Aldehydes and Ketones Test

Result

Aldehydes and Ketones Test - 11

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Question 1
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Which of the following fail(s) to reduce Fehling ’s solution?

A
H₂ CO

B
C₆ H₅ CHO

C

D
Option B,C are Correct

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
Question 2
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Which of the following reagents can be used for the separation of a mixture containing CH₃ CHO and CH₃ CN?

A
NaHSO₃

B
Schiff`s reagent

C
2, 4-dinitrophenyl hydrazine

D
Option A,C are Correct

Solution

Both NaHSO₃ and 2,4-dinitrophenyl hydrazine form precipitate with aldehydes and ketones, from which carbonyls can be recovered back easily.
Question 3
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One mole of a symmetrical alkane on ozonolysis gives two moles of an aldehyde having molecular mass of 44u. The alkene is:

A
Ethene

B
Propene

C
1-butene

D
2-butene

Solution

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.
Question 4
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Which compound(s) given below, upon acid hydrolysis gives a compound that forms a yellow precipitate with KOH/I₂ solution?

A

B

C

D
Option B,C are Correct

Solution

In acid hydrolysis will give a compound which forms a yellow precipitate with KOH/I₂ solution.
Question 5
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Comprehension Type

Direction (Q. Nos. 5 - 7) 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(C₁₀ H₂₀ O₂ ) 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 (C₂ H₅ O)₃ Al solution gives back A.

Q.

How many different stereoisomers exist for A ?

A
2

B
4

C
5

D
6

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 CH₃ CH(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 6
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A neutral organic compound A(C₁₀ H₂₀ O₂ ) 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 (C₂ H₅ O)₃ Al solution gives back A.

Q.

What ist rue regarding D ?

A
It forms a yellow precipitate with KOH/I₂

B
It forms an orange precipitate with 2, 4-dinitrophenyl hydrazine

C
On oxidation with K₂ Cr₂ O₇ /H₂ SO₄ , it forms B

D
On treatment with NaHSO₃ , it forms salt which cannot be separated by fractional crystallisation

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 CH₃ CH(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 7
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A neutral organic compound A(C₁₀ H₂₀ O₂ ) 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 (C₂ H₅ O)₃ Al solution gives back A.

Q.

The correct statement regarding the following reaction is

A
Product is achiral

B
Product is a racemic mixture

C
Product is a pair of diastereomer

D
No α-halogenation occur with 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 CH₃ CH(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 8
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CH₃ CHO and C₆ H₅ CH₂ CHO can be distinguished chemically by:

A
Tollen 's reagent test

B
Fehling solution test

C
Benedict test

D
Iodoform test
Question 9
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How many different isomers of C₅ H₈ O, on treatment with 2, 4-dinitrophenyl hydrazine gives orange precipitate?

A
9

B
5

C
4

D
1

Solution

Conjugated aldehydes and ketones give orange precip itate with 2, 4-dinitrophenyl hydrazine
Question 10
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One mole of an organic compound 'A 'with the formula C₃ H₈ O 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 ______.

A
Propan-1-ol

B
Propan-2-ol

C
ethoxyethane

D
methoxyethane
Question 11
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Matching List Type

Direction (Q. No. 11) 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.

A
a

B
b

C
c

D
d

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 Cu₂ O. (iv) NaHSO₃ forms white precipitate of bisulphite with all carbonyls.