t-Pentyl alcohol is reacted with HCl to make t-Pentyl chloride. Some 2-methyl-2-butene may be produced in the reaction as a by-product. Give a mechanism for its production.
(SN1) C5H12O + HCl —-> C5H11Cl
The Correct Answer and Explanation is :
The reaction of t-pentyl alcohol (C5H12O) with HCl to form t-pentyl chloride (C5H11Cl) typically follows an SN1 (unimolecular nucleophilic substitution) mechanism, due to the formation of a stable carbocation intermediate. In this case, the by-product, 2-methyl-2-butene, is produced via an elimination pathway (E1 mechanism), which competes with the nucleophilic substitution process. Here’s a detailed explanation of the mechanism:
Step 1: Protonation of Alcohol
The reaction begins with the protonation of the alcohol (t-pentyl alcohol) by HCl. The alcohol oxygen is quite electronegative, so it readily accepts a proton from the HCl, turning into a good leaving group (water). This step is essential for the subsequent carbocation formation. The structure of t-pentyl alcohol, which has a tertiary carbon (C3) attached to the hydroxyl group, makes it particularly favorable for this reaction.
Step 2: Departure of Water and Carbocation Formation
The protonated alcohol then undergoes the departure of water (H2O), which forms a tertiary carbocation at the C3 position. The stability of this carbocation is crucial in the SN1 mechanism, and the tertiary carbocation is highly stabilized due to inductive effects and hyperconjugation from adjacent alkyl groups.
Step 3: Rearrangement to Form a More Stable Carbocation (Optional)
In some cases, carbocations may undergo rearrangement to increase their stability. However, in the case of t-pentyl alcohol, there is no need for further rearrangement because the initial carbocation is already tertiary and highly stable.
Step 4: Elimination (Formation of 2-methyl-2-butene)
Instead of the nucleophilic chloride ion attacking the carbocation to form t-pentyl chloride (C5H11Cl), an elimination reaction (E1) can occur. The leaving group (water) departs, and a hydrogen is abstracted from the adjacent carbon (C2). This results in the formation of a double bond between C2 and C3, producing 2-methyl-2-butene as a by-product.
Step 5: Nucleophilic Attack (SN1)
If the reaction proceeds via substitution (SN1), the chloride ion (Cl-) will attack the carbocation, leading to the formation of t-pentyl chloride.
Conclusion
In summary, the reaction of t-pentyl alcohol with HCl involves both SN1 and E1 pathways. The SN1 pathway leads to the formation of t-pentyl chloride, while the E1 pathway can lead to the formation of the alkene by-product, 2-methyl-2-butene.