Aldehydes Ketones And Carboxylic Acids

The correct answer is (C) :

The correct answer is (A) :

The correct answer is (B) : CrO₃ / acetic anhydride, H₃O⁺

The correct answer is (C) : MnO
Benzoic acid can be converted to benzaldehyde in presence of MnO.

The correct answer is (C) :

To determine the major product 'B' of the given reaction sequence, we need to analyze each step:

1. The initial compound is an alkene, which is treated with in methanol . This is a halogenation reaction in which the alkene reacts with bromine to form a vicinal dihalide.
2. Methanol acts as a nucleophile here and attacks the more substituted carbon, leading to the formation of a bromomethoxy compound.
3. The second step involves the treatment of this bromomethoxy compound with hydrogen iodide .
4. is a strong acid and also provides the iodide nucleophile. It reacts with the bromomethoxy compound to replace the methoxy group with an iodide, primarily through an mechanism.
5. Thus, the major product 'B' formed will have iodine attached in place of the methoxy group.

Considering these steps, the major product 'B' is an iodo-alkane. Below is the correct option representing the major product:

This represents the structure of 'B' correctly as the iodo-alkane formed after the reaction steps described above.

The correct answer is (B)

To solve this question, we need to match chemical reactions in List-I with their corresponding names in List-II. Let's analyze each reaction one by one:

1. Reaction (A): Benzoyl chloride (C₆H₅COCl) is being reduced to benzaldehyde (C₆H₅CHO) using H₂ over palladium-barium sulfate (Pd-BaSO₄). This is known as the Rosenmund reaction. This matches with List-II (IV).
2. Reaction (B): The conversion of a nitrile to an aldehyde using SnCl₂ and HCl followed by hydrolysis is the characteristic of the Stephen reaction. This matches with List-II (III).
3. Reaction (C): Oxidation of a methyl group to an aldehyde using CrO₂Cl₂ (chromyl chloride) is known as the Etard reaction. This matches with List-II (II).
4. Reaction (D): Formation of benzaldehyde (C₆H₅CHO) from benzene in the presence of CO, HCl, and anhydrous AlCl₃ is an example of the Gattermann Koch reaction. This matches with List-II (I).

Thus, the correct match is:

• A (IV): Rosenmund reaction
• B (III): Stephen reaction
• C (II): Etard reaction
• D (I): Gattermann Koch reaction

Therefore, the correct answer is: A(IV), B(III), C(II), D(I)

So, the correct option is (B).

so, the correct option is (D) : cyclohexa-1, 4-diene

Step 1: Analyze the Structure of Compound ‘A’
Given the molecular formula , compound ‘A’ gives a positive neutral FeCl test, indicating the presence of a phenolic ( ) group. Compound ‘A’ also decolourises alkaline KMnO , indicating the presence of a double bond.
Step 2: Chemical Reactions of ‘A’
- Treatment with NaOH and CH Br forms , confirming the presence of a second hydroxyl group.
- Hydroiodic acid treatment yields methyl iodide, confirming the presence of an ether bond ( ).
- Hot NaOH treatment forms compound ‘B’, , which retains a double bond.
Step 3: Determine the Number of Bonds
The structure of compound ‘A’ contains:
One aromatic benzene ring with three bonds.
One aliphatic double bond outside the aromatic ring.
Thus, the total number of bonds in compound ‘A’ is:

Conclusion:
The number of bonds present in compound ‘A’ is .

Statement I:
Clemmensen reduction involves the use of Zn(Hg) and HCl, which reduces ketones or aldehydes to their corresponding alkanes under acidic conditions.
In the given compound, , the ketone group ( ) is reduced, and the intermediate carboxylic acid product ( ) is correct under these conditions.
Hence, Statement I is true.
Statement II:
Wolff-Kishner reduction involves hydrazine ( ) and a strong base such as KOH, which reduces carbonyl groups (ketones or aldehydes) to their corresponding alkanes.
However, in the compound , the chlorine atom may react with the basic medium, leading to a side reaction (dehydrohalogenation) instead of simple reduction. The final product will not necessarily be , as suggested.
Hence, Statement II is false.

The correct answer is (A) : A is false but R is true

The acid sensitive alcohol group reacts with HCl, hence Clemmenson reduction is not suitable for above conversion.

