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No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for $\mathrm{E2}$ elimination. And $\mathrm{E1}$ elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to $\mathrm{S_N2}$ reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to S_N2 reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to $S_{N}2$ reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for $\mathrm{E2}$ elimination. And $\mathrm{E1}$ elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to $\mathrm{S_N2}$ reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrangment, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to $S_{N}2$ reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to $S_{N}2$ reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.