Consider the following reaction and identify the product (P):
3-Methylbutan-2-ol (CH3-CH(CH3)-CH(OH)-CH3) reacts with HBr.
Possible products include:
1) CH3-CH(CH3)-CH(Br)-CH3
2) CH3-C(CH3)2-CH2Br
3) Br-CH3-C(CH3)-CH2-CH3
4) CH3CH=CH-CH3
Identify the correct product (P).
Answer (2)
In this question, the reaction involves 3-Methylbutan-2-ol, which is an alcohol, reacting with hydrobromic acid (HBr). This is a typical example of a substitution reaction, specifically reacting an alcohol with HBr to form an alkyl halide.
Let's break this down step-by-step:
Identify the Alcohol and the Leaving Group: The alcohol we are starting with is 3-Methylbutan-2-ol. The hydroxyl group (-OH) from the alcohol is the part that will be replaced in this reaction.
React with HBr: When alcohols react with hydrobromic acid, the -OH group gets protonated by HBr and then leaves as a water molecule, forming a carbocation.
Carbocation Intermediate Formation: In this case, a 3° carbocation can be formed since the molecule can rearrange to stabilize the positive charge on the most substituted carbon. This rearrangement happens as 3-Methylbutan-2-ol undergoes hydrogen shift from the methyl group adjacent to the carbocation.
Formation of the Bromoalkane: The bromide ion from HBr attacks the carbocation, resulting in the formation of a bromoalkane.
Considering the possible products listed:
CH3-CH(CH3)-CH(Br)-CH3
CH3-C(CH3)2-CH2Br
Br-CH3-C(CH3)-CH2-CH3
CH3CH=CH-CH3
The correct product is (2) CH3-C(CH3)2-CH2Br . This product is formed due to the rearrangement which results in the stable 3° carbocation, allowing bromide to attack the positively charged carbon thereby replacing the hydroxyl position with a bromide.
Thus, the answer is option (2).
The reaction of 3-Methylbutan-2-ol with HBr produces CH3-C(CH3)2-CH2Br as the primary product. This occurs through the formation of a carbocation that rearranges to a more stable tertiary form, allowing for bromide substitution. Therefore, the correct answer is option (2).
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