CH ENGR 104C Lecture Notes - Lecture 9: Tetrahedral Carbonyl Addition Compound, Hemiacetal, Deprotonation

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Carbonyl Chemistry:
Survey of Reactions and
Mechanisms
Course Notes
Chemistry 14D
Images and sample reactions
taken from the Chemistry 14D
Thinkbook for Fall 2004, and
Organic Chemistry
by Paula Yurkanis Bruice 4th edition
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Carbonyl Chemistry:
Survey of Reactions and Mechanisms:
When dealing with Carbonyls, we consider two general mechanism types:
Carbonyls
Have X as a Leaving Group
Don’t have X as a Leaving Group
Carboxylic Acid (RCO2H)
Aldehyde (RCHO)
Acyl Halide (ROX)
Ketone (RCOR’)
Acid Anhydride (RCO2COR’)
Ester (RCO2R’)
Amide (RCONR’2)
Nitrile (RCN) (doesn’t look like a Carbonyl because it doesn’t
have a C=O but it reacts very similarly) H3C—C=N
2)… we then get a
tetrahedral
intermediate whose
fate is determined by
the presence of a
leaving group
1) we start with
the carbonyl...
which is attacked
by the nucleophile
3b) …If X is NOT a leaving group, the O- accepts a
“H+” and the result is an ADDITION REACTION
3a) …If X is a leaving group, then you kick out the leaving
group and the result is a SUBSTITUTION REACTION.
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Another tetrahedral
intermediate is formed.
The HOCH3+ is
deprotonated by the :B
and ultimately, the ketal
is formed, as shown in the
pink box.
C= O Survey of Reactions
I. HEMIACETAL/ACETAL and
HEMIKETAL/KETAL FORMATION
Aldehyde Hemiacetal Acetal
Ketone Hemiketal Ketal
Ketone/aldehyde + alcohol = hemiacetal/hemiketal after one equivalent of alcohol
= acetal/ketal after two equivalents of alcohol
*hemiacetal, hemiketal, and half all start with the letter h, this is a
reminder that when you produce the hemiacetal or hemiketal, you are
halfway to the acetal/ketal
A hemiaceta/hemiketal has a carbon attached to an “ether” on one end and an alcohol on the
other.
o i.e. RO—C—OH
An acetal/ketal has a carbon that is attached to two “ethers”. If you take a hemiacetal and
replace the OH with an OR’ group then you get an acetal.
o i.e. RO—C—OR’
First, let us look at a generic mechanism for the formation of an acetal or ketal. Here, we start with the
ketone, so we will be forming a hemiketal, and then a ketal.
In step 1) the oxygen is protonated. This makes the carbonyl carbon more electrophilic by
giving it a greater partial positive charge
Then the first equivalent of alcohol attacks the electrophile in step 2)
Then, the tetrahedral intermediate is formed, when this is deprotonated by the :B, the hemiketal
is formed as seen inside the orange box
The mechanism doesn’t stop here… the remaining alcohol group on the molecule is protonated
by H—B+, to form H2O as a leaving group, with this, a double bond is formed.
Then, the second equivalent of alcohol attacks the electrophilic carbonyl carbon
1)
2)
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