Test: Tests And Distinctions


21 Questions MCQ Test Chemistry Class 12 | Test: Tests And Distinctions


Description
This mock test of Test: Tests And Distinctions for Class 12 helps you for every Class 12 entrance exam. This contains 21 Multiple Choice Questions for Class 12 Test: Tests And Distinctions (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Tests And Distinctions quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this Test: Tests And Distinctions exercise for a better result in the exam. You can find other Test: Tests And Distinctions extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.
QUESTION: 1

Only One Option Correct Type

Direction (Q. Nos. 1-8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

Q. 

Tollen’s reagent used for the distinction of aldehydes with ketones is

Solution:

Tollen’s reagent is ammoniacal solution of silver nitrate in which silver remains as [Ag(NH3)2]+.

QUESTION: 2

A hydrocarbon X on treatment with O3 followed by the reduction of ozonide with Zn-H2O gives Y. Y gives both Tollen's test as well as yellow precipitate with NaOH/I2 solution. Which is a possible structure of X ?

Solution:


The above product has both —CHO and —COCH3 groups required for Tolien's test and iodoform test respectively.

QUESTION: 3

Which compound given below does not form red precipitate with ammoniacal solution of Cu(lI) tartarate ?

Solution:

Fehling's test is not given by benzaldehyde (aromatic aldehyde).

QUESTION: 4

Which reagent can differentiate between benzaldehyde and acetophenone?

Solution:

Benzaldehyde gives Tolien’s test but acetophenone does not. On the other hand, acetophenone (Ph—COCH3) gives iodoform test but benzaldehyde does not.

QUESTION: 5

Which reagent can be used to separate a mixture of ethanol and butanone into components? 

Solution:

Butanone forms precipitate with 2, 4-dinitrophenyl hydrazine, which after separation, can be hydrolysed to obtain back the ketone. 

QUESTION: 6

Reagent that can differentiate 2-propanoi from acetone is

Solution:

Cerric nitrate forms coloured solution with alcohols but does not react with ketone.

QUESTION: 7

Which reagent given below can differentiate propanal from propanone?

Solution:

Schiff`s test is given by aldehydes but not by ketones

QUESTION: 8

Which of the following would produce an orange coloured precipitate with 2, 4-dinitrophenyl hydrazine?

Solution:

Conjugated aldehydes and ketones form orange coloured precipitate with 2, 4-dinitrophenyl hydrazine. Non-conjugated carbonyls form yellow precipitate in the same reaction.

*Multiple options can be correct
QUESTION: 9

One or More than One Options Correct Type

Direction (Q. Nos. 9-14) This section contains 6 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.  Which of the following form(s) yellow precipitate with NaOH/I2 solution?

Solution:

Carbonyls with  group or an alcohol with a methyl group and a “H” at α-carbon gives iodoform test. 

*Multiple options can be correct
QUESTION: 10

Which of the following compounds reduce(s) ammoniacal silver nitrate solution?

Solution:

Both aliphatic aldehydes and aromatic aldehydes give Tolien’s test. Besides, α-hydroxy ketones are also oxidised by Tollen’s reagent.

*Multiple options can be correct
QUESTION: 11

Which of the following fail(s) to reduce Fehling’s solution?

Solution:

Aromatic aldehydes are not oxidised by Fehling's solution. Acid derivatives like ester, anhydride and amide are not oxidised by this reagent. However, α-hydroxy ketones are oxidised by both Fehling’s solution and Tollen’s reagent

*Multiple options can be correct
QUESTION: 12

Which of the following reagents can be used for the separation of a mixture containing CH3CHO and CH3CN?

Solution:

Both NaHSO3 and 2,4-dinitrophenyl hydrazine form precipitate with aldehydes and ketones, from which carbonyls can be recovered back easily.

QUESTION: 13

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

Solution:
*Multiple options can be correct
QUESTION: 14

Which compound(s) given below, upon acid hydrolysis gives a compound that forms a yellow precipitate with KOH/I2 solution?

Solution:

In acid hydrolysis will give a compound which forms a yellow precipitate with KOH/I2 solution.

