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How many moles of alkyl halides will react with a mole of ethylamine converting it into its ammonium salt?

Short Answer

Expert verified
1 mole of alkyl halides will react with 1 mole of ethylamine.

Step by step solution

01

Understand the Reaction

The reaction in question involves ethylamine, which is a nucleophile, and an alkyl halide, which typically acts as an electrophile. Ethylamine will donate its lone pair to the alkyl halide, leading to the formation of an ammonium salt.
02

Identify the Stoichiometry

For this type of reaction, one mole of ethylamine reacts with one mole of an alkyl halide to produce one mole of the corresponding ammonium salt. The reaction can be depicted as: \[ \text{R-X} + \text{C}_2\text{H}_5\text{NH}_2 \rightarrow \text{C}_2\text{H}_5\text{NH}_3^+ \text{X}^- \] where \( \text{R-X} \) represents the alkyl halide.
03

Conclusion Based on Stoichiometry

The balanced equation indicates a 1:1 molar ratio between the alkyl halide and ethylamine. This means that for every mole of ethylamine, one mole of alkyl halide is required to fully react and form the ammonium salt.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alkyl Halides
Alkyl halides, also known as haloalkanes, are organic compounds that consist of carbon atoms bonded to a halogen atom. The general structure can be represented as R-X, where R is the alkyl group and X is the halogen (e.g., fluoride, chloride, bromide, or iodide). These halides play a crucial role in organic chemistry due to their ability to undergo a variety of reactions.

Alkyl halides are classified based on the halogen and the type of carbon to which it is attached:
  • Primary alkyl halides: The halogen is attached to a carbon that is connected to only one other carbon.
  • Secondary alkyl halides: The halogen is attached to a carbon that is connected to two other carbons.
  • Tertiary alkyl halides: The halogen is attached to a carbon that is connected to three other carbons.
These compounds serve as electrophiles in many reactions due to the polar nature of the C-X bond, where the carbon is slightly positive, and the halogen is slightly negative.
This polarity makes alkyl halides susceptible to attack by nucleophiles, facilitating various substitution and elimination reactions.
Nucleophile and Electrophile Reactions
In organic chemistry, nucleophiles and electrophiles are essential players in chemical reactions. A nucleophile is an electron-rich species that seeks out the positively charged or electron-deficient sites of other molecules. In contrast, electrophiles are electron-deficient and tend to accept electrons.

The reaction between ethylamine (a nucleophile) and an alkyl halide (an electrophile) exemplifies this dynamic. Ethylamine has a lone pair of electrons on the nitrogen atom, making it ready to attack the electrophilic carbon atom of the alkyl halide. The process involves the nucleophile donating its lone pair to form a new bond while displacing the halogen atom originally bonded to carbon.
This results in the substitution reaction:
  • The nucleophilic attack by ethylamine on the alkyl halide.
  • Displacement of the halogen, forming a new bond.
  • Formation of an ammonium salt where the nitrogen of ethylamine gains an additional proton.
Such reactions are fundamental in organic chemistry, forming bonds that build complex molecules from simpler ones.
Ammonium Salt Formation
Ammonium salt formation is a specific type of reaction that occurs when an amine reacts with an alkyl halide. This process is significant because it transforms basic amines into more stable ionic compounds. When ethylamine reacts with an alkyl halide, the lone pair on the nitrogen atom bonds with the carbon atom of the halide, resulting in the formation of an ammonium ion.

The overall equation can be depicted as follows:
  • Ethylamine ( \( \text{C}_2\text{H}_5\text{NH}_2 \)) acts as a nucleophile, donating an electron pair.
  • The alkyl halide (\(\text{R-X}\)) contains an electrophilic carbon atom.
  • Upon reaction, the newly formed bond results in \( \text{C}_2\text{H}_5\text{NH}_3^+ \text{X}^- \).
The product is an ammonium salt where the ethylammonium ion is paired with a halide ion.
This transformation is important in many synthetic pathways as it makes the properties of a compound more predictable and manageable, often enhancing solubility and stability in various conditions.

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