CLASS XII SET A

(ii)Hydrolysis (Y) In the above ... Formic acid > Acetic acid > Propionic acid (ACID STRENGTH) (b) Cyclohexanol < Phenol < Benzoic acid ... CLASS XII ...

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CLASS XII

SET A CHEMISTRY

31.

Which of the following is not the mechanism of Cannizaro reaction? 2PhCHO →

⊝ OH

PhCH2 OH + PhCOO⊝

(a) The attack of OH− at the (C=O) group (b) The transfer of H ⊝ ion to the (C=O) group (c) The abstraction of H ⊕ ion from carboxylic acid (d) The deprotonation of PhCH2 OH Ans: (D) The first three options are a part of the mechanism of Canizzaro’s reaction in that order. 32.

HgSO4

CH ≡ CH → H2 SO4

LiAIH4

(X) →

(a) Ethylene bromide (c) Ethyl bromide

P4 /Br2

(Y) →

(Z) In this sequence of reaction, (Z) is:

(b) Ethanol (d) Methyl bromide

Ans: (C) X is acetaldehyde, Y is ethyl alcohol, Z is (C) Mild

33.

(i)CH3 MgI

(CH3 )2 CHOH → (X) → (Y) (ii)Hydrolysis Oxidation[O] In the above sequence of reaction, (Y) is:

(a) Isobutyl alcohol (b) 𝑛 −Butyl alcohol (c) Tertiary butyl alcohol (d) Isobutylene Ans: (C) X is acetone, which then undergoes Grigrnard addition to give (C) 34.

Which of the following ketone will not give yellow precipitate with NaOH/I2 ?

(a) 2-Pentanone (b) Benzophenone Ans: (B) No keto-methyl or 𝛼 − 𝐻 in (B) 35.

(c) Acetone

(d) 2-Hexanone

2-Propanone (acetone) can be obtained by Wacker’s process starting with:

(a) Propyne Ans: (A)

(b) Propan-2-ol

(c) Propene

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(d) Propanol-1-ol

CLASS XII 36.

SET A

A compound has molecular formula C6 H12 O. It does not reduce Tollens or Fehling’s reagent, but gives a crystalline derivative with 2,4-dinitro-phenyl hydrazine. With alkali and I2 , it gives yellow solid with a medicinal odour. Clemmensen reduction converts it to 2-methylpentane. The structural formula of the compound is most likely to be: (a) CH3 − COCH2 − CH − (CH3 )2 (c) CH3 CH2 CH2 − CO − CH2 CH3

(b) CH3 − CH2 − CO − CH − (CH3 )2 (d) (CH3 )2 − CH − CO − CH − (CH3 )2

Ans: (A) The compound must have a keto-methyl group. Assertion-Reasoning Type In each of the following questions, a statement of Assertion (A) is given. It is followed by a corresponding statement of Reason (R). Of the statements, mark the correct answers as: (a) If both (A) and (R) are true and (R) is the correct explanation of (A). (b) If both (A) and (R) are true but (R) is not the correct explanation of (A). (c) If (A) is true but (R) is false. (d) If (A) is false but (R) is true. 37.

Assertion (A) : The following intramolecular aldol reaction occurs.

Reason (R) 38.

Ans: D Assertion (A) : The equilibrium constant for cyanohydrin formation is 1013 times greater for cyclohexanone than for cyclopentanone. Reason (R)

39.

: For a five-membered ring, reactions are more favorable when a ring changes from 𝑠𝑝3 to 𝑠𝑝2 because eclipsing interactions are removed.

Ans: A Assertion (A) : Pentan-3-one on reaction with NaHSO3 gives sodium bisulphite adduct product. Reason (R)

40.

: The carbanion does not add to C-4 because a strained fourmembered ring would result.

: (C=0) group of pentan-3-one is sterically hindered and nucleophilic addition reaction does not occur.

Ans: D Assertion (A) : The following intramolecular aldol reaction occurs.

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CLASS XII Reason (R) Ans: D

41.

SET A

: (CHO) is a better acceptor than (C=0).

PhCONH2 (B) → → (C). Product (C) is: (A) KOBr MeOH (a) PhNH2 (b) PhNHCOOMe (c) PhNHCOOPh (d) None Ans: (B) Hoffman bromamide reaction. 𝑃ℎ − 𝑁 = 𝐶 = 𝑂 reacts with MeOH to give urethane (B)

42.

Mark the correct order or increasing reactivity. (a) CH3 CONH2 < CH3 COOC2 H5 < CH3 COCl (b) CH3 COOC2 H5 < CH3 COCl < CH3 CONH2 (c) CH3 COCl < CH3 CONH2 < CH3 COOC2 H5 (d) CH3 COOC2 H5 < CH3 CONH2 < CH3 COCl Ans: (A)

43. Here, (X) is: (a) Glycollic acid (c) Succinic acid Ans: (D)

(b) α-Hydroxypropionic acid (d) Malonic acid

44. (a) C2 H5 I

(b) C2 H5 OH

(c) CHI3

Ans: (C)

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(d) CH3 CHO

CLASS XII 45.

