Chapter 4 Solutions to Problems – An Overview of Organic Reactions

Kirsten Kramer; Cassandra Lilly

4 Chapter 4 Solutions to Problems – An Overview of Organic Reactions

Chapter 4 – An Overview of Organic Reactions

Solutions to Problems

4.1	(1)	Find the longest chain containing the double bond and name it. In (a), the longest chain is a pentene.
	(2)	Identify the substituents. There are three methyl groups in (a).
	(3)	Number the substituents, remembering that the double bond receives the lowest possible number. The methyl groups are attached to C3 and C4 (two methyl groups).
	(4)	Name the compound, remembering to use the prefix “tri-” before “methyl” and remembering to use a number to signify the location of the double bond. The name of the compound in (a) is 3,4,4-trimethyl-1-pentene.
	(a)
	(b)
	(c)
	(d)

4.2	It is much easier to draw a structure from a given name than it is to name a structure. First, draw the carbon chain, placing the double bond or bonds in the designated locations. Then attach the cited groups in the proper positions.
	(a)
	(b)
	(c)
	(d)

4.3	(a)
	(b)
	(c)

4.4	In the new naming system, the bond locant appears directly before –ene or –diene.
	(a)
	(b)

4.5	The rules for naming alkynes are almost the same as the rules for naming alkenes. The suffix –yne is used, and compounds containing both double bonds and triple bonds are –enynes, with the double bond taking numerical precedence.
	(a)
	(b)
	(c)
	(d)
	(e)

4.6

4.7	Compounds (c), (e), and (f) can exist as cis–trans isomers.
	(c)
	(e)
	(f)

4.8	(a)
	(b)

4.9

Review the sequence rules presented in Section 4.4. A summary:

Rule 1:

An atom with a higher atomic number has priority over an atom with a lower atomic number.

Rule 2:

If a decision can’t be reached by using Rule 1, look at the second, third, or fourth atom away from the double-bond carbon until a decision can be made.

Rule 3:

Multiple-bonded atoms are equivalent to the same number of single-bonded atoms.

	Higher	Lower	Rule
(a)	–CH₃	–H	1
(b)	–Cl	–CH₂Cl	1
(c)	–CH=CH₂	–CH₂CH₂Br	3
(d)	–OCH₃	–NHCH₃	1
(e)	–CH=O	–CH₂OH	3
(f)	–CH=O	–CH₂OCH₃	3

4.10	Highest priority ⎯⎯⎯→ Lowest Priority
	(a)	–Cl, –OH, –CH₃, –H
	(b)	–CH₂OH, –CH=CH₂, –CH₂CH₃, –CH₃
	(c)	–CO₂H, –CH₂OH, –C≡N, –CH₂NH₂
	(d)	–CH₂OCH₃, –C≡N, –C≡CH, –CH₂CH₃

4.11	(a)	First, consider the substituents on the right side of the double bond. –Cl ranks higher than –CH₂OH by Rule 1 of the Cahn–Ingold–Prelog rules. On the left side of the double bond, –CH₂CH₃ ranks higher than –CH₃. The isomer is Z when the two higher priority groups lie on the same side of the double bond. Otherwise, the isomer is E.
	(b)
	(c)	Notice that the upper substituent on the left side of the double bond is of higher priority because of the methyl group attached to the ring.
	(d)

4.12

4.13

(a) CH₃Br + KOH → CH₃OH + KBr substitution

(b) CH₃CH₂Br → H₂C = CH₂ + HBr elimination

(c) H₂C = CH₂ + H₂ → CH₃CH₃ addition

4.14

Keep in mind:

(1) An electrophile is electron-poor, either because it is positively charged, because it has a functional group that is positively polarized, or because it has a vacant orbital.

(2) A nucleophile is electron-rich, either because it has a negative charge, because it has a functional group containing a lone electron pair, or because it has a functional group that is negatively polarized.

(3) Some molecules can act as both nucleophiles and electrophiles, depending on the reaction conditions.

(a) The electron-poor carbon acts as an electrophile.

(b) CH₃S^– is a nucleophile because of the sulfur lone-pair electrons and because it is negatively charged.

(c) C₄H₆N₂ is a nucleophile because of the lone-pair electrons of nitrogen. (Only one of the nitrogens is nucleophilic, for reasons that will be explained in a later chapter.)

(d) CH₃CHO is both a nucleophile and an electrophile because of its polar C=O bond.

(a)

(b)

(c)

(d)

4.15

BF₃ is likely to be an electrophile because the electrostatic potential map indicates that it is electron-poor (blue). The electron-dot structure shows that BF₃ lacks a complete electron octet and can accept an electron pair from a nucleophile.

4.16

Reaction of cyclohexene with HCl or HBr is an electrophilic addition reaction in which halogen acid adds to a double bond to produce a haloalkane.

4.17

The mechanism is pictured in Figure 4.7.

