Practice Homework for Chapters 9, 10, and 11:
Draw the structure of…
a) ethylisopropylacetylene
b) cis-pent-2-en-4-yne
Name the following structures using IUPAC nomenclature (new or old acceptable)
HO
HS
OH
Give the potential product/s of each reaction.
1) NaNH2
2) acetone
3) H3O+
HgSO4
H2SO4,H2O
1)KMnO4 ,KOH, 
2) H3O+
1) Ozone,-78°C
2) Zn, Acetic Acid
CH3
1) Hg(OAc)2, H2O,THF
2) NaBH4, NaOH, H2O
CH2CH CH2
1) BH3*THF, diglyme
2) H2O2, H2O, NaOH
Br
1) Mg, (CH3CH2)2O
2) Acetone
3) H3O+
Cl
1) Mg, (CH3CH2)2O
2) O
Cl
3) H3O+
OH OH
H2SO4
100 C
1)NaBH4
2) H3O+
O
Raney Ni, H2
HO
O
1)LiAlH4
2) H3O+
OH
CrO3*pyridine*HCl
CH2Cl2
OH
HO
SOCl2
dioxane
1) TsCl, pyridine
2) NaCN, CH3OH
Design a multi-step synthesis of the desired product from the starting material provided. You can use
any reactions you have learned in organic I so far.
A handy guide to retrosynthetic analysis can be found starting on page 376 in chapter 8 and page 416 in
chapter 9.
Starting Materials
Desired Product
OH
a)
Br
H
Multistep
H
CH3
Br
b)
A new drug has been produced by the Jonesco Pharmaceutical Company. It is used to treat Organic
chemistry students who, due to hyperactivity, cannot pay attention in class. The drug is Attentionall A.
Affordable starting materials for Attentionall A can be extracted from the San Antonion Lion. Propose a
reaction scheme for the synthesis of Attentionall A. (HINT 1: the scheme will be more than one step.
HINT 2: All required reactions have been covered in chapters 10, and 11.)
O
O
HO
Multiple
Steps
H
H
Cl
O
Attentionall A
Practice Homework for Chapters 9, 10, and 11:
Draw the structure of…
a) Ethylisopropylacetylene
b) cis-pent-2-en-4-yne
Name the following structures using IUPAC nomenclature (new or old acceptable)
HO
1-hexen-3-ol or hex-1-en-3-ol
HS
OH
2-cylcopropyl-3-mercaptopropan-1-ol
2-cylcopropyl-3-mercapto-1-propanol
Give the potential product/s of each reaction.
1) NaNH2
2) acetone
3) H3O+
 
Na+
1) NaNH2
O
O
+ NH3
H H
H O
O
OH
HgSO4
H2SO4,H2O
H
H2O
Hg+
HO
Tautomerization
see pg 403 for
mechanism
-H+
HO
H
HO
H
H
Hg+
H
+
Hg2+
1)KMnO4 ,KOH, 
2) H3O+
H2C
OH O
O C O +
O
OH
O
O
HO
HO
H
Hg+
Tautomerization mechanisms are on pages 411-413
1)KMnO4 ,KOH, 
2) H3O+
H
Hg+
enol- unstable product
ketone - stable product
H
H
H
H
Hg2+
HgSO4
+ H2O
H2SO4
O
H
O
O
OH
H H
H O
When an alkene is reacted with potassium permanganate (excellent oxidizer) at high temperatures or
high concentrations, the alkene will initially be oxidized to a glycol (-OH groups on adjacent carbons).
However, the glycol will be further oxidized resulting in oxidative cleavage of the alkene (breaking of the
carbon-carbon bonds). A tertiary carbon will be oxidized to a ketone, a secondary carbon will be
oxidized to a carboxylic acid (second and third products), and a primary carbon will be oxidized to
carbon dioxide (first product). (See mechanism pgs 366).
