Question

WTF AM I DOING WRONG http://gyazo.com/c310d30080f91c65d190bf0e0dd80840.png

Answer 1

the NaOH is there to deprotonate the ketone to produce a nucleophile that can attack via substitution reaction. Does that make sense?

Answer 2

Do you know the general pKa for an aldehyve vs ketone? That might help you with these.

Answer 3

Not too much. I'm just frustrated becuase I'm super close but I don't know what to change.

Answer 4

Attacch it to the ketone on the cyclopentanone

Answer 5

Attach what to what?

Answer 6

the \(\sf \color{blue}{ketone}\)

Answer 7

What about it?

Answer 8

the \(\sf \color{blue}{ketone}\)

Answer 9

WHAT ABOUT IT!

Answer 10

looooooooooooooooooooool

Answer 11

Can you not yell at me, please. This hurts my feelings.

Answer 12

wait, I didn't type that twice. da funk?

Answer 13

lmao

Answer 14

what he is drawn ifs right :)

Answer 15

All I know is that NaOH is a catalyst in this reaction. I dont know anything.

Answer 16

If you mix the benzyladehyde with NaOH, you won't really get a reaction....the product would be too unstable. Thus, NaOH reacts with the cyclopentanone (deprotonating the cyclopentanone) to make it more reactive toward benzaldehyde.

Answer 17

Guys I'll look back at that in a sec. Can you help me with this one? http://gyazo.com/a58a420c9eb11373d6f6a0731736cf91.png

Answer 18

The last one is gringard reaction. Very important reaction in organic chemistry.

Answer 19

What do I change it to?

Answer 20

Note that the oxygen on the ketone carries a partial negative charge \(\delta^-\), while the carbon has a partial positive charge \(\delta^+\).
Similarly, the magnesiumbromoethane, the MgBr carries a \(\delta^+\), the carbon \(\delta^-\). Etc.
Basically, + goes with - and so forth.

Answer 21

Then, water protinates the organometalic structure. Organometallics are pretty unstable and highly reactive. So, you get an alcohol.

Answer 22

The CrO\(_3\) promotes a jones oxidation reaction. I am not really sure what it does, because it tends to form alcohols \(\rightarrow\) ketones. But I find it hard to form one since it is already a 3º alcohol. Maybe a very unstable product forms.