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In terms of electrophilic character, carboxyl groups are not as reactive as carbonyl groups. Examination of the resonance structures reveals that the carbonyl carbon bears a higher degree of positive charge than the carboxyl carbon, and is therefore a better (more reactive) electrophile.
carbonyl group
carboxyl group
Although the above example uses a carboxylic acid as the instance of carboxyl group, acid chlorides and esters behave similarly. You should be able to draw the resonance structures for both of these groups as well.
The difference in reactivity between the two groups means that the carbonyl group can be reduced with both high reactivity reducing agents such as lithium aluminum hydride, and less reactive agents such as sodium borohydride. The carboxyl group, on the other hand, will respond only to lithium aluminum hydride and will not be affected by sodium borohydride. This is illustrated by the following example.
21.05: Reduction of Carboxyl Groups to Aldehydes Using Modified Hydride Reagents
It was stated before that carboxyl groups get reduced all the way to primary alcohols when full reactivity reducing agents are used. The mechanism of reduction goes through an aldehyde stage, but it cannot stop there because the aldehyde gets further reduced to alcohol. So the question is, can we stop at the aldehyde stage by using modified hydride reagents that have bulky groups in the structure and are capable of delivering only one hydride per equivalent. The answer is yes, because those reactions are slower and we can control the number of hydride ions delivered, so that by limiting this parameter we prevent the aldehyde from undergoing further reduction. Such reductions can be accomplished using DIBAL-H or LiAlH(OtBu)3
SEE NEXT PAGE FOR A SUMMARY OF REDUCTIONS OF CARBONYL AND CARBOXYL GROUPS
21.06: Hydride Reductions of Carbonyl and Carboxyl Groups Chart
All reactions with LiAlH4 assume treatment with water or dilute acid as the last step
21.07: Reduction of Carbonyl Compounds and Acid Chlorides Through Catalytic Hydrogenation
Another way to reduce carbonyl groups and acid chlorides is through the catalytic addition of hydrogen. Just like the C=C bond, the C=O bond is capable of adding one mole of hydrogen. The catalyst typically used to accomplish this is called Raney Nickel.
If there are any C=C bonds present in the molecule, obviously they will also take up hydrogen. If selective reduction of the carbonyl group is desired, use NaBH4 instead.
As with the case of hydride reductions, the above reactions also go through the aldehyde stage, but cannot stop due to the high reactivity of the H2 /catalyst mixture. However, just as was the case in the addition of hydrogen to triple bonds, the process can be stopped at the aldehyde stage by the use of a reduced reactivity version of the H2 /catalyst mixture. This is accomplished by the addition of a “poison,” just as it was done with alkynes to stop at the alkene stage. It turns out that Lindlar’s catalyst works in this case as well.
"Poisoned" catalysts for hydrogenation
Pd / BaSO4 / S Pd / BaSO4 / quinoline (Lindlar's catalyst)
EXAMPLE:
21.09: Summary of Methods to Synthesize Alcohols From Carbonyl Compounds
1. TREATMENT OF CARBONYL COMPOUNDS WITH CARBON NUCLEOPHILES SUCH AS GRIGNARD REAGENTS. Excellent method for the synthesis of primary, secondary, and tertiary alcohols with high degree of specificity.
2. TREATMENT OF CARBONYL COMPOUNDS WITH HYDRIDE REDUCING AGENTS. Good method for synthesizing primary or secondary alcohols with the same number of carbons as the starting material.
3. CATALYTIC HYDROGENATION OF CARBONYL COMPOUNDS. Same applications as number 2 above, but will also reduce pi bonds to alkanes if they are present.
22.01: Study Guide for Chapters 8 10
1. SECTIONS TO COVER
Chapter 8: All, except for sections 11 and 16.
Chapter 9: All, except for sections 4, 8, and 9 (D and E).
Chapter 10: All, except for sections 5, 10B, and 12.
2. RECOMMENDED PROBLEMS FROM THE CHAPTERS. Most of these have been suggested in the relevant notes. In general, work any problems relevant to the topics covered in class.
3. RECOMMENDED PROBLEMS FROM THE END OF THE CHAPTERS
Chapter 8: 47, 49(a-f), 50(c, e, g, h), 59, 63.
Chapter 9: 27(a–d), 28, 30, 32, 33, 34(h and i), 36, 37, 39.
Chapter 10: 31(a–e), 33(a–e), 34, 37, 38(a–h and k–n), 39(a–g), 49. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/21%3A_Hydride_Reactions/21.04%3A_Electrophilicity_of_Carbonyl_Groups_Versus_Carboxyl_Groups.txt |
1. The atomic number of boron is 5. The correct electronic configuration of boron is:
A. 1s22s3 B. 1s22p3 C. 1s22s22p1 D. 2s22p3 E. 1s22s23s1
2. How many distinct p orbitals exist in the second electron shell, where n = 2?
3. An oxygen atom has __________ valence electrons.
4. Draw a correct Lewis structure for boric acid, B(OH)3.
5. Draw a correct Lewis structure for t-butanol, (CH3)3COH.
6. Draw a correct Lewis structure for acetonitrile, CH3CN.
7. The formal charge on nitrogen in the compound below is __________.
8. Which of the following are correct Lewis structures for nitric acid, HNO3?
9. The formal charge on oxygen in dimethyl ether, CH3OCH3, is ________.
10. Which of the following choices represent(s) a pair of resonance forms?
11. Draw an acceptable line-angle formula for the molecule shown
12. Orbitals which are equal in energy are referred to as __________.
A. degenerate B. polar C. nodes D. filled E. nonpolar
13. When filling two or more orbitals of the same energy with electrons, the electrons will go into different orbitals rather than pair up in the same orbital. ___True ___False
14. One or more of the atoms in the structure shown should have nonzero formal charges. Redraw the structure and indicate any such charges.
15. One or more of the atoms in the structure shown should have nonzero formal charges. Redraw the structure and indicate any such charges.
16. When a molecule can best be represented as a series of resonance forms, each of these forms always contributes to the same degree in the hybrid.
___True ___False
17. The __________ tells us that each orbital can hold a maximum of 2 electrons.
A. Aufbau principle
B. Pauli exclusion principle
C. Hund's rule principle
D. LeChatelier principle
E. Uncertainty principle
18. Draw a line-angle formula for (CH3)2CHCH2CH2NH2
19. In the compound sodium methoxide (NaOCH3), there is __________ bonding.
A. ionic
B. polar covalent
C. nonpolar covalent
D. a mixture of ionic and covalent
E. resonance stabilized
20. A carbon-hydrogen bond in ethane (CH3CH3) is best described as __________.
A. highly polar
B. essentially nonpolar
C. ionic
D. a multiple bond
E. resonance stabilized
21. What kind of orbitals result when orbitals of different atoms interact?
22. What kind of molecular orbital (sigma or pi) results when the two atomic orbitals shown below approach each other head to head as indicated?
23. What kind of molecular orbital (sigma or pi) results when the two atomic orbitals shown below approach each other sideways as indicated?
24. Which of the following statements about pi molecular orbitals is/are correct?
A. They are cylindrical.
B. Most of their electron density is centered above and below the internuclear axis.
C. When two atoms are connected by a double bond, both of those bonds are pi bonds.
D. Both B and C.
E. A, B, and C.
25. The HCN bond angle in hydrogen cyanide (HCN) is __________ degrees.
26. The HCH bond angle in propane is __________ degrees.
27. The CCO bond angle in acetone (CH3COCH3) is __________ degrees.
28. In boron trifluoride (BF3) the boron atom is ____ hybridized and the FBF bond angle is ______ degrees.
29. In triethylamine [(CH3CH2)3N] the N atom is ____ hybridized and the geometry of the N atom is ________.
30. Structures which differ only in rotations about a single bond are called __________.
A. Structural isomers B. Geometric isomers C. Conformers D. Resonance structures
31. What is the hybridization of the oxygen atom in CH3CHO?
32. What is the hybridization of the ntrogen atom in (CH3)2CHCN?
33. How many pi bonds are present in the following molecule?
34. Which of the molecules below can be properly called an ether?
A. CH3CH2CH(CH3)2
B. CH3OCH2CH2CH3
C. CH3COOH
D. CH3COOCH3
E. HCCCH3
35. Which of the molecules below has the higher boiling point? Briefly explain your choice.
CH3CH2CH2OH or CH3CH2OCH3
36. Which of the molecules below has the higher boiling point? Briefly explain your choice.
(CH3)3N or CH3CH2CH2NH2
37. Which compound is more soluble in water? Briefly explain your choice.
CH3OCH3 or CH3CH2OH
38. Which compound is more soluble in water? Briefly explain your choice.
(CH3)2NH or CH3CH2CH3
39. Are the two compounds shown best described as geometric isomers, structural isomers, or not isomeric?
40. Are the two compounds shown best described as geometric isomers, structural isomers, or not isomeric?
41. Are the two compounds shown best described as geometric isomers, structural isomers, or not isomeric?
42. Are the two compounds shown best described as geometric isomers, structural isomers, or not isomeric?
43. Are the two compounds shown best described as geometric isomers, structural isomers, or not isomeric?
44. Are the two compounds shown best described as geometric isomers, structural isomers, or not isomeric?
45. Which one of the molecules shown below has no net molecular dipole moment?
A. CH3Cl B. H2C=CH2 C. CH2O D. CH2Cl2 E. CH3OH
46. Choose the term below which best describes the geometry of acetylene (HCCH).
A. trigonal bipyramidal
B. trigonal
C. tetrahedral
D. square planar
E. linear
47. Which of the functional groups below contain a hydroxyl group as a part of their structure?
A. aldehyde B. alcohol C. carboxylic acid D. amine E. B and C
48. What is the name of the functional group found in the molecule CH3CH2COOH?
49. How many sigma bonds and how many pi bonds are present in the molecule shown?
50. How many sigma bonds and how many pi bonds are present in acrylonitrile, CH2=CHCN?
51. Which of the following statements concerning the cyclic molecule shown is not true?
A. It contains a pi molecular orbital formed by the overlap of a carbon p orbital with an oxygen p orbital.
B. It contains a sigma molecular orbital formed by the overlap of two carbon sp2 orbitals.
C. It contains a sigma molecular orbital formed by the overlap of two carbon sp3 orbitals.
D. It contains a pi molecular orbital formed by the overlap of two carbon p orbitals.
E. It contains a sigma molecular orbital formed by the overlap of a carbon p orbital with an oxygen sp3 orbital.
52. Give structures for the 3 isomers with molecular formula C5H12 and provide the common name of each.
53. Provide an acceptable IUPAC name for the alkane shown below.
54. Provide an acceptable IUPAC name for the alkane shown below.
55. Provide an acceptable IUPAC name for the alkane shown below.
56. Provide an acceptable IUPAC name for the alkane shown below.
57. Draw an acceptable structure for 4-ethyl-6-(1,2-dimethylpropyl)decane.
58. Draw an acceptable structure for 3-ethyl-3-methylhexane.
59. Draw an acceptable structure for 6-ethyl-2,6,7-trimethyl-5-propylnonane.
60. Give the IUPAC name for the cycloalkane shown below.
61. Provide an acceptable name for the compound shown below.
62. Provide an acceptable IUPAC name for CH3CH2CH2C(CH3)2I.
63. Provide an acceptable IUPAC name for the compound shown below.
64. Provide the structure of isopropyl iodide.
65. Provide the structure of 1-bromo-3-methylhexane.
66. Which of the following is a secondary alkyl halide?
A. methyl bromide
B. isopropyl chloride
C. t-butyl iodide
D. propyl bromide
E. isobutyl chloride
67. Which of the following is a primary alkyl halide?
A. methyl bromide
B. isopropyl bromide
C. t-butyl iodide
D. cyclohexyl bromide
E. isobutyl chloride | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.1%3A_Chapter_1-3_Exercises.txt |
PART A - Molecules (a) through (m) below have been drawn in a way that makes their symmetry apparent if they are in fact symmetric. All the molecules labeled chiral can exist as enantiomeric pairs. All molecules with two chiral carbons and a plane of symmetry represent meso compounds, namely (e), (k), and (m). Chiral carbons have been marked with an asterisk.
dichloromethane achiral
1-bromo-1-chloroethane one chiral carbon chiral
2-bromopropane achiral
2-chlorobutane one chiral carbon chiral
cis-1,2-dichlorocyclopropane 2 chiral carbons achiral
trans-1,2-dichlorocyclopropane 2 chiral carbons chiral
trans-1-bromo-3-chlorocyclobutane achiral
trans-1-bromo-3-chlorocyclobutane achiral
cis-1-bromo-2-chloroethene achiral (planar)
trans-1-bromo-2-chloroethene achiral (planar)
(2S,3R)-2,3-dibromobutane 2 chiral carbons achiral
(2R,3R)-2,3-dibromobutane 2 chiral carbons chiral
meso-1,3-dimethylcyclohexane 2 chiral carbons achiral
PART B - Before assigning configuration to carbons, make sure they are chiral!
a) Chiral molecule and its mirror image - enantiomers.
b) R-isomer on the left, S-isomer on the right - enantiomers.
c) -OH and -Br are in the same positions, but -H and -CH3 have been exchanged - enantiomers.
d) R-isomer on the left, R-isomer on the right - same molecule.
e) Both molecules are chiral, but they do not have the same groups attached to the chiral carbon - unrelated.
f) Each molecule is chiral (no plane of symmetry) and they are mirror images - enantiomers.
g) The easiest way to approach this one is to assign configurations to the chiral carbons. The molecules have the same molecular formula and the same connectivities, but their 3D arrangement is different. They are stereoisomers. The configurations of their chiral centers mirror each other, which makes them enantiomers.
h) The molecule is not chiral (easiest to see in top view) and they’re both cis-isomers - same molecule.
cis-1-bromo-3-methylcyclobutane.
i) A pair of cis/trans isomers - diastereomers
j) Two trans-isomers (chiral) and mirror images - enantiomers.
k) Rotating the molecule on the left as shown leads to the molecule on the right - same molecule.
l) Cis-isomers, same substituents on the same carbons (1 and 3), but different conformations - same molecule.
m) Each molecule has two chiral carbons and a plane of symmetry. Although they mirror each other, they are the same molecule (a meso compound).
n) Both structures represent the S-isomer - same molecule.
o) S-isomer on the left, R-isomer on the right - enantiomers.
p) Both molecules represent 2,3-dibromobutane, but the molecule on the left has all the groups (-H, -Br, -CH3)
anti to each other. One can rotate the front carbon until all the groups eclipse and match each other. That is not the case with the molecule on the right. Both molecules have two chiral carbons. The one on the left has a plane of symmetry, which makes it a meso compound. The molecule on the right has no symmetry and therefore it is chiral. They are diastereomers. For added clarity turn the molecules around to view them from the side.
1.11: Practice Problems for Bronsted-Lowry Acid-Base Chemistry
1. For each of the species below, identify the most acidic proton and provide the structure of the corresponding conjugate base. You might want to draw detailed Lewis formulas in some cases.
2. For each of the species below, identify the basic atom and provide the structure of the corresponding conjugate acid. You might want to draw detailed Lewis formulas in some cases.
