Q 18P Finish Solved Problem 7-3 by sho... [FREE SOLUTION]

Chapter 7: Q 18P (page 359)

Finish Solved Problem 7-3 by showing how the rearranged carbocations give the four products shown in the problem. Be careful when using curved arrows to show deprotonation and/or nucleophilic attack by the solvent. The curved arrows always show movement of electrons, not movement of protons or other species.

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Step by step solution

Step by step solution: Step 1: Formation of E1 products

E1 mechanism or E1 elimination involves two steps鈥攊onization and deprotonation. During ionization, intermediate carbocation is formed; and in deprotonation, a proton is lost by carbocation. This happens in the presence of a base which leads to formation of double bond. Carbocation formation is the slow and rate-determining step of this reaction.

Methanol can act as base as well as nucleophile. Thus, when it acts as a base, it abstracts the proton from carbon which is adjacent to the carbocation, and on charge neutralisation of carbocation, a double bond or E1 products are formed.

Formation of E1 products

Step-2. Formation of products:

reaction mechanism follows three steps. Firstly a carbocation is formed from the removal of leaving group, halogen. Then, carbocation is attacked by the nucleophile. Lastly, deprotonation takes place which gives the required product.

Methanol acts as nucleophile and attacks the carbocation which leads to formation of product. There is no rearrangement of carbocation as formed carbocations are stable.