CUNY College of Staten Island Organic Chemistry Carbon Bonded Question

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7.1 Classify the alkyl halide as 1°, 2°, or 3° by counting the number of carbons bonded directly to the carbon bonded to the halogen. C bonded to 1 C 1° alkyl halide a. J CH3CH 2 CH 2 CH 2 CH 2 -Br b.(k C bonded to 2 C's 2° alkyl halide ?H3 l c. CH3-C-CHCH3 I C bonded to 3 C's 3° alkyl halide I CH 3 Cl \ } d. ~ C bonded to 3 C's 3° alkyl halide 7.2 Use the directions from Answer 7 .1. 3°, neither 2°, neither 2°, neither ! J a, b. ~'::a' Cl 3o, njther telfairine Br~ vinyl ! Cl Br - 1 °, neither ,,,Cl-3°, allylic Br l 2°, neither halomon Cl -vinyl 7.3 Draw a compound of molecular formula C6 H13 Br to fit each description. Br a. Br~ b. 1° alkyl halide one stereogenic center C. 2° alkyl halide two stereogenic centers 3° alkyl halide no stereogenic centers 7.4 To name a compound with the IUPAC system: [I] Name the parent chain by finding the longest carbon chain. [2] Number the chain so the first substituent gets the lower number. Then name and number all substituents, giving like substituents a prefix (di, tri, etc.). To name the halogen substituent, change the -ine ending to -o. [3] Combine all parts, alphabetizing substituents, and ignoring all prefixes except iso. a. (CH3hCHCH(Cl)CH2CH 3 l re-draw [1] ~ 2-methyl [2)~ f [3] 3-chloro-2-methylpentane 2 Cl-3-chloro Cl 5 carbon alkane = pentane b. [1] 1tz>, e> (I) C: w Reaction coordinate 7,23 To draw the products of SN2 reactions, replace the leaving group by the nucleophile, and then draw the stereochemistry with inversion at the stereogenic center. CH3CH2 q D CH CH ': a. C-Br + -OCH 2CH 3 I :_/ - CH 3CH 20-C \ H 2 3 H 7.24 Increasing the number of R groups increases crowding of the transition state and decreases the rate of an SN2 reaction. Cl a. or 2° alkyl halide ~Cl I 1° alkyl halide faster reaction b . Q-sr or 2° alkyl halide faster reaction 3° alkyl halide 7.25 loss of --- a proton IV \__/'Br ,,,Q O N H CH3 nicotine 7.26 ln a first-order reaction, the rate changes with any change in [RX]. The rate is independent of any change in [:Nu-]. a. [RX] is tripled, and [:Nu-] stays the same: rate triples. b. Both [RX] and [:Nu-] are tripled: rate triples. c. [RX] is halved, and [:Nu-] stays the same: rate halved. d. [RX] is halved, and [:Nu-] is doubled: rate halved. 7 27 [n Sr,1 reactions,_ racemization_ always occurs at a stereogenic center. Draw two products, · with the two possible configurations at the stereogenic center. leaving group ?H3 l nucleophile l / C,,.,Br a. (CH3)2CH \H2CH3 CH 3 H20 I + (CH3hCH ,......-C\ 'OH + HBr CH 2CH 3 enantiomers nucleophile . l CH3Coo- H,° ' C H , 3 b CH 3CH{ ,,,I dlastereomers leaving group ?.28 Carbocations are classified by the number of R groups bonded to the carbon: oR groups= methyl, 1 R group= 1°, 2 R groups= 2°, and 3 R groups= 3°. + a. + b. (CH3l3CCH2 2 R groups 2° carbocation 1 R group 1° carbocation c.(y d.(X 3 R groups 3° carbocation 2 R groups 2° carbocation 7.29 For carbocations: Increasing number of R groups= Increasing stability. + CH 3-C-CH3 I 1° carbocation least stable CH 3 2° carbocation intermediate stability 3° carbocation most stable 7.30 For carbocations: Increasing number of R groups= Increasing stability. + 1° carbocation least stable 2° carbocation intermediate stability 3° carbocation most stable 7.31 The rate of an SNl reaction increases with increasing alkyl substitution. o- CH3Br a. (CH 3lJCBr or (CH 3)JCCH 2Br 3° alkyl halide 1° alkyl halide faster SN1 reaction slower SN1 reaction b. or 3° alkyl halide faster SN1 reaction 2° alkyl halide slower SN1 reaction 7·32 • For methyl and 1° alkyl halides, only SN2 will occur. • For 2° alkyl halides, SNI and SN2 will occur. • For 3° alkyl halides, only SNI will occur. CH 3 H I I a. CH 3-C-C - Br I / ' y Br b. ~ I CH 3 CH 3 c. 0 - Br Br / \ 3° alkyl halide SN1 2° alkyl halide SN1 and SN2 1° alkyl halide SN2 2° alkyl halide SN1 and SN2 d. 7.33 • Draw the product of nucleophilic substitution for each reaction. • For methyl and 1° alkyl halides, only SN2 will occur. • For 2° alkyl halides, SNI and SN2 will occur and other factors determine which mechanism operates. • For 3° alkyl halides, only SNI will occur. Strong nucleophile favorsi5N2. \ / a. ~ c. CH 30H Cl - ~ OCH3 U ' CH,CH,0- U OCH,CH: I + HCI 3° alkyl halide only SN1 2° alkyl halide Both SN 1 and SN2 are possible. Weak nucleophile favors SN1. ( " rsr b.v ( " rsH+ V j Br- CH30 H d.~ Br 1° alkyl halide only SN2 I + HBr OCH 3 2° alkyl halide Both SN1 and SN2 are possible. 