CBSE 12th Chemistry - Aldehydes , Ketones and Carboxylic Acids Model Question Paper | QB365

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CBSE 12th Chemistry - Aldehydes , Ketones and Carboxylic Acids Model Question Paper

Published on: 04/12/2019

Aldehydes , Ketones and Carboxylic Acids

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Std: 12th Standard CBSE - Chemistry

chapter: - Model Question paper

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Questions:

Generic 2

3 x 2 = 6

How will you carry out the following conversions?
(i) Acetylene to acetic acid
(ii) Toluene to m-nitro benzoic acid

What product will be formed on reaction of propanal with 2-methylpropanal in the presence of NaOH? What products will be formed? Write the name of the reaction also.

Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Pentan-2-one and Pentan-3-one
(ii) Ethanal and Propanal

Generic 3

4 x 3 = 12

An aromatic compound 'A' (Molecular formula C₈H₈O) gives positive 2, 4-DNP test. It gives a yellow precipitates of compound 'B' on treatments with iodine and sodium hydroxide solution. Compound 'A' does not give Tollen's or Fehling's test. On drastic oxidation with potassium permanganate it forms a carboxylic acid 'C' (Molecular formula C₇HO)₆O₂), which is also formed along with the yellow compound in the above reaction. Identify A, B and C write all the reacitons involved.

An alkene 'A' (Mol. formula C₅H₁₀) on ozonolysis gives a mixture of two compounds 'B' and 'C'. Compound 'B' gives positive Fehling's test and also forms iodoform on treatment with I₂ and NaOH. Compound 'C' does not give Fehling's test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C.

Give reasons for the following:
(a) Carboxylic acids do not give characteristic reactions of carbonyl group.
(b) Treatment of benzaldehyde with HCN gives a mixture of two isomers which cannot be separated even by careful fractional distillation.
(c) Sodium bisulphite is used for the purification of aldehydes and ketones.

How are the following conversions carried out?
(i) Ethylcyanide to ethanoic acid
(ii) Butan-1-ol to butanoic acid
(iii) Benzoic acid to m-bromobenzoic acid.

Generic 5

3 x 5 = 15

A compound A on oxidation gives B (C₂H₄O₂). A reacts with dil. NaOH and on subsequent heating forms C. C on catalytic hydrogenation gives D. Identify A, B, C, D and write down the reactions involved.

A compound with molecular formula, C₄H₁₀O₃ on acetylation with acetic anhydride gives a compound with molecular weight 190. Find out the number of hydroxyl groups present in the compound.

10.

Write down functional isomers of a carbonyl compound with molecular formula C₃H₆O. Which isomer will react faster with HCN and why? Explain the mechanism of the reaction also. Will the reaction lead to the completion with the conversion of whole reactant into product at reaction condition. If a strong acid is added to the reaction mixture what will be the effect on concentration of the product and why?

Multiple Choice Question

5 x 1 = 5

11.

Which of the following is the industrial method of preparation of acetaldeyde ?

(a)

CH3CN \(\xrightarrow [ HCI ]{ Sn{ CI }_{ 2 } } \) CH3CH = NH \(\xrightarrow {H_3O^{+}}\)

(b)

CH3COCI + H2 \(\xrightarrow [ BaSO_4 ]{Pd } \) CH3CHO + HCI

(c)

CH2 = CH2 + H2O \(\xrightarrow {Pd ^{2+}}\)

(d)

All of the above.

12.

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is

(a)

2-butene

(b)

ethene

(c)

propane

(d)

1-butene

13.

The product formed in Aldol condensation is

(a)

an alpha, beta unsaturated ester

(b)

a beta hydroxy acid

(c)

a beta hydroxyaldehyde or ketone

(d)

an alpha hydroxy aldehyde or ketone

14.

Which of the following organic compounds answers both iodoform test and Fehling's test?

(a)

Ethanal

(b)

Propanone

(c)

Ethanol

(d)

Methanol

15.

In a reaction, RCHO is reduced to RCH₃ using amalgamated zinc and concentrated HCl and warming the solution. The reaction is known as

(a)

Meerwein-Ponndorf reaction

(b)

Clemmensen reduction

(c)

Wolff-Kishner reduction

(d)

Schiff's reaction

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Answers Key:

Generic 2

3 x 2 = 6

(i) \(CH\equiv CH\overset { dil{ H }_{ 2 }{ SO }_{ 4 } }{ \underset { { Hg }^{ 2+ } }{ \longrightarrow } } { CH }_{ 2 }=CHOH\overset { Tautomerisation }{ \longrightarrow } { CH }_{ 3 }CHO\overset { { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 } }{ \underset { { H }_{ 2 }{ SO }_{ 4 } }{ \longleftarrow } } \)
(ii)

(i) Add I₂ and NaOH. P-2- one will give yellow ppt of iodoform, whereas pentan-3- one will not react.
(ii) Add I2 and NaOH. Ethanal will yellow precicipate of iodoform, whereas proponal does not react.

Generic 3

4 x 3 = 12

(a) It is due to resonance.

(b) It is because we get two optical isomers which have same physical properties but differ in sign of optical rotation therefore, cannot be separated by fractional distillation.

(c) Aldehydes and ketones from addition compounds with NaHSO3 whereas impurities do not.
On hydrolysis, we get pure aldehydes and ketone back.

Generic 5

3 x 5 = 15

(i) Since compound A on oxidation gives compound 8 with M.F. C₂H₄O₂, therefore, compound 8 may be acetic acid, CH₃COOH and A may be acetaldehyde, CH₃CHO.
(ii) Since compound A, i.e., acetaldehyde reacts with dil. NaOH, therefore, it undergoes aldol condensation to afford an aldol.Further since this aldol on heating gives compound (C), therefore, (C) must be an α, β- unsaturated aldehyde, i.e., but-2-en-I-al (crotonaldehyde).
(iii) Since compound (C) on catalytic hydrogenation gives compound D, therefore, D may be either I-butanal or I-butanol depending upon the extent of hydrogenation.
AIl the reactions involved in this question are explained below:

During acetylation, one H-atom (at. mass = Iamu) of the OH group is replaced by an acetyl group, i.e., CH₃CO (mol. mass = 43 amu). .
-OH + (CH₃CO)₂O ⇾ -O-COCH₃ + CH₃COOH
In other words, acetylation of each OH group increases the mass by 43 - I = 42 amu. Now the mol. mass of C₄H₁₀O = 106 amu while that of the acetylated product is 190 amu, therefore, the number of OH groups III the compound\(={190-106\over 42}=2\)

10.

Compound I will react faster with HCN because it is less stearically hindered and has more positive charge on carbonyl compounds.
Mechanism of the reaction:

The reaction will not lead to completion.It is reversible reaction, hence equilibrium established.
Addition of strong acid will will inhibit the reaction because dissociation of HCN will be reduced due to common ion effect and formation of CN^- is prevented.