Sodium hydride is a very strong base. In this reaction, one equivalent of sodium hydride is reacting with the thioalcohol shown with a 1-to-1 stoichiometry. Predict the products of this reaction (include both conjugate acid and base), draw the mechanistic arrows and explain why this product is formed. H Na :H .S: 1 equivalent エーの: エ-O:

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Sodium hydride is a very strong base. In this reaction, one equivalent of sodium hydride is reacting with the thioalcohol shown with a 1-to-1 stoichiometry. Predict the products of this reaction (include both conjugate acid and base), draw the mechanistic arrows and explain why this product is formed.

**Diagram Explanation:**

- The reactant is a thioalcohol containing an OH group and an SH group.
- Sodium hydride (Na⁺ ⁻H) is shown as the other reactant, with one equivalent indicated.
- An arrow points from the reactants toward the products, illustrating the reaction pathway.

To complete the reaction:

1. Identify the acidic hydrogen in the thioalcohol that will be deprotonated by hydride.
2. Show the flow of electrons from the hydride ion attacking the hydrogen atom, leading to the formation of H₂ gas and a thiolate ion.
3. The product side should include sodium thiolate (where the sulfur anion pairs with the sodium cation) and hydrogen gas formed.
Transcribed Image Text:Sodium hydride is a very strong base. In this reaction, one equivalent of sodium hydride is reacting with the thioalcohol shown with a 1-to-1 stoichiometry. Predict the products of this reaction (include both conjugate acid and base), draw the mechanistic arrows and explain why this product is formed. **Diagram Explanation:** - The reactant is a thioalcohol containing an OH group and an SH group. - Sodium hydride (Na⁺ ⁻H) is shown as the other reactant, with one equivalent indicated. - An arrow points from the reactants toward the products, illustrating the reaction pathway. To complete the reaction: 1. Identify the acidic hydrogen in the thioalcohol that will be deprotonated by hydride. 2. Show the flow of electrons from the hydride ion attacking the hydrogen atom, leading to the formation of H₂ gas and a thiolate ion. 3. The product side should include sodium thiolate (where the sulfur anion pairs with the sodium cation) and hydrogen gas formed.
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