Silberberg7e solution manual ch 15

If the ethyl group is attached to the second carbon from the left, the longest chain becomes 6 carbons instead of 5 and the structure is the same as 3.. With the triple bond between the

CHAPTER 15 ORGANIC COMPOUNDS AND THE ATOMIC PROPERTIES OF CARBON

FOLLOW–UP PROBLEMS

15.1A a) Plan: For compounds with seven carbons, first start with 7 C atoms in a straight chain Then make all

arrangements with 6 carbons in a straight chain and 1 carbon branched off the chain Examine the structures to make sure they are all different Continue in the same manner with 5 C atoms in chain and 2 branched off the chain, then 4 C atoms in chain and 3 branched off, and 3 C atoms in chain and 4 branched off Examine all structures to guarantee there are no duplicates

Six carbons in chain:

H

CH

as (1)

Five carbons in chain:

C

H

HH

H

HH

HH

Starting with a methyl group on the second carbon from the left, add another methyl group in a systematic pattern Structure (4) has the second methyl on the second carbon, (5) on the third carbon, (6) on the 4th carbon These are all the unique possibilities with the first methyl group on the second carbon Any other variation will produce a structure identical to a structure already drawn Next, move the first methyl group to the third carbon where the only unique placement for the second methyl group is on the third carbon as shown in structure (7) Which structure above would be the same as placing the first methyl group on the third carbon and the second methyl group on the fourth carbon? Are there any more arrangements with a 5-carbon chain? Think of attaching the two extra carbons as an ethyl group (—CH2CH3) If the ethyl group is attached to the second carbon from the left, the longest chain becomes 6 carbons instead of 5 and the structure is the same as (3) If the ethyl group is instead attached to the third carbon, the longest chain is still 5 carbons and the ethyl group is a side chain to give structure (8)

CH

b) Plan: For a five-carbon compound start with 5 C atoms in a chain and place the triple bond in as many

unique places as possible

HH

HH

4 C atoms in chain: There are two unique placements for the triple bond: one between the first and second

carbons and one between the second and third carbons in the chain

The fifth carbon is added as a methyl group branched off the chain With the triple bond between the first and second carbons, the methyl group is attached to the third carbon to give structure (3)

(3)H

15.1B a) Plan: Start with a ring consisting of 4 atoms Make sure to include the double bond in the ring Then work with

a ring consisting of 3 atoms and 1 carbon branched off the ring Move the position of the double bond relative to the branch to draw the 3 remaining structures The double bond can occur within the ring or between a ring carbon and the branch carbon

Solution:

Four carbons in the ring:

Three carbons in the ring:

b) Plan: For a four-carbon compound start with 4 C atoms in a chain and place the two double bonds in as many unique combinations as possible (there are 2) It will not be possible to draw a molecule with 3 C atoms in a chain, 1 branch , and 2 double bonds without a C atom having more than 4 bonds

Solution:

15.2A Plan: Examine the structure for chain length and side groups; then use the structure to write the name In part d),

examine the structure for carbons that are bonded to four different groups Those are the chiral carbons

Solution:

a) 3,3-diethylpentane There are 5 single-bonded carbons in the main chain and two ethyl groups attached to

carbon #3 of that chain The end of the name, pentane, indicates a 5 C chain (pent- represents 5 C) with only single bonds between the carbons (-ane represents alkanes, only single bonds) 3,3-diethyl- means there are two ethyl groups and that each ethyl group is attached to carbon #3

b) 1-ethyl-2-methylcyclobutane The main “chain” is a four-membered ring with an ethyl group and a methyl

group attached as branches The name cyclobutane describes the 4 C ring (cyclobut-) with only single bonds (-ane) The 1-ethyl- indicates that an ethyl group (–CH2CH3) is attached to carbon #1 The 2-methyl- indicates that

a methyl group (–CH3) is attached to carbon #2 The lower number carbon is assigned to the ethyl group because

it precedes methyl alphabetically

c) trans-3-methyl-3-hexene There are 6 carbons in the main chain, with a double bond between carbons #3 and

