ORGANIC SYNTHESIS:
O
CH3CH3 Æ Æ Æ
H3C
NH
CH3
To produce an amide, we need an amine plus either an ester or an acid halide. To produce the amine from ethane: 1) CH3CH3 + Cl2/hν → CH3CH2Cl + HCl (ethane) (chloroethane)
(substitution reaction or halogenation)
2) CH3CH2Cl + NH2– → CH3CH2NH2 + Cl– (chloroethane) (ethanamine/ethylamine)
(substitution reaction)
To produce an ester from ethane: 1) CH3CH3 + Cl2/hν → CH3CH2Cl + HCl (ethane) (chloroethane)
(substitution reaction or halogenation)
2) CH3CH2Cl + OH– → CH3CH2OH + Cl– (chloroethane) (ethanol)
(nucleophilic substitution reaction)
3) CH3CH2OH + Cr2O72–/H+ → CH3COOH (ethanol) (ethanoic acid)
(oxidation)
4) CH3COOH + CH3CH2OH + H+ → CH3C(O)OCH2CH3 + H2O (substitution reaction or (ethanoic acid) (ethyl ethanoate) condensation reaction) To produce an acid halide: 1) CH3CH3 + Cl2/hν → CH3CH2Cl + HCl (ethane) (chloroethane)
(substitution reaction or halogenation)
2) CH3CH2Cl + OH– → CH3CH2OH + Cl– (chloroethane) (ethanol)
(nucleophilic substitution reaction)
3) CH3CH2OH + Cr2O72–/H+ → CH3COOH (ethanol) (ethanoic acid)
(oxidation)
4) CH3COOH + SOCl2 → CH3C(O)Cl + HCl + SO2 (ethanoic acid) (acetyl chloride)
(substitution reaction)
To produce the amide: O
CH3C(O)OCH2CH3 + CH3CH2NH2 → H3C
NH
CH3
O
CH3C(O)Cl + CH3CH2NH2 → H3C
NH
CH3
+ HCl
+ HOCH2CH3
(substitution reaction) (substitution reaction) 1
Review Questions from Last Class: 1. CH3CH(CH3)CHCH2 + ?? → CH3CH(CH3)CH(Br)CH3 2. What is the most likely major product when 2-methylpentane is treated with Br2/hv? 3. What is the product of the oxidation of 2-propanol? 4. What is the product of the reaction of the following with H2O/H2SO4? CH2
i) 1-methylcyclopentene
ii)
5. What is the product of 1-bromo-3-methylbutane + NaOH?
6. i) ii) benzaldehyde + ethanol →
7. Which of the following are chiral? 8. Draw the Lewis structure for C22– and give the formal charges for each carbon atom. 9. Draw the Lewis structure for nitrite ion, give the formal charges and predict the shape. 10. How many sigma and pi bonds are present in H3CSCN? 11. A solution of bleach, sodium hypochlorite, labeled “0.030M NaClO”, had a pH of 10.00. Ka for HClO.
Calculate
12. Write NIE’s for the reactions that occur when perchloric acid and sodium hydroxide are added to a C2H5NH3Cl/C2H5NH2 buffer. 13. Calculate the pH after the addition of 0.0250 moles of HCl to 1.00 L of a buffer solution of 0.0425 M acetic acid and 0.0596 M sodium acetate. Assume no change in volume. 14. For the titration curve below, determine the pKb for the base. A) B) C) D) E)
2.0 3.0 5.3 7.0 8.7
15. a) What is the pH at the stoichiometric point of 0.130 M HCOOH (aq) titrated with 0.130 M KOH (aq) b) Which indicator should be used for the titration above? C) bromothymol blue (pKIn = 7.1) A) methyl red (pKIn = 5.0) B) alizarin yellow R (pKIn = 11.1) D) cresol red (pKIn = 7.9) 2
Answers for the Review Questions:
1.
CH3
CH3
H3C
+ HBr → H3C
H3C
CH2
Reactant - alkene (3-methyl-1-butene)
CH3
CH3
+
H3C
→
Intermediate (2o carbocation)
Br
Product – alkyl halide (2-bromo-3-methylbutane)
This is an addition reaction. The electrophile, H+, opens up the pi bond of the alkene by bonding with two electrons. For the major product, the H goes to the side with the most H’s (to produce a methyl group on the end). We now have a secondary carbocation intermediate due to the formal charge left on the carbon atom. The nucleophile, Br– , is then attracted to the 2o carbocation to complete the structure.
