An alkyl iodide on standing darkens, due to

Question

An alkyl iodide on standing darkens, due to (A) hydrolysis (B) conversion into ether (C) liberation of iodine (D) formation of alkanes.

Tardigrade Admin · Accepted Answer

Since C-I bond is a weak bond thus, iodides decompose on standing in presence of light. 2 R- I xrightarrowh v R-R+ I 2 This liberated I 2 dissolves in alkyl iodide to impart it a dark (violet or brown) colour.

Q. An alkyl iodide on standing darkens, due to

hydrolysis

conversion into ether

liberation of iodine

formation of alkanes.

Since $C - I$ bond is a weak bond thus, iodides decompose on standing in presence of light.
$2 R - I h v R - R + I_{2}$

This liberated $I_{2}$ dissolves in alkyl iodide to impart it a dark (violet or brown) colour.

Q. An alkyl iodide on standing darkens, due to

hydrolysis

conversion into ether

liberation of iodine

formation of alkanes.

Since C−I bond is a weak bond thus, iodides decompose on standing in presence of light. 2R−Ihv​R−R+I2​ This liberated I2​ dissolves in alkyl iodide to impart it a dark (violet or brown) colour.

Since $C - I$ bond is a weak bond thus, iodides decompose on standing in presence of light.
$2 R - I h v R - R + I_{2}$

This liberated $I_{2}$ dissolves in alkyl iodide to impart it a dark (violet or brown) colour.