chapter 6comsizo.com.br/resolucoes/organicchemistrychapter6.pdfchapter 6 6.1 calculate the degree of...

Chapter 6 6.1 Calculate the degree of the unsaturation in the following hydrocarbons:

(a) C8H14; (b) C5H6 (c) C12H20 (d) C20H32 (e) C40H56

Solution: (a)△=2 (b)△=3 (c)△=3 (d)△=5 (e)△=13

6.2 Calculate the degree of the unsaturation in the following formula, and the draw as many structures

as you can for each: (a) C4H5; (b) C4H6; (c) C3H4;

Solution: (a)△=1

H2C CH

H2C CH3

H3C CH

CH

CH3

H2C C CH3

CH3 (b)△=2

H2C CH

CH

CH2

HC CH2C CH3

H3C C C CH3

H2C C CH

CH3

(c)△=2

HC C CH3 H2C C CH2

6.3 Calculate the degree of unsaturation in the following formulas: (1) C6H5N 5 (2) C6H5NO2 5 (3) C8H9Cl3 3 (4) C9H10Br2 1 (5) C10H12N2O3 6 (6) C20H32ClN 5 6.4 Give the IUPAC name for the following compounds: (1)

3,4,4-trimethyl-1-pentene

(2)

3-methyl-3-hexene

(3)

4,7-dimethyl-2,5-octadiene

6.5 Draw structures corresponding to the following IUPAC names: (a)2-Methyl-1,5-hexadiene (b)3-Ethyl-2,2-dimethyl-3-heptene (c)2,3,3-Trimethyl-1,4,6-octatriene (d)3,4-diisopropyl-2,5-dimethyl-3-hexene (e)4-tert-Butyl-2-methylheptane Solution:

(a)

H2C CH2C

H2C C

HCH2

CH3 (b)

H3C C C CH

H2C CH3

CH3

CH3

CH2

CH3

(c)

H2C C C CH

CH

CH

CH

CH3

CH3

CH3

CH3 (d)

H3CHC C C C

HCH3

CH3

CH3

CH

H3C CH3

CH

H3C CH3

(e)

H3C CH

H2C

HC

H2C CH3

CH3

CH3C CH3

CH3

6.6 Name the following cycloalkenes:

(a)

CH3

CH3 (b)

CH3

CH3

(c)

CH(CH3)2

Solution: (a) 1,2-Dimethyl-cyclohexene (b) 4,4-Dimethyl-cycloheptene (c) 3-Isopropyl-cyclopentene 6.7 Which of the following compounds can exist as pairs of cis-trans isomers? Draw each cis-trans pair,

and indicate the geometry of each isomer. (a) H3CHC CH2 (b) (H3C)2C CHCH3 (c) H3CH2CHC CHCH3 (d) (H3C)2C C(CH3)CH2CH3 (e) ClHC CHCl (f) BrHC CHCl Solution: (c)

cis:

C C

H3CH2C

H H

CH3

trans:

C C

H3CH2C

H CH3

H

(e)

cis:

C C

Cl

H H

Cl

trans:

C C

Cl

H Cl

H

(f)

cis:

C C

Br

H H

Cl

trans:

C C

Br

H Cl

H

6.8 Name the following alkenes, including the cis or trans designation:

(a)

C C

CH

H H

CH3

CH3

CHH3C

H3C

(b)

C C

H

CH2 H

CH2CH3

HCH3C

H3C

Solution: (a) cis-4,5-Dimethyl-2-hexene (b) trans-6-Methyl-3-heptene 6.9 Which member in each of the following sets has higher priority? (a) –H or –Br (b) –Cl or –Br (c) –CH3 or –CH2CH3 (d) –NH2 or –OH (e) –CH2OH or –CH3 (f) –CH2OH or –CH=O Solution: (a) –Br (b) –Br (c) –CH2CH3

(d) –OH (e) –CH2OH (f) –CH=O 6.10 Rank the following sets of substituents in order of Cahn-Ingold-Prelog priorities. (a) –CH3, -OH, -H, –Cl (b) –CH3, –CH2CH3, -CH=CH2 (c) –CO2H, -CH2OH, -C≡N, -CH2NH2 (d) –CH2CH3, -C≡CH, -C≡N, -CH2OCH3

Solution: (From High to low) (a) –Cl, -OH, –CH3, -H (b) -CH=CH2, –CH2CH3, –CH3 (c) –CO2H, -CH2OH, -C≡N, -CH2NH2 (d) -CH2OCH3, -C≡N, -C≡CH, –CH2CH3

6.11 Assign E or Z configuration to the following alkenes: Solution:

C C

Cl

CH2OHH3C

H3CH2C

C C

C C

C C

Cl

H3CO CH2CH2CH3

CH2CH3

CN

CH2NH2H3C

H

CO2H

CH2OH

H3C

Z E

Z E 6.12 Assign stereochemistry to the following alkene, and convert the drawing into a skeletal structure: Solution:

CH2OH

COOCH3

Z

6.13 Name the following alkenes, and tell which compounds in each of the following pairs are more

stable: (a)

H2C CHCH2CH3H2C CCH3

CH3or

1-butene 2-methyl-1-propene 2-methyl-1-propene is more stable because it is more substituents. (b)

orC C

H

H3C

H

CH2CH2CH3 H3C

CH2CH2CH3

H

H

cis-2-hexene trans-2-hexene trans-2-hexene is more stable because it is trans isomer. (c)

orC C

CH3 CH3

1-methyl-1-cyclohexene 3-methyl-1-cyclohexene

1-methyl-1-cyclohexene is more stable because the carbon has only sp2-sp2 bonds while the other has sp2-sp3 bonds.

6.14 Predict the products of the following reactions:

(a)

+ HCl

Cl

(b)

C CH

H3C

H3C H2C CH3

HBr

BrCH3

(c) CH3CH2CH2CH CH2

H3PO4KI

I

(d)

CH2

+ HBr

CH3

Br

6.15 What alkenes would you start with to prepare the following alkyl halides? (a) Bromocyclopentane (b) 1-Ethyl-1-iodocyclohexane (c)

CH3CH2CHCH2CH2CH3

Br

Solution: (a) Cyclopentene (b) 1-Ethylcyclohexene

(c) H3CH2CHC CHCH2CH3 6.16 Show the structures of carboations you would expect in the following reactions: (a)

H3CH2CC CHCHCH3

CH3CH3

+ ?BrH

(b)

CHCH3 + HI ?

Solution: (a)

H3CH2CC CH2CH2CH3

CH3 CH3

(b)

CH2CH3 +

6.17 Draw a skeletal structure of the following carbocation. Identify it as primary, secondary, or tertiary, and identify the hydrogen atoms that are involved in hyperconjugation in the conformation shown.

Solution:

C C

CH3

H

CH3H

CH3

It is secondary carbocation. Only the methyl-group C—H that is parallel to the carbocation p orbital can show hyperconjugation. 6.18 What about the second step in the electrophilic addition of HCl to an alkene—the reaction of

chloride ion with the carbocation intermediate? Is this step exergonic or endergonic? Does the transition state for this second step resemble the reactant (carbocation) or product (alkyl chloride)? Make a rough drawing of what the transition-state structure might look like.

Solution: This step is exergonic. The transition state for this second step resemble carbocation.

C C

H

R

RR

R

Br

Intermediate

transition state

Product

6.19 On treatment with HBr, vinylcyclohexane undergoes addition and rearrangement to yield 1-bromo-1-ethylcyclohexane. Using curved arrows, propose a mechanism to account for this result.

Solution:

vinylcyclohexane

+ H Br C

CH3

+ Br

H

C

CH3

+ Br

CH2CH3

Br 6.20 Name the following alkenes, and convert each drawing into a skeletal structure.

(a) 2,4,5-trimethyl-2-hexene

(b)

1-ethyl-3,3-dimethyl-1-cyclohexene 6.21. Assign stereochemistry (E or Z) to each of the following alkenes, and convert each drawing into a

skeletal structure:

(a) (b)

SOLUTION:

Cl

O

E(a) (b)

O

O

HO

Z

6.22. The following carbocation is an intermediate in the electrophilic addition reaction of HCl with

two different alkenes. Identify both, and tell which C H bonds in the carbocation are aligned for maximum hyperconjugation with the vacant p orbital on the positively charged carbon.

SOLUTION:

1

2

3

6.23 Calculate the degree of unsaturation in the following formulas, and draw five possible for each: (a) C10H16 (b) C8H8O (c) C7H10Cl2

(d) C10H16O2 (e) C5H9NO2 (f) C8H10ClNO Solution” a) It has 3 unsaturation degrees.

Possible structures:

b) It has 5 unsaturation degrees. Possible structures:

O

OOH

OH OH

c) It has 2 unsaturation degrees.


Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl Cl

d) It has 3 unsaturation degrees.


OH

OH

OH

OH

OH

OH

OHHO OH

HO

e) it has 2 unsaturation degrees.


N

H

N

H

OH

OH

OH

OHN

H

OH

OH

N

H

OH

OH N

H

OHHO

f) it has 4 unsaturation degrees.


N

H

OH

N

H

OH

N

H

OH

N

H

OH

N

H

OH

6.24 A compound of formula C10H14 undergoes catalytic hydrogenation but absorbs only 2 molar

equivalents of hydrogen. How many rings does the compound have? Solution: It has 4 unsaturation degrees. Two of them are rings. 6.25 A compound of formula C12H13N contains two rings. How many molar equivalents of hydrogen

does it absorb if all the remaining unsaturations are double bonds? Solution: C12H13N=C12H12

Degree of Unsaturation: 72

)122212(=

−+×

It can absorb 7-2=5 molar equivalents of hydrogen. 6.26 Name the following alkenes:

(a)

C C

H3C

H CHCH2CH3

H

H3C

(b)

CH3CHCH2CH2CH

CH3 CH2CH3

C C

H H

CH3

(c) H2C CCH2CH3

CH2CH3

(d)

H2C CHCHCH

CH3

CH3

C C

H CH3

H

(e)

C C

H3CH2CH2C

H3C C

CH3

C

H H

CH3

(f) H2C C CHCH3 Solution: (a).trans-4-Methyl-2-hexene or (E)-4-Methyl-2-hexene (b).cis-4-Ethyl-7-methyl-2-octene or (Z)-4-Ethyl-7-methyl-2-octene (c).2-Ethyl-1-butene (d).trans-3,4-Dimethyl-1,5-heptadiene or (E)-3,4-Dimethyl-1,5-heptadiene (e).(2-cis,4-trans)-4,5-Dimethyl-2,4-octadiene or (2Z,4E)-4,5-Dimethyl-2,4-octadiene (f).1,2-Butadiene 6.27 Ocimene is a triene found in the essential oils of many plants. What is its IUPAC name, including

stereochemistry? Solution:

(3E)3,7-dimethyl-octa-1,3,6-triene

6.28 α-Farnesene is a constituent of the natural wax found on apples. What is its IUPAC name, including stereochemistry?

Solution:

α-Farnesene

(3E,6E)3,7,11-trimethyl-dodeca-1,3,6,10-tetraene

6.29 Draw structures corresponding to the following systematic names:

(a) (4E)-2,4-Dimethyl-1,4-hexadiene

Solution:

C C

H

H3C

H2C

CH3

C C

H3C

H

H

(b) cis-3,3-Dimethyl-4-propyl-1,5-octadiene

Solution:

C C

C2H5

H H

CH

C

C C

H

HHC3H7

H3C CH3

(c) 4-Methyl-1,2-pentadiene

Solution:

C CH

HCH

CHCH3

CH3

(d) (3E,5Z)-2,6-Dimethyl-1,3,5,7-octatetraene.

Solution:

C C

H

H

C

CH3

C

H C

H

C

H CH3

C C

H

H

H

(e) 3-Butyl-2-heptene

Solution: H3C

H C4H9

C4H9 (f) trans-2,2,5,5-Tetramethyl-3-hexene

Solution:

C C

H

CH3C

CCH3

H

H3C CH3

H3C CH3

6.30 Menthene, a hydrocarbon found in mint plants, has the systematic name

1-isopropyl-4-methylcyclohexhene. Draw its structure.

Solution:

6.31 Draw and name the 6 pentene isomers, C5H10, including E, Z isomers.

1-Pentene (Z)-2-Pentene (E)-2-Pentene

2-Methyl-1-butene 2-Methyl-2-butene 3-Methyl-1-butene

6.32 Draw and name the 17 hexene isomers, C6H12, including E, Z isomers.

2,3-Dimethyl-1-butene 2,3-Dimethyl-2-butene

(E)-3-Methyl-2-pantene (Z)-3-Methyl-2-pantene 3,3-Dimethyl-1-butene

4-Methyl-1-pantene 3-Methyl-1-pantene 2-Ethyl-1-butene

1-Hexene (E)-2-Hexene (Z)-2-Hexene

(E)-3-Hexene (Z)-3-Hexene 2-Methyl-1-pantene

2-Methyl-2-pantene (E)-4-Methyl-2-pantene (Z)-4-Methyl-2-pantene

6.33 trans-2-Butene is more stable than cis-2-butene by only 4 KJ/mol, but trans-2, 2, 5,

5-tetramethyl-3-hexene is more stable than cis-2, 2, 5, 5-tetrameethyl-3-hexene by 39 KJ/mol. Explain.

Alkene ΔΗohydrog

(KJ/mol) (kcal/mol) cis-2-butene -119.7 -28.6 trans-2-Butene -115.5 -27.6 cis-2, 2, 5, 5-tetrameethyl-3-hexene -151.5 -36.2 trans-2,2,5,5-tetrameethyl-3-hexene -112.6 -26.9 Solution: Between cis-2-butene and trans-2-Butene, the cis-2-butene has steric strain, so the

trans-2-Butene is more stable. But the steric strain is not large, the difference between them is not very large. And the cis-2, 2, 5, 5-tetrameethyl-3-hexene has large steric strain, so the

difference between cis-2, 2, 5, 5-tetrameethyl-3-hexene and trans-2, 2, 5, 5-tetrameethyl-3-hexene is larger.

6.34 Cyclodecene can exist in both cis and trans forms, but cyclohexene cannot. Explain.(Making

molecular models is helpful.) Solution: the cyclodecene can be

But the cyclohexene can only be:

6.35 Normally, a trans alkene is more stable than its cis isomer. trans-Cyclooctene, however, is less stable than cis-cyclooctene by 38.5 kJ/mol. Explain.

Build models of the two cyclooctenes and notice that the large amount of torsional strain in trns-cyclooctene relative to cis-cyclooctene. This strain causes the trans isomer ot be of higher energy.

6.36 Trans-Cyclooctene is less stable than cis-cyclooctene by 38.5 kJ/mol, but trans-cyclononene is less stable than cis-cyclononene by only 12.2 kJ/mol. Explain.

trans-Cyclononene has more carbon than trans-Cyclooctene, and the ring is bigger, so the steric strain is smaller.

6.37 Allene(1.2-propadiene), C CH2H2C ,has two adjacent double bonds. What kind of hybridization must the central carbon have? Sketch the bonding π orbitals in allene. What shape do you predict for allene.

Solution: The central carbon is sp hybridized. The bonding πorbitals in allene is showed as follow:

C CC

H

HHH

C C C

H

H

HH

The structure of allene is showed as follow:

C CHH

C

H

H

6.38 The heat of hydrogenation for allene(Problem 6.37) to yield propaneis -298 kJ/mol, and the heat

of hydrogenation for a typical monosubsituted alkene such as propene is -126kJ/mol. Is allene more or less stable then you might except for diene? Explain.

Solution: Less stable. Because allene give out more heat than diene in hydrogenation which means that it contains more energy. 6.39 Predict the major product in each of the following reactions: (a)

CH3CH2CH CCH2CH3

CH3

+ HCl CH3CH2CH2 CCH2CH3

CH3

Cl (b)

+Br

HBr

1-Ethylcyclopentene (c)

2,2,4-Trimethyl-3-hexene

+ HI CH3CCH2CCH2CH3

CH3

CH3

CH3

I

(d)

1,6-Heptadiene+ 2HCl

Cl Cl

(e)

CH3

+ HBr

CH3

Br

6.40 Predict the major product from addition of HBr to each of following alkenes: (a)

+ HBr

CH2 CH3

Br

(b)

+ HBr

Br

(c)

+ HBrH3CHC CHCHCH3

CH3

Br 6.41 Rank the following sets of substituents in order of priority according to the Cahn-Ingold-Prelog

sequence rules: (a) CH3, Br, H, I Solution: 1. I, 2. Br, 3. CH3, 4. H. (b) OH, OCH3, H, COOH Solution: 1. OCH3, 2. OH, 3. COOH, 4. H.