#4 (counting from the right hand side) There is also a methyl group attached to carbon #3 The name 3-hexene describes the main chain (hex- indicates that there are 6 carbons in the chain, -ene indicates that there is a double bond in the chain, 3- indicates that the double bond starts at carbon #3) The name 3-methyl- indicates that there is

a methyl group on carbon #3 Finally, because there is a double bond, we have to determine whether the groups

bonded to the double bond are in a cis- (same side of the double bond) or trans- (opposite sides of the double

bond) configuration In this structure, the longer carbon chains attached to the double bond are on opposite sides

of the double bond, so we add the prefix trans- to the name of the compound

d) 1-methyl-2-propylcyclopentane The main “chain” in the molecule is a 5 C ring with only single bonds A

methyl group and a propyl group are attached to the ring The name cyclopentane describes a 5-membered ring (cyclopent-) with all single bonds (-ane) The name 1-methyl- indicates that a methyl group is attached to the ring

at the #1 position The name 2-propyl- indicates that a propyl group is attached to the ring at the #2 position The lower number carbon is assigned to the methyl group because it precedes propyl alphabetically The chiral carbons (those bonded to four different groups) are marked with asterisks

15.2B Plan: Analyze the name for chain length and side groups; then draw the structure

Solution:

a) 3-ethyl-3-methyloctane The end of the name, octane, indicates an 8 C chain (oct- represents 8 C) with only single bonds between the carbons (-ane represents alkanes, only single bonds) 3-ethyl means an ethyl group (–CH2CH3) attached to carbon #3 and 3-methyl means a methyl group (–CH3) attached to carbon #3

H3C CH2 C CH2 CH2 CH2 CH2 CH3

CH3

CH2

CH3b) 1-ethyl-3-propylcyclohexane The hexane indicates 6 C chain (hex-) and only single bonds (-ane) The cyclo- indicates that the 6 carbons are in a ring The 1-ethyl indicates that an ethyl group (–CH2CH3) is attached to carbon #1 Select any carbon atom in the ring as carbon #1 since all the carbon atoms in the ring are equivalent The 3-propyl indicates that a propyl group (–CH2CH2CH3) is attached to carbon #3

c) 3,3-diethyl-1-hexyne The 1-hexyne indicates a 6 C chain with a triple bond (-yne) between C #1 and C #2 The 3,3-diethyl means two (di-) ethyl groups (–CH2CH3) both attached to C #3

d) trans-3-methyl-3-heptene The 3-heptene indicates a 7 C chain (hept-) with one double bond (-ene) between the

3rd and 4th carbons The 3-methyl indicates a methyl group (–CH3) attached to carbon #3 The trans indicates that the arrangement around the two carbons in the double bond gives the two smaller groups on opposite sides of the double bond Therefore, the smaller group on the third carbon (which would be the methyl group) is above the double bond while the smaller group, H, on the fourth carbon is below the double bond

15.3A Plan: In an addition reaction, atoms are added to the carbon(s) in a double bond Atoms are removed in an

elimination reaction, resulting in a product with a double bond In a substitution reaction, an atom or group of atoms substitutes for another one in the reactant

Solution:

a) In this reaction, the Br on the carbon chain is replaced with a hydroxyl group while the carbons maintain the

same number of bonds This is a substitution reaction

b) The methyl group on the left-hand side of the structure is replaced by a hydrogen atom while the carbons

maintain the same number of bonds This is a substitution reaction

c) Water (in the form of a –H group and an –OH group) is added to the double bond, resulting in the product having three more atoms Additionally, there is a second O bonded to the C in the product compared to the

reactant This is an addition reaction

15.3B Plan: a) An addition reaction involves breaking a multiple bond, in this case the double bond in 2-butene, and

adding the other reactant to the carbons in the double bond The reactant Cl2 will add –Cl to one of the carbons and –Cl to the other carbon

b) A substitution reaction involves removing one atom or group from a carbon chain and replacing it with another atom or group For 1-bromopropane the bromine will be replaced by hydroxide

c) An elimination reaction involves removing two atoms or groups, one from each of two adjacent carbon atoms, and forming a double bond between the two carbon atoms For 2methyl2propanol, the –OH group from the center carbon and a hydrogen from one of the terminal carbons will be removed and a double bond formed between the two carbon atoms