3. Fill in the reactants or products for the following acid-base reactions. Keep in mind that in line-angle formulas hydrogens are not shown. When in doubt, write complete Lewis structures. Abbreviations used: Ph=phenyl (a benzene ring attached to a carbon chain)
4. Arrange the substances by order of acidity or basicity as indicated. You may use pKa tables, periodic tables, or any other tools available to you (except for your cell phone).
5. Circle the side favored by equilibrium in the following acid-base reactions.
6. The conjugate acid of ammonia, NH3, is:
A) NH4 + B) NH2OH C) NH2 - D) none of the above
7. Methanesulfonic acid, CH3SO3H, has a pKa of -7 while ethanol, CH3CH2OH, has a pKa of 15.9. Which is the stronger acid and what accounts for this large difference in relative acidity?
8. Would you predict trifluoromethanesulfonic acid, CF3SO3H, to be a stronger or weaker acid than methanesulfonic acid, CH3SO3H? Explain your reasoning.
9. Consider the species CH3O-, NH2 -, and CH3COO-. Rank these ions in order of increasing basicity, and explain your rationale.
10. The Ka of formic acid is 1.7 x 10-4. The pKa of formic acid is __________.
A) 1.7 B) 10.3 C) 4.0 D) 3.8 E) -2.3
11. Provide the structure of the conjugate base of phenol (shown below) and all its resonance forms.
12. Rank the following in order of increasing acidity: CH3OH, HCl, NH3, and CH4.
13. Rank the following in order of increasing basicity: CH3O-, H2N-, H2O, and NH3.
14. When methanol (CH3OH) acts as a base, its conjugate acid is __________. A) CH4OH B) CH3OH2+ C) CH4O+ D) CH3O- E) -CH2OH
15. Which of the following pairs of bases lists the stronger base first?
A) H2O > HO-
B) H2N- > CH3COO-
C) CH3COO- > HO-
D) I- > Cl-
E) HO- > H2N-
16. Draw the structure of the conjugate acid of acetone (CH3COCH3).
1.12: Practice Problems for Bronsted-Lowry Acid-Base Chemistry Answers
1.
2.
3.
4.
5. The numbers represent approximate pKa values for the substances acting as acids
6. A
7. Methanesulfonic acid is the stronger acid. The lower the pKa, the stronger the acid. A lower pKa is associated with a larger Ka which signifies greater dissociation. The large relative difference in acidity in this case can be most easily seen by gauging the relative basicities of the conjugate bases. The weaker the base, the stronger the corresponding conjugate acid. Methanesulfonate, CH3SO3-, is considerably stabilized by resonance delocalization which is not found in ethoxide, CH3CH2O-. This effect greatly reduces the basicity of methanesulfonate relative to ethoxide. Draw the Lewis formula for methanesulfonate and the resonance forms for practice.
8. Trifluoromethanesulfonic acid is a stronger acid. Compare the strengths of the conjugate bases and remember that the weaker the base, the stronger the conjugate acid. Both bases are stabilized by resonance, but in the case of the trifluoro derivative, the presence of the highly electronegative fluorine atoms serves to delocalize the negative charge to an even greater extent by the inductive effect of fluorine. This additional delocalization makes trifluoromethanesulfonate a weaker base.
9. CH3COO- < CH3O- < NH2-
first factor to consider is the nature of the atom which bears the negative charge. The more electronegative the atom that bears the negative charge, the more stable the anion. Stable anions are less reactive and are hence weaker bases. Since O is more electronegative than N, the NH2- is the strongest base in the set. In the remaining two species, the negative charge is on the O, but in the case of CH3COO-, the negative charge is also delocalized by resonance.
10. D
11. See class notes, or p. 417 of the Wade textbook, for all the resonance structures of the phenoxide ion shown below.
12. CH4 < NH3 < CH3OH < HCl (periodic trend)
13. H2O < NH3 < CH3O- < H2N- Negatively charged ions are stronger bases than neutral counterparts. A table of pKa values will further aid in deciding the final order of basicity.
14. B
15. B
16. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.10%3A_Stereochemistry_Exercises_Part_2_Answers.txt |
1) Which of the following best represents the carbon-chlorine bond of methyl chloride?
2) Provide a detailed, stepwise mechanism for the reaction below.
3) Rank the species below in order of increasing nucleophilicity in hydroxylic solvents:
CH3CO2- CH3S- HO- H2O
4) Give a stereochemical structure of the product from the reaction between (S)-2-iodopentane and KCN in DMF (dimethyl formamide, a good polar solvent for ionic reagents).
5) Consider the reaction of (CH3)3CO- with iodomethane. Will the reaction rate increase, decrease, or remain the same if the concentration of iodomethane is increased? Explain.
6) Which of the following compounds will undergo an Sn2 reaction most readily?
A) (CH3)3CCH2I
B) (CH3)3CCl
C) (CH3)2CHI
D) (CH3)2CHCH2CH2CH2I
E) (CH3)2CHCH2CH2CH2Cl
7) What is the major organic product in the following reaction?
8) Would 2-chloropropane or 1-chloro-2,2-dimethylpropane undergo substitution faster with Na+ -CCH? Give the structure of the substitution product.
9) t-butyl chloride undergoes solvolysis in 70% water/30% acetone at a rate slower than in 80% water/20% acetone. Explain.
10) Provide the major organic product of the reaction below and a detailed, stepwise mechanism which accounts for its formation.
11) Sn2 reactions involving chiral electrophiles usually proceed with: A) inversion of configuration B) slightly more inversion than retention. C) slightly more retention then inversion. D) retention of configuration. E) equal amounts of inversion and retention of configuration
12) Which compound undergoes solvolysis in aqueous ethanol most rapidly and why? Remember: solvolysis refers to ionization of the molecule aided by the solvent.
cyclohexyl bromide
isopropyl chloride
methyl iodide
3-chloropentane
3-iodo-3-methylpentane
13) Why does CH2=CHCHBrCH3 undergo solvolysis much more rapidly than 2-bromobutane?
14) Provide the structure of the major organic products which result in the reaction below.
15) What combination of reactants would be best to prepare CH3OCH(CH3)2 by an Sn2 reaction?
16) The reaction between 2-iodohexane and ethanol to give a substitution product most likely follows an ______ mechanism.
17) Which of the following alkyl halides is most likely to undergo rearrangement in an Sn1 reaction?
A) 3-bromopentane
B) 2-chloro-3,3-dimethylpentane
C) 3-chloropentane
D) bromocyclohexane
E) 1-bromo-4-methylcyclohexane
18) Which compound is most nucleophilic? A) CH3SH B) CH3OH C) H2O D) CH3CO2H E) BF3
19) Which halide has the smallest dipole moment?
A) CH3F B) CH3Cl C) CH2I2 D) CH2Cl2 E) CF4
20) When 2,2-dimethylbutane is subjected to free-radical chlorination, ________ distinct monochlorinated products are possible and ________ of these contain asymmetric carbon atoms.
A) 4, 2 B) 5, 0 C) 3, 0 D) 5, 2 E) 4, 0
21) Arrange the substrates in order of increasing Sn2 reactivity with NaCN: Bromoethane, 1-chloro-3,3-dimethylpentane, 1-chloro-2,2-dimethylpentane, and 2-bromo-2-methylpentane.
22) Arrange the following compounds in order of increasing reactivity toward ethanol solvolysis: t-butyl bromide, t-butyl iodide, isopropyl chloride, and methyl iodide.
1.14: Sn1 and Sn2 Reactions Answers
1) I
2)
3) H2O < CH3CO2- < HO- < CH3S-
4)
5) This is an Sn2 reaction. Rate \(= k[(CH_{3})_{3}CO^{-}][CH_{3}I]\). The rate increases as [CH3I] increases.
6) D
7)
The reaction is Sn2
8) 2-Chloropropane. The reaction is Sn2, and even though 1-chloro-2,2-dimethylpropane is a primary chloride, it is more sterically hindered than 2-chloropropane, which is secondary.
9) This is purely a solvent effect. The greater the percentage of water in this solvent mixture, the more polar the solvent. The more polar the solvent, the more easily solvated the developing carbocation and the more rapidly it is formed.
10)
1)
2) carbocation rearrangement from secondary to tertiary by means of a hydride shift
3)
11) A
12) 3-Iodo-3-methylpentane. This molecule forms the most stable cation (tertiary).
13) The intermediate carbocation is resonance stabilized (allylic). Can you draw the resonance structures?
14)
weak nucleophile and secondary substrate promote Sn1 mechanism
15)
The following combination might also work, but less effectively. The substrate is more sterically hindered than methyl iodide.
16) Sn1 - secondary substrate plus weak nucleophile.
17) B
18) A
19) E
20) A
21) 2-bromo-2-methylpentane < 1-chloro-2,2-dimethylpentane < 1-chloro-3,3-dimethylpentane < bromoethane
22) methyl iodide < isopropyl chloride < t-butyl bromide < t-butyl iodide
Again, solvolysis means ionization aided by the solvent. The substrate that can make the most stable cation will be most reactive. In this case there are two tertiary substrates. The one with the best leaving group (iodide) will be most reactive. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.13%3A_Sn1_and_Sn2_Reactions_Exercises.txt |
1) One of the products that results when 1-bromo-2,2-dimethylcyclopentane is heated in ethanol is shown below. Give a mechanism by which it is formed and give the name of this mechanism.
2) Provide the structure of the major organic product in the following reaction
3) Provide the structure of the major organic product from following reaction
4) Which diastereomer of 1-bromo-4-t-butylcyclohexane, the cis or the trans, undergoes elimination more rapidly when treated with sodium ethoxide? Explain your answer.
5) Provide the structure of the major organic product from the following reaction.
6) When 1-iodo-1-methylcyclohexane is treated with NaOCH2CH3 as the base, the more highly substituted alkene product predominates. When KOC(CH3)3 is used as the base, the less highly substituted alkene predominates. Give the structures of the two products and offer an explanation.
7) Which of the following statements apply to E1 reactions of alkyl halides? Choose as many as necessary.
I. Rate = k[base]
II. Rate = k[base][RX]
III. Rate = k[RX]
IV. The reactions occur in two or more distinct steps.
V. Rearrangements are sometimes seen.
8) What is Saytzeff's rule?
9) What major product results when 2-bromo-2-methylbutane is treated with sodium ethoxide.
10) How many distinct alkenes can result from E2 elimination of the compound below? Give their structures and IUPAC names.
11) Give the major product and the mechanism of the following reaction
12) Predict the most likely mechanism and the product for the reaction below.
13) Predict the most likely mechanism and the product from the reaction between 2-chloro-2-methylpentane and sodium ethoxide in ethanol.
14) The major product which results when 2-chloro-2-methylpentane is heated in ethanol is an ether. Show and name the mechanism by which this ether forms.
15) Which of the following mechanisms feature carbocation intermediates?
A) SN1 only
B) SN2 only
C) E1 only
D) E2 only
E) both SN1 and E1
16) Which mechanism(s) give(s) alkenes as the major products, Sn1, Sn2, E1, or E2?
17) Which compound produces only one alkene when treated with sodium methoxide?
A) 2-chloro-2-methylpentane
B) 3-chloro-3-ethylpentane
C) 3-chloro-2-methylpentane
D) 2-chloro-4-methylpentane
E) 2-chloro-3-ethylpentane
18) When 3-iodo-3-ethylpentane is treated with sodium methoxide in methanol, the major organic product is an ______ that is generated through an ______ mechanism.
A) ether, SN1 B) ether, SN2 C) ether, E1 D) alkene, E2 E) alkene, E1
19) Provide the structure of the major alkene product of the reaction below
20) Based on Saytzeff's rule, select the most stable alkene.
A) 1-methylcyclohexene
B) 3-methylcyclohexene
C) 4-methylcyclohexene
D) They are all of equal stability
21) Based on Saytzeff's rule, select the most stable alkene.
A) 1,2-dimethylcyclohexene
B) 1,6-dimethylcyclohexene
C) cis-3,4-dimethylcyclohexene
D) They are all of equal stability
FOR QS. 22-24, DRAW ALL ALKENE PRODUCTS AND CIRCLE THE PREDOMINANT ONE
22)
23)
24)
25) Propose a detailed, step-by-step mechanism for the reaction shown below.
26) Draw all likely products of the following reaction and circle the product you expect to predominate.
27) Draw all likely products of the following reaction and circle the product you expect to predominate.
28) Which base, ammonia (NH3) or triethylamine [(CH3CH2)3N], would be more effective to use for the following conversion?
29) Which compound would undergo dehydrohalogenation with strong base to give the alkene shown below as the only alkene product?
A) 1-chloropentane
B) 2-chloropentane
C) 3-chloropentane
D) 1-chloro-2-methylbutane
E) 1-chloro-3-methylbutane
FOR SYNTHESES # 30-32 GIVE THE MISSING REAGENTS AND STRUCTURES.
30)
31)
32)
33) Which of the following statements applies to the E2 mechanism?
A) It occurs with inversion of stereochemistry.
B) It occurs with racemization of stereochemistry.
C) It proceeds through the more stable carbocation intermediate.
D) The C-H and C-X bonds that break must be anti.
E) Use of a bulky base gives the more highly substituted alkene product. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.15%3A_Elimination_Reactions_and_Alkene_Synthesis_Exercises.txt |
1)
E1 mechanism with carbocation rearrangement
2) In questions 2 and 3, only the proton trans to the leaving group can eliminate.
3)
4)
Due to the presence of the bulky t-butyl group, the ring is practically locked up in the most stable conformation with the bulky group being equatorial.
Of the two isomers, the cis is the only one that fulfills the anticoplanar arrangement for E2, where the leaving group and adjacent proton must be anti to each other and in the same plane.
The atoms shown in red fulfill the anticoplanar requirement for E2. Elimination is possible and fast
The atoms shown in red cannot fulfill the anticoplanar requirement. Elimination is slower or not possible
5)
6) The small, unhindered base ethoxide yields the more stable alkene (Saytzeff’s product, i.e. the more highly substituted alkene). When the bulky t-butoxide base is used, the most accessible hydrogen is removed. This results in the least highly substituted alkene (Hoffman’s product).
7) III, IV, and V.
8) In elimination reactions, the most highly substituted alkene predominates.
9)
10)
11) E2. The molecule must rotate around the central cabon-carbon bond to aquire the anticoplanar arrangement required for E2. This is a stereospecific reaction that results in formation of the product where the phenyl groups are cis to each other.
12) Strong base and bulky substrate favor E2. Only carbon 6 has protons trans to the leaving group. The pi bond can only form between carbons 1 and 6. Can you name the product by IUPAC rules?
13) Strong base and bulky substrate favor E2 with preferential formation of Saytzeff’s product.
14) Weak nucleophile (ethanol, the solvent) and bulky substrate favor Sn1 as shown below. Can you tell which is the rate-determining step? Can you tell what type of reaction is involved in the last step?
15) E
16) E1 and E2
17) B
18) D
19) The reacton is E2. See question 12 for a similar case
20) A
21) A
22)
23)
24)
25) First step is protonation of the alcohol by the strong acid to form a potential water molecule as a leaving group. Next is departure of the leaving group with formation of a carbocation. Next is a carbocation rearrangement from secondary to tertiary. The last step is an elimination step where the water abstracts the acidic proton next to the positive charge to form the alkene.
