Beginning with how pi bonds are formed within a carbonyl bond (C=O) to distinguishing between aldehydes and ketones, these series of notes will help you go the extra mile in your studies. These I've crafted myself helped me go from a C in year 12 to an A by the end of year 13, and they can help you...
Describe how the i-bond forms in a care
mys group
:
-
c =
0
/ -
-
sideways overlapping of adjacent
R2 -
C -
O p-orbital
&
it above and below the C-0
.
-
& the =0 bond does not react readily with electrophiles :
-
the (is f* +
the C 2 =
becomes palaried
-the it bind elections are unavailable due to
being polarized
-
electrophiles are repelled
+
-
nuclephiles are attracted to the Cf
Reduction of Carbonyls to form alcinols.
aldehydes
+
ketones can be reduced to
-
an alcohol
[]
-Reducing Agent :
NaBH4(an)
represented as [H]
in equasions.
, example 1 :
Reduction of Propanal
H H H
5
H
I
j c
I I
+
2[H] > c OH
-
-
H
-
H
- - -
-
-
11
It H It H
Propanal Propan-1-01
example 2 : Ketone
H H
-
It It It
It 0 It
1 I
I 1
- - -
+ +
2) - -
- - -
- -
-
I & I I & I
I I
H H H (t H H H H It
Pentone Pentan-1-01
Mechanism :
Nucleophilic Addition
BHG acts a source of : H-
-
as
Example :
Ethanal (Aldenyde) st
jo+St
H
,
f- H
I 8 H
It
H-c- I
I
I
> H - -
H-c-C-O
-
-
H
iH
·
intermediate I
products :
ethanol +
hydroxide ion-
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