CHEM 233 Lab 6 Study Questions PDF

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CHEM%233%–%Lab%6%Study%Questions! ! pp.!167'171:!Questions!in!black.!!Answers!in!blue.! ! 1)!Using!Equation! 5.9,!show!that!three!extractions!with!5'mL!portions! of!a! solvent!give!better! recovery! than!a!single!extraction!with!15!mL!of!solvent!when!K"=!0.5.! n ! Vo Error:"Equation"5.9"should"read:!! FA = KVx + Vo

3 ! Vo F = A For!the!3!extractions,!the!equation!would!look!like:!! (0.5)(5) + Vo ! ! 1 ! Vo FA = For!the!1!extraction,!the!equation!would!look!like:!! (0.5)(15) + Vo ! ! ! The!cubed!denominator!(3!extractions)!will!be!larger!than!the!single!denominator!(1!extraction):!! 1 3 ! Vo Vo < FA = FA = ! (0.5)(15) + Vo (0.5)(5) + Vo ! (If!you!have!trouble!conceptualizing!this,!pick!a!value!for!Vo!and!sub!it!in!and!solve!both!equations!–!you! will!see!that!the!fraction!remaining!(FA)!is!smaller!with!the!3!extractions!than!with!1!extraction).! ! 5)! Assume! that! the! partition! coefficient,! K,! for! partitioning! of! compound! A! between! diethyl! ether! and! water!is!3;!that!is,!A"preferentially!partitions!into!ether.! ! ! a)! Given! 400mL! of! an! aqueous!solution! containing! 12! g! of!compound! A,! how! many!grams! of!A" could!be!removed!from!the!solution!by!a!single"extraction!with!200!mL!of!diethyl!ether?! ! ! !

FA =

!

1

400

= 0.4 (40%) remaining

(3)(200) + 400

! !

Fraction removed = 1 - 0.4 = 0.6 (60%)

!

Mass removed = (0.6)(12 g) = 7.2 g compound A

!

% ! ! ! ! ! ! ! !

b)!How!many!total!grams!of!A"can!be!removed!from!the!aqueous!solution!with!three!successive! extractions!of!67!mL!each?! !

! FA =

400 (3)(67) + 400

3 = 0.295 (29.5%) remaining

Fraction removed = 1 - 0.295 = 0.705 (70.5%) Mass removed = (0.705)(12 g) = 8.46 g compound A

6)!Based!on!the!principle!of!“like!dissolves!like”,!indicate!by!placing!an!“x”!in!the!space!those!compounds! listed!below!that!are!likely!to!be!soluble!in!an!organic!solvent!like!diethyl!ether!or!dichloromethane!and! those!that!are!soluble!in!water.! %% Organic%soluble:%naphthalene,!anthracene,!phenol,!aniline,!CH3CO2H! % Water%soluble:%C6H5CO2Na,!C6H5ONa,!NaCl,!CH3CO2H! % 7)! Benzoic! acid! (5)! is! soluble! in! diethyl! ether! but! not! water;! however,! benzoic! acid! is! extracted! from! diethyl!ether!with!aqueous!sodium!hydroxide.! ! ! a)!Complete!the!acid!base!reaction!below!by!writing!the!products!of!the!reaction.! ! b)!In!the!reaction!in!Part!a),!label!the!acid,!the!base,!the!conjugate!acid,!and!conjugate!base.! ! c)!Indicate!the!solubility!of!benzoic!acid!and!its!conjugate!base!in!diethyl!ether!and!water.! ! O O ! - + + HO ! OH + NaOH (aq) O Na 2 ! ! Water Sodium benzaote Benzoic acid Sodium hydroxide ! Acid Conjugate Acid Base Conjugate Base ! Soluble in ether Insoluble in ether ! Insoluble in water Soluble in water ! ! 8)! Aniline! (17),! an! amine,! is! soluble! in! diethyl! ether! but! not! water;! however,! aniline! is! extracted! from! diethyl!ether!with!aqueous!hydrochloric!acid.! ! ! a)!Complete!the!acid!base!reaction!below!by!writing!the!products!of!the!reaction.! ! b)!In!the!reaction!in!Part!a),!label!the!acid,!the!base,!the!conjugate!acid,!and!conjugate!base.! ! c)!Indicate!the!solubility!of!aniline!and!its!conjugate!acid!in!diethyl!ether!and!water.! ! + NH2 NH3 ! + HCl (aq) + Cl ! ! Chloride ion Hydrochloric acid Anilinium ion Aniline ! Base Conjugate Base Acid Conjugate Acid ! Soluble in ether Insoluble in ether ! Insoluble in water Soluble in water ! ! 10)!There!are!three!common!functional!groups!in!organic!chemistry!that!are!readily!iozined!by!adjusting! the! pH! of! the! aqueous! solution! during! an! extraction.! Name! and! write! the! chemical! structure! of! these! three!functional!groups,!and!show!each!of!them!in!both!their!neutral!and!ionized!forms.! ! ! ! ! !

! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !

O R

-H OH

R

Carboxylic acid NH2

O

+

O

-

Carboxylate ion +H

NH3

+

R

+

R Anilinium ion

Aniline OH

-H

R

O

+

-

R Phenoxide ion

Phenol

+

'

12)!Consider!the!base,!monosodium!phosphate,!Na (OH)2P(=O)O .!!! ! ! a)!Write!the!structure!of!the!conjugate!acid!of!this!base.! ! O O ! + HO P O Na HO P OH ! OH OH ! Monosodium Phosphoric acid ! Phosphate (conj. acid) ! ! b)! Given! that! the! pKa! of! this! conjugate! acid! is! 2.1,! explain! why! an! aqueous! solution! of! ! monosodium!phosphate!would!be!ineffective!for!extracting!benzoic!acid!(5)!from!a!diethyl!ether! ! solution.! ! The! pKa! of! benzoic! acid! is! 4.2! and! the! pKa! of! monosodium! phosphate! is! 2.1.! ! Therefore,! ! ! monosodium! phosphate! is! more% acidic" than! benzoic! acid,! and! as! such! cannot! act! as! a! base! to! ! deprotonate! benzoic!acid.! ! If!benzoic! acid!is! not! deprotonated!to! give! benzoate,!then! it!cannot! ! be! extracted! into! an! aqueous! solution! from! ether,! since! neutral! organic! compounds! are! ! generally!not!soluble!in!water.! ! 14)! The! equilibrium! for! phenol! (16),! sodium! phenoxide! (18),! sodium! bicarbonate,! and! carbonic! acid! is! shown!below:! OH O ! + + NaHCO3 H2CO3 ! ! Phenol (16) Sodium Sodium Carbonic ! Bicarbonate Acid Phenoxide (18) ! ! ! a)! The! pKas! for! phenol! and! carbonic! acid! are! 10.0! and! 6.4,! respectively.! ! Determine! the! Keq! for! ! this!reaction.!

pKeq = pKa (PhOH) - pKa (H2CO3) ! ! ! pKeq = 10 - 6.4 = 3.6 ! Keq = 10-pKeq = 10-3.6 = 2.5 x 10-4 ! ! ! b)! Based! upon! your! answer! to! Part! a),! would! sodium! bicarbonate! be! a! suitable! base! for! ! separating!phenol!from!a!neutral!organic!compound!via!an!aqueous!extraction?! ! ! When! Keq! ! 1,! the! ! equilibrium! lies! to! the! right! (i.e.! toward! the! products).! ! Because! the! Keq! for! this! particular! ! reaction!is!2.5!x!10'4!(i.e.!...