Question

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54 CHAPTER 1 Structure and Bonding 1-54 Predict the hybridization and geometry of the carbon and nitrogen atoms in the following molecules and ions. (Hint: Resonance.) (a) CH, C-CH (b) HN-CH-CH-CH, (c) CH-CN 9 1-55 1-56 Draw orbital pictures of the pi bonding is the following compounds CH,COCH, | b) HCN (d) CHCOCHO CH.CHC=CHCH, (a) Draw the structure of cir-CH-CH=CH-CH.CH. S ches (b) Circle the six coplanar atoms in this compound e) Draw the transisomer, and y still six? (d) Circle the coplanar atoms in the following structure: CH (c) CH=CH-CHCHCN (1) CH-CH=N-CH=C=0 r geometry. i 1-57 1.58 In pent-2-yne (CH,COCH, CH), there are four stoms in a straight line. Use dashed lines and wedges to draw a three-dimensional representation of this molecule, and circle the four atoes that are in a straight line. Which of the following compounds show.cis-trans Isomerism? Draw the cis and trans isomers of the ones that do (a) CH,CH=CHCH, | (b) CH-CH-CH, (6) CHCICH), (d) cyclopentene, (e) CH-GH-C-CH-CH, in CH-CH-N-CH, CH,CH.CH Give the relationships between the following pairs of structures. The possible relationships are as follows: same compound, cis-trans isomers, constitutional structural) isomers, and not isomers (different molecular formula) W CH CH CH.CH, and (CH),CH (b) CH-CH-CH2Cl and CHCI-CH-CH, CH, CH, CH, (d) CH, CH, CH, and 1.59 D CH, CH, | CH-CH, and CH.CH, CH-CH, CH.CH Сн, 1-60 Dimethyl sulfoxide (DMSO) has been used as an anti-inflammatory rub for race horses. DMSO and acetone appear to have similar structures, but the C O carbon atom in acetone is planar, while the S- O sulfur atom in DMSO is pyramidal Draw Lewis structures for DMSO and acetone, predict the hybridizations, and explain these observations CH-S-CH, DMSO CH-C-CH, cetone

Answer 1

INTRODUCTION CIS AND TRANS

Alkene R-CH=CH-R can exist cis and trans isomers. If two R efroups are same side af (C=C). : CSACR R same side cio Trans Two R enroups are on side of the le=c) carbon carbon opposite double bond R !

56 ѕ c H2 -CH=CH-CH2 – сна нс СН 2 -СН3 се с нн тас - I RAroup same side cis If its trans structure become н сH-CH3 c=c . H₂CH CR erroup opposite side) Dristorent Trans 58 = а, сна -СН = cн - сн3 сіз нс с = с = с сна cis But a ene Trans & But qene

b) But 2 yne not exhibit cis trans Isomerism. Because Linear exeometry for both of the triple bonded Carbon. CH3 C-CH3 Nocis hans form CH2=C(CH3)2 CH2= this ring not large enough to accommodate a trans double bond e z Bethyl hexe 2 ene CH2-CH2-CHI сн. -СН 3 cZCHEH.-C. in form I CH2-CH₂-CH3 -CH2-CH3 at E form