7.34 First decide whether the reaction will proceed via an SNI or SN2 mechanism. Then draw the products with stereochemistry. a. + ~ H Br 2° alkyl halide SN1 and SN2 H20 l + HBr - Weak nucleophile favors SN1. HOH HOH enantiomers HC=C)("' b. ~Cl H D 1° alkyl halide SN2 only SN1 = racemization at the stereogenic C D H + CI- SN2 = inversion at the stereogenic C . , . , , ..... , ,~vu 7.3 5 c ompounds wi a. cH th 3CH 2CH2CI l 't U \, l t;U t,JIIII IV V\..l"-,1 -.J u , ...... ,- I hi larz~;t~:or ei Q) CH~ fH 3 C=C I H \ H cis-2-butene C: w + H2 -- CH3CH2CH,CH3 111-f' = -120 kJ/mol --- ______ ii~~er in energy butane smaller 2 H' for cis-2-butene lower in energy, more stable 8.35 a. CH CH=CHCH2CH2CH (CH3}i + 3 (loss of P1 H) monosubslituted (loss of P2 H) major product disubstituted Oo~ DBU b. c. only product + CH 3CH 2C(CH 3)=C(CH 3)CH2CH2CH3 cH CH 2CH(CH 3)C(CH3)=CHCH2CH3 3 (loss of P2 H) (loss of P1 H) major product telrasubstituted + lriwbst'"I•~ (loss of P3 H) disubstituted p d. Cl - OC(CH3)J only product P2 e. -OH ~ I CH3CH2CH2CH2CH=CHCH3 P1 f. + (loss of P2 H) monosubstituted (loss of P1H) major product disubstituted +t"'' - OH 1 JJ fl + (loss of P1 H) disubstituted (loss of P2 H) major product trisubstituted 8.36 To give only one alkene as the product of elimination, the alkyl halide must have either: • only one pcarbon with a hydrogen atom • all identical pcarbons, so the resulting elimination products are identical CH 3 9H3 a. CH3-9-9-H H b. Cl 0---{I H ' I c=cI CH3 \ H - - Q-cH=CH 2 9H3~ CH3 - C-C - H I Cl I H o - - rCI Alkyl Halides and Elimination Reactions 8-17 c cl - 00 -- ~ Cl s.37 Draw the products of the E2 reaction and compare the number of C's bonded to the C=C. ~(!) 1 Br ~@] P1 ~2 I½ + (CH3)iCHCH=CHCH 3 a major product trisubstituted disubstituted a (CH3)iCHCH=CHCH 3 P1 major product disubstituted + monosubstituted A yields a t~isubst_ituted alkene as the ma)or product and a disubstituted alkene as minor product. 8 yields a d1subst1tuted alkene as the maJor product and a monosubstituted alkene as minor product. Since the major and minor products formed from A have more alkyl groups on the C=C (making them more stable) than those formed from B, A reacts faster in an elimination reaction. 8.38 a. Mechanism: ~ . I-..! 0 by-products .. 81/-=\?C(CH3)J HJ (CH3)JCOH b. Rate= k[R-Br]LOC(CH3)3] [ l] Solvent changed to DMF (polar aprotic) = rate increases [2] tOC(CH3)J] decreased = rate decreases [3] Base changed to -oH = rate decreases (weaker base) [4] Halide changed to 2° = rate increases (More substituted RX reacts faster.) [5] Leaving group changed to i- = rate increases (better leaving group) 8.39 [>-Q 2° halide Cl CH3CH2CH=CH 2 -OC(CH 3)J ~I Cf + (cis and trans) E1 1° halide d. CH 3CH=CHCH 3 weak base 2° halide CH3CH2CH=CH2 + (cis and trans) E2 2° halide C. CH 3CH=CHCH 3 strong base Br strong base E2 OH strong base E2 CrCHCH2CH3 + CH 3 q-CH2CH 2CH3 CH 3 >-0 >-0 + >--0 + q- CH2CH 2CH 3 CH 3 g,4J The order of reactivity is the same for both E2 and EI : I < 20 < 3o. O CH3 b--c1 Q s, QcH3 CH3 Cl 2° halide 3° halide 3° halide+ better leaving group Increasing reactivity in E1 and E2 s.44 Cl ('r(CH3 a. v H2O b. , CI 3° halide-faster reaction -OH C. (CH3)JCCI I c=rBr -OH l s,,oog base-E2 DMSO C) cis-cyclodecene bromocyclodecane H2O CH3 Cl / 3° halide-faster reaction 8.45 (CH3)JCCI 1 polar aprolic solvent faster reaction In a ten-membered ring, the cis isomer is more stable and, therefore, the preferred elimination product. The trans isomer is less stable because strain is introduced when two ends of the double bond are connected in a trans arrangement in this medium-sized ring. 8.46 With the strong base -0CH2CH3, the mechanism is E2, whereas with dilute base, the mechanism is E1. E2 elimination proceeds with anti periplanar arrangement of H and X. In the EI mechanism there is no requirement for elimination to proceed with anti periplanar geometry. In this case the major product is always the most stable, more substituted alkene. Thus, C is the major product under El conditions. (In Chapter 9, we will learn that additional elimination products may form in the El reaction due to carbocation rearrangement.) OCH2CH3 A strong base E2 rO B .. . Since this is an E2 mechanism, dehydrohalogenalion needs an anti periplanar H to form the double bond. There is only one H trans to Cl, so the disubstituted alkene B must form . CH 3OH + weak base A E1 B disubstituted alkene C trisubslituted alkene more stable Chapter B-20 8.47 + CH 3CH=CHCHi .,p'v 2-butene 1-butene from loss of from loss of P2H P, H Na' -ocH2CH3 19% 81 % K'-OC(CH 3lJ 33% 67% The H's on the Cl-1 2 group of the ~2carbon are more sterically hindered than the ~~son the C~ i group of the~, carbon . Since K•-oqCHi)i is a much bulkier base than Na OCH2CH 3, 1115 easier to remove the more accessible H on ~1, giving it a higher percentage of !