1559T_ch03_38-5410/22/052:59Page43 ⊕ EQA Solutions to Problems.43 Cbond participating (b)In naming these,remember that the radical carbon is alwaysCl(just like the"point of CH-CHz CH3-CH2 CH-CH-CH-CH CH-CH-C-CH 2-Ethylbutyl radical 1-Ethyl-1-mthypropyl radica Primary,less stable ⊕ (c)Left to right:1.2-dimethylpropyl radical,secondary.intermediate in stability:1.1-dimethylpropyl radical,tertiary,most stable;3-methylbutyl radical,primary,least stable Hyperconjugation in the 1.1-dimethylpropyl radical is the same as in 1-ethyl-1-methylpropyl [(b)above]: in your picture.an H should replace one of the end CH groups. aCH,C,CH,一CH,cH,·+·CH,c-Cond c Then there are three possible recombinations: (d)CH3 CH3CH2-CHaCH2CH3 (Reverse of first step) Propane Solutions to Problems • 43 COH bonds overlapping with the radical p orbital, and one with a COC bond participating instead of a COH: (b) In naming these, remember that the radical carbon is always C1 (just like the “point of attachment” carbon in alkyl groups). The parent name is based on the longest carbon chain beginning at C1, and all appendages are named as substituents: (c) Left to right: 1,2-dimethylpropyl radical, secondary, intermediate in stability; 1,1-dimethylpropyl radical, tertiary, most stable; 3-methylbutyl radical, primary, least stable. Hyperconjugation in the 1,1-dimethylpropyl radical is the same as in 1-ethyl-1-methylpropyl [(b) above]; in your picture, an H should replace one of the end CH3 groups. 15. Work problems like this “mechanistically”: Proceed via general reaction steps as you have previously seen illustrated in the text until you reach stable molecules. Pyrolysis of propane starts as follows: (a) Then there are three possible recombinations: (b) (c) (d) Propane CH3
CH3CH2 CH3CH2CH3 (Reverse of first step) Butane 2 CH3CH2 CH3CH2CH2CH3 Ethane 2 CH3 CH3CH3 CH3CH2 CH3 CH3CH2
CH3 C—C bond cleavage H H H C H2C CHCH3 CH3HC and H H H H C C H2C CH3 H H H H C C H2C CH3 1559T_ch03_38-54 10/22/05 2:59 Page 43