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Neopentyl iodide is treated with aq. AgN...

Neopentyl iodide is treated with aq. `AgNO_3` solution, a yellow precipitate is formed along with other compound which is

A

`(CH_3)_3C CH_2ONO_2`

B

`(CH_3)_2oversetoverset(OH)(|)C-CH_2CH_3`

C

`(CH_3)_3C CH_2OH`

D

`(CH_3)_2oversetoverset(ONO_2)(|)C-CH_2CH_3`

Text Solution

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To solve the problem of what compound is formed when neopentyl iodide is treated with aqueous AgNO3 solution, we can follow these steps: ### Step-by-Step Solution: 1. **Identify Neopentyl Iodide Structure**: Neopentyl iodide has the structure: \[ \text{C(CH}_3\text{)}_4\text{CH}_2\text{I} \] This can also be represented as: \[ \text{(CH}_3\text{)}_4\text{C-CH}_2\text{I} \] 2. **Reaction with Aqueous AgNO3**: When neopentyl iodide reacts with aqueous AgNO3, the iodide ion (I-) is replaced by a hydroxyl group (OH) due to the nucleophilic substitution reaction. The silver ion (Ag+) reacts with iodide to form a yellow precipitate of silver iodide (AgI). 3. **Formation of the Carbocation**: The reaction mechanism involves the formation of a carbocation after the iodide leaves: \[ \text{C(CH}_3\text{)}_4\text{C}^+ \text{ (carbocation)} \] 4. **Methyl Shift**: A methyl shift occurs to stabilize the carbocation, leading to the formation of a more stable tertiary carbocation: \[ \text{C(CH}_3\text{)}_3\text{C}^+\text{ (new carbocation)} \] 5. **Nucleophilic Attack by Water**: Water (H2O) acts as a nucleophile and attacks the carbocation: \[ \text{C(CH}_3\text{)}_3\text{C}^+\text{ + H}_2\text{O} \rightarrow \text{C(CH}_3\text{)}_3\text{C(OH)}\text{H}_2 \] 6. **Deprotonation**: The final step involves the removal of a proton (H+) from the hydroxyl group, leading to the formation of the final alcohol product: \[ \text{C(CH}_3\text{)}_3\text{C(OH)H}_2 \rightarrow \text{C(CH}_3\text{)}_3\text{C(OH)CH}_2\text{ (neopentyl alcohol)} \] 7. **Final Products**: The final products of the reaction are: - Yellow precipitate of silver iodide (AgI) - Neopentyl alcohol (C(CH3)3CH2OH) ### Conclusion: The other compound formed along with the yellow precipitate (AgI) when neopentyl iodide is treated with aqueous AgNO3 is neopentyl alcohol.

To solve the problem of what compound is formed when neopentyl iodide is treated with aqueous AgNO3 solution, we can follow these steps: ### Step-by-Step Solution: 1. **Identify Neopentyl Iodide Structure**: Neopentyl iodide has the structure: \[ \text{C(CH}_3\text{)}_4\text{CH}_2\text{I} ...
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