Name the following compounds (include stereochemistry, cis/trans, E/Z when appropriate). Cl E- 6 chloro, 5 ethyl, 4 methyl 3-octene

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Problem Set 6 Name the following compounds (include stereochemistry, cis/trans, E/Z when appropriate). E- 6 chlo, 5 ethyl, 4 methyl 3-octene 5 methyl, 3 vinyl cyclohexene 7Z- 7 bromo 4 ethyl, 6 methyl deca 1,7 diene Allyl bromide 3 bromo 1propene 3 ethyl, 1 methyl cyclopentene -trans, 4 cis hepta,4 diene Calculate the degrees of unsaturation (DBs) f the following molecular fmulas. C 8 8 DB = n -#s -#s #Ns = x8-8 = 5 C 1 19 N DB = x1-19 1 = 4 C 13 1 N DB = x13-1 -1 = 4 1

β-eliminations that follow an E mechanism have stereochemical implications that need to be addressed. Elimination MUST occur from a confmation with an anti periplanar between the leaving group and β-hydrogen. This cresponds to a staggered confmation in acyclic systems and trans-diaxial in cyclohexane chairs. A. Examples of acyclic systems 3 C C 3 NaC 3 C 3 C 3 C 3 t-bu rotate Na 3 C t-bu eliminate t-bu C 3 rotate Na two 180 o apart eliminate trans product

B. Examples of cyclic systems (give all products and label maj/min). f there is no reaction likely, simply write N EACTN. Note: recall that large bulky groups (i.e. tbu) D NT occupy the axial position... NaC 3 maj min NaC 3 NaC 3 no reaction proceeds due to t-bu group, chair cannot adopt a confmation with LG axial 3

Draw ALL possible elimination products of the following reactions involving alkyl halides. f there is me than one product possible, CCLE the maj product (Zaitsev's Law). NaC 3 C 3 (C 3 ) 3 CK Draw the E Elimination Mechanism f the fmation of the MAJ product of the reaction shown below. emember: LG and eliminated- must be "anti-periplanar"! C 3 (C 3 ) 3 CK Draw the E1 Elimination Mechanism f fmation of the MAJ product of the reaction shown below. 4

E Elimination in cyclohexanes: Draw the product(s) of the following reactions. f there are two products possible, circle the maj product. f no elimination products are possible, simply write N.. C 3 C 3 C 3 C 3 N When cis-1-iodo--methylcyclohexane is treated with potassium ter, the maj product is 1-methylcyclohexene, a trisubstituted alkene. owever, when the cresponding trans isomer is subjected to the same conditions, the reaction progresses at a slower rate and the maj product is 3-methylcyclohexene, a disubstituted alkene. Explain these observations. The E elimination is an anti elimination that requires that the leaving group and the β-hydrogen to be 180 apart. As seen from the perspective drawing of the most stable chair (C 3 ) 3 C K confmation of the trans isomer, the leaving group is not anti to any β- (C 3 ) 3 C hydrogens. A ring flip is needed to bring the molecule to the requisite Δ maj confmation leading to elimination. This confmation has two axial substituents and is thus much less stable. This explains why the reaction is so (C 3 ) 3 C K sluggish. Note also that in this confmation the leaving group can only be (C 3 ) 3 C Δ anti to one of the C(6) hydrogens. This explains the fmation of the less maj substituted alkene. me stable C 3 ring flip base C 3 less stable n examination of the most stable chair confmation of the cis isomer, it is clear that the molecule is perfectly set f anti elimination. Two β- hydrogens iented anti to the leaving group are now present; the one attached to C() is removed preferentially, as this gives rise to the me substituted alkene product. base C 3 5

Both vicinal and geminal dihalides can undergo two consecutive elimination reactions produce the two π bonds of an alkyne. A very strong base (NaN ) is needed to facilitate this reaction. C C vicinal dihalide C C geminal dihalide Draw the mechanism f two sequential E eliminations of the following vicinal dihalide to give the alkyne product: NaN DMS NaN Na N 3 6

Provide the maj ganic products f the following reactions. Be sure to include all stereoisomers and consider rearrangements. syn-epoxide mcpba addition S 4 addition addition KMn4 3 anti-addition syn-epoxide mcpba addition ydride Shift anti- syn-addition 1. B 3. 1. g(ac),. NaB 4, non syn/anti selectivity 1. 3. Zn, 3 anti-addition C 3 C 3 Possible Methyl Shift addition C 3, Pd/C syn-addition non syn/anti selectivity NBS 1. s 4, pyridine.nas 3 syn-addition 7