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Neopentyl alcohol on treatment with HCl ...

Neopentyl alcohol on treatment with HCl and anhydrous `ZnCl_(2)` gives

A

2-Chloro-3-methylbutane

B

2-Chloro-2-methylbutane

C

1-Chloro-2,2-dimethylpropane

D

None of these

Text Solution

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The correct Answer is:
To solve the problem of what product is formed when neopentyl alcohol is treated with HCl and anhydrous ZnCl₂, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Structure of Neopentyl Alcohol**: - Neopentyl alcohol is a branched alcohol with the formula \(C_5H_{12}O\). Its structure can be represented as: \[ \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}_2-\text{OH} \] - This shows that the hydroxyl group (-OH) is attached to a carbon that is also connected to two other methyl groups. 2. **Reaction with HCl**: - When neopentyl alcohol reacts with HCl, the hydroxyl group (-OH) is protonated, making it a better leaving group. This leads to the formation of a carbocation after the loss of water (H₂O) and chloride ion (Cl⁻): \[ \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}_2^+ \] - The carbocation formed here is a primary carbocation (1°) because it is attached to only one alkyl group. 3. **Carbocation Rearrangement**: - Primary carbocations are not stable. To stabilize the carbocation, a methyl shift occurs. This means that one of the methyl groups from the adjacent carbon shifts to the carbocation carbon, resulting in a more stable tertiary carbocation: \[ \text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \] - This new carbocation is now a tertiary carbocation (3°) and is more stable. 4. **Nucleophilic Attack by Cl⁻**: - The chloride ion (Cl⁻) acts as a nucleophile and attacks the positively charged carbon of the tertiary carbocation, leading to the formation of the final product: \[ \text{CH}_3-\text{C}(\text{Cl})(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \] - This compound is 2-chloro-2-methylbutane. 5. **Final Product**: - The final product of the reaction is identified as: \[ \text{2-chloro-2-methylbutane} \] ### Conclusion: The product formed when neopentyl alcohol is treated with HCl and anhydrous ZnCl₂ is **2-chloro-2-methylbutane**.

To solve the problem of what product is formed when neopentyl alcohol is treated with HCl and anhydrous ZnCl₂, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Structure of Neopentyl Alcohol**: - Neopentyl alcohol is a branched alcohol with the formula \(C_5H_{12}O\). Its structure can be represented as: \[ \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}_2-\text{OH} ...
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Knowledge Check

  • Neopentyl alcohol is:

    A
    `1^(@)` alcohol
    B
    `2^(@)` alcohol
    C
    `3^(@)` alcohol
    D
    an ether
  • Alkyl chloride is formed when alcohol is treated with HCl in presence of anhydrous ZnCl_(2) . The order of reactivity with respect to alcohol is

    A
    `3^(@) gt 2^(@) gt 1^(@)`
    B
    `1^(@) gt 2^(@) gt 3^(@)`
    C
    `2^(@) gt 1^@ gt 3^(@)`
    D
    `1^(@) gt 3^@ gt 2^(@)`
  • The hydroxyl compound that gives a precipitate immediately when treated with concentrated HCl and anhydrous ZnCl_(2) is :

    A
    3- methyl-2-butanol
    B
    3-methyl -1-butanol
    C
    1-butanol
    D
    2-methyl-2-butanol
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    From amongst the following alcohols, the one that would react fastest with conc. HCl and anhydrous ZnCl_(2) , is

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