Problem 75
Question
\(\mathrm{R}-\mathrm{Cl}+\mathrm{AgCN} \longrightarrow \mathrm{A} \stackrel{\text { reductions }}{\longrightarrow} \mathrm{B}\) \(A\) and \(B\) respectively are (a) \(\mathrm{RCN}, \mathrm{RCH}_{2} \mathrm{NH}_{2}\) (b) \(\mathrm{RNC}, \mathrm{RNHCH}_{3}\) (c) \(\mathrm{RCN}, \mathrm{RNHCH}_{3}\) (d) \(\mathrm{RNC}, \mathrm{RCH}_{2} \mathrm{NH}_{2}\)
Step-by-Step Solution
Verified Answer
The correct answer is (b) \( \mathrm{RNC}, \mathrm{RNHCH}_{3} \).
1Step 1: Understanding the Reaction with AgCN
The reaction of an alkyl halide, \( \text{R-Cl} \), with silver cyanide, \( \text{AgCN} \), leads to the formation of an isocyanide. This happens because \( \text{AgCN} \) acts as a nucleophile, and the \( \text{CN}^- \) ion from \( \text{AgCN} \) tends to bond through the nitrogen atom, forming \( \text{RNC} \). Therefore, the compound \( A \) is \( \text{RNC} \) (an isocyanide).
2Step 2: Reduction of the Isocyanide
The reduction of an isocyanide (\( \text{RNC} \)) typically involves the addition of hydrogen atoms to the nitrogen and carbon atoms. This reduction results in a secondary amine \( \text{RNHCH}_{3} \). The compound \( B \), therefore, is \( \text{RNHCH}_{3} \).
3Step 3: Comparing Options with Final Products
With \( A \) as \( \text{RNC} \) and \( B \) as \( \text{RNHCH}_{3} \), compare these findings with the options provided: - Option (a): \( \text{RCN}, \text{RCH}_{2} \text{NH}_{2} \) - Option (b): \( \text{RNC}, \text{RNHCH}_{3} \) - Option (c): \( \text{RCN}, \text{RNHCH}_{3} \) - Option (d): \( \text{RNC}, \text{RCH}_{2} \text{NH}_{2} \)The pair that matches our findings is option (b): \( \text{RNC}, \text{RNHCH}_{3} \).
Key Concepts
Nucleophilic SubstitutionReduction ReactionsIsocyanidesAlkyl Halides
Nucleophilic Substitution
Nucleophilic substitution is a vital reaction in organic chemistry. It involves a nucleophile, a species rich in electrons, attacking an electron-poor carbon, usually found in a molecule with a leaving group like a halogen. In our exercise, \(\text{R-Cl}\), an alkyl halide, is a classic example of having a good leaving group, chloride (Cl). When nucleophilic substitution takes place,
- The nucleophile replaces the leaving group.
- In our reaction, silver cyanide (\(\text{AgCN}\)) acts as the nucleophile.
- The \(\text{CN}^-\) ion attacks the carbon bonded to chlorine, eventually replacing the chlorine.
Reduction Reactions
Reduction reactions in organic chemistry involve the addition of hydrogen or the removal of oxygen from a molecule. In reduction, electrons are transferred to the molecule, decreasing its oxidation state. For isocyanides, like \(\text{RNC}\), the reduction is achieved typically by adding hydrogen. During this process:
- The nitrogen and carbon receive hydrogen atoms.
- This act transforms the structure, converting a linear isocyanide to a more complex organic molecule.
Isocyanides
Isocyanides are unique organic compounds characterized by the presence of the \(-\text{NC}\) group, where nitrogen is bonded directly to carbon. They are the less common form of isomers compared to nitriles. The formation of isocyanides is quite intriguing:
- The nucleophile \(\text{CN}^-\) in \(\text{AgCN}\) prefers to bond through nitrogen rather than carbon.
- This preference arises due to the complex bonding environment and the nature of silver cyanide.
Alkyl Halides
Alkyl halides, also known as haloalkanes, are molecules where a halogen atom is bonded to an alkyl group. They serve as foundational compounds in organic chemistry, acting as a reactive center for numerous reactions like substitutions and eliminations.In our given problem:
- The alkyl halide used is \(\text{R-Cl}\), where R is an alkyl group, and Cl is the halogen.
- The presence of a halogen makes these compounds reactive, often serving as a good leaving group.
- During nucleophilic substitution, the chloride ion is replaced by another group or atom through a reaction with a nucleophile.
Other exercises in this chapter
Problem 73
Compound \(\mathrm{A}\) on reduction gives \(\mathrm{B}\), which on further reaction with \(\mathrm{CHCl}_{3}\) and alcoholic KOH gives compounds \(\mathrm{C}\)
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\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl} \stackrel{\mathrm{NaCN}}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{Ni} / \mathrm{H}_{2}}{\longrightarrow} \mat
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In a set of reactions propionic acid yielded a compound D. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \stackrel{\mathrm{SOCl}_{2}}{\longrightarrow} \mathrm
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Identify \(\mathrm{X}\) and \(\mathrm{Y}\) in the following sequence. \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br} \stackrel{\mathrm{X}}{\longrightarrow}\) produ
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