Carboxylic Acids Test

Result

Carboxylic Acids Test - 4

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Question 1
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Consider the reaction given below,

Q.

The correct observation regarding the above reaction is/are

A
Precipitate formation takes place

B
Racemic mixture of salts are formed

C
Salts on acid hydrolysis gives back the original reactant

D
All are correct

Solution

Aldehydes and ketones form bisulp hite salt with NaHSO₃ which is insoluble in concentrated NaHSO₃ solution due to common ion effect. Hence, if NaHSO₃ is in limited quantity, precipitation may not take place.
Question 2
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Consider the reaction given below,

Q.

The correct deduction(s) regarding the above reagent is/are

A
Zn(Hg)-HCI can bring about the above reaction

B
If N₂ H₄ /NaOH/Heat is used, chloride may also be substituted by the hydroxide group

C
NaBH₄ followed by acid work-up can efficiently bring about the above reaction

D
Option A,B are Correct

Solution

Reducing agent (Zn(Hg)-HCI) in Clemmensen reduction does not affect the halide group. If W olff-Kishner (N₂ H₄ - OH^- ) reduction is done, OH^- may bring about S_N 2 and E2 reaction with halide group.
Question 3
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Consider the following reaction,

Q.

The correct deduction(s) regarding mechanism of the above reaction is/are

A
Reaction involves nucleophilic addition at carbonyl carbon

B
If carbonyl carbon turns chiral, racemic mixture of products is formed provided no other chiral carbon is present

C
If the intermediate aluminium salt is hydrolysed with DCI-D₂ O deuterated alcohol (R —OD) is formed

D
All are correct

Solution

H^- (hydride) ion undergoes nu cle ophilic ad dition at sp² carbon of carbonyls. Aldehydes are more reactive than ketones due to less steric hindrance in former case. Since, nucleophilic attack of hydride ion occur at planar carbonyl carbon, there is equal probability of attack from both side of plane giving racemic mixture if chiral carbon is generated.
Question 4
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Comprehension Type

Direction : 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

An organic compound A (C₇ H₁₆ O) shows both enantiomerism and diastereomerism. Treatment of a pure enantiomer of A with Na₂ CrO₄ /Dil. H₂ SO₄ gives B (C₇ H₁₄ O) - Also A on dehydration with concentrated H₂ SO₄ gives a single alkene C (C₇ H₁₄ ). Ozonolysis of C followed by work-up with Zn-H₂ O gives D (C₅ H₁₀ O) as one of the product which gives racemic mixture on reduction with NaBH₄ .

Q.

The structure of A satisfying the above criteria is

A

B

C

D

Solution

Since, A is completely saturated, it must contain more than one chiral carbon atoms in order to show both enantiomerism and diastereomerism. Also, C on ozonolysis gives D (C₅ H₁₀ O) as one product, other product must be CH₃ SHO. Hence, C must be
Question 5
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An organic compound A (C₇ H₁₆ O) shows both enantiomerism and diastereomerism. Treatment of a pure enantiomer of A with Na₂ CrO₄ /Dil. H₂ SO₄ gives B (C₇ H₁₄ O) - Also A on dehydration with concentrated H₂ SO₄ gives a single alkene C (C₇ H₁₄ ). Ozonolysis of C followed by work-up with Zn-H₂ O gives D (C₅ H₁₀ O) as one of the product which gives racemic mixture on reduction with NaBH₄ .

Q.

If B is reduced with LiAIH₄ followed by acid hydrolysis will give

A
a pure enantiomer

B
a racemic mixture

C
a pair of diastereomers

D
an achiral alcohol

Solution

Since, A is completely saturated, it must contain more than one chiral carbon atoms in order to show both enantiomerism and diastereomerism. Also, C on ozonolysis gives D (C₅ H₁₀ O) as one product, other product must be CH₃ SHO. Hence, C must be
B is enantiomeric. If a pure enantiomer of B is reduced with LiAIH₄ , pair of diastereomers would be formed.
Question 6
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An organic compound A (C₇ H₁₆ O) shows both enantiomerism and diastereomerism. Treatment of a pure enantiomer of A with Na₂ CrO₄ /Dil. H₂ SO₄ gives B (C₇ H₁₄ O) - Also A on dehydration with concentrated H₂ SO₄ gives a single alkene C (C₇ H₁₄ ). Ozonolysis of C followed by work-up with Zn-H₂ O gives D (C₅ H₁₀ O) as one of the product which gives racemic mixture on reduction with NaBH₄ .

Q.

The correct statement regarding the compound D is

A
With CH₃ MgBr followed by acid hydrolysis gives racemic mixture

B
With CH₃ CH₂ CH₂ MgBr followed by acid hydrolysis gives racemic mixture

C
With CH₃ CH₂ MgBr followed by acid hydrolysis gives racemic mixture

D
Both (b) and (c) are correct

Solution

Since, A is completely saturated, it must contain more than one chiral carbon atoms in order to show both enantiomerism and diastereomerism. Also, C on ozonolysis gives D (C₅ H₁₀ O) as one product, other product must be CH₃ SHO. Hence, C must be
Question 7
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One integer Value Correct Type

Direction : This section contains 6 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

Q.

If all the aldehyde isomers of C₅ H₁₀ O is independently treated with HCN/NaCN solution, how many of them will give racemic mixture of cyanohydrin?

A
3

B
4

C
2

D
1

Solution

IV is enantiomeric. Its pure enantiomer, with HCN/NaCN, would produce pair of diastereomers.
Question 8
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If one molecule of trioxane is heated, how many molecules of formaldehyde would be formed?

A
1

B
2

C
3

D
4

Solution
Question 9
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A mixture containing all isomeric cyclohexanedione is treated with excess of NaHSO₃ solution. How many different disulphite salts are formed?

A
1

B
3

C
5

D
8

Solution
Question 10
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Consider the following reaction,

Q .

How many deuterium would be incorporated in the final product?

A
1

B
3

C
2

D
5

Solution
Question 11
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If glyoxal glycol is treated with a mixture of CH₃ MgBr and C₂ H₅ MgBr in diethyl ether followed by acid hydrolysis, how many different diols would be formed, which are simultaneously optically active?

A
8

B
3

C
5

D
1

Solution

One pair of enantiomers for each (I) and (II) while two pairs of enantiomers for (III).
Question 12
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Consider the reaction given below,

Q.

X is formed in which a new ring is formed. How many atoms are present in this new ring?

A
1

B
5

C
8

D
3

Solution
Question 13
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Matching List Type

Direction : Choices for the correct combination o f 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 reactants from Column I with the reagents and expected outcomes from Column II. Mark the correct option form the codes given below.

A
a

B
b

C
c

D
d

Solution

(i) It has a carbonyl carbon that turns chiral on reduction with hydrides. Hence, with LiAIH₄ or aluminium isopropoxide, gives racemic mixture. Also, it has no other functional groups, either Clemmensen reduction or Wolff-Kishner reduction can be used.
(ii) It has an olefinic double bond, Clemmensen reduction would not be suitable for selective reduction of carbonyl group.
(iii) LiAlH₄ also reduces primary halide but aluminium isopropoxide does not. Wolff-Kishner reduction would not be suitable because HO^- reacts with halide group (E_N 2 or E2).
(iv) Same reasons as with (i).
Question 14
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Match the reactions from Column I with the properties of products from Column II. Mark the correct option form the codes given below.

A
a

B
b

C
c

D
d

Solution