26)
27)
28) Triethylamine. Amines can be nucleophiles or bases. Increasing their steric bulk near the nitrogen atom diminishes their nucleophilicity while retaining the basicity. Since the substrate is sterically accessible to nucleophilic attack, a bulky base is needed to promote elimination.
29) C
30) This is a typical example of a simple, multistep synthesis (in this case only two steps). This tests your ability to use previously learned reactions (e.g. from ch. 4) to design a synthesis towards a particular product, in this case cyclopentene. The last step is an elimination reaction. Any strong base combination will serve the same purpose as NaOH and acetone.
31) Similar to the previous problem, but this time Hoffman’s product is desired. A bulky base must be used in the last step, such as t-butoxide ion.
32) Same as above, but this time Sayteff’s product is desired. A small base must be used in the last step.
33) D
1.2: Chapter 1-3 Answers
1. C
2. Three: 2px, 2py, 2pz
3. Six
4.
5.
6.
7. +1
8. B
9. Zero
10. B and C only. Atoms have been moved in A, which is not allowed
11.
12. A
13. TRUE
14.
15.
16. False
17. B
18.
19. D
20. B
21. Molecular orbitals
22. Sigma
23. Pi
24. B
25. 180 degrees
26. 109.5 degrees
27. 120 degrees
28. sp2 /120
29. sp3/tetrahedral
30. C
31. sp2
32. sp
33. Six. The triple bond contains two of them.
34. B
35. The alcohol, CH3CH2CH2OH, has the higher boiling point, since it is capable of intermolecular hydrogen bonding. The ether is not.
36. CH3CH2CH2NH2 has the higher boiling point, since it is capable of intermolecular hydrogen bonding.
37. The alcohol, CH3CH2OH, is more soluble in water since it can form a hydrogen bond to water and accept a hydrogen bond from water. The ether, CH3OCH3, can only accept a hydrogen bond from water.
38. The amine, (CH3)2NH, is more soluble in water since it can hydrogen bond with water. Alkanes are not capable of hydrogen bonding with water.
39. Structural isomers: same molecular formulas, but different atom connectivities.
40. Structural isomers
41. Geometric (cis-trans) isomers: They owe their existence to hindered rotation around the pi-bond.
42. Structural isomers
43. Structural isomers
44. Geometric (cis-trans) isomers:
45. B
46. E
47. E
48. Carboxylic acid:
49. 14 sigma bonds and 3 pi bonds. You must count every C-C and C-H bond. The CN bond is a triple bond, which alone accounts for two pi and one sigma bond.
50. 6 sigma bonds and 3 pi bonds.
51. E. There is no sp3 oxygen in the molecule.
52.
53. 2,5-dimethylheptane
54. 2,2-dimethyl-5-(1-methylpropyl)nonane
55. 4-(1-methylethyl)decane
56. 3-ethyl-7-methyl-5-propylnonane
57.
58.
59.
60. trans-1-ethyl-2-methylcyclopentane
61. 1-bromo-4,4-dimethylcyclohexane
62. 2-iodo-2-methylpentane
63. cis-1, 2-dichlorocyclopentane
64. (CH3)2CH-I
65. CH3CH2CH2CH(CH3)CH2CH2Br
66. B (Isoproyl chloride). Chlorine is attached to a secondary carbon.
67. E (Isobutyl chloride). Chlorine is attached to a primary carbon: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.16%3A_Elimination_Reactions_and_Alkene_Synthesis_Exercises_Answers.txt |
1) Draw a Newman projection of the most stable conformation of 2-methylpropane.
2) The structures below are:
A) not isomers.
B) conformational isomers.
C) cis-trans isomers.
D) structural isomers.
E) both B and D
3) Define the term conformation.
4) View a butane molecule along the C2-C3 bond and provide a Newman projection of the lowest energy conformer.
5) Provide a representation of the gauche conformer of butane.
6) Among the butane conformers, which occur at energy minima on a graph of potential energy versus dihedral angle?
A) gauche only
B) eclipsed and totally eclipsed
C) gauche and anti
D) eclipsed only
E) anti only
7) Which of the following cycloalkanes exhibits the greatest molar heat of combustion per CH2 group?
A) cyclooctane B) cycloheptane C) cyclohexane D) cyclobutane E) cyclopropane
8) Which of the following correctly ranks the cycloalkanes in order of increasing ring strain per methylene (CH2 group)?
A) cyclopropane < cyclobutane < cyclohexane < cyclopentane
B) cyclohexane < cyclopentane < cyclobutane < cyclopropane
C) cyclohexane < cyclobutane < cyclopentane < cyclopropane
D) cyclopentane < cyclopropane < cyclobutane < cyclohexane
E) cyclopropane < cyclopentane < cyclobutane < cyclohexane
9) Describe the source of angle strain and torsional strain present in cyclopropane.
10) Which of the following statements is a correct description of the most stable conformation of 1,1,3- trimethylcyclohexane?
A) The methyl group at C-3 is equatorial. B) C-1 is a tertiary carbon and C-3 is a primary carbon. C) C-1 is a quaternary carbon and C-3 is a secondary carbon. D) C-1 is a tertiary carbon and C-3 is a secondary carbon. E) Both methyl groups at C-1 are equatorial.
11) Draw the most stable conformation of trans-1,2-dimethylcyclohexane.
12) Draw the most stable conformation of cis-1,2-dimethylcyclohexane.
13) Which of the statements below correctly describes the chair conformations of trans-1,4- dimethylcyclohexane?
A) The two chair conformations are of equal energy.
B) The higher energy chair conformation contains one axial methyl group and one equatorial methyl group.
C) The lower energy chair conformation contains one axial methyl group and one equatorial methyl group.
D) The higher energy chair conformation contains two axial methyl groups.
E) The lower energy chair conformation contains two axial methyl groups.
14) Which of the statements below correctly describes the chair conformations of trans-1,3- diethylcyclohexane.
A) The two chair conformations are equal in energy.
B) The higher energy chair conformation contains two axial ethyl groups.
C) The higher energy chair conformation contains two equatorial ethyl groups.
D) The lower energy chair conformation contains two axial ethyl groups.
E) The lower energy chair conformation contains two equatorial ethyl groups.
15) Draw the most stable conformation of trans-1-tert-butyl-3-ethylcyclohexane.
16) Which of the following correctly lists the conformations of cyclohexane in order of increasing energy?
A) chair < boat < twist < half-chair
B) half-chair < boat < twist < chair
C) chair < twist < half-chair < boat
D) chair < twist < boat < half-chair
E) half-chair < twist < boat < chair
17) The energy difference between the axial and equatorial conformers of methylcyclohexane is:
A) < 0.1 kcal/mol B) 0.9 kcal/mol C) 1.7 kcal/mol D) 2.5 kcal/mol E) > 5.0 kcal/mol
18) Draw the most stable conformation of cis-1-ethyl-3-methylcyclohexane.
19) Draw the most stable conformation of cis-1-ethyl-4-isopropylcyclohexane.
20) From the perspective of viewing down the C2-C3 bond, draw the Newman projection of the most stable conformation of 2,3-dimethylbutane.
21) In the lowest energy chair conformation of cis-1,3-dimethylcyclohexane, how many axial positions are occupied by hydrogen atoms?
A) 2 B) 3 C) 4 D) 5 E) 6
22) Arrange the following conformers of butane in order of energy, lowest to highest: eclipsed, totally eclipsed, gauche, and anti.
23) In the lowest energy conformation of the compound below, how many alkyl substituents are axial?
A) 0 B) 1 C) 2 D) 3 E) 6
1.4: Conformational Analysis Answers
1)
2) D
3) Conformations are different arrangements of the same molecule formed by rotations about single bonds.
4)
5)
6) C
7) E – cyclopropane, due to its high ring strain.
8) B
9) The angle strain arises from the compression of the ideal tetrahedral bond angle of 109.5o to 60o. The large torsional strain occurs since all C-H bonds on adjacent carbons are eclipsed.
10) A
11)
12)
13) D
14) A
15)
16) D
17) C (see table 3-6 in the textbook)
18)
19)
20)
21) E
22) anti < gauche < eclipsed < totally eclipsed
23) A – all alkyl groups are equatorial.
The most stable conformation has to be a chair and the bulkiest group (isopropyl) has to be equatorial. The ethyl group has to be cis to isopropyl, and the methyl group has to be trans to ethyl. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.3%3A_Conformational_Analysis_Practice_Exercises.txt |
1) Write an equation to describe the initiation step in the chlorination of methane.
2) Reaction intermediates that have unpaired electrons are called __________.
3) When light is shined on a mixture of chlorine and chloromethane, methylene chloride (dichloromethane) is one of the products. Write the propagation steps that explain the formation of dichloromethane from chloromethane under these conditions.
4) Which of the following is NOT a possible termination step in the free radical chlorination of methane? Hint: see textbook, p. 128.
A) $\cdot CH_{3} + Cl_{2} \rightarrow CH_{3}Cl + Cl \cdot$
B) $\cdot CH_{3} + \textrm{wall of container} \rightarrow CH_{3}-wall$
C) $\cdot CH_{3} + \cdot CH_{3} \rightarrow CH_{3}CH_{3}$
D) $\cdot CH_{3} + Cl \cdot \rightarrow CH_{3}Cl$
E) $Cl \cdot + \textrm{wall of container} \rightarrow Cl-wall$
5) Chlorination of methane can result in a mixture of chlorinated products. What experimental conditions should be used to favor the production of chloromethane over the other chlorinated products?
6) Write a detailed, stepwise mechanism (i.e. initiation and propagation steps) for the following reaction.
7) Provide a definition of bond dissociation energy.
8) Energy is __________ when bonds are formed and is __________ when bonds are broken; therefore, bond dissociation energies are always __________.
A) released / consumed / endothermic
B) consumed / released / exothermic
C) consumed / released / endothermic
D) consumed / released / isothermic
E) released / consumed / exothermic
9) Consider the bond dissociation energies listed below in kcal/mol.
CH3-Br = 70 CH3CH2-Br = 68 (CH3)2CH-Br = 68 (CH3)3C-Br = 65
These data show that the C-Br bond is weakest when bromine is bound to a ______.
A) methyl carbon B) quaternary carbon C) primary carbon D) 3o carbon E) 2o carbon
10) If stronger bonds are formed and weaker bonds are broken in a chemical reaction, then the reaction is
A) exothermic
B) an equilibrium reaction
C) endothermic
D) all of the above
E) none of the above
11) Given the bond dissociation energies below (in kcal/mol), estimate the ΔHo for the propagation step $(CH_{3})_{2}CH \cdot + Cl_{2} \rightarrow (CH_{3})_{2}CHCl + Cl \cdot$
CH3CH2CH2-H 98
(CH3)2CH-H 95
Cl-Cl 58
H-Cl 103
CH3CH2CH2-Cl 81
(CH3)2CH-Cl 80
A) -40 kcal/mol B) +22 kcal/mol C) -22 kcal/mol D) +45 kcal/mol
12) Using your table of bond dissociation energies (in kcal/mol), calculate the overall ΔHo for the following reaction: (CH3)3CH + Br2 ---------> (CH3)3CBr + HBr
13) What term describes the highest-energy state in a molecular collision which leads to reaction?
A) exothermic
B) endothermic
C) reaction intermediate
D) transition state
E) free radical
14) The difference in energy between reactants and the transition state is known as __________.
15) Consider the conversion of C to D via a one-step mechanism. The activation energy of this conversion is 3 kcal/mol. The energy difference between D and the transition state of the reaction is 7 kcal/mol. Estimate ΔHo for the reaction C ----> D.
6) Consider the reaction: $CH_{3}CH_{2} \cdot + Br_{2} \rightarrow CH_{3}CH_{2}Br + Br \cdot$
If this reaction has an activation energy of +6 kcal/mol and a ΔHo of -22 kcal/mol, sketch a reaction-energy diagram for this reaction, label the axes, and show Ea and ΔHo on your drawing.
17) Consider the one-step conversion of F to G. Given that the reaction is endothermic by 5 kcal/mol and that the energy difference between G and the transition state for the process is 15 kcal/mol, sketch a reaction-energy diagram for this reaction. Make sure to show how the given energy differences are consistent with your sketch.
18) The major monobrominated product which results when ethylcyclohexane is subjected to free radical bromination is:
A) bromomethane
B) a primary bromide
C) a quaternary bromide
D) a secondary bromide
E) a tertiary bromide
19) How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination?
20) What is the relative reactivity of secondary vs. primary hydrogens in the free radical bromination of n-butane if the ratio of 1-bromobutane to 2-bromobutane formed is 7:93?
21) Predict the major monobromination product in the following reaction, and name it by IUPAC rules.
$(CH_{3})_{3}CCH_{2}CH_{3} + Br_{2} \rightarrow$
22) Predict the major monobromination product in the following reaction, and name it by IUPAC rules
23) What C5H12 isomer will give only a single monochlorination product?
24) Which of the following statements is the best statement of the Hammond Postulate?
A) In an exothermic reaction, the transition state is closer in energy to the products.
B) The structure of the transition state in an organic reaction is always modeled on the structure of the reactants leading to that transition state.
C) In an endothermic reaction, the transition state is closer to the reactants in structure.
D) Transition states are molecular species of finite lifetime whose properties can be probed through free radical reactions.
E) Related species that are similar in energy are also similar in structure.
25) Use the Hammond Postulate to explain why free radical brominations are more selective than free radical chlorinations.
26) List the following radicals in order of increasing stability
$(CH_{3})_{3}C \cdot$ $CH_{2}=CHCH_{2} \cdot$ $CH_{3}CH_{2} \cdot$ $CH_{3} \cdot$ $(CH_{3})_{2}CH \cdot$
27) When Bromine reacts with 1-butene (CH3CH2CH=CH2), the hydrogen atom which is preferentially abstracted is the one which produces a resonance stabilized radical. Draw the major resonance contributing forms of this radical.
28) How do alkyl substituents stabilize a carbocation?
A) Through an inductive donation of electron density to the cationic center.
B) Through an inductive removal of electron density from the cationic center.
C) Through hyperconjugation.
D) both A and C
E) both B and C
29) Describe the hybridization of the cationic center and predict the CCC bond angle in (CH3)3C.
30) Free radical bromination of pentane results in poor yields of 1-bromopentane, while cyclopentane can be readily brominated under similar conditions to yield bromocyclopentane. Offer an explanation.
31) What is the name of the major monobrominated product which results when 3-methylpentane is subjected to Br2/hν conditions?
32) What is the name of the major monobrominated product which results when methylcyclohexane is subjected to Br2/hν conditions?
33) Provide the two propagation steps in the free-radical chlorination of ethane.
34) In an exothermic reaction, are stronger bonds broken and weaker bonds formed or are weaker bonds broken and stronger bonds formed?
35) How many secondary hydrogens are present in the hydrocarbon below?
A) 16 B) 7 C) 8 D) 2 E) 6
36) How many tertiary hydrogens are present in the hydrocarbon below?
