CHAPTER 14 ORGANOMETALLIC COMPOUNDS SOLUTIONS TO TEXT PROBLEMS 14.1 (b) Magnesium bears a cyclohexyl substituent and a chlorine. Chlorine is named as an anion. The compound is cyclohexylmagnesium chloride. 14.2 (b) The alkyl bromide precursor to sec-butyllithium must be sec-butyl bromide. 14.3 (b) Allyl chloride is converted to allylmagnesium chloride on reaction with magnesium. (c) The carbon–iodine bond of iodocyclobutane is replaced by a carbon–magnesium bond in the Grignard reagent. Mg diethyl ether Iodocyclobutane I Cyclobutylmagnesium iodide MgI Allyl chloride CHCH 2 ClH 2 C Allylmagnesium chloride CHCH 2 MgClH 2 C Mg diethyl ether 2LiH11001 LiBrH11001CH 3 CHCH 2 CH 3 Br 2-Bromobutane (sec-butyl bromide) CH 3 CHCH 2 CH 3 Li 1-Methylpropyllithium (sec-butyllithium) 342 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Bromine is attached to an sp 2 -hybridized carbon in 1-bromocyclohexene. The product of its reaction with magnesium has a carbon–magnesium bond in place of the carbon–bromine bond. 14.4 (b) 1-Hexanol will protonate butyllithium because its hydroxyl group is a proton donor only slightly less acidic than water. This proton-transfer reaction could be used to prepare lithium 1-hexanolate. (c) The proton donor here is benzenethiol. 14.5 (b) Propylmagnesium bromide reacts with benzaldehyde by addition to the carbonyl group. (c) Tertiary alcohols result from the reaction of Grignard reagents and ketones. (d) The starting material is a ketone and so reacts with a Grignard reagent to give a tertiary alcohol. diethyl ether H 3 O H11001 CH 3 CH 2 CH 2 H 3 C C OMgBr CH 3 CH 2 3-Methyl-3-hexanol CH 3 CH 2 CH 2 COH CH 2 CH 3 CH 3 CH 3 CH 2 CH 2 MgBr O CH 3 CH 2 C H 3 C Propylmagnesium bromide H11001 2-butanone 1. diethyl ether 2. H 3 O H11001 H11001CH 3 CH 2 CH 2 MgBr O CH 2 CH 2 CH 3 OH 1-Propylcyclohexanol diethyl ether H 3 O H11001 C 6 H 5 CO H CH 3 CH 2 CH 2 C 6 H 5 C OMgBr H 1-Phenyl-1-butanol C 6 H 5 CHCH 2 CH 2 CH 3 OH CH 3 CH 2 CH 2 MgBr H11001H11001C 6 H 5 SH Benzenethiol C 6 H 5 SLi Lithium benzenethiolate CH 3 CH 2 CH 2 CH 2 Li Butyllithium CH 3 CH 2 CH 2 CH 3 Butane H11001H11001CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH 1-Hexanol CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OLi Lithium 1-hexanolate CH 3 CH 2 CH 2 CH 2 Li Butyllithium CH 3 CH 2 CH 2 CH 3 Butane Mg diethyl ether 1-Bromocyclohexene Br 1-Cyclohexenylmagnesium bromide MgBr ORGANOMETALLIC COMPOUNDS 343 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 344 ORGANOMETALLIC COMPOUNDS 14.6 Ethyl anion reacts as a Br?nsted base to remove a proton from the alkyne. The proton at C-1 is removed because it is the most acidic, having a pK a of approximately 25. 14.7 (b) The target alcohol is tertiary and so is prepared by addition of a Grignard reagent to a ketone. The retrosynthetic transformations are: Because two of the alkyl groups on the hydroxyl-bearing carbon are the same (methyl), only two, not three,different ketones are possible starting materials: 14.8 (b) Recall that the two identical groups bonded to the hydroxyl-bearing carbon of the alcohol arose from the Grignard reagent. That leads to the following retrosynthetic analysis: Thus, the two phenyl substituents arise by addition of a phenyl Grignard reagent to an ester of cyclopropanecarboxylic acid. 14.9 (b) Of the three methyl groups of 1,3,3-trimethylcyclopentene, only the one connected to the double bond can be attached by way of an organocuprate reagent. Attachment of either of H11001H11001 1. diethyl ether 2. H 3 O H11001 (C 6 H 5 ) 2 C OH Cyclopropyl- diphenylmethanol CH 3 OH Methanol O COCH 3 Methyl cyclopropanecarboxylate 2C 6 H 5 MgBr Phenylmagnesium bromide (C 6 H 5 ) 2 C OH O COR 2C 6 H 5 MgXH11001 MgBr H11001 CH 3 CCH 3 O Phenylmagnesium bromide Acetone CCH 3 1. diethyl ether 2. H 3 O H11001 OH CH 3 2-Phenyl-2-propanol CCH 3 CH 3 MgI CCH 3 H11001 1. diethyl ether 2. H 3 O H11001 OHO CH 3 Methylmagnesium iodide Acetophenone 2-Phenyl-2-propanol H11001H11001CH 3 H11002 C O CH 3 C OH CH 3 CH 3 H11002 CH 3 CCH 3 O H11001H11001CH 3 CH 3 Ethane1-Hexyne CCH 2 CH 2 CH 2 CH 3 HC Ethyl anion CH 3 CH 2 H11002 Conjugate base of 1-hexyne CCH 2 CH 2 CH 2 CH 3 C H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website the other methyls would involve a tertiary carbon, a process that does not occur very efficiently. 14.10 (b) Methylenecyclobutane is the appropriate precursor to the spirohexane shown. 