May you could do a video about methylparaben like of the mechanism from a benzoe acid to like the paraben? Bcs I have to write a documentary of 12 Pages and it‘s much more complicated than I thought it would be. Stupid mistake of my chemistry abilities😂 could you please help out?
Hey netspirit79, So as stated in the video, I wanted to show the effect of the reduction on the left side of the molecule (to show that NaBH4 and the EtOH workup is alkene/alkyne safe). The left side of the molecule (the carbonyl carbon) is first attacked by H:-, an addition elimination mechanism follows, and the right side of the molecule is ejected as the leaving group when the tetrahedral intermediate collapses (which is a great LG because the carboxylate is resonance stabilized). The result of the first attack is an aldehyde, which is then attacked AGAIN by ANOTHER H:-. This, after workup from the EtOH, produces the product shown, which still has 4 carbons, as seen in the left side of the reactant. Did that make sense?
nice
With the last example LiAlH4 would convert it to alcohol so why didn't you rather use a protecting group and add H2 in raney nickel
May you could do a video about methylparaben like of the mechanism from a benzoe acid to like the paraben? Bcs I have to write a documentary of 12 Pages and it‘s much more complicated than I thought it would be. Stupid mistake of my chemistry abilities😂 could you please help out?
in the third example (excess NaBH4) is there actually an oxygen between the 2 carbonyls? i wonder how it gets replaced with a carbon
Hey netspirit79,
So as stated in the video, I wanted to show the effect of the reduction on the left side of the molecule (to show that NaBH4 and the EtOH workup is alkene/alkyne safe). The left side of the molecule (the carbonyl carbon) is first attacked by H:-, an addition elimination mechanism follows, and the right side of the molecule is ejected as the leaving group when the tetrahedral intermediate collapses (which is a great LG because the carboxylate is resonance stabilized).
The result of the first attack is an aldehyde, which is then attacked AGAIN by ANOTHER H:-. This, after workup from the EtOH, produces the product shown, which still has 4 carbons, as seen in the left side of the reactant.
Did that make sense?
@@jOeCHEM Oh got it. Sorry, all clear!
@@netspirit79 glad it made sense! And also, thanks for watching