A) 2 B) 3 C) 4 D) 1 E) 5 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.5%3A_Practice_Questions_for_Chapter_4.txt |
1) $Cl-Cl + photon (h \nu) \rightarrow 2 Cl \cdot$ (a photon is a unit of light energy)
2) radicals or free radicals
3) $CH_{3}Cl + Cl \cdot \rightarrow \cdot CH_{2}Cl + HCl$
$\cdot CH_{2}Cl + Cl-Cl \rightarrow CH_{2}Cl_{2} + Cl \cdot$
4) A (termination steps involve combination of free radicals without generation of new ones)
5) Use a large excess of methane.
6)
1) Initiation
2) Propagation Ph = Phenyl group (benzene ring attached to a carbon chain)
7) Amount of energy required to break a covalent bond homolytically.
8) A
9) D
10) A
11) C
12) -16 kcal/mol. Hint: remember to look up the correct BDEs for the type of bond, i.e., primary, secondary, tertiary, etc
13) D
14) the activation energy or Ea
15) -4 kcal/mol
16)
17)
18) E (1-Bromo-1-ethylcyclohexane)
19) Three:
20) The 2o hydrogens are 20 times more reactive than the 1o ones (Refer to p. 156 of your textbook for a similar example).
21) (CH3)3CCHBrCH3 (2-Bromo-3,3-dimethylbutane). The secondary hydrogens are the most reactive in the reactant. There are no tertiary hydrogens.
22) 1-Bromo-1-phenylpropane. The benzylic hydrogens are the most reactive in the reactant because the free radical that leads to their formation is resonance stabilized (can you draw it and provide all the resonance structures?) Refer to section 17-4B of your textbook for a very similar example, complete with resonance structures. You should be familiar with these concepts for the test. Don’t let the fact that this example is in ch.17 scare you. You know or have the tools to understand most of the concepts discussed in this section ;-)
23) (CH3)4C (neopentane or 2,2-dimethylpropane).
24) E
25) The first propagation step in free radical bromination is endothermic while the analogous step in free radical chlorination is exothermic. From the Hammond Postulate, this means that the transition state for the bromination is product-like (i.e., radical-like) while the transition state for the chlorination is reactant-like. The product-like transition state for bromination has the C-H bond nearly broken and a great deal of radical character on the carbon atom. The energy of this transition state reflects most of the energy difference of the radical products. This is not true in the chlorination case where the transition state possesses little radical character.
26) $CH_{3} \cdot < CH_{3}CH_{2} \cdot < (CH_{3})_{2}CH \cdot < (CH_{3})_{3}C \cdot < CH_{2}=CHCH2 \cdot$ (resonance-stabilized)
27)
28) D (refer to section 4-16A, p. 163 for an explanation of concepts such as hyperconjugation and inductive effect).
29) Cationic center is sp2 hybridized, therefore the CCC bond angle is 120o.
30) In the bromination of pentane, the lowest energy reaction pathways go through secondary free radical intermediates to produce secondary alkyl bromides (2-bromopentane and 3- bromopentane). Thus 1-bromopentane is a very minor product.
In cyclopentane, on the other hand, all of the hydrogen atom abstractions lead to the same secondary radical which eventually leads to bromocyclopentane.
31) 3-bromo-3-methylpentane
32) 1-bromo-1-methylcyclohexane
33) $CH_{3}CH_{3} + Cl \cdot \rightarrow CH_{3}CH_{2} \cdot + H-Cl$
$CH_{3}CH_{2} \cdot + Cl-Cl \rightarrow CH_{3}CH_{2}Cl + Cl \cdot$
34) In an exothermic reaction, weaker bonds are broken and stronger bonds are formed.
35) E
36) B | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.6%3A_Ch._4_Answers.txt |
1) Is the molecule shown below chiral or achiral?
2) Is the molecule shown below chiral or achiral?
3) Is the molecule shown below chiral or achiral?
4) Is the molecule shown below chiral or achiral?
5) Is the molecule shown below chiral or achiral?
6) Which of the following terms best describes the pair of compounds shown: enantiomers, diastereomers, or the same compound?
7) Which of the following terms best describes the pair of compounds shown: enantiomers, diastereomers, or the same compound?
8) Label each asymmetric carbon in the compound below as R or S.
9) Label each asymmetric carbon in the compound below as R or S.
10) Label each assymetric carbon in the compound below as R or S.
11) Draw the structure of (2R,3S)-2,3-dichloropentane. Take particular care to indicate three-dimensional stereochemical detail properly.
12) Draw the structure of (S)-1-bromo-1-chloropropane. Take particular care to indicate three-dimensional stereochemical detail properly.
13) Draw the structure of a meso form of 1,3-dichlorocyclopentane. Take particular care to indicate three-dimensional stereochemical detail properly.
14) How many asymmetric carbons are present in the compound below?
15) How many asymmetric carbons are present in the compound below?
16) How many asymmetric carbons are present in the compound below?
17) How many asymmetric carbons are present in the compound below? -ethyl-2,2,4-trimethylpentane
18) Can the molecule shown below be properly described as a meso compound?
19) Can the molecule shown below be properly described as a meso compound?
20) Can the molecule shown below be properly described as a meso compound?
21) Draw the structure of (1R, 2R)-1-bromo-2-chlorocyclobutane. Take particular care to indicate stereochemistry properly.
22) Stereoisomers which are not mirror image isomers are __________.
23) Is it theoretically possible to separate the pair of compounds below by distillation? Explain briefly.
24) Draw the Fischer projection of (S)-2-hydroxybutanoic acid, CH3CH2CH(OH)COOH.
25) How many enantiomers are there of the molecule shown below?
A. 6 B. 2 C. 0 D. 1 E. 3
26) Which of the following terms best describes the pair of compounds shown: enantiomers, diastereomers, or the same compound?
27) Which of the following terms best describes the pair of compounds shown: enantiomers, diastereomers, or the same compound?
28) Label each asymmetric carbon in the molecule below as having the R or S configuration
1.8: Ch. 5 Stereochemistry Exercises Part 1 Answers
1) achiral
2) achiral
3) achiral
4) achiral
5) chiral
6) the same compound
7) enantiomers
8)
9)
10)
11)
12)
13)
14) 1
15) 5
16) 2
17) 1
18) No
19) Yes
20) No
21)
22) diastereomers
23) Yes. The molecules are related as diastereomers and hence have different boiling points.
24)
25) C
26) the same compound
27) enantiomers
28)
1.9: Ch. 5 Stereochemistry Exercises Part 2
PART A
Consider the following molecules and answer the questions.
a) dichloromethane
b) 1-bromo-1-chloroethane
c) 2-bromopropane
d) 2-chlorobutane
e) cis-1,2-dichlorocyclopropane
f) trans-1,2-dichlorocyclopropane
g) trans-1-bromo-3-chlorocyclobutane
h) trans-1-bromo-2-chlorocyclobutane
i) cis-1-bromo-2-chloroethene
j) trans-1-bromo-2-chloroethene
k) (2S, 3R)-2,3-dibromobutane
l) (2R, 3R)-2,3-dibromobutane
m) meso-1,3-dimethylcyclohexane
1. Which of these molecules are chiral (i.e. asymmetric)?
2. Which of these molecules contain chiral carbons? In your drawings label them with an asterisk.
3. Which of these molecules can exist as enantiomeric pairs?
4. Which of these molecules represent meso compounds?
PART B
Indicate whether the following pairs of compounds represent the same molecule, pairs of enantiomers, diastereomers, meso compounds, or stereochemically unrelated molecules.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
o)
p) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/Problems/1.7%3A_Stereochemistry_Practice_Problems_Part_1.txt |
Thumbnail: Four sp3 orbitals. (CC BY SA 3.0; Jfmelero via Wikipedia)
01: Basic Concepts in Chemical Bonding and Organic Molecules
To summarize simply, a chemical bond is the attractive force holding atoms or ions together. Such attractive interaction leads to a more stable state for the whole system comparing to individual atoms.
Valence electrons play a fundamental role in chemical bonding. In the electron configuration of an atom, the outermost shell is called the valence shell, and the electrons in the valence shell (outermost shell) are known as valence electrons. Take the carbon atom for example: the electron configuration of carbon is 1s22s22p2. The outermost shell is the 2nd principal shell, so there are 4 valence electrons in carbon. Valence electrons are the electrons that are the furthest away from the nucleus, and thus experience the least attraction from the nucleus and therefore are most reactive. They play the most important role in chemical bonding.
Exercises 1.1
Determine the number of valence electrons for following elements: B, N, O, Cl, Mg.
Answers to Practice Questions Chapter 1
Ionic Bond and Covalent Bond
There are two major types of chemical bonding: ionic bonds and covalent bonds. An ionic bond is a bond that results from the electrostatic attraction (force) between ions of opposite charges. Ionic bonds apply to ionic compound, such as sodium chloride (NaCl).
In simple ionic compounds, the metal element loses valence electron(s) to form the cation and the non-metal element gains electron(s) to form the anion. With the proper number of electron(s) lost or gained, both the cation and the anion achieve a full outer shell that contains eight electrons, as in the following examples of Na+, Ca2, Cland O2. According to Lewis’s Theory, an atom is most stable if its outer shell is filled or contains eight electrons. This is also called the octet rule.
Na (atom) → Na+ + e Ca (atom) → Ca2++ 2e
Cl (atom) + e→ Cl O (atom) + 2e→ O2-` `
A covalent bond is a bond formed through the sharing of electron pairs between the two bonding atoms. The shared electron pairs are mutually attracted by the nuclei of both atoms. By sharing the electron pairs, both atoms also gain a filled outer shell, or an octet. Almost all of the bonds involved in organic compounds are covalent bonds.
Covalent bond can be non-polar or polar.
For covalent bonds formed between two identical atoms, the electron pairs are shared equally between the two nuclei. Electron density is distributed evenly through the bond, making the bond a non-polar bond. Examples include all homonuclear molecules, such as H-H, Cl-Cl, O=O, N≡N.
For heteronuclear bonds (the bond formed between two different atoms), the electron pairs are not shared evenly, and the bond is polar. The electron pairs are more attracted to the atom that has the stronger ability to pull the electron pairs towards itself. This ability is measured with electronegativity. The relative values of electronegativity (EN) are listed using the scale devised by Linus Pauling, as summarized in the following table:
Notes about electronegativity values for Organic Chemistry purposes:
• It is much more important to know the trend of electronegativity than to memorize the values. The trend is that EN values decrease along the group from top to bottom and increase along the period from left to right (the trend mainly works for Main Group elements, not transition metal elements).
• It is very useful (although not mandatory) to know the EN values of a few select elements: F (4.0, highest), O (3.5), N (3.0), C (2.5) and H (2.1).
• The EN of C (2.5) and H (2.1) is rather close, which makes the C-H bond (the bond involved in all organic compounds) technically non-polar.
With the introduction to the concept of electronegativity, bond polarity can be represented with the electronegativity difference between the two bonding atoms, which is known as ΔEN. For non-polar bonds, ΔEN equals to zero, and for polar bonds, ΔEN is not zero. The greater the ΔEN, the more polar the bond is.
Exercises 1.2
1. Identify the following bonds as “polar” or “non-polar”: C-C, C-H, B-F, O-O, C=N
2. Rank the following bonds in order of increasing bonding polarity: C—S, C—O, C—F (referring to the trend of EN, you do not need to use the exact EN values).
Answers to Practice Questions Chapter 1
Because of the electronegativity difference, the atom with the higher EN attracts the shared electron pairs more strongly, therefore bearing a slightly negative charge (δ-). The other atom with a lower EN bears a slightly positive charge (δ+). The direction of the bond polarity can be indicated with an arrow, with the head of the arrow pointing to the negative end and a short perpendicular line near the tail of the arrow marking the positive end. The following example of an H-Cl molecule indicates how to show the bond polarity and partial charges of the polar bond. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/01%3A_Basic_Concepts_in_Chemical_Bonding_and_Organic_Molecules/1.01%3A_Chemical_Bonding.txt |
The Lewis structure is a structure that shows the bonding between atoms as short lines (some books use pairs of dots), and non-bonding valence electrons as dots.
1.2.1 Lewis Structure of Diatomic Molecules
To learn about Lewis structures, we will start with the Lewis symbol. The Lewis symbol is the chemical symbol of an element with valence electrons represented as dots. The Lewis symbols of some elements are shown here:
For simple diatomic molecules, combining the Lewis symbols of each element gives its Lewis structure.
H2 example: (H only needs two electrons; usually referred to as a duet.)
F2 example:
Terms used in Lewis structures (see example of F2):
• Bonding pair: The pair of valence electrons involved in a covalent bond. The covalent bonds are drawn as short lines in this book, and one covalent bond means one pair of bonding electrons, that is 2 electrons. Single bonds and multiple bonds (double or triple bonds) may be involved.
• Lone pair: The pairs of valence electrons not involved in the covalent bond. Lone pair electrons can also be called non-bonding electron pairs.
Special note: Non-bonding electrons can also be unpaired (single) electrons. A species with one or more unpaired (single) electrons is called a radical (free radical). More examples of radicals with single electrons will be in section 1.2.5 and Chapter 9.
HCl example:
O2 example:
Exercises 1.3
Draw the Lewis structure of the N2 molecule.
Answers to Practice Questions Chapter 1
1.2.2 Lewis Structures of Polyatomic Molecules or Ions
For more complicated polyatomic molecules and ions, the Lewis structures cannot be obtained by simply combing Lewis symbols. A specific procedure with certain steps have to be followed. It is very important that you follow the following procedure in order to get the correct Lewis structures for polyatomic molecules and ions.
Lewis Structure Drawing Procedure for Polyatomic Molecules and Ions
1. Calculate the total number of valence electrons. For ions, make sure charges are properly included in the calculation. For example of NH4+ cation:
the total number of electrons = 5 (N atom) + 4×1 (four H atoms) -1 (minus the charge for cation) = 8 valence electrons
2. Write a plausible skeletal structure using the following steps:
a) Write atomic symbols for the central and terminal atoms.
• Hydrogen atoms are always terminal
• Central atoms are generally those with the lowest electronegativity
• Carbon atoms are always central
b) Connect the central atom with each of the terminal atom by drawing a single bond.
3. For each single bond, subtract two electrons from the total number of valence electrons.
4. Using the remaining valence electrons, complete the octets of the terminal atoms first, and then complete as many as possible for the central atoms.
5. If you have used up all of the valence electrons to complete octets for all of the atoms, you are done.
6. If not, then complete the octets of all central atoms by moving lone-pairs from terminal atoms to form multiple bonds.
7. Calculate the Formal Charges on all atoms and label the non-zero formal charges in the structure:
Formal Charge on an atom = No. of valence electrons in free atom–No. of lone pair electrons –½ (No. of bonding electrons)
Formula 1.1
Examples: Here we will take CO2 molecule as an example to explain the procedure step by step:
1. Total number of valence electrons: 4 (C atom) + 2×6 (2 O atoms) = 16
Always DOUBLE CHECK: In the correct Lewis structure, the total number of electrons involved (bonding plus non-bonding electrons) must be equal to this number, less or more are both incorrect!!
2. Write a plausible skeletal structure:
Carbon atoms are always central, so the skeletal structure is: O — C — O
3. Four electrons are used so far, and there are 16 – 4 = 12 electrons remained.
4. The remaining 12 electrons must be used to complete the octet for both terminal O atoms first, and no electrons left after that.
