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In the reaction:
$CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3 + HI \xrightarrow{\text{Heated}}$
Which of the following compounds will be formed?
$CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3 + HI \xrightarrow{\text{Heated}}$
Which of the following compounds will be formed?
A
$CH_3-CH(CH_3)-CH_3 + CH_3CH_2OH$
B
$CH_3-CH(CH_3)-CH_2OH + CH_3CH_3$
C
$CH_3-CH(CH_3)-CH_2OH + CH_3-CH_2-I$
D
$CH_3-CH(CH_3)-CH_2-I + CH_3CH_2OH$
Detailed Solution
In the cleavage of mixed ethers having two different alkyl groups, the alcohol and the alkyl iodide that form depend on the nature of the alkyl groups.
Step 1: the ether oxygen is protonated by HI to give an oxonium ion.
Step 2: the iodide ion attacks one of the alkyl carbons by the $S_N2$ mechanism.
When primary or secondary alkyl groups are present, the iodide ion attacks the smaller, less hindered alkyl group, so it is the lower alkyl group that forms the alkyl iodide.
Here the two groups are ethyl and isobutyl (2-methylpropyl). The ethyl carbon is less hindered than the isobutyl carbon, which has a branched $\beta$-carbon.
$CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3 + HI \xrightarrow{\Delta} CH_3-CH(CH_3)-CH_2OH + CH_3CH_2I$
The products are 2-methylpropan-1-ol and ethyl iodide.
Step 1: the ether oxygen is protonated by HI to give an oxonium ion.
Step 2: the iodide ion attacks one of the alkyl carbons by the $S_N2$ mechanism.
When primary or secondary alkyl groups are present, the iodide ion attacks the smaller, less hindered alkyl group, so it is the lower alkyl group that forms the alkyl iodide.
Here the two groups are ethyl and isobutyl (2-methylpropyl). The ethyl carbon is less hindered than the isobutyl carbon, which has a branched $\beta$-carbon.
$CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3 + HI \xrightarrow{\Delta} CH_3-CH(CH_3)-CH_2OH + CH_3CH_2I$
The products are 2-methylpropan-1-ol and ethyl iodide.