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 yield cyanohydrins. The reaction mechanism proceeds as follows:

1. The base deprotonates HCN, generating the cyanide ion (CN^-) and a proton (H⁺). The base is essential for generating the nucleophile, CN^-.
2. The nucleophilic cyanide ion (CN^-) attacks the electrophilic carbon of the aldehyde's carbonyl group, forming a tetrahedral intermediate.
3. The oxygen within the tetrahedral intermediate abstracts a proton (H⁺), resulting in the formation of a cyanohydrin.

This reaction is classified as a nucleophilic addition, where the cyanide ion (CN^-) adds across the aldehyde's carbonyl group. Aldehydes are typically more reactive than ketones due to steric and electronic considerations, leading to preferential cyanohydrin formation with aldehydes.

Therefore, the product of this reaction is Cyanohydrin (Option C).

Answer 2

The decreasing order of basic strength of amines in aqueous solution is determined by electronic effects and steric hindrance. Basicity is affected by the electron-donating capacity of alkyl groups attached to nitrogen and the steric environment around it.

Factors to Consider:

• Electronic Effects: Electron-donating alkyl groups increase electron density on nitrogen, facilitating proton (H⁺) acceptance and thus enhancing basicity.
• Steric Hindrance: An increasing number of alkyl groups can impede proton access to nitrogen, thereby reducing basicity.

Analysis of each amine:

• (CH₃)₂NH (Dimethylamine): Possesses strong electron donation from two methyl groups and less steric hindrance than trimethylamine, making it the most basic.
• CH₃NH₂ (Methylamine): With one methyl group, it exhibits good electron donation and less steric hindrance than dimethylamine, ranking second in basicity.
• (CH₃)₃N (Trimethylamine): Despite having three electron-donating methyl groups, significant steric hindrance considerably lowers its basicity compared to the other two amines.
• NH₃ (Ammonia): Lacking alkyl groups, its basicity is the lowest due to the absence of electron-donating groups that would augment nitrogen's proton-accepting capability.

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

Answer 3

The objective is to identify reactions that result in the formation of a new carbon-carbon (C-C) bond. The following reactions are analyzed:

• Cannizzaro reaction: This reaction involves the disproportionation of an aldehyde lacking an α-hydrogen in a basic medium, producing an alcohol and a carboxylic acid. This reaction does not form a new C-C bond.
• Friedel-Crafts alkylation: This reaction is employed to append alkyl groups to an aromatic ring, establishing a C-C bond between the aromatic ring and the attached alkyl or acyl moiety. C-C bond formation is observed.
• Clemmensen reduction: This process reduces aldehydes or ketones to their corresponding hydrocarbons using zinc amalgam and hydrochloric acid. No new C-C bond is generated.
• Riemer-Tiemann reaction: This reaction typically introduces a formyl group onto a phenolic substrate. It involves the creation of new C-C bonds between the phenol and the formyl group. C-C bond formation occurs.

Consequently, new C-C bond formation is observed in Friedel-Crafts alkylation and the Riemer-Tiemann reaction. Therefore, the reactions capable of forming new C-C bonds 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 options are: (A), (B) and (D) only.

Answer 4

Tollen's test identifies aldehydes using ammoniacal silver nitrate (Ag(NH₃)₂⁺), a mild oxidizing agent. A positive result is indicated by a silver mirror forming in the test tube. Of the carboxylic acids, only methanoic acid (formic acid) reacts with Tollen’s reagent.

• Ethanoic acid (CH₃COOH), Propanoic acid (C₂H₅COOH), and Butanoic acid (C₃H₇COOH) are standard carboxylic acids lacking an aldehyde group, thus they do not react with Tollen’s test.
• Methanoic acid (HCOOH) possesses an aldehyde-like structure due to the hydrogen atom bonded to the carbonyl carbon. This characteristic enables its oxidation to carbon dioxide by Tollen’s reagent, resulting in a positive test.

Methanoic acid's reactivity stems from its electronic structure, where the carbonyl carbon’s bonded hydrogen atom renders it susceptible to oxidation by Tollen’s reagent, mimicking aldehyde behavior.

Answer 5

The order of haloalkane reactivity with nucleophiles is determined by the carbon-halogen bond strength. Higher bond strength correlates with decreased reactivity in nucleophilic substitution reactions.

The carbon-halogen bond strength is affected by the halogen's size and electronegativity. Larger halogens create weaker bonds with carbon due to greater distances and less orbital overlap.

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

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