Assign 15-2 - 3 orbitals (a hybrid orbital with more s...

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Assignment 15 -2 1. Give an acceptable name for each compound: (a) C C CH CH 2 CH 3 H H (b) C C CH 3 H CH 2 CH 2 C H C H H CH 3 CH 2 2. Indicate the more (most) stable (lower energy) species in each of the following sets: vs. CH CH CH CH 3 CH CH 2 CH CH 2 CH 2 CH 2 3. Why is the C 2 -C 3 bond of 1,3-butadiene (1.47 Å ) shorter than the C_C bond of ethane (1.54 Å)? 4. In the addition of HCl to 1,3-butadiene, why does the 1,2 addition product form faster than the 1,4 addition product? _______________________________ ANSWERS 1. (a) (3E)-1,3-pentadiene (b) (2Z, 6Z)-2,6-nondiene 2. CH CH CH CH 3 CH 2 has the lower energy. Conjugated C=C’s are more stable than nonconjugated C=C’s. 3. Two reasons: (1) The C 2 -C 3 bond in 1,3 butadiene is formed from the overlap of two relatively short sp 2 orbitals. This contrasts with the C-C bond of ethane which is formed from the overlap of two longer sp
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Unformatted text preview: 3 orbitals (a hybrid orbital with more s character is shorter). (2) Because of electron delocalization, there is partial double bond character in the C 2-C 3 bond, making it shorter. 4. The addition of HC1 to l,3-butadiene involves the bonding of Cl -to a (+) carbon of an allylic cation intermediate. A 1,2 addition product forms when the Cl -bonds to C-2 of the allylic cation; a 1,4 addition product forms when Cl -bonds to C-4. In the allylic cation, C-2 has a larger positive charge than C-4 (C-2 is 2 o cation while C-4 is a 1 o cation). This makes it easier for Cl -to bond to C-2 than C-4. Thus, the 1,2 addition product forms faster....
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This note was uploaded on 04/07/2009 for the course CHE 312 taught by Professor Drake during the Spring '08 term at American International.

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