Q20E

Question


Predict the major product (s) from the addition of 1 equivalent of HX and show the mechanism for each reaction below.



Step-by-Step Solution

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Answer

Step-by-Step -Solution

1Step1: Addition of HX

Electrophilic addition of HX to a conjugated diene takes place via the formation of allylic carbocation intermediates. In the first step, protonation of ends of diene occurs to form allylic carbocations (two types are possible), each of the two carbocations then reacts with CI2 to generate four possible structures.

2Step 2: Addition of HCl


Electrophilic addition of HCl to the given conjugated diene takes place via the formation of allylic carbocation intermediates. In the first step, protonation of C-2 and C-3 occurs to form allylic carbocation that can react further to generate 1,2 and 1,4 adducts. Here in this case three different carbocations are formed when protonation occurs, so three different products are possible 4-Chloro-2,3-Hexa diene, 2-chloro-Hexa 3,4-diene, and 5-chloro-hexa-3,4-diene.

3Step 3: Addition of HBr


When addition of HBr takes place to the diene, three types of carbocations are possible (as two carbocations are the same) as shown. Out of the three carbocations, carbocation 1 is the most stable one, as it is a tertiary allylic carbocation. Therefore the major product formed is 1-Bromo-1-methyl-1,2-penta-diene.

4Step 3: Addition of HCl

When addition of HCl takes place to the cyclic diene, two types of carbocations are possible, one is primary and another is tertiary, as shown. Out of the two carbocations, carbocation 1 is the most stable one, as it is a tertiary allylic carbocation. Therefore the major product formed is the one obtained from tertiary carbocation.