hemistry 000 (Spring 014) Prolem Set #9: Rections nd Acid/Bse hemistry Solutions Ansers to Questions in Silererg (only those /out nsers t the ck of the ook) 1.1 () ddition () elimintion (c) sustitution 1. ond 1.3 Yes. Addition of cross ond (giving n lkne from n lkene, for emple) is redo rection in hich the hydrogen toms of ere oidized (oidtion stte ent from 0 to 1) nd the cron toms of the lkene ere reduced (oidtion stte of ech loered y 1). Addition of Br cross ond (giving hlolkne from n lkene, for emple) is redo rection in hich the romine toms of Br ere reduced (oidtion stte ent from 0 to -1) nd the cron toms of the lkene ere oidized (oidtion stte of ech rised y 1). There re mny more emples, mny of hich you ill encounter in EM 600. Note tht orgnic chemists only tend to clssify rection s redo rection if the verge oidtion sttes of the cron (or sulfur) toms chnges. So, they don t clssify ddition of l cross ond s redo ecuse the verge oidtion stte of cron remins the sme (one cron increses y1; one cron decreses y 1); lso, the oidtion sttes of nd l do not chnge. An orgnic chemist could lso provide emples of elimintion rections hich re lso redo rections. There re even fe sustitution rections hich re lso redo rections (though those re less common). You hve proly not yet seen ny emples of either. 1.5 () () ddition sustitution (techniclly, this is ctully to step process: n ddition rection folloed y n elimintion rection ut you cn t tell tht from the informtion provided, nd you ouldn t e epected to kno tht) 1.17 An electrophile is n electron pir cceptor. () is the est electrophile of these three choices. (c) cn esily e ruled out on the sis of the negtive chrge of the orgnic component ( 3 O ). While the electrophilic cron tom is slightly more positive in () (since l is more electronegtive thn Br), the other requirement of n electrophile is the ility to form ne ond to ccept the electron pir. For () nd (), the only y to form this ond is for the hlide ion (l or Br ) to leve. Br is more stle nion thn l so it is etter t leving ( etter leving group ), mking 3 Br etter electrophile thn 3 l. 1.19 Note tht you should e compring the nions not the slts. N is neutrl specttor ion. (d) is the most nucleophilic ecuse it is the lest stle of these nions nd therefore the most rective. When compring nions ith negtively chrged toms of similr size (e.g. vs. N vs. O ), the est nucleophile ill e the one ith the lest electronegtive negtively chrged tom (in this cse, the of 3 ). While lrger toms tend to e more nucleophilic thn smller toms (e.g. l is more nucleophilic thn F ), tht effect is primrily oserved hen the toms re in the sme Group of the periodic tle. It is not enough to overcome the instility of.
1.1 The nucleophilic tom in thiophenol is sulfur. The nucleophilic tom in phenol is oygen. Both nucleophilic toms re neutrl, nd they re in the sme Group of the periodic tle. As such, their reltive rectivity cn e ttriuted to the size (nd therefore the polrizility) of the toms. Sulfur is lrger thn oygen, so its electrons re held less tightly nd it is sttisticlly more likely tht they ill e unevenly distriuted (hich ould creted temporry induced dipole). As such, the sulfur tom is etter electron pir donor thn n equivlent oygen tom. 1.43 () is the most sic then () then (c) then (d) A strong se hs ek conjugte cid (nd vice vers). The conjugte cid of (d) is 3 O. The conjugte cid of (c) is phenol. The conjugte cid of () is 3 O. The conjugte cid of () is ( 3 ) 3 O. The strength of n cid is determined y the stility of its conjugte se. So, 3 O is the strongest cid ecuse (d) is the most stle of the conjugte ses. Similrly, ( 3 ) 3 O is the ekest cid ecuse () is the lest stle of the conjugte ses. To evlute the stility of these ses, strt y compring the toms ering the negtive chrge. In this question, they re ll oygen toms so neither electronegtivity nor size is relevnt. Net, compre the mount of chrge on ech oygen tom. If the negtive chrge is deloclized (i.e. shred) onto other toms, tht stilizes