We show in Chapter 12 that C― Br bonds can break to give a carbocation and a bromide anion. For which of the organohalides (A or B) would you expect this process to be fastest? Explain.

Verified step by step guidance

Identify the structure of the two organohalides (A and B) and determine the type of carbon atom (primary, secondary, or tertiary) to which the bromine atom is attached. This is crucial because the stability of the resulting carbocation will influence the rate of bond cleavage.

Recall that the rate of C―Br bond cleavage to form a carbocation and bromide anion depends on the stability of the carbocation formed. Tertiary carbocations are more stable than secondary carbocations, which are more stable than primary carbocations, due to inductive effects and hyperconjugation.

Analyze the structure of organohalide A: Determine whether the carbon bonded to bromine is primary, secondary, or tertiary. If it forms a tertiary carbocation, it will likely undergo bond cleavage faster.

Analyze the structure of organohalide B: Similarly, determine whether the carbon bonded to bromine is primary, secondary, or tertiary. Compare its carbocation stability to that of organohalide A.

Conclude which organohalide (A or B) will undergo the C―Br bond cleavage process faster based on the relative stabilities of the carbocations formed. The organohalide that forms the more stable carbocation will have the faster bond cleavage rate.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Carbocation Stability

Carbocations are positively charged carbon species that can form when a bond breaks, such as a C―Br bond. The stability of a carbocation is influenced by the degree of substitution; tertiary carbocations are more stable than secondary, which are more stable than primary. This stability affects the rate of reaction, as more stable carbocations form more readily, leading to faster reactions.

Leaving Group Ability

The ability of a leaving group to depart from a molecule is crucial in determining reaction rates. Bromide (Br⁻) is a good leaving group due to its ability to stabilize the negative charge after bond cleavage. The better the leaving group, the faster the reaction will proceed, as it can facilitate the formation of the carbocation more efficiently.

Reaction Mechanism

Understanding the reaction mechanism is essential for predicting the speed of the process. In this case, the reaction likely follows an SN1 mechanism, where the bond breaks to form a carbocation before the nucleophile attacks. The rate of this reaction depends on the stability of the carbocation formed and the efficiency of the leaving group, which together dictate how quickly the reaction occurs.