1559rch21373-38810/18/0518:42Pag0378 EQA 378.chapter 21 AMINES AND THEIR DERIVATIVES:FUNCTIONAL GROUPS CONTAINING NITROGEN 一一4一 :NH2 Guanidine Resonance stabilization of conjugate acids enhances base strengths canes are not po method would work well for a five-or six-membered ring.(i)Not at all.Reaction shown is for a benzene,not a cyclohexane compound.(j)Well. 35.(CH,CH)O solvent and H'.HO work-up are understood for Grignard and LiAlH reactions. (a)1.NaN3.DMSO.2.LiAH(b)You need to be devious.Add a carbon and then take it out again (CH.).CCOH 0 0 (CH).Ho A,(ma .CH.CH. es imine Not howCHddformaldeyde,rection (e Sane mpk可e0snwh○一r d的ing Be〈○一N.ne15a will slowly hydrogenate the ring. 36.CH.CH,NH2.(CH,CH2)2NH.(CH,CH2),N.and(CH,CH2)4N" 37Make pseudoephedrine from phenylpropanolamine using reductive amination.As in Problem 35(d).use one equivalent of HC=O to avoid dimethylation. ture rr'nh NaOH.H2O.4.CH;CHO.H*.5.NaBH;CN,CH;CH2OH 39(a)Lots of ways!Bromobutane+azide,then LiAlH reduction.r bromopropane+cyanide,then LiAlHor bromobutane+phthalimide salt.then hydrolysis
Resonance stabilization of conjugate acids enhances base strengths. 34. (a) Not at all. This process adds a carbon (the CN group), making 1-pentanamine. (b) Not at all. SN2 reactions with tertiary haloalkanes are not possible. (c) Well. (d) Poorly. Further alkylation can occur, making (e) Poorly. The haloalkane, although primary, is highly branched and will not react well in SN2 reactions. (f), (g) Well. (h) Poorly. Four-membered rings are strained and difficult to form. The method would work well for a five- or six-membered ring. (i) Not at all. Reaction shown is for a benzene, not a cyclohexane compound. (j) Well. 35. (CH3CH2)2O solvent and H, H2O work-up are understood for Grignard and LiAlH4 reactions. (a) 1. NaN3, DMSO, 2. LiAlH4. (b) You need to be devious. Add a carbon and then take it out again! (d) 1. NaN3, 2. LiAlH4 (makes primary amine), 3. H2CPO [makes imine; use one equivalent only, to avoid dimethylation (Problem 50)], 4. NaBH3CN, CH3CH2OH (completes reductive amination). Note how CH3 group is introduced as formaldehyde, with a subsequent reduction step. (e) Same as (b). (h) No simple way to improve the situation. (i) Start with and do reactions shown, making The H2, Pt (Section 15-2) will slowly hydrogenate the ring. 36. CH3CH2NH2, (CH3CH2)2NH, (CH3CH2)3N, and (CH3CH2)4N. 37. Make pseudoephedrine from phenylpropanolamine using reductive amination. As suggested in Problem 35(d), use one equivalent of H2CPO to avoid dimethylation. 38. Secondary (general structure RRNH). (a) 1. CH3CH2NH2, H, 2. NaBH3CN, CH3CH2OH; (b) 1. NaN3, DMF, 2. LiAlH4, THF, 3. CH3CHO, H, 4. NaBH3CN, CH3CH2OH; (c) 1. SOCl2, 2. NH3, 3. Br2, NaOH, H2O, 4. CH3CHO, H, 5. NaBH3CN, CH3CH2OH 39. (a) Lots of ways! Bromobutane azide, then LiAlH4 reduction, or bromopropane cyanide, then LiAlH4, or bromobutane phthalimide salt, then hydrolysis. H2C O, NaBH3CN RNH2 RNHCH3 Br NH2. Br 1. Mg 2. CO2 1. SOCl2 2. NH3 (CH3)3CCl (CH3)3CCOH O Br2, NaOH, H2O (Hofmann rearrangement) (CH3)3CCNH2 (CH3)3CNH2 O ( ( NCH3. 2 NH2 C NH2 NH2 NH2 C NH2 NH2 NH2 C NH2 NH2 NH2 C NH2 NH2 Guanidine 378 • Chapter 21 AMINES AND THEIR DERIVATIVES: FUNCTIONAL GROUPS CONTAINING NITROGEN 1559T_ch21_373-388 10/18/05 18:42 Page 378
1559T_ch21_373-38810/18/0518:42Pa9e379 ⊕ EQA Solutions to Problems.379 (b)First make methar thalimide sat(similar to CHs (e)First cycle: CH2=CH(CH2)N(CH)2CH,CH-CH(CH2)2N(CH): N(CH)z CH CHCH-CH-CH Second cvclez CH2=CHCH:CH-CH CH,CH-CHCH-CH2 CH=CH, (e)First cycl O CH Second eycle ◇C◇ (CH3)2 (CH3) CH1 (CH3) Third 41.The reaction takes place under acidic conditions. CH.NH.+.COH CH.NH-CH-OH CH.+ product Enol of aldehyd
Solutions to Problems • 379 (b) First make methanamine from iodomethane and either azide or phthalimide salt (similar to butanamine syntheses in (a), and then carry out reductive amination using the methanamine and butanal: (c) Make butanamine (a) and then go with double reductive amination using excess formaldehyde and NaBH3CN. Best way to make any N,N-dimethyl tertiary amines (see Problem 50). 40. (a) (b) (c) First cycle: CH2PCH(CH2)3N(CH3)2 CH3CHPCH(CH2)2N(CH3)2 N(CH3)2 A CH3CHCH2CHPCH2 Second cycle: CH2PCHCH2CHPCH2 CH3CHPCHCHPCH2 (d) (e) First cycle: Second cycle: Third cycle: 41. The reaction takes place under acidic conditions. CH3NH CH2 H3C H3C OH H C C product H Enol of aldehyde CH3NH2 CH3NH2 CH2 OH CH3NH CH2 OH2 H2C OH (CH3)2 N N N N (CH3)2 (CH3)2 (CH3)2 N (CH3)2 CH3 N CH3 N CH3 N CH3 N N(CH3)2 CH CH2 CH2 CH3 and CH ( CHCH Z and E) 3 CH3I CH3NHCH2CH2CH2CH3 1. CH3CH2CH2CHO 2. NaBH3CN, pH 3 1. N3 2. LiAlH4 CH3NH2 1559T_ch21_373-388 10/18/05 18:42 Page 379