The reaction sequence involves multiple steps:

Step 1: Cyanohydrin Formation (Intermediate P) - The aldehyde group ( ) in chlorobenzaldehyde reacts with in the presence of , leading to the formation of a cyanohydrin intermediate. - Cyanohydrin structure ( ): .

Step 2: Hydrolysis of Cyanohydrin (Intermediate Q) - The cyanohydrin ( ) undergoes hydrolysis with ethanol and acidic conditions ( ) to produce a -hydroxy ketone intermediate ( ). - Structure of ( ): .

Step 3: Grignard Reaction (Formation of Final Product R) - The intermediate ( ) reacts with methyl magnesium bromide ( ) to add two groups to the carbonyl carbon, forming a tertiary alcohol. - This is followed by hydrolysis with , leading to the final product . - Structure of : with groups on the same carbon as the hydroxyl group.

Conclusion: The correct product matches Option (1).

The correct answer is :

The correct answer is (B) :

The correct option is (D): B only

To determine the second homologue in the homologous series of monocarboxylic acids, it's essential to understand the general formula for monocarboxylic acids. The general formula for a monocarboxylic acid is where is a positive integer.

Monocarboxylic acids, also known as aliphatic carboxylic acids, consist of a carboxyl group ( ) attached to an alkyl chain. The series starts with formic acid (HCOOH) where .

To find the second homologue:

1. Formic acid (first compound in this series): (equivalently written as ) corresponds to .
2. Acetic acid (second compound in the series), substituting in the formula: .

Thus, the correct molecular formula of the second homologue in the homologous series of monocarboxylic acids is .

Correction: The initial provided correct answer seems to have been misunderstood as . Upon re-evaluation, using the formula explained above, is indeed the correct formula for acetic acid, the second homologue.

• Option 1 ( ) does not fit the general formula for monocarboxylic acids.
• Option 2 ( ) lacks the carboxyl group.
• Option 3 ( ) is the correct molecular formula for acetic acid.
• Option 4 ( ) corresponds to propionic acid, the third homologue.

Thus, the correct answer is Option C.

To determine the acidity order of the –OH group in the given compounds, we need to understand the effect of substituents on the phenol ring.

1. Phenol Acidity: Phenols are acidic because they can donate a hydrogen ion (proton) from the hydroxyl group, resulting in a phenoxide ion. The stability of the phenoxide ion determines the acidity of phenols.
2. Effect of Substituents:
   • Electron-withdrawing groups (EWG) like increase acidity by stabilizing the negative charge on the phenoxide ion through resonance or induction.
   • Electron-donating groups (EDG) like decrease acidity by destabilizing the phenoxide ion.
3. Analysis of Each Compound:
   • (A) Bu – OH: A simple butanol with no resonance stabilizing effects, therefore, least acidic.
   • (B) : Contains a strong electron-withdrawing group, significantly increasing acidity.
   • (C) : Contains an electron-donating group, decreasing acidity.
   • (D) Phenol: Standard reference for acidic strength in phenols.
   • (E) : The presence of two groups greatly enhances the acidity due to strong electron-withdrawing effects.
4. Conclusion: Based on the effects explained above, the correct order of acidity is:

To determine which sugar does not give a reddish-brown precipitate with Fehling's reagent, we need to understand the chemistry behind Fehling's test.

Fehling's solution is used to test for reducing sugars. It is an alkaline solution containing copper(II) sulfate. When a reducing sugar is present, it reduces the copper(II) ions to copper(I) oxide, which precipitates as a reddish-brown solid. Here's the reaction:

Now, let's consider each option:

1. Lactose - It is a reducing sugar because it has a free anomeric carbon, allowing it to reduce Fehling's solution, producing a reddish-brown precipitate.
2. Maltose - Similarly, maltose is also a reducing sugar due to its free anomeric carbon, and it gives a reddish-brown precipitate with Fehling's reagent.
3. Glucose - Glucose is a reducing sugar known for its ability to give a positive Fehling's test.
4. Sucrose - This is a non-reducing sugar because the glycosidic bond in sucrose involves the anomeric carbon of both glucose and fructose, leaving no free anomeric carbon to act as a reducing agent. Therefore, sucrose does not reduce Fehling's solution and does not produce a reddish-brown precipitate.

Hence, the sugar which does not give a reddish-brown precipitate with Fehling's reagent is: Sucrose