Click here👆to get an answer to your question ️ och3 group is

2017年10月15日 · An alkoxide molecule has more steric hindrance and less distinct polarization, making it less attracted to the positive carbon center. So "OCH3"^- is a weaker nucleophile than the hydroxide ion in a protic solvent though "OCH3"^- is a stronger base. "C"_2"H"_5"O" is more bulky than (more steric hindrance)"OCH3"^- so reasonably it is less ...

− N H 2, − O C H 3, − C 6 H 5 are all ortho-para directing group. These groups when present on benzene ring, direct the second or incomming elctrophile to the ortho-para positions.

Which of the following is a hemiacetal? OCH3 HO-CH-CH3 HO-CH-CH3 HO-C-CH3 OCH он OCH3 он D, CH3O-CH-CH3 The hemiacetal shown below is formed from the carbonvl addition between which compounds? OCH(CH 3)2 HO-CH- CH2CH3 A. propanal and 2-propanol C. acetone and 1-propanol E. None of these can form the hemiacetal shown.

Click here👆to get an answer to your question ️ ch3och2 ch2 och3

OCH3 Na, CH3OH NH3 OCH3 OCH3 OCH3 OCH3 OCH3 . Show transcribed image text. There are 2 steps to solve this ...

Question: Draw the Lewis structure of (CHO)OCH3 and then choose the appropriate set of molecular geometries of the three central atoms. Your answer choice is independent of the orientation of your drawn structure.

Question: Pick the solvent that gives the fastest SN2 reaction between CH2CH2Br and OCH3. 18:43 сезон H20 CH2CH2OH Sta DMSO tep < Prey - !! 0 Bie 2 of 25 O Next > Show transcribed image text There’s just one step to solve this.

The conversion of acetyl chloride to methyl acetate occurs via the following two-step mechanism: What is the rate equation for this reaction if the first step is rate determining? Rate =k [acetyl chloride ] Rate =k [acetyl chloride] [OCH3]2 Rate =k [acetyl chloride] [OOCH3] Rate =k[OCH3]

excess CH2OH -OCH3 + HCI trityl chloride methoxy triphenyl methane The intermediate in the above reaction is the trityl cation o | trityl cation Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.