question about destroyer road maps

[phamdmd]

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hi guys, hope you had a merry xmas. i just needed help on a few questions:

for the road map 3:
how do you know where the double bonds go for
(CH3)3CO-Na+ with (CH3)3COH versus C2H5O-K+ with C2H5OH ?
Is there a rule or is this just straight rememorization?

for road map 4:
how come the NaBH4 with CH3OH produces no reaction? while 1)LiAlH4,Et20 2)H2O produces that product? Is it because LiAlH4 is stronger? Can someone give me an explanation for this one why NaBH4 doesn't work while LiAlH4 works?

Finally, i have a question about grignards. i know that they basically add X with MgCl and the X goes on the halide or whatever... but i was wondering where it adds on? like to most substituated or what? also, if you use a base or acid with it, how does it change?

ok thanks all.

 

[joonkimdds]

For road map4 (which I don't even know what that road map look like).
If it has something to do with OCHEM1(I took it about 1 month ago and the last chapter was about reducing agent NABH4 vs. LiAlH4), the only difference that I learned is that LiAlH4 is a stronger reducing agent than NaBH4.
Sorry, that's all I know... 😀

Edit: Ok here is what I just read from my textbook.
Sodium borohydride is a less powerful reducing agent than lithium aluminum hydride. Lithium aluminum hydride reduces acids, esters, aldehydes, and ketones, but sodium borohydride reduces only aldehydes and ketones.


I think what I wrote above is everything that my textbook has. It's only 2 pages in my textbook.
Basically it's explaning that LiAlH4 is a stronger reducing agent and used to reduce carboxylic acid or ester into alcohol.
NaBH4 is a weaker reducing agent and used to reduce aldehyde and ketone into alcohols.