the nion. This is relevnt for oth (c) nd (d). In (d), the negtive chrge is shred eqully eteen to oygen toms so ech hs chrge of -½. In (c), the negtive chrge is shred eteen one oygen tom nd three cron toms. Since oygen is much more electronegtive thn cron, the oygen tom ers most (ut not ll) of the -1 chrge. So, (d) is the most stle conjugte se nd (c) is the second most stle conjugte se. There is no delocliztion of negtive chrge in () or () so compre inductive effects. The only difference eteen these to nions is tht to of the in () hve een replced y 3 in (). Alkyl groups (like 3 ) re electron donors y induction since they hve lrger electron clouds thn. Dontion of electrons tord negtive chrge destilizes the nion. So, () is less stle thn (). Note tht the inductive effect here is ek reltive to the resonnce (delocliztion of electrons) effect. This is clerly illustrted y the p vlues for the four conjugte cids of these ses. The p of cetic cid ( 3 O ;(conjugte cid of (d)) is 4.74. The p of phenol (conjugte cid of (c)) is 10. The p of ethnol ( 3 O; conjugte cid of ()) is 16. The p of tert-utnol (( 3 ) 3 O; conjugte cid of ()) is 17. 1.44 Note tht you should e compring the nions not the slts. N is neutrl specttor ion. () is the most sic. ompre their conjugte cids. The conjugte cid of O is O (p = 14). The conjugte cid of N is N 3 (p 35). The conjugte cid of I is I (p -10). The strongest cid (I) hs the ekest conjugte se (I ) hile the ekest cid (N 3 ) hs the strongest conjugte se (N ). You should e le to nser this question even ithout ccess to p vlues. See elo. I is strong cid hile O nd N 3 re ek cids. So, the conjugte se of I (I ) is the ekest of the ses. Since N nd O re in the sme Period, the reltive stility of their nions is determined y their electronegtivity. O is more electronegtive thn N, so O is more stle thn N. As such, N is less stle thn O, mking N 3 eker cid thn O nd N stronger se thn O,
1.5 Mrkovnikov s rule sttes tht in n electrophilic ddition rection, the product is formed vi the most stle croction intermedite. (It is often misstted s the hydrogen ill dd to the cron ith more hydrogens. While this is usully true, it is not true in 100% of cses.) e.g. When l is dded to ut-1-ene (four cron chin ith doule ond t one end), the product is -chloroutne (chlorine tom ttched to the second cron in the chin): l l l more stle thn In this prticulr emple, the secondry croction (shon in lue in the rection mechnism) is more stle thn the lterntive primry croction (shon in green elo the rection mechnism). The chloride nion ill ttck the positively chrged cron tom of the more stle croction to give the mjor product of this rection. 1.76 The conjugte se of croylic cid is more stle thn the conjugte se of n lcohol. As such, croylic cid is more cidic thn n lcohol (since the strength of n cid depends on the stility of its conjugte se more stle conjugte se corresponds to stronger cid). See the nser to 1.43 for n eplntion of hy the conjugte se of croylic cid is more stle thn the conjugte se of n lcohol. 1.96 () () (c) sustitution This rection ill proceed until equilirium is reched. If X = Br nd Y = I then I is the etter nucleophile (compred to Br ) ut R I is the etter electrophile (compred to R Br). So, the rection ill proceed ut it ill not go to completion unless rection conditions re mnipulted to drive the rection forrd (y removing one or oth products). In prctice, this rection ill go to completion if the solvent used is cetone ecuse NI is solule in cetone ut NBr is not. noing those specific rection conditions is eyond the scope of this course. elimintion ddition The product shon is not the Mrkovnikov product, so it ould not form under the rection conditions shon. While there re ys to produce the so-clled nti-mrkovnikov ddition product, they re eyond the scope of this course. 16.5 A conjugte cid-se pir is pir of compounds tht differ y. When is removed from the conjugte cid, the product is the conjugte se. When is dded to the conjugte se, the product is the conjugte cid.
16.6 () Rections proceed tord the side ith loer energy species. Loer energy products men tht the equilirium lies tord the products; loer energy rectnts men tht the equilirium lies tord the rectnts. The strength of n cid is determined y the stility of the conjugte se. Therefore, the stronger cid ill hve the eker conjugte se (nd vice vers). So, the side of the rection ith the eker cid nd eker se ill e loer energy thn the side of the rection ith the stronger cid nd stronger se. As such, the equilirium lies tord tht loer energy side. () The eker contins 5 A : A : B : 5 B. Thus, A hs een 9% dissocited [/(5) = 0.857] hile B hs een 71% dissocited [5/(5) = 0.7143]. Since B dissocites to greter etent in the sme solution, it is the stronger cid. Since A is the eker of the to cids, its conjugte cse (B ) is the stronger of the to ses. 16.43 () () O As such, hen [O ] increses y fctor of 10, [ ] decreses y fctor of 10 (ssuming idel conditions so tht = c / 1 M). p log Therefore, if [ ] decreses y fctor of 10, p increses y 1 unit. p p log therefore 10 When p decreses y 3 units, [ ] increses y fctor of 10 3 = 1000 (ssuming idel conditions so tht = c / 1 M). 16.57 Scene A contins purple dots. Scene B contins 4 purple dots. So, Scene B contins 1 times s much 3 O s Scene A. This question cn either e pproched stepise or using proportions. A stepise pproch involves: 1. clcultion of for Scene A from the p (10-4.8 = 10-5 ). clcultion of for Scene B (1 ( 10-5 ) = 10-4 ) 3. clcultion of p for Scene B (-log( 10-4 ) = 3.7) Alterntely, log rules cn e used to determine the impct on p if increses 1-fold: p B log log 1 p log1 log A B A log1 log A log1 p A B 1.08 4.8 3. 7 16.59 All Brønsted-Lory ses hve pir of electrons tht cn e used to mke ond to the they remove from Brønsted-Lory cid. This pir of electrons is frequently lone pir (though it cn e pir of onding electrons. 16.60 Most conjugte ses re strong enough ses tht significnt frction of molecules of the se cn remove from ter molecules. When ter is deprotonted, O is formed. As such, Ois higher thn nd the solution is sic. onjugte ses of strong cids re not sic. These include l, Br, I, NO 3,lO 4 nd SO 4.
16.61 () () 3 O (q), 3 O (q), (q) (or, if you prefer, 3 O (q)) nd O (q) In neutrl queous solutions (nd in pure ter), the concentrtions of (q) nd O (q) re equl. In 0.1 M 3 O, some of the 3 O rects ith ter to give 3 O nd (q). The solution then contins more (q) thn O (q), mking it cidic. In 0.1 M 3 O, some of the 3 O rects ith ter to give 3 O nd O (q). The solution then contins more O (q) thn (q), mking it sic. 16.6 () () 16.63 () () 5 5 N (q) O (l) 5 5 N (q) O (q) O 3 (q) O (l) O 3 (q) O (q) 6 5 O (q) O (l) 6 5 O (q) O (q) ( 3 ) 3 N (q) O (l) ( 3 ) 3 N (q) O (q) 5 N 5 N ( q) 5 5 ( q) O O 6 6 ( 3( q) 3( q) O 5 5 ( q) O 3 ) 3 ( q) N ( q) ( 3 ) 3N ( q) O O O O ( q) ( l ) ( q) ( l ) O ( q) O ( l ) O O ( q) ( l ) 16.66 here is for the conjugte cid nd is for the conjugte se of the pir () of cetic cid ( 3 O ) is 1.8 10-5 14 110 10 A 5.6 10 5 A 1.8 10 () of niline ( 6 5 N ) is 4.0 10-10 B B 14 110 5.5 10 10 4.0 10 16.68 () of chlorous cid (lo ) is 1.1 10-14 110 13 A 9.1 10 A 1.1 10 () of dimethylmine (( 3 ) N) is 5.9 10-4 B B 14 110 11 1.7 10 4 5.9 10 16.70 (c) In very dilute solution of ek cid, the cid ill e fully or lmost fully dissocited. One could rgue tht the concentrtion of cid is very lo oth efore nd fter dissocition, ut there is significnt difference eteen very lo nd negligile.
