CBSE Class 12 Chemistry Aldehyde Ketones And Acids Worksheet Set 02

Assertion Reason type Question

a statement of Assertion followed by a statement reason is given. Choose the correct options out of the following choices.
(A) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
(B) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
(C) Assertion is correct, but reason is wrong statement.
(D) Assertion is wrong, but reason is correct statement

Question. Assertion: Compounds containing –CHO group are easily oxidised to corresponding carboxylic acids.
Reason: Aldehydes cantains C-H bond cannot be easily broken.
Answer: c

Question. Assertion: aromatic aldehyde & formaldehyde undergo cannizaro reaction.
Reason: aromatic aldehydes are as reactive as formaldehyde
Answer: d

VSA type questions

Question. what is popoff’s rule?
Answer. when asymmetric ketone is oxidised to carboxylic acid, then rupturing of the C-C bond takes place in such a way that CO group goes with the less number of carbon. It is known as popoff’s rule.

Question. Why p-nitro benzoic acid has higher Ka value than benzoic acid?
Answer. higher Ka value stronger is the acid.as nitro group has –R as well as –I effect which makes the O-H more polar.hence p-nitro benzoic acid is stronger acid. So it has higher Ka value.

Question. What are acetal?
Answer. when aldehydes react with Alcohols in the presence of dry HCl gas it form alkoxyalcohol (Hemiacetal) then with another molecule of alcohol ot form gem-dialkoxy compound known as acetal.

Question. Higher carboxylic acid are insoluble in water why?
Answer. this is due to increased hydrophobic part, H bonding does not takes place.

Question. what is the role of acid catalysts in nucleophillic addition reaction?
Answer. it increases the +ve charge on carbonyl carbon and make it more electrophillic as a result weak nucleophile can easily attack.

Question. Arrange the following in order of increasing reactivity towards nucleophillic addition reaction Ehanal, propanal, propanone, butanone
Answer. due to steric hindrance & +I effect the reactivity decreases . so order is butanone, propanone, propanal, ethanol.

Question. Why th pH should be around 3.5 during the addition of ammonia derivative compounds to aldehyde/ ketone?
Answer. when the medium is too acidic ammonia derivative get protonated, and they can not act as nucleophile.
But when the medium is slightly acidic the protonation of carbonyl cabon will not takes place, hence the pH shoul be around 3-5.

Question. what is Tollen,s reagent?
Answer. it is an ammoniacal solution of silver nitrate. Ag[ (NH₃)₂]OH. It give silver mirror test with aldehydes. Ketones do not give this test, it is used to distinguish between aldehydes& ketones.

Question. write the IUPAC name of CH₃CH(OH) CH₂ CH₂CHO
Answer. 4- Hydroxypentanal

Question. What is the similiarity between alkene and carbonyl compounds?
Answer. i In alkene and carbonyl compounds C is sp² hybridised.
ii. C is planer.
iii both undergoes addition reaction.

Question. What is formalin?
Answer. It is 40% aq. Solution of formaldehyde , used to preserve the biological specimens.

Question. What is the dissimiliarity between alkene and carbonyl compounds ?
Answer. i. carbonyls under nucleophillic addition reaction whereas alkenes undergoes electrophillic addition reaction.
ii. C-C bond in alkene is longer than C-O bond of carbonyl group.

Question. Which is more reactive towards nucleophillic addition reaction: p-nitro benzaldehyde or benzaldehyde?
Answer.. p-nitrobenzaldehyde. As withdrawing group increases the reactivity towards nucleophillic addition reaction.

Long Answer type question:

Question. How would you distinguished between: propanal & propanone
Answer. As propanone contain CH₃CO- group it will give positive iodoform test, propanal do not give iodoform test.

Question. Complete the reaction
Answer. C₆H₅CHO + H₂NCONHNH₂ →C₆H₅CHO= NCONHNH₂ (Benzaldehyde semicabazide)

Question. Give plausible explanation: cyclohexanone forms cynohydrins in good yield but 2,2,6-trimetheylcyclohexanone does not:
Answer. Due to steric hindrance 2,2,6-trimetheylcyclohexanone does not form cyanohydrins as compared to cyclohexanone.

Question. Give simple chemical tests to distinguish between the following pairs of compounds :
(i) Benzaldehyde and benzoic acid
(ii) Propanal and propanone.
Answer. (i) Benzaldehyde and benzoic acid can be distinguished by sodium bicarbonate test.
Benzoic acid will give effervescence with NaHCO₃ but benzaldehyde will not react.
(ii) Propanal and propanone can be distinguished by their reactions with Tollens’ reagent.
Propanal will form the silver mirror, but propanone does not react.

Question. Account for the following :
CH₃CHO is more reactive than CH₃COCH₃
towards reaction with HCN.
Answer. It is a nucleophilic addition reaction, in which CN – acts as a nucleophile. CH₃CHO undergoes nucleophilic addition reactions faster than CH₃COCH₃ as in CH₃COCH₃ there are two electron releasing methyl groups attached to the carbonyl carbon that hinders the approach of nucleophile to carbonyl carbon and reduce the electrophilicity of the carbonyl group while in CH₃CHO, there is only one methyl group attached to carbonyl carbon.

Question. Distinguish between the following :
(i) C₆H₅—COCH₃ and C₆H₅—CHO
Answer. Benzaldehyde and acetophenone can be distinguished by Tollens’ test.
Benzaldehyde will form silver mirror, on treatment with Tollens’ reagent whereas acetophenone will not show Tollens’ Test.

Question. Aldehydes and Ketones have lower boiling points than corresponding alcohols. Why?
Answer. The boiling points of aldehydes and ketones are lower than that of corresponding alcohols and acids due to absence of intermolecular H–bonding in aldehydes and ketones.

Question. Arrange the following in the increasing order of their boiling points.
CH₃CHO, CH₃COOH, CH₃CH₂OH
Answer. Increasing order of boiling point :
CH₃ — CHO < C₂H₅OH < CH₃ — COOH

Question. Give simple chemical tests to distinguish between the following pair of compounds.
Propanal and Butan-2-one.
Answer. Propanal and Butan-2-one can be distinguised by their reactions with tollen’s reagent. Propanal will form the silver mirror, but Butan-2-one does not react.

Question. Give simple chemical tests to distinguish between the following pair of compounds :
Ethanal and Propanal
Answer.Ethanal and propanal can be distinguished by iodoform test.
Yellow precipitate of iodoform will be formed from ethanal on heating with iodine and sodium hydroxide solution.

Question. Given reason :
pH of reaction should be carefully controlled while preparing ammonia derivatives of carbonyl compounds.
Answer. In strongly acidic medium ammonia derivatives being basic will react with acids and will not react with carbonyl compound. In basic medium, OH– will attack carbonyl group. Therefore, pH of a reaction should be carefully controlled.

Question. An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Answer. The compound forms 2,4-DNP derivative. It shows that it is a carbonyl compound. Further it reduces Tollens’ reagent which shows that it contains aldehydic group. It undergoes Cannizzaro reaction indicating that aldehyde group is without any a-hydrogen. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid which shows that there are two carbon residues on benzene ring. Since the molecular formula is C₉H₁₀O, it fits into the structure, 2-ethylbenzaldehyde.