On subjecting mesityl oxide to the iodoform reaction, one of the products is the sodium salt of an organic acid. Which acid is obtained?

Question

On subjecting mesityl oxide to the iodoform reaction, one of the products is the sodium salt of an organic acid. Which acid is obtained? (A) ( CH 3)2 C = CH - CH 2 COOH (B) ( CH 3)2 C = CH - CO - COOH (C) ( CH 3)2 C = CH - COOH (D) ( CH 3)2 CH - COOH

Tardigrade Admin · Accepted Answer

Mesityl oxide undergoes iodoform reaction. CH 3- overset underset|CH3C = CH - overset underset||OC - CH 3 xrightarrow NnOH +I2CH 3- overset underset|CH3C = CH - COONa + CHI 3

Q. On subjecting mesityl oxide to the iodoform reaction, one of the products is the sodium salt of an organic acid. Which acid is obtained?

$(C H_{3})_{2} C = C H - C H_{2} COO H$

$(C H_{3})_{2} C = C H - CO - COO H$

$(C H_{3})_{2} C = C H - COO H$

$(C H_{3})_{2} C H - COO H$

Mesityl oxide undergoes iodoform reaction.
$C H_{3} - C ∣ C H_{3} = C H - C ∣∣ O - C H_{3} N n O H + I_{2} C H_{3} - C ∣ C H_{3} = C H - COON a + C H I_{3}$

Q. On subjecting mesityl oxide to the iodoform reaction, one of the products is the sodium salt of an organic acid. Which acid is obtained?

(CH3​)2​C=CH−CH2​COOH

(CH3​)2​C=CH−CO−COOH

(CH3​)2​C=CH−COOH

(CH3​)2​CH−COOH

Mesityl oxide undergoes iodoform reaction. CH3​−C∣CH3​​=CH−C∣∣O​−CH3​NnOH+I2​​CH3​−C∣CH3​​=CH−COONa+CHI3​

$(C H_{3})_{2} C = C H - C H_{2} COO H$

$(C H_{3})_{2} C = C H - CO - COO H$

$(C H_{3})_{2} C = C H - COO H$

$(C H_{3})_{2} C H - COO H$

Mesityl oxide undergoes iodoform reaction.
$C H_{3} - C ∣ C H_{3} = C H - C ∣∣ O - C H_{3} N n O H + I_{2} C H_{3} - C ∣ C H_{3} = C H - COON a + C H I_{3}$