14.11 Syn addition of dibromocarbene to cis-2-butene yields a cyclopropane derivative in which the methyl groups are cis. Conversely, the methyl groups in the cyclopropane derivative of trans-2-butene are trans to one another. 14.12 Iron has an atomic number of 26 and an electron configuration of [Ar]4s 2 3d 6 . Thus, it has 8 valence electrons and requires 10 more to satisfy the 18-electron rule. Five CO ligands, each providing two electrons, are therefore needed. The compound is Fe(CO) 5 . 14.13 (a) Cyclopentyllithium is It has a carbon–lithium bond. It satisfies the requirement for classification as an organo- metallic compound. (b) Ethoxymagnesium chloride does not have a carbon–metal bond. It is not an organometallic compound. (c) 2-Phenylethylmagnesium iodide is an example of a Grignard reagent. It is an organometallic compound. CH 2 CH 2 MgI CH 3 CH 2 OMgCl CH 3 CH 2 O H11002 Mg 2H11001 Cl H11002 or HLi CC H H 3 C CH 3 H trans-2-Butene trans-1,1-Dibromo-2,3- dimethylcyclopropane CHBr 3 KOC(CH 3 ) 3 H 3 C CH 3 BrBr H H CC H H 3 H 3 H cis-2-Butene cis-1,1-Dibromo-2,3- dimethylcyclopropane H 3 CCH 3 BrBr HH CHBr 3 KOC(CH 3 ) 3 CH 2 I 2 Zn(Cu), ether CH 2 Methylenecyclobutane Spiro[3.2]hexane (22%) H11001LiCu(CH 3 ) 2 diethyl ether 1,3,3-Trimethylcyclopentene1-Bromo-3,3- dimethylcyclopentene Lithium dimethylcuprate CH 3 CH 3 Br H 3 C CH 3 CH 3 ORGANOMETALLIC COMPOUNDS 345 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Lithium divinylcuprate has two vinyl groups bonded to copper. It is an organometallic compound. (e) Sodium carbonate, Na 2 CO 3 can be represented by the Lewis structure. There is no carbon–metal bond, and sodium carbonate is not an organometallic compound. ( f ) Benzylpotassium is represented as It has a carbon–potassium bond and thus is an organometallic compound. 14.14 The two alkyl groups attached to aluminum in [(CH 3 ) 2 CHCH 2 ] 2 AlH are isobutyl groups. The hy- drogen bonded to aluminum is named in a separate word as hydride. Thus, “dibal” is a shortened form of the systematic name diisobutylaluminum hydride. 14.15 (a) Grignard reagents such as pentylmagnesium iodide are prepared by reaction of magnesium with the corresponding alkyl halide. (b) Acetylenic Grignard reagents are normally prepared by reaction of a terminal alkyne with a readily available Grignard reagent such as an ethylmagnesium halide. The reaction that takes place is an acid–base reaction in which the terminal alkyne acts as a proton donor. (c) Alkyllithiums are formed by reaction of lithium with an alkyl halide. (d) Lithium dialkylcuprates arise by the reaction of an alkyllithium with a Cu(I) salt. 14.16 The polarity of a covalent bond increases with an increase in the electronegativity difference be- tween the connected atoms. Carbon has an electronegativity of 2.5 (Table 14.1). Metals are less elec- tronegative than carbon. When comparing two metals, the less electronegative one therefore has the more polar bond to carbon. (a) Table 14.1 gives the electronegativity of lithium as 1.0, whereas that for aluminum is 1.5. The carbon–lithium bond in CH 3 CH 2 Li is more polar than the carbon–aluminum bond in (CH 3 CH 2 ) 3 Al. H11001 CuX (X H11005 Cl, Br, or I)Pentyllithium, from part (c) 2CH 3 CH 2 CH 2 CH 2 CH 2 Li H11001 LiX Lithium dipentylcuprate LiCu(CH 2 CH 2 CH 2 CH 2 CH 3 ) 2 H11001 2Li 1-Halopentane (X H11005 Cl, Br, or I) CH 3 CH 2 CH 2 CH 2 CH 2 X H11001 LiX Pentyllithium CH 3 CH 2 CH 2 CH 2 CH 2 Li H11001 diethyl ether CH 3 CH 2 MgI Ethylmagnesium iodide H11001 CH 3 CH 3 Ethane1-Butyne CH 3 CH 2 CCH 1-Butynylmagnesium iodide CH 3 CH 2 CCMgI MgH11001 diethyl ether CH 3 CH 2 CH 2 CH 2 CH 2 I 1-Iodopentane CH 3 CH 2 CH 2 CH 2 CH 2 MgI Pentylmagnesium iodide orCH 2 K CH 2 K H11001 H11002 Na H11001 CO O O H11002 H11002 Na H11001 Li H11001 (H 2 C CH CH 2 )Cu CH H11002 346 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The electronegativity of magnesium (1.2) is less than that of zinc (1.6). (CH 3 ) 2 Mg therefore has a more polar carbon–metal bond than (CH 3 ) 2 Zn. (c) In this part of the problem two Grignard reagents are compared. Magnesium is the metal in both cases. The difference is the hybridization state of carbon. The sp-hybridized carbon in HC>CMgBr is more electronegative than the sp 3 -hybridized carbon in CH 3 CH 2 MgBr, and HC>CMgBr has a more polar carbon–magnesium bond. 14.17 (a) (b) (c) (d) (e) ( f ) (g) (h) (i) ( j) 1. diethyl ether 2. H 3 O H11001 Propyllithium CH 3 CH 2 CH 2 Li H11001 1-Butanol CH 3 CH 2 CH 2 CH 2 OHHCH O D 2 O DCl Isopropyllithium CH 3 CHCH 3 Li 2-Deuteriopropane CH 3 CHCH 3 D Propylmagnesium bromide CH 3 CH 2 CH 2 MgBr 1-Deuteriopropane CH 3 CH 2 CH 2 D D 2 O DCl H11001 Lithium dipropylcuprate Iodobenzene Propylbenzene (CH 3 CH 2 CH 2 ) 2 CuLi I CH 2 CH 2 CH 3 H11001 Heptane CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 Br 1-BromobutaneLithium dipropylcuprate (CH 3 CH 2 CH 2 ) 2 CuLi H11001 CuI Propyllithium 2CH 3 CH 2 CH 2 Li Lithium dipropylcuprate (CH 3 CH 2 CH 2 ) 2 CuLi H11001 diethyl ether Mg 2-Iodopropane Isopropylmagnesium iodide MgI CH 3 CHCH 3 CH 3 CHCH 3 I H11001 diethyl ether 2Li 2-Iodopropane CH 3 CHCH 3 H11001 LiI Isopropyllithium Li CH 3 CHCH 3 I H11001 diethyl ether Mg 1-Bromopropane CH 3 CH 2 CH 2 Br Propylmagnesium bromide CH 3 CH 2 CH 2 MgBr H11001 diethyl ether 2Li 1-Bromopropane CH 3 CH 2 CH 2 Br H11001 LiBr Propyllithium CH 3 CH 2 CH 2 Li ORGANOMETALLIC COMPOUNDS 347 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (k) (l) (m) (n) (o) (p) (q) (r) 14.18 In the solutions to this problem, the Grignard reagent butylmagnesium bromide is used. In each case the use of butyllithium would be equally satisfactory. 1-Pentene CHCH 2 CH 2 CH 3 H 2 C CHBr 3 KOC(CH 3 ) 3 1,1-Dibromo-2-propylcyclopropane H CH 2 CH 2 CH 3 H H BrBr cis-1-Ethyl-2-hexyl- cyclopropane CH 2 I 2 Zn(Cu), diethyl ether CC H CH 3 CH 2 H (CH 2 ) 5 CH 3 (Z)-3-Decene CH 3 CH 2 (CH 2 ) 5 CH 3 H H trans-1-Heptyl-2- methylcyclopropane CH 2 I 2 Zn(Cu), diethyl ether CC H H 3 C (CH 2 ) 6 CH 3 H (E)-2-Decene H 3 C H H (CH 2 ) 6 CH 3 CH 2 I 2 1-Octene Zn(Cu), diethyl ether 1-Cyclopropylhexane H 2 C CH 2 CH(CH 2 ) 5 CH 3 CH(CH 2 ) 5 CH 3 H 2 C H11001H110012CH 3 CH 2 CH 2 MgBr Propylmagnesium bromide Methyl benzoate C 6 H 5 COCH 3 O 1. diethyl ether 2. H 3 O H11001 4-Phenyl-4-heptanol CH 3 OHC 6 H 5 C(CH 2 CH 2 CH 3 ) 2 OH Methanol H11001 2-Butanone CH 3 CCH 2 CH 3 O 1. diethyl ether 2. H 3 O H11001 2,3-Dimethyl-3-pentanol CH 3 CH CCH 2 CH 3 OH CH 3 CH 3 Isopropyl- magnesium iodide CH 3 CHCH 3 MgI H11001 Isopropyllithium CH 3 CHCH 3 Li 1. diethyl ether 2. H 3 O H11001 Cycloheptanone O 1-Isopropylcycloheptanol OH(CH 3 ) 2 CH H11001 Propylmagnesium bromide CH 3 CH 2 CH 2 MgBr Benzaldehyde CH O 1-Phenyl-1-butanol CHCH 2 CH 2 CH 3 OH 1. diethyl ether 2. H 3 O H11001 348 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (a) 1-Pentanol is a primary alcohol having one more carbon atom than 1-bromobutane. Retrosyn- thetic analysis suggests the reaction of a Grignard reagent with formaldehyde. An appropriate synthetic scheme is (b) 2-Hexanol is a secondary alcohol having two more carbon atoms than 1-bromobutane. As re- vealed by retrosynthetic analysis, it may be prepared by reaction of ethanal (acetaldehyde) with butylmagnesium bromide. The correct reaction sequence is (c) 1-Phenyl-1-pentanol is a secondary alcohol. Disconnection suggests that it can be prepared from butylmagnesium bromide and an aldehyde; benzaldehyde is the appropriate aldehyde. (d) The target molecule 3-methyl-3-heptanol has the structure C OH CH 3 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 1. ether 2. H 3 O H11001 Benzaldehyde CH O 1-Phenyl-1-pentanol CHCH 2 CH 2 CH 2 CH 3 OH CH 3 CH 2 CH 2 CH 2 MgBr Butylmagnesium bromide H11001 H11001 Butylmagnesium halide CH 3 CH 2 CH 2 CH 2 MgX 1-Phenyl-1-pentanol CH 3 CH 2 CH 2 CH 2 CH OH Benzaldehyde CH O CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane Butylmagnesium bromide CH 3 CH 2 CH 2 CH 2 MgBr 2-Hexanol CH 3 CH 2 CH 2 CH 2 CHCH 3 OH Mg diethyl ether 2. H 3 O H11001 1. CH 3 CH O H11001 2-Hexanol Butylmagnesium halide Ethanal (acetaldehyde) CH 3 CH 2 CH 2 CH 2 MgX CH 3 CH O CH 3 CH 2 CH 2 CH 2 CHCH 3 OH CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane Mg diethyl ether Butylmagnesium bromide CH 3 CH 2 CH 2 CH 2 MgBr CH 3 CH 2 CH 2 CH 2 CH 2 OH 1-Pentanol 2. H 3 O H11001 1. HCH O H11001CH 3 CH 2 CH 2 CH 2 1-Pentanol Butylmagnesium halide Formaldehyde CH 3 CH 2 CH 2 CH 2 MgX H 2 CCH 2 OH O ORGANOMETALLIC COMPOUNDS 349 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website By retrosynthetically disconnecting the butyl group from the carbon that bears the hydroxyl substituent, we see that the appropriate starting ketone is 2-butanone. Therefore (e) 1-Butylcyclobutanol is a tertiary alcohol. The appropriate ketone is cyclobutanone. 14.19 In each part of this problem in which there is a change in the carbon skeleton, disconnect the phenyl group of the product to reveal the aldehyde or ketone precursor that reacts with the Grignard reagent derived from bromobenzene. Recall that reaction of a Grignard reagent with formaldehyde (H 2 C?O) yields a primary alcohol, reaction with an aldehyde (other than formaldehyde) yields a secondary alcohol, and reaction with a ketone yields a tertiary alcohol. (a) Conversion of bromobenzene to benzyl alcohol requires formation of the corresponding Grignard reagent and its reaction with formaldehyde. Retrosynthetically, this can be seen as Therefore, (b) The product is a secondary alcohol and is formed by reaction of phenylmagnesium bromide with hexanal. 1-Phenyl-1-hexanolHexanal MgBr Phenylmagnesium bromide H11001 CH 3 CH 2 CH 2 CH 2 CH 2 CH OH CHCH 2 CH 2 CH 2 CH 2 CH 3 O 1. diethyl ether 2. H 3 O H11001 Phenylmagnesium halide Hexanal MgX HC(CH 2 ) 4 CH 3 1-Phenyl-1-hexanol H11001 O CH(CH 2 ) 4 CH 3 OH Br Bromobenzene Phenylmagnesium bromide MgBr Benzyl alcohol CH 2 OH 2. H 3 O H11001 1. HCH O Mg diethyl ether CH 2 OH MgX H 2 CH11001 O H11001 Butylmagnesium bromide CH 3 CH 2 CH 2 CH 2 MgBr 1-Butylcyclobutanol OH CH 2 CH 2 CH 2 CH 31. diethyl ether 2. H 3 O H11001 Cyclobutanone O CH 3 CH 2 CH 2 CH 2 MgBr CH 3 CH 2 CH 2 CH 2 CCH 2 CH 3 CH 3 CCH 2 CH 3 H11001 Butylmagnesium bromide 2-Butanone 3-Methyl-3-heptanol 1. diethyl ether 2. H 3 O H11001 O OH CH 3 H11001 Butylmagnesium halide CH 3 CH 2 CH 2 CH 2 MgXCH 3 CH 2 CH 2 CH 2 CCH 2 CH 3 OH CH 3 2-Butanone CCH 2 CH 3 CH 3 O 350 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The desired product is a secondary alkyl bromide. A reasonable synthesis would be to first prepare the analogous secondary alcohol by reaction of phenylmagnesium bromide with benz- aldehyde, followed by a conversion of the alcohol to the bromide. Retrosynthetically this can be seen as (d) The target molecule is a tertiary alcohol, which requires that phenylmagnesium bromide react with a ketone. By mentally disconnecting the phenyl group from the carbon that bears the hydroxyl group, we see that the appropriate ketone is 4-heptanone. The synthesis is therefore (e) Reaction of phenylmagnesium bromide with cyclooctanone will give the desired tertiary alcohol. ( f ) The 1-phenylcyclooctanol prepared in part (e) of this problem can be subjected to acid- catalyzed dehydration to give 1-phenylcyclooctene. Hydroboration–oxidation of 1-phenyl- cyclooctene gives trans-2-phenylcyclooctanol. H 2 SO 4 , heat 1. B 2 H 6 2. H 2 O 2 , HO H11002 OH 1-Phenylcyclooctanol 1-Phenylcyclooctene H trans-2-Phenylcyclooctanol H H OH O OH H11001 1. diethyl ether 2. H 3 O H11001 1-PhenylcyclooctanolCyclooctanone MgBr Phenylmagnesium bromide 1. diethyl ether 2. H 3 O H11001 4-Phenyl-4-heptanol4-Heptanone MgBr OH Phenylmagnesium bromide H11001 CH 3 CH 2 CH 2 CCH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CCH 2 CH 2 CH 3 O 4-Heptanone O CH 3 CH 2 CH 2 CCH 2 CH 2 CH 3 4-Phenyl-4-heptanol OH CH 3 CH 2 CH 2 CCH 2 CH 2 CH 3 H11001 H11002 1. diethyl ether 2. H 3 O H11001 Benzaldehyde Diphenylmethanol CH OH MgBr Bromodiphenylmethane CH Br Phenylmagnesium bromide H11001 HC O HBr or PBr 3 OBr C 6 H 5 CH H11001C 6 H 5 OH C 6 H 5 CH C 6 H 5 MgX C 6 H 5 CHC 6 H 5 ORGANOMETALLIC COMPOUNDS 351 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 14.20 In these problems the principles of retrosynthetic analysis are applied. The alkyl groups attached to the carbon that bears the hydroxyl group are mentally disconnected to reveal the Grignard reagent and carbonyl compound. (a) (b) (c) (d) 6-Methyl-5-hepten-2-ol CHCH 2 CH 2 (CH 3 ) 2 CCH 3 CH OH (1) (2) H11001 Methylmagnesium halide XMgCH 3 5-Methyl-4-hexenal O CHCH 2 CH 2 CH(CH 3 ) 2 CH11001 Ethanal HCCH 3 O 4-Methyl-3-hexen-1-ylmagnesium halide CHCH 2 CH 2 MgX(CH 3 ) 2 C (2)(1) H11001(CH 3 ) 3 CCH 2 OH 2,2-Dimethyl-1-propanol (CH 3 ) 3 CMgX tert-Butylmagnesium halide HCH O Formaldehyde CH 1-Cyclopropyl-1-(p-anisyl)methanol OCH 3 (1) (2) OH Cyclopropane- carbaldehyde p-Anisylmagnesium halide CH O H11001 XMg OCH 3 Cyclopropyl- magnesium halide p-Anisaldehyde O MgX H11001 HC OCH 3 (2)(1) 5-Methyl-3-hexanol CH 3 CH 2 CH CH 2 CH(CH 3 ) 2 OH (1) (2) (2)(1) H11001 Propanal Isobutylmagnesium halide CH 3 CH 2 