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Which group gets higher priority and why?

$\ce{R-CH_2OCH_3}$

$\ce{R-CH=O}$

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    $\begingroup$ Aldehyde over ether, oxygen in the $\ce{C=O}$ double bond counts twice. $\endgroup$ – Klaus-Dieter Warzecha Feb 20 '15 at 20:47
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As mentioned by Yomen , the double bond counts for two atoms with a single bond between them . They are called ghost atoms . This makes aldehyde group at higher priority than the ether .

Wikedia has a very nice explanation on it :

If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a ghost atom that has the same atomic number as B but is not attached to anything except A. In turn, when B is replaced with a list of attached atoms, A itself is excluded in accordance with the general principle of not doubling back along a bond that has just been followed, but a ghost atom for A is included so that the double bond is properly represented from both ends.

A triple bond is handled in a similar manner .

Wikipedia link : Cahn–Ingold–Prelog priority rules

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$\ce{R-CH=O}$ has higher priority over $\ce{R-CH_2OCH_3}$. Because, if $\ce{R}$ group is the same, then we have to compare between the next substituents: For $\ce{R-CH=O}$ you have one carbon atom connected to 2 oxygen atoms (a double bond is equivalent to two simple bonds) and one hydrogen atom. For $\ce{R-CH_2OCH_3}$, you have one carbon atom connected to one oxygen atom and two hydrogen atoms. So, $\ce{R-CH=O}$ has higher priority.

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