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Chapter 15: Problem 16

Show how spiro[2.2]pentane can be prepared in one step from organic compounds containing three carbons or less and any necessary inorganic reagents or solvents. Spiro [2.2] pentane

Short Answer

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Question: Describe a one-step synthesis of spiro[2.2]pentane from an organic compound containing three carbons or less, along with any necessary inorganic reagents or solvents. Answer: Spiro[2.2]pentane can be synthesized in one step from 1,3-dibromopropane using sodium amide (NaNH2) and n-butyllithium (n-BuLi). The process involves deprotonation of 1,3-dibromopropane by NaNH2, formation of an organolithium species with n-BuLi, and an intramolecular nucleophilic attack to form the spiro[2.2]pentane structure.

Step by step solution

01

1. Identify the starting organic compound and necessary reagents

The starting organic compound for this synthesis should have the potential to form cyclopropane rings. An appropriate candidate is 1,3-dibromopropane as it contains the necessary three carbon atoms for the synthesis. The necessary reagents for this synthesis are NaNH2 (sodium amide) for deprotonation, and n-BuLi (n-butyllithium) for the formation of an organolithium species.
02

2. Deprotonation of 1,3-dibromopropane

First, NaNH2 (sodium amide) is used to deprotonate 1,3-dibromopropane at its central carbon atom to form a carbanion intermediate. NaNH2 is a strong base which facilitates the removal of a proton, leaving a negative charge on the central carbon atom: \[ C_{2}H_{5}Br - C_{H} - BrC_{2}H_{5} + NaNH_{2} \rightarrow C_{2}H_{5}BrC^{-}BrC_{2}H_{5} + Na^{+}NH_{2}^{-} \]
03

3. Formation of an organolithium species

Next, the obtained carbanion intermediate is treated with n-BuLi (n-butyllithium). The carbanion reacts with n-BuLi, replacing one of the bromine atoms with a lithium atom to form an organolithium species: \[ C_{2}H_{5}BrC^{-}BrC_{2}H_{5} + BuLi \rightarrow C_{2}H_{5}BrCLiC_{2}H_{5} + BuBr \]
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4. Intramolecular nucleophilic attack

Finally, the organolithium species undergoes an intramolecular nucleophilic attack, with the nucleophilic lithium-bound carbon attacking the electrophilic carbon attached to the bromine atom. This forms a cyclopropane ring and changes the lithium to a bromine atom, resulting in the formation of spiro[2.2]pentane: \[ C_{2}H_{5}BrCLiC_{2}H_{5} \rightarrow \mathrm{Spiro[2.2]pentane} + LiBr \] So, spiro[2.2]pentane can be synthesized in one step from 1,3-dibromopropane using NaNH2 and n-BuLi.

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