OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
Chapter 6 – An Overview of Organic Reactions
Solutions to Problems
6.1 (a) CH3Br + KOH ⎯⎯→ CH3OH + KBr substitution
(b) CH3CH2Br ⎯⎯→ H2C = CH2 + HBr elimination
(c) H2C = CH2 + H2 ⎯⎯→ CH3CH3 addition
6.2 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)
6.3 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.
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6.4 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.
6.5 The mechanism is pictured in Figure 6.4. 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.
6.6 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.
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(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.
6.7 This mechanism will be studied in a later chapter.
6.8
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Chapter 6 – An Overview of Organic Reactions
Solutions to Problems
6.1 (a) CH3Br + KOH ⎯⎯→ CH3OH + KBr substitution
(b) CH3CH2Br ⎯⎯→ H2C = CH2 + HBr elimination
(c) H2C = CH2 + H2 ⎯⎯→ CH3CH3 addition
6.2 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)
6.3 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.
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, OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
6.4 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.
6.5 The mechanism is pictured in Figure 6.4. 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.
6.6 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.
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(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.
6.7 This mechanism will be studied in a later chapter.
6.8
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