$4$-Pentenoic acid when treated with $I_2$ and $NaHCO_3$ gives:

  • A
    $4, 5$-diiodopentanoic acid
  • B
    $5$-iodomethyl-dihydrofuran-$2$-one
  • C
    $5$-iodo-tetrahydropyran-$2$-one
  • D
    $4$-pentenoyl iodide

Explore More

Similar Questions

Identify the product of the following reaction:

Which one of the following esters cannot undergo self Claisen condensation?

Difficult
View Solution

The most acidic of the following is

Which of the following compounds reacts with methyl magnesium iodide to produce ethyl acetate?

When formic acid is heated with conc. $H_{2}SO_{4}$,the gas evolved is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo