Chapter 20
There are several sets of answers; one is:
(a) C
5
H
12
(b) C 5 H 10
(c) C 5 H 8
Both reactions result in bromine being incorporated into the structure of the product. The difference is the way in which that incorporation takes place. In the saturated hydrocarbon, an existing C–H bond is broken, and a bond between the C and the Br can then be formed. In the unsaturated hydrocarbon, the only bond broken in the hydrocarbon is the π bond whose electrons can be used to form a bond to one of the bromine atoms in Br 2 (the electrons from the Br–Br bond form the other C–Br bond on the other carbon that was part of the π bond in the starting unsaturated hydrocarbon).
Unbranched alkanes have free rotation about the C–C bonds, yielding all orientations of the substituents about these bonds equivalent, interchangeable by rotation. In the unbranched alkenes, the inability to rotate about the bond results in fixed (unchanging) substituent orientations, thus permitting different isomers. Since these concepts pertain to phenomena at the molecular level, this explanation involves the microscopic domain.
They are the same compound because each is a saturated hydrocarbon containing an unbranched chain of six carbon atoms.
(a) C 6 H 14
(b) C 6 H 14
(c) C 6 H 12
(d) C 6 H 12
(e) C 6 H 10
(f) C 6 H 10
(a) 2,2-dibromobutane; (b) 2-chloro-2-methylpropane; (c) 2-methylbutane; (d) 1-butyne; (e) 4-fluoro-4-methyl-1-octyne; (f) trans -1-chloropropene; (g) 4-methyl-1-pentene
(a) 2,2,4-trimethylpentane; (b) 2,2,3-trimethylpentane, 2,3,4-trimethylpentane, and 2,3,3-trimethylpentane:
In the following, the carbon backbone and the appropriate number of hydrogen atoms are shown in condensed form:
In acetylene, the bonding uses sp hybrids on carbon atoms and s orbitals on hydrogen atoms. In benzene, the carbon atoms are sp 2 hybridized.
(a)
(b)
65.2 g
9.328 10 2 kg
(a) ethyl alcohol, ethanol: CH 3 CH 2 OH; (b) methyl alcohol, methanol: CH 3 OH; (c) ethylene glycol, ethanediol: HOCH 2 CH 2 OH; (d) isopropyl alcohol, 2-propanol: CH 3 CH(OH)CH 3 ; (e) glycerine, l,2,3-trihydroxypropane: HOCH 2 CH(OH)CH 2 OH
(a) 1-ethoxybutane, butyl ethyl ether; (b) 1-ethoxypropane, ethyl propyl ether; (c) 1-methoxypropane, methyl propyl ether
HOCH 2 CH 2 OH, two alcohol groups; CH 3 OCH 2 OH, ether and alcohol groups
(a)
(b) 4.593 10 2 L
(a)
(b)
(a)
(b)
(c)
A ketone contains a group bonded to two additional carbon atoms; thus, a minimum of three carbon atoms are needed.
Since they are both carboxylic acids, they each contain the –COOH functional group and its characteristics. The difference is the hydrocarbon chain in a saturated fatty acid contains no double or triple bonds, whereas the hydrocarbon chain in an unsaturated fatty acid contains one or more multiple bonds.
(a) CH 3 CH(OH)CH 3 : all carbons are tetrahedral; (b) the end carbons are tetrahedral and the central carbon is trigonal planar; (c) CH 3 OCH 3 : all are tetrahedral; (d) CH 3 COOH: the methyl carbon is tetrahedral and the acid carbon is trigonal planar; (e) CH 3 CH 2 CH 2 CH(CH 3 )CHCH 2 : all are tetrahedral except the right-most two carbons, which are trigonal planar
(a)
(b)
Trimethyl amine: trigonal pyramidal, sp 3 ; trimethyl ammonium ion: tetrahedral, sp 3
CH 3 C H = C HCH 3 ( sp 2 ) + Cl CH 3 C H(Cl)H(Cl)CH 3 ( sp 3 ); 2 C 6 H 6 ( sp 2 ) + 15O 2 12 C O 2 ( sp ) + 6H 2 O
The carbon in CO 3 2− , initially at sp 2 , changes hybridization to sp in CO 2 .