CH 3 CHCH CH
4. Isomerization
4.5 Summary of Alkane Chemistry
SUPPLEMENTARY PROBLEMS
Problem 4.24 Assign numbers, ranging from (1) for LOWESTto (3) for HIGHEST, to the boiling points of the fol-lowing hexane isomers: 2,2-dimethylbutane, 3-methylpentane, and n-hexane. Do not consult any tables for data.
Hexane has the longest chain, the greatest intermolecular attraction, and, therefore, the highest boiling point, (3). 2,2-Dimethylbutane is (1), since, with the most spherical shape, it has the smallest intermolecular contact and attraction. This leaves 3-methylpentane as (2).
Problem 4.25 Write structural formulas for the five isomeric hexanes, and name them by the IUPAC system.
The isomer with the longest chain is hexane, CH3CH2CH2CH2CH2CH3. If we use a five-carbon chain, a CH3 may be placed on either C2 or C4 to produce 2-methylpentane, or on C3 to give another isomer, 3-methylpentane.
With a four-carbon chain, either a CH3CH2or two CH3’s must be added as branches to give a total of 6 C’s. Plac-ing CH3CH2anywhere on the chain is ruled out because it lengthens the chain. Two CH3’s are added, but only to central C’s to avoid extending the chain. If both CH3’s are introduced on the same C, the isomer is 2,2-di-methylbutane. Placing one CH3on each of the central C’s gives the remaining isomer, 2,3-dimethylbutane.
Problem 4.26 Write the structural formulas for (a) 3,4-dichloro-2,5-dimethylhexane; (b) 5-(1,2-dimethyl-propyl)-6-methyldodecane. (Complex branch groups are usually enclosed in parentheses.)
(a)
(b) The group in parentheses is bonded to the fifth C. It is a propyl group with CH3’s on its first and second C’s (denoted 1′ and 2′) counting from the attached C.
Problem 4.27 Write the structural formulas and give the IUPAC names for all monochloro derivatives of (a) isopentane, (CH3)2CHCH2CH3; (b) 2,2,4-trimethylpentane, (CH3)3CCH2CH(CH3)2.
(a) Since there are four kinds of equivalent H’s:
there are four isomers:
(b) There are four isomers because there are four kinds of H’s (CH13)3CCH22CH3(CH43)2.
Problem 4.28 Give topological structural formulas for (a) propane, (b) butane, (c) isobutane, (d) 2,2-di-methylpropane, (e) 2,3-dimethylbutane, (f ) 3-ethylpentane, (g) 1-chloro-3-methylbutane, (h) 2,3-dichloro-2-methylpentane, (i) 2-chloro-2,4-trimethylpentane.
In this method, one writes only the C⎯C bonds and all functional groups bonded to C. The approximate bond angles are used
Problem 4.29 Synthesize (a) 2-methylpentane from CH2CHCH ⎯ CH(CH3)2, (b) isobutane from isobutyl chloride, (c) 2-methyl-2-deuterobutane from 2-chloro-2-methylbutane. Show all steps.
(a) The alkane and the starting compound have the same carbon skeleton.
(b) The alkyl chloride and alkane have the same carbon skeleton.
(c) Deuterium can be bonded to C by the reaction of D2O with a Grignard reagent.
Problem 4.30 RCl is treated with Li in ether solution to form RLi. RLi reacts with H2O to form isopentane.
Using the Corey-House method, RCl is coupled to form 2,7-dimethyloctane. What is the structure of RCl?
To determine the structure of a compound from its reactions, the structures of the products are first consid-ered and their formation is then deduced from the reactions. The coupling product must be a symmetrical mol-ecule whose carbon-to-carbon bond was formed between C4and C5of 2,7-dimethyloctane. The only RCl which will give this product is isopentyl chloride:
This alkyl halide will also yield isopentane.
Problem 4.31 Give steps for the following syntheses: (a) propane to (CH3)2CHCH(CH3)2, (b) propane to 2-methylpentane, (c)14CH3Cl to 14CH314CH214CH214CH3.
(a) The symmetrical molecule is prepared by coupling an isopropyl halide. Bromination of propane is pre-ferred over chlorination because the ratio of isopropyl to n-propyl halide is 96%/4% in bromination and only 56%/44% in chlorination.
Problem 4.32 Synthesize the following deuterated compounds: (a) CH3CH2D, (b) CH2DCH2D.
Problem 4.33 In the dark at 150º C, tetraethyl lead, Pb(C2H5)4, catalyzes the chlorination of CH4. Explain in terms of the mechanism.
Pb(C2H5)4readily undergoes thermal homolysis of the Pb⎯ C bond.
The CH3CH2 then generates the Cl that initiates the propagation steps.
Problem 4.34 Hydrocarbons are monochlorinated with tert-butyl hypochlorite, t-BuOCl.
t-BuOCl RH RCl t-BuOH Write the propagating steps for this reaction if the initiating step is
t-BuOCl t-BuO Cl
The propagating steps must give the products and also form chain-carrying free radicals. The formation of t-BuOH suggests H-abstraction from RH by t-BuO, not by Cl. The steps are:
R and t-BuO are the chain-carrying radicals.
Problem 4.35 Calculate the heat of combustion of methane at 25ºC. The bond energies for C⎯H, OO, CO, and O ⎯H are, respectively, 413.0, 498.3, 803.3, and 462.8 kJ/mol.
First, write the balanced equation for the reaction.
The energies for the bonds broken are calculated. These are endothermic processes and ΔH is positive.
Next, the energies are calculated for the bonds formed. Bond formation is exothermic, so the ΔH values are made negative.
The enthalpy for the reaction is the sum of these values:
+1652 0 996 6 1606 6 1851 2. + . − . − . = −809 2. kJ/mol (Reaaction is exothermic)
Problem 4.36 (a) Deduce structural formulas and give IUPAC names for the nine isomers of C7H16. (b) Why is 2-ethylpentane not among the nine?
(b) Because the longest chain has six C’s, and it is 3-methylhexane.
Problem 4.37 Singlet methylene, :CH2(Section 3.2), may be generated from diazomethane, CH2N2; the other product is N2. It can insert between C⎯H bonds of alkanes:
Determine the selectivity and reactivity of :CH2from the yields of products from methylene insertion in pentane:
Name, Kind of CH insertion Hexane, CH1 2-Methylpentane, CH2 3-Methylpentane, CH3
Yield 48% 35% 17%
Calculate the theoretical % yield based only on the probability factor and then compare the theoretical and observed % yields.
The almost identical agreement validates the assumption that methylene is one of the most reactive and least se-lective species in organic chemistry.
Hexane
2-Methylpentane 3-Methylpentane Total
1 2 3
6 4 2 12
6/12 = 50 4/12 = 33.3 2/12 = 16.7
48 35 17
PRODUCT KIND NUMBER
H’s OF PENTANE
PROPORTION × 100% = CALC. % YIELD OBSERVED % YIELD
Problem 4.38 How would the energy-conformation diagrams of ethane and propane differ?
They would both have the same general appearance showing minimum energy in the staggered form and max-imum energy in the eclipsed form. But the difference in energy between these forms would be greater in propane than in ethane (13.8 versus 12.5 kJ/mol). The reason is that the eclipsing strain of the Me group and H is greater than that of two H’s.
Problem 4.39 1,2-Dibromoethane has a zero dipole moment, whereas ethylene glycol, CH2OHCH2OH, has a measurable dipole moment. Explain.
1,2-Dibromoethane exists in the anti form, so that the C哫Br dipoles cancel and the net dipole moment is zero. When the glycol exists in the gauche form, intramolecular H-bonding occurs. Intramolecular H-bonding is a stabilizing effect which cannot occur in the anti conformer.
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