Find the major product for the reaction shown | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction is an iodolactonization followed by dehydroiodination.$1$. Treatment of the unsaturated carboxylic acid with $I_2/NaHCO_3$ leads to t

Vedclass · Accepted Answer

Vedclass · Answer

(A) The reaction is an iodolactonization followed by dehydroiodination.$1$. Treatment of the unsaturated carboxylic acid with $I_2/NaHCO_3$ leads to the formation of an iodolactone intermediate. The carboxylate oxygen attacks the iodonium ion formed across the double bond, resulting in a bicyclic lactone with an iodine atom at the bridgehead position.$2$. Subsequent treatment with pyridine and heat $(\Delta)$ promotes an $E2$ elimination reaction, removing the iodine atom and a hydrogen atom from the adjacent carbon to form a double bond.$3$. The final product is a bicyclic lactone containing a double bond in the six-membered ring, which corresponds to option $A$.

Find the major product for the reaction shown below. The reactant is $2-(cyclohex-1-en-1-yl)acetic acid$. The reaction conditions are: $(1) I_2, NaHCO_3$ and $(2) \text{pyridine}, \Delta$.

Similar Questions

Allyl chloride on dehydrochlorination gives:

The alkene $C_6H_{10}$ producing $OHC-(CH_2)_4-CHO$ on ozonolysis is

The reaction conditions used for converting $1, 2-$dibromoethane to ethene are

How many products are formed in the following reaction?
$1,4$-dimethylcyclohex-$1$-ene + $HBr \rightarrow$ ?

The major product of the following chemical reaction is: $CH_3-CH(CH_3)-CH=CH_2 + HBr \xrightarrow{(C_6H_5CO)_2O_2} ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module