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When pyrrolidine (C₄H₈NH) is dissolved in water, the following reaction occurs:

a. C₄H₈NH + H₂O → C₄H₈N⁺ + OH⁻
b. C₄H₈NH + H₂O → C₄H₈NH2⁺ + OH⁻
c. C₄H₈NH + H₂O → C₄H₈N⁺ + H₂O⁻
d. C₄H₈NH + H₂O → C₄H₈NH₂⁺ + H₂O⁻

1 Answer

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Final answer:

The ionization of pyrrolidine in water involves the nitrogen compound accepting a proton from water, forming pyrrolidinium and hydroxide ions. The correct answer is option b: C₄H₈NH + H₂O → C₄H₈NH2⁺ + OH−.

Step-by-step explanation:

When pyrrolidine (C₄H₈NH) is dissolved in water, a base ionization reaction typically occurs, where the base accepts a proton from a water molecule. This is comparable to other reactions involving similar compounds such as pyridine or aniline, where the base (the nitrogen-containing organic compound) accepts a hydrogen ion (H+) from water (H₂O), forming a positively charged ion and releasing hydroxide ions (OH−) into solution. The correct reaction for pyrrolidine becoming ionized in water is:

b. C₄H₈NH + H₂O → C₄H₈NH2+ + OH−

This shows pyrrolidine gaining a hydrogen ion to form pyrrolidinium (C₄H₈NH2+) and leaving behind a hydroxide ion, thereby increasing the pH of the solution. This type of reaction is known as a base ionization or sometimes hydrolysis, which is common with bases that are also weak acids in water.

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