Identify the following compounds 'A' to 'J' (write only structural formula): (i) Phthalic acid (benzene ring w

Question 53 of 56

Q. Identify the following compounds 'A' to 'J' (write only structural formula):

(i) Phthalic acid (benzene ring with two -COOH groups) --NH3--> A --strong heat(delta)--> B.

(ii) CH3COOH --SOCl2--> C --(i) CH2N2(excess),

(ii) Ag2O, H2O, delta--> D.

(iii) Toluene (ring-CH3) --(i) CrO2Cl2,

(ii) H2O--> E.

(iv) CH3COONa + NaOH --CaO, delta--> F.

(v) CH3-C(triple bond)CH --20% H2SO4, 1% Hg+2, 80C--> G.

(vi) Phenyl acetate (ring-OCOCH3) --anhydrous AlCl3, CS2, delta--> H + I.

(vii) CH3CHO --NaHSO3--> J. [1/2 x 10 = 5] OR

(i) An organic compound produces acetic acid and ethyl alcohol on acid hydrolysis. Write the structural formula of the compound. How can you prepare the compound from acetaldehyde in one step?

(ii) Give an example of Cannizzaro reaction.

(iii) How would you convert (Give only arrow head reaction):

(x) Acetone -> Mesitylene? (y) Benzaldehyde -> Cinnamaldehyde? [2+1+2]

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 5mImportance★★★★★

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Each lettered product follows from a standard named reaction — ammonolysis/cyclisation, Arndt–Eistert homologation, Etard oxidation, soda-lime decarboxylation, Markovnikov alkyne hydration, Fries rearrangement, and bisulphite addition.

(i) Phthalic acid $+ NH_3 \rightarrow$ A = ammonium phthalate ( $C_6H_4(COONH_4)_2$ ) $\xrightarrow{\text{strong heat}}$ B = phthalimide ( $C_6H_4(CO)_2NH$ , loses 2 H $_2$ O and cyclises to the imide).

(ii) $CH_3COOH \xrightarrow{SOCl_2}$ C = acetyl chloride ( $CH_3COCl$ ) $\xrightarrow{\text{(i) excess }CH_2N_2\text{, (ii) }Ag_2O,H_2O,\Delta}$ D = propanoic acid ( $CH_3CH_2COOH$ ) — the Arndt–Eistert homologation, adding one carbon via a diazoketone/Wolff rearrangement.

(iii) Toluene $\xrightarrow{\text{(i) }CrO_2Cl_2\text{ (ii) }H_2O}$ E = benzaldehyde ( $C_6H_5CHO$ ) — the Etard reaction.

(iv) $CH_3COONa + NaOH \xrightarrow{CaO,\Delta}$ F = methane ( $CH_4$ ) — decarboxylation (soda-lime reaction).

(v) $CH_3\text{-}C{\equiv}CH \xrightarrow{20\%\,H_2SO_4,\,1\%\,HgSO_4,\,80^\circ C}$ G = acetone ( $CH_3COCH_3$ ) — Markovnikov (mercuric-ion-catalysed) hydration of the terminal alkyne via an unstable enol that tautomerises to the ketone.

(vi) Phenyl acetate $\xrightarrow{\text{anhydrous }AlCl_3,\,CS_2,\,\Delta}$ H = o-hydroxyacetophenone + I = p-hydroxyacetophenone — the Fries rearrangement, giving a mixture of ortho and para isomers.

(vii) $CH_3CHO + NaHSO_3 \rightarrow$ J = $CH_3\text{-}CH(OH)\text{-}SO_3Na$ — the bisulphite addition compound of acetaldehyde.

OR:

(i) A compound giving acetic acid and ethanol on acid hydrolysis is the ester ethyl acetate, $CH_3COOC_2H_5$ . Prepared in one step from acetaldehyde via the Tishchenko reaction:

$2CH_3CHO \xrightarrow{Al(OC_2H_5)_3} CH_3COOC_2H_5$

(a disproportionation where one molecule is oxidised to the acid part and another reduced to the alcohol part, combining directly as the ester.)

(ii) Cannizzaro reaction example: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$_g$
(a) methanal
(b) ethanal
(c) propanal
(d) propanone
›Reveal solutionSolution

Hg $^{2+}$ /H $_2$ SO $_4$ hydration of propyne (Markovnikov) gives propanone via the enol.

