Organic Chemistry 1 · Bonding and formal charge
Methylamine takes a proton from hydrogen chloride
Conditions: HCl, diethyl ether
Draw this problem yourself
Drag from the electrons that move to where they go, or tap one and then the other. Every arrow is checked; Hint helps when you're stuck. No account needed.
The mechanism, step by step
Next plays each step's curved arrows and explains them. The written steps are below.
Read all 3 written steps
The nitrogen lone pair pulls off the proton
Methylamine is a base because its nitrogen carries a lone pair, and HCl is a strong acid with a weak, polar H–Cl bond. Two arrows move in one step: one makes the N–H bond and one breaks the H–Cl bond.
The curved arrows
- The nitrogen lone pair reaches for the partly positive hydrogen and becomes the new N–H bond.
- The H–Cl bonding pair moves onto chlorine, the more electronegative atom, so chlorine leaves as chloride.
Charge on the ammonium nitrogen
Count the bonds and lone pairs on nitrogen after the proton transfer.
Methylammonium chloride
Nitrogen now has four bonds, no lone pair and a +1 formal charge, and its geometry stays tetrahedral (sp3). Chloride keeps the former bonding pair as its fourth lone pair and carries −1, so the salt is neutral overall.
What you'll practice
Draw the proton transfer from hydrogen chloride to methylamine and assign the formal charge on nitrogen in methylammonium.
You just pushed every arrow of this one.
Every one of the 5349 Orgo 1 and Orgo 2 problems works like this: feedback on each arrow, hints when you're stuck, a no-hint quiz mode. Free with an account, no card.
Stuck on your own homework?
Snap it — first scan free, no sign-up
Scanning every week? Scan Standard: for about problems