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A.) p-nitrotoluene
B.) o-nitroanisole
C.) phthalic anhydride
D.) phenyl benzoate
E.) p-nitrobenzophenone
F.) p-t-butyltoluene

2007-03-04 07:28:39 · 2 answers · asked by augusta 1 in Science & Mathematics Chemistry

2 answers

Just so you actually learn something: look for the ortho or para positions next to a strong electron donator, such as an amine in anisole, a methyl in toluene, or a hydroxyl.

A) 2,4-dinitrotoluene
B) 2,4-dinitroanisole
C) don't know how to number it, but the nitro group would be two carbons away from the carbonyl carbons, and I'm not absolutely certain about this.
D) 3-nitro phenylbenzoate
E) no idea how to name it, but the nitro group would come ortho to the carbonyl on one of the rings
F) 4-t-butyl-2-nitrotoluene (the t-butyl is more electron donating, but very sterically bulky so it would hinder the reaction)

2007-03-04 07:37:32 · answer #1 · answered by Some Body 4 · 0 0

ok for section a you may desire to have a octane molecule so draw 7 lines zigzagged. the 1st end could have Cl linked to it and the different end could have an I linked. the molecule will become an octane while u upload a dashed line on the comparable C the place the Cl is linked and a wedged line on the comparable C because of the fact the I. b) this could help u visualize the 1st section much greater and perhaps help u make it. Its named (R)-2-chloro-7-iodo-2-methyloctane because of the fact the chloro and the dashed methyl is on the C2 and iodine on C7. oh and for section d) because of the fact the Br will react and replace the I, you will get a million-bromobutane

2016-09-30 04:51:48 · answer #2 · answered by ? 4 · 0 0

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