16.71 No. The strength of n cid refers to the rediness ith hich it dissocites. It is mesured y the (or p ) of the cid. Even though these to solutions hve similr p vlues due to their different concentrtions, tht does not chnge the fct tht l is strong cid (negtive p vlue) hile 3 O is ek cid (positive p vlue). 16.7 In neutrl ter t 5 (i.e. efore ny cid dissocites), [ 3 O ] = 1 10-7 M As such, fter the cid (A) dissocites, [ 3 O ] [A ] (1 10-7 M) (This is not perfect equlity ecuse the equilirium eteen 3 O, O nd ter is lso relevnt hen [ 3 O ] is lo). Therefore, if [A ] >> 1 10-7 M then [ 3 O ] [A ] (1 10-7 M) [A ] () Yes. This solution of 3 O is concentrted enough tht the mount of cid hich dissocites to give 3 O ill e significntly greter thn 1 10-7 M. () No. Even though the cid in this solution fully dissocites, only 1 10-7 M 3 O ill e produced. This lso produces 1 10-7 M 3 O in ddition to the initil 1 10-7 M 3 O from the ter. Tht gives totl of 10-7 M 3 O efore ccounting for the equilirium eteen 3 O, O nd ter. (c) No. A solution contining oth 0.1 M 3 O nd 0.1 M 3 O hs p equl to the p of 3 O. Since p = 4.74 then [ 3 O ] = 10-4.74 = 1.8 10-5. This is not close to [ 3 O ] hich is 0.1 M. 16.73 Removl of ech successive proton forms conjugte se ith greter negtive chrge. As such, ech successive conjugte se is less stle thn the lst. e.g. 3 PO 4 is stronger cid thn PO 4 hich is stronger cid thn PO 4. This is ecuse PO 4 is more stle thn PO 4 hich is more stle thn PO 4 3. While the negtive chrge of ll three nions is resonnce stilized, the verge negtive chrge on ech oygen tom increses s the overll negtive chrge increses. 16.75 Step 1: Write lnced chemicl eqution for this rection nd set up n IE tle. A A ( q) ( q) ( q) initil (M) 0.035 1 10-7 0 chnge (M) equilirium (M) 0.035 (1 10-7 ) Step : Mke resonle ssumptions to simplify rithmetic. Rememer to check these ssumptions lter! (1 10-7 ) good ssumption for most solutions unless very dilute 0.035 0.035 good ssumption for most ek cids Step 3: Write the equilirium constnt epression. ( q) A( q) A ( q) Step 4: Determine hich ctivities re knon (or cn e redily clculted from dt in question). p 4.88 5 7 5 10 10 1.3 10 therefore 1 10 1.3 10 therefore 1.3 10 A 1.3 10 5 1.3 10 5 0. 035 A 0.035 0.035 Note tht e hve no checked oth ssumptions mde in Step. In this prticulr question, e could hve skipped Step, ut it is usully orth the effort so it is good hit to get into. 5
Step 5: lculte 5.3 10 1.3 10 0.035 5 1 9 5.0 10 16.105 Incresing the size of nonmetl tom increses the stility of the conjugte se (since the negtive chrge ill e distriuted over lrger surfce re; chrge density of the conjugte se is loer for lrger toms). As such, inry hydrides of lrger nonmetls re more cidic thn those of smller nonmetls. e.g. l is stronger cid thn F. 16.156 l is strong cid so e cn ssume 100% dissocition of the cid. lo ( = 1.1 10 - ) nd lo ( =.9 10-8 ) re oth ek cids so e cnnot ssume 100% dissocition of either cid. () p l log log 0.10 1. i. 0 ii. lo ( q) ( q) lo ( q) initil (M) 0.10 1 10-7 0 chnge (M) equilirium (M) 0.10 (1 10-7 ) 0.10 ( q) lo( q) 1.1 10 0.10 0.10 lo ( q) 1.1 10 1.1 10 3 1.1 10 3.3 10 3 heck ssumptions!!! 