Original post by evelync12
i dont understand the mechanism for this question:
A cation is formed when methylamine reacts with a large excess of bromoethane. Name the mechanism involved in the reaction and draw the structure of the cation formed
A cation is formed when methylamine reacts with a large excess of bromoethane. Name the mechanism involved in the reaction and draw the structure of the cation formed
Due to the polar bond between bromine and carbon, the carbon is slightly positive. The lone pair on the nitrogen is attracted to the slightly positive charge on the carbon and donates it's lone pair to the carbon. The bromine takes the electron pair and leaves the molecule. Now because the methylamine has donated an electron pair via the nitrogen, it has a positive charge. To be more stable it takes an electron pair from one of the hydrogen's attached to it. H+ leaves the molecule and another molecule of methylamine donates it's lone pair to it. so now you have a molecule of ammonium bromide (NH4+) and a molecule of N-methylethanamine.
(edited 4 years ago)
The mechanism taking place is nucleophilic substitution, i.e. the bromine atom in bromoethane is being substituted by a nucleophile, in this case methylamine
Furthermore, the specific mechanism taking place is amine alkylation, and in this instance it undergoes via SN2 as bromoethane is a primary alkyl halide
What you said in your comment is generally fine, but this part is incorrect
If the proton is being accepted by another molecule of methylamine, then you won’t be forming ammonium bromide; you’ll be forming methylammonium bromide, and thus the structure will be slightly different
Furthermore, the specific mechanism taking place is amine alkylation, and in this instance it undergoes via SN2 as bromoethane is a primary alkyl halide
Original post by Harrybeld
...H+ leaves the molecule and another molecule of methylamine donates it's lone pair to it. so now you have a molecule of ammonium bromide (NH4+)...
What you said in your comment is generally fine, but this part is incorrect
If the proton is being accepted by another molecule of methylamine, then you won’t be forming ammonium bromide; you’ll be forming methylammonium bromide, and thus the structure will be slightly different
(edited 4 years ago)
Original post by Kian Stevens
The mechanism taking place is nucleophilic substitution, i.e. the bromine atom in bromoethane is being substituted by a nucleophile, in this case methylamine
Furthermore, the specific mechanism taking place is amine alkylation, and in this instance undergoes via SN2 as bromoethane is a primary alkyl halide
What you said in your comment is generally fine, but this part is incorrect
If the proton is being accepted by another molecule of methylamine, then you won’t be forming ammonium bromide; you’ll be forming methylammonium bromide, and thus the structure will be slightly different
Furthermore, the specific mechanism taking place is amine alkylation, and in this instance undergoes via SN2 as bromoethane is a primary alkyl halide
What you said in your comment is generally fine, but this part is incorrect
If the proton is being accepted by another molecule of methylamine, then you won’t be forming ammonium bromide; you’ll be forming methylammonium bromide, and thus the structure will be slightly different
Yes. you're right apologies. Usually used to drawing this mechanism with NH3.
Original post by Kian Stevens
The mechanism taking place is nucleophilic substitution, i.e. the bromine atom in bromoethane is being substituted by a nucleophile, in this case methylamine
Furthermore, the specific mechanism taking place is amine alkylation, and in this instance it undergoes via SN2 as bromoethane is a primary alkyl halide
What you said in your comment is generally fine, but this part is incorrect
If the proton is being accepted by another molecule of methylamine, then you won’t be forming ammonium bromide; you’ll be forming methylammonium bromide, and thus the structure will be slightly different
Furthermore, the specific mechanism taking place is amine alkylation, and in this instance it undergoes via SN2 as bromoethane is a primary alkyl halide
What you said in your comment is generally fine, but this part is incorrect
If the proton is being accepted by another molecule of methylamine, then you won’t be forming ammonium bromide; you’ll be forming methylammonium bromide, and thus the structure will be slightly different
thanks so much!!
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