-butene. 8 48 · Han_d Br must be anti during the E2 elimination . Rotate if necessary to make them anti ; then eliminate. CH 3 a. C5H5~ C6H5 H - f y cH,cH, E2 CH3 CH 3 CH,CH3 CH 3 Br b. ",Q)'"' C5H5 $ CH3 rotate CH3 CH 2CH 3 C, Hs CH 2CH 3 CH, CH3 CH, CH 3 C, Hs CH2CH 3 H C6H5 ~ '"'~" Br CH, H E2 CH 3 C6H5 C, c , Hs rotate ~", C H E2 CH,CH 3 CH 3 CH 3 CH 3 8.49 a. ('y Cl two chair H H axial _C',Ql)f9_1'!1)~~9~~-ma V. •,,CH H 3 : CH(CH 3), CIH (CH 3>,CH CH (CH 3),CH A3H H_ B Choose this conformation. axial Cl H (CH,l,CH H 0 one axial H H (CH3),CH ~ CH 3 H 0.. : only product 'CH 3 CH(CH 3)2 b. ex Cl two chair H _c_~~f_or~~tio~~ H _ CH, ~ C CH(CH,l:, (CH3),CH A H (CH, )2CCHH rY~i, H Choose this conformation . - axial Cl H H B Pi H (CH3)2CH ~ CH3~ H I!) (CH,l:,CH ~ H axial CH P1 H n H (CH3),CH ~ i : : \ CH 3~ (loss of P, H) two axial H's I re-draw I re-draw Q : CH 3 CH(CH3)2 U CH3 CH(CH 3l, (loss of P2 H) D 60 D~ H H D l This conformation reacts. axial Cl H (loss of P2 H) major product trisubstituted enantiomers D o--fi-o H = Q D (loss of P1 H) (loss of P, D) H p~l o~Pi I IL o~ HffZH ...,___----1\ -~O.,_,_ H_ _ _ __ _ _ This conformation reacts . axial Cl H- - f i - H D (loss of P1 D) enantiomers Q =0 Chapter 8-22 8.50 enantiomers a. enantiomers 9H3 •.H 6 CJH 3- ~~(CH C .•c-c•:--CH3 3 CH 3CH{ I CH3 CJ \9J B l A 2-chloro-3-methylpentane ll -HCI H and Cl are arranged anti in each stereoisomer, for anti periplanar elimination . CH 3CH~ H I 1 \ CH2CH3 \ I C=C CH3 CH3 c=c CH3 -HCI CH3 ----, identical b. Two different alkenes are fom,ed as products. . .d • I products A and B . . (A and B) c. Tl1e pro ducts are diastereomers: Two enant1omers . give .I ent1ca lk Thus· are diastereomers of C and D. Each pair of enantiomers gives a sing 1e a ene. ' diastereomers give diastereomeric products. 8.51 NaNH2 a. (2 equiv) b. c. 9H3 CH3CH2-9-9HcH2Br CH 3 Br Cl CH3 I CH3-9-CH2CH 3 Cl d. (2 equiv) 9H3 CH 3CH 2-9-C=CH (excess) i\.ti-o Cl Cl ;\__C=C-0 (2 equiv) \_F 8.52 H H Br I I CH3 - 9-CH2CH3 or Br CH3 b. CH3-C - C=CH I CH3 CH3 Br I I CH 3-9-9-CH3 CH3 Br or I CH3-9 - 9-CH3 Br Br 9H3~r CH3-9-cH-CH2Br or 9H3 CH3-9-cH2CHBr2 CH3 CH 3 0-cH-cH-0 Br Br or I I \ /; IJ.53 H I H I CH3 - 9 - 9 - CH 3 CH, - C =' C - CH 3 Br Br 11 spsp 2.3-dibromobutane CH 3 - CH =C=CH 2 CH 2=C H - C H =CH 2 1 C sp B A S.54 Use the "S ummary chart on the four mechanisms: SN I, SN2, EI , or E2" on p. 8- 2 to answer the quest10ns. a. Both SN I and EI involve carbocation intermediates. b. Both SN I and E 1 have two steps. c. SN I, SN2, E 1, and E2 a ll have increased reaction rates with better leav ing groups. d. Both SN2 and E2 have increased rates when changing from CH 30H (a protic solvent) to (CH 3 )zSO (an aprotic solvent). e. Jn SN I and EI reactions, the rate depends on on ly the alkyl halide concentration . f Both SN2 and E2 are concerted reactions. g. cH 3 CH 2 B~ and NaO~ react by an SN2 mechanism. h. Racemizat1on occurs in SN 1 reactions. _ i. In SNI, E I , and E2 mechanisms, 3° alkyl halides react faster than 1° or 2° hahdes. · E2 and SN2 reactions follow second-order rate equations. J· 8.55 a. - OC(CHah Br sterically 11° halide hindered base SN2 or E2 b. ~ - OCH2CH3 I E2 ~ OCH2CH3 strong nucleophile 1 ° halide SN2 or E2 Cl I c. CH3-?-CH3 Cl dihalide d. ( ) Br 1° halide SN2 or E2 HC=C-CH3 equiv) strong base (2 DBU sterically hindered base er e. ( XCH2CH 3 - OC(CH3}J Br 2° halide SN1, SN2, E1, E2 sterically hindered base E2 ( YCH2CH3 V + major product E2 Chapter 8-2 4 f. r"r(Br V., CH2CH3 weak base 3• halide no SN2 g, E1 products 2 NaNH2 (CH3)iCH-9HcH 2Br V ('YCHCH3 H OCH2C \ CH3CH 20H (CH )iCH - C= CH 3 dihalide Br h. KOC(CH 3lJ 9H3 CH 3- 9-c = CH (2 equiv) CH 3 DMSO dihalide I OCH 2CH 3 i. CH3CH 20H 2° halide SN1,SN2,E1,E2 weak base - H20 j. 3• halide no SN2 weak base + SN1 product % + E1 product + CH 3CH = CHCH3 (cis and trans) E1 product CH 3CH 2C(CH 3)= CHCH3 + (cis and trans) E1 product SN1 product E1 product 8.56 [ 1] NaOCOCH3 is a good nucleophile and weak base, and substitution is favored. [3] KOC(CH3)J is a strong, bulky base that reacts by E2 elimination when there is a~ hydrogen in the alkyl halide. a. CH3CI [1] Na0C0CH 3 [2] Na0CH 3 CH30C0CH 3 C. [1] Na0C0CH3 O-ococH 3 O-c1 [2] Na0CH3 CH30CH3 0 - 0CH3 SN2 [3] K0C(CH3)J b. CH30C(CH3)J ~ C l [1] Na0C0CH 3 ~OCOCH3 [3] KOC(CH3)J d. d c i [1] Na0C0CH 3 ·O 0 dococH3 [2] Na0CH3 ~ O C H3 [2] Na0CH 3 (J [3] K0C(CH3)J O= [3] K0C(CH3)J (J E2 8.s7 two enantiomers: a. (CH3lJC -o (CH3lJC " " A Q ' B b. The bulky tert-butyl group anchors the cyclohexane ring and occupies the more roomy equatorial po~ition. The cis isomer has the Br atom axial, while the trans isomer has the Br atom equatonal. For dehydrohalogenation to occur on a halo cyclohexane, the halogen must be axial to afford trans diax.ial elimination of Hand X. The cis isomer readily reacts since the Br atom is axial. The only way for the trans isomer to react is for the six-membered ring to flip into a highly unstable conformation having both (CH 3)3C and Br axial. Thus, the trans isomer reacts much more slowly. ~ ~ t r a n s diaxial (CH3'3C~ / HH/ (CH 3l J C ~ B r cis-1-bromo-4-tert-butylcyclohexane lrans-1-bromo-4-lert-butylcyclohexane c. two products: OCH3 D= (CH 3) J C ~ d. cis-l-Bromo-4-tert-butylcyclohexane reacts faster. With the strong nucleophile -0CH3, backside attack occurs by an SN2 reaction, and with the cis isomer, the nucleophile can approach from the equatorial direction, avoiding 1,3-diaxial interactions. 1,3-diaxial interactions Br Hi (CH 3l J C ~ ~ C H 3 ~QCH3 (CH 3)Jc--MBr equatorial approach preferred cis-1-bromo-4-lert-butylcyclohexa ne axial approach lrans-1 -bromo-4-lert-butylcyclohexane e. The bulky base -0C(CH3)J favors elimination by an E2 mechanism, affording a mixture of two enantiomers, A and B. The strong nucleophile -oCH 3 favors nucleophilic substitution by an SN2 mechanism. Inversion of configuration results from backside attack of the nucleophile. 8.58 Cl H a. ~ 2' halide SN1, SN2, E1, E2 Cl b. OH strong base SN2 and E2 ~+ + + SN2 product major E2 product inversion at stereogenic center minor E2 product H ~ H~ weak base 2' halide SN1, SN2, E1, E2 SN1 and E1 + ~ + SN 1 products + major E1 product + minor E1 product ;=v minor E1 product minor E2 product CH 3 CH3 c tCH 3 Cl ···csHs C. c tCH3 OCH3 C6Hs CH30H weak base SN1 and E1 3° halide no SN2 e. CH 3COO ~Br f. ~ ···o b. C(c1 strong base SN2 and E2 CH30H weak base SN1 and E1 major E2 product • halide 3 minor E2 product >; u., D u + SN2 product inversion at stereogenic center (trans diaxial elimination of D, :·,·; E2 product cc +C(+C(+C( ex ex + cc 'OCH3 E1 SN1 --X- E1 + 8.60 a. C5H5 achiral SN1 product aOCH; •, SN1 KOH strong base E2 + -OC(CH3h .# CH3 ·00CCH3 x , ,OH 8.59 C(c1 0 KOH 2• halide SN1,SN2,E1 , E2 a. ( :t°"' E1 product OH SN2 product weak base good nucleophile SN1 3° halide no SN2 ct/ + C6Hs 0CH3 ~+ Cl strong base • halide SN2 and E2 2 SN1,SN2,E1,E2 0CH3 'Br + SN 1 products NaOH d. CH3 CHC (J CH3 OC(CH3h strong bulky base _ _ _ _ _ _ __ E2 [QJ U CH3 + c rCH2 major product more substituted alkene No substitution occurs with a strong bulky base and a 3° RX. The C with the leaving group is too crowded for an SN2 substitution to occur. Elimination occurs instead by an E2 mechanism. E1 Alkyl Halides and Elimination Reactions 8-27 ~ b. + - OCH 3 Br 1• halide ---)(- strong nucleophile I SN2 - - - - - - - - [ ~OCH 3 All elimination reactions are slow with 1• halides -----~ The strong nucleophile reacts by an SN2 mecha~ism instead. a- ct? c. ('Y strong base V OCH minor prodru_ ct_o_nl.cy___ _ _ _ _ __ 2 E --------- 3° halide \ 1 alkene is favored. 1 d. 3 More substituted minor product only good nucleophile, weak base - - - - - - - - SN2 favored I major product Cl 2° halide I The 2° halide can react by an E2 or SN2 reaction with a negatively charged nucleophile or base . Since 1- is a weak base, substitution by an SN2 mechanism is favored. 8.61 3• halide, weak base: SN1 and E1 CH 3CH 2OH a. \Cl The steps: l overall ~ reaction + ~ + HCI SN1 CH 3CH29H or E1H ~ ) :ci": E1 +HCI + or ~ +HCI + ~ +HCI Any base (such as CH 3CH 2OH or cI-) can be used to remove a proton to form an alkene. If cI- is used, HCI is formed as a reaction by-product. If CH 3CH 2OH is used, (CH 3CH 2OH 2)' is formed instead. Chapter 8-28 o -CH Cl b. 3 _ OH overall reaction 3° halide strong base E2 Each product: U CH3 a CH, 'H,O' + ceJ 1 : cf "' _.. one step L---9.H + H29 + or y c.CI Cl ct .CH - i ; ----:qH a Cl -one step 8·62 Draw the products of each reaction with the I alkyl hal ide. O DBU a. ~ NaOCH 2CH 3 Cl b. ~ c. strong nucleophile SN2 H ·. KCN Cl l·.--. Cl H ·. sterically hindered base E2 ~ strong nucleophile SN2 H ·. ,~ CN H ·. 8.63 ci-H\ H H - H20: . + Hi): above - : H ·Br· ·.. · H ' : : re-draw ' + - ::B,~ : H H 0: 2 below .. :OH ctf'"~. _ Cv•• , H ~ H + HBr + HBr : H H +p-H\ .· \l H \ :Br:~ ~ ~·· H HBr :5:iH .