It is very important to keep in mind that the remaining electrons should be used to give the octet of terminal atoms first!
5. The central C atom does not get octet yet, we should do next step.
6. Moving one lone pair from each terminal O atom, the following structure is obtained.
this is the complete Lewis structure of CO2.
For Lewis structure purposes, the lone-pairs can only be moved from terminal atoms to the central atom to form multiple bonds, not the other way around.
7. Formal charges check: all atoms have formal charges equals to 0 in this structure.
FC (C) = 4 -½×(4×2) = 0
FC (O) = 6 -4-½×(2×2) = 0
Since the two oxygen atoms have the same bonding, one calculation is enough for both oxygen atoms.
1.2.3 Guidelines about Formal Charges in Lewis Structures
The purpose of formal charges is to compare the difference between the number of valence electrons in the free atom and the number of electrons the atom “owns” when it is bonded. The smaller the difference the “happier” (more stable) the atom is. The atom owns all of the lone pair (non-bonding) electrons and half of the bonding (shared) electrons, which is why the formula is in the way given in Formula 1.1.
Formal charges can be used as guidelines to determine the plausibility of Lewis structures by comparing the stability of non-equivalent resonance structures, which is particularly important for organic species. The rules about formal charges are:
• The sum of the formal charges must equal to the total charge on the molecule or ion.
• Formal charges should be as small as possible (comparing the absolute value of formal charges for such purposes).
• “-” FC usually appears on the most electronegative atoms (with the stronger ability to pull the shared electrons; this atom is “winning” electrons in the sharing).
• “+” FC usually appears on least electronegative atoms (with the weaker ability to pull the shared electrons; this atom is “losing” electrons in the sharing).
• Structures having formal charges of the same sign on adjacent atoms are unlikely.
There is a derived way for calculating formal charge: since each bond contains 2 electrons, half of the bonding electrons simply equals to the number of bonds. So, the formal charge can also be calculated based on the derived version of the formula:
Formal Charge on an atom = No. of valence electrons in free atom–No. of lone pair electrons – No. of covalent bonds around the atom Formula 1.2
Double bonds count as 2 and triple bond count as 3 in Formula 1.2. Both Formula 1.1 and 1.2 work for counting the formal charge; you can choose either one for your convenience. While almost all of the other textbooks show Formula 1.1 as the official way, Formula 1.2 is easier to use and can be regarded as the practical one based on experience.
Exercises 1.4
Why is the following structure not the best way to show the Lewis structure of CO2 ?
Answers to Practice Questions Chapter 1
1.2.4 Kekulé Structures vs Lewis Structures
The complete Lewis structure always has to include all the bonding electrons and lone pair electrons. However, organic species are usually shown as KeKulé structures (more discussion will be in Chapter 2) with all the lone pair electrons completely omitted (with exceptions to the lone pairs that are shown to highlight special properties). Therefore, when viewing Kekulé structures, it is very helpful to keep in mind that atoms other than C and H should have a certain number of lone pairs. Examples of Kekulé structures of some compounds are given here:
To count how many lone pairs should be involved on a certain atom, apply the octet rule. All of the atoms (except H) should have 8 electrons around it, therefore, N usually has 1 lone pair, O has 2 lone pairs and halogens have 3 lone pairs.
1.2.5 Exceptions to Octet Rule in Lewis Structure
So far we have always been applying the octet rule in Lewis structures, however there are some cases in which the rule does not apply. For example, H only needs 2 electrons. Here we will see some other cases where the octet rule is compromised.
• Odd number of electrons
If the total number of valence electron is an odd number, the octet rule can not be applied to all atom in the species. The examples could include NO (nitrogen monoxide or nitric oxide), NO2 (nitrogen dioxide) and alkyl radicals.??
NO molecule: Although NO is a diatomic molecule, it is possible to draw the Lewis structure by following the procedure. Depending on which atom is given the octet first in Step 4, you may get two possible structures. By applying the formal charge guideline, we can decide that the first structure is the better choice with zero formal charges.
NO2 molecule: The Lewis structure of NO2 molecule is shown below.
For above molecules, they all contain unpaired (single) electrons. The neutral species that contain an unpaired electron is called radical (or free radical). When the carbon atom of a alkyl group has an unpaired electron, the species is the alkyl radical.
Alkyl radicals: The simplest example of alkyl radical is •CH3, with the total number of valence electron as 7. The structure is:
More discussions about the properties and reactions of radicals will be included in Chapter 9.
• Incomplete Octet
An incomplete octet means that the atom has less than 8 electrons involved. This could be because the total number of valence electrons is less than 8, or due to formal charge concerns.
BH3 molecule: The total number of valence electrons is 6, so the central boron atom does not get an octet.
BF3 molecule: Even though all of the atoms do have a chance to get octets in the structure of BF3, the actual structure of BF3 keeps the incomplete octet. Applying the FC guideline explains why the first structure is the better choice. Similar examples include BeF2, AlCl3.
F is the atom with the highest electronegativity, therefore F never has “+” formal charge in any plausible Lewis structure!
CH3+: This is another reactive intermediate in organic reactions (more discussions in Chapter 8). FC calculations indicate that the “+” charge lies on the C atom, so such a species is also called a carbocation . Carbon has an incomplete octet.
• Expanded Valence Shell
For elements in Period 3 or higher, they can have more than 8 electrons if that helps to lower the formal charges. Common examples involve the species with P, S or Cl, etc as central atoms. Sometimes multiple double bonds are necessary to minimize the formal charge of the central atom. The structure of the phosphate anion, PO43-, is given here as an example.
Elements in Period 3 (or higher) have 3 (or more than 3) principle shells, so the d orbital is available in the valence shell. That is why they can accommodate more than 8 electrons.
Key Takeaways
For elements in 2nd period, C, N, O, F and Ne, the maximum number of electrons involved in Lewis structure is eight!!! | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/01%3A_Basic_Concepts_in_Chemical_Bonding_and_Organic_Molecules/1.02%3A_Lewis_Structure.txt |
In the case that more than one reasonable (plausible) Lewis structure can be drawn for a species, these structures are called resonance structures or resonance contributors. Resonance structures can be either equivalent or non-equivalent.
Equivalent Resonance Structures
Let’s consider the example of carbonate anion, CO32-:
By following Step 6 in the Lewis structure drawing procedure, the double bond can be built between the central C and any of the terminal O’s to generate three structures, and they all look “the same”. However, they are not really identical (or same), they are just equivalent. Each structure is called a resonance structure, and they can be connected by the double-headed resonance arrow. There are total three equivalent resonance structures for CO32-, and the actual structure of CO32-is the hybrid of the three resonance contributors.
The arrows used here to connect between resonance structures is the “resonance arrow“, which has double arrow heads. Resonance structures have to be connected using resonance arrows.
Since the resonance structures are equivalent, they are all in the same level of energy and have the same stability, so they make the same contributions to the actual structure of CO32-. This is supported by the experimental evidence that all the carbon-oxygen bonds in CO32- are the same bond length, which is longer than a regular double bond but shorter than a single bond. As a result of the resonance structures, the two negative charges in CO32- are not localized on any oxygen atoms, but are spread evenly among all three oxygen atoms, and is called charge delocalization. Because of charge delocalization, each oxygen atom has two-thirds of a full negative charge. Charge delocalization helps to stabilize the whole species. The stability a species gains from having charge delocalization through resonance contributors is called resonance stabilization effect. The greater the number of resonance contributors, the greater the resonance stabilization effect, and the more stable the species is.
The actual structure of the carbonate anion is a combination of all the three equivalent resonance structures, that can be called a hybrid. What does the actual structure look like, and can we draw one structure on paper to show the actual structure? The actual structure can not be shown with a conventional Lewis structure, because the regular Lewis structures do not include partial charges, and there is two-thirds of a full negative charge on each oxygen atom in CO32-. An attempt to show the hybrid structure can be by using dashed lines to show that the bond between carbon and oxygen is somewhere between a single and double bond, and each oxygen atom has partial charges.
The delocalized charges can also be represented by the calculated electrostatic potential map of the electron density in the CO32- anion. In an electrostatic potential map, regions with different charges are shown in different colours. More specifically, colours trending towards red means higher negative charges, while colours trending toward blue means more positive charge (the colour system generated by different softwares might not be same, but will follow the same trend). In the electrostatic potential map of carbonate anion below, the same shade of red of all three oxygen atoms indicate the equal charge distribution at the three oxygen atoms.
Exercises 1.5
Draw all the equivalent resonance structures for following species. Include any non-zero formal charges in the structures.
• O 3 molecule
• nitrate anion NO 3
• chlorate anion ClO 3 .
Answers to Practice Questions Chapter 1
Non-equivalent Resonance Structures
Resonance structures can also be non-equivalent. For the example of OCN, there are three non-equivalent resonance structures, depending on how the multiple bonds are formed in Step 6 of the Lewis structure drawing procedure.
For non-equivalent resonance structures, the bonding and charge distributions are different, so they are in different energy levels. Some are more stable (better) resonance structures than others. The guidelines for comparing the relative stability between non-equivalent resonance structures are (the lower the energy, the more stable the structure is and vice versa):
• The structure with complete octets is usually more stable, except in the cases in section1.2.4 “Exceptions to Octet Rule”.
• The structure involving the smaller formal charges is more stable.
• Negative charges should be preferentially located on atoms with greater electronegativity, and positive charges should be preferentially located on atoms with less electronegativity
• Charge separation decreases the stability (increases the energy).
By applying the rules above, we can predict that for OCN, structure 3 is the least stable one since it has the highest formal charges. For both structures 1 and 2, the formal charge is “-1”. It is more preferable for negative formal charges to be on oxygen, the more electronegative atom; therefore structure 2 is the most stable one.
Exercises 1.6
Draw all of the resonance structures for azide anion, N3, and indicate the most stable one.
Answers to Practice Questions Chapter 1 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/01%3A_Basic_Concepts_in_Chemical_Bonding_and_Organic_Molecules/1.03%3A_Resonance_Structures.txt |
Resonance stabilization effect (also known as resonance effect), as briefly mentioned in Section 1.3, is one of the fundamental concepts of Organic Chemistry and has broad applications. The discussion of resonance effect heavily relies on the understanding of resonance structures. Here we will focus on how to draw resonance structures (or resonance contributors) for organic chemistry species, and how to compare the relative stabilities between the structures.
According to resonance effect, the greater the number of resonance contributors, the greater the resonance stabilization effect, and the more stable the species is. Therefore, to predict whether the resonance effect applies or not, we usually need to construct “new” resonance structures (contributors) based on the “original” one that is available. There are some very important rules we need to follow for such purposes.
Guidelines for Drawing Resonance Structures:
• All resonance structures must be valid Lewis structures. (Keep in mind that all the rules applied to Lewis structures still apply here!)
• All resonance structures must have the same atom connectivity, and only differ in the electron arrangement. (Atoms NEVER move, only electrons move.)
• All resonance structures have the same number of electrons and net charge. (Formal charges on individual atom could be different, but net charge, that is the sum of all the charges, must be the same.)
• To move electrons, only π electrons and lone-pair electrons (NEVER move σ bonds!) can be moved from the higher electron density area to lower electron density area by following one of the three transformations:
• π bond forms another π bond;
• π bond forms the lone pair electrons;
• lone pair electrons forms a π bond.
• Use curved arrows to indicate the electron movement in the “original” resonance structure. The “new” resonance structure should be a “product” automatically obtained by following the arrows.
• Calculate the formal charge in the “new” structure and label any non-zero formal charges.
The way to use curved arrows show electron transfer is also called arrow pushing, it is a very important fundamental skill you need to master in organic chemistry. For the purpose of constructing “new” resonance structures, arrows have to be shown in the “original” structure.
Examples: Draw another resonance structure based on the given one.
1.
Approach: There is only one π bond in this example, and no any lone pairs, so only the π electrons can be moved around. There is a carbocation beside the π bond, which is the low electron density spot. Therefore it is reasonable to move the π electrons to the position beside carbocation to form another π bond, and that gives the “new” structure. The two resonance structures here are equivalent.
Solution
2.
Approach: More electrons available for movement in this example: several lone pairs and one π bond. The guideline of “move electrons from the higher electron density area to the lower electron density area” provides a useful hint about where to start. The nitrogen atom has a “-” formal charge, meaning it has relatively high electron density, higher than other neutral spots. So it is reasonable to move the lone pair on nitrogen away to form a π bond (keep in mind that lone pair can only form π bond, not another lone pair). However, when the new π bond is formed around the carbon atom, there are 5 bonds (10 electrons) on that carbon, which is not allowed. So, another electron pair has to be moved away, and the only available electron pair to be moved is the π electrons in C=O bond. It can be moved onto the oxygen atom and become another lone pair on the oxygen atom.
Solution:
The two resonance structures in this example are non-equivalent, so one is more stable than the other. By applying the formal charge guideline, the “-“ formal charge is more preferable on oxygen, which is more electronegative than nitrogen, so the 2nd structure is the more stable one with lower energy, and makes more contribution to the actual structure in this species. The more stable structure can also be called as the major resonance contributor.
Comparing the relative stability of different resonance contributors:
• Structures with a maximum of octets are most important.
• Charge separation usually decrease the stability (increase the energy of the contributor).
• Negative charges should be preferentially located on atoms with greater electronegativity, and positive charges should be preferentially located on atoms with less electronegativity.
Common errors for drawing resonance structures:
1. σ bond is moved
2. Atom is moved
3. More than eight electrons located around C, N or O
4. Arrows are not shown in the proper way
5. Electron pairs are moved too far away, they should only be moved to the next position/atom.
Exercises 1.7 Draw new resonance structure and compare the relative stability, show arrows in the original structure.
2.
Answers to Practice Questions Chapter 1 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/01%3A_Basic_Concepts_in_Chemical_Bonding_and_Organic_Molecules/1.04%3A_Resonance_structures_in_Organic_Chemistry.txt |
The Valence-Shell Electron-Pair Repulsion (VSEPR) theory helps us to understand and predict the geometry (shape) of molecules or ions. The theory is:
• Electron pairs repel each other whether they are in chemical bonds or lone pairs.
• Valence electron pairs are oriented to be as far apart as possible to minimize repulsions.
Based on this theory, depending on the number of electron pairs (both bonding pairs and lone pairs) around the central atom, a certain shape is adopted to minimize the repulsion between election pairs, as summarized in the table below:
Total number of electron groups (electron pairs) around central atom
Geometry (Shape) of electron groups (electron pairs)
2
linear
3 trigonal planar
4 tetrahedral
5 trigonal bipyramidal
6 octahedral
Table 1.1 Basic VSEPR Shapes
Notes:
• For VSEPR purpose, the terms “shape” and “geometry” are interchangeable; “electron pair” and “electron group” are also interchangeable.
• Multiple bonds (double or triple bond) are regarded as one electron group for VSEPR purpose.
For species that do not have any lone pair electrons (LP), the geometry (shape) of the species is just the same as the geometry of the electron groups.