1) Ozone,-78°C
2) Zn, Acetic Acid
H2C
1) Ozone,-78°C
2) Zn/Acetic Acid
H
O C
H
O
H
+
H
O
O
O
H
H
O
H
Ozone can be used to oxidatively cleave alkene bonds. The oxidation that occurs during ozonolysis is
less intense than that using potassium permanganate. Tertiary carbons are oxidized to a ketone. Both
secondary and primary carbons are oxidized to an aldehyde, without over-oxidizing to carbon dioxide or
a carboxylic acid. Step 2 involves the reduction of the intermediate ozonide by either zinc metal (a good
reducing agent) or dimethyl sulfide (CH3)2S (can be oxidized to dimethyl sulfoxide (CH3)2SO). (See
mechanism, pg 367)
Ozonolysis can be used in helping to identify the location of a carbon-carbon double bond in an
unknown compound by identifying the products of the oxidative cleavage.
CH3
1) Hg(OAc)2, H2O,THF
2) NaBH4, NaOH, H2O
CH3
OAc
Hg 
1) Hg(OAc)2, H2O, THF
2) NaBH4, NaOH, H2O

H
CH3
O
H
H
OAc
-H+
Hg CH3
NaBH4
NaOH(aq)
H
OH
H CH3
H
OH
CH2CH CH2
1) BH3*THF, diglyme
2) H2O2, H2O, NaOH
H
BH2
CH2C
H
CH2

H
BH
CH2C  CH2
H
H
BH2
CH2C
H
CH2
Repeat H2O2
twice
NaOH
H
CH2C
H
OH
CH2
Hydroboration can be run in an inert solvent such as tetrahydrofuran (THF) and diglyme
(CH3OCH2CH2OCH2CH2OCH3).
Br
1) Mg, (CH3CH2)2O
2) Acetone
3) H3O+
Alkylhalide in the presence of Magnesium metal in an ether solvent will result in the formation of a
Grignard reagent. This grignard reagent will then attack the carbonyl (acetone) via a nucleophilic attack.
The addition of weak acid will protonate the alkoxide formed in step 2.
Br
Mg, diethylether
Cl
O
H3C C
MgBr
CH3
1) Mg, (CH3CH2)2O
2) O
Cl
3) H3O+
H
CH3
C O
CH3
O
H
H
CH3
C OH
CH3
Alkylhalide in the presence of Magnesium metal in an ether solvent will result in the formation of a
Grignard reagent. This grignard reagent will then attack the acid chloride via a nucleophilic attack. As
chlorine is a good leaving group, two R groups will be added in successive nucleophilic attacks. The
addition of weak acid will protonate the alkoxide formed in step 2.
Cl
Mg, diethylether
MgCl
MgCl
O
O
Cl
Cl
O
H3O+
OH
O
OH OH
H2SO4
100 C
This is another example of a pinacol type rearrangement. One of the 3º alcohols is protonated, making it
a good leaving group. Upon leaving, a 3º carbocation is formed. Upon formation of the carbocation,
methyl migration occurs following by donation of an oxygen lone pair to form a oxygen stabilized cation.
Deprotonation then occurs leaving an expanded ring with a ketone.
H O H
H
H
O H
OH
OH
OH
OH
H
O
O H
H
O
OH
The following problem displays the varying reducing abilities of three reagents. NaBH4 will only reduce
ketones and aldehydes to the corresponding alcohols. LiAlH4 is a stronger reducing agent and will
reduce all of the carbonyl containing functional groups to the corresponding alcohols. Raney nickel will
reduce all carbonyls to alcohols, as well as, alkenes to alkanes.
HO
O
1)NaBH4
2) H3O+
HO
HO
O
O
HO
O
1)LiAlH4
2) H3O+
HO
HO
O
O
HO
Raney Ni, H2
HO
O
HO
O
CrO3*pyridine*HCl, better known as pyridinium chlorochromate or PCC is an oxidizing reagent that will
convert 2º alcohols to ketones and 1º alcohols to aldehydes. To further convert the aldehyde to a
carboxylic acid, a stronger oxidizing agent would be needed, such as, Na2Cr2O7/H2SO4.
OH
CrO3*pyridine*HCl
CH2Cl2
O
H
Thionyl chloride is a good chlorinating agent for converting 1º and 2º alcohols to alkychlorides.
