Ans_First_Review_Prob_Set

Ans_First_Review_Prob_Set - P1 OXT/SRB P2 xxx...

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P1: OXT/SRB P2: xxx Printer: Hamilton JWDD052-ANSWER JWDD052-Solomons-v2 April 24, 2007 19:5 ANSWERS TO FIRST REVIEW PROBLEM SET 1. (a) OH CH 3 CH 3 CH 3 A HA H 2 O +H 2 O CH 3 CH 3 CH 3 OH 2 + + CH 3 + 2 ° Carbocation 3 ° Carbocation methanide shift H CH 3 CH 3 A CH 3 + CH 3 CH 3 CH 3 HA CH 3 CH 3 (b) + + Br + Br Br then, + Br Cl Br Cl Br (c) The enantiomer of the product given would be formed in an equimolar amount via the following reaction: + Br Cl Cl Br The trans -1,2-dibromocyclopentane would be formed as a racemic form via the reaction of the bromonium ion with a bromide ion: + + Br Br + Br Racemic trans -1,2-dibromocyclopentane Br Br Br 231
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P1: OXT/SRB P2: xxx Printer: Hamilton JWDD052-ANSWER JWDD052-Solomons-v2 April 24, 2007 19:5 232 ANSWERS TO FIRST REVIEW PROBLEM SET And, trans -2-bromocyclopentanol (the bromohydrin) would be formed (as a racemic form) via the reaction of the bromonium ion with water. + OH 2 Br + + Br H 2 O + + OH 2 Br A + + OH OH Br Br HA Racemic trans -2-bromocyclopentanol 2. (a) CHCl 3 (b) The cis isomer (c) CH 3 Cl 3. This indicates that the bonds in BF 3 are geometrically arranged so as to cancel each others’ polarities in contrast to the case of NF 3 . This, together with other evidence, indicates that BF 3 has trigonal planar structure and NF 3 has trigonal pyramidal structure. 4. (a) sp 3 (b) They are much smaller (60 ) than the expected 109.5 . (c) That the bent bonds in cyclopropane, like compressed springs, contain stored energy that is available to assist in reactions that open the 3-membered ring. 5. All of these differences can be explained by the contribution to the CH 2 CHCl molecule made by the following A and B resonance structures. CC H H H Cl Cl H H H + AB (a) Because of the contribution made to the hybrid by B, the C Cl bond of CH 2 CH Cl has some double-bond character and is, therefore, shorter than the “pure” single bond of CH 3 CH 2 Cl. (b) The contribution made to the hybrid by B imparts some single-bond character to the carbon-carbon double bond of CH 2 CH Cl, causing it to be longer than the “pure” double bond of CH 2 CH 2 . (c) Electronegativity differences would cause a carbon-chlorine bond to be polarized as follows: l δ + δ
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P1: OXT/SRB P2: xxx Printer: Hamilton JWDD052-ANSWER JWDD052-Solomons-v2 April 24, 2007 19:5 ANSWERS TO FIRST REVIEW PROBLEM SET 233 And this effect accounts, almost entirely, for the dipole moment of CH 3 CH 2 Cl. Cl CH 3 CH 2 δ + δ μ = 2.05 D With CH 2 CH Cl, however, the resonance contribution of B tends to oppose the polariza- tion of the C Cl bond caused by electronegativity differences. That is, the resonance effect partially cancels the electronegativity effect, causing the dipole moment to be smaller. 6. B = CH 3 (CH 2 ) 11 CH 2 C CNa C = CH 3 (CH 2 ) 11 CH 2 C CCH 2 (CH 2 ) 6 CH 3 A = CH 3 (CH 2 ) 11 CH 2 C CH Muscalure = CH 3 (CH 2 ) 11 CH 2 CC HH CH 2 (CH 2 ) 6 CH 3 7.
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This note was uploaded on 09/10/2008 for the course CHE 321 taught by Professor Fowler/sampson during the Spring '08 term at SUNY Stony Brook.

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Ans_First_Review_Prob_Set - P1 OXT/SRB P2 xxx...

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