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The major organic product in the reaction,
$CH_3-O-CH(CH_3)_2 + HI \rightarrow$ Product is :
$CH_3-O-CH(CH_3)_2 + HI \rightarrow$ Product is :
A
$CH_3OH + (CH_3)_2CHI$
B
$ICH_2OCH(CH_3)_2$
C
$CH_3O-C(I)(CH_3)_2$
D
$CH_3I + (CH_3)_2CHOH$
Detailed Solution
Step 1: the ether oxygen is protonated by HI to give an oxonium ion: $CH_3-\overset{+}{O}(H)-CH(CH_3)_2$
Step 2: the iodide ion, a good nucleophile, attacks one of the alkyl carbons by the $S_N2$ mechanism.
In an $S_N2$ attack the nucleophile prefers the less sterically hindered carbon.
The methyl carbon is much less hindered than the isopropyl (secondary) carbon, so $I^-$ attacks the methyl group.
$CH_3-O-CH(CH_3)_2 + HI \xrightarrow{S_N2} CH_3I + (CH_3)_2CH-OH$
So the smaller alkyl group forms the alkyl iodide and the larger group forms the alcohol.
The products are methyl iodide and propan-2-ol: $CH_3I + (CH_3)_2CHOH$.
Step 2: the iodide ion, a good nucleophile, attacks one of the alkyl carbons by the $S_N2$ mechanism.
In an $S_N2$ attack the nucleophile prefers the less sterically hindered carbon.
The methyl carbon is much less hindered than the isopropyl (secondary) carbon, so $I^-$ attacks the methyl group.
$CH_3-O-CH(CH_3)_2 + HI \xrightarrow{S_N2} CH_3I + (CH_3)_2CH-OH$
So the smaller alkyl group forms the alkyl iodide and the larger group forms the alcohol.
The products are methyl iodide and propan-2-ol: $CH_3I + (CH_3)_2CHOH$.