Terminal alkynes undergo Markovnikov (mercuric-ion catalysed) hydration with dilute $H_2SO_4$ in presence of 1% $HgSO_4$ , adding water across the triple bond to give an enol which tautomerises to a ketone (following Markovnikov's rule).

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Ozonolysis of 2, 3 dimethyl but-2-ene, followed by decomposition by Zn dust and water gives _______.$
(a) acetaldehyde
(b) propionaldehyde and acetone
(c) acetone
(d) acetaldehyde and butyraldehyde
›Reveal solutionSolution

2,3-Dimethylbut-2-ene, $(CH_3)_2C=C(CH_3)_2$ , is symmetric, so ozonolysis cleaves it into two identical acetone molecules.

2,3-Dimethylbut-2-ene has the structure $(CH_3)_2C=C(CH_3)_2$ — both carbons of the double bond bear two identical methyl substituents. Ozonolysis (formation of the ozonide, followed by reductive cleavage with Zn dust/water) cleaves the $C=C$ bond, converting each doubly-bonded carbon into a carbonyl carbon:

…

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CBSE 2022Set E1 markMCQ
$Q. When acetylene is passed through dil. H2SO4 in the presence of HgSO4, the compound formed is$
(a) Ethyl alcohol
(b) Acetone
(c) Acetaldehyde
(d) Carbide of Hg
›Reveal solutionSolution

Hydration of acetylene over HgSO4/dil. H2SO4 gives acetaldehyde.

When acetylene (HC≡CH) is passed through dilute sulphuric acid in the presence of mercuric sulphate as catalyst, water adds across the triple bond (Markovnikov):

HC≡CH + H2O --(HgSO4/dil.H2SO4)--> [CH2=CH-OH] -> CH3-CHO

…

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CBSE 2022Set ANNUAL1 mark
$Q. Write the reagent required (denoted '?') for the following reaction: benzonitrile (ring-CN) --?--> benzaldehyde (ring-CHO).$
›Reveal solutionSolution

A nitrile can be reduced selectively to an aldehyde by the Stephen reduction (SnCl2/HCl) or by DIBAL-H, both followed by hydrolysis of the intermediate imine.

Identifying the reaction: Nitrile (-CN) converting to aldehyde (-CHO) is a partial (controlled) reduction — the nitrile must be stopped at the imine stage rather than going all the way to the primary amine.

Method 1 — Stephen reduction: Benzonitrile is treated with stannous chloride (SnCl2) and HCl gas, which reduces it to an imine-stannic chloride complex; this is then hydrolysed with water to release the aldehyde.

$C_6H_5CN \xrightarrow{SnCl_2/HCl} C_6H_5CH=NH\cdot HCl \xrightarrow{H_2O} C_6H_5CHO$

…

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CBSE 2021Set A1 markMCQ
$Q. Dry distillation of calcium formate gives$
(a) HCHO
(b) HCOOH
(c) CH3COOH
(d) CH3CHO
›Reveal solutionSolution

Dry distillation of the calcium salt of formic acid (calcium formate) yields formaldehyde, HCHO.

Dry distillation of the calcium salt of a carboxylic acid is a classic route to aldehydes/ketones. Calcium formate, (HCOO)2Ca, on dry distillation gives formaldehyde:

(HCOO)2Ca ---> HCHO + CaCO3

…

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CBSE 2016Set ANNUAL1 markMCQ
$Q. Which of the following compounds is obtained when calcium acetate is dry distilled?$
(a) Formic acid
(b) Formaldehyde
(c) Acetone
(d) Butanone
›Reveal solutionSolution

Dry distillation of the calcium salt of a carboxylic acid is a classic laboratory route to a ketone — for calcium acetate, this gives acetone.

When the calcium salt of acetic acid (calcium acetate) is heated strongly in the absence of air (dry distillation), it undergoes ketonic decarboxylation: two acetate groups combine, releasing calcium carbonate and forming a ketone — in this case, the simplest ketone, acetone (propan-2-one):

$Ca(CH_3COO)_2 \xrightarrow{\Delta,\,dry\ distillation} CaCO_3 + CH_3COCH_3\ (acetone)$

…

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Q.Identify the following compounds 'A' to 'J' (write only structural formula):

Concept understanding — Preparation of Aldehydes and Ketones

Q.Identify the following compounds 'A' to 'J' (write only structural formula):

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