0.10 = 0.10 0.033 = 0.067 (not 0.10; ssumption fils) (1 10-7 ) =(1 10-7 )0.033 = 0.033 (ssumption psses) Since ssumption tht 0.10 filed, repet clcultion ithout tht ssumption. This requires use of the qudrtic eqution. ( q) lo( q) 1.1 10 3 1.1 10 1.1 10 0 0.10 lo ( q) 0.10 1.1 10 3 1.1 10 1.1 10
3 1.1 10 1.1 10 41 1.1 10 4c 0.011 0.067 0.08 or 0.039 Since ctivities cnnot e negtive, cnnot e -0.039 therefore = 0.08. 7 7 1 10.8 10 1 10.8 10 This re-checks the ssumption tht s re-used in the second ttempt to clculte. p lo log log.8 10 1. 55 1 iii. lo ( q) ( q) lo ( q) initil (M) 0.10 1 10-7 0 chnge (M) equilirium (M) 0.10 (1 10-7 ) 0.10.9 10 ( q) 8 0.10 0.10 lo lo ( q) 8.9 10.9 10 ( q) 9.9 10 5.4 10 5 9 heck ssumptions!!! 0.10 = 0.10 0.000054 = 0.10 (ssumption psses) (1 10-7 ) = (1 10-7 )(5.4 10-5 ) = 5.4 10-5 (ssumption psses) 5.4 10 5 p lo log log5.4 10 5 4. 7 () By dding ter, the p of ny solution cn e rought closer to neutrlity. So, im to mke the p of the solutions of the to stronger cids equl to the p of the ekest cid. (i.e. Pln to dd enough ter tht ll three solutions hve p 4.7. p 4.7 5 i. For n cid to hve p 4.7, it needs 10 10 5.4 10. Since l is strong cid, it is fully dissocited t ll concentrtions, so e do not hve to consider equilirium effects.
ii. For idel solutions, ctivity is equl to concentrtion divided y 1 M, so [ ] = 5.4 10-5 M nd therefore (since l is strong cid) e need to find the volume of ter tht ould hve given n initil [l] = 5.4 10-5 M. To chieve this, ter must e dded until the finl volume is: c V c V 1 1 0.10M 100. ml c 1V 1 5 V 1.9 10 ml 1.9 10 L 5 c 5.4 10 M To chieve finl volume of 186 L of solution, dd 186 L of ter to the 100. ml of 0.10 M l. p 4.7 5 For n cid to hve p 4.7, it needs 10 10 5.4 10. Since lo is ek cid, e should consider equilirium effects. lo lo ( q) ( q) ( q) initil (M) y 1 10-7 0 chnge (M) equilirium (M) y (1 10-7 ) 5 7 5 5.4 10 therefore lo lo 5.4 10 5 y y 5.4 10 110 5.4 10 therefore 5 5.4 10 5 y ( q) lo lo ( q) ( q) 5 5 5.4 10 5.4 10 1.1 10 5 y 5.4 10 5 5 5 5.4 10 5.4 10 5.4 10 1.1 10 5 5 5.4 10 5.4 10 5 y 5.4 10 1.1 10 7 5 y.6 10 5.4 10 5 y 5.4 10 For idel solutions, ctivity is equl to concentrtion divided y 1 M, so e need to find the volume of ter tht ould hve given n initil [l] = 5.4 10-5 M. To chieve this, ter must e dded until the finl volume is: c V c V 1 1 c 1V 1 V c 0.10M 100. ml 5 1.9 10 ml 1.9 10 L 5 5.4 10 M To chieve finl volume of 186 L of solution, dd 186 L of ter to the 100. ml of 0.10 M lo. Note tht lo is one of the stronger ek cids. While it turned out tht e needed to dd the sme volume of ter to ech solution (l nd lo ), tht ould not hve een the cse for eker cid. (This is lso hy one ssumption filed in prt () of this question.)
