What happens in this reaction? butan-1-amine + CH3I -->
1 answer:
Here we have to get the product between the reaction of butane-1-amine with methyl iodide (CH₃I).
The reaction between 1 mole of butan-1-amine and 1 mole of methyl iodide produces Methyl-butamine which is a secondary amine.
However, In presence of 2 moles of methyl iodide the reaction proceed to N, N-di-methylbutamine. The reaction is shown in the figure.
This is one of the effective reaction method to generate secondary and tertiary amine from primary amine.
The primary amine reacts with alkyl iodide to form secondary to tertiary amine. The final product depends upon the quantity of the alkyl iodide present in the reaction.
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The organic product formed when 1−hexyne is treated with H₂O, H₂SO₄, and HgSO₄ will be 2-hexanone (structure attached).
This reaction is an example of an oxymercuration reaction of the organic product 1−hexyne.
Oxymercuration is shown in three steps to the right. The nucleophilic double bond attacks the mercury ion, releasing an acetoxy group. The mercury ion's electron pair attacks carbon on the double bond, generating a positive-charged mercuronium ion. Mercury's dxz and 6s orbitals give electrons to the double bond's lowest unoccupied molecular orbitals.
In the second stage, the nucleophilic H₂O attacks the highly modified carbon, freeing its mercury-bonding electrons. Electrons neutralize mercury ions by collapsing. Water molecules have positive-charged oxygen.
In the third stage, the negatively charged acetoxy ion released in the first step attacks the hydrogen of the water group, generating the waste product HOAc. The two electrons in the oxygen-hydrogen link collapse into oxygen, neutralizing its charge and forming alcohol.
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