what do the martyed army man need from poets?
Answers
Answer:
HERE YOU GO SISTER
Explanation:
A tertiary alcohol would result after workup.
(
H
3
C
)
2
C
=
O
+
R
M
g
X
→
(
H
3
C
)
2
R
C
−
O
−
→
(
H
3
C
)
2
R
C
O
H
Explanation:
The Grignard reagent features a more or less direct bond between a carbon centre and a metal centre:
R
H
2
C
δ
−
+
δ
M
g
X
, or even as a full-blown carbanion,
R
H
2
C
−
+
M
g
X
. Given this formulation, its interaction with a carbonyl species may be rationalized.
The Grignard will react with an aldehyde to form (after workup) a
2
∘
alcohol, with a ketone to give a
3
∘
alcohol, with carbon dioxide, and ethylene oxide, to give a carboxylic acid, or a primary alcohol respectively. All of these are examples of prized
C
−
C
bond forming reactions.
Of course, the ipso carbon on each substrate is somewhat electron-poor, i.e. electrophilic, viz.,
δ
−
O
=
C
+
δ
=
O
δ
−
. Upon addition of the Grignard, (which you can usually do by pouring the ethereal solution of
R
M
g
X
directly onto dry ice) a direct
C
−
C
bond is formed between the Grignard residue and the carbonyl. This gives a carboxylate salt, and after workup, a carboxylic acid that is 1 carbon longer than the original Grignard residue.
With a ketone, acetone, a
3
∘
alcohol is formed:
R
3
C
δ
−
+
δ
M
g
X
+
R
'
2
C
(
=
O
)
→
R
'
2
(
−
O
)
C
−
C
R
3
If it were acetone, then the alcohol product would be
(
H
3
C
)
2
C
(
O
H
)
R
.
All of this demands fast-working, being on the ball, and reasonably dry solvents. Today, you can even buy pre-prepared solutions of Grignard and lithium reagents for direct use in synthesis.
Can you predict the product between acetone and methyl magnesium chloride?