15.4A Plan: Determine the functional group(s) of the organic reactant(s) and then examine any inorganic reactant(s) to

decide on the reaction type

Solution:

a) This reaction is an oxidation (elimination) reaction because Cr2O72– and H2SO4 are oxidizing agents An alcohol group (–OH) is oxidized to a ketone group and a single bond between C and O is converted to a double bond

b) This is a substitution reaction because the bromine atoms on the organic reactant (an alkyl halide) can be

replaced with the cyanide groups from the inorganic reactant

15.4B Plan: Examine any inorganic compounds and the organic product to determine the organic reactant Look for

differences between the reactants and products, if possible

product, change the carbonyl group, =O, to an –OH group In part b) ethyllithium, CH3CH2–Li, and H2O react to

convert a ketone or aldehyde to form the alcohol and to add an ethyl group to the carbonyl carbon To form the

product, find the carbonyl carbon, add an ethyl group (from CH3CH2–Li) to the carbon and change the carbonyl group, =O, to a –OH bond

Solution:

a)

b)

15.5B Plan: The oxidizing agents in reaction a) indicate that the ketone group (=O) has been oxidized from an alcohol

To form the reactant, replace the C=O with an alcohol group, (C–OH) In reaction b) the reactants CH3CH2–Li and H2O indicate that a ketone or aldehyde reacts to form the alcohol To form the reactant find the carbon with the alcohol group, remove the ethyl group that came from CH3CH2–Li and change the –OH bond to a carbonyl group, =O

Solution:

a) In the reactant, a hydrogen atom is lost from the oxygen and another hydrogen is lost from the adjacent carbon

A double bond forms between the carbon and the oxygen

CH3OH

b) In the reactant, the first carbon off the ring will have a carbonyl group (–C=O) in place of the alcohol group (–OH) and the ethyl group (–CH2CH3)

HC

O

Check: When the reactants are combined, the products are identical to those given

15.6A Plan: Determine the functional group(s) of the organic reactant(s) and then examine any inorganic reactant(s) to

identify the reaction type Then draw the structure of the product(s) based on the reaction type

Solution:

a) The reactant is an ester because it contains the unit:

Reacting an amide with a reducing agent like LiAlH4 converts an amide to an amine

15.6B Plan: a) The product is an ester because it contains the unit:

b) The product is an amide because it contains the unit:

15.7A Plan: Examine the structures for known functional groups: alkenes (C=C), alkynes (CC), haloalkanes (C–X,

where X is a halogen), alcohols (C–OH), esters (–COOR), ethers (C–O–C), amines (NR3), carboxylic acids (–COOH), amides (–CONR2), aldehydes (O=CHR), and ketones (O=CR2 where R cannot be H)

b) The structure contains an amide and a haloalkane

where X is a halogen), alcohols (C–OH), esters (–COOR), ethers (C–O–C), amines (NR3), carboxylic acids (–COOH), amides (–CONR2), aldehydes (O=CHR), and ketones (O=CR2 where R cannot be H)

Solution:

a) The structure contains an aromatic ring, a haloalkane, a tertiary amine, and an aldehyde group

b) The structure contains an amide, a nitrile group, a ketone group, and an aromatic ring