(c) CO2H, CO2CH3, CH2OH, CH3 Solution: 1. CO2CH3, 2. CO2H 3. CH2OH, 4. CH3.

(d) CH3, CH2CH3, CH2CH2OH, CCH3

O

Solution: 1. CCH3

O

, 2. CH2CH2OH, 3. CH2CH3, 4. CH3.

(e) CH

CH2, CN, CH2NH2, CH2Br

Solution: 1. CH2Br, 2. CN, 3. CH2NH2, 4. CH

CH2.

(f) CH

CH2, CH2CH3, CH2OCH3, CH2OH

Solution: 1. CH2OCH3 , 2. CH2OH , 3. CH

CH2 , 4.

CH2CH3. 6.42 Assign E or Z configuration to each of the following alkenes:

(a)

C C

H3C

HOH2C CH3

H

Z

(b)

C C

Cl

HOOC H

OCH3

Z

(c)

C C

NC

H3CH2C CH2OH

CH3

E

(d)

C C

HO2C

H3CO2C HC

CH2CH3

CH2Z

6.43 Name the following cycloalkenes:

（a）

CH3

(b) (c)

(d) (e) (f)

Solution: (a) 3-Methyl-1-cyclohexene (b) 2,3-Dimethyl-1-cyclopentene (c) 1-Ethyl-1,3-cyclobutadiene (d) 1,2-Dimethyl-1,4-cyclohexadiene (e) 5-Methyl-1,3-cyclohexadiene

(g) 1,5- cyclooctadiene

6.44 Which of the following E,Z designations are correct, and which are incorrect?

(a)

C

H3C

CCO2H

HZ (b)

C CH

H3C

CH2CH

CH2CH(CH3)2

CH2

E

(c)

C CBr

H

CH2NH2

CH2NHCH3Z (d)

C CNC

(H3C)2NH2C

CH3

CH2CH3E

(e) CC

Br

H Z (f)

C CCO2H

COCH3

HOH2C

H3COH2CE

Solution: (a) correct (b) Correct (c) incorrect, it should be E (d) correct (e) Doesn’t show E / Z isomerism. (f) correct

6.45 tert-Butyl esters [RCO2C(CH3)3] are converted into carboxylic acids (RCO2H) by reaction with

trifluoroacetic acid, a reaction useful in protein synthesis (see Section 26.10). Assign E,Z designation to the double bonds of both reactant and product in the following scheme, and explain why there is an apparent change of double-bond stereochemistry:

C C

H

H3C

C OCH3

O

C OC(CH3)3O

CF3CO2HC C

H

H3C

C OCH3

O

C OHO

Solution: The reactant is Z designation, while the product is E designation. Because C OC(CH3)3

O

receives higher priority than C OCH3

O

, while C OCH3

O

receives higher priority than

C OH

O

, and in the reaction C OH

O

substitute for C OC(CH3)3

O

.

6.46 Use the bond dissociation energies in Table 5.4 to calculate ΔH0 for the reaction of ethylene with

HCl, HBr, and HI. Which reaction is most favorable?

H2C CH2 + H Cl

235 KJ / mol 432 KJ / mol

HH2C

H2C Cl


H o = -91 KJ / mol

H2C CH2 + H Br


HH2C

H2C Br


H o = -104 KJ / mol

H2C CH2 + H I


HH2C

H2C I


H o = -109 KJ / mol

6.47: Each of the following carboncations can rearrange to a more stable ion. Propose structure for the

likely rearranges for the likely rearrangement product.

(a) (b) (c)

Solution: (a) (b) (c)

6.48 Addition of HCl to 1-isopropyl-cyclohexane yields a rearranged product. Propose a mechanism, showing the structures of the intermediates and using curved arrows to indicate electron flow in each step.

