Chem 226 — Problem Set #9 —
“Fundamentals of Organic Chemistry,” 4th edition, John McMurry.
Chapter 9
2.
Name the following aldehydes and ketones.
(a)
O
(b)
CH2CH2CHO
CH3CH2CCH(CH3)2
H CH3
O
(c) O
CH3CCH2 CH2 CH2CCH2CH3
(d)
H
C O
H
(a) 2-methyl-3-pentanone, (b) 3-phenylpropanal, (c) 2,6-octanedione,
(d) (1R,2R)-2-methylcyclohexanecarbaldehyde
4.
How could you prepare pentanal from the following starting materials?
(a) 1-pentanol, (b) CH3CH2CH2CH2COOH, (c) 5-decene
(a)
CH3CH2CH2 CH2 CH2OH
PCC
CH2Cl2
CH3CH2CH2CH2 CHO
O
CH3CH2CH2 CH2C OH
(b)
CH3CH2CH2 CH2CH2OH
1. LiAlH4
+
2. H3O
PCC
(c)
CH3CH2 CH2 CH2C OH
CH3CH2 CH2 CH2CH2OH
CH3CH2CH2 CH2 CHO
CH2Cl2
CH3CH2 CH2 CH2CH CHCH2CH2CH2 CH3
O
CH3CH2CH2CH2 CH2 OH
1. LiAlH4
+
2. H3 O
PCC
CH2Cl2
KMnO4
H3O
+
CH3CH2CH2CH2CH2 OH
CH3CH2CH2 CH2 CHO
5.
How could you prepare 2-hexanone from the following starting materials?
(a) 2-hexanol, (b) 1-hexyne, (c) 2-methyl-1-hexene
OH
O
(a) CH3 CH2CH2CH2CHCH3
K 2CrO7
CH3CH2CH2CH2CCH3
O
(b) CH3 CH2CH2CH2C CH
CH3
(c) CH3 CH2CH2CH2C CH2
H2SO 4
HgSO4
H2O
CH3CH2CH2CH2CCH3
O
KMnO 4
H3O
CH3CH2CH2CH2CCH3
+
(b) The enol that initially forms here undergoes keto-enol tautomerism. Terminal alkynes give methyl
ketones; they follow Markovnikov’s rule.
(c) Potassium permanganate under basic conditions would hydroxylate the alkene to form a vicinal diol,
but under acidic conditions it completely cleaves the double bond. The =CH2 group would become
carbon dioxide.
6.
How would you carry out the following transformations? More than one step may be required.
(a) 3-hexene
3-hexanone
(b) benzene
1-phenylethanol
H
(a) CH3CH2CH CHCH2CH3
H2O
CH3CH2CH CHCH2CH3
H2 SO 4
H
K 2Cr2 O7
CH3CCl
AlCl3
NaBH4
O
CH3CH2CH CCH2CH3
O
(b)
OH
C CH3
O
H
C CH3
OH
9.
The reduction of a ketone to a secondary alcohol on treatment with NaBH4 is a nucleophilic addition
reaction in which the nucleophile hydride ion (:H-) adds to the carbonyl group, and the alkoxide ion
intermediate is then protonated. Show the mechanism of this reduction.
H
H B H
H
C
H
H
C
C
O
O H
O
H O CH2CH3
O CH2CH3
Ethanol is frequently used as a solvent in these
reactions. In any event a proton must be
supplied by something in the solution.
11.
When dissolved in water, trichloroacetaldehyde (chloral, CCl3CHO) exists primarily as the gem diol,
chloral hydrate. Show the structure of chloral hydrate.
CCl3CH(OH)2
14.
Show the mechanism of the acid-catalyzed formation of a cyclic acetal from ethylene glycol and
acetone.
H2 C
CH2
H2C
OH OH
H3 C
C
CH3
H3 C
O
C
CH3
H3C
O H
CH2
H2C
CH2
OH OH
O
OH
C
CH3
H3C
C
O H
O H
H
H+
H2C
CH2
O
H3C
O
C
CH3
H3 C
CH3
O H
C
CH3
H2C
CH2
H2C
CH2
O
OH
O
OH
H3C
C
CH3
H3 C
O
C
H3C
CH3
C
CH3
+ H2O
17.
Show the products obtained from addition of CH3MgBr to the following compounds.
(a) cyclopentanone, (b) benzophenone, (c) 3-hexanone
One assumes that McMurry means the alcohol one would get upon acidification after adding the
Grignard reagent to the ketone.
HO
C
(b)
(a)
OH
H3 C
CH3
OH
(c)
CH3CH2CCH2 CH2 CH3
CH3
18.
How might you use a Grignard addition reaction to prepare the following alcohols?
(a) 2-methyl-2-propanol, (b) 1-methylcyclohexanol, (c) 3-methyl-3-pentanol
To make a tertiary alcohol, we can react a ketone with the Grignard reagent.