CH2C

N

C

H2

HNCO

TOOLS OF THE LABORATORY AND CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B15.1 Plan: Protons (hydrogen atoms) that have identical environments produce one peak, while protons

in different environments produce different peaks Examine the structure of each compound for

different types (different environments) of hydrogen atoms

B15.2 Plan: Convert the frequency in MHz to frequency in Hz Use the relationship between energy and frequency,

E = hν, to find the energy associated with the given frequency

B15.3 Plan: A compound with two peaks in the NMR spectrum must have hydrogen atoms in two different

environments Since the two peaks are in a 3:1 ratio, the ratio of hydrogen atoms in the two environments must be 3:1 Look at the structure of each isomer and identify the number of environments of hydrogen atoms in each one

Solution:

Isomer A has a total of four hydrogen atoms that are each bonded to a C atom which is bonded to a Br atom and another C atom These would produce one peak Another four hydrogen atoms are bonded to the two C atoms in the middle of the molecule; these produce one peak Since there are four hydrogen atoms in two different

environments, two peaks are produced with a ratio of 1:1 and the two peaks would be the same size

Isomer B has six hydrogen atoms in one environment, bonded to a C atom that is bonded to another C atom that is bonded to a Br atom These six hydrogen atoms produce one peak Then there are two hydrogen atoms bonded to

a C atom that is also bonded to a Br atom These produce a second peak Since there are six hydrogen atoms in one environment and two hydrogen atoms in another environment, two peaks are produced with a size ratio of 6:2

or 3:1 Isomer B is characterized by a spectrum of two peaks in a 3:1 ratio

Isomer C has hydrogen atoms in three different environments Four hydrogen atoms are at each end of the chain, bonded to a C atom that is also bonded to a Br atom; three hydrogen atoms are bonded to a C atom which is bonded to the middle C atom; one hydrogen atom is bonded to the middle C atom Since there are three types of hydrogen atoms, three peaks are expected in the NMR spectrum

B15.4 Plan: When a ddNTP is incorporated into the growing DNA chain, polymerization stops since no more

phosphor-diester bonds can be formed When ddATP is added, the chain stops as each base T is encountered (base A pairs with base T); when ddCTP is added, the chain stops as each base G is encountered (base C pairs with base G)

Solution:

With ddATP, the chain stops at each base T:

TACAGGTTCAGT

ddATP will give four complementary chain pieces: A, AGTCA, AAGTCA, ATGTCCAAGTCA

With ddCTP, the chain stops at each base G:

TACAGGTTCAGT

ddCTP will give three complementary chain pieces: CA, CAAGTCA, CCAAGTCA

B15.5 Plan: The longest complementary chain piece can be used to find the sequence of bases in the

DNA fragment Bases A and T pair and bases C and G pair

Solution:

CATATG is the longest complementary chain piece; matching G for C, C for G, A for T, and T for A results in a

DNA fragment of GTATAC

15.1 Organic: Methane (natural gas) CH4 Acetic acid (in vinegar) C2H4O2

Inorganic: Calcium carbonate CaCO3 Sodium bicarbonate NaHCO3

15.2 a) Carbon’s electronegativity is midway between the most metallic and nonmetallic elements of Period 2 To attain

a filled outer shell, carbon forms covalent bonds to other atoms in molecules (e.g., methane, CH4), network covalent solids (e.g., diamond), and polyatomic ions (e.g., carbonate, CO32–)

b) Since carbon has four valence shell electrons, it forms four covalent bonds to attain an octet

c) Two noble gas configurations, He and Ne, are equally near carbon’s configuration To reach the He

configuration, the carbon atom must lose four electrons, requiring too much energy to form the C4+ cation This is confirmed by the fact that the value of the ionization energy for carbon is very high To reach the Ne

configuration, the carbon atom must gain four electrons, also requiring too much energy to form the C4– anion The fact that a carbon anion is unlikely to form is supported by carbon’s electron affinity The other possible ions would not have a stable noble gas configuration

d) Carbon is able to bond to itself extensively because carbon’s small size allows for closer approach and greater orbital overlap The greater orbital overlap results in a strong, stable bond

e) The C–C bond is short enough to allow the sideways overlap of unhybridized p orbitals on neighboring C atoms The sideways overlap of p orbitals results in double and triple bonds