Cl

ClH+

Solution:

(E)

Cl

ClH

H

Cl

Cl

++

+

6.49 Addition of HCl to 1-isopropenyl-1-methylcyclopentane yields 1-chloro-1, 2, 2-trimethylcyclohexane. Propose a mechanism, showing the structures of the intermediates and using curved arrows to indicate electron flow in each step.

CH3

+ H Cl

Cl

CH3

CH3

CH3

Solution:

CH3

+ H Cl

CH3

Cl

CH3

CH3

CH3

CH3

CH3

CH3

Cl

6.50 Vinylcyclopropane reacts with HBr to yield a rearranged alkyl bromide. Follow the flow of

electrons as represented by the curved arrows, show the structure of the carbocation intermediate in brackets, and show the structure of the final product.

H Br

? ?Br

Solution:

H Br

BrBr

6.51: Calculate the degree of unsaturation in each of the following formulas. Solution:

(a) C27H46O D=5 (b) C14H9Cl5 D=8 (c) C20H34O5 D=4

(d) C8H10N4O2 D=6 (e) C21H28O5 D=8 (f) C17H23 NO3 D=7

6.52: Is the rearrangement exergonic or endergonic? Draw the transition state. Solution: It is a exergonic. State

H3C CH2

H3CH

6.53 Draw a reaction energy diagram for the addition of HBr to 1-pentene. Let one curve on your diagram show the formation of 1-bromopentane product and another curve on the same diagram show the formation of 2-bromopentane product. Label the postions for all reactants, intermediates, and products. Which curve has higher-energy carbocation intermediate? Which curve has the higher-energy first transition state?

Solution:

Br

H2C

CH2

H2C

CH2

CH3

The vertical axial presents energy, and the horizontal one shows the progress of the reaction. The red one stands for 2-bromopentane, and the blue one stands for 1-bromopentane. The blue one has higher-energy cabocation intermediate and the higher-energy first transition state. 6.54 Make sketches of the transition state structures involved in the reaction of HBr with 1-pentene

(Problem 6.53). Tell whether each structure resembles reactant or product. Solution: The first one is for forming 1-bromopentane, and the second one is for forming 2-bromopentane.

H3C CH

Br

H2C

H2C CH3

Each structure resembles carbocation. 6.55 Aromatic compounds such as benzene react with alkyl chlorides in the presence of AlCl3 catalyst

to yield alkylbenzenes. The reaction occurs through a carbocation intermediate, formed by reaction of the alkyl chloride with AlCl3 (R－Cl+ AlCl3 R+＋AlCl4

-). How can you explain the observation that reaction of benzene with 1-chloropropane yields isopropylbenzene as the major product?

H3CH2CH2C ClAlCl3

Solution:

C CH

H

δH

R

H BrBr

C CH

R

δH

H

H BrBr

δδ

δδ

H3CH2CH2C Cl AlCl3 H C

H

C

H

H

CH3

H3C CH CH3

AlCl4

H

AlCl4

HCl AlCl3

6.56 Alkenes can be converted into alcohols by acid-catalyzed addition of water. Assuming that Markonikov’s rule is valid, predict the major product from each of the following alkenes.

(a) H3CH2CC CH2

CH3

(b)

CH2

(c) H3C C

H

CH3

CH2CH CH2

Solution:

(a)

H3CH2CC CH2

CH3

H2O C CH3

OH

CH3

H3CH2C

(b)

CH2

H2O OH

(c) H3C C

H

CH3

CH2CH CH2H2O

HC CH3

OH

(H3C)2H2C

6.57 Reaction of 2,3-dimethyl-1-butene with HBr leads to an alkyl bromide C6H13Br. On treatment of this alkyl bromide with KOH in methanol, elimination of HBr occurs and a hydrocarbon that is isomeric with the starting alkene is formed. What is the structure of this hydrocarbon, and how do

you think is formed from the alkyl bormide? Solution:

H2C C CH

CH3

CH3 CH3

Br H

H3C C CH

CH3

CH3 CH3

Br

H3C C C CH3

CH3 CH3

Br H

OH

H3C C C CH3

CH3 CH3

chapter 6comsizo.com.br/resolucoes/organicchemistrychapter6.pdfchapter 6 6.1 calculate the degree of...

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