CH XMgCH 2 CH(CH 3 ) 2 O Ethylmagnesium halide 3-Methylbutanal H11001CH 3 CH 2 MgX HCCH 2 CH(CH 3 ) 2 O 352 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) 14.21 (a) Meparfynol is a tertiary alcohol and so can be prepared by addition of a carbanionic species to a ketone. Use the same reasoning that applies to the synthesis of alcohols from Grignard reagents. On mentally disconnecting one of the bonds to the carbon bearing the hydroxyl group we see that the addition of acetylide ion to 2-butanone will provide the target molecule. The alternative, reaction of a Grignard reagent with an alkynyl ketone, is not acceptable in this case. The acidic terminal alkyne C@H would transfer a proton to the Grignard reagent. (b) Diphepanol is a tertiary alcohol and so may be prepared by reaction of a Grignard or organo- lithium reagent with a ketone. Retrosynthetically, two possibilities seem reasonable: and N(C 6 H 5 ) 2 CCH (C 6 H 5 ) 2 C OH CH 3 NCHO CH 3 H11002 H11001 N(C 6 H 5 ) 2 CCH C 6 H 5 OH CH 3 NC 6 H 5 CCH O CH 3 H11002 H11001 HC CH 3 CH 2 CCH 3 Sodium acetylide 2-Butanone Meparfynol (94%) CNa H11001 CH 3 CH 2 CC CH OH CH 3 O 1. NH 3 2. H 3 O H11001 O OH CH 3 CH 3 CH 3 CH 2 CC CH 3 CH 2 CCH CHC H11002 OH (1) (2)(3) 4-Ethyl-4-octanol H11001 O 4-Octanone XMg Ethylmagnesium halide H11001O 3-Hexanone XMg Butylmagnesium halide H11001MgX Propylmagnesium halide O 3-Heptanone (3) (2) (1) ORGANOMETALLIC COMPOUNDS 353 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In principle either strategy is acceptable; in practice the one involving phenylmagnesium bromide is used. (c) A reasonable last step in the synthesis of mestranol is the addition of sodium acetylide to the ketone shown. Acetylide anion adds to the carbonyl from the less sterically hindered side. The methyl group shields the top face of the carbonyl, and so acetylide adds from the bottom. 14.22 (a) Sodium acetylide adds to ketones to give tertiary alcohols. (b) The substrate is a ketone, which reacts with ethyllithium to yield a tertiary alcohol. (c) The first step is conversion of bromocyclopentene to the corresponding Grignard reagent, which then reacts with formaldehyde to give a primary alcohol. MgBr 1-Cyclopentenylmagnesium bromide CH 2 OH 1-Cyclopentenylmethanol (53%) Br 1-Bromocyclopentene Mg THF 2. H 3 O H11001 1. HCH O O H11001 CH 3 CH 2 Li 1. diethyl ether 2. H 3 O H11001 OH CH 2 CH 3 2-Ethyl-2-adamantanol (83%)2-Adamantanone H11001 NaC CHC O 1. liquid ammonia 2. H 3 O H11001 C CCH OH 1,1-Diphenyl-2-propyn-1-ol (50%) Benzophenone O H 3 C H 3 C CH 3 O Shields top face 1. NaC 2. H 3 O H11001 CH, NH 3 CCH OH CH 3 O Mestranol N(C 6 H 5 ) 2 CCHC 6 H 5 MgBr OH CH 3 NC 6 H 5 CCH O CH 3 H11001 1. diethyl ether 2. H 3 O H11001 Phenylmagnesium bromide Diphepanol 354 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) The reaction is one in which an alkene is converted to a cyclopropane through use of the Simmons–Smith reagent, iodomethylzinc iodide. (e) Methylene transfer using the Simmons–Smith reagent is stereospecific. The trans arrangement of substituents in the alkene is carried over to the cyclopropane product. ( f ) Lithium dimethylcuprate transfers a methyl group, which substitutes for iodine on the iodoalkene. Even halogens on sp 2 -hybridized carbon are reactive in substitution reactions with lithium dialkylcuprates. (g) The starting material is a p-toluenesulfonate ester. p-Toluenesulfonates are similar to alkyl halides in their reactivity. Substitution occurs; a butyl group from lithium dibutylcuprate replaces p-toluenesulfonate. 14.23 Phenylmagnesium bromide reacts with 4-tert-butylcyclohexanone as shown. The phenyl substituent can be introduced either cis or trans to the tert-butyl group. The two alcohols are therefore stereoisomers (diastereomers). 4-tert-Butylcyclohexanone C(CH 3 ) 3 O 1. C 6 H 5 MgBr, diethyl ether 2. H 3 O H11001 4-tert-Butyl-1-phenylcyclohexanol C 6 H 5 C(CH 3 ) 3 HO H11001 CH 2 CH 2 CH 2 CH 2 CH 3 O 3-Pentylfuran LiCu(CH 2 CH 2 CH 2 CH 3 ) 2 Lithium dibutylcuprate(3-Furyl)methyl p-toluenesulfonate CH 3 CH 2 OS O O O H11001 LiCu(CH 3 ) 2 I CH 3 O 2-Iodo-8-methoxybenzonorbornadiene CH 3 CH 3 O 8-Methoxy-2-methylbenzonorbornadiene (73%) (E)-1-Phenyl-2-butene CH 2 I 2 Zn(Cu), ether CC CH 2 H H CH 3 trans-1-Benzyl-2-methylcyclopropane (50%) CH 2 H CH 3 H CH 2 I 2 Zn(Cu) diethyl ether CH 2 CH CH 2 Allylbenzene Benzylcyclopropane (64%) CH 2 ORGANOMETALLIC COMPOUNDS 355 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Dehydration of either alcohol yields 4-tert-butyl-1-phenylcyclohexene. 