15.3 a) The elements that most frequently bond to carbon are other carbon atoms, hydrogen, oxygen, nitrogen,

phosphorus, sulfur, and the halogens, F, Cl, Br, and I

b) In organic compounds, heteroatoms are defined as atoms of any element other than carbon and hydrogen The elements O, N, P, S, F, Cl, Br, and I listed in part a) are heteroatoms

c) Elements more electronegative than carbon are N, O, F, Cl, and Br Elements less electronegative than carbon

are H and P Sulfur and iodine have the same electronegativity as carbon

d) The more types of atoms that can bond to carbon, the greater the variety of organic compounds that are

possible

15.4 Atomic and bonding properties produce three crucial differences between C and Si Si is larger, forms

weaker bonds, and unlike C, has d orbitals available

15.5 Oxidation states of carbon range from –4 to +4 In carbon dioxide (CO2) carbon has a +4 oxidation state In methane (CH4), carbon has a –4 oxidation state

15.6 Plan: Chemical reactivity occurs when unequal sharing of electrons in a covalent bond results in regions of high

and low electron density

Solution:

The C–H, C–C, and C–I bonds are unreactive because electron density is shared equally between the two atoms

The C=O bond is reactive because oxygen is more electronegative than carbon and the electron rich pi bond is above and below the C–O bond axis, making it very attractive to electron-poor atoms The C–Li bond is also

reactive because the bond polarity results in an electron-rich region around carbon and an electron-poor region around Li

15.7 a) An alkane is an organic compound consisting of carbon and hydrogen in which there are no multiple bonds

between carbons, only single bonds A cycloalkane is an alkane in which the carbon chain is arranged in a ring

An alkene is a hydrocarbon with at least one double bond between two carbons An alkyne is a hydrocarbon with

at least one triple bond between two carbons

b) The general formula for an alkane is CnH2n+2

The general formula for a cycloalkane is CnH2n Elimination of two hydrogen atoms is required to form the additional bond between carbons in the ring

For an alkene, assuming only one double bond, the general formula is CnH2n When a double bond is formed in an alkane, two hydrogen atoms are removed

For an alkyne, assuming only one triple bond, the general formula is CnH2n–2 Forming a triple bond from a double bond causes the loss of two hydrogen atoms

c) For hydrocarbons, “saturated” is defined as a compound that cannot add more hydrogen An unsaturated hydrocarbon contains multiple bonds that react with H2 to form single bonds The alkanes and cycloalkanes are saturated hydrocarbons since they contain only single C–C bonds

15.8 a) Constitutional isomers are those with different sequences of bonded atoms

b) Geometric isomers have different orientation of groups around a double bond or a cyclic structure

c) Optical isomers are a type of stereoisomerism that arises when a molecule and its mirror image cannot be superimposed on each other

Constitutional and geometric isomers are not stereoisomers

15.9 Alkynes are linear about the triple bond Aromatics can only assume one (planar) orientation in space

15.10 Plan: An asymmetric molecule has no plane of symmetry

Solution:

a) A circular clock face numbered 1 to 12 o’clock is asymmetric Imagine that the clock is cut in half, from 12 to

6 or from 9 to 3 The one-half of the clock could never be superimposed on the other half, so the halves are not identical Another way to visualize symmetry is to imagine cutting an object in half and holding the half up to a mirror If the original object is “re-created” in the mirror, then the object has a plane of symmetry

b) A football is symmetric and has two planes of symmetry — one axis along the length and one axis along the fattest part of the football

c) A dime is asymmetric Either cutting it in half or slicing it into two thin diameters results in two pieces that

cannot be superimposed on one another

d) A brick, assuming that it is perfectly shaped, is symmetric and has three planes of symmetry at right angles to

e) A hammer is symmetric and has one plane of symmetry, slicing through the metal head and down through the handle

f) A spring is asymmetric Every coil of the spring is identical to the one before it, so a spring can be cut in half

and the two pieces can be superimposed on one another by sliding (not flipping) the second half over the first However, if the cut spring is held up to a mirror, the resulting image is not the same as the uncut spring