The steps:

(1) Attack of the π electrons of the double bond on HBr, forming a carbocation

(2) Formation of a C–Br bond by electron pair donation from Br^– to form the neutral addition product

4.18

A reaction with ΔG^‡ = 45 kJ/mol is faster than a reaction with ΔG^‡ = 70 kJ/mol because a larger value for ΔG^‡ indicates a slower reaction.

4.19

4.20

For curved arrow problems, follow these steps:

(1) Locate the bonding changes. In (a), a bond from nitrogen to chlorine has formed, and a Cl-Cl bond has broken.

(2) Identify the nucleophile and electrophile (in (a), the nucleophile is ammonia and the electrophile is one Cl in the Cl₂ molecule), and draw a curved arrow whose tail is near the nucleophile and whose head is near the electrophile.

(3) Check to see that all bonding changes are accounted for. In (a), we must draw a second arrow to show the unsymmetrical bond-breaking of Cl₂ to form Cl^–.

(a)

(b)

A bond has formed between oxygen and the carbon of bromomethane. The bond between carbon and bromine has broken. CH₃O^– is the nucleophile and bromomethane is the electrophile.

(c)

A double bond has formed between oxygen and carbon, and a carbon–chlorine bond has broken. Electrons move from oxygen to form the double bond and from carbon to chlorine.

4.21

This mechanism will be studied in a later chapter.

Additional Problems

Visualizing Chemistry

4.22	(a)
	(b)

4.23	(a)
	(b)

4.24

4.25

4.26	(a)	The electrostatic potential map shows that the formaldehyde oxygen is electron-rich, and the carbon–oxygen bond is polarized. The carbon atom is thus relatively electron- poor and is likely to be electrophilic.
	(b)	The sulfur atom is more electron-rich than the other atoms of methanethiol and is likely to be nucleophilic.

4.27

(a) ΔG° is positive.

(b) There are two steps in the reaction.

(c) There are two transition states, as indicated on the diagram.

4.28

(a) The reaction involves four steps, noted above.

(b) Step 1 is the most exergonic because the energy difference between reactant and product (ΔG°) is greatest.

(c) Step 2 is slowest because it has the largest value of ΔG‡.

Naming Alkenes

4.29	(a)
	(b)
	(c)
	(d)
	(e)
	(f)

4.30	(a)
	(b)
	(c)
	(d)
	(e)
	(f)

4.31	(a)
	(b)
	(c)
	(d)
	(e)
	(f)

4.32

Because the longest carbon chain contains 8 carbons and 3 double bonds, ocimene is an octatriene. Start numbering at the end that will give the lower number to the first double bond (1,3,6 is lower than 2,5,7). Number the methyl substituents and, finally, name the compound.

4.33

(3E,6E)-3,7,11-Trimethyl-1,3,6,10-dodecatetraene

4.34

4.35

Naming Alkynes

4.36	(a)
	(b)
	(c)
	(d)
	(e)
	(f)

4.37	(a)
	(b)
	(c)
	(d)
	(e)
	(f)
	(g)
	(h)

4.38

(a)

CH₃CH=CHC≡CC≡CCH=CHCH=CHCH=CH₂.

1,3,5,11-Tridecatetraen-7,9-diyne

Using E–Z notation: (3E,5E,11E)-1,3,5,11-Tridecatetraen-7,9-diyne

The parent alkane of this hydrocarbon is tridecane.

(b)

CH₃C≡CC≡CC≡CC≡CC≡CCH=CH₂. 1-Tridecen-3,5,7,9,11-pentayne This hydrocarbon also belongs to the tridecane family.

Alkene Isomers and Their Stability

4.39	Highest priority ⎯⎯⎯→ Lowest Priority
	(a)	–I, –Br, –CH₃, –H
	(b)	–OCH₃, –OH, –CO₂H, –H
	(c)	–CO₂CH₃, –CO₂H, –CH₂OH, –CH₃
	(d)	–COCH₃, –CH₂CH₂OH, –CH₂CH₃, –CH₃
	(e)	–CH₂Br, –C≡N, –CH₂NH₂, –CH=CH₂
	(f)	–CH₂OCH₃, –CH₂OH, –CH=CH₂, CH₂CH₃

4.40	(a)
	(b)
	(c)
	(d)

4.41	(a)
	(b)
	(c)
	(d)
	(e)
	(f)

4.42

Energy Diagrams and Reaction Mechanisms

4.43

A transition state represents a structure occurring at an energy maximum. An intermediate occurs at an energy minimum between two transition states. Even though an intermediate may be of such high energy that it cannot be isolated, it is still of lower energy than the transition states surrounding it.

4.44

(a) Mechanism:

(b) Mechanism:

(c) Mechanism:

Polar Reactions

4.45

(a)

(b)

(c)

(d)

(e)

(f)

4.46	(a)	The reaction between bromoethane and sodium cyanide is a substitution because two reagents exchange parts.
	(b)	This reaction is an elimination because two products (cyclohexene and H₂O) are produced from one reactant
	(c)	Two reactants form one product in this addition reaction.
	(d)	This is a substitution reaction.