Chapter 6 Solutions to Problems – Organohalides

Kirsten Kramer; Cassandra Lilly

6 Chapter 6 Solutions to Problems – Organohalides

Chapter 6 – Organohalides

Solutions to Problems

6.1

The rules that were given for naming alkanes in Section 1.15 are used for alkyl halides. A halogen is treated the same as an alkyl substituent but is named as a halo group.

(a)

(b)

(c)

(d)

(e)

(f)

6.2

(a)

(b)

(c)

(d)

(e)

(f)

6.3

Remember that halogen acids are used for converting tertiary alcohols to alkyl halides. PBr₃ and SOCl₂ are used for converting secondary and primary alcohols to alkyl halides. (CH₃CH₂)₂NSF₃ and HF in pyridine can be used to form alkyl fluorides.

(a)

(b)

(c)

(d)

6.4

Table 5.2 shows that the pK_a of CH₃–H is 60. Since CH₄ is a very weak acid, ^–:CH₃ is a very strong base. Alkyl Grignard reagents are similar in base strength to ^–:CH₃, but alkynyl Grignard reagents are somewhat weaker bases. Both reactions (a) and (b) occur as written.

(a)

(b)

6.5

(a)

The oxidation level of a compound can be found by adding the number of C–O, C–N, and C–X bonds and subtracting the number of C–H bonds. Cyclohexane, the first compound shown, has 12 C–H bonds, and has an oxidation level of –12. Cyclohexanone has 2 C–O bonds (from the double bond) and 10 C–H bonds, for an oxidation level of –8. 1 – Chlorocyclohexene has one C–Cl bond and 9 C–H bonds, and also has an oxidation level of –8. Benzene has 6 C–H bonds, for an oxidation level of –6.

In order of increasing oxidation level:

(b)

6.6

(a)	The aldehyde carbon of the reactant has an oxidation level of 1 (2 C–O bonds minus 1 C–H bond). The alcohol carbon of the product has an oxidation level of –1 (1 C–O bond minus 2 C–H bonds). The reaction is a reduction because the oxidation level of the product is lower than the oxidation level of the reactant.
(b)	The oxidation level of the upper carbon of the double bond in the reactant changes from 0 to +1 in the product; the oxidation level of the lower carbon of the double bond changes from 0 to –1. The total oxidation level, however, is the same for both product and reactant, and the reaction is neither an oxidation nor a reduction.

Additional Problems

Visualizing Chemistry

6.7

(a)

(b)

6.8

The name of the compound is (R)-2-bromopentane. Reaction of (S)-2-pentanol with PBr₃ to form (R)-2-bromopentane occurs with a change in stereochemistry because the configuration at the chirality center changes from S to R.

Mechanism Problems

6.9

Naming Alkyl Halides

6.10

(a)

(b)

(c)

(d)

(e)

6.11

(a)

(b)

(c)

(d)

6.12

(a)

2-methylbutane

There are six different products. Four of the compounds are chiral and optically active.

(b)

methylcyclopropane

There are six different products. All of the compounds whose names include the absolute configuration (R and/or S) in their name are chiral and optically active.

(c)

2,2-dimethylpentane

There are six different products. All of the compounds whose names include the absolute configuration (R or S) in their name are chiral and optically active.

Synthesizing Alkyl Halides

6.13

(a)

(b)

Hydroboration/oxidation can also be used to form cyclopentanol.

6.14

(a)

(b)

(c)

(d)

Oxidation and Reduction

6.15

Remember that the oxidation level is found by subtracting the number of C–H bonds from the number of C–O, C–N, and C–X bonds. The oxidation levels are shown beneath the structures.

In order of increasing oxidation level:

(a)

but-2-ene, but-1-ene < butanal < butanoic acid

(b)

6.16

All of the compounds except 3 (which is more oxidized) have the same oxidation level.

6.17

(a)	This reaction is an oxidation.
(b)	The reaction is neither an oxidation nor a reduction because the oxidation level of the reactant is the same as the oxidation level of the product.
(c)	This reaction is a reduction.

General Problems

6.18

(a)

(b)

(c)

6.19

Abstraction of a hydrogen atom from the chirality center of (S)-3-methylhexane produces an achiral radical intermediate, which reacts with bromine to form a 1:1 mixture of R and S enantiomeric, chiral bromoalkanes. The product mixture is optically inactive.

6.20

Abstraction of a hydrogen atom from carbon 4 yields a chiral radical intermediate. Reaction of this intermediate with chlorine does not occur with equal probability from each side, and the two diastereomeric products are not formed in a 1:1 ratio. The first product is optically active, and the second product is a meso compound.

6.21

Fluoroalkanes do not usually form Grignard reagents.

6.22

A Grignard reagent cannot be prepared from a compound containing an acidic functional group because the Grignard reagent is immediately quenched by the proton source. For example, the –CO₂H, –OH, –NH₂, and RC≡CH functional groups are too acidic to be used for preparation of a Grignard reagent. BrCH₂CH₂CH₂NH₂ is another compound that does not form a Grignard reagent.

6.23

Reaction of the ether with HBr can occur by either path A or path B. Path A is favored because its cation intermediate can be stabilized by resonance.

6.24

Tertiary carbocations (R₃C⁺) are more stable than either secondary or primary carbocations, due to the ability of the three alkyl groups to stabilize a positive charge. If the substrate is also allylic, as in the case of H₂C=CHC(CH₃)₂Br, positive charge can be further delocalized. Thus, H₂C=CHC(CH₃)₂Br should form a carbocation faster than (CH₃)₃CBr because the resulting carbocation is more stable.

6.25

Carboxylic acids are more acidic than alcohols because the negative charge of a carboxylate ion is stabilized by resonance. This resonance stabilization favors formation of carboxylate anions over alkoxide anions, and increases K_a for carboxylic acids.

Alkoxide anions are not resonance stabilized.

6.26

(a)

The primary alcohol would go exclusively through an S_N2 mechanism. This secondary alcohol will undergo a mixture of S_N1 and S_N2 substitution. However, formation of a secondary carbocation is relatively slow, as is the S_N2 reaction compared to the primary alcohol. Thus, the primary alcohol will be the faster reaction.

(b)

Tertiary alcohols undergo the fastest of S_N1 reactions through a tertiary carbocation, while a secondary alcohol is relatively slow for S_N2 because of the sterics of the substrate. Thus, the tertiary alcohol would be expected to react faster.

(c)

Tertiary alcohols undergo the fastest of S_N1 reactions through a tertiary carbocation, while the primary alcohol of neopentyl alcohol is one of the slowest S_N2 substrates for substitution because of the sterics on the tert-butyl group. Thus, the tertiary alcohol would be expected to react faster.

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Solutions Manual Fundamentals of Organic Chemistry-Openstax Adaptation Copyright © by Kirsten Kramer and Cassandra Lilly is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.