16.158 () m N 4( m) N 3( l ) N ( m) () The strongest cid tht cn eist in N 3 is its conjugte cid, N 4. The strongest se tht cn eist in N 3 is its conjugte se, N. (c) NO 3 nd O re oth stronger cids thn N 4 (nd therefore N 3 is stronger se thn either of their conjugte ses). As such, the folloing to rections re product-fvoured: NO N N NO O 3( m) 3( l ) 4( m) 3( m) ( m) N 3( l ) N 4( m) O ( m) (d) In pure N 3(l), (e) 16.181 N 4( m) N ( m) We lso kno tht 1. m m N m) 4( N 3( l ) 1 Therefore, N N ( m) 3( l ) 5.1 10 7.1 10 7 14 m 14 14 7.1 10 therefore N4( m N 4 7.1 10 M ) SO4( l ) 3SO4( sulf ) SO4( sulf ) sulf SO 3 4( sulf ) SO 4( l ) In pure SO 4(l), SO SO 3 4( sulf ) 4( sulf ) SO We lso kno tht 1. sulf 1 Therefore, sulf SO 4( l ).7 10 4 4( sulf ) 1.6 10 1.6 10 therefore SO4( sulf SO4 1.6 10 M ) t 33. t 0. Aqueous solution of propnoic cid (A (q) ): m A = 7.500 g M A = 74.079 g/mol V totl = 100.0 ml for propnoic cid = 1.3 10-5 (The tet seems to epect tht you on t look this up. Tht seems kind of silly. Looking up this vlue mkes the question much esier nd it s in Appendi. They used n lterntive much more comple route clculting mole frction of ter in the solution from the freezing point of the solution then using tht to clculte the moles of A in solution) Solution is mostly ter; freezing point is -1.890. Freezing point of pure ter is 0.
() Step 1: lculte moles of A in solution 1mol n A 7.500g 0. 101mol 74.079g Step : lculte molrity of solution na 0.101mol 1000mL mol c A 1.01 1. 01M L Vtotl 100.0mL 1L () Step 1: Write lnced chemicl eqution for this rection nd set up n IE tle. A A ( q) ( q) ( q) initil (M) 1.01 1 10-7 0 chnge (M) equilirium (M) 1.01 (1 10-7 ) Step : Mke resonle ssumptions to simplify rithmetic. Rememer to check these ssumptions lter! (1 10-7 ) good ssumption for most solutions unless very dilute 1.01 1.01 good ssumption for most ek cids Step 3: Write the equilirium constnt epression. ( q) A( q) A ( q) Step 4: Use to clculte 5 1.3 10 1.01 5 1.3 10 1.01 3 3.6 10 heck ssumptions!!! 1.01 = 1.01 0.0036 = 1.009 (ssumption fils) (rely) (1 10-7 ) = (1 10-7 )(3.6 10-3 ) = 3.6 10-3 (ssumption psses) Since ssumption tht 1.01 filed, repet clcultion ithout tht ssumption. This requires use of the qudrtic eqution. ( q) A( q) 1.3 10 5 1.01 5 1.01 1.3 10 5 5 1.3 10 1.3 10 5 5 1.3 10 1.3 10 0 A ( q)
5 5 5 1.3 10 1.3 10 41 1.3 10 4c 1 5 1.3 10 0.0073 0.0036 or -0.0036 Since ctivities cnnot e negtive, cnnot e -0.0036 therefore = 0.0036. It turns out tht e get the sme nser s e did mking the ssumption. This is not overly surprising given tht the ssumption only filed on the fourth sig. fig. of the checked nser hile this nser only hs to sig. fig. The molrity of the propnote nion (i.e. of A ) is therefore 0.0036 mol/l = 0.0036 M. (c) M A 0.0036M % dissocition 100% 100% 0.36% M M 1.01M A A 16.184 Imge for X shos 6 X : 3 O : X Imge for Y shos Y : 6 3 O : Y Imge for Z shos 4 Z : 4 3 O : 4 Z Therefore, Y is stronger cid thn Z hich is stronger cid thn X (since Y is more dissocited thn Z hich is more dissocited thn X). () is higher for stronger cids, so (X) < (Z) < (Y) () p is smller (nd/or more negtive) for stronger cids, so p (Y) < p (Z) < p (X) (c) Strength of the conjugte se is inversely relted to strength of the cid. p is smller (nd/or more negtive) for stronger ses, so p (X ) < p (Z ) < p (Y ) N X (d) % dissociti on 100% 100% 5% N N 6 (e) X X Since Z is the strongest of the cids, Z is the ekest of the conjugte ses. Therefore, the solution of NZ ould hve the highest po. (Just s high p = less cidic, high po = less sic.) It ould lso hve the loest p. Additionl Prctice Prolems 1. For ech of the rections elo, dr the mjor orgnic product(s). () 3 Br 3 Br = Br
() 3 3 l 3 l = l 3 l l (c) 3 O O SO 4 = O (d) 3 ecess l 3 3 l l = l l. There re si isomers (including oth structurl isomers nd stereoisomers) ith the moleculr formul 4 8. () Dr ll si isomers of 4 8.
() Identify hich isomers ill rect ith l. For those rections hich proceed, dr the product. 3 l l 3 3 only possile product 3 3 l l 3 3 mjor product 3 3 l l 3 3 mjor product 3 3 l l 3 3 3 mjor product (c) Identify hich isomers ill rect ith Br. For those rections hich proceed, dr the product. 3 Br Br Br 3 3 3 Br Br 3 3 Br trns ddition product 3 3 Br Br 3 3 Br trns ddition product 3 3 Br Br 3 Br 3
3. () Which of the folloing ses cn eist in queous solution? 3 O O S N 3 N nd 3 re stronger ses thn O - (the conjugte se of ter), so they cnnot eist in queous solution. Insted, they rect ith ter to give O nd either N 3 or 4 (their *very ek* conjugte cids). S is stronger cid thn O, so its conjugte se (S ) is eker se thn O. () Which of the folloing cids cn eist in queous solution? 3 O N 4 F Br Br is stronger cid thn 3 O (the conjugte cid of ter), so it cnnot eist in queous solution. Insted, it rects ith ter to give solvted proton ( (q) or 3 O (q)) nd Br. F is eker cid thn 3 O (due to the high chrge density of F ) so it only prtilly dissocites (rects ith ter). 4. In order for phrmceuticl to rech its trget site, it is often necessry for it to e (t lest temporrily) ter solule. One strtegy tht cn e used to improve the ter soluility of drug is to convert n lcohol group to sulfte ester: O R O R O S O onsider the distriution curves for the lcohol nd sulfte ester shon ove. (Sketching them ould e good ide.) Assume tht the lcohol hs p vlue of 17. In this cse, it eists primrily in protonted form (R-O) t p vlues elo 16 nd primrily in deprotonted form (R-O ) t p vlues ove 18. As such, the lcohol ill e neutrl under physiologicl conditions. The strongest intermoleculr forces eteen the drug nd ter ill therefore e dipoledipole ttrctions; hoever, these my not e sufficient to soluilize the drug if it hs lrge nonpolr regions. The sulfte ester hs p vlue of -3. In this cse, it eists primrily in protonted form (R-OSO 3 ) t p vlues elo -4 nd primrily in deprotonted form (R-OSO 3 - ) t p vlues ove -. As such, the sulfte ester ill e negtively chrged under physiologicl conditions. The strongest intermoleculr forces eteen the drug nd ter ill therefore e ion-dipole ttrctions. These re stronger thn dipole-dipole ttrctions nd ill therefore increse the soluility of the drug in ter. O
5. When sodium hydroide is dded to hydrogen peroide ( O ), the folloing rection occurs: O (q) O - (q) O (l) O - (q) () Rerite this eqution using Leis structures nd sho the movement of electrons using curly rros......... O O O -1......... O.... -1.. O... O () Which is stronger cid, ter or hydrogen peroide? Suggest one reson hy this might e so. Since the rection is fvoured in the forrd direction, O must e stronger cid thn ter (nd O stronger se thn O ). This implies tht O (the conjugte se of O ) must e more stle thn O (the conjugte se of O). This is likely ecuse of the electronegtive oygen tom onded to the oygen ering the negtive chrge in O. The second oygen tom ill pull electron density y from the O, stilizing it y lessening the chrge density on tht negtive oygen tom. This is n emple of n inductive effect. (c) If the equilirium constnt for the rection ove is 00 t 5, clculte the nd p of O. int: If you dd to rections (1 nd ), the overll equilirium constnt ill e 1. O (q) (q) O - (q) = ( O ) (q) O - (q) O (l) = 1 O (q) O - (q) O (l) O - (q) = ( O ) rn ( ( O O ) rn ) 14 1 0010 10 p log( ) log 10 1 11. 7 (d) lculte the nd p of O (t 5). 14 10 3 510 1 10 p log( ) log5 10 3. 3