Orgo Ch. 4 Practice Problems

Give IUPAC names for the following compounds:

Draw structures corresponding to the following IUPAC names:

(a) Methyl-1,5-hexadiene

(b) Ethyl-2,2-dimethyl-3-heptene

(c) 2,3,3-Trimethyl-1,4,6-octatriene

(d) 3,4-Diisopropyl-2,5-dimethyl-3-hexene

Name the following cycloalkenes:

Change the following old names to new, post-1993 names, and draw the structure of each compound:

(a) 2,5,5-Trimethyl-2-hexene

(b) 2,3-Dimethyl-1,3-cyclohexadiene

Alkyne nomenclature follows the general rules for hydrocarbons discussed in Section 2.4 and 4.1. The suffix -yne is used, and the position of the triple bond is indicated by giving the number of the first alkyne carbon in the chain. Numbering the main chain begins at the end nearer the triple bond so that the triple bond receives as low a number as possible.

Compounds with more than one triple bond are called diynes, triynes, and so forth; compounds containing both double and triple bonds are called enynes (not ynenes). Numbering of an enyne chain starts from the end nearer the first multiple bond, whether double or triple. When there is a choice in numbering, double bonds receive lower numbers than triple bonds. For example:

As with alkyl and alkenyl substituents derived from alkanes and alkenes, respectively, alkynyl groups are also possible.

Name the following alkynes:

There are seven isomeric alkynes with the formula C₆H₁₀. Draw and name them.

Which of the following compounds can exist as pairs of cis–trans isomers? Draw each pair, and indicate the geometry of each isomer.

(a) CH₃CH=CH₂

(b) (CH₃)₂C=CHCH₃

(c) CH₃CH₂CH=CHCH₃

(d)(CH₃)₂C=C(CH₃)CH₂CH₃

(e) ClCH=CHCl

(f) BrCH=CHCl

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

Which member in each of the following sets ranks higher?

(a)−H or −CH₃

(b)−Cl or −CH₂Cl

(c)–CH₂CH₂Br or –CH=CH₂

(d)−NHCH₃ or −OCH₃

(e)−CH₂OH or −CH=O

(f)−CH₂OCH₃ or −CH=O

Rank the substituents in each of the following sets according to the sequence rules:

(a)−CH₃, −OH, −H, −Cl

(b)−CH₃, −CH₂CH₃, −CH=CH₂, −CH₂OH

(c)−CO₂H, −CH₂OH, −C≡N, −CH₂NH₂

(d)−CH₂CH3, −C≡CH, −C≡N, −CH₂OCH₃

Assign E or Z configuration to the following alkenes:

Assign stereochemistry (E or Z) to the double bond in the following compound, and convert the drawing into a skeletal structure (red = O):

Classify each of the following reactions as an addition, elimination, substitution, or rearrangement:

(a) CH₃Br + KOH ⟶ CH₃OH + KBr

(b) CH₃CH₂Br ⟶ H₂C=CH₂ + HBr

(c) H₂C=CH₂ + H₂ ⟶ CH₃CH₃

Which of the following species are likely to be nucleophiles and which electrophiles? Which might be both?

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.

An electrostatic potential map of boron trifluoride is shown. Is BF₃ likely to be a nucleophile or an electrophile? Draw a Lewis structure for BF₃, and explain your answer.

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.

What product would you expect from reaction of cyclohexene with HBr? With HCl?

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.

Reaction of HBr with 2-methylpropene yields 2-bromo-2-methylpropane. What is the structure of the carbocation formed during the reaction? Show the mechanism of the reaction.

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

Which reaction is faster, one with ΔG^‡ = +45 kJ/mol or one with ΔG^‡ = +70 kJ/mol?

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.

Sketch an energy diagram for a two-step reaction in which both steps are exergonic and in which the second step has a higher-energy transition state than the first. Label the parts of the diagram corresponding to reactant, product, intermediate, overall ΔG^‡, and overall ΔG°.

Add curved arrows to the following polar reactions to indicate the flow of electrons in each:

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^–.

Predict the products of the following polar reaction, a step in the citric acid cycle for food metabolism, by interpreting the flow of electrons indicated by the curved arrows:

This mechanism will be studied in a later chapter.