QUESTION: 15

Comprehension Type

Direction (Q. Nos. 15-17) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Passage

A neutral organic compound A(C10H20O2) neither reduces Tollen’s reagent nor forms precipitate with 2, 4-dinitrophenyl hydrazine, but can be resolved into enantiomers. A on acid hydrolysis forms two compounds B and C, both are enantiomeric. C neither reduces Fehling’s solution nor forms iodoform with alkaline iodine solution. C on oxidation with CrO3/HCI /pyridineforms D which is still resolvable into enantiomers. D on further treatment with aqueous (C2H5O)3 Al solution gives back A.

Q. 

 How many different stereoisomers exist for A ?

Solution:

From the given information, it appears that A is an ester. Since, D on Tischenko reaction gives back A indicates that in ester, both acid and alcohol fragments (B and C respectively) has five carbon atoms each. Also, C is an alcohol which has a chiral carbon and no CH3CH(OH)— grouping (not giving iodoform). Controlled oxidation of C gives D, a carbonyl which is still chiral. Hence, D must be an aldehyde with a chiral carbon.


The above ester has two chiral carbons and no symmetry hence, four stereoisomers.

QUESTION: 16

A neutral organic compound A(C10H20O2) neither reduces Tollen’s reagent nor forms precipitate with 2, 4-dinitrophenyl hydrazine, but can be resolved into enantiomers. A on acid hydrolysis forms two compounds B and C, both are enantiomeric. C neither reduces Fehling’s solution nor forms iodoform with alkaline iodine solution. C on oxidation with CrO3/HCI /pyridineforms D which is still resolvable into enantiomers. D on further treatment with aqueous (C2H5O)3 Al solution gives back A.

Q. 

 What ist rue regarding D ?

Solution:

From the given information, it appears that A is an ester. Since, D on Tischenko reaction gives back A indicates that in ester, both acid and alcohol fragments (B and C respectively) has five carbon atoms each. Also, C is an alcohol which has a chiral carbon and no CH3CH(OH)—grouping (not giving iodoform). Controlled oxidation of C gives D, a carbonyl which is still chiral. Hence, Dmust be an aldehyde with a chiral carbon.

QUESTION: 17

A neutral organic compound A(C10H20O2) neither reduces Tollen’s reagent nor forms precipitate with 2, 4-dinitrophenyl hydrazine, but can be resolved into enantiomers. A on acid hydrolysis forms two compounds B and C, both are enantiomeric. C neither reduces Fehling’s solution nor forms iodoform with alkaline iodine solution. C on oxidation with CrO3/HCI /pyridineforms D which is still resolvable into enantiomers. D on further treatment with aqueous (C2H5O)3 Al solution gives back A.

Q. 

The correct statement regarding the following reaction is

Solution:

From the given information, it appears that A is an ester. Since, D on Tischenko reaction gives back A indicates that in ester, both acid and alcohol fragments (B and C respectively) has five carbon atoms each. Also, C is an alcohol which has a chiral carbon and no CH3CH(OH)—grouping (not giving iodoform). Controlled oxidation of C gives D, a carbonyl which is still chiral. Hence, D must be an aldehyde with a chiral carbon.

QUESTION: 18

CH3CHO and C6H5CH2CHO can be distinguished chemically by:

Solution:
*Answer can only contain numeric values
QUESTION: 19

How many different isomers of C5H8O, on treatment with 2, 4-dinitrophenyl hydrazine gives orange precipitat ?


Solution:

Conjugated aldehydes and ketones give orange precip itate with 2, 4-dinitrophenyl hydrazine


QUESTION: 20

One mole of an organic compound 'A' with the formula C3H8O reacts completely with two moles of HI to form X and Y. When 'Y' is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound 'A' is ______.

Solution:
QUESTION: 21

Matching List Type

Direction (Q. No. 21) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

Q.

Match the qualitative tests listed in Column I with compounds from Column II that gives positive response to these tests and mark the correct option from the codes given below.

Solution:

(i) Benzaldehyde forms orange precipitate with 2, 4-dinitrophenyl hydrazine while all others form yellow precipitate.
(ii) Benzaldehyde, being conjugated, forms orange precipitate with 2, 4-dinitrophenyl hydrazine.
(iii) Aldehydes (aliphatic) and a-hydroxy ketones are oxidised by Fehling's solution giving red precipitate of Cu2O. (iv) NaHSO3 forms white precipitate of bisulphite with all carbonyls. 

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