Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

The rearrangement which most likely occurs in the following reaction can be described as a

HBr
(CH3)2CHCH=CH2 →

a. hydride shift from C-2 to C-1.
b. hydride shift from C-3 to C-2.
c. proton shift from C-2 to C-1.
d. methyl group shift from C-3 to C-2.

Sagot :

pstnonsonjoku

Answer:

b. hydride shift from C-3 to C-2.

Explanation:

Markovnikov's rule states that *in the addition of a protic acid HX or other polar reagent to an asymmetric alkene, the acid hydrogen (H) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (X) group or electronegative part gets attached to the carbon with more alkyl substituents* (wikipedia).

This rule implies that the hydrogen of HBr will be attached to C-1 and the carbocation will be on C-2. Remember that the order of stability of carbocations is tertiary > secondary > primary > methyl. A hydride shift can yield a tertiary carbocation.

C-3 is a tertiary carbon atom. If the hydride on carbon 3 shifts to carbon 2, a tertiary and more stable carbocation is formed. This accounts for the major product in the reaction.

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.