14.24 (a) By working through the sequence of reactions that occur when ethyl formate reacts with a Grignard reagent, we can see that this combination leads to secondary alcohols. This is simply because the substituent on the carbonyl carbon of the ester, in this case a hydrogen, is carried through and becomes a substituent on the hydroxyl-bearing carbon of the alcohol. (b) Diethyl carbonate has the potential to react with 3 moles of a Grignard reagent. The tertiary alcohols that are formed by the reaction of diethyl carbonate with Grignard reagents have three identical R groups attached to the carbon that bears the hydroxyl substituent. 14.25 If we use the 2-bromobutane given, along with the information that the reaction occurs with net inversion of configuration, the stereochemical course of the reaction may be written as The phenyl group becomes bonded to carbon from the opposite side of the leaving group. Applying the Cahn–Ingold–Prelog notational system described in Section 7.6 to the product, the order of decreasing precedence is C 6 H 5 H11022 CH 3 CH 2 H11022 CH 3 H11022 H LiCu(C 6 H 5 ) 2 C H CH 3 CH 2 CH 3 C 6 H 5 CH CH 3CH 3 CH 2 Br H11001 CH 3 CH 2 OMgXH11001RMgX Grignard reagent CH 3 CH 2 OCOCH 2 CH 3 O Diethyl carbonate RCOCH 2 CH 3 O Ester RCR O Ketone RCR R OH Tertiary alcohol RMgX 1. RMgX 2. H 3 O H11001 CH 3 CH 2 OMgXH11001 H11001 CH 3 CH 2 OMgXH11001RMgX Grignard reagent HCOCH 2 CH 3 O Ethyl formate RCH O Aldehyde RCHR OH Secondary alcohol 1. RMgX, diethyl ether 2. H 3 O H11001 C(CH 3 ) 3 C 6 H 5 H OH C(CH 3 ) 3 C 6 H 5 H HO H H11001 , heat or C(CH 3 ) 3 C 6 H 5 H 4-tert-Butyl-1-phenylcyclohexene 356 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Orienting the molecule so that the lowest ranked substituent (H) is away from us, we see that the order of decreasing precedence is clockwise. The absolute configuration is R. 14.26 The substrates are secondary alkyl p-toluenesulfonates, and so we expect elimination to compete with substitution. Compound B is formed in both reactions and has the molecular formula of 4-tert-butylcyclohexene. Because the two p-toluenesulfonates are diastereomers, it is likely that compounds A and C, especially since they have the same molecular formula, are also diastereomers. Assuming that the substitution reactions proceed with inversion of configuration, we conclude that the products are as shown. Inversion of configuration is borne out by the fact given in the problem that compound C is more stable than compound A. Both substituents are equatorial in C; the methyl group is axial in A. 14.27 We are told in the statement of the problem that the first step is conversion of the alcohol to the cor- responding p-toluenesulfonate. This step is carried out as follows: Alkyl p-toluenesulfonates react with lithium dialkylcuprates in the same way that alkyl halides do. Treatment of the preceding p-toluenesulfonate with lithium dibutylcuprate gives the desired compound. O OTs 3,8-Epoxyundecyl p-toluenesulfonate O 4,9-Epoxypentadecane LiCu(CH 2 CH 2 CH 2 CH 3 ) 2 p-Toluenesulfonyl chloride (TsCl) CH 3 SO 2 Cl pyridine H11001 3,8-Epoxy-1-undecanol O OH 3,8-Epoxyundecyl p-toluenesulfonate O OTs (CH 3 ) 3 C CH 3 cis-4-tert-Butylcyclohexyl p-toluenesulfonate trans-1-tert-Butyl-4-methylcyclohexane (compound C, C 11 H 22 ) Compound B LiCu(CH 3 ) 2 H11001 (CH 3 ) 3 C OTs (CH 3 ) 3 C (CH 3 ) 3 C OTs trans-4-tert-Butylcyclohexyl p-toluenesulfonate cis-1-tert-Butyl-4-methylcyclohexane (compound A, C 11 H 22 ) 4-tert-Butylcyclohexene (compound B, C 10 H 18 ) LiCu(CH 3 ) 2 (CH 3 ) 3 C CH 3 H11001 (CH 3 ) 3 C C 6 H 5 H 3 C CH 2 CH 3 ORGANOMETALLIC COMPOUNDS 357 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website As actually performed, a 91% yield of the desired product was obtained in the reaction of the p-toluenesulfonate with lithium dibutylcuprate. 