, and so increases acidity.

In $$p$$-$$\mathrm{H_3C-C_6H_4-COOH}$$ ($$p$$-toluic acid) the $$\mathrm{-CH_3}$$ group is electron-releasing ($$+I$$, hyperconjugation). It intensifies the negative charge on the carboxylate, destabilises it, and so decreases acidity.

Hence $$p$$-(trifluoromethyl)benzoic acid, $$p$$-$$\mathrm{F_3C-C_6H_4-COOH}$$, is the stronger acid.

This is the conversion of a primary alcohol into an aldehyde. An ordinary oxidising agent would carry the oxidation further to the carboxylic acid; a mild, selective oxidant is needed that stops at the aldehyde stage.

The reagent is pyridinium chlorochromate (PCC), $$\mathrm{C_5H_5N{\cdot}HCrO_3Cl}$$ (a complex of $$\mathrm{CrO_3}$$ with pyridine and HCl), used in dichloromethane.

$$\mathrm{CH_3(CH_2)_4CH_2OH \xrightarrow{PCC} CH_3(CH_2)_4CHO}$$

Cyclohexanol is a secondary alcohol; oxidation of a secondary alcohol gives a ketone, and the ketone is not oxidised further under ordinary conditions. So an ordinary oxidising agent may be used.

The reagent is anhydrous chromium trioxide ($$\mathrm{CrO_3}$$) (a chromium(VI) oxidant; acidified $$\mathrm{K_2Cr_2O_7}$$ also works).

$$\mathrm{Cyclohexanol \xrightarrow{CrO_3} Cyclohexanone}$$

A ring methyl group is to be oxidised to $$\mathrm{-CHO}$$ without going on to $$\mathrm{-COOH}$$. Two selective methods are used:

• $$\mathrm{CrO_3}$$ in the presence of acetic anhydride. The aldehyde is trapped as a geminal diacetate (which protects it from further oxidation) and is then released by hydrolysis.
• Chromyl chloride, $$\mathrm{CrO_2Cl_2}$$ (Étard reaction). The methyl group forms a chromium complex which is hydrolysed by $$\mathrm{H_2O}$$ to the aldehyde.

$$\mathrm{p\text{-}F\text{-}C_6H_4\text{-}CH_3 \xrightarrow[2.\ H_2O]{1.\ CrO_2Cl_2} p\text{-}F\text{-}C_6H_4\text{-}CHO}$$

A nitrile ($$\mathrm{CH_3C{\equiv}N}$$) is to be partially reduced to an aldehyde. A full reduction would give the amine, so a controlled reducing agent is needed that delivers only one hydride and stops at the imine stage, which hydrolyses to the aldehyde.

The reagent is diisobutylaluminium hydride (DIBAL-H), used at low temperature, followed by hydrolysis.

$$\mathrm{CH_3C{\equiv}N \xrightarrow[2.\ H_2O]{1.\ DIBAL\text{-}H} CH_3CHO}$$

Allyl alcohol, $$\mathrm{CH_2=CH-CH_2OH}$$, is a primary alcohol that also contains a $$\mathrm{C=C}$$ double bond. It is to be oxidised to propenal (acrolein), $$\mathrm{CH_2=CH-CHO}$$, without touching the double bond and without over-oxidation to the acid.

The reagent is pyridinium chlorochromate (PCC), which selectively oxidises the $$\mathrm{-CH_2OH}$$ group to $$\mathrm{-CHO}$$ and leaves the $$\mathrm{C=C}$$ bond intact.

$$\mathrm{CH_2=CH-CH_2OH \xrightarrow{PCC} CH_2=CH-CHO}$$

But-2-ene, $$\mathrm{CH_3-CH=CH-CH_3}$$, is symmetrical about the double bond. Cleaving the $$\mathrm{C=C}$$ bond and capping each carbon with an oxygen gives two molecules of ethanal.

This is done by ozonolysis: treatment with ozone ($$\mathrm{O_3}$$) to form an ozonide, followed by reductive work-up with zinc dust and water.

$$\mathrm{CH_3CH=CHCH_3 \xrightarrow[2.\ Zn,\ H_2O]{1.\ O_3} 2\,CH_3CHO}$$

Butan-1-ol is a primary alcohol. Vigorous oxidation of a primary alcohol gives a carboxylic acid with the same number of carbon atoms. Acidified potassium permanganate (or potassium dichromate) is used.

$$\mathrm{CH_3CH_2CH_2CH_2OH \xrightarrow[\Delta]{KMnO_4/H^+} CH_3CH_2CH_2COOH}$$

Phenylethanoic acid, $$\mathrm{C_6H_5CH_2COOH}$$, has one more carbon than benzyl alcohol, $$\mathrm{C_6H_5CH_2OH}$$. The extra carbon (as $$\mathrm{-COOH}$$) is introduced through a nitrile.

First the $$\mathrm{-OH}$$ group is replaced by $$\mathrm{-Cl}$$ using $$\mathrm{SOCl_2}$$; the chloride is displaced by cyanide; finally the nitrile is hydrolysed to the acid.

$$\mathrm{C_6H_5CH_2OH \xrightarrow{SOCl_2} C_6H_5CH_2Cl \xrightarrow{KCN} C_6H_5CH_2CN \xrightarrow[\Delta]{H_3O^+} C_6H_5CH_2COOH}$$

The straightforward Grignard route used in the other parts cannot be applied here. A Grignard reagent is a very strong nucleophile and base, whereas the nitro group ($$\mathrm{-NO_2}$$) is strongly electrophilic and easily reduced; the two are incompatible. The organomagnesium species attacks (and reduces) the nitro group, so an aryl Grignard reagent simply cannot be prepared from a nitro-substituted aryl halide.

Instead the bromine is replaced by a cyano group and the nitrile is then hydrolysed. Heating 3-nitrobromobenzene with cuprous cyanide ($$\mathrm{CuCN}$$) replaces $$\mathrm{-Br}$$ by $$\mathrm{-CN}$$; acidic hydrolysis of the resulting nitrile gives the carboxylic acid, the nitro group remaining untouched.

$$\mathrm{m\text{-}O_2N\text{-}C_6H_4\text{-}Br \xrightarrow[\Delta]{CuCN} m\text{-}O_2N\text{-}C_6H_4\text{-}CN}$$

$$\mathrm{m\text{-}O_2N\text{-}C_6H_4\text{-}CN \xrightarrow[\Delta]{H_3O^+} m\text{-}O_2N\text{-}C_6H_4\text{-}COOH + NH_4^+}$$

The product is 3-nitrobenzoic acid. (3-Nitrobenzoic acid is also conveniently prepared by direct nitration of benzoic acid, since the $$\mathrm{-COOH}$$ group is meta-directing and sends the incoming $$\mathrm{-NO_2}$$ mainly to the 3-position.)

4-Methylacetophenone, $$\mathrm{p\text{-}CH_3\text{-}C_6H_4\text{-}COCH_3}$$, carries two side groups on the ring: a methyl group and an acetyl group. On vigorous oxidation, both are degraded to $$\mathrm{-COOH}$$ groups.

$$\mathrm{p\text{-}CH_3\text{-}C_6H_4\text{-}COCH_3 \xrightarrow[\Delta]{KMnO_4,\ KOH} p\text{-}KOOC\text{-}C_6H_4\text{-}COOK \xrightarrow{H_3O^+} p\text{-}HOOC\text{-}C_6H_4\text{-}COOH}$$

The product is benzene-1,4-dicarboxylic acid (terephthalic acid).

Cyclohexene is a six-membered ring with one $$\mathrm{C=C}$$ double bond. Hot, concentrated $$\mathrm{KMnO_4}$$ cleaves the double bond oxidatively. Since both alkene carbons bear a hydrogen, each is oxidised to a $$\mathrm{-COOH}$$ group; the ring opens to give a straight-chain dicarboxylic acid.

$$\mathrm{Cyclohexene \xrightarrow[\Delta]{KMnO_4\ (conc.)} HOOC-CH_2-CH_2-CH_2-CH_2-COOH}$$

The product is hexane-1,6-dioic acid (adipic acid).

An aldehyde is very easily oxidised to the corresponding carboxylic acid with the same number of carbon atoms. Even a mild oxidising agent suffices; common reagents are $$\mathrm{KMnO_4}$$, $$\mathrm{K_2Cr_2O_7}$$, or Tollens' reagent.

$$\mathrm{CH_3CH_2CH_2CHO \xrightarrow{[O]} CH_3CH_2CH_2COOH}$$

(For example, with mild $$\mathrm{KMnO_4}$$, $$\mathrm{CH_3CH_2CH_2CHO \xrightarrow{KMnO_4} CH_3CH_2CH_2COOH}$$.)

A cyanohydrin is the product formed when hydrogen cyanide ($$\mathrm{HCN}$$) adds across the carbonyl group of an aldehyde or a ketone. It contains a hydroxyl group ($$\mathrm{-OH}$$) and a cyano group ($$\mathrm{-CN}$$) on the same carbon atom; i.e. it is an $$\alpha$$-hydroxynitrile.