OH
SOCl2
dioxane
Cl
The following is a two-step reaction. The first step involves converting a poor leaving group –OH to a
better leaving group –OTs. The tosylate is then displaced by the good nucleophile, CN-.
2) NaCN, CH3OH
1) TsCl, pyridine
HO
TsO
N
C
Design a multi-step synthesis of the desired product from the starting material provided. You can use
any reactions you have learned in organic I so far.
A handy guide to retrosynthetic analysis can be found starting on page 376 in chapter 8 and page 416 in
chapter 9.
Starting Materials
Desired Product
OH
a)
You are given two starting materials to achieve your final product. You notice that your alkyne will
eventually bond to the methylene carbon on the cyclohexane ring. You will need to think of what types
of reactions can add an existing terminal alkyne to a carbon atom. We went over two types previously,
the nucleophilic substitution of an acetylide anion with an alkylhalide (pg 389), and the addition of the
acetylide anion to a carbonyl carbon (pg 401). As an alcohol is produced in the reaction with a carbonyl,
and an -OH group is present on your final product, this is the probable reaction.
O
OH
+
You now must compare your ketone to your starting material. This reaction must involve the oxidation
of a carbon-carbon double bond to a carbon-oxygen double bond. This can be achieved by the oxidative
cleavage of an alkene by either ozone or potassium permanganate.
O
You can now piece together your synthesis.
1) O3
2) H2O
O
or
KMnO4, KOH
H2O, heat
NaNH2, NH3
O 1) Na+
2)H3O+
Na+
OH
Br
H
Multistep
H
CH3
Br
b)
Beware of stereochemistry.
H
Multistep
Br
Br
H
CH3
The product is drawn in a way as to emphasis the anti-addition of 2 bromine atoms. As your starting
material is an alkyne, it appears that the total synthesis will be short: a hydrogenation then a
bromination. With the stereochemistry shown, the bromination occurs with a trans-substituted alkene.
The trans-substituted alkene can be formed via the metal-ammonia reduction of an alkyne (p 398-399).
Bromination occurs via the addition of bromine across a C-C double bond (p 342-345).
H
Na/NH3
H
H
H
Br2
H
Br
Br
H
CH3
A new drug has been produced by the Jonesco Pharmaceutical Company. It is used to treat Organic
chemistry students who, due to hyperactivity, cannot pay attention in class. The drug is Attentionall A.
Affordable starting materials for Attentionall A can be extracted from the San Antoinion Lion. Propose a
reaction scheme for the synthesis of Attentionall A. (HINT 1: the scheme will be more than one step.
HINT 2: All required reactions have been covered in chapters 10, and 11.)
O
O
HO
Multiple
Steps
H
H
Cl
O
Attentionall A
The synthesis of Attentionall A needs to be approached by looking at the two reactive sites of the
molecule, the carboxylic acid and the alkene. The carboxylic acid needs to be converted to a chloride.
The alkene needs to be cleaved into two aldehydes. The alkene cleavage can be achieved in two
methods, either by the synthesis of a glycol followed by periodic acid cleavage, or by ozonolysis.
The conversion of the carboxylic acid to a chloride must be done, in two steps. The first step involves
hydride reduction to an alcohol, then the replacement of an alcohol with a chloride.
The carboxylic acid has to be reduced by LAH. A stronger reducing agent like Raney nickel will reduce
the alkene to an alkane, and a weaker agent like NaBH4 will not reduce a stable carbonyl like a carboxylic
acid. If the alkene is cleaved first, the aldehydes formed would be reduced to alcohols as well.
Keeping this all in mind, the carboxylic acid must be reduced to an alcohol via LAH. The next step can be
achieved by any number of good chlorinating agents for primary alcohols: PCl3, PCl5, or SOCl2. The
cleavage of the alkene via glycolysis and periodic acid can then be done to obtain Attentionall A.
O
1)LiAlH4
2)H3O+
PCl3
PCl5
or SOCl2
HO
OsO4
H2O2
Cl
Cl
HO
O3, CH3SCH3
O
H
H
OH
HIO4
Cl
Cl
O
Attentionall A
OH