"Identify A/B/C" & Mixed Reasoning

CHEExam Practice & Reasoning

These questions chain reactions together (e.g. $CH_3COCl \xrightarrow{H_2/Pd\text{-}BaSO_4} A$ ) and ask you to name each intermediate, or they pose one-line "give reasons" prompts.

Success depends on knowing each reagent's signature transformation and the standard reactivity/acidity reasoning the chapter is built on.

Rosenmund (acyl chloride to aldehyde)

RCOCl \xrightarrow{H_2,\ Pd\text{-}BaSO_4} RCHO + HCl

Pd

partially poisoned with

S

or quinoline so reduction stops at the aldehyde

Stephen reduction (nitrile to aldehyde)

RCN \xrightarrow{SnCl_2/HCl} RCH{=}NH \cdot HCl \xrightarrow{H_3O^+} RCHO + NH_4^+

imine salt is hydrolysed on work-up to give the aldehyde

Grignard adds one carbon

R'MgX + RCHO \rightarrow RCH(OMgX)R' \xrightarrow{H_3O^+} RCH(OH)R'

HCHO \rightarrow 1^\circ

alcohol; aldehyde

\rightarrow 2^\circ

; ketone

\rightarrow 3^\circ

Reagent signatures to memorise: Rosenmund $H_2/Pd\text{-}BaSO_4$ gives an aldehyde from acyl chloride; Stephen reduction $SnCl_2/HCl$ then $H_3O^+$ turns a nitrile into an aldehyde; $RMgX$ then $H_3O^+$ adds one carbon and gives an alcohol; PCC oxidises $1^\circ$ alcohol to aldehyde, $2^\circ$ to ketone (no over-oxidation).
Diagnostic test logic: 2,4-DNP (orange ppt) confirms a carbonyl (aldehyde OR ketone); Tollens'/Fehling's positive means aldehyde; iodoform positive means a $CH_3CO\text{-}$ or $CH_3CH(OH)\text{-}$ group. Combine them to pin down a structure.
$C_3H_6O$ + iodoform + 2,4-DNP but no Tollens' mirror $\rightarrow$ propanone (methyl ketone, no $\text{-}CHO$ ); $C_3H_6O$ that is also Tollens'/Fehling's positive would instead be propanal.
Aldehydes > ketones in nucleophilic addition: ketones have two electron-donating $+I$ alkyl groups (less electrophilic carbonyl C) plus more steric crowding around the attack site.
HCHO has no $\alpha\text{-}H$ , so it cannot self-condense (aldol) and instead undergoes Cannizzaro (disproportionation) with conc. alkali; $CH_3CHO$ has $\alpha\text{-}H$ and gives aldol.
Carboxylic acids resist nucleophilic addition because the $\text{-OH}$ lone pair donates into the carbonyl by resonance, lowering its electrophilicity; they react instead by nucleophilic acyl substitution.
Acidity order: a carboxylic acid is more acidic than phenol, which is more acidic than ethanol, because the carboxylate spreads charge over two equivalent O atoms while phenoxide delocalises over a ring and alkoxide has no resonance stabilisation.
Alcohols boil higher than carbonyls of similar mass (e.g. propan-1-ol vs propanal) because alcohols H-bond intermolecularly, whereas aldehydes/ketones lack $\text{O-H}$ and show only weaker dipole-dipole attraction.
Build the molecular formula and degree of unsaturation first: one ring or one $\pi$ bond each count as one degree; a lone $\text{C=O}$ in $C_nH_{2n}O$ accounts for that one degree and signals a simple aldehyde/ketone.
Track the carbon count through a chain: Grignard, $CO_2$ carbonation and cyanohydrin/nitrile routes each ADD one carbon, whereas decarboxylation (soda lime) and haloform reactions REMOVE carbon - use this to cross-check A/B/C.
If a $C_3H_6O$ isomer is Tollens'-positive AND iodoform-negative it must be propanal; iodoform-positive but Tollens'-negative makes it propanone - the two tests together fix the isomer unambiguously.
For "give reason" pairs, always name the controlling effect explicitly: resonance/ $-I$ for acidity, $+I$ /steric for carbonyl reactivity, and H-bonding (and dimerisation in acids) for boiling-point comparisons.

Where the marks go

Reading 2,4-DNP as proof of an aldehyde - it only confirms a carbonyl; you still need Tollens'/Fehling's to decide aldehyde vs ketone.
Forgetting the acidic $H_3O^+$ work-up arrow after Grignard or Stephen reductions - leaving it out gives the magnesium alkoxide/imine salt, not the named product A.
Calling benzaldehyde iodoform-positive or Fehling's-positive: it has no $CH_3CO\text{-}$ group and gives NO red $Cu_2O$ with Fehling's (though it does reduce Tollens').
Mixing up the alcohol class from Grignard: $HCHO$ gives a $1^\circ$ alcohol, an aldehyde a $2^\circ$ , and a ketone a $3^\circ$ - students routinely shift this by one.
Stating ethanol is more acidic than water or that phenol is more acidic than a carboxylic acid - the correct order is carboxylic acid > carbonic acid > phenol > water > alcohol.

How the board asks it

Identify / classifydiagnostic test logic and carbon-count tracking
An organic compound $A$ ( $C_3H_6O$ ) gives a positive iodoform test and an orange precipitate with 2,4-DNP, but does not give a silver mirror with Tollens' reagent. On reaction with $HCN$ , $A$ forms $B$ , which on acid hydrolysis gives $C$ . Identify $A$ , $B$ and $C$ and write the equations involved.
Predict the productnamed-reagent signatures in a reaction chain
Identify $A$ and $B$ in the sequence $CH_3COCl \xrightarrow{H_2/Pd\text{-}BaSO_4} A \xrightarrow{\text{dil. } NaOH} B$ , and write the structure of $B$ .
Give reasons $+I$ /steric, resonance and $\alpha$ -H reasoning
Give reasons: (i) Aldehydes are more reactive than ketones towards nucleophilic addition. (ii) Carboxylic acids are more acidic than phenols. (iii) Formaldehyde ( $HCHO$ ) undergoes the Cannizzaro reaction but not the aldol condensation.
Conversionrosenmund and stephen one-carbon routes
How will you convert: (i) acetyl chloride ( $CH_3COCl$ ) into acetaldehyde; (ii) methyl cyanide ( $CH_3CN$ ) into acetaldehyde?
Distinguishtollens'/fehling's diagnostic logic
How will you distinguish between propanal and propanone by a chemical test? Give the observation in each case.

PreviousConversions & Roadmaps

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.