Question
Compound ' $X$ ' with a molecular formula $C _4 H _{11} N$ did not react with Hinsberg's reagent, but reacted with one mole of $CH _3$ l to form a salt. What is the structure of ' $X$ '?
✓
Answer
The structure of compound ‘X’ is :
Since the compound ' $X$ ' does not react with $NaN02$ and $HC$ 1 i.e. nitrous acid ( $HO - N = O$ ), it must be a tertiary amine. The tertiary amine reacts with one mole of $CH 3$ I to give a quaternary ammonium salt.
$\underset{\text { ethyldimethylamine }}{\left( CH _3\right)_2 N - C _2 H _5}+\underset{\text { methyliodide }}{ CH _3 I \xrightarrow{\Delta}} \underset{\text { ethyltrimethyl ammonium iodide }}{\left[\left( CH _3\right)_3 \stackrel{+}{ N }- C _2 H _5\right] I ^{-}}$
Since the compound ' $X$ ' does not react with $NaN02$ and $HC$ 1 i.e. nitrous acid ( $HO - N = O$ ), it must be a tertiary amine. The tertiary amine reacts with one mole of $CH 3$ I to give a quaternary ammonium salt.
$\underset{\text { ethyldimethylamine }}{\left( CH _3\right)_2 N - C _2 H _5}+\underset{\text { methyliodide }}{ CH _3 I \xrightarrow{\Delta}} \underset{\text { ethyltrimethyl ammonium iodide }}{\left[\left( CH _3\right)_3 \stackrel{+}{ N }- C _2 H _5\right] I ^{-}}$
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