[NewtoAtoms]

Which one of the following would furnish the highest yield of methyl ether on reaction with NaOCH3 under conditions of the Williamson synthesis?

a.  p-nitrophenol               NO - This doesn't come anywhere close to yielding a methyl ether.
b.  bromobenzene              NO - This is a benzene with OCH₃ attached Carbon #1
c.  tert-butyl bromide         MAYBE - This becomes (CH₃)₃COCH₃
d.  1-butene                     This can't undergo Williamson synthesis as there is no alkoxide ion
e.  benzyl chloride              NO - This becomes C₆H₅CH₂OCH₃ + NaCl

My problem with this question is that NONE give simply methyl ether.  C comes close but it isn't simply methyl ether.

Can anyone shed some light on this question?

Thank you

[ARGOS++]

[NewtoAtoms]

Hello Argos+ +

Thank you for your reply. I looked at the website that you suggested and under 'OTHER NAMES' I do recognize that a molecule of C₈H₁₀O is called a Benzyl methyl ether.  Therefore in this reaction, benzyl chloride would indeed furnish the highest yield of methyl ether. 

In comparison tert-butyl bromide would only yield dimethyl ether -  which is not correct either.

Am I correct in my thinking?

[ARGOS++]

[azmanam]

b doesn't give product because you can't do SN2 reactions on sp2 carbon atoms.

c doesn't give product because you can't do SN2 reactions on tertiary halides (if it was SN1...)

d doesn't because the reaction is not acid catalyzed (if it was with methanol and acid, it would work)

e is the right answer, because benzyllic halides are good leaving groups.

Like Argos said, the question is looking for a methyl ether.