# Etymology of alanine

Oxford dictionary online gives etymology of alanine as:

Coined in German as Alanin, from aldehyde + -an (for ease of pronunciation) + -ine.

But I see no resemblance to the aldehyde structure in alanine. Is this etymology incorrect, or am I mistaken about the resemblance?

In the original German paper [1] Adolf Strecker used Aldehyd-Ammoniak or aldehyde-ammonia as a precursor, that's where the name derives from:

Vor einigen Jahren habe ich gezeigt, daſs Aldehyd-Ammoniak und Blausäure beim Erwärmen mit verdünnter Chlorwasserstoffsäure sich zu einer schwachen Basis, Alanin genannt, vereinigen [...]: $$\ce{\underset{\text{Aldehyd-Ammoniak}}{C4H4O2 * NH3} + HCl + \underset{Blausäure}{C2NH} +2 HO = \underset{Alanin}{C6H7NO4} + NH4Cl}$$

As David Richerby mentioned in the comments, Strecker's brutto-formula ($\ce{C6H7NO4}$) deviates from the moden one ($\ce{C3H7NO2}$), also the reaction scheme is a bit different.

1. Strecker, A. Annalen der Chemie und Pharmacie 1854, 91 (3), 349–351. DOI 10.1002/jlac.18540910309.
• @andselisk Easy-to-understand answers to easy-to-understand questions always get lots of votes. Think of it as compensation for the times when you've spent an hour writing a detailed answer to a difficult question and it scores +2 because nobody can be bothered to read it. – David Richerby Aug 27 '17 at 10:56
• Having said that, I'm a bit confused. Isn't alanine $\ce{C3H7NO2}$? – David Richerby Aug 27 '17 at 11:10
• @DavidRicherby I'm not 100% sure, but most likely it's brutto-formula was established incorrectly at that time. Many other sources around that period of time (mid 19th century) refer to alanine as to $\ce{C6H7NO4}$, or, to meet the era's spirit, $\ce{C^6H^7NO^4}$. – andselisk Aug 27 '17 at 16:10
• Ah, I see. Might be worth adding a note about that to the answer. – David Richerby Aug 27 '17 at 18:30

User364914 may be confused by the "odd" formulas in Strecker's 1854 paper. At that time there was no universal agreement on the masses of atoms. Was carbon 6 or 12? Was oxygen 8 or 16? Strecker used 6 for carbon and 8 for oxygen.

The aldehyde is acetaldehyde, $$\ce{NH3}$$ is OK, blausaure is hydrogen cyanide and $$\ce{OH}$$ is water. The corrected structures should have half the number of carbons and oxygens. The modern equation should read:

$$\ce{C2H4O.NH3 + HCl + HCN + H2O -> C3H7NO2 + NH4Cl}$$