- H.64 good nucleophile 0 II CH;(c ' o - CH3- HCH 3 7 o , c , CH 3 II CH3COQ- is a good nucleophile and a weak base and so it favors substitution by SN2 · 0 (only) strong base CH3 ~ 7HCH3 + CH 3CH = CH The strong base gives both SN2 and 2 OCH,CH 3 E2 products. but since the 2• RX is 20% 8.65 80% somewhat hindered to substitution , the E2 product is favored. c5 6 3• halide weak base SN1 and E1 HCI + HCI HCI or + HCI AJJ:,vvc;- 1 .:, 9.1 Lv, ,v._,,,._,,,--> 0 • Alcohols are classified as I , 2°, or 3°, depending on the number of carbon atoms bonded to the carbon with the OH group. • Symmetrical ethers have two identical R groups, and unsymmetrical ethers have R groups that are different. OH .._____,,,o'-./ CHa , O ~ 2' alcohol symmetrical ether unsymmetrical ether ~OH CHa, ~ O unsymmetrical ether ~OH 1" alcohol XH 3' alcohol 1' alcohol Chapter 9-6 9.2 Use the definitions in Answer 9.1. CH 20 H - 10 .oH-30 ·•·CH3 0 9.3 To name an alcohol: bstituent. Name the molecule as a [I] Find the longest chain that has the OH gro~p a:h: ~; ending of the alkane to the suffix ·of. derivative of that number of carbons by changing the tower number. When the OH [2] Number the carbon chain to give the OH group. ·ng with the OH group, and the "I'' group is bonded to a ring, the ring is numbered beginni is usually omitted. h name [3] Apply the other rules of nomenclature to complete t e · 1 a. [1] IZ X S JOH [2] ~ t t 5 carbons = pentanol b. [1] [3] 3,3-dimethyl-1-pentanol OH ~ C H3 [ 2] MoH [RCH 3-2-methyl [3] cis-2-methylcyclohexanol ~OH 6 carbon ring = cyclohexanol 1 -6-methyl , . l1] ~ 0H 121 ~ : [3] 5-ethyl-6-methyl-3-nonanol t 9 carbons = nonanol 5-ethyl 9.4 To work backwards from a name to a structure: [I] Find the parent name and draw its structure. [2] Add the substituents to the long chain. OH 7 , . 7,Mmothy1·4-octaool q+ 3~ o 2"erl·bo\y1·3•methylcyoloh"'ool 4 b. 5-methyl-4-propyl-3-heptanol 2 1OH / 5 d. trans-1,2-cyclohexanediol cr 1 OH 'OH ( \ ' ,,OH or ~OH To name simple ethers: 9,5 [ 1] Name both alkyl groups bonded to the oxygen. [2] Arrange these names alphabetically and add the word ether. For symmetrical ethers, name the alkyl group and add the prefix di. To name ethers using the IUPAC system: [I] Find the two alkyl groups bonded to the ether oxygen. The smaller chain becomes the substituent, named as an alkoxy group. [2] Number the chain to give the lower number to the first substituent. a. common name: IUPAC name: CH 3-0-CH2CH2CH2CH 3 I ~ C H2CH 2CH 2CH 3 -larger group- 4 C's I methyl methoxy butyl methyl ether 1-methoxybutane IUPAC name: b. common name: y (YOCH3 (YOCH3 --substituent - V m1thyl methoxy t larger group - 6 C's cyclohexane cyclohexyl cyclohexyl methyl ether methoxycyclohexane IUPAC name: c. common name: ICH3CH2CH2-ofCH2CH 2CH3 I 1 propoxy propane CH 3CH 2CH 2-0-CH2CH 2CH 3 I propyl butane Isubstituent: butyl I propyl dipropyl ether 1-propoxypropane 9.6 Name the ether using the rules from Answer 9.5 . 1 CH 3 I CH 3 C I 2-methyl CH3 0-CH 3 I ) 2-methoxy propane 2-methoxy-2-methylpropane 9.7 Three ways to name epoxides: [ l] Epoxides are named as derivatives of oxirane, the simplest epoxide. [2] Epoxides can be named by considering the oxygen as a substituent called an epoxy group, bonded to a hydrocarbon chain or ring. Use two numbers to designate which two atoms the oxygen is bonded to. [3] Epoxides can be named as alkene oxides by mentally replacing the epoxide oxygen by a double bond. Name the alkene (Chapter I0) and add the word oxide. Three possibilities: [1] methyloxlrane [2] 1,2-epoxypropane [3] propene oxide p 0 a. Three possibilities: [ ) cis-2-methyl-3-propyloxiran 1 [2) cis-2,3-epoxyhe~ane e [ ) cis-2-hexene oxide 3 CH 3 1 (;o- \ CH3- 1 •methyl b. epoxy group 2 Two possibilities: ne [1] 6 carbons = cyclohh~~~clohexane 1,2-epoxy-1-met y oxide [2] 1-methylcycloheXene 9.8 Two rules for boiling point: [ 1] The stronger the forces the higher the bp, d' [2] Bp increases as the extent of the hydrogen b_on same number of carbon atoms: hydrogen bondtng 1° ROH . a. 0 CH t vow lowest bp 3 a CH3 t 0 OH vow intermediate bp hydrogen bonding highest bp DD b. ~ OH ~ Oii t t 3' ROH lowest bp DD yv OH t t vow increases. For alcohols with the bp's increase: 3° ROH < 2° ROH HO (1R,2SJ-2-isobutylcyclopentanol d. constitutional isomer i'()-oH TsCI (1S,3S)-3-isobutylcyclopentanol [6]~ HO pyridine e. constitutional isomer with an ether butoxycyclopentane 9.37 HO H HBr a. Br H + s HO R H H = racemization SN2 = inversion. R HO 2° alcohol SN1 PBr3 follows b. C. H Br HCI Cl H 2° alcohol + s ' R SN1 = racemization Alcohols, Ethers, and Epoxldes 9-17 HO H ~ ~ \ .Cl pyridine ,,,,. d. ~ soc12 follows SN2 =Inversion. R 9 38 Draw the structure of each alcohol using the d. ti . . , OH a. [>-I ' A 10 . c m111ons m Answer 9. 1. OH ~ 2' OH b. H 3' enol 9.39 Use the directions from Answer 9.3. (CH 3),CHCH2CH2CH2OH a. [1] H [2] [ CH, -?-CH2CH2CH20 B] 1 [3] @:Ha- y- CH, CH, CH, oBJ CH3 s carbons = pentanol 4-methyl-1-pentanol CH, -4-methyl b. (CH 3),CHCH,CH(CH2CH3)CH(OH)CH CH 2 3 [1] H ?H [2] [GH3- 9 CH2-y CH CH2C':fil H H O.!:!.