14.28 (a) The desired 1-deuteriobutane can be obtained by reaction of D 2 O with butyllithium or butyl- magnesium bromide. Preparation of the organometallic compounds requires an alkyl bromide, which is synthesized from the corresponding alcohol. (b) In a sequence identical to that of part (a) in design but using 2-butanol as the starting mate- rial, 2-deuteriobutane may be prepared. An analogous procedure involving sec-butyllithium in place of the Grignard reagent can be used. 14.29 All the protons in benzene are equivalent. In diphenylmethane and in triphenylmethane, protons are attached either to the sp 2 -hybridized carbons of the ring or to the sp 3 -hybridized carbon between the rings. The large difference in acidity between diphenylmethane and benzene suggests that it is not a ring proton that is lost on ionization in diphenylmethane but rather a proton from the methylene group. The anion produced is stabilized by resonance. It is a benzylic carbanion. C H11002 H etc. C H11002 H (C 6 H 5 ) 2 CH H H11001 H11001 H11002 (C 6 H 5 )CH 2 Diphenylmethane PBr 3 Mg ether D 2 O 2-Deuteriobutane CH 3 CHCH 2 CH 3 D sec-Butylmagnesium bromide CH 3 CHCH 2 CH 3 MgBr 2-Bromobutane CH 3 CHCH 2 CH 3 Br 2-Butanol CH 3 CHCH 2 CH 3 OH CH 3 CH 2 CH 2 CH 2 Li Butyllithium CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane Li ether CH 3 CH 2 CH 2 CH 2 MgBr Butylmagnesium bromide Mg ether CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CH 2 Br 1-Butanol 1-Bromobutane PBr 3 or HBr H11001CH 3 CH 2 CH 2 CH 2 Li Butyllithium CH 3 CH 2 CH 2 CH 2 D 1-Deuteriobutane Deuterium oxide D 2 O H11001 Butylmagnesium bromide CH 3 CH 2 CH 2 CH 2 MgBr D 2 O or 358 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Both rings are involved in delocalizing the negative charge. The anion from triphenylmethane is stabilized by resonance involving all three rings. Delocalization of the negative charge by resonance is not possible in the anion of benzene. The pair of unshared electrons in phenyl anion is in an sp 2 hybrid orbital that does not interact with the H9266 system. 14.30 The titanium-containing compound is a metallocene. (It has cyclopentadienyl rings as ligands.) With an atomic number of 22, titanium has an electron configuration of [Ar]4s 2 3d 2 . As the following accounting shows, this titanium complex is 2 electrons short of satisfying the 18-electron rule. 1,3-Butadiene(tricarbonyl)iron satisfies the 18-electron rule. The electron configuration of iron is [Ar]4s 2 3d 6 . 14.31 Using 1-decene as an example, we can see from the following schematic that the growing polymer will incorporate a C 8 side chain at every point where 1-decene replaces ethylene. 14.32–14.36 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. 5 Ethylene molecules H11001 1 1-decene molecule Section of linear low-density polyethylene Fe: 8 electrons 1,3-Butadiene ligand: 4 electrons Three CO ligands: 6 electrons Total: 18 electrons Fe CO CO OC HH HH Ti: 4 electrons Two cyclopentadienyl rings: 10 electrons Two chlorine atoms: 2 electrons Total: 16 electrons Ti Cl Cl Not delocalized into H9266 system C H11002 C H11002 etc. ORGANOMETALLIC COMPOUNDS 359 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SELF-TEST PART A A-1. Give a method for the preparation of each of the following organometallic compounds, using appropriate starting materials: (a) Cyclohexyllithium (b) tert-Butylmagnesium bromide (c) Lithium dibenzylcuprate A-2. Give the structure of the product obtained by each of the following reaction schemes: (a)(d) (b e) (c) A-3. Give two combinations of an organometallic reagent and a carbonyl compound that may be used for the preparation of each of the following: (a) (b) A-4. Gives the structure of the organometallic reagent necessary to carry out each of the following: (a) (b) (c) A-5. Compounds A through F are some common organic solvents. Which ones would be suitable for use in the preparation of a Grignard reagent? For those that are not suitable, give a brief reason why. CH 3 CH 2 CH 2 CH 2 OCH 2 CH 2 CH 2 CH 3 A CH 3 OCH 2 CH 2 OCH 3 B HOCH 2 CH 2 OH C D CH 3 COCH 2 CH 3 O F CH 3 COH O E O ? CH 3 CH 3 CH 3 CH 3 1. ? 2. H 3 O H11001 C 6 H 5 CH 2 CO 2 CH 3 (CH 3 ) 2 CHCCH(CH 3 ) 2 OH CH 2 C 6 H 5 ? CH 3 I CH 3 CH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CHCH 2 CH 2 CH 2 CH 3 OH C 6 H 5 CHC(CH 3 ) 3 OH H 3 CBr ? 1. Mg 3. H 3 O H11001 2. H 2 C O KOC(CH 3 ) 3 (CH 3 ) 3 COH ? H11001 CHBr 3 (CH 3 ) 2 CHCH 2 Li D 2 O ? O 1. CH 3 CH 2 Li 2. H 3 O H11001 ? CH 3 CO 2 CH 2 CH 3 ? 1. 