Which will form lactone on treatment with NaO | vedclass

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Question

(D) Lactones are cyclic esters formed by the intramolecular reaction of hydroxy acids or halo acids. $\delta$-Bromo acids (or $\delta$-hydroxy acids)

Vedclass · Accepted Answer

$\delta$-Bromo acid

Vedclass · Answer

(D) Lactones are cyclic esters formed by the intramolecular reaction of hydroxy acids or halo acids. $\delta$-Bromo acids (or $\delta$-hydroxy acids) undergo intramolecular nucleophilic substitution in the presence of a base like $NaOH$ to form a stable five- or six-membered lactone ring. Specifically,$\delta$-bromo acids form a six-membered lactone ring,which is thermodynamically stable due to low ring strain.

Which will form lactone on treatment with $NaOH$?

Similar Questions

The correct decreasing order for acidic strength is

The product $(A)$ is:

The decreasing order of the rate of alkaline hydrolysis for the following esters is:
$I: C_6H_5COOC_2H_5$
$II: p-Cl-C_6H_4COOC_2H_5$
$III: p-NO_2-C_6H_4COOC_2H_5$
$IV: p-CH_3O-C_6H_4COOC_2H_5$

Toluene $\xrightarrow[\Delta ]{KMnO_4} X$ $\xrightarrow{Br_2/H_2O} Y$; $Y$ is

$CH_3COOH + C_2H_5^{18}OH \xrightarrow{Conc. H_2SO_4} (X)$
$CH_3COOH + (CH_3)_3C^{18}OH \xrightarrow{Conc. H_2SO_4} (Y)$
In the above reaction $(X)$ and $(Y)$ are respectively:

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