-- 3 9 CH, 9 -c:H-CH CH2J 2 - ; ;cl:H ::-,~-± c-:c H-, c,-H,------''--.::J f 3 @, cH, CH2CH, 7 carbons = heptanol 6-methyl [3) 4-ethyl-6-methyl-3-heptanof 4-ethyl c. [1] [2] ~ :thyl [3] s carbons =octanol 4-ethyl-5-methyl-3-octanol OH 4-ethyl M 3 d. [1] 4 2 [ ] HO-bLOH HO-OOH cyclohexanediol [1] [2] [2] OH 1 [2] 00 5 carbons =cyclopentanof [3] OH 5-methyl-2,3,4-heptanetriol OH 5-methyl g, HO° °'CH(CH3)2 (2R,3R)-2,3-butanediol 2R,3R 7 carbons =heptanetriol [1] [3] HOH =butanediol ¼ OH cis-1 ,4-cyclohexanediol HOH HOH f. [3] cis et1J 4 carbons 1 .G [3] HO' ' ~CH(CH3)2 I 3-isopropyl trans-3-lsopropylcyclopentanol Chapter g_ 18 97 9 .4o Use the rules from Answers 9.5 and · · a. c. x .2-rnethylhexane 1.2~b~tJ1-2-rnethylox_irane or 2 thylhexene oxide or 2-rne cH 3 GH3 I I c - o - c - GH3 cH, - , , d. cH, GH3 O o-0 dlcycloheJ1yl ether ~ 4-fk 3 I I ~ CH3 h. 1-ethoxy-3-ethylheptane . R )_ 3.jsopropyl-2-hexanol I. (2 ,35 2 d. 6 -sec-butyl-7.7-diethyl-4-decanol OH j. (ZS)-z-ethoxy- 1,1-dirnethylcyclopen~ane e. 3-chloro-1 ,2-propanedlol 1 3 HO~ CI OH Alcohols , Ethers, and Epoxldes 9-19 9,42 Eight constitutional isomers of mo 1ecular formula C H s 120 containing an OH group: OH 1-pentanol ~ 1° alcohol ® OH 2-pentanol 2° alcohol [zspJ OH 2 3-pentanol ~ 1 OH J-me th Y1·1-butanol 1° alcohol 1° alcohol H 2,2-dimethyl-1-propanol 1° alcohol 0H 20 alcohol OH 2-methyl-1-butanol ~ OH 3-methyl-2-butanol 2° alcohol 2-methyi-2-butanol 2 3° alcohol 9.43 Use the boiling point rules from Answer 9_8_ a. CH 3CH20CH3 (CH3hCHOH I ether no hydrogen bonding lowest bp I 2° alcohol hydrogen bonding intermediate bp b. CH3CH2CH2CH2CH2CH3 no OH group lowest water solubility CH3CH2CH 20H I 1° alcohol hydrogen bonding highest bp CH 3CH 2CH 2CH CH cH 0H 2 2 2 HOCH2CH2CH2CH2CH 2CH 20H one OH group two OH groups intermediate water solubility highest water solubility 9.44 Melt_ing po_ints depend on intennolecular forces and symmetry. (CHihCHCHiOH has a lower melting pomt than CH1CH2CH2CH20H because branching decreases surface area and makes (CH1)2CHCH20H less symmetrical so it packs less well. Although (CH 3) 3COH has the most branch in~ ~nd _least s~rface area,_it is the most symmetrical, so it packs best in a crystalline lattice, g1vmg 11 the highest melting point. ~OH -108 °C lowest melting point -90°C intermediate melting point 26 ' C highest melting point 9.45 Stronger intermolecular forces increase boiling point. All of the compounds can hydrogen bond, but both dials have more opportunity for hydrogen bonding since they have two OH groups, making their bp 's higher than the bp of 1-butanol. 1,2-Propanediol can also intramolecularly hydrogen bond. lntramolecular hydrogen bonding decreases the amount of intermolecular hydrogen bonding, so the bp of 1,2-propanediol is somewhat lower. Increasing boiling point HOT 1-butanol 118 °C OH 1,2-propanediol 187°C HO~OH 1,3-propanediol 215 °C 9.46 NaH HCI ZnCl 2 HBr 8• CH3CH2CH 20H f. CH3CH 2CH 20H g. CH3CH 2CH 20H SOCl2 pyridine PBr3 CH 3CH2CH2Br TsCI CH 3CH 2CHz0TS pyridine [1] NaH [1] TsCI (Y, 9.47 a. CfoH b. CfoH c. CfoH d. CfoH NaH C fo H ~ NaCl N.R. HBr C fs, HCI e. C foH ( fo - Na ' + H2 Cfc1 g. (foe __I'!""'-- U +H20 a,N.R. 1,1ce,oa,a, Cfo-ce,ce, h. ( fo H ~ U pynd1ne • u 9.48 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major product. a. Jx OH / ~/T Jy tetrasubstituted major product TsOH b. c. U disubstituted CH 2C: 3 CX ·CX 0 trisubstituted disubstituted major product CHCH 3 d. CH 3CH2CH2CH20H OH -~, ,..,,..,, '- ll l t:,1..:, 1 C I I U - ~~UAIU ~t; 'd- TsOH - TsOH e. tetrasubslituted disubstituted major product Two products formed by carbocation rearrangement 9.49 The most stable alkene is the major product. OH + trans and disubstituted monosubstituted major product cis and disubstituted 9,50 OTs is a_go?d leaving group_an? will easily be replaced by a nucleophile. Draw the products by substttutrng the nucleophlle in the reagent for OTs in the starting material. NaOH SN2 K+ -oc(CH3l3 E2 9.51 a. ~ - HO H b. ~ OH H D _!::!fL. ZnCl2 l'-0 pyridine HO H d. / ·,_ H Br Cl ~ - H Cl pyridine HO H Configuration is maintained. C-O bond is not broken. 2° Alcohol will undergo SN 1. racemization 1° Alcohol will undergo SN2· inversion SOCl2 C. + ~ - Br H Ts0 H SOCl 2 always implies SN2. inversion _!SL_ SN2 inversion H I ' Chapter 9-22 -oH -- TsO H c HP 9.52 A= ~ NaH (b) HO H TsCI pyridine . "'- (c~ (2R) -2-hexanol C = ,,~ E= ~ H Br CH3I B= cHP o= --- F= . . entical products, Routes (a) and (c) given rd . or 9.53 Acid-catalyzed dehydration follow s an El mechan1sm . make a good leaving group. The three steps ~re. p [ l] Protonate the oxygen to make a good leaving grou · [2] Break the C-0 bond to form a carbocation. [3) Remove a phydrogen to form the re bond. H OCH 3 labeled Band F. , 2° and 3° ROH with an added step t o :~ A / CH3 H.L6so3H V overall The steps~ I reaction ~ U cH 3 + I (_:0.