Download Chapter Twenty-One

Chapter
Twenty-One
Organic and
Biochemical
Molecules
Question
A student gave a molecule the following name:
3-methyl-4-isopropylpentane
The teacher pointed out that, although the
molecule could be correctly drawn from this
name, the name violates the IUPAC rules. What
is the correct (IUPAC) name of the molecule?
a)
b)
c)
d)
e)
4-Isopropyl-3-methylpentane
2-Isopropyl-3-methylpentane
1,1,2,3-Tetramethylpentane
2,3,4-Trimethylhexane
3,4-Dimethylheptane
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Answer
d) 2,3,4-Trimethylhexane
Section 21.1, Alkanes: Saturated Hydrocarbons
The molecule would have six carbons in the
longest chain and three methyl groups. The
correct name is 2,3,4-trimethylhexane.
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Question
Which of the following names is a correct
one?
a) 3,4-Dichloropentane
b) cis-1,3-Dimethylpropane
c) 2-Bromo-1-chloro-4,4-diethyloctane
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Answer
c) 2-Bromo-1-chloro-4,4-diethyloctane
Section 21.1, Alkanes: Saturated
Hydrocarbons
Choice (a) should be 2,3-dichloropentane.
Choice (b) should be pentane.
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Question
Which of the following has the lowest
boiling point?
a)
b)
c)
d)
Butane
Ethane
Propane
Methane
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Answer
d) Methane
Section 21.1, Alkanes: Saturated
Hydrocarbons
The smallest of the saturated
hydrocarbons will have the lowest boiling
point.
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Question
How many isomers are there with the
formula C2H2Br2? Include both structural
and geometric isomers.
a)
b)
c)
d)
e)
2
3
4
5
6
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Answer
b) 3
Section 21.2, Alkenes and Alkynes
The formula given could be any of these:
cis-1,2-Dibromoethene
trans-1,2-Dibromoethene
1,1-Dibromoethene
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Question
Which of the following is an incorrect
name?
a)
b)
c)
d)
trans-1,2-Dichloroethene
Propylene
Ethylene
cis-1,2-Dichloroethane
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Answer
d) cis-1,2-Dichloroethane
Section 21.1, Alkanes: Saturated Hydrocarbons;
Section 21.2, Alkenes and Alkynes
Given that 1,2-dichloroethane is a saturated
hydrocarbon, no cis or trans designation is
necessary.
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Question
How many different “tetramethylbenzenes” are possible?
a)
b)
c)
d)
e)
2
3
4
5
6
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Answer
b) 3
Section 21.3, Aromatic Hydrocarbons
Here are the three possibilities:
1,2,3,4-Tetramethylbenzene
1,2,3,5-Tetramethylbenzene
1,2,4,5-Tetramethylbenzene
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Question
H2CCHCH2N(CH3)2 is
a)
b)
c)
d)
an alkyne and a secondary amine.
an alkene and a primary amine.
an alkene and a tertiary amine.
an alkyne and a tertiary amine.
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Answer
c) an alkene and a tertiary amine.
Section 21.2, Alkenes and Alkynes;
Section 21.4, Hydrocarbon Derivatives
This species is a tertiary amine because
three carbons are bonded to the nitrogen
and the molecule contains a C=C bond.
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Question
For which of the following compounds are
cis and trans isomers possible?
a)
b)
c)
d)
2,3-Dimethyl-2-butene
3-Methyl-2-pentene
4,4-Dimethylcyclohexanol
ortho-Chlorotoluene
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Answer
b) 3-Methyl-2-pentene
Section 21.2, Alkenes and Alkynes; Section 21.3,
Aromatic Hydrocarbons; Section 21.4, Hydrocarbon
Derivatives
Choices (c) and (d) do not contain different groups
across a double bond. For choice (a), the second
carbon contains two methyl groups, so the cis-trans
designation is not necessary.
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Question
Which of the following types of
compounds may lack an sp2-hybridized
carbon center?
a)
b)
c)
d)
Aldehydes
Ketones
Alcohols
Alkenes
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Answer
c) Alcohols
Section 21.4, Hydrocarbon Derivatives
Since aldehydes and ketones have C=O bonds
and alkenes and benzene have C=C bonds,
these derivatives have sp2-hybridized carbon
centers. Alcohols have C–O–H bonds and can
have sp3-hybridized carbon centers.
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Question
For which of the following choices do all of the
functional groups listed have a C=O bond?
a)
b)
c)
d)
e)
Ester, aldehyde, secondary alcohol, ketone
Alcohol (any), ether, ester
Secondary alcohol, ketone, aldehyde
Ester, aldehyde, ketone
Carboxylic acid, ether, tertiary alcohol
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Answer
d) Ester, aldehyde, ketone
Section 21.4, Hydrocarbon Derivatives
Table 21.4 lists the functional groups.
Carboxylic acids, esters, aldehydes, and
ketones all contain C=O bonds. Alcohols
contain the –OH group. Ethers contain the –O–
group.
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Question
Oxidation of secondary alcohols results in
a)
b)
c)
d)
e)
ketones.
tertiary alcohols.
aldehydes.
esters.
ethers.
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Answer
a) ketones.
Section 21.4, Hydrocarbon Derivatives
Ketones may be prepared from the
oxidation of secondary alcohols.
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Question
What might be the product of the oxidation
of 2-methyl-1-butanol?
a)
b)
c)
d)
e)
2-Methyl-2-butanone
2-Methylbutanal
2-methylbutanoic acid
Both b and c
Both a and c
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Answer
d) Both b and c
Section 21.4. Hydrocarbon Derivatives
The primary alcohol could be oxidized to
an aldehyde or a carboxylic acid, so the
answers are 2-methyl-1-butanal and 2methylbutanoic acid.
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Question
Which of the following is optically active
(i.e., chiral)?
a)
b)
c)
d)
e)
HN(CH3)2
CH2Cl2
2-Chloropropane
2-Chlorobutane
3-Chloropentane
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Answer
d) 2-Chlorobutane
Section 21.4, Hydrocarbon Derivatives
For a molecule to be optically active, it must
have a carbon atom bonded to four different
species. For 2-chlorobutane, the carbon at the 2
position is bonded to a methyl group, a chlorine,
a hydrogen, and an ethyl group.
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Question
Oxidation of primary alcohols results in
a)
b)
c)
d)
e)
ketones.
tertiary alcohols.
aldehydes.
esters.
ethers.
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Answer
c)
aldehydes.
Section 21.4, Hydrocarbon Derivatives
An aldehyde results from the oxidation of
a primary alcohol.
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Question
The boiling point of methanol is much higher
than that of ethane. This is primarily due to
a) the significant difference in the molar masses
of methanol and ethane.
b) the hydrogen bonding in methanol and the lack
of hydrogen bonding in ethane.
c) the significant difference in the molecular sizes
of methanol and ethane.
d) the carbon–oxygen bond in the methanol.
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Answer
b)
the hydrogen bonding in methanol and
the lack of hydrogen bonding in ethane.
Section 21.4, Hydrocarbon Derivatives
Methanol is a polar molecule with hydrogen
bonding; ethane is a nonpolar molecule with
London dispersion forces. The boiling point will
be much higher for polar molecules with
hydrogen bonding.
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Question
No atoms are lost from the starting
material in making which kind of polymer?
a) Condensation polymer
b) Polyester polymer
c) Addition polymer
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Answer
c) Addition polymer
Section 21.5, Synthetic Polymers
When monomers add to form polymers,
no atoms are lost.
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Question
The structure of the polymer used in a freezer
wrap can mainly be described as follows:
[CCl2 – CH2 – CCl2 – CH2 – CCl2 – CH2 – CCl2 – CH2]n
What is the structure of chief monomer of this
wrap?
a) CCl2 CH2
b) Cl2C–CH2
c) Cl2C CH2 CCl2
d) CCl2
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Answer
a)
CCl2CH2
Section 21.5, Synthetic Polymers
For the polymer given, the monomeric
unit is CCl2=CH2, which adds to form the
polymer.
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Question
Which of the following steps will increase
the rigidity of a polymer?
a)
b)
c)
d)
Use shorter polymer chains
Make chains more branched
Decrease cross-linking
Introduce the possibility of hydrogen
bonding between chains
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Answer
d) Introduce the possibility of hydrogen
bonding between chains
Section 21.5, Synthetic Polymers
Increasing hydrogen bonding in a polymer
can give it greater strength and rigidity.
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