Alkyl Halides and Aryl HalidesMedium
Question
CH3OC2H5 and (CH3)3 C - OCH3 are treated with hydriodic acid. the fragments obtained after reactions are
Options
A.CH3I + HOC2H5; (CH3)3CI + HOCH3
B.CH3OH + C2H5I;(CH3)3CI + HOCH3
C.CH3OH + C2H5I;(CH3)3COH + CH3I
D.CH3I+ HOC2H5; CH3I(CH3)3COH.
Solution
In case of unsymmetrical ethers,the site of cleage depends on the nature of alkyl group.
e.g., CH3 - O - C2H5 + HI
CH3I + C2H5OH
The alkyl haide is formed from the smaller alkyl group.
However, in case of teritiary alkyl ether following reaction occurs.
CH3- O -(CH3)3 + HI (CH3)3 C - I + CH3 OH
ter-butyl methyl ether
Tnhe alkyl halide is formed from the tertiary alkyl group and the cleavage of such ethers occurs by SN1mechanism as the product is controlled by the formation of more stable intermedied ether.
e.g., CH3 - O - C2H5 + HI
CH3I + C2H5OHThe alkyl haide is formed from the smaller alkyl group.
However, in case of teritiary alkyl ether following reaction occurs.
CH3- O -(CH3)3 + HI
(CH3)3 C - I + CH3 OHter-butyl methyl ether
Tnhe alkyl halide is formed from the tertiary alkyl group and the cleavage of such ethers occurs by SN1mechanism as the product is controlled by the formation of more stable intermedied ether.
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