-H CH3 H-KY + CH3 -- - + HS04 H H ~_) .. _ + H20 .. HS0 4 2° carbocation and &~H3 2° carbocation 1,2Hsh< cr·~1Cr°"'\. 3° carbocation and ~H + HS04 OCH3_ 9.54 With POCh (pyridine), elimination occurs by an E2 mechanism. Since only one carbon has a p hydrogen, only one product is formed. With H2S04, the mechanism of elimination is E1. A 2° carbocation rearranges to a 3° carbocation, which has three pathways for elimination. 0 0 ,, :oH cb 0 ~( r - - - , 11/l Cl , IP' CI Cl N"" \_ H t±> cb + I/ POCl2 Cl + ~ .o +N + -0PHOCl2 0 +N H () ~ H- OS03H :oH cb I/ + COH2 cb + HS04- cb + H20 3° carbocation 2° carbocation Cb 3° carbocation Hso,- 3° carbocation j j j m co m 9.55 To (I] [2] (3] [4] Cb 1,2-CH3 shift y draw the mechanism : Protonate the oxygen to make a good leaving group. Break the C-O bond to form a carbocation. Look for possible rearrangements to make a more stable carbocation. Remove a p hydrogen to form the 7t bond. Dark and light circles are meant to show where the carbons in the starting material appear in the product. ~-H 1+ H + Hso.- - : - a - 2° carbocation + HS04 - J 3° carbocation C( + H2S04 Chapter 9-24 The 2' alcohol reacts b an SN1 mechanism 1/ form a carbocation that rearranges. er 9.56 - OH l-methyl-2-butanol 2-malhyl-1-propanol ~ ~ J-. {1}HBr , carbocalion 3 [2]-H 20 "/'-y HCl 2' carbocat1on HBr \- H ,q7T ---:li'r. The 1° alcohol reacts with HBr by an SN2 mechanism. no carbocatlon Intermediate = no rearrangement possible er~ • H,Q. ·5N2 no carbocation \H-Br " 9.57 + :d: - 9.58 CH,CH,CH,CH,OH H,so,, NaBr overall reaction two resonance structures tor the carbocation CH CH CH CH 2Br + CH 3CH 2CH=CH 2 3 2 2 + CH CH CH 2CH 20CH 2CH 2CH 2CH 3 3 2 • ormat,on of a good leaving group • step 111for all product5 • F r--.HQ,S0 H CH,CH 2CH 2CH,-OH 3 CH,CH2CH 2CH,-?;-H Formation of CH3CH2CH=CH2: CH,CH CH-CH ~0-H _ _ _ _ _ _ r - -- - 2 2 H Hso,-rH, J E2cH,cH,cH=cH,I ,+ + HS04 H I • H,9. + H,so, Ether formation (from the protonated alcohol): CH3CH2CH2CH2~?,.- CH2 CH2CH2 CH3 + H 2Q rH HSO,- j 1,2-shift l,) ..) 9 + Hso,- +Hso,- + -CY 9 + H2SO, 9,60 ~QisoaH H QH ~ ,:;H2 _ ~QH + Hso,- 9.61 a. /t+ Br~ 2° halide b. I /'-Br + --0~ 1° halide less hindered RX preferred path 2 2 3 / a j H CH CH a l o- + BrCH2CH2CH3 cY'' . V 1' halide less hindered RX preferred path 2' halide _JH,CH,CH, Chapter 9-26 C. TH,CH,CH, CH 3CH.f)CH2CH 2CH~ 1 CH3CH20- + BrCH2CH2CH3 CH 3CH 2Br + --OCH2CH2CH3 1• halide 1° halide Neither path preferred. 9.62 A tertiary halide is loo hindered and an aryl halide too unreactive to undergo a Williamson ether synthesis. . sets of start·m9 materials: Two possible 9H3 O- C- CH I cr 3 CH 3 Br + yH3 -o-c-CH3 yH3 Br-C-CH3 + er ~I o- I I CH3 CH3 3° alkyl halide too sterically hindered for SN2 aryl halide unreactive in SN2 9.63 B. b. (CH3l3COCH 2CH 2CH 3 Q-o--Q ( ~:~iv) 2 9.64 (CH 3lJCBr + BrCH 2CH 2CH3 (2 equiv) + H,O C. 0 - - 0CH3 __!!L 0-sr + (2 equiv) CH, 3: + H,o 2 Q -sr + H,o -~\ :3 overall reaction / \ CH3 a. (_____)'( l H-1 The steps: / \ CH3 .. \o~CH3 + :_v ·1 H :i j ~,b. Cl ~ ~H 0.p. + H2 + NaCl q,66 0 0 HBr I \ a. ,,c - C" •H H/ 'H H 0 I \ c. e. 0 H I \ CH3CH20CH2CH 20H [2] H20 f. HJ c-c,.H H H 9,67 0 a. a b CH30H CH3CH20H CH{ A · H CH3 H I I I I c. CH 3-C-C-H H2S04 ~H3 CH3CH20 (1) CH3CH20- Na+ a H CH3 OH d. (2] H20 OCH 2CH 3 9.68 a. H Cl CH3~ . / c-c.. (I I \'''CH 3H- 0 : H \~' HOCH2CH20H [2] H20 (1] CH3CH20 - I \ .c-c ... H H'/ 'H HC c: C- CH 2- CH20H [2] H20 (1) OH I \ HOCH 2CH 20H H2S0 4 0 H 0 H20 ' H H [1]HC = C- d. H ,c - C·,, H b H/ I \ BrCH 2CH 20H ct ct H Cl.,_ CH3-.:,. / ,../ c-c,, \ ''CH 3:q: H [1] CH3S- CH 3SCH2CH20H [2] H20 ~ HBr r OH OH (1)NaCN C rCN (2) H20 The 2 CH 3 groups are anti in the starting material, making them trans in the product. /_J Na'H: CH3 ,CH 3 The 2 CH 3 groups are gauche in the H.::·c-c·::.H starting material, making them b. '}f: cis in the product. C4H80 Cl c. \ OH I ,C-C,, CH{ ' \'''CH3 H H rotate CI /C~3 C \c-c·· ,' ,•'JI CH:i H \ '-- :V backside attack CH3 ,H H.::•c-c·::.cH 3 \; :o : C4H 80 Chapter 9-28 . t rial and both products. Because I starting Jl1a e d n the epoxide oxygen, an one must t coille 1rOI 9,69 First, use the names to draw the structures oft ic ' the product has two OH groups, one OH mus come from the nuclcophile, either -oH or H20 . OH With -oH, the nucleophile attacks at the less o __.substituted end of the _ [2] H20 epoxide to form the R isomer of the product. (2R)-2-ethyl-2-methyloxirane ~c@J
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7.1 Categorize the following alkyl halide as 1o, 2o, or 3o depending on the number of carbon
atoms bonded to the carbon directly bonded to the halogen