The cyanide ion attacks the carbonyl carbon and the alkoxide intermediate then picks up a proton:

$$\mathrm{CH_3CHO + HCN \rightarrow CH_3-CH(OH)-CN}$$

The product (acetaldehyde cyanohydrin, i.e. 2-hydroxypropanenitrile) is a cyanohydrin.

An acetal is a gem-dialkoxy compound, $$\mathrm{RCH(OR')_2}$$, in which two alkoxy groups ($$\mathrm{-OR'}$$) are attached to the same (terminal) carbon. It is obtained when an aldehyde is treated with two equivalents of a monohydric alcohol in the presence of dry hydrogen chloride.

Addition of one molecule of alcohol first gives a hemiacetal, which then reacts with a second molecule of alcohol (loss of water) to give the acetal:

$$\mathrm{CH_3CHO + 2\,C_2H_5OH \xrightarrow{dry\ HCl} CH_3CH(OC_2H_5)_2 + H_2O}$$

The product, 1,1-diethoxyethane, is an acetal.

A semicarbazone is the product formed when an aldehyde or ketone reacts with semicarbazide, $$\mathrm{H_2N-NH-CO-NH_2}$$. It is a nucleophilic addition-elimination reaction in which the $$\mathrm{>C=O}$$ group is converted into the $$\mathrm{>C=N-NH-CO-NH_2}$$ group with loss of water.

$$\mathrm{CH_3CHO + H_2N-NH-CO-NH_2 \rightarrow CH_3CH=N-NH-CO-NH_2 + H_2O}$$

The product, acetaldehyde semicarbazone, is a sharp-melting crystalline solid useful for the identification of carbonyl compounds.

An aldol is a $$\beta$$-hydroxy aldehyde (or $$\beta$$-hydroxy ketone) formed when two molecules of an aldehyde (or ketone) having at least one $$\alpha$$-hydrogen combine in the presence of dilute alkali. The name 'aldol' comes from aldehyde + alcohol, the two functional groups present in the product.

The dilute base removes an $$\alpha$$-hydrogen to give a carbanion (enolate), which attacks the carbonyl carbon of a second molecule:

$$\mathrm{2\,CH_3CHO \xrightarrow{dil.\ NaOH} CH_3-CH(OH)-CH_2-CHO}$$

The product, 3-hydroxybutanal, is the aldol.

A hemiacetal is the compound formed by the addition of one molecule of a monohydric alcohol to an aldehyde. It has both a hydroxyl group ($$\mathrm{-OH}$$) and an alkoxy group ($$\mathrm{-OR}$$) on the same carbon atom.

$$\mathrm{CH_3CHO + C_2H_5OH \xrightarrow{dry\ HCl} CH_3-CH(OH)(OC_2H_5)}$$

The hemiacetal is an intermediate in acetal formation; with a further molecule of alcohol it goes on to give the acetal.

An oxime is the product formed when an aldehyde or ketone reacts with hydroxylamine ($$\mathrm{H_2N-OH}$$) in the presence of a weak acid. By nucleophilic addition-elimination the $$\mathrm{>C=O}$$ group is converted into the $$\mathrm{>C=N-OH}$$ group with loss of water.

$$\mathrm{CH_3CHO + H_2N-OH \xrightarrow{H^+} CH_3CH=N-OH + H_2O}$$

The product, acetaldehyde oxime (ethanal oxime), is an oxime.

A ketal is a gem-dialkoxy compound, $$\mathrm{R_2C(OR')_2}$$, derived from a ketone. It is obtained when a ketone reacts with two equivalents of a monohydric alcohol (or with a diol such as ethylene glycol) in the presence of dry HCl.

$$\mathrm{CH_3COCH_3 + 2\,C_2H_5OH \xrightarrow{dry\ HCl} (CH_3)_2C(OC_2H_5)_2 + H_2O}$$

The product, 2,2-diethoxypropane, is a ketal. (In current usage, ketals are also classed simply as acetals.)

An imine is a compound containing the $$\mathrm{>C=N-}$$ group. It is formed when an aldehyde or ketone reacts with a primary amine ($$\mathrm{R-NH_2}$$): the amine adds to the carbonyl group and water is then eliminated.

$$\mathrm{CH_3CHO + C_2H_5NH_2 \xrightarrow{H^+} CH_3CH=N-C_2H_5 + H_2O}$$

The product, a substituted imine ($$N$$-ethylethanimine), is also called a Schiff's base.

A 2,4-DNP derivative (a 2,4-dinitrophenylhydrazone) is the product formed when an aldehyde or ketone reacts with 2,4-dinitrophenylhydrazine, $$\mathrm{H_2N-NH-C_6H_3(NO_2)_2}$$ (Brady's reagent). The $$\mathrm{>C=O}$$ group is converted into $$\mathrm{>C=N-NH-C_6H_3(NO_2)_2}$$ with loss of water.

$$\mathrm{CH_3CHO + H_2N-NH-C_6H_3(NO_2)_2 \rightarrow CH_3CH=N-NH-C_6H_3(NO_2)_2 + H_2O}$$

2,4-DNP derivatives are yellow, orange or red crystalline solids; their formation is used as a test for the carbonyl group and (from their sharp melting points) for identifying aldehydes and ketones.

A Schiff's base is a substituted imine — a compound containing the $$\mathrm{>C=N-R}$$ group — obtained by the condensation of an aldehyde or ketone with a primary amine, with loss of water.

$$\mathrm{C_6H_5CHO + C_6H_5NH_2 \xrightarrow{H^+} C_6H_5CH=N-C_6H_5 + H_2O}$$

The product, benzylideneaniline ($$N$$-phenylbenzaldimine), is a Schiff's base.

The principal group is $$\mathrm{-CHO}$$, so this carbon is C-1. Numbering the longest chain from the $$\mathrm{-CHO}$$ end:

$$\mathrm{\overset{1}{C}HO-\overset{2}{C}H_2-\overset{3}{C}H_2-\overset{4}{C}H(CH_3)-\overset{5}{C}H_3}$$

The chain has 5 carbons (parent: pentanal) with a methyl branch on C-4.

IUPAC name: 4-methylpentanal.

The principal group is the ketone $$\mathrm{>C=O}$$. The longest chain that contains the carbonyl carbon runs through six carbons. Numbering so that the carbonyl carbon gets the lowest locant:

$$\mathrm{\overset{1}{C}H_3-\overset{2}{C}H_2-\overset{3}{C}O-\overset{4}{C}H(C_2H_5)-\overset{5}{C}H_2-\overset{6}{C}H_2Cl}$$

Parent: hexan-3-one. Substituents: an ethyl group on C-4 and a chloro group on C-6.

IUPAC name: 6-chloro-4-ethylhexan-3-one.

The principal characteristic group is the aldehyde group $$\mathrm{-CHO}$$. It has the highest priority, so the carbon chain is numbered to give the $$\mathrm{-CHO}$$ carbon the lowest possible locant — here it is C-1.

Numbering from the $$\mathrm{-CHO}$$ end:

$$\mathrm{\overset{1}{C}HO-\overset{2}{C}H=\overset{3}{C}H-\overset{4}{C}H_3}$$

The chain has 4 carbons (parent: butenal) with a carbon-carbon double bond between C-2 and C-3.

IUPAC name: but-2-enal (common name: crotonaldehyde).

The molecule has a five-carbon chain with two ketonic $$\mathrm{>C=O}$$ groups:

$$\mathrm{\overset{1}{C}H_3-\overset{2}{C}O-\overset{3}{C}H_2-\overset{4}{C}O-\overset{5}{C}H_3}$$

Parent: pentane; two oxo groups at C-2 and C-4 (a diketone).

IUPAC name: pentane-2,4-dione (common name: acetylacetone).

The principal group is the ketone $$\mathrm{>C=O}$$. The longest chain containing the carbonyl carbon has six carbons. Numbering to give the carbonyl carbon the lowest locant:

$$\mathrm{\overset{1}{C}H_3-\overset{2}{C}O-\overset{3}{C}(CH_3)_2-\overset{4}{C}H_2-\overset{5}{C}H(CH_3)-\overset{6}{C}H_3}$$

Parent: hexan-2-one. Substituents: two methyl groups on C-3 and one methyl group on C-5.

IUPAC name: 3,3,5-trimethylhexan-2-one.

The principal group is $$\mathrm{-COOH}$$ (C-1). The longest chain containing the carboxyl carbon is four carbons long (one of the three methyl groups of the tert-butyl part is counted into the main chain):

$$\mathrm{\overset{1}{C}OOH-\overset{2}{C}H_2-\overset{3}{C}(CH_3)_2-\overset{4}{C}H_3}$$

Parent: butanoic acid; the remaining two methyl groups are substituents on C-3.

IUPAC name: 3,3-dimethylbutanoic acid.

The compound is a benzene ring carrying two $$\mathrm{-CHO}$$ groups para to each other (positions 1 and 4). When $$\mathrm{-CHO}$$ groups are attached directly to a ring, the suffix used is carbaldehyde; two such groups give dicarbaldehyde.

IUPAC name: benzene-1,4-dicarbaldehyde (common name: terephthalaldehyde).

Butanal is the four-carbon aldehyde $$\mathrm{\overset{1}{C}HO-\overset{2}{C}H_2-\overset{3}{C}H_2-\overset{4}{C}H_3}$$, the $$\mathrm{-CHO}$$ carbon being C-1. The prefix 3-methyl places a methyl group on C-3.

$$\mathrm{CH_3-CH(CH_3)-CH_2-CHO}$$, i.e. $$\mathrm{(CH_3)_2CH-CH_2-CHO}$$.

Propiophenone is ethyl phenyl ketone, $$\mathrm{C_6H_5-CO-CH_2CH_3}$$ (1-phenylpropan-1-one). The ring carbon bearing the $$\mathrm{-COCH_2CH_3}$$ group is C-1, so 'p-nitro' places a nitro group at the para (C-4) position.

Structure: a benzene ring with a $$\mathrm{-COC_2H_5}$$ group at one position and an $$\mathrm{-NO_2}$$ group para to it:

$$\mathrm{p\text{-}O_2N\text{-}C_6H_4\text{-}CO\text{-}CH_2CH_3}$$

Benzaldehyde is $$\mathrm{C_6H_5-CHO}$$. The ring carbon bearing the $$\mathrm{-CHO}$$ group is C-1; 'p-methyl' places a methyl group at the para (C-4) position.

Structure: a benzene ring with a $$\mathrm{-CHO}$$ group at one position and a $$\mathrm{-CH_3}$$ group para to it:

$$\mathrm{p\text{-}CH_3\text{-}C_6H_4\text{-}CHO}$$

Pent-...-2-one is a five-carbon chain with the keto group at C-2. The locants tell us: double bond between C-3 and C-4, and a methyl branch on C-4.

$$\mathrm{\overset{1}{C}H_3-\overset{2}{C}O-\overset{3}{C}H=\overset{4}{C}(CH_3)-\overset{5}{C}H_3}$$

Structure: $$\mathrm{CH_3-CO-CH=C(CH_3)-CH_3}$$ (common name: mesityl oxide).

Pentan-2-one is the five-carbon ketone with the keto group at C-2. The prefix 4-chloro places a chlorine atom on C-4.

$$\mathrm{\overset{1}{C}H_3-\overset{2}{C}O-\overset{3}{C}H_2-\overset{4}{C}HCl-\overset{5}{C}H_3}$$

Structure: $$\mathrm{CH_3-CO-CH_2-CHCl-CH_3}$$.

Pentanoic acid is the five-carbon acid with $$\mathrm{-COOH}$$ as C-1. The prefixes place a bromine atom on C-3 and a phenyl group on C-4.

$$\mathrm{\overset{1}{C}OOH-\overset{2}{C}H_2-\overset{3}{C}HBr-\overset{4}{C}H(C_6H_5)-\overset{5}{C}H_3}$$

Structure: $$\mathrm{HOOC-CH_2-CHBr-CH(C_6H_5)-CH_3}$$.

Benzophenone is diphenyl ketone, $$\mathrm{C_6H_5-CO-C_6H_5}$$. The labels $$p$$ and $$p'$$ refer to the para positions of the two separate rings; '$$p,p'$$-dihydroxy' places one $$\mathrm{-OH}$$ group at the para position of each ring (para to the $$\mathrm{-CO-}$$ link).

Structure: $$\mathrm{(p\text{-}HO\text{-}C_6H_4)-CO-(C_6H_4\text{-}OH\text{-}p)}$$, i.e. a carbonyl group joining two $$p$$-hydroxyphenyl rings.

The parent is a six-carbon acid (hex- + -oic acid), with $$\mathrm{-COOH}$$ as C-1. The locants indicate a double bond between C-2 and C-3 (en) and a triple bond between C-4 and C-5 (yn).

$$\mathrm{\overset{1}{C}OOH-\overset{2}{C}H=\overset{3}{C}H-\overset{4}{C}\equiv\overset{5}{C}-\overset{6}{C}H_3}$$

Structure: $$\mathrm{HOOC-CH=CH-C\equiv C-CH_3}$$.

Expanding, $$\mathrm{CH_3CO(CH_2)_4CH_3 = CH_3-CO-CH_2CH_2CH_2CH_2-CH_3}$$. The longest chain through the ketone carbon has seven carbons; numbering to give the $$\mathrm{>C=O}$$ the lowest locant puts it at C-2.

IUPAC name: heptan-2-one. Common name: methyl pentyl ketone (methyl n-amyl ketone).

The $$\mathrm{-CHO}$$ carbon is C-1. Numbering the longest chain from that end:

$$\mathrm{\overset{1}{C}HO-\overset{2}{C}H(CH_3)-\overset{3}{C}H_2-\overset{4}{C}HBr-\overset{5}{C}H_2-\overset{6}{C}H_3}$$

Parent: hexanal. Substituents: a methyl group on C-2 and a bromo group on C-4.

IUPAC name: 4-bromo-2-methylhexanal. (No common name.)

The compound $$\mathrm{CH_3(CH_2)_5CHO}$$ is a straight-chain aldehyde with seven carbon atoms in all (one $$\mathrm{-CHO}$$ + five $$\mathrm{-CH_2-}$$ + one $$\mathrm{-CH_3}$$).

IUPAC name: heptanal. Common name: heptaldehyde (oenanthaldehyde).

The $$\mathrm{-CHO}$$ carbon is C-1. The chain bearing it is $$\mathrm{\overset{1}{C}HO-\overset{2}{C}H=\overset{3}{C}H-}$$ (three carbons, prop-2-enal), with a phenyl group on C-3.

IUPAC name: 3-phenylprop-2-enal. Common name: cinnamaldehyde.

When a $$\mathrm{-CHO}$$ group is attached directly to a ring, the carbon of the ring cannot be included in the chain ending in '-al'; instead the suffix carbaldehyde is used. Here a $$\mathrm{-CHO}$$ group is bonded to a cyclopentane ring.

IUPAC name: cyclopentanecarbaldehyde. (No common name in general use.)

$$\mathrm{PhCOPh}$$ is $$\mathrm{C_6H_5-CO-C_6H_5}$$ — a carbonyl group flanked by two phenyl groups. It is named as a substituted methanone (the $$\mathrm{>C=O}$$ being the 'methanone' carbon).

IUPAC name: diphenylmethanone. Common name: benzophenone.

Benzaldehyde is $$\mathrm{C_6H_5CHO}$$. Its 2,4-dinitrophenylhydrazone is formed by replacing the carbonyl oxygen with the group $$\mathrm{=N-NH-C_6H_3(NO_2)_2}$$ (loss of $$\mathrm{H_2O}$$):

$$\mathrm{C_6H_5-CHO + H_2N-NH-C_6H_3(NO_2)_2 \rightarrow C_6H_5-CH=N-NH-C_6H_3(NO_2)_2 + H_2O}$$

Structure: $$\mathrm{C_6H_5-CH=N-NH-C_6H_3(NO_2)_2}$$, where the dinitrophenyl ring carries $$\mathrm{-NO_2}$$ groups at positions 2 and 4.

Cyclopropanone is a three-membered ring in which one ring carbon bears a $$\mathrm{>C=O}$$ group. Its oxime is obtained by replacing the carbonyl oxygen with $$\mathrm{=N-OH}$$ (reaction with hydroxylamine, loss of $$\mathrm{H_2O}$$).

Structure: a cyclopropane ring ($$\mathrm{-CH_2-CH_2-}$$ joined to the third carbon) in which the third ring carbon carries a $$\mathrm{=N-OH}$$ group, i.e. $$\mathrm{(CH_2)_2C=N-OH}$$ (ring).

Acetaldehyde is $$\mathrm{CH_3CHO}$$. Its dimethyl acetal is formed by addition of two molecules of methanol across the carbonyl group (with loss of water); both methoxy groups end up on the same carbon:

$$\mathrm{CH_3CHO + 2\,CH_3OH \xrightarrow{dry\ HCl} CH_3CH(OCH_3)_2 + H_2O}$$

Structure: $$\mathrm{CH_3-CH(OCH_3)_2}$$ (1,1-dimethoxyethane).

Cyclobutanone is a four-membered ring with one ring carbon bearing a $$\mathrm{>C=O}$$ group. Its semicarbazone is formed by reaction with semicarbazide ($$\mathrm{H_2N-NH-CO-NH_2}$$), in which the carbonyl oxygen is replaced by $$\mathrm{=N-NH-CO-NH_2}$$ (loss of $$\mathrm{H_2O}$$).

Structure: a cyclobutane ring whose one ring carbon carries a $$\mathrm{=N-NH-CO-NH_2}$$ group, i.e. $$\mathrm{(CH_2)_3C=N-NH-CO-NH_2}$$ (ring).

Hexan-3-one is $$\mathrm{CH_3CH_2-CO-CH_2CH_2CH_3}$$ (an ethyl propyl ketone). Its ethylene ketal is formed by reaction with ethylene glycol ($$\mathrm{HO-CH_2-CH_2-OH}$$) in the presence of dry HCl; the carbonyl oxygen is replaced by a cyclic $$\mathrm{-O-CH_2-CH_2-O-}$$ unit, giving a five-membered 1,3-dioxolane ring (loss of $$\mathrm{H_2O}$$).

Structure: the former carbonyl carbon now bears an ethyl group, a propyl group and two oxygens that are joined to each other through a $$\mathrm{-CH_2-CH_2-}$$ bridge:

$$\mathrm{C_2H_5-\underset{\displaystyle (O{-}CH_2{-}CH_2{-}O)}{C}-C_3H_7}$$ (a 1,3-dioxolane).

Formaldehyde is $$\mathrm{HCHO}$$. Its methyl hemiacetal is formed by the addition of one molecule of methanol across the carbonyl group; one $$\mathrm{-OH}$$ and one $$\mathrm{-OCH_3}$$ end up on the same carbon:

$$\mathrm{HCHO + CH_3OH \rightarrow H_2C(OH)(OCH_3)}$$

Structure: $$\mathrm{CH_2(OH)(OCH_3)}$$ (methoxymethanol).

Cyclohexanecarbaldehyde is $$\mathrm{C_6H_{11}-CHO}$$ (a cyclohexyl group attached to $$\mathrm{-CHO}$$). A Grignard reagent adds to the carbonyl group: the phenyl carbanion-like group attacks the carbonyl carbon, and acidic work-up protonates the alkoxide. Addition of a Grignard reagent to an aldehyde gives a secondary alcohol.

$$\mathrm{C_6H_{11}CHO \xrightarrow{C_6H_5MgBr} C_6H_{11}CH(OMgBr)C_6H_5 \xrightarrow{H_3O^+} C_6H_{11}CH(OH)C_6H_5}$$

The product is cyclohexyl(phenyl)methanol.

Tollens' reagent (ammoniacal silver nitrate, containing the $$\mathrm{[Ag(NH_3)_2]^+}$$ ion) oxidises an aldehyde to the corresponding carboxylate ion (carboxylic acid), while $$\mathrm{Ag^+}$$ is reduced to metallic silver, which deposits as a bright silver mirror.

$$\mathrm{C_6H_{11}CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow C_6H_{11}COO^- + 2Ag\!\downarrow + 4NH_3 + 2H_2O}$$

The organic product is cyclohexanecarboxylic acid ($$\mathrm{C_6H_{11}COOH}$$ after acidification), accompanied by a silver mirror.

Semicarbazide ($$\mathrm{H_2N-NH-CO-NH_2}$$) reacts with the carbonyl group by nucleophilic addition-elimination; in the presence of a weak acid the carbonyl oxygen is replaced by the group $$\mathrm{=N-NH-CO-NH_2}$$, with loss of water.

$$\mathrm{C_6H_{11}CHO + H_2N-NH-CO-NH_2 \xrightarrow{H^+} C_6H_{11}CH=N-NH-CO-NH_2 + H_2O}$$

The product is the semicarbazone of cyclohexanecarbaldehyde.

An aldehyde reacts with an alcohol in the presence of dry HCl (acid). One molecule of ethanol first gives a hemiacetal; with a second molecule of ethanol (excess alcohol) the hemiacetal is converted into the acetal (a gem-diethoxy compound), with loss of water.

$$\mathrm{C_6H_{11}CHO + 2\,C_2H_5OH \xrightarrow{dry\ HCl} C_6H_{11}CH(OC_2H_5)_2 + H_2O}$$

The product is the diethyl acetal of cyclohexanecarbaldehyde.

Zinc amalgam with concentrated/dilute hydrochloric acid brings about the Clemmensen reduction: the carbonyl group ($$\mathrm{>C=O}$$) of an aldehyde or ketone is reduced all the way to a $$\mathrm{-CH_2-}$$ (methylene) group.

$$\mathrm{C_6H_{11}CHO \xrightarrow{Zn(Hg),\ HCl} C_6H_{11}CH_3 + H_2O}$$

The $$\mathrm{-CHO}$$ group is reduced to $$\mathrm{-CH_3}$$; the product is methylcyclohexane.

Methanal (formaldehyde), $$\mathrm{HCHO}$$, has no $$\alpha$$-hydrogen atom — there is no carbon adjacent to the carbonyl. Hence it cannot undergo aldol condensation. An aldehyde with no $$\alpha$$-hydrogen, on warming with concentrated alkali, undergoes the Cannizzaro reaction (self oxidation-reduction / disproportionation).

$$\mathrm{2\,HCHO \xrightarrow{conc.\ NaOH} CH_3OH + HCOO^-Na^+}$$

One molecule is reduced to methanol and the other is oxidised to sodium formate (methanoate).

2-Methylpentanal is $$\mathrm{CH_3CH_2CH_2-CH(CH_3)-CHO}$$ (molecular formula $$\mathrm{C_6H_{12}O}$$). Its $$\alpha$$-carbon (C-2) carries a methyl group and one hydrogen atom. Because it has an $$\alpha$$-hydrogen it undergoes the aldol reaction; it cannot give the Cannizzaro reaction, which requires an aldehyde with no $$\alpha$$-hydrogen.

In dilute alkali one molecule loses its single $$\alpha$$-hydrogen to form the enolate (carbanion) — the nucleophile — which attacks the carbonyl carbon of a second molecule — the electrophile. The $$\alpha$$-carbon of the first molecule becomes bonded to the carbonyl carbon of the second, and the carbonyl oxygen becomes an $$\mathrm{-OH}$$ group:

$$\mathrm{2\,CH_3CH_2CH_2CH(CH_3)CHO \xrightarrow{dil.\ NaOH} CH_3CH_2CH_2C(CH_3)(CHO)CH(OH)CH(CH_3)CH_2CH_2CH_3}$$

The aldol product is therefore $$\mathrm{C_3H_7-C(CH_3)(CHO)-CH(OH)-CH(CH_3)-C_3H_7}$$ (where $$\mathrm{C_3H_7-}$$ is the $$n$$-propyl group). It is a $$\beta$$-hydroxy aldehyde of molecular formula $$\mathrm{C_{12}H_{24}O_2}$$ — exactly the addition product of two $$\mathrm{C_6H_{12}O}$$ molecules.

Numbering from the $$\mathrm{-CHO}$$ carbon, the parent chain is heptanal; it carries methyl and propyl groups on C-2, an $$\mathrm{-OH}$$ group on C-3 and a methyl group on C-4. Hence the aldol is 3-hydroxy-2,4-dimethyl-2-propylheptanal.

Note that in this aldol the $$\alpha$$-carbon (C-2, which bears the $$\mathrm{-CHO}$$) carries no hydrogen, so the usual dehydration to an $$\alpha,\beta$$-unsaturated aldehyde cannot occur; the aldol itself is the final product.

Benzaldehyde, $$\mathrm{C_6H_5CHO}$$, has the $$\mathrm{-CHO}$$ group attached directly to the benzene ring; there is no $$\alpha$$-hydrogen. Hence it cannot undergo aldol condensation but readily undergoes the Cannizzaro reaction with concentrated alkali.

$$\mathrm{2\,C_6H_5CHO \xrightarrow{conc.\ NaOH} C_6H_5CH_2OH + C_6H_5COO^-Na^+}$$

One molecule is reduced to benzyl alcohol and the other is oxidised to sodium benzoate.

Benzophenone, $$\mathrm{C_6H_5-CO-C_6H_5}$$, is a ketone in which the carbonyl group is flanked by two phenyl groups. It has no $$\alpha$$-hydrogen, so it cannot undergo aldol condensation. The Cannizzaro reaction is given only by aldehydes lacking $$\alpha$$-hydrogen, not by ketones.

Hence benzophenone undergoes neither aldol condensation nor the Cannizzaro reaction.

Cyclohexanone has $$\alpha$$-hydrogen atoms (on the two ring carbons adjacent to the $$\mathrm{>C=O}$$ group). Hence it undergoes aldol condensation (it cannot give the Cannizzaro reaction, which is for aldehydes).

In dilute alkali the $$\alpha$$-carbon of one cyclohexanone molecule attacks the carbonyl carbon of a second molecule, giving the $$\beta$$-hydroxy ketone (ketol): 1-(2-oxocyclohexyl)cyclohexan-1-ol. On heating this loses water to give the $$\alpha,\beta$$-unsaturated ketone:

$$\mathrm{2\,(Cyclohexanone) \xrightarrow{dil.\ NaOH} ketol \xrightarrow{-H_2O} 2\text{-}(cyclohex\text{-}1\text{-}en\text{-}1\text{-}yl)cyclohexan\text{-}1\text{-}one}$$

So the final aldol condensation product is 2-(cyclohex-1-en-1-yl)cyclohexan-1-one — a cyclohexanone ring bearing, on its $$\alpha$$-carbon, a cyclohex-1-enyl group.

1-Phenylpropan-1-one (propiophenone) is $$\mathrm{C_6H_5-CO-CH_2-CH_3}$$. The $$\mathrm{-CH_2-}$$ group adjacent to the carbonyl carries $$\alpha$$-hydrogen atoms, so it undergoes aldol condensation.

In dilute alkali the $$\alpha$$-carbon ($$\mathrm{CH_2}$$) of one molecule attacks the carbonyl carbon of a second molecule. The aldol (ketol) then loses water to give the $$\alpha,\beta$$-unsaturated ketone:

$$\mathrm{2\,C_6H_5COCH_2CH_3 \xrightarrow{dil.\ NaOH} C_6H_5CO-CH(CH_3)-C(OH)(C_6H_5)-CH_2CH_3 \xrightarrow{-H_2O} C_6H_5CO-C(CH_3)=C(C_6H_5)-CH_2CH_3}$$

The condensation product is 2-methyl-1,3-diphenylpent-2-en-1-one.

Phenylacetaldehyde is $$\mathrm{C_6H_5-CH_2-CHO}$$. The $$\mathrm{-CH_2-}$$ group between the ring and the $$\mathrm{-CHO}$$ is an $$\alpha$$-carbon bearing hydrogen, so the compound undergoes aldol condensation.

In dilute alkali the $$\alpha$$-carbon of one molecule attacks the carbonyl carbon of a second molecule to give the aldol; on heating, water is lost to give the $$\alpha,\beta$$-unsaturated aldehyde:

$$\mathrm{2\,C_6H_5CH_2CHO \xrightarrow{dil.\ NaOH} C_6H_5CH_2-CH(OH)-CH(C_6H_5)-CHO \xrightarrow{-H_2O} C_6H_5CH_2-CH=C(C_6H_5)-CHO}$$

The condensation product is 2,4-diphenylbut-2-enal.

Butan-1-ol, $$\mathrm{CH_3CH_2CH_2CH_2OH}$$, is an alcohol — it contains no carbonyl ($$\mathrm{>C=O}$$) group at all. Both the aldol condensation and the Cannizzaro reaction are reactions of aldehydes/ketones.

Since butan-1-ol is neither an aldehyde nor a ketone, it undergoes neither reaction.

2,2-Dimethylbutanal is $$\mathrm{CH_3CH_2-C(CH_3)_2-CHO}$$. The $$\alpha$$-carbon (C-2) is bonded to the $$\mathrm{-CHO}$$ group, two methyl groups and an ethyl group — it carries no hydrogen. With no $$\alpha$$-hydrogen, aldol condensation is not possible; instead the aldehyde undergoes the Cannizzaro reaction with concentrated alkali.

$$\mathrm{2\,(CH_3CH_2)(CH_3)_2C-CHO \xrightarrow{conc.\ NaOH} (CH_3CH_2)(CH_3)_2C-CH_2OH + (CH_3CH_2)(CH_3)_2C-COO^-Na^+}$$

One molecule is reduced to 2,2-dimethylbutan-1-ol and the other is oxidised to sodium 2,2-dimethylbutanoate.

Two molecules of ethanal first undergo the aldol reaction in dilute alkali to give 3-hydroxybutanal (an aldol). The aldehyde group of the aldol is then reduced (e.g. with $$\mathrm{NaBH_4}$$, or $$\mathrm{H_2}$$/Ni) to a $$\mathrm{-CH_2OH}$$ group, giving the diol.

$$\mathrm{2\,CH_3CHO \xrightarrow{dil.\ NaOH} CH_3CH(OH)CH_2CHO}$$

$$\mathrm{CH_3CH(OH)CH_2CHO \xrightarrow{NaBH_4} CH_3CH(OH)CH_2CH_2OH}$$

The product is butane-1,3-diol.

Two molecules of ethanal undergo aldol condensation. In dilute alkali they first give the aldol 3-hydroxybutanal; on warming, this loses a molecule of water (dehydration) to give the $$\alpha,\beta$$-unsaturated aldehyde.

$$\mathrm{2\,CH_3CHO \xrightarrow{dil.\ NaOH} CH_3CH(OH)CH_2CHO \xrightarrow[\Delta]{-H_2O} CH_3CH=CHCHO}$$

The product is but-2-enal (crotonaldehyde).

First obtain but-2-enal from ethanal by aldol condensation (as in the previous part). The $$\mathrm{-CHO}$$ group of but-2-enal is then mildly oxidised to a $$\mathrm{-COOH}$$ group; a mild oxidant such as Tollens' reagent is used so that the $$\mathrm{C=C}$$ double bond is not attacked.

$$\mathrm{2\,CH_3CHO \xrightarrow{dil.\ NaOH} CH_3CH(OH)CH_2CHO \xrightarrow[\Delta]{-H_2O} CH_3CH=CHCHO}$$

$$\mathrm{CH_3CH=CHCHO \xrightarrow{[O]\ (mild)} CH_3CH=CHCOOH}$$

The product is but-2-enoic acid (crotonic acid).

Addition of HCN to a carbonyl compound is a nucleophilic addition. Reactivity falls as the groups on the carbonyl carbon become larger and more electron-releasing (greater steric hindrance and greater +I effect, both of which oppose the attack of $$\mathrm{CN^-}$$).

• Acetaldehyde ($$\mathrm{CH_3CHO}$$): only one small methyl group on the carbonyl carbon — most reactive.
• Acetone ($$\mathrm{CH_3COCH_3}$$): two methyl groups — more hindered than acetaldehyde.
• Methyl tert-butyl ketone ($$\mathrm{CH_3CO\,C(CH_3)_3}$$): one bulky tert-butyl group — strongly hindered.
• Di-tert-butyl ketone ($$\mathrm{(CH_3)_3C\,CO\,C(CH_3)_3}$$): two bulky tert-butyl groups — the most hindered, least reactive.

Hence the increasing order of reactivity towards HCN is as shown.

Acid strength rises when the conjugate base (carboxylate) is stabilised. The electron-withdrawing $$\mathrm{-Br}$$ ($$-I$$ effect) increases acidity, and its effect is stronger the closer it is to the $$\mathrm{-COOH}$$ group. An electron-releasing group ($$+I$$) decreases acidity.

• $$\mathrm{(CH_3)_2CHCOOH}$$ (2-methylpropanoic acid): an extra electron-releasing methyl group, no Br — weakest.
• $$\mathrm{CH_3CH_2CH_2COOH}$$ (butanoic acid): no Br, plain straight chain — weak, but slightly stronger than the branched one above.
• $$\mathrm{CH_3CH(Br)CH_2COOH}$$ (3-bromobutanoic acid): Br on C-3 ($$\beta$$-carbon) — moderately strong.
• $$\mathrm{CH_3CH_2CH(Br)COOH}$$ (2-bromobutanoic acid): Br on C-2 ($$\alpha$$-carbon, nearest $$\mathrm{-COOH}$$) — strongest.

Hence the increasing order of acid strength is as shown.

In substituted benzoic acids, electron-withdrawing groups (like $$\mathrm{-NO_2}$$) stabilise the carboxylate ion and increase acidity, while electron-releasing groups (like $$\mathrm{-OCH_3}$$) destabilise it and decrease acidity. More $$\mathrm{-NO_2}$$ groups mean greater acidity.

• 4-Methoxybenzoic acid: $$\mathrm{-OCH_3}$$ is electron-releasing (it strongly donates electrons by resonance from the para position) — weakest acid.
• Benzoic acid: no substituent — reference.
• 4-Nitrobenzoic acid: one electron-withdrawing $$\mathrm{-NO_2}$$ group — stronger than benzoic acid.
• 3,4-Dinitrobenzoic acid: two electron-withdrawing $$\mathrm{-NO_2}$$ groups — strongest acid.

Hence the increasing order of acid strength is as shown.

Propanal is an aldehyde; propanone is a ketone. Aldehydes are easily oxidised and so reduce mild oxidising reagents, whereas ketones do not.

Tollens' test: Warm each compound with Tollens' reagent (ammoniacal $$\mathrm{AgNO_3}$$).

• Propanal reduces it, depositing a bright silver mirror: $$\mathrm{CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag\!\downarrow + 4NH_3 + 2H_2O}$$.
• Propanone gives no silver mirror.

(Fehling's test serves equally well: propanal gives a red precipitate of $$\mathrm{Cu_2O}$$, propanone does not.)

Acetophenone, $$\mathrm{C_6H_5COCH_3}$$, is a methyl ketone; benzophenone, $$\mathrm{C_6H_5COC_6H_5}$$, is not.

Iodoform test: Warm each compound with iodine and sodium hydroxide ($$\mathrm{I_2/NaOH}$$).

• Acetophenone contains a $$\mathrm{CH_3CO-}$$ group, so it gives a yellow precipitate of iodoform: $$\mathrm{C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3\!\downarrow + 3NaI + 3H_2O}$$.
• Benzophenone has no $$\mathrm{CH_3CO-}$$ group and gives no yellow precipitate.

Benzoic acid is a carboxylic acid, strong enough to liberate $$\mathrm{CO_2}$$ from sodium hydrogencarbonate; phenol is only weakly acidic and cannot.

Sodium hydrogencarbonate test: Add a little $$\mathrm{NaHCO_3}$$ solution to each.

• Benzoic acid gives brisk effervescence due to the evolution of $$\mathrm{CO_2}$$: $$\mathrm{C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2\!\uparrow}$$.
• Phenol gives no effervescence.

(Alternatively, the neutral $$\mathrm{FeCl_3}$$ test: phenol gives a violet colouration, benzoic acid gives a buff/dull-yellow precipitate.)

Benzoic acid has a free $$\mathrm{-COOH}$$ group and is acidic; ethyl benzoate is an ester and is neutral.

Sodium hydrogencarbonate test: Add $$\mathrm{NaHCO_3}$$ solution to each.

• Benzoic acid gives brisk effervescence of $$\mathrm{CO_2}$$: $$\mathrm{C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2\!\uparrow}$$.
• Ethyl benzoate ($$\mathrm{C_6H_5COOC_2H_5}$$), being an ester, gives no effervescence.

(Benzoic acid also dissolves in $$\mathrm{NaOH}$$ giving a clear solution; the ester is insoluble.)

Pentan-2-one, $$\mathrm{CH_3CO\,CH_2CH_2CH_3}$$, is a methyl ketone (it has a $$\mathrm{CH_3CO-}$$ group); pentan-3-one, $$\mathrm{CH_3CH_2\,CO\,CH_2CH_3}$$, is not.

Iodoform test: Warm each with $$\mathrm{I_2/NaOH}$$.

• Pentan-2-one gives a yellow precipitate of iodoform: $$\mathrm{CH_3COC_3H_7 + 3I_2 + 4NaOH \rightarrow C_3H_7COONa + CHI_3\!\downarrow + 3NaI + 3H_2O}$$.
• Pentan-3-one gives no yellow precipitate.

Benzaldehyde, $$\mathrm{C_6H_5CHO}$$, is an aldehyde; acetophenone, $$\mathrm{C_6H_5COCH_3}$$, is a methyl ketone.

Tollens' test: Warm each with Tollens' reagent.

• Benzaldehyde, being an aldehyde, reduces it and gives a bright silver mirror.
• Acetophenone, a ketone, gives no silver mirror.

(Alternatively, the iodoform test distinguishes them the other way round: acetophenone, a methyl ketone, gives a yellow precipitate of $$\mathrm{CHI_3}$$, whereas benzaldehyde does not.)

Both are aldehydes, so an oxidation test (Tollens', Fehling's) cannot tell them apart — both are positive. They differ in that ethanal, $$\mathrm{CH_3CHO}$$, has a $$\mathrm{CH_3CO-}$$-type grouping (a methyl group attached to the carbonyl carbon) needed for the iodoform reaction, whereas propanal, $$\mathrm{CH_3CH_2CHO}$$, does not.

Iodoform test: Warm each with $$\mathrm{I_2/NaOH}$$.

• Ethanal gives a yellow precipitate of iodoform: $$\mathrm{CH_3CHO + 3I_2 + 4NaOH \rightarrow HCOONa + CHI_3\!\downarrow + 3NaI + 3H_2O}$$.
• Propanal gives no yellow precipitate.

Benzene is first converted to toluene by Friedel-Crafts alkylation (a one-carbon organic reagent, $$\mathrm{CH_3Cl}$$). The methyl group is oxidised to a $$\mathrm{-COOH}$$ group, and the benzoic acid is finally esterified with methanol.

$$\mathrm{C_6H_6 \xrightarrow[anhyd.\ AlCl_3]{CH_3Cl} C_6H_5CH_3 \xrightarrow[\Delta]{KMnO_4,\ KOH;\ then\ H_3O^+} C_6H_5COOH}$$

$$\mathrm{C_6H_5COOH + CH_3OH \xrightarrow[\Delta]{conc.\ H_2SO_4} C_6H_5COOCH_3 + H_2O}$$

The product is methyl benzoate.

The two substituents needed are $$\mathrm{-COOH}$$ and $$\mathrm{-NO_2}$$, meta to each other. The $$\mathrm{-COOH}$$ group is meta-directing, so it must be introduced first; nitration then occurs at the meta position.

$$\mathrm{C_6H_6 \xrightarrow[anhyd.\ AlCl_3]{CH_3Cl} C_6H_5CH_3 \xrightarrow[\Delta]{KMnO_4,\ KOH;\ then\ H_3O^+} C_6H_5COOH}$$

$$\mathrm{C_6H_5COOH \xrightarrow[conc.\ H_2SO_4]{conc.\ HNO_3} m\text{-}O_2N\text{-}C_6H_4\text{-}COOH}$$

The product is $$m$$-nitrobenzoic acid.

Here $$\mathrm{-COOH}$$ and $$\mathrm{-NO_2}$$ must be para to each other. A $$\mathrm{-CH_3}$$ group is ortho/para-directing, so toluene is nitrated first (giving mainly $$p$$-nitrotoluene); the methyl group is then oxidised to $$\mathrm{-COOH}$$.

$$\mathrm{C_6H_6 \xrightarrow[anhyd.\ AlCl_3]{CH_3Cl} C_6H_5CH_3 \xrightarrow[conc.\ H_2SO_4]{conc.\ HNO_3} p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3}$$

$$\mathrm{p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3 \xrightarrow[\Delta]{KMnO_4,\ KOH;\ then\ H_3O^+} p\text{-}O_2N\text{-}C_6H_4\text{-}COOH}$$

The product is $$p$$-nitrobenzoic acid.

Phenylacetic acid, $$\mathrm{C_6H_5CH_2COOH}$$, has one carbon more than the benzylic position of toluene. Benzene is first converted to toluene, then the methyl group is chlorinated in the side-chain (free-radical conditions) to give benzyl chloride; the chloride is displaced by cyanide and the nitrile is hydrolysed to the acid.

$$\mathrm{C_6H_6 \xrightarrow[anhyd.\ AlCl_3]{CH_3Cl} C_6H_5CH_3 \xrightarrow[(light)]{Cl_2,\ hv} C_6H_5CH_2Cl}$$

$$\mathrm{C_6H_5CH_2Cl \xrightarrow{KCN} C_6H_5CH_2CN \xrightarrow[\Delta]{H_3O^+} C_6H_5CH_2COOH}$$

The product is phenylacetic acid.

Benzene is converted to toluene, which is nitrated; since $$\mathrm{-CH_3}$$ is ortho/para-directing, $$p$$-nitrotoluene is the main product. The methyl group of $$p$$-nitrotoluene is then oxidised selectively to a $$\mathrm{-CHO}$$ group with chromyl chloride (Étard reaction); chromyl chloride stops the oxidation at the aldehyde stage.

$$\mathrm{C_6H_6 \xrightarrow[anhyd.\ AlCl_3]{CH_3Cl} C_6H_5CH_3 \xrightarrow[conc.\ H_2SO_4]{conc.\ HNO_3} p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3}$$

$$\mathrm{p\text{-}O_2N\text{-}C_6H_4\text{-}CH_3 \xrightarrow[2.\ H_3O^+]{1.\ CrO_2Cl_2} p\text{-}O_2N\text{-}C_6H_4\text{-}CHO}$$

The product is $$p$$-nitrobenzaldehyde.

Step 1 — Reduction: Propanone is reduced to propan-2-ol with a hydride reagent such as $$\mathrm{NaBH_4}$$ (or $$\mathrm{LiAlH_4}$$).

$$\mathrm{CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_3}$$

Step 2 — Dehydration: Propan-2-ol is dehydrated with hot concentrated sulphuric acid to give propene.

$$\mathrm{CH_3CH(OH)CH_3 \xrightarrow[\Delta]{conc.\ H_2SO_4} CH_3CH=CH_2 + H_2O}$$

Step 1: Benzoic acid is converted to benzoyl chloride with thionyl chloride.

$$\mathrm{C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl + SO_2 + HCl}$$

Step 2 — Rosenmund reduction: Benzoyl chloride is hydrogenated over palladium poisoned with $$\mathrm{BaSO_4}$$, which stops the reduction cleanly at the aldehyde stage.

$$\mathrm{C_6H_5COCl + H_2 \xrightarrow{Pd\text{-}BaSO_4} C_6H_5CHO + HCl}$$

Step 1 — Oxidation: Ethanol is oxidised to ethanal with a mild/controlled oxidant such as PCC (or by passing its vapour over hot copper).

$$\mathrm{CH_3CH_2OH \xrightarrow{PCC} CH_3CHO}$$

Step 2 — Aldol reaction: Two molecules of ethanal combine in dilute alkali to give the aldol, 3-hydroxybutanal.

$$\mathrm{2\,CH_3CHO \xrightarrow{dil.\ NaOH} CH_3CH(OH)CH_2CHO}$$

Step 1 — Friedel-Crafts acylation: Benzene is acetylated with acetyl chloride and anhydrous $$\mathrm{AlCl_3}$$ to give acetophenone.

$$\mathrm{C_6H_6 + CH_3COCl \xrightarrow{anhyd.\ AlCl_3} C_6H_5COCH_3 + HCl}$$

Step 2 — Nitration: The $$\mathrm{-COCH_3}$$ group is electron-withdrawing and meta-directing, so nitration of acetophenone with conc. $$\mathrm{HNO_3}$$/conc. $$\mathrm{H_2SO_4}$$ introduces the $$\mathrm{-NO_2}$$ group at the meta position.

$$\mathrm{C_6H_5COCH_3 \xrightarrow[conc.\ H_2SO_4]{conc.\ HNO_3} m\text{-}O_2N\text{-}C_6H_4\text{-}COCH_3}$$

Step 1 — Grignard addition: Benzaldehyde is treated with phenylmagnesium bromide and then hydrolysed with acid. Addition of the Grignard reagent to the aldehyde, followed by protonation, gives the secondary alcohol diphenylmethanol (diphenylcarbinol).

$$\mathrm{C_6H_5CHO \xrightarrow[2.\ H_3O^+]{1.\ C_6H_5MgBr} C_6H_5CH(OH)C_6H_5}$$

Step 2 — Oxidation: The secondary alcohol is oxidised to the ketone benzophenone (e.g. with PCC or $$\mathrm{CrO_3}$$).

$$\mathrm{C_6H_5CH(OH)C_6H_5 \xrightarrow{PCC} C_6H_5COC_6H_5}$$

Step 1 — Grignard formation: Bromobenzene is converted to phenylmagnesium bromide with magnesium in dry ether.

$$\mathrm{C_6H_5Br \xrightarrow{Mg,\ dry\ ether} C_6H_5MgBr}$$

Step 2 — Addition to ethanal: The Grignard reagent adds to ethanal; acidic hydrolysis then gives the secondary alcohol 1-phenylethanol.

$$\mathrm{C_6H_5MgBr \xrightarrow{CH_3CHO} C_6H_5CH(CH_3)OMgBr \xrightarrow{H_3O^+} C_6H_5CH(OH)CH_3}$$

Step 1 — Cross aldol condensation: Benzaldehyde (no $$\alpha$$-H) is condensed with ethanal in dilute alkali; loss of water gives the $$\alpha,\beta$$-unsaturated aldehyde cinnamaldehyde.

$$\mathrm{C_6H_5CHO + CH_3CHO \xrightarrow[\Delta]{dil.\ NaOH} C_6H_5CH=CHCHO + H_2O}$$

Step 2 — Reduction: Catalytic hydrogenation reduces both the $$\mathrm{C=C}$$ double bond and the $$\mathrm{-CHO}$$ group, giving the saturated primary alcohol.

$$\mathrm{C_6H_5CH=CHCHO \xrightarrow{H_2,\ Ni} C_6H_5CH_2CH_2CH_2OH}$$

The product is 3-phenylpropan-1-ol.

Step 1 — Cyanohydrin formation: Benzaldehyde adds HCN across its carbonyl group to give the cyanohydrin (an $$\alpha$$-hydroxynitrile).

$$\mathrm{C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN}$$

Step 2 — Hydrolysis: The nitrile group of the cyanohydrin is hydrolysed with dilute acid to a $$\mathrm{-COOH}$$ group.

$$\mathrm{C_6H_5CH(OH)CN \xrightarrow[\Delta]{H_3O^+} C_6H_5CH(OH)COOH}$$

The product is $$\alpha$$-hydroxyphenylacetic acid (mandelic acid).

Step 1 — Nitration: The $$\mathrm{-COOH}$$ group is electron-withdrawing and meta-directing, so nitration of benzoic acid gives $$m$$-nitrobenzoic acid.

$$\mathrm{C_6H_5COOH \xrightarrow[conc.\ H_2SO_4]{conc.\ HNO_3} m\text{-}O_2N\text{-}C_6H_4\text{-}COOH}$$

Step 2 — Reduction: The $$\mathrm{-COOH}$$ group is reduced to a $$\mathrm{-CH_2OH}$$ group with diborane ($$\mathrm{B_2H_6}$$), which reduces the carboxyl group but leaves the $$\mathrm{-NO_2}$$ group untouched.

$$\mathrm{m\text{-}O_2N\text{-}C_6H_4\text{-}COOH \xrightarrow[2.\ H_3O^+]{1.\ B_2H_6} m\text{-}O_2N\text{-}C_6H_4\text{-}CH_2OH}$$

The product is $$m$$-nitrobenzyl alcohol.

Acetylation is the introduction of an acetyl group, $$\mathrm{CH_3CO-}$$, into a molecule — usually in place of the hydrogen of an $$\mathrm{-OH}$$, $$\mathrm{-NH_2}$$ or $$\mathrm{-NHR}$$ group. It is carried out by treating the compound with acetyl chloride ($$\mathrm{CH_3COCl}$$) or acetic anhydride [$$\mathrm{(CH_3CO)_2O}$$], generally in the presence of a base such as pyridine (which removes the HCl/acetic acid formed).

For example, the acetylation of an alcohol or amine:

$$\mathrm{C_2H_5OH + CH_3COCl \xrightarrow{pyridine} CH_3COOC_2H_5 + HCl}$$

$$\mathrm{C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH}$$

Acetylation lowers the reactivity of the $$\mathrm{-OH}$$ or $$\mathrm{-NH_2}$$ group (it is used, for instance, as a protecting step).

The Cannizzaro reaction is the self oxidation-reduction (disproportionation) undergone by an aldehyde that has no $$\alpha$$-hydrogen when it is warmed with a concentrated solution of an alkali.

One molecule of the aldehyde is reduced to the corresponding primary alcohol while a second molecule is oxidised to the salt of the carboxylic acid.

$$\mathrm{2\,HCHO \xrightarrow{conc.\ NaOH} CH_3OH + HCOO^-Na^+}$$

$$\mathrm{2\,C_6H_5CHO \xrightarrow{conc.\ NaOH} C_6H_5CH_2OH + C_6H_5COO^-Na^+}$$

Cross (mixed) aldol condensation is an aldol condensation carried out between two different aldehydes and/or ketones.

If both partners contain $$\alpha$$-hydrogen atoms, each can act as the nucleophile and as the electrophile, so a mixture of four products is formed and the reaction has little synthetic value. It becomes useful when one of the carbonyl compounds has no $$\alpha$$-hydrogen (e.g. benzaldehyde, formaldehyde); such a compound can only act as the electrophile, so a single major product is obtained.

For example, benzaldehyde (no $$\alpha$$-H) with acetaldehyde:

$$\mathrm{C_6H_5CHO + CH_3CHO \xrightarrow[\Delta]{dil.\ NaOH} C_6H_5CH=CHCHO + H_2O}$$

giving cinnamaldehyde as the main product.

Decarboxylation is the removal of a molecule of carbon dioxide ($$\mathrm{CO_2}$$) from a carboxylic acid (or its salt).

When the sodium salt of a carboxylic acid is heated with soda lime (a mixture of $$\mathrm{NaOH}$$ and $$\mathrm{CaO}$$), the $$\mathrm{-COOH}$$ group is lost as carbonate and a hydrocarbon with one carbon atom fewer is obtained.

$$\mathrm{R{-}COONa + NaOH \xrightarrow[\Delta]{CaO} R{-}H + Na_2CO_3}$$

For example: $$\mathrm{CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4 + Na_2CO_3}$$.

(Aromatic acids behave similarly, e.g. sodium benzoate gives benzene. Decarboxylation also occurs at the anode in Kolbe's electrolysis of carboxylate solutions.)

Hot alkaline $$\mathrm{KMnO_4}$$ oxidises any alkyl side-chain on a benzene ring (regardless of its length) all the way down to a single $$\mathrm{-COOH}$$ group attached to the ring. The ethyl group of ethylbenzene is therefore degraded to a carboxyl group; with $$\mathrm{KOH}$$ present the product is the potassium salt.

$$\mathrm{C_6H_5CH_2CH_3 \xrightarrow[KOH,\ \Delta]{KMnO_4} C_6H_5COOK\ (\xrightarrow{H_3O^+} C_6H_5COOH)}$$

The product is benzoic acid (isolated as potassium benzoate, then acidified).

Thionyl chloride ($$\mathrm{SOCl_2}$$) converts a $$\mathrm{-COOH}$$ group into an acid chloride ($$\mathrm{-COCl}$$), the by-products ($$\mathrm{SO_2}$$ and $$\mathrm{HCl}$$) being gases. Phthalic acid has two $$\mathrm{-COOH}$$ groups, so both are converted to $$\mathrm{-COCl}$$.

$$\mathrm{o\text{-}C_6H_4(COOH)_2 + 2\,SOCl_2 \rightarrow o\text{-}C_6H_4(COCl)_2 + 2\,SO_2 + 2\,HCl}$$

The product is benzene-1,2-dicarbonyl dichloride (phthaloyl chloride).

$$\mathrm{H_2N{-}CO{-}NH{-}NH_2}$$ is semicarbazide. It reacts with the carbonyl group of benzaldehyde by nucleophilic addition-elimination: the $$\mathrm{-CHO}$$ group is converted into $$\mathrm{-CH=N-NH-CO-NH_2}$$ with loss of a molecule of water.

$$\mathrm{C_6H_5CHO + H_2N{-}CO{-}NH{-}NH_2 \rightarrow C_6H_5CH=N-NH-CO-NH_2 + H_2O}$$

The product is benzaldehyde semicarbazone.

Benzophenone is a diaryl ketone, made by introducing a benzoyl group ($$\mathrm{C_6H_5CO-}$$) onto a benzene ring. This is a Friedel-Crafts acylation: benzene is treated with benzoyl chloride in the presence of anhydrous aluminium chloride.

$$\mathrm{C_6H_6 + C_6H_5COCl \xrightarrow{anhyd.\ AlCl_3} C_6H_5-CO-C_6H_5 + HCl}$$

So the missing reagent is benzoyl chloride, $$\mathrm{C_6H_5COCl}$$, with anhydrous $$\mathrm{AlCl_3}$$.

$$\mathrm{[Ag(NH_3)_2]^+}$$ is the active species of Tollens' reagent — a mild oxidising agent. It oxidises only the easily-oxidised aldehyde group to a $$\mathrm{-COOH}$$ group; the ketone group is not affected.

So in 4-formylcyclohexanone only the $$\mathrm{-CHO}$$ group is oxidised, while the ring ketone remains:

$$\mathrm{4\text{-}formylcyclohexanone \xrightarrow{[Ag(NH_3)_2]^+} 4\text{-}oxocyclohexane\text{-}1\text{-}carboxylic\ acid}$$

(A silver mirror is also deposited.) The product is 4-oxocyclohexane-1-carboxylic acid.

$$\mathrm{NaCN}$$ with $$\mathrm{HCl}$$ furnishes $$\mathrm{HCN}$$, which adds across the aldehyde carbonyl group to form a cyanohydrin. The much less reactive $$\mathrm{-COOH}$$ group does not add HCN.

The starting material is $$o$$-formylbenzoic acid, $$o$$-$$\mathrm{HOOC-C_6H_4-CHO}$$. The $$\mathrm{-CHO}$$ group becomes $$\mathrm{-CH(OH)CN}$$:

$$\mathrm{o\text{-}HOOC\text{-}C_6H_4\text{-}CHO \xrightarrow{NaCN/HCl} o\text{-}HOOC\text{-}C_6H_4\text{-}CH(OH)CN}$$

The product is the cyanohydrin of $$o$$-formylbenzoic acid, i.e. $$o$$-(carboxyphenyl) carrying a $$\mathrm{-CH(OH)CN}$$ group.

This is a cross aldol condensation. Benzaldehyde has no $$\alpha$$-hydrogen, so it can act only as the electrophile (the carbonyl component). Propanal, $$\mathrm{CH_3CH_2CHO}$$, has $$\alpha$$-hydrogen; its $$\alpha$$-carbon ($$\mathrm{-CH_2-}$$) provides the nucleophile (enolate).

The $$\alpha$$-carbon of propanal attacks the carbonyl carbon of benzaldehyde; the aldol then loses water on heating to give the $$\alpha,\beta$$-unsaturated aldehyde:

$$\mathrm{C_6H_5CHO + CH_3CH_2CHO \xrightarrow[\Delta]{dil.\ NaOH} C_6H_5CH=C(CH_3)CHO + H_2O}$$

The product is 2-methyl-3-phenylprop-2-enal ($$\alpha$$-methylcinnamaldehyde).

Sodium borohydride ($$\mathrm{NaBH_4}$$) is a mild reducing agent. It reduces an aldehyde or ketone carbonyl to an alcohol, but it is not strong enough to reduce an ester group.

Ethyl acetoacetate, $$\mathrm{CH_3COCH_2COOC_2H_5}$$, has a ketone group and an ester group. Only the keto $$\mathrm{>C=O}$$ is reduced, to a $$\mathrm{>CH-OH}$$ group; the $$\mathrm{-COOC_2H_5}$$ ester group survives.

$$\mathrm{CH_3COCH_2COOC_2H_5 \xrightarrow[(ii)\ H^+]{(i)\ NaBH_4} CH_3CH(OH)CH_2COOC_2H_5}$$

The product is ethyl 3-hydroxybutanoate.

Cyclohexanol is a secondary alcohol. Oxidation of a secondary alcohol gives a ketone, and the ketone is not oxidised further under these conditions. Chromium trioxide ($$\mathrm{CrO_3}$$) effects this oxidation.

$$\mathrm{Cyclohexanol \xrightarrow{CrO_3} Cyclohexanone}$$

The product is cyclohexanone.

Methylenecyclohexane has an exocyclic $$\mathrm{C=CH_2}$$ double bond. Cyclohexanecarbaldehyde has a $$\mathrm{-CHO}$$ group on the ring carbon. So the terminal $$\mathrm{=CH_2}$$ must become $$\mathrm{-CHO}$$, and a hydrogen must add to the ring carbon.

Step 1 — Hydroboration-oxidation: Treatment with diborane ($$\mathrm{B_2H_6}$$) followed by alkaline hydrogen peroxide adds $$\mathrm{H}$$ and $$\mathrm{OH}$$ across the double bond in an anti-Markovnikov manner, placing the $$\mathrm{-OH}$$ on the terminal carbon. This gives the primary alcohol cyclohexylmethanol ($$\mathrm{C_6H_{11}-CH_2OH}$$).

Step 2 — Controlled oxidation: The primary alcohol is oxidised with PCC to the aldehyde, cyclohexanecarbaldehyde ($$\mathrm{C_6H_{11}-CHO}$$).

$$\mathrm{C_6H_{10}{=}CH_2 \xrightarrow[2.\ H_2O_2/OH^-]{1.\ B_2H_6} C_6H_{11}CH_2OH \xrightarrow{PCC} C_6H_{11}CHO}$$

The reagents $$\mathrm{O_3}$$ followed by $$\mathrm{Zn/H_2O}$$ describe ozonolysis: a $$\mathrm{C=C}$$ double bond is cleaved and each doubly-bonded carbon becomes a carbonyl carbon.

To get two molecules of cyclohexanone, both carbons of the cleaved double bond must each end up as the carbonyl carbon of a cyclohexanone ring. This means the starting alkene has a $$\mathrm{C=C}$$ double bond joining the carbonyl-position carbons of two cyclohexane rings — i.e. the two rings share a double bond between them.

$$\mathrm{(Cyclohexylidene)cyclohexane \xrightarrow[(ii)\ Zn/H_2O]{(i)\ O_3} 2\ Cyclohexanone}$$

The starting material is cyclohexylidenecyclohexane (bicyclohexylidene), in which one ring carbon of each cyclohexane is joined by a double bond, $$\mathrm{C_6H_{10}{=}C_6H_{10}}$$.

Cyanohydrin formation is a nucleophilic addition: the cyanide ion ($$\mathrm{CN^-}$$) must approach and bond to the carbonyl carbon.

In cyclohexanone the carbons next to the $$\mathrm{>C=O}$$ group carry only hydrogen atoms, so the carbonyl carbon is open to attack — $$\mathrm{CN^-}$$ adds easily and the cyanohydrin is formed in good yield.

In 2,2,6-trimethylcyclohexanone there are three methyl groups on the two carbons flanking the carbonyl group (two on C-2, one on C-6). These bulky methyl groups crowd around the carbonyl carbon and sterically hinder the approach of $$\mathrm{CN^-}$$. The attack is therefore blocked, and little or no cyanohydrin is formed.

Semicarbazide is $$\mathrm{H_2N-NH-CO-NH_2}$$. Of its two $$\mathrm{-NH_2}$$ groups, one is attached to the carbonyl carbon (it is part of the $$\mathrm{-CO-NH_2}$$ amide unit) and the other (the terminal $$\mathrm{H_2N-NH-}$$) is not.

The $$\mathrm{-NH_2}$$ group that is joined to the $$\mathrm{>C=O}$$ has its lone pair of electrons drawn into resonance (conjugation) with the carbonyl group:

$$\mathrm{-\overset{..}{N}H_2-C{=}O \leftrightarrow -\overset{+}{N}H_2{=}C-O^-}$$

Because this lone pair is delocalised, that $$\mathrm{-NH_2}$$ group is much less nucleophilic and is not available to attack a carbonyl compound.

The other $$\mathrm{-NH_2}$$ group is not in conjugation with the $$\mathrm{>C=O}$$; its lone pair is fully available, so it is the reactive nucleophile that condenses with the aldehyde/ketone. Hence only one $$\mathrm{-NH_2}$$ group takes part in semicarbazone formation.

Esterification (Fischer esterification) is a reversible reaction:

$$\mathrm{R{-}COOH + R'{-}OH \underset{}{\overset{H^+}{\rightleftharpoons}} R{-}COOR' + H_2O}$$

At equilibrium, an appreciable amount of the acid and alcohol remains unreacted, so the yield of ester is limited.

By Le Chatelier's principle, removing one of the products as soon as it is formed disturbs the equilibrium; the system responds by making more product to partly restore the balance. So removing the water (or distilling off the ester) continually shifts the equilibrium to the right, driving the reaction towards completion and giving a much higher yield of ester.