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Chapter 6: Problem 2

1-Bromo-3-chloropropane reacts with one molar equivalent of sodium cyanide in aqueous ethanol to give a single organic product. What is this product?

Short Answer

Expert verified
The product is 3-chloropropanenitrile.

Step by step solution

01

Understand the Reaction Type

This reaction involves a nucleophilic substitution, specifically an S\(_N\)2 reaction because sodium cyanide (NaCN) is a nucleophile and will attack the electrophilic carbon bonded to the bromine in 1-Bromo-3-chloropropane.
02

Identify the Leaving Group

In 1-Bromo-3-chloropropane, both bromine and chlorine are potential leaving groups. However, since bromine is a better leaving group than chlorine due to its larger size and weaker bond with carbon, bromine will leave to form the product.
03

Determine the Nucleophile

The nucleophile in this reaction is the cyanide ion (CN\(^-\)) from sodium cyanide. The CN\(^-\) ion will attack the carbon from which the bromine leaves.
04

Predict the Organic Product

The CN\(^-\) ion replaces the bromine in the compound, resulting in the formation of 3-chloropropanenitrile. The structure is: \( \text{ClCH}_2\text{CH}_2\text{CH}_2\text{CN} \). This is because the CN\(^-\) substitutes at the location where bromine was attached.

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Key Concepts

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

S_N2 reaction
An S\(_N\)2 reaction, or bimolecular nucleophilic substitution, is a fundamental type of nucleophilic substitution. In an S\(_N\)2 reaction, the bond-breaking and bond-forming steps occur simultaneously. This means that the nucleophile attacks the substrate at the same time as the leaving group departs. The term "bimolecular" refers to the transition state of the reaction, where both the nucleophile and the substrate are involved.
  • The S\(_N\)2 reaction is characterized by one step, involving a single transition state.
  • Unlike S\(_N\)1 reactions, where the reaction rate depends only on the substrate concentration, S\(_N\)2 reactions depend on the concentration of both the nucleophile and the substrate.
  • The reaction proceeds with inversion of configuration, often referred to as a "backside attack." This is because the nucleophile approaches the opposite side of the leaving group, leading to a product that is a mirror image of the original configuration.
Thus, in the reaction involving 1-Bromo-3-chloropropane with sodium cyanide, the S\(_N\)2 mechanism ensures a direct substitution of the leaving group by the nucleophile.
leaving group
The leaving group in a nucleophilic substitution reaction is the atom or group of atoms that is replaced by the nucleophile. Its stability as a free species in the reaction medium determines its ability to leave. A good leaving group typically forms a stable ion after leaving, and therefore, it should typically be the weaker base.
  • Bromine is a preferred leaving group compared to chlorine due to its larger atomic size and the weaker bond it forms with the carbon atom in an organic substrate.
  • Larger atoms like bromine are better at stabilizing the negative charge when they leave, as they have greater polarizability.
In the case of 1-Bromo-3-chloropropane, bromine acts as the leaving group since it can better stabilize once it is dissociated from the carbon atom, leading to the formation of the desired product.
nucleophile
A nucleophile is a chemical species that donates an electron pair to form a chemical bond in reaction. In nucleophilic substitution reactions, the nucleophile targets an electron-deficient carbon.
  • In the given exercise, the nucleophile is the cyanide ion (CN\(^-\)) provided by sodium cyanide (NaCN).
  • Cyanide ions are strong nucleophiles due to the electron density on the carbon atom, which seeks to share its lone pair with an electrophilic target.
In the reaction with 1-Bromo-3-chloropropane, CN\(^-\) efficiently attacks the carbon atom from which the bromine leaves, resulting in the successful substitution to form 3-chloropropanenitrile.
organic product prediction
Predicting the organic product of a nucleophilic substitution involves understanding both the reaction mechanism and the components involved. Here, sodium cyanide reacts via an S\(_N\)2 mechanism and replaces the bromine atom in the substrate.
  • The bromine leaves as Br\(^-\), and CN\(^-\) takes its place directly on the carbon chain.
  • This direct substitution effectively converts 1-Bromo-3-chloropropane into 3-chloropropanenitrile, represented as \(\text{ClCH}_2\text{CH}_2\text{CH}_2\text{CN}\).
  • The chlorine atom remains untouched in this process, as it is not the preferable leaving group under the given conditions.
Thus, by examining the nature of the reactants and the S\(_N\)2 mechanism, one can confidently predict that the resultant organic product is 3-chloropropanenitrile.

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