Disassemble a ballpoint pen and cut the spring inside to verify this explanation

15.11 A polarimeter is used to measure the angle that the plane of polarized light is rotated A beam of light consists of

waves moving in all planes A polarizing filter blocks all waves except those in one plane, so the light emerging through the filter is plane-polarized An optical isomer is said to be optically active because it rotates the plane of

the polarized light The dextrorotatory isomer (designated d or +) rotates the plane of light to the right; the

levorotatory isomer (designated l or –) is the mirror image of the first and rotates the plane to the left

15.12 Aromatic hydrocarbons have carbon in sp2 hybridization while cycloalkanes have carbon in sp3 hybridization

Aromatic hydrocarbons are planar while most cycloalkanes assume puckered ring structures

15.13 Plan: To draw the possible skeletons, it is useful to have a systematic approach to make sure no structures are

missed Draw the chain or ring and then draw structures with branches or a double bond at different points along the chain

Solution:

a) Since there are seven C atoms but only a six-carbon chain, there is one C branch off of the chain First, draw the skeleton with the double bond between the first and second carbons and place the branched carbon in all possible positions starting with C #2 Then move the double bond to between the second and third carbon and place the branched carbon in all possible positions Then move the double bond to between the third and fourth carbons and place the branched carbon in all possible positions The double bond does not need to be moved further in the chain since the placement between the second and third carbon is equivalent to placement between the fourth and fifth carbons and placement between the first and second carbons is equivalent to placement between the fifth and sixth carbons The other position to consider for the double bond is between the branched carbon and the six-carbon chain

Double bond between first and second carbons:

C

The total number of unique skeletons is eleven To determine if structures are the same, build a model of one skeleton and see if you can match the structure of the other skeleton by rotating the model and without breaking any bonds If bonds must be broken to make the other skeleton, the structures are not the same

b) The same approach can be used here with placement of the double bond first between C #1 and C #2, then between C #2 and C #3 Since there are seven C atoms but only five C atoms in the chain, there are two C branches

Double bond between first and second carbons:

C

CC

C

C

CC

15.14 To draw the possible skeletons it is useful to have a systematic approach to make sure no structures are missed a)

15.15 Plan: Add hydrogen atoms to make a total of four bonds to each carbon

c) The second carbon in the chain has five bonds, so move the ethyl group from the second carbon to the third

To do this, a hydrogen atom must be removed from the third carbon atom

CH2

CH3d) Structure is correct

d)

CH3 CH2 CH

CH3

CH2 CH2 CH315.19 Plan: The longest chain is named Then we find the lowest branch numbers by counting C atoms from the end

closer to a branch Name each branch (root- + -yl) and put the names alphabetically before the chain name

6

5

4

2 1

4 and 5 Since the goal is to obtain the lowest numbering position for a side group, the correct name is

3,4-dimethylheptane Note that the prefix “di” is used to denote that two methyl side groups are present in this

molecule

d) This molecule is a 4–carbon chain, with two methyl groups (dimethyl) located at the position 2 carbon The

correct name is 2,2-dimethylbutane

15.21 Plan: The longest chain is named Then we find the lowest branch numbers by counting C atoms from the end

closer to a branch Name each branch (root- + -yl) and put the names alphabetically before the chain name

CH2 CH3

6c) 2-methylcyclohexane means a 6 C ring with a methyl group on carbon #2:

