Confusion on syn and anti

I came across the following compound syn acetaldoxime
However why is it syn because the lone pair having lowest priority is anti to H which is lowest priority on other side.
Please clarify 


[![enter image description here](https://i.stack.imgur.com/Bw3Rc.jpg)](https://i.stack.imgur.com/Bw3Rc.jpg)




In an oxime, the syn or anti configuration is determined by the position of the lone pair on nitrogen with respect to the alkyl or aryl group. That is, syn configuration is one in which the lone pair on nitrogen and the alkyl or aryl group are in the same side, while, in anti configuration, they are on opposite sides.


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By traditional convention, the drawing displays aldoxime in (*syn*)-configuration because both OH (of oxime) and remaining H (from the once aldehyde) are on the same side of the double bond.


Your right, on first sight (even on subsequent ones ...) it is puzzling.


An obvious alternative is to adhere again to the [CIP-rules](https://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules), where you check each side of the double bond for the substitutent of highest priority; which yields in the example provided by you the (*E*)-configuration. Applying the CIP-rules comes with the advantage to be consistent with the naming around other C=C double bonds, and being applicable for ketoximes, too.


Side note: Do not forget, often reactions affording the oxime yield a mixture of (*E*)- and (*Z*)-configurated product. Albeit the (*E*)-aldoxime is thermodynamically favoured (sugar chemistry, for example), you may then recognise the patterns of the two isomers in the 1H NMR spectrum.


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