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Test: Tests & Distinctions - NEET MCQ


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21 Questions MCQ Test Topic-wise MCQ Tests for NEET - Test: Tests & Distinctions

Test: Tests & Distinctions for NEET 2024 is part of Topic-wise MCQ Tests for NEET preparation. The Test: Tests & Distinctions questions and answers have been prepared according to the NEET exam syllabus.The Test: Tests & Distinctions MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Tests & Distinctions below.
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Test: Tests & Distinctions - 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

Detailed Solution for Test: Tests & Distinctions - Question 1

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

Test: Tests & Distinctions - 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 ?

Detailed Solution for Test: Tests & Distinctions - Question 2


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

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Test: Tests & Distinctions - Question 3

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

Detailed Solution for Test: Tests & Distinctions - Question 3

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

Test: Tests & Distinctions - Question 4

Which reagent can differentiate between benzaldehyde and acetophenone?

Detailed Solution for Test: Tests & Distinctions - Question 4

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

Test: Tests & Distinctions - Question 5

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

Detailed Solution for Test: Tests & Distinctions - Question 5

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

Test: Tests & Distinctions - Question 6

Reagent that can differentiate 2-propanoi from acetone is

Detailed Solution for Test: Tests & Distinctions - Question 6

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

Test: Tests & Distinctions - Question 7

Which reagent given below can differentiate propanal from propanone?

Detailed Solution for Test: Tests & Distinctions - Question 7

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

Test: Tests & Distinctions - Question 8

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

Detailed Solution for Test: Tests & Distinctions - Question 8

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
Test: Tests & Distinctions - 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?

Detailed Solution for Test: Tests & Distinctions - Question 9

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

*Multiple options can be correct
Test: Tests & Distinctions - Question 10

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

Detailed Solution for Test: Tests & Distinctions - Question 10

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
Test: Tests & Distinctions - Question 11

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

Detailed Solution for Test: Tests & Distinctions - Question 11

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
Test: Tests & Distinctions - Question 12

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

Detailed Solution for Test: Tests & Distinctions - Question 12

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

Test: Tests & Distinctions - Question 13

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

Detailed Solution for Test: Tests & Distinctions - Question 13

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 g/mol. The alkene is 2-butene.
Both the cis and trans form of 2-Butene on ozonolysis give 2 moles of acetaldehyde with the molecular mass of 44 g/mol. In this process of ozonolysis, ozone reacts with alkenes to break the double bond and forms two carbonyl groups.

*Multiple options can be correct
Test: Tests & Distinctions - Question 14

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

Detailed Solution for Test: Tests & Distinctions - Question 14

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

Test: Tests & Distinctions - 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 ?

Detailed Solution for Test: Tests & Distinctions - Question 15

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.

Test: Tests & Distinctions - 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 ?

Detailed Solution for Test: Tests & Distinctions - Question 16

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.

Test: Tests & Distinctions - 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

Detailed Solution for Test: Tests & Distinctions - Question 17

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.

Test: Tests & Distinctions - Question 18

CH3CHO and C6H5CH2CHO can be distinguished chemically by:

*Answer can only contain numeric values
Test: Tests & Distinctions - Question 19

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


Detailed Solution for Test: Tests & Distinctions - Question 19

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


Test: Tests & Distinctions - 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 ______.

Test: Tests & Distinctions - 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.

Detailed Solution for Test: Tests & Distinctions - Question 21

(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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