CH3

In a ring structure, whichever carbon is bonded to the methyl group is automatically assigned as carbon #1 Since

this is automatic, it is not necessary to specify 1-methyl in the name Correct name is methylcyclohexane

d) 3,3-methyl-4-ethyloctane means an 8 C chain with 2 methyl groups attached to the 3rd carbon and one ethyl group to the 4th carbon

CH3 CH2 C CH CH2 CH2 CH2 CH3

CH3

CH3

CH2 CH3Numbering is good for this structure, but the fact that there are two methyl groups must be indicated by the

prefix di- in addition to listing 3,3 The branch names appear in alphabetical order Correct name is

4-ethyl-3,3-dimethyloctane

15.22 a) 3,3-dimethlybutane should be 2,2-dimethylbutane

3

4 5

CH3

CH3H

Clchiral carbon chiral carbon

15.24 A carbon atom is chiral if it is attached to four different groups The circled atoms below are chiral

Hchiral carbon

HCCC

HH

H

H

HH

Both can exhibit optical activity

15.25 Plan: The longest chain is named Then we find the lowest branch numbers by counting C atoms from the end

closer to a branch Name each branch (root- + -yl) and put the names alphabetically before the chain name An

optically active compound contains at least one chiral center, a carbon with four distinct groups bonded to it Solution:

a) This compound is a six-carbon chain with a Br on the third carbon 3-bromohexane is optically active because carbon #3 has four distinct groups bonded to it: 1) –Br, 2) –H, 3) –CH2CH3, 4) –CH2CH2CH3

Br

chiral carbon

b) This compound is a five-carbon chain with a Cl and a methyl (CH3) group on the third carbon

3-chloro-3-methylpentane is not optically active because no carbon has four distinct groups The third carbon has three distinct groups: 1) –Cl, 2) –CH3, 3) two –CH2CH3 groups

15.27 Plan: Geometric isomers are defined as compounds with the same atom sequence but different arrangements of

the atoms in space The cis-trans geometric isomers occur when rotation is restricted around a bond, as in a

double bond or a ring structure, and when two different groups are bonded to each atom in the restricted bond Solution:

a) Both carbons in the double bond are bonded to two distinct groups, so geometric isomers will occur The double bond occurs at position 2 in a five-carbon chain

b) Cis-trans geometric isomerism occurs about the double bond The ring is named as a side group (cyclohexyl)

occurring at position 1 on the propene main chain

b) No, geometric isomers occur because the right carbon participating in the double bond is attached to two identical methyl (–CH3) groups

c) Cis-trans geometric isomerism occurs about the double bond

CH2 CH2 CH3

CH2 CH2 CH3

15.29 Plan: Geometric isomers are defined as compounds with the same atom sequence but different arrangements of

the atoms in space The cis-trans geometric isomers occur when rotation is restricted around a bond, as in a

double bond or a ring structure, and when two different groups are bonded to each atom in the restricted bond Solution:

a) The structure of propene is CH2=CH–CH3 The first carbon that is involved in the double bond is bonded to two of the same type of group, hydrogen Geometric isomers will not occur in this case

b) The structure of 3-hexene is CH3CH2CH=CHCH2CH3 Both carbons in the double bond are bonded to two distinct groups, so geometric isomers will occur

15.30 a) The structure of 1-pentene is CH2=CH–CH2–CH2–CH3 The first carbon that is involved in the double bond is

bonded to two of the same type of group, hydrogen Geometric isomers will not occur in this case

d) There are no geometric isomers because the first carbon has two hydrogens attached to it

15.31 Plan: Benzene is a planar, aromatic hydrocarbon It is commonly depicted as a hexagon with a circle in the middle

to indicate that the  bonds are delocalized around the ring and that all ring bonds are identical With two groups

attached to the ring, number the C atoms so that a group is attached to ring C-1 Alternatively, the ortho (o-), meta (m-), and para (p-) naming system is used to denote the location of attached groups in benzene compounds

only, not other ring structures like the cycloalkanes