For the exampleof the PCl5molecule, there are five electron groups on the central phosphorous, and they are all bonding pairs (BP). The shape of the electron groups is trigonal bipyramidal, and the shape of the PCl5 molecule is trigonal bipyramidal as well. The trigonal bipyramidal shape can be drawn on paper using solid and dashed wedges: the three bonds lie within the paper plane are shown as ordinary lines, the solid wedge represent a bond that points out of the paper plane, and the dashed wedge represent a bond that points behind the paper plane.
However, for the species that has lone pair electrons on the central atom, the shape of the species will be different to the shape of the electron groups. The reason is that even though the lone pairs occupy the space, there are no terminal atoms connected with lone pair, so the lone pair become “invisible” for the shape of the species.
For the example of the water (H2O) molecule, the central oxygen atom has two BPs and two LPs, and the shape of all the electron groups is tetrahedral. The shape of a water molecule is bent because only the atoms are counted towards the molecular shape, not the lone pair electrons.
The VSEPR shapes can be rather diverse, considering the different numbers of total electron pairs together with the different numbers of lone pairs involved. The most common shapes are summarized in the following table (Table 1.2). To describe a certain shape, the specific name has to be used properly, and the bond angle information is important as well.
Total number of e-groups Geometry (shape) ofall the electron groups # of Bonding Pairs (BP) and Lone Pairs (LP) Geometry (shape) of the species Angles (°)
2 linear 2BP linear 180
3 trigonal planar 3BP trigonal planar 120
2BP, 1LP bent <120
4 tetrahedral 4BP tetrahedral 109.5
3BP, 1LP trigonal pyramidal <109.5
2BP, 2LP bent <109.5
5 trigonal bipyramidal 5BP trigonal bipyramidal 120, 90, 180
4BP, 1LP see-saw <120, 90, 180
3BP, 2LP T-shape 90, 180
2BP, 3LP linear 180
6 octahedral 6BP octahedral 90, 180
5BP, 1LP square pyramidal 90, 180
4BP, 2LP square planar 90, 180
Table 1.2 Summary of specific VSEPR shapes
The website https://phet.colorado.edu/sims/html/molecule-shapes/latest/molecule-shapes_en.html provides good resources for visualizing and practicing VSEPR topics.
We will see more applications of VSEPR in organic compounds in next section. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/01%3A_Basic_Concepts_in_Chemical_Bonding_and_Organic_Molecules/1.05%3A_Valence-Shell_Electron-Pair_Repulsion_Theory_%28VSEPR%29.txt |
1.6.1 Valence Bond Theory
We have talked about how covalent bonds are formed through the sharing of a pair of electrons; here we will apply the valence bond theory to explain in more detail how the sharing happens. The valence bond theory describes the covalent bond formed from the overlap of two half-filled atomic orbitals on different atoms.
Let’s start with the simple molecule H2. The atomic electron configuration of a hydrogen atom is 1s1, meaning that there is one electron (which is also the valence electron) in the sphere-shaped 1s orbital.
When two hydrogen atoms are approaching each other, the two 1s orbitals overlap, allowing the two electrons (each H donates 1 electron) to pair up for the bonding with the overlapping orbitals. The shared pair of electrons are under the attraction of both hydrogen nuclei simultaneously, resulting in them serving as a “glue” that holds the two nuclei together.
The overall energy changes of the system versus the distance between the two hydrogen nuclei can be summarized in the energy diagram below.
When the two atoms are separate, there is no overlap and no interaction. As they are getting closer, orbitals start to overlap, and there is attraction between the nucleus of one atom and the electron of the other atom, so the total energy of the system lowers. The energy lowers to its minimum level when the two atoms approach the optimal distance. The optimal distance is also defined as the bond length. H2 molecules have a bond length of 74 pm (often referred to as 0.74 Å, 1Å= 10-10m). The energy difference between the most stable state (lowest energy state with optimum distance) and the state in which the two atoms are completely separated is called the bond (dissociation) energy. The bond energy is 7.22×10-19 J for one H-H bond, or 435kJ/mol.
When the two atoms get closer than the optimal distance, the repulsion between the two nuclei become predominant, and the energy of the system becomes even higher.
Another important character of the covalent bond in H2 is that the two 1s orbitals overlap in a way that is referred to as head-to-head. The bond formed by head-to-head overlap is called σ (sigma) bond. σ bonds are cylindrically symmetrical, meaning if a cross-sectional plane is taken of the bond at any point, it would form a circle.
The valence bond theory works well to explain the bonding in HF as well, with the 2p orbital of fluorine atom involved in the overlapping.
The fluorine atom has the valence electron configuration of 2s22p5 as shown in the orbital diagram.
For the three 2p orbitals, two of them are filled and the other one is half-filled with one single electron. The filled orbital cannot form bonds, so only the half-filled 2p is available for overlap. Therefore, the 1s orbital of the hydrogen atom overlaps head-to-head with the half-filled 2p orbital of the fluorine atom to form the H-F σ bond, as shown below.
A σ bond can also be formed through the overlap of two p orbitals. The covalent bond in molecular fluorine, F2, is a σ bond formed by the overlap of two half-filled 2p orbitals, one from each fluorine atom as shown here.
However, when the valence bond theory is applied to organic molecules, for instance CH4, it does not work. The valence electron configuration of carbon atom is 2s22p2 as shown in the orbital diagram.
Based on the valence bond theory, with two half-filled orbitals available, the carbon atom should be able to form two bonds. However, carbon always has four bonds in any stable organic compound. To explain the bonding of carbon and other atoms that cannot fit into the simple valence bond theory, a new theory called orbital hybridization will be introduced as a supplement to the valence bond theory.
1.6.2 Hybridization and the Structure of CH4
Simply speaking, hybridization means the mathematical combination of several orbitals to generate a set of new hybrid orbitals.
In the hybridization for CH4, the 2s and three 2p orbitals are combined to give a new set of four identical orbitals, that are called sp3 hybrid orbitals. The symbol sp3 here identify the numbers and types of orbitals involved in the hybridization: ones and three p orbitals. For the hybridization process,
number of hybrid orbitals = the total number of atomic orbitals that are combined
It means that with total four orbitals combined, four new hybrid orbitals are generated, and they all named as sp3 hybrid orbitals. These new hybrid orbitals are all in the same energy level that is between those of 2s and 2p orbitals, and are directed in a tetrahedral shape overall with the angle between any two orbitals as 109.5°. Each sp3 hybrid orbital has two lobes that are very different in size. The lobe with the larger size is in the positive phase and is responsible for bonding.
Since there are four sp3 hybrid orbitals available, each of the four valence electrons occupies one of them, so there are four half-filled sp3 orbitals in the carbon atom that are able to form four bonds. Therefore, the C-H bond of CH4 is formed by the overlapping between the 1s orbital in the hydrogen atom and the sp3 orbital in the carbon atom.
Because the arrangement of the four sp3 hybrid orbitals is in a tetrahedral, the shape of the CH4 molecule is also a tetrahedral, which is consistent with the shape predicted by VSEPR. The tetrahedral shape of the sp3 carbon can usually be drawn using the solid and dashed wedges. Out of the fours bonds, the two bonds that lie within the paper plane are shown as ordinary lines, the solid wedge represent a bond that point out of the paper plane, and the dashed wedge represent a bond that point behind the paper plane. These perspective drawings that show the 3D tetrahedral shape is particularly important in the discussion of stereochemistry in Chapter 5.
1.6.3 Hybridization and VSEPR
Other than sp3 hybridization, there are also other types of hybridization that include sp, sp2, sp3d and sp3d2. Usually the hybridization on a certain atom can simply be determined by counting the total number of electron groups (bonding pairs and lone pairs). The total number of electron groups just equals the total number of orbitals involved in the certain hybridization. For example, in a CH4 molecule, the central carbon atom has four 4 bonding pairs, so the hybridization of carbon is sp3 (one s and three p orbitals, 1+3=4). If a central atom has total five 5 electron groups (bonding pairs and lone pairs all together) around, then the hybridization is sp3d (ones, three p and one d orbitals, 1+3+1=5).
This correlation may remind you of VSEPR. Hybridization and VSEPR are two separate concepts, however they can be correlated together via the number of electron groups in common. The following table is very useful in correlating the hybridization and VSEPR shape/bond angles around the central atom and the total number of electron groups together.
Hybridization on central atom
Total number of electron pairs (BP and LP) around central atom
Geometry (Shape) of electron groups (electron pairs)
sp
2
linear
sp2 3 trigonal planar
sp3 4 tetrahedral
sp3d 5 trigonal bipyramidal
sp3d2 6 octahedral
Table 1.3 Correlation between Hybridization and VSEPR
Exercises 1.8
1. What is the hybridization of the oxygen atom in H2O molecule?
2. What is the hybridization of the xenon atom in XeF4 molecule, and what is the shape of the whole molecule?
Answers to Practice Questions Chapter 1
1.6.4 The Hybridization and VSEPR in Organic Molecules
Organic molecules usually contain more than one central atom, so it is not practical to name the shape of the whole molecule; instead we can talk about the shape/bond angle about each central atom individually. For such purposes, make sure to include the lone pairs that are usually left out in the organic structures (refer to section 1.2.4). The different structural formulas of ethanol, acetic acid and ethanenitrile molecules are shown in the table below. The 3D molecular model for each compound is shown as well to help you visualize the spatial arrangement. We can see that the hybridization and VSEPR shapes need to be indicated for each internal atom separately. Taking the oxygen atom in the OH group of ethanol as an example, since there two pairs of lone pair electrons on the oxygen atom as well (omitted in the structures in the table though), the oxygen has sp3 hybridization and is in the tetrahedral shape.
Ethene (C2H4)
We will take Ethene (C2H4) as an example for understanding the structure of a double bond.
According to the structure formula of C2H4, there are three electron groups around each carbon. Through referring to Table 1.3 it is determined that both carbons are in sp2hybridization, with the trigonal planar shape and a 120° bond angle. What does sp2hybridization mean to the carbon atom in this compound? It means that only three orbitals are involved in the hybridization (one 2s and two of 2p orbitals) out of the total four, and there is one 2p orbital left out, or not included in the hybridization, which is called the unhybridized 2p.
The three new sp2 hybrid orbitals and the unhybridized 2p are directed in the following arrangement: the three sp2 hybrid orbitals are in the trigonal planar shape, and the unhybridized 2p is in the position that is perpendicular to the plane. Each orbital has one single electron, so all the orbitals are half-filled and are available for bonding. Both carbon atoms have the same set of orbitals (three sp2hybrid orbital and one unhybridized 2p) as shown below.
When the two carbons approach each other, the sp2 on the x axis overlaps head-to-head to form the C-C σ sigma bond, and the “unhybridized” 2p overlaps side-by-side to form another new bond. The side-by-side orbital overlapping forms the π (pi) bond.
So now we understand that the C=C double bond contains two different bonds: σ (sigma) bond from sp2 –sp2 orbital overlapping and π (pi) bond from 2p–2p overlapping. Because of the π bond, the overall shape of the whole C2H4 molecule is co-planar.
The other sp2 hybrid orbitals on each carbon atom overlap with 1s orbital of H atoms and give total four C-H σ (sigma) bonds.
Ethyne (C2H2)
Ethyne C2H2 (common name is acetylene) has a C≡C triple bond. Generally, triple bonds involve one σ sigma bond and two π (pi)bonds. Both carbon atom is in sp hybridization and in linear shape. With sp hybridization, each carbon has two sp hybrid orbitals and two unhybridized 2p orbitals. Each carbon uses one sp hybrid orbital to overlap head-to-head and gives the C-C the σ sigma bond, meanwhile the 2p orbitals overlap side-by-side to give two π bonds as shown in the diagram below. The other sp orbitals are used for overlapping with 1s of hydrogen atoms to form C-H σ bonds.
Image Descriptions
Table 1.4 image description: Ethanol’s CH3, CH2, and OH are all in a sp3 tetrahedral shape. Acetic acid’s CH3, and OH are in a sp3 tetrahedral shape and CO is in a sp2 trigonal planar. Lastly, ethanenitrile’s (acetonitrile) CH3 in a sp3 tetrahedral shape, and CN is in a sp linear shape. [Return to Table 1.4]
1.07: Answers to Practice Questions Chapter 1
Xin Liu
1.1 Number of valence electrons:
B: 3 valence electrons
N: 5 valence electrons
O: 6 valence electrons
Cl: 7 valence electrons
Mg: 2 valence electrons
1.2
• Identify the following bond is “polar” or “non-polar”?
C-C: non-polar C-H : non-polar (very close electronegativity for C and H)
B-F : polar. O-O : non-polar C=N : polar
• Rank the following bonds in the order of increasing bonding polarity: C—S, C—O, C—F (referring to the trend of EN, no need to use the exact EN values).
bonding polarity: C—S < C—O < C—F
1.3 Draw the Lewis structure of N2 molecule:
1.4 Why following structure is not the best way to show the Lewis structure of CO2?
Because the formal charges are not minimized in above structure. The formal charge in the best Lewis structure of CO2 are all zero, and the best Lewis structure of CO2 is shown here:
1.5 Draw all the equivalent resonance structures for following species. Include any non-zero formal charges in the structures.
• O3 molecule
• nitrate anion NO3
• chlorate anion ClO3
1.6 Draw all the resonance structures for azide anion, N3, and indicate the most stable one.
1.7 Draw new resonance structure and compare the relative stability, show arrows in the original structure.
1.8
• What is the hybridization of oxygen atom in H2O molecule?
• What is the hybridization of xenon atom in XeF4 molecule, and what is the shape of the whole molecule? | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/01%3A_Basic_Concepts_in_Chemical_Bonding_and_Organic_Molecules/1.06%3A_Valence_Bond_Theory_and_Hybridization.txt |
2.1.1 Structures and Different Structure Formulas
Alkane is the simplest hydrocarbon with only C-C single bonds. The chain alkane fits the general formula of CnH2n+2 (n: positive integer), and the number of H atoms reaches the maximum level in chain alkanes. The names and structures of straight-chain alkanes up to ten carbons are listed in the table below.
Number of Carbons Name
Formula (CnH2n+2)
Condensed Structure
1 methane CH4 CH4
2 ethane C2H6 CH3CH3
3 propane C3H8 CH3CH2CH3
4 butane C4H10 CH3CH2CH2CH3
5 pentane C5H12 CH3CH2CH2CH2CH3
6 hexane C6H14 CH3CH2CH2CH2CH2CH3
7 heptane C7H16 CH3CH2CH2CH2CH2CH2CH3
8 octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3
9 nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3
10 decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
Table 2.1 Names and Structures of Straight-Chain Alkanes
The primary sources of alkanes are natural gas and petroleum. Natural gas contains mainly methane (70 –90%) and some ethane. Petroleum refining separates crude oil into different fractions and each fraction consists of alkanes of similar number of carbons. Propane and butane are common fuels in propane gas burners and cigarette lighters. Alkanes with 5 to 8 carbons are the major components of gasoline, while diesel contains alkanes ranging from 9 to 16 carbons. As the number of carbons increase, the boiling point and viscosity of alkanes increase.
There are a variety of formats to show the structural formulas of organic compounds, it is important to be able to recognize different formula drawings, and use them correctly to represent the structures.