2.
H3C
CH3
H3C
+ Br2/hv → CH3
• CH3
Reactant - alkane (2-methylpentane)
H3C
CH3
CH3
⋅
+ HBr + Br →
Intermediate (3o free radical)
Br
CH3
Major Product – alkyl halide (2-bromo-2-methylpentane)
This is a substitution reaction. Light (or heat) breaks the bromine molecule into two free radicals of bromine. One radical removes a H atom from the alkane. The preference is a tertiary hydrogen over a secondary or primary. This produces HBr and a tertiary free radical. The second bromine radical then collides with the tertiary free radical to produce the major product. Other products are formed in this reaction, but because the tertiary free radical is the most stable form, this will then produce the major product 2-bromo-2-methylpentane. (Other monobrominated species would include: 1-bromo-2methylpentane (intermediate = 1o free radical), 3-bromo-2-methylpentane (intermediate = 2o free radical), 2bromo-4-methylpentane (intermediate = 2o free radical), and 1-bromo-4-methylpentane (intermediate = 1o free radical), – a total of 5 possible monobrominated species)
3. Reaction (oxidation of 2-propanol):
O
OH
+ Cr2O7 2–/H+→ H3C H3C
CH3
Reactant – 2o alcohol (2-propanol)
CH3
Product - ketone (propanone or acetone)
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4. i)
H3C H3C
OH
+ H2O/H2SO4 → Product – cycloalkanol (1-methylcyclopentanol)
Reactant – cycloalkene (1-methylcyclopentene) ii) H2C
H3C
OH
+ H2O/H2SO4 → Product – cycloalkanol (1-methylcyclopentanol)
Reactant – cycloalkane + double bond (methylenecyclopentane)
5.
Br
HO
CH3
+ NaOH →
CH3
CH3
+ NaBr Product – alcohol (3-methyl-1-butanol)
Reactant – alkyl halide (1-bromo-3-methylbutane)
6. i) Step #1:
CH3
OH
Na O CH3 + 2Na(s) →
2 H3C
CH3
2 H3C
+ H2(g) Product – nucleophile (Na+OR–)
Reactant – alcohol (2-butanol) Step #2:
H3C O
CH3I
+
Reactant – alkyl halide (methyl iodide)
Na CH3
O H3C CH3 H3C
→
+ NaI Product – ether (2-methoxybutane)
nucleophile
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ii) benzaldehyde + ethanol → hemiacetal O
H
H HO
O
+ CH3CH2OH →
CH3
Reactant – aldehyde (benzaldehyde)
Product – hemiacetal (ethoxy(phenyl)methanol - just for interest sake)
7. Which of the following are chiral? 1,2-dichlorbutane H3C
H
OH
H3C
2-aminopropanoic acid
Cl
Cl
*
CH3
HO
Achiral
CH3
CH3
*
CH3
Chiral
CH3
H
N
H H
* H
O
H H
Chiral
H
O
Chiral
1,4,4-trimethylcyclohexanol is achiral (no carbon has four different groups due to the symmetry within this molecule – subsitutents are directly across from one another). 2-methylcyclopentanol contains one asymmetric carbon which is bonded to –OH, –CH3, –CH2– (left hand side of ring) and –C(CH3)2 (right hand side of ring).
8. Lewis structure: [:C≡C:]2– ; formal charge for each carbon atom = 4 – (2 + 6/2) = –1. 9. Nitrite ion, NO2– :
..
N
Formal Charges: Single bonded oxygen: 6 – (6 +2/2) = –1 Nitrogen: 5 – (2 + 2/2 + 4/2) = 0 Double bonded oxygen: 6 – (4 + 4/2) = 0
..
O
-
:..O: 10. Every bond contains one sigma bond.
Double bonds .. are made up of one sigma and one pi bond. Triple bonds are made up of one sigma and two pi bonds. S
H3C
N
This molecule contains 6 sigma and 2 pi.
11. A solution of bleach, sodium hypochlorite, labeled “0.030M NaClO”, had a pH of 10.00. Calculate Ka for HClO. NaClO → Na+ (a neutral) + ClO–(a weak base) Rxn: ClO–(aq) + H2O(l) ↔ HClO(aq) + OH–(aq); Kb(ClO–) At equilibrium: 0.030 – x x x ; Kb = x2/(0.030 – x) The pH of the solution gives us the concentration of hydroxide ions, i.e., pH = 10.00, so pOH = 14.00 – pH = 4.00; [OH–] = 10–4.00 = 1.0×10–4 M = x Kb = x2/(0.030 – x) = (1.0×10–4)2/(0.030 – 1.0×10–4) = 3.3×10–7 Ka = Kw / Kb = 1.0×10–14 / 3.3×10–7 = 3.0×10–8
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12. Write NIE’s for the reactions that occur when perchloric acid and sodium hydroxide are added to a C2H5NH3Cl/C2H5NH2 buffer. Buffer contains: C2H5NH2 - weak base & C2H5NH3Cl – acidic salt (C2H5NH3+ – weak acid) Addition of perchloric acid (HClO4 → H3O+ + ClO4– – strong acid + spectator ion): NIE: C2H5NH2(aq) + H3O+(aq) ↔ C2H5NH3+(aq) + H2O(l); Keq = 1/Ka = Kb/Kw = LARGE!! Because water appears on the product side, the equilibrium constant for this reaction is one over a K value. Because water is with the weak acid, then the value is the Kafor C2H5NH3+. Since we won’t be able to find this value, we need to calculate it from the Kb value. This equilibrium constant would then be: Keq = Kb/Kw = 4.7×10–4/1.0×10–14 = 4.7×1010 Addition of sodium hydroxide (NaOH → Na+ + OH– – spectator ion + strong base): NIE: C2H5NH3+(aq) + OH–(aq) ↔ C2H5NH2(aq) + H2O(l); Keq = 1/Kb = 2100
13. Calculate the pH after the addition of 0.0250 moles of HCl to 1.00 L of a buffer solution of 0.0425 M acetic acid and 0.0596 M sodium acetate. Assume no change in volume. Initial buffer solution: Weak acid = 0.0425 M acetic acid (CH3COOH or HAc) Weak base = 0.0596 M acetate (CH3COO– or Ac– – sodium is a spectator ion) Since the volume is 1L, then the concentration values are the same as the moles values. (Based on the H-H equation, the initial pH of this buffer is 4.89 – more basic than its pKa) Addition of HCl (HCl → H3O+ + Cl– – strong acid + spectator ion): I C E
H3O+ + CH3COO–(aq) ↔ CH3COOH(aq) + H2O(l); 0.0250 0.0596 0.0425 moles – 0.0250 – 0.0250 +0.0250 0 0.0346 0.0675
H3O+ is the limiting reagent
pH = pKa + log {CH3COO–/CH3COOH} To find the pKa, look up the Ka value from the tables (Ka = 1.8×10–5) and take the negative log of this value, i.e., pKa = – log Ka = 4.74 pH = 4.74 + log {0.0346/0.0675} = pH = 4.74 – 0.29 = 4.45
14. From the titration curve, find the equivalence point (where the curvature changes from concaved down to concaved up). The volume at the equivalence point is 25 mL. At the half–way point (i.e., 25 mL/2 = 12.5 mL) pH is equal to the pKa (i.e., 5.25). To determine the pKb for the base, subtract 5.25 from 14.00 = 8.75.
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15. At the stoichiometric point, the amount of moles of HCOOH and KOH must be equal (based on the stoichiometry). Therefore, if 1L (or 10 mL) of HCOOH is titrated, then 1L (or 10 mL) of KOH would be needed. The total volume would then be 2L (or 20 mL). HCOOH(aq) + OH–(aq) ↔ HCOO–(aq) + H2O(l) I 0.130 0.130 0 moles C – 0.130 – 0.130 +0.130 E 0 0 0.130 moles ∴[HCOO–] = 0.130 moles/2L = 0.0650 M Based on Le Chatalier’s Principle, the equilibrium will respond to having no HCOOH and OH– and will move in the reverse direction, i.e., I C E
HCOO–(aq) + H2O(l) ↔ HCOOH + OH–(aq); Kb = Kw/Ka = 1.0×10–14/1.7×10–4 = 5.9×10–11 0.0650M 0 0 –x +x +x 0.0650 – x x x
Kb = 5.9×10–11= x2/(0.0650 – x) ; x = √(5.9×10–11 × 0.0650) = 2.0×10–6 = [OH–] pOH = –log (2.0×10–6) = 5.71; pH = 14.00 – pOH = 8.29 b) When selecting the proper indicator, choose one which has a pKIn value which falls within the range of pH at equivlance point ± 1, i.e., 8.29 ± 1 = 7.29 – 9.29. Based on the list provided, cresol red would be the most appropriate indicator.
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