2C 6 H 5 MgBr 2. H 3 O H11001 360 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-6. Show by a series of chemical equations how you would prepare octane from 1-butanol as the source of all its carbon atoms. A-7. Synthesis of the following alcohol is possible by three schemes using Grignard reagents. Give the reagents necessary to carry out each of them. A-8. Using ethylbenzene and any other necessary organic or inorganic reagents, outline a synthe- sis of 3-phenyl-2-butanol. A-9. Give the structure of the final product of each of the following sequences of reactions. (a) (b) (c) PART B B-1. Which (if any) of the following would not be classified as an organometallic substance? (a) Triethylaluminum (b) Ethylmagnesium iodide (c) Potassium tert-butoxide (d) None of these (all are organometallic compounds) B-2. Rank the following species in order of increasing polarity of the carbon–metal bond (least B most polar): (a)3H11021 1 H11021 2(b)2H11021 1 H11021 3(c)1H11021 3 H11021 2(d)2H11021 3 H11021 1 B-3. Which sequence of reagents would carry out the following conversion? (a)H 2 SO 4 , heat; then B 2 D 6 ; then H 2 O 2 , HO H11002 (b)H 2 SO 4 , heat; then D 2 , Pt (c)CH 3 MgBr; then D 2 O (d) HBr; then Mg; then D 2 O B-4. Arrange the following intermediates in order of decreasing basicity (strongest B weakest): (a)2H11022 1 H11022 4 H11022 3(c)3H11022 4 H11022 1 H11022 2 (b)4H11022 1 H11022 2 H11022 3(d)3H11022 2 H11022 4 H11022 1 H 2 C CHNa HC CNaCH 3 CH 2 Na CH 3 CH 2 ONa 12 4 CH 3 CH 2 CHCH 3 OH CH 3 CH 2 CHCH 3 D ? CH 3 CH 2 MgCl 1 CH 3 CH 2 Na 2 (CH 3 CH 2 ) 3 Al 3 NaNH 2 H 3 O H11001 ?CH 3 CCH O HCl Mg H 3 O H11001 ?1-Butene CH 3 CH O Br 2 FeBr 3 Mg CH 3 CCH 2 CH 3 O H 3 O H11001 ? (CH 3 ) 2 CHC(CH 3 ) 2 OH ORGANOMETALLIC COMPOUNDS 361 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-5. Which, if any, of the following pairs of reagents could be used to prepare 2-phenyl- 2-butanol? (a) (b) (c) (d) (e) None of these combinations would be effective. B-6. Which of the following reagents would be effective for the following reaction sequence? (a) Sodium ethoxide (c) Butyllithium (b) Magnesium in diethyl ether (d) Potassium hydroxide B-7. What is the product of the following reaction? B-8. Which of the following combinations of reagents will yield a chiral product after hydrolysis in aqueous acid? (d) Both (a) and (c) O H11001(b) CH 3 MgBr H11001(c) 2CH 3 MgBrCH 3 CH 2 COCH 3 O O H H11001(a) CH 3 MgBr HOCH 2 CH 2 CH 2 CH 2 CHOCH 3 (d) CH 3 HOCH 2 CH 2 CH 2 CH 2 COH(b) CH 3 CH 3 CH 3 OCH 2 CH 2 CH 2 CH 2 CHCH 3 (c) OH HOCHCH 2 CH 2 CH 2 CHOH(a) CH 3 CH 3 O O H11001 2CH 3 MgBr 1. diethyl ether 2. H 3 O H11001 C 6 H 5 CCH C 6 H 5 C CCH 2 OH 3. H 3 O H11001 1. ? 2. H 2 C O C 6 H 5 MgCl CH 3 CCH 2 CH 2 CH 3 H11001 O CH 3 MgI C 6 H 5 CH 2 CCH 3 H11001 O CH 3 CH 2 MgBr C 6 H 5 CH 2 CHH11001 O CH 3 CH 2 MgBr C 6 H 5 CH 2 CCH 3 H11001 O CH 3 CH 2 CC 6 H 5 CH 3 OH 2-Phenyl-2-butanol 362 ORGANOMETALLIC COMPOUNDS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-9. Which sequence of steps describes the best synthesis of 2-phenylpropene? (a) Benzene H11001 2-chloropropene, AlCl 3 (b) Benzene H11001 propene, H 2 SO 4 (c) 1. Benzaldehyde (C 6 H 5 CH?O) H11001 CH 3 CH 2 MgBr, diethyl ether 2. H 3 O H11001 3. H 2 SO 4 , heat (d) 1. Bromobenzene H11001 Mg, diethyl ether 2. Propanal (CH 3 CH 2 CH?O) 3. H 3 O H11001 4. H 2 SO 4 , heat (e) 1. Bromobenzene H11001 Mg, diethyl ether 2. Acetone [(CH 3 ) 2 C?O)] 3. H 3 O H11001 4. H 2 SO 4 , heat B-10. What sequence of steps represents the best synthesis of 4-heptanol (CH 3 CH 2 CH 2 ) 2 CHOH? (a)CH 3 CH 2 CH 2 MgBr (2 mol) H11001 formaldehyde (CH 2 ?O) in diethyl ether followed by H 3 O H11001 (b)CH 3 CH 2 CH 2 MgBr H11001 butanal (CH 3 CH 2 CH 2 CH?O) in diethyl ether followed by H 3 O H11001 (c)CH 3 CH 2 CH 2 CH 2 MgBr H11001 acetone [(CH 3 ) 2 C?O] in diethyl ether followed by H 3 O H11001 (d) (CH 3 CH 2 CH 2 ) 2 CHMgBr H11001 formaldehyde (CH 2 ?O) in diethyl ether followed by H 3 O H11001 (e)CH 3 CH 2 CH 2 MgBr H11001 ethyl acetate ( ) in diethyl ether followed by H 3 O H11001 B-11. All of the following compounds react with ethylmagnesium bromide. Alcohols are formed from four of the compounds. Which one does not give an alcohol? B-12. Give the major product of the following reaction: (a) cis-1-Ethyl-2-methylcyclopropane (b) trans-1-Ethyl-2-methylcyclopropane (c) 1-Ethyl-1-methylcyclopropane (d) An equimolar mixture of products (a) and (b) (E)-2-pentene ? CH 2 I 2 , Zn(Cu) CH(a) O COH(b) O COCH 3 (c) O CH 2 OCCH 3 (e) O CCH 3 (d) O CH 3 COCH 2 CH 3 O ORGANOMETALLIC COMPOUNDS 363 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website