The compound which reacts with Fehling’s solution is: (a) C6 H5 COOH Ans: (B)

46.

SET A (b) HCOOH

(c) C6 H5 CHO

(d) CH2 ClCH3

Which of the following order is incorrect w.r.t. the property indicated? (a) Formic acid > Acetic acid > Propionic acid (ACID STRENGTH) (b) Cyclohexanol < Phenol < Benzoic acid (ACID STRENGTH) (c) Benzamide < Cyclohexylamine < Aniline (BASIC STRENGTH) (d) FCH2 COOH > ClCH2 COOH > BrCH2 COOH (ACID STRENGTH) Ans: (C)

For Questions (47-49), refer to the reaction map below:

47.

The compound (B) is:

Ans: (B) 48.

The compound (C) is:

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CLASS XII

SET A

Ans: (C) 49.

The compound (D) is:

Ans: (C) Solution:

50.

In Rosenmund reduction, which of the following does not poison the catalyst Pd?

(a) BaSO4 (b) S (c) Quinoline (d) Xylene Ans: (D) Xylene is used as a solvent for this reaction. The rest are poisons for the catalyst (by reducing the surface area of Pd), thereby reducing the reductive nature of the reaction.

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CLASS XII

SET A

51.

Ans: B

52.

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CLASS XII

SET A

Ans: D

Read the paragraph below to answer Questions (53 to 55) The “OLIVINE” series of minerals consists of crystals in which Fe2+ and Mg 2+ ions may substitute for each other causing substitutional impurity defects without changing the volume of unit cell. In “OLIVINE” series of minerals, O2− ions exist as FCC with Si4+ occupying one-fourth of Octahedral Voids and divalent metal ions occupying one-fourth of Tetrahedral Voids. The density of “forsterite” (magnesium silicate) is 3.21g cm−3 and that of “fayalite” (ferrous silicate) is 4.34g cm−3. 53.

The formula of “fayalite mineral” is: (a) Fe2 SiO4 Ans: (A)

54.

(c) Fe2 SiO6

(d) FeSiO3

If in “forsterite mineral” bivalent Mg 2+ ions are to be replaced by unipositive Na⊕ ions, and if Na⊕ ions are occupying half of Tetrahedral Voids in FCC lattice, the arrangement of rest of the constituents is kept same, then the formula of the new solid is: (a) Na2 SiO4 Ans: (C)

55.

(b) FeSiO4

(b) Na2 SiO3

(c) Na4 SiO4

(d) Na2 Si2 O6

The percentage of “fayalite” in an “OLIVINE” with density 3.88 g cm−3 approximately is:

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CLASS XII (a) 75% Ans: (B)

SET A (b) 59%

(c) 35%

Paragraph for Problems (56 to 59) Extraction of aluminium can be understood by:

Electrolytic reduction of Al2 O3 : Electrolyte: Al2 O3 + Cryolite + CaF2 Cathode: Carbon inside the Fe container Anode: Graphite rods 56.

The purpose of adding cryolite is (a) To remove the impurities as slag (b) To lower the melting point of Al2 O3 (c) To decrease the electrical conductivity of pure aluminium (d) To increase the Al percentage in the yield Ans: (B)

57.

Coke powder is spread over the molten electrolyte to (a) Prevent the corrosion of graphite anode (b) Prevent the heat radiation from the surface (c) Prevent the oxidation of molten aluminium by air (d) Both (a) and (b) Ans: (D)

58.

The function of fluorspar (CaF2 ) is (a) To increase the melting point of electrolyte (b) To increase electrolytic conductivity power (c) To remove the impurities as slag (d) All of these Ans: B

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(d) 29%

CLASS XII 59.

SET A

The molten electrolytes contain Na⊕ , Al3+ and Ca2+ but only Al gets deposited at cathode because, (a) Standard reduction potential of Al is more than that of Na and Ca (b) Standard oxidation potential of Al is more than that of Na and Ca (c) Graphite reacts only with Al3+ and not with Na⊕ and Ca2+ (d) Discharge potential of Al3+ is higher than Na⊕ and Ca2+ Ans: A

60.

Consider the following metallurgical processes: 1. 2. 3.

Heating impure metal with CO and distilling the resulting volatile carbonyl (Boiling Point 43°C) and finally decomposing at 150°C to 230°C to get the pure metal Heating the sulphide ore in air until a part is converted to oxide and then further heating in the absence of air to let the oxide react with unchanged sulphide Electrolysing the molten electrolyte containing CaCl2 to obtain the metal

The processes used for obtaining sodium, nickel, and copper are, respectively (a) 1, 2 and 3

(b) 2, 3 and 1

(c) 3, 1 and 2

(d) 2, 1 and 3

Ans: (C) Sodium = Electrolytic method, Nickel = Mond’s process, Copper = Auto-reduction

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