Question

Which among the following compound is formed when aldehyde reacts with HCN in presence of base?
(A) Cyanide
(B) Isocyanide
(C) Cyanohydrin
(D) Hydrogen cyanide
Choose the correct answer from the options given below:

• A. Cyanide
• B. Isocyanide
• C. Cyanohydrin
• D. Hydrogen cyanide
• A. CH₃NH₂ > (CH₃)₂NH > NH₃ > (CH₃)₃N
• B. CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N > NH₃​
• C. NH₃ > (CH₃)₃N > (CH₃)₂NH > CH₃NH₂​
• D. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
• A. (B) and (D) only
• B. (A), (B) and (D) only
• C. (B), (C) and (D) only
• D. (A), (B), (C) and (D)
• A. Ethanoic acid
• B. Methanoic acid
• C. Propanoic acid
• D. Butanoic acid
• A. R-I>R-Br>R-Cl
• B. R-Cl>R-Br>R-I
• C. R-Br>R-Cl>R-I
• D. R-Br>R-I>R-Cl

Answer 1

The Correct Option is C

Aldehydes react with hydrogen cyanide (HCN) in the presence of a base to form cyanohydrins. Let's explore how this reaction occurs:

The reaction mechanism can be described as follows: 

1. In the presence of a base, HCN dissociates into CN^- (cyanide ion) and H⁺.
   The base helps to generate the nucleophile, CN^-, which plays a critical role in the reaction.
2. Cyanide ion (CN^-) acts as the nucleophile and attacks the electrophilic carbon in the carbonyl group of the aldehyde, resulting in a tetrahedral intermediate.
3. The oxygen in the tetrahedral intermediate takes up a proton (H⁺) to form the final product: a cyanohydrin.

The formation of cyanohydrins is a type of nucleophilic addition reaction, where the nucleophile (CN^-) adds across the carbonyl group of the aldehyde. Due to both steric and electronic factors, aldehydes are generally more reactive than ketones, which is why this reaction predominantly yields cyanohydrins when aldehydes are used.

Thus, the correct choice for the compound formed is Cyanohydrin (Option C).

Answer 2

To determine the correct decreasing order of basic strength of the given amines in an aqueous solution, we need to consider both electronic effects and steric hindrance. In general, the basicity of amines is influenced by the electron-donating ability of the alkyl groups attached to the nitrogen atom and the steric hindrance around the nitrogen. 

Key Considerations:

• Electronic Effects: Alkyl groups are electron-donating and can enhance the basicity of the amine by increasing the electron density on the nitrogen atom, making it more available to accept a proton (H⁺).
• Steric Hindrance: As the number of alkyl groups increases, steric hindrance can reduce the accessibility of nitrogen to protons, affecting the basicity.

Analyzing each amine:

• (CH₃)₂NH (Dimethylamine): This compound benefits from both a strong electron-donating effect from two methyl groups and relatively low steric hindrance compared to (CH₃)₃N, which makes it the most basic in the group.
• CH₃NH₂ (Methylamine): With one methyl group, it has a good electron-donating effect but less steric hindrance than dimethylamine, making it the next in basicity.
• (CH₃)₃N (Trimethylamine): Although it has three electron-donating methyl groups, the steric hindrance significantly decreases its basicity compared to the other two amines.
• NH₃ (Ammonia): Having no alkyl groups, its basicity is the lowest due to lack of electron-donating groups enhancing the nitrogen's ability to accept protons.

Thus, the correct decreasing order of basic strength is: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

Answer 3

The given question asks us to identify the reactions where a new C-C bond formation is possible. Let's evaluate each reaction: 

• Cannizzaro reaction: This is a reaction where an aldehyde without an α-hydrogen atom undergoes disproportionation under base conditions, resulting in an alcohol and a carboxylic acid. No C-C bond is formed in this reaction.
• Friedel-Crafts alkylation: This reaction is used to add alkyl groups to an aromatic ring. It involves the formation of a C-C bond between the aromatic ring and the alkyl or acyl group. C-C bond formation occurs.
• Clemmensen reduction: This reaction reduces aldehydes or ketones to hydrocarbons using zinc amalgam and hydrochloric acid. No new C-C bond is formed.
• Riemer-Tiemann reaction: This reaction typically involves the formylation of phenols to introduce an aldehyde group to the aromatic ring. It involves the formation of new C-C bonds between the phenolic substrate and the formyl group. C-C bond formation occurs.

Based on this analysis, new C-C bond formation occurs in the Friedel-Crafts alkylation and Riemer-Tiemann reaction. Thus, the reactions in which a new C-C bond formation is possible are: Friedel-Crafts alkylation and Riemer-Tiemann reaction.

Reaction	New C-C Bond Formation
Cannizzaro	No
Friedel-Crafts Alkylation	Yes
Clemmensen Reduction	No
Riemer-Tiemann	Yes

The correct answer is: (A), (B) and (D) only.

Answer 4

The Tollen’s test is used to detect the presence of aldehydes. It involves a mild oxidizing agent, specifically ammoniacal silver nitrate (Ag(NH₃)₂⁺), also known as Tollen’s reagent. This test produces a silver mirror on the inside of the test tube if an aldehyde is present. Among carboxylic acids, only methanoic acid (formic acid) reacts with Tollen’s reagent. 
While examining the given options:

• Ethanoic acid (CH₃COOH), Propanoic acid (C₂H₅COOH), and Butanoic acid (C₃H₇COOH) are typical carboxylic acids and do not contain an aldehyde group. Therefore, they do not respond to Tollen’s test.
• Methanoic acid (HCOOH) contains an aldehyde-like structure because it has a hydrogen attached to the carbonyl carbon, allowing it to be oxidized to carbon dioxide in the presence of Tollen’s reagent, and hence, it responds to Tollen's test.

The reason methanoic acid reacts is that, electronically, it behaves like an aldehyde, as the carbon in the acid group (COOH) has a bonded hydrogen atom making it susceptible to oxidation by Tollen’s reagent.

Answer 5

To determine the order of reactivity of haloalkanes towards nucleophiles, we need to consider the bond strength between the carbon and the halogen atom. The reactivity of haloalkanes in nucleophilic substitution reactions generally decreases as the bond strength between carbon and the halogen increases.

The carbon-halogen bond strength is influenced by the size and electronegativity of the halogen atom. Larger halogen atoms form weaker bonds with carbon due to longer bond lengths and reduced overlap between orbitals.

From iodine to chlorine, the size of the halogen decreases (I > Br > Cl), while the bond strength increases (C-Cl > C-Br > C-I).

Therefore, the order of reactivity of haloalkanes is: R-I > R-Br > R-Cl