7.2 Refer to 7.1 as a guide

7.3 Illustrate the skeletal structures of C6 H13Br that will best fit the given description

7.4 Steps in naming a compound using the IUPAC system
a. Name the longest chain. This will be the parent chain
b. Number the carbon atoms in the chain such that the first substituent will get the lowest
possible number.
c. Name the substituents using prefixes di-, tri-, tetra-, etc. when applicable
d. Name the halongen by changing the -ine to -o.
e. Combine the answers/names generated from steps a to d. Remember to alphabetize the
substituents (including their prefixes except for iso-)

7.5 How to draw a structure given the name
a. Identify the parent chain. The parent chain is the alkane and thus, is usually found at the
end of the name. This will always end in -ane
b. Assign arbitrary numbers to the carbon chain. Ensure that these are numbers in
chronological order from left to right or vice versa

7.6 Recall that an sp2 carbon has higher percent s-character than sp3 carbons. This means that
they hold their electrons closer and thus, away from the halogen, in this case chlorine. This
makes the Csp2-Cl bond less polar than the Csp3-Cl bond .

7.7 The structure of chondrocole A,10 C's and only one functional group capable of hydrogen
bonding to water (an ether), makes it rather insoluble in water. However, since it is organic, it is
soluble in CH2Cl2.

7.8 Drawing products of nucleophilic substitution reactions
a. Locate the sp3 hybridized electrophilic carbon. This should be bonded to a leaving
group.
b. Identify the nucleophile. These usually contain electrons or lone pairs in pi bonds
c. As the leaving group leaves, substitute this with the nucleophile. This will now be bonded
to the electrophilic carbon.

7.9 Refer to 7.8 as a guide. Show the proton transfer reaction

7.10 Illustrate the structure of CPC. Use 7.8 as a guide

7.11 Compare the leaving groups based on the following parameters
a. Weaker bases are better leaving groups
b. Uncharged leaving groups are more favorable than their conjugate bases

7.12 Examples of good leaving groups are Cl-, Br-,I-, and H2O.

7.13 Equilibrium will always favor the weaker base and thus, would proceed in that direction.
Always compare the strengths of the nucleophile and the leaving group.

7.14 CH3CH2CH2OH cannot be converted to CH3CH2CH2Cl through nucleophilic substitution
with NaCl. This is because -oH is a stronger base and poorer leaving group than the chloride
ion. The equilibrium will thus favor the reactants, not the products.

7.15 How to find the stronger nucleophile
a. Compare nucleophiles with the same nucleophilic atom. The stronger nucleophile is the
stronger base
b. Negatively charged nucleophiles are always more preferred and thus stronger than their
conjugate acids. .
c. When comparing nucleophiles with similar charges, remember that nucleophilicity
decreases across a row

7.16 Since polar protic solvents contain a H bonded to an O or N, it can exhibit H-bonding. Polar
aprotic solvents on the other hand, do not have H bonded to O or N and thus, will not exhibit H
bonding

7.17
a. When looking at polar protic solvents, remember that nucleophilicity is opposite to the
trend in ba...


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