To make a secondary alcohol we can react an aldehyde with the Grignard reagent.
To make a primary alcohol we can react formaldehyde with the Grignard reagent.
OH
CH3CCH3
CH3
H3C
1. CH3MgBr
+
2. H3O
O
CH3CCH3
O
OH
1. CH3MgBr
+
2. H3O
OH
CH3CH2 CCH2CH3
CH3
1. CH3MgBr
+
2. H3O
O
CH3CH2CCH2 CH3
19.
Identify the different kinds of carbonyl groups in the following molecules.
carboxylic acid
HO
C O
(a)
ester
(carboxylic)
O
CH3
ester
N
COCH3
OCCH3 (b)
OCC 6H5
H
O
H
ester
O
ester (cyclic
CH2OH esters are sometimes
called lactones)
HOHC
O
O
(c)
O
OH
ketone
We usually call the esters of carboxylic acids just esters since this type of ester is so common.
However, other acids – sulfonic acids, for example – can also form esters.
22.
Draw and name the seven aldehydes and ketones with the formula C5H10O.
O
O
CH3CH2CH2CH2CH
pentanal
CH3
O
CH3CHCH2CH
3-methylbutanal
23.
O
CH3 CH2CH2CCH3
2-pentanone
CH3
CH3 CHCCH3
CH3 CH2CCH2CH3
3-pentanone
CH3
HCCHCH2CH3
CH3
HCCCH3
O
O
O CH3
3-methyl-2-butanone 2-methylbutanal 2,2-dimethylpropanal
Which of the compounds you identified in question 22 are chiral?
Our approach here will be to look for stereocenters. If a molecule has one, and only one stereocenter,
it will be chiral. If a molecule does not have any stereocenters it is unlikely to be chiral. [ In this course
it is very unlikely we will see chiral molecules that do not have stereocenters.] If a molecule has two or
more stereocenters it will be chiral unless it is a meso structure.
2-Methylbutanal is the only molecule above that has has a stereocenter, and it has just one (C-2). It is
chiral.
32.
Starting from 2-cyclohexenone and any other reagents needed, how would you prepare the following
substances? (More than one step may be required.)
(a) 1,3-cyclohexadiene, (b) 1-methylcyclohexanol, (c) cyclohexanol,
(d) 1-phenyl-2-cyclohexen-1-ol
(a)
OH
H2SO 4
heat
OH
H3C
1. CH3MgBr
O
NaBH4
2. H3O
O
(b)
OH
H2
Pt
+
NaBH4
MgBr
(c)
+
H3O
34.
(d)
OH
Use a Grignard reaction on an aldehyde or ketone to synthesize the following compounds.
(a) 2-pentanol, (b) 2-phenyl-2-butanol, (c) 1-ethylcyclohexanol, (d) diphenylmethanol
When undertaking a Grignard synthesis of an alcohol keep in mind that two of the groups (including all
hydrogens) attached to the carbon bearing the OH group will come from the carbonyl compound and
one of them (not a hydrogen) will come from the Grignard reagent.
OH
H3C C CH2 CH2 CH3
H
+
H3 O
O
H3O
+
O
H3C C
C CH2CH2 CH3
+
H
CH3MgBr
+
H
BrMgCH2CH2 CH3
(a) Viewed one way, we are trying to
make a secondary alcohol, so we
employ an aldehyde and a Grignard
reagent. The Grignard reagent can
provide either of the boxed alkyl
groups, while the aldehyde will provide
the other. The hydrogen must be
provided by the aldehyde – there is no
HMgX that can react with a ketone to
give a secondary alcohol.
(b) Since we are trying to make a tertiary
alcohol, we will employ a ketone and a
CH3CCH2CH3
CH3C + BrMgCH2CH3 Grignard reagent. The ketone will provide
two of the groups attached to the carbon
H3O+
holding the OH (as well as that carbon
itself) and the Grignard reagent will provide
the third group. In this case I have
arranged for the Grignard to provide the
boxed ethyl group and the acetophenone will provide the methyl and phenyl groups. Two other
combinations could have been used. What are they?
OH
O
(c) We are, again, trying to make a
+ BrMgCH2CH3 tertiary alcohol, so we will use a ketone
and the Grignard. The carbon holding
the OH and two of the groups attached
+
H3O
to it must come from the ketone and the
third group must come from the
Grignard. The situation here is made a
little interesting by virtue of the ring, since the ring constitutes two of the groups attached to the OH
bearing carbon, but is “one piece” and cannot come from two different sources. Consequently, the
only choice for the ketone is cyclohexanone and this forces the Grignard to be ethylmagnesium halide.
HO
CH2CH3
O
H
O
C OH
C H
+
H3O +
MgBr
(d) A secondary alcohol requires an
aldehyde and Grignard reagent.
There is no choice here since two of
the groups are phenyls. The
aldehyde must be benzaldehyde and
the Grignard must provide another
phenyl group.