Question

3. (3 pts) Provide a reasonable arrow-pushing mechanism for the following transformation. Why does the more substituted cyclopentanone not form? Me OEt 1. NaOET, ETOH X 2. NH4Cl (aq soln) Eto EIO Me

Answer 1

Mechanism for the reaction is as given below:

Description:

Step 1: Deprotonation of more acidic proton using the nucleophillic base 'ethoxide' to generate an enolate.

Step 2: Negatively charged carbon nucleophillically attacks on the electrophillic carbonyl carbon to generate a tetrahedral intermediate containing a negatively charged oxygen.

Step 3: Negative charge on the oxygen moves back to form double bond with the carbon and simultaneously the ethoxide group leaves in order to maintain the tetravalency of the carbon and the required product is obtained.

​​​​​​Reason why more substituted product is not formed:

Det or H И .O Et a b? Eto- 7) O Me

The given reactant had two sites for deprotonation marked as a and b. The site of deprotonation depends upon the stability of the enolate/carbanion formed after deprotonation. If any electron withdrawing group is present on on the negatively charged carbon then its its acidic strength increases beacuse electron withdrawing group can help to stabilise the negative charge after deprotonation. However the case is exactly opposite if an electron releasing group is present because such group will intensify the negative charge on the carbon making it less stable.  

In the given molecule among the two sites available for deprotonation, the one marked as 'b' has an extra methyl group as compared to 'a'. Methyl group can intensify the negative charge on carbon via electron releasing inductive effect thereby decreasing the acidic strength on 'b'.

As acidic strength of 'b' is less than that of 'a' that's why the more substituted product will not be formed.