Kekul é Structure
We have had some discussions on Kekulé structures in section 1.2.4. They are similar to Lewis structures with all the bonding electrons shown in short lines and all the atoms included as element symbols. However, the lone-pair electrons are left out in Kekulé structures, which is the major difference between Kekulé structures of organic compound and Lewis structures.
Condensed Structure Formula
In condensed structure formulas, the C-H bonds are omitted and all the H atoms attached to a certain carbon (or other atoms) are usually shown as a group like CH3, CH2, NH2, OH. The structures in Table 2.1 are shown as condensed structures. The C-C bond sometimes can be omitted as well (as for 2-methylpropane and 2-hexanol in the examples below). Usually, if the structure has a branch, the bonding between the parent structure to the branch needs to be shown with a short line. It is faster to draw a structure with condensed structure formula, and the structure does not look as bulky as Kekulé structures.
Short-Line Structure Formula
The structure drawing can be further simplified by short-line structure (or “bond-line structure”, “skeletal formula” in other books) with most atoms omitted, it is also the very common type of structure formula used in Organic Chemistry because of its simplicity. To apply and interpret the short-line structures correctly, it is very important to understand the conventions of this type of drawing clearly.
• Each short line represents a bond.
• The carbon chains are shown in a zig-zag way.
• No carbon atoms are shown (as an exception, it is optional to show the CH3 group at the end of the chain, or as a branch); each bend in a line or terminus of a line represents a carbon atom, unless another atom is shown explicitly.
• Hydrogen atoms bonded to carbons are not shown; hydrogen atoms bonded to other atoms are shown explicitly.
• Atoms other than C and H, for example N, O, Cl, need to be shown explicitly.
In short-line structures, the number of hydrogen atoms attached to each carbon can be calculated by applying the octet rule and checking formal charges involved.
Perspective Formula of 3D Structure
When it is necessary to highlight the spatial arrangement of groups around a tetrahedral sp3 carbon for conformation (Chapter 4) or stereochemistry (Chapter 5) purposes, the perspective formula with solid and dashed wedges are used. Out of the four bonds on a tetrahedral carbon, two bonds lie within the paper plane and are shown as ordinary lines, the solid wedge represents a bond that points out of the paper plane, and the dashed wedge represents a bond that points behind the paper plane.
2.1.2 Constitutional Isomers
For methane, ethane and propane, there is only one way of carbon arrangement. As the number of carbon increases to 4 carbons, there are two ways for the carbon atoms to be connected, one as a straight-chain (blue structure below), and the other one as a branch on the chain (red structure below).
Two Constitutional Isomers with Formula C4H10
Butane
Isobutane (i-butane)
“iso” means “isomeric”
b.p. = 0 °C
b.p. = -12 °C
density: 0.622 g/mL density: 0.604 g/mL
As we can see, these two different structures represent two different compounds, with different names and different physical properties; however, they both have the same formula of C4H10, and they are called Constitutional (Structural) isomers. Constitutional (Structural) isomers are different compounds with the same molecular formula, but their atoms arranged in a different order. (i.e. the atoms are bonded in different ways.)
Let’s see more examples of constitutional isomers.
For alkanes with 5 carbons, there are a total of three constitutional isomers. Check the notes besides for the strategy to build constitutional isomers.
For alkanes with 6 carbons, there are a total of five constitutional isomers.
Exercises 2.1
Draw all the constitutional isomers with a formula of C7H16.
Answers to Practice Questions Chapter 2
The constitutional isomers we have so far have different lengths of carbon “backbones”, and are also called skeletal constitutional isomers. The other possible situations include positional and functional constitutional isomers that we will encounter later.
As the number of carbons increase, the number of constitutional isomers increases dramatically. For the example of alkanes with 20 carbons, that is C20H42, there are 366,319 constitutional isomers. While there is no simple formula allowing us to predict the total number of isomers for a certain amount of carbons, the phenomena of constitutional isomers partially explains the high diversity of organic structures.
2.1.3 Recognition of 1 °, 2°, 3°, 4°carbons
The carbon atoms in organic structure can be categorized as primary (1°), secondary (2°), tertiary (3°) and quaternary (4°), depending on how many other carbons it connects with. Specifically:
• Primary (1°) carbon: attached directly to only one other C atom;
• Secondary (2°) carbon: attached directly to two other C atoms;
• Tertiary (3°) carbon: attached directly to three other C atoms;
• Quaternary (4°) carbon: attached to four other C atoms.
The hydrogen atoms attached on 1°, 2° and 3° carbon, are labeled as 1°, 2° and 3° hydrogen respectively.
In one compound, carbons (or hydrogens) that belong to different category show different structural and reactive properties. This concept has a lot more applications in later sections. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/02%3A_Fundamental_of_Organic_Structures/2.01%3A_Structures_of_Alkenes.txt |
As we have realized that the number of constitutional isomers increases dramatically as the number of carbons increases, it is impossible to give each structure its own common name, like isobutane. So, a systematic method with certain rules is necessary when it comes to naming organic compounds. In this book, we will learn about IUPAC nomenclature; it is also the systematic nomenclature that has been widely adopted internationally. IUPAC nomenclature was initially designed by a commission for the International Union of Pure and Applied Chemistry in 1892, and it has been continually revised by the commission since then.
IUPAC NOMENCLATURE of ALKANES
1. Identify the longest continuous carbon chain as the parent chain. This chain determines the parent name (or last name) of the alkane.
• If there are two choices of the same length, then the parent chain is the longest chain with the greatest number of “branches”. The term substituent will be used from now on as the official name for “branch”.
1. Number the chain beginning at the end that is closest to any substituents, thus ensuring the lowest possible numbers for positions of substituents.
2. Use these numbers to designate the location of the substituent groups, whose names are obtained by changing the “-ane” suffix to “-yl“.
The substituents derived from alkane are also called alkyl groups.
1. If an alkyl substituent group appears more than once, use the prefixes di, tri, tetra, penta, hexa (meaning 2, 3, 4, 5, 6 respectively) for each type of alkyl group.
2. List the substituent groups alphabetically (use the substituent group name from step 3, ignore the prefixes from 4, but include “iso” and “cyclo”).
3. Write the name as a single word. Numbers are separated from letters by “-“; numbers are separated by “,”.
Alkane Naming Examples:
More notes about the branched alkyl groups: The common names of the branched alkyl groups have been used broadly, and are adopted as part of the IUPAC system. Understanding the origin of these common names is very helpful in distinguishing and memorizing the names.
Three-carbon branched alkyl groups
Both of the two 3-carbon branched alkyl groups come from propane. Since propane has two types of hydrogens, primary (1°) and secondary (2°), so there are two alkyl groups depends on which H is removed. Four-carbon branched alkyl groups Out of the four 4-carbon branched alkyl groups, two come from butane and the other two come from isobutane (or 2-methylpropane).
IUPAC name of branched alkyl groups
The branched alkyl groups can also be named by the IUPAC rules. To do that, they are treated as if it were a compound itself. Begin numbering at the point of attachment to the parent chain, and number the branches the same as before to avoid confusion. The complex substituent name is put in parentheses when the name of the complete molecule is written.
For the example of isobutyl below, the part that connect directly onto the parent chain has 3 carbons, so it is “propyl”. There is another CH3 on the 2nd carbon of propyl, therefore the whole group is called “2-methylpropyl”.
Naming of Cycloalkanes
Cycloalkanes are alkanes that contain a ring(s) as part of the structure. For the cycloalkane that contain one ring, there are two fewer hydrogens than the non-cyclic alkane, so the general formula of cycloalkanes with one ring is CnH2n.
IUPAC NOMENCLATURE of CYCLOALKANES
1. The parent name is “cycloalkane”.
2. Number the ring to provide the lowest possible numbering sequence (when two such sequences are possible, cite substituents in alphabetical order and the No.1 position is given to first cited substituent).
Example:
3. When both ring and chain are included in the structure, compare the number of carbons in ring vs chain, and select the one with more carbons as the parent structure; the other is treated as a substituent.
Example:
4. When higher priority functional groups are present (more in section 2.2), parent structure will contain that functional group.
Example: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/02%3A_Fundamental_of_Organic_Structures/2.02%3A_Nomenclature_of_Alkanes.txt |
Functional groups are the most reactive parts in organic compounds, and determine the major properties of compounds. The summary of common functional groups is included in Table 2.2. Knowing the functional groups well is one of the fundamental skills required for this course. It is required in order for students to quickly identify and name the functional groups included in molecules, as well as to understand, interpret and draw the specific structure of each functional group clearly. The IUPAC naming of compounds containing a couple of functional groups is required as well.
Alkene and alkynes are hydrocarbon functional groups; the π bond in multiple bonds accounts for the reactivity of alkenes and alkynes.
Benzene rings (C6H6) are a special type of hydrocarbon. Historically, because of the special aroma (sweet smelling) that benzene and its derivatives release, they are called aromatic compounds. The structure of benzene can be represented as three C=C double bonds alternate with single bonds, however, the actual structure of benzene has nothing to do with alkenes. Detailed discussions on the structure of benzene, which is a big conjugation system, and the chemistry definitions of aromatic/aromaticity will be a topic of Organic Chemistry II. Benzene rings can be shown with any of the following structure drawings.
When a halogen is connected with carbon, the group is called alkyl halide (or haloalkane). The halide can be categorized as a primary (1°), secondary (2°) or tertiary (3°) halide, depending on what category the carbon connected with the halogen is in.
Alcohol is a functional group that you probably are familiar with. In organic chemistry, the term alcohol refers to a compound containing the OH (hydroxy) group. Depending on the position of the OH group, alcohols can also be categorized as primary (1°), secondary (2°) or tertiary (3°).
Another functional group that contains the oxygen atom in single bonds is ether. In ether, the O atom connects with two carbon-containing R groups through two C-O σ bonds. For compounds with ether as the only functional group, it is usually named with the common name “alkyl alkyl ether”. When the two alkyl groups are the same, they can be combined as “dialkyl”.
Ether can be in cyclic structure as well. It may not be that intuitive to recognize the following structure as ether, and labelling the carbon atom will be helpful for identification.
Both nitrile and nitro groups contain nitrogen atom, and it might be easy to get them mixed up. Nitrile has a C≡N triple bond, and therefore can only be at the end of a structure, while nitro (NO2) can be in any position on the carbon chain or ring.
Amine is the organic derivative of ammonia, NH3. When the hydrogen atom(s) in NH3 is replaced with R groups, it produces amine. The amine can be primary (1°), secondary (2°) or tertiary (3°) depending on how many R groups are connected with nitrogen. The amines can also be named with common names.
For the functional groups on the 2nd part of Table 2.2, they all have a common structural unit of a carbonyl group C=O; the different structure of “W” in the general formula determines the nature of the functional group. It is usually more challenging to identify and draw these functional groups correctly, because they are kind of similar. More practice is needed.
Aldehyde and ketone are similar in terms of their structures and properties. Aldehyde can be regarded as a special case of ketone since “H” can be regarded as an R with zero carbon. Because H has to be connected on one side of the C=O group in aldehyde, aldehyde can only be at the end of a structure. Ketone, on the other hand, must be in the middle position to ensure both sides of the C=O groups are connected with R groups. Ketone can also be in a cyclic structure.
The last four functional groups are related in terms of structures and chemical properties. When an OH group is connected with C=O, the whole COOH is called a carboxylic acid functional group. The other three, ester, anhydride and amide, are all derivatives of carboxylic acid, meaning they can be prepared with carboxylic acid as the starting material. For these three functional groups, it is important to remember that the “W” part has to be considered together with the C=O, and overall it determines the functional group correctly. For example, the COOR is ester; it can not be recognized as a “ketone” plus an “ether”. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/02%3A_Fundamental_of_Organic_Structures/2.03%3A_Functional_Groups.txt |
With the ability to identify functional groups, next we will learn how to give IUPAC names to compounds containing a few functional groups, by following a set of rules.
IUPAC NOMENCLATURE of COMPOUNDS with FUNCTIONAL GROUPS
1. Find the longest carbon chain containing the functional group with highest priority (see Table 2.3). This chain determines the parent name of the compound.
2. Change the ending of the parent alkane/alkene/alkyne to the suffix of the highest priority group, which gives the parent name of the compound (usually, drop the last letter “e” before adding the suffix, except for nitrile where the “e” is kept).
3. Number the chain from the end closest to the highest functional group.
4. The other groups are named as substituents by using the appropriate prefixes.
5. Assign stereochemistry, E/Z or R/S, as necessary (details in Chapter 5).
For naming purposes, the functional groups are assigned with priorities (Table 2.3). If the compound includes more than one functional groups, the one with the highest priority is the “parent structure” and determines the “parent name”; the other groups will be regarded as “substituents”. “Suffix” is used to indicate the name of the parent structure, and “prefix” is for the substituent. The order of the groups listed in Table 2.3 is based on the decreasing order of the priority, where carboxylic acid group is in the highest priority. The groups in the subordinate table have no difference in terms of priority, and they are usually listed in the alphabetic order.
We will go through several examples for more details about the naming rules.
1.
The parent structure is the 6-carbon carboxylic acid with a double bond, so the last name comes from “hexene”. To add the suffix, the last letter “e” will be dropped, so the parent name is “hexeneoicacid”. A number is necessary to indicate the position of the double bond, so the name is “4-hexenoic acid”. The carboxylic acid group is always on the #1 position, so it is NOT necessary to include that number for the position.
2.
This is a ketone based on a cycloalkane, so the last name comes from “cyclohexane’. By adding the suffix, it become “cyclohexanone”, and the complete name is “3-ethylcyclohexanone”.
3.
With the multiple groups involved, the ketone has the highest priority, so it decides the last name. The 8-carbon alkene chain with ketone should be name as “octenone”. The numbers on the chain should start from the left side to ensure that ketone has the lowest number. When the OH group is regarded as a substituent, it is indicated by the prefix “hydroxy”. So the complete name is “5-bromo-7-chloro-6-hydroxy-2,2,5-trimethyl-7-octen-4-one”.
4.
It is not difficult to find the parent structure for this compound, which is a cyclic alcohol, so the last name is “cyclopropanol”. The naming of the substituent with the benzene ring is bit challenging. When benzene is a “substituent”, it is called “phenyl”; and since there is an isopropyl group on the “phenyl”, the whole substituent is called “3-isopropylphenyl”, and the complete name of the compound is “2,2-dimethyl-3-(3-isopropylphenyl)cyclopropanol”.
5.
In ester, an OR group replaces the OH group of a carboxylic acid. When naming the ester, the name of the R in the OR group is stated first, followed by the name of the acid, with “oic acid” replaced by “oate”. As a net result, the R in the OR is regarded as the “substituent”, even though it is not. So, the complete name of the ester above is “tert-butyl propanoate”.
Naming of substituted benzene and benzene derivatives
For substituted benzene, the benzene ring is regarded as the parent structure, and the positions and names of substituents are added to the front.
For di-substituted benzene, there is another unique way to indicate the relative position of the two substituents by using ortho-, meta- and para-. Although this o-, m-, p- system is the common naming system for benzene derivatives, they have been applied broadly in books and literatures.
