Q.Which of the following compounds do not undergo aldol condensation? (Two or more options may be correct.)

Concept and Intuition

Aldol condensation joins two carbonyl compounds bearing $\alpha$-hydrogens using a base catalyst; the initial aldol (β-hydroxy ketone/aldehyde) then dehydrates on heating to give an $\alpha,\beta$-unsaturated carbonyl compound. For acetone, the base-catalysed self-condensation (classically with $Ba(OH)_2$) gives diacetone alcohol, which on heating loses water to form mesityl oxide.

Step-by-Step Solution

1. Mesityl oxide structure: $(CH_3)_2C=CH-COCH_3$ — an unsaturated ketone with 6 carbons, consistent with joining two 3-carbon acetone units and losing one water molecule.
2. Option (A): two acetaldehyde molecules under dil. NaOH/Δ give crotonaldehyde, not mesityl oxide.
3. Option (B): two acetone molecules under $Ba(OH)_2$/Δ — this is exactly the known route: acetone → diacetone alcohol → (–H2O) → mesityl oxide. …

Concept and Intuition

Vicinal dihalides undergo double elimination with hot alcoholic KOH to give alkynes. A terminal alkyne can be hydrated with dilute acid and $\mathrm{Hg^{2+}}$ catalyst via Markovnikov addition, producing an enol that immediately tautomerises (isomerises) to the more stable keto form — for propyne this gives acetone. Acetone, being a simple dialkyl ketone with no aldehyde group, cannot be oxidised by the mild oxidants used in Tollens'/Fehling's tests, which specifically detect the aldehyde (–CHO) functional group.

Step-by-Step Solution

1. $\mathrm{CH_3CHBrCH_2Br}$ (a vicinal dibromide) $\xrightarrow{\text{Alc. KOH}, \Delta}$ double dehydrohalogenation gives propyne, $\mathrm{CH_3-C\equiv CH}$.
2. $\mathrm{NaNH_2}$ acts on the terminal alkyne (deprotonating/confirming the terminal triple bond position); X remains propyne (a terminal alkyne) after aqueous workup.
3. X $\xrightarrow[333K]{H_2O,\ Hg^{2+}/H^+}$ [y]: Markovnikov hydration of the terminal alkyne gives the unstable enol $\mathrm{CH_2=C(OH)CH_3}$ (bracketed as an unstable intermediate, [y]).
4. [y] $\xrightarrow{\text{Isomerisation}}$ z: keto-enol tautomerisation converts the enol into the stable keto form, z = acetone, $\mathrm{CH_3COCH_3}$. …

Concept and Intuition

Ozonolysis (O3 then Zn/H2O) cleaves a C=C bond into two carbonyl fragments — each alkene carbon becomes a carbonyl carbon. A crossed aldol condensation between a ketone (which has $\alpha$-hydrogens) and formaldehyde (which has none) always proceeds cleanly: the ketone supplies the enolate nucleophile, formaldehyde is the electrophile, and mild base + heat drives addition then dehydration to the enone.

Step-by-Step Solution

1. Ozonolysis of $\mathrm{(CH_3)_2C=CH_2}$: the double bond carbons are $\mathrm{(CH_3)_2C=}$ and $\mathrm{=CH_2}$. Reductive workup converts each into a carbonyl: $\mathrm{(CH_3)_2C=O}$ (acetone) and $\mathrm{H_2C=O}$ (formaldehyde). So $X$ = acetone, $Y$ = formaldehyde (order interchangeable).
2. Dilute NaOH generates the enolate of acetone at its $\alpha$-carbon ($\mathrm{CH_3}$ group), which attacks the electrophilic carbonyl carbon of formaldehyde (aldol addition step): $\mathrm{CH_3-CO-CH_3} + \mathrm{HCHO} \to \mathrm{CH_3-CO-CH_2-CH_2OH}$ (4-hydroxybutan-2-one).
3. Heating ($\Delta$) causes base-catalysed dehydration (E1cb, loss of the $\beta$-hydroxyl as water) to form the conjugated enone: $\mathrm{CH_3-CO-CH=CH_2}$. …

Concept and Intuition

Benzaldehyde ($\mathrm{C_6H_5CHO}$) has no $\alpha$-hydrogens, so it cannot form its own enolate — it can only act as the electrophile. Acetophenone ($\mathrm{C_6H_5COCH_3}$) has $\alpha$-hydrogens on its methyl group, so under $OH^-$ it forms an enolate that attacks the benzaldehyde carbonyl carbon (crossed/Claisen–Schmidt aldol). The initial $\beta$-hydroxy ketone readily loses water because the resulting C=C is conjugated with two aromatic rings — this conjugated enone (chalcone) is thermodynamically very stable, so it forms as the major product even without deliberately heating. $NaBH_4$ is a mild hydride donor: it reduces ketone/aldehyde carbonyls (1,2-reduction) but does not touch isolated or conjugated C=C double bonds under normal conditions, so it converts the ketone in chalcone to a secondary allylic alcohol while leaving the alkene intact.

Step-by-Step Solution

1. Acetophenone's $\mathrm{CH_3}$ loses an $\alpha$-H to $OH^-$, forming the enolate $\mathrm{C_6H_5C(O^-)=CH_2}$.
2. This enolate attacks the carbonyl carbon of benzaldehyde, giving the aldol adduct $\mathrm{C_6H_5CH(OH)-CH_2-CO-C_6H_5}$. …

Concept and Intuition

This question chains four classic reactions: Markovnikov hydrohalogenation, nucleophilic substitution (hydrolysis), catalytic dehydrogenation of a secondary alcohol to a ketone, and finally base-catalyzed aldol condensation of the resulting ketone with itself. The last step is the standard example (from NCERT) of acetone's self-condensation: two molecules combine (aldol addition) to give diacetone alcohol, which dehydrates on heating to give the enone mesityl oxide.

Step-by-Step Solution

1. $CH_3-CH=CH_2$ (propene) + $HBr$: Markovnikov addition puts Br on the more substituted carbon, giving 2-bromopropane, $CH_3-CHBr-CH_3$.
2. $+ NaOH$: nucleophilic substitution (hydrolysis) replaces Br with OH, giving isopropanol (propan-2-ol), $CH_3-CH(OH)-CH_3$.
3. $Cu$, $573\ K$: catalytic dehydrogenation of a secondary alcohol gives a ketone (removes $H_2$ across the C-OH/C-H), giving acetone, $CH_3-CO-CH_3$. …

Concept and Intuition

Aldol condensation proceeds by base (or acid) generating an enolate/enol from a carbonyl compound with $\alpha$-hydrogens; this nucleophile attacks the electrophilic carbonyl carbon of a second (possibly different) carbonyl compound, giving a $\beta$-hydroxy carbonyl that can dehydrate to an $\alpha,\beta$-unsaturated carbonyl. For a crossed reaction to be classified as an aldol condensation, at least one partner must supply enolizable $\alpha$-H and the other must present a sufficiently electrophilic carbonyl carbon.

Step-by-Step Solution

1. (A) Two different aldehydes, both with $\alpha$-H: each can act as the enolate donor and the carbonyl acceptor for the other — a crossed aldol condensation occurs (though a statistical mixture of four products results).
2. (B) Two different ketones, both with $\alpha$-H: similarly, crossed aldol condensation occurs, though ketone carbonyls are less electrophilic than aldehydes, so yields/selectivity are poorer — but the reaction does occur.
3. (C) An aldehyde with $\alpha$-H and one without (e.g., an aromatic aldehyde like benzaldehyde): the one with $\alpha$-H forms the enolate and attacks the other's (very electrophilic, unhindered) carbonyl — this is the classic Claisen–Schmidt condensation, which occurs efficiently and selectively. …

Concept and Intuition

Three classic named transformations are chained here: (i) $LiAlH_4$ is a strong reducing agent that reduces carboxylic acids all the way to primary alcohols;

(ii) passing an alcohol vapour over copper catalyst at $573\,K$ dehydrogenates it (removes $H_2$) to the corresponding carbonyl compound — a primary alcohol gives an aldehyde;

(iii) an aldehyde with alpha-hydrogens, treated with dilute base, undergoes aldol condensation/addition, where one molecule's enolate attacks a second molecule's carbonyl carbon, giving a $\beta$-hydroxy carbonyl compound ("aldol").

Step-by-Step Solution

1. $CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH$: acetic acid is reduced fully to ethanol. This is X.
2. $CH_3CH_2OH \xrightarrow[573K]{Cu} CH_3CHO$: a primary alcohol over Cu at $573\,K$ dehydrogenates to the aldehyde, acetaldehyde. This is Y. …