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.
Answer:
because it covers a large area as there are more then 1 lakh compounds of organic chemistry.
Answer:

Explanation:
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In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

Thus, we solve for the molarity of the acid to obtain:

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Answer:
The central and most important part of an object, movement, or group, forming the basis for its activity and growth.
Explanation:
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C. Protons, neutrons, electrons