• ortho- (o-): 1,2- (next to each other in a benzene ring)
• meta- (m): 1,3- (separated by one carbon in a benzene ring)
• para- (p): 1,4- (across from each other in a benzene ring)
For the following mono-substituted benzene derivatives, phenol, benzoic acid and benzaldehyde, their common names are adopted in the IUPAC system.
When other substituents are introduced into those benzene derivatives, the common name will be used as the parent name of the compound with the base functional group (OH for phenol, COOH for benzoic acid and CHO for benzaldehyde) given the #1 position. For example:
When benzene is the connected with a carbon chain that has six or more carbons, the carbon chain should be regarded as the parent structure, and the benzene ring becomes the substituent and will be indicated with the prefix “phenyl”. An example is given here: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/02%3A_Fundamental_of_Organic_Structures/2.04%3A_IUPAC_Naming_of_Organic_Compounds_with_Functional_Groups.txt |
Now with lots functional groups introduced, the extent of constitutional isomers will be expanded a lot. To further explore the phenomena of constitutional isomers, we will need to understand the concept of Degree of Unsaturation (or: Index of Hydrogen Deficiency/IHD).
Let’s compare three compounds first: pentane, 1-pentene and cyclopentane
The formula for pentane is C5H12. For a compound containing 5 carbons, the maximum number of hydrogens is 12, so the structure of pentane is saturated (no more hydrogen atoms can be added in), or we can say that pentane has zero degree of unsaturation.
For 1-pentene C5H10, there are two less hydrogens than the saturated level (pentane), which means the 1-pentene has one degree of unsaturation. With a ring introduced, cyclopentane (C5H10) also has to sacrifice two hydrogens, so cyclopentane also has one degree of unsaturation. The trend is that when a double bond (essentially a π bond), or a ring, is involved in the structure, it leads to one degree of unsaturation of the compound.
Formula
Degree of Unsaturation/
Index of Hydrogen Deficiency (IHD)*
Structure Unit Involved
CnH2n+2
0
chain alkane only
CnH2n
1
1 double bond
or 1 ring
CnH2n-2
2
2 double bonds
or 2 rings
or 1 double bond plus 1 ring
or 1 triple bond
Table 2.5 Summary of degree of unsaturation/IHD vs structure unit involved
The degree of unsaturation could be accumulated, and Table 2.5 summarizes the situations up to two degrees. As we can see, adding 1 ring or 1 π bond contributes to one degree of unsaturation. Therefore, the essential meaning of degree of unsaturation is the “number of rings plus π bonds” in a structure.
If the structure of a compound is available to us, the total degrees of unsaturation can simply be counted through inspecting the structure.
Example:
If the formula of a compound is given, we can also calculate the degree of unsaturation by comparing the number of hydrogens vs the saturated level, by using the equation:
(n: number of carbons; X = number of H + number of Halogen – number of N)
This is a general equation that accounts for the presence of heteroatoms as well. Please note that oxygen atoms are ignored in this calculation.
For example, for a compound with a formula given as C4H7NO, it is calculated that the degree of unsaturation is 2 for this compound:
Now we are ready to solve constitutional isomer questions with the application of degrees of unsaturation. Usually, the formula information is available to us for such questions, and we will need to build constitutional isomers based on the given formula together with other requirements. To solve this type of question, it is very helpful to do it strategically by following certain steps:
• Calculate the degree of unsaturation based on the given formula.
• With the value of this specific unsaturation degree, how many double bonds or rings might be included in the structure?
• Combine your knowledge of functional groups with the degree of unsaturation, as well with certain atoms included in the formula, to see what functional group(s) may be possible.
• Build constitutional isomers according to the above information (separate the isomers by different functional group).
Examples: Draw and name all the constitutional isomers with the molecular formula C4H10O.
Approach: Answering the following questions lead you to the solution.
• What is the degree of unsaturation for the formula C4H10O? 0
• How many double bonds, or rings, could be involved? none
• What are the possible functional groups that matche with that degree of unsaturation, and include one oxygen atom? alcohol or ether
• With these hints, we can try to “build” the constitutional isomers for each functional group separately. total seven structures
Solutions:
alcohols:
ethers:
Exercises 2.2
Draw all the constitutional isomers that include a C=O bond with formula the C5H10O.
Answers to Practice Questions Chapter 2 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/02%3A_Fundamental_of_Organic_Structures/2.05%3A_Degree_of_Unsaturation_Index_of_Hydrogen_Deficiency.txt |
2.6.1 Intermolecular Forces
In Organic Chemistry, the understanding of physical properties of organic compounds, for instance boiling point (b.p.), molecular polarity and solubility, is very important. It provides us with helpful information about dealing with a substance in the proper way. Those physical properties are essentially determined by the intermolecular forces involved. Intermolecular forces are the attractive force between molecules and that hold the molecules together; it is an electrical force in nature. We will focus on three types of intermolecular forces: dispersion forces, dipole-dipole forces and hydrogen bonds.
Dispersion Forces
Dispersion Forces (also called London Forces) result from the instantaneous dipole and induced dipole of the molecules. For nonpolar molecules, the constant shifting and distortion of electron density leads to a weak short-lived dipole at a given moment, which is called an instantaneous dipole. Such temporary dipoles will induce the electrons in a neighbouring molecule to get distorted as well, and to develop a corresponding transient dipole of its own, which is the induced dipole. At the end, all nonpolar molecules are attracted together via the two types of temporary dipoles as shown in Fig. 2.6a. The dispersion force is weak in nature, and is the weakest intermolecular force. However, since it applies to all types of molecules (it is the only intermolecular force for nonpolar molecules), dispersion forces are also the most fundamental intermolecular force.
The magnitude of dispersion forces depends on two factors:
• The relative polarizability of electrons. The simple understanding of polarizability is how easily the electrons get distorted. For larger atoms, there are more electrons in a larger space, therefore the electrons are more loosely held and more easily polarized, so the dispersion force is stronger. Generally, the larger the molar mass of the molecule, the stronger the dispersion force.
• The relative surface area of the molecule. Molecules with longer, flatter or cylindrical shapes have a greater surface area compared to the bulky, branched molecules, and therefore have a stronger dispersion force. Taking the two constitutional isomers of C4H10 (section 2.1.2), butane and isobutane as an example, the dispersion force of butane is stronger than that of isobutane.
Dipole-Dipole Force
For polar molecules, molecules are attracted to each other because of a permanent dipole, and this type of attractive force is called a dipole-dipole force. As shown below in the electrostatic potential map of acetone, one end of acetone has a partial negative charge (red) and the other end has a partial positive charge (blue). The dipole-dipole force is an attraction force between the positive end of one molecule and the negative end of the neighbouring molecule.
Hydrogen Bonds
First of all, do not let the name mislead you! Although it is called a “bond”, a hydrogen bond is not a covalent bond, it is a type of intermolecular force. The hydrogen bond is the force between a H atom that is bonded to O, N or F (atoms with high electronegativity) and the neighbouring electronegative atom,. It can be shown in a general way as:
The most common example of hydrogen bonding is for water molecules. Water has two O-H bonds, and both are available as hydrogen bond donors for neighbouring molecules. This explains the extraordinarily high b.p. of water (100 °C), considering the rather small molar mass of 18.0 g/mol. As a comparison, the methane molecule CH4 with a similar size has a b.p. of -167.7 °C.
For organic compounds, hydrogen bonds play important roles in determining the properties of compounds with OH or NH bonds, for example alcohol (R-OH), carboxylic acid (R-COOH), amine (R-NH2) and amide RCONH2.
The three major types of intermolecular forces are summarized and compared in Table 2.6.
Polar vs Non-Polar molecules
As indicated in Table 2.6, the nature of molecular polarity determines the types of force(s) applied to a certain substance. So here we will have discussions about how to tell whether a molecule is polar or non-polar.
The polarity of the compound can be determined by its formula and shape.
For diatomic molecules, the molecular polarity is the same as the bonding polarity. That means all homonuclear molecules, like H2, N2, O2, F2, are non-polar because of their non-polar bond, while all heteronuclear molecules, like HF, HCl, are polar.
For polyatomic molecules, the molecular polarity depends on the shape (refer to VSEPR in Section 1.5) of the molecule as well. Let’s see the examples of H2O and CO2.
Both H2O and CO2 have two polar bonds. H2O is in the bent shape, so the bond polarities of the two O-H bonds add up to give the molecular polarity of the whole molecule (shown above), therefore H2O is polar molecule. On the other hand, the shape of CO2 is linear, and the bond polarities of the two C=O bonds cancel out, so the whole CO2 molecule is non-polar.
There are other examples of non-polar molecules where the bond polarity cancels out, such as BF3, CCl4, PCl5, XeO4 etc.
For organic compounds, the hydrocarbons (CxHy) are always non-polar. This is mainly because of the small electronegativity difference between carbon atoms and hydrogen atoms, making C-H bonds technically non-polar bonds.
For other organic compounds that contain functional groups with heteroatoms, like R-O-R, C=O, OH, NH, they are all polar molecules.
The diagram here (Fig. 2.6g) provides a summary of all the discussions about molecular polarities.
Other than the three types of intermolecular forces, there is another interaction that is very important for understanding the physical property of a compound, which is the ion-dipole force.
Ion-Dipole Force
Ion-dipole force is not categorized as an intermolecular force, however it is a type of important non-covalent force that is responsible for the interaction between ions and other polar substance. A simple example is the dissolving of an ionic solid, or salt, in water. When table salt (NaCl) is dissolved in water, the interactions between the ions and water molecules are strong enough to overcome the ionic bond that holds the ions in the crystal lattice. As a result, the cations and anions are separated apart completely, and each ion is surrounded by a cluster of water molecules. This is called a solvation process. The solvation occurs through the strong ion-dipole force. Lots salts, or ionic compounds, are soluble in water because of such interactions.
2.6.2 Physical Properties and Intermolecular Forces
The comprehension of intermolecular forces helps us to understand and explain the physical properties of substances, since it is intermolecular forces that account for physical properties such as phases, boiling points, melting points, viscosities, etc. For organic chemistry purposes, we will focus on boiling point (b.p.) and solubility.
Boiling point (b.p):
The boiling point trend of different substance directly correlates with the total intermolecular forces. Generally speaking, the stronger the overall intermolecular force applied to a certain substance, the higher the boiling point of the substance. Boiling point is the temperature at which the liquid phase of the substance vaporizes to become a gas. In order to vaporize a liquid, the intermolecular forces that hold the molecules together must be overcome. The stronger the forces, the more energy is needed to overcome the forces, and a higher temperature is required, thus leading to a higher boiling point.
Example:
All three compounds here have similar Molar Masses, so the dispersion forces are at a similar level. However, the three compounds have different molecular polarities. Butane is a non-polar substance that only has dispersion forces, propanal is a polar molecule with both dispersion forces and dipole-dipole forces, and propanol is a polar molecule with an OH bond, so all three types of forces apply to. Therefore, the overall amount of intermolecular forces is strongest for propanol, and weakest for butane, which is in the same order as their boiling points.
Solubility:
A general rule for solubility is summarized by the expression “like dissolves like”. This means that one substance can dissolve in another with similar polarity, and as a result, with similar intermolecular forces. More specifically:
• Nonpolar substances are usually soluble in nonpolar solvents.
• Polar and ionic substances are usually soluble in polar solvents.
• Polar and nonpolar substances are insoluble to each other.
Determining the polarity of a substance has already been summarized in an earlier part of this section (Fig. 2.6g). Water, methanol and ethanol are examples of very polar solvents that can form Hydrogen bonds. Ether, ketone, halide and esters are polar solvents as well, but not as polar as water or methanol. Non-polar solvents include hydrocarbons like hexane, benzene, toluene etc.
For some organic compounds, however, it may not be that easy to simply call it polar or non-polar, because part of the compound may be polar, and the another part may be nonpolar. This is often described as hydrophilic or hydrophobic.
• Hydrophobic (hydro, water; phobic: fearing or avoiding) meaning it does not like water, or is insoluble in water;
• Hydrophilic (hydro, water; philic: loving or seeking) meaning it likes water, or is soluble in water.
The hydrocarbon part of the organic compound is hydrophobic, because it is nonpolar and therefore does not dissolve in polar water. The functional group of OH, COOH, NH2etc is polar and is therefore hydrophilic. With both hydrophobic and hydrophilic parts present in an organic compound, the overall polarity depends on whichever part is the major one. If the carbon chain is short (1~3 carbons), the hydrophilic effect of the polar group is the major one, so the whole compound is soluble in water; with carbon chains of 4~5 carbons, the hydrophobic effect begins to overcome the hydrophilic effect, and water solubility is lost.
The solubility differences of different alcohols demonstrates this trend clearly; as the length of the carbon chain increases, the solubility of alcohol in water decreases dramatically (Table 2.7):
Alcohol
Solubility in water
(g/100mL)
methanol, ethanol, propanol
(CH3OH, CH3CH2OH, CH3CH2CH2OH)
miscible
(dissolve in all proportions)
1-butanol (CH3CH2CH2CH2OH)
9
1-pentanol (CH3CH2CH2CH2CH2OH)
2.7
1-octanol (CH3CH2CH2CH2CH2CH2CH2CH2OH)
0.06
Table 2.7 Solubility of different alcohols in water
For organic compounds that are water insoluble, they can sometimes be converted to the “salt derivative” via a proper reaction, and thus can become water soluble. This method is used commonly in labs for the separation of organic compounds.
Example:
Applying acid-base reactions is the most common way to achieve such purposes. As shown in the above example, by adding a strong base to the benzoic acid, an acid-base reaction occurs and benzoic acid is converted to its salt, sodium benzoate, which is water soluble (because of the ion-dipole force as we learned earlier). The benzoic acid can therefore be brought into water (aqueous) phase, and separated from other organic compounds that do not have similar properties.
2.07: Answers to Practice Questions Chapter 2
2.1 Draw all the constitutional isomers with a formula of C7H16
2.2 Draw all the constitutional isomers that include C=O bond with formula C5H10O. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/02%3A_Fundamental_of_Organic_Structures/2.06%3A_Intermolecular_Force_and_Physical_Properties_of_Organic_Compounds.txt |
15
The most commonly applied definition of acids and bases is the Brønsted-Lowry definition:
• Brønsted-Lowry Acid: a substance that can donate a proton (H+);
• Brønsted-Lowry Base: a substance that can accept a proton (H+).
Therefore, according to the Brønsted-Lowry definition, an acid-base reaction is a proton transfer process in which the acid gives away a proton and base accepts a proton as shown in the general equation:
General equation for acid-base reaction
The species that forms when an acid loses its proton is called the conjugate base of that acid; similarly, the species that forms when a base accepts a proton is called the conjugate acid of that base. In the general equation above, HA is the conjugate acid of A, and A is the conjugate base of HA. HA and A can also be called a conjugate acid-base pair; another pair is HB+ and B.
A strong acid donates the proton completely, and the arrow “→” can be used in the reaction equation to indicate that the reaction goes to completion. The dissociation reaction of the strong acid HCl in water is used as an example here:
HCl (g) + H2O (l) →H3O+(aq) + Cl(aq)
For weak acids (HA is used as a general formula), the proton is only donated partially and the reaction stays at equilibrium. The equilibrium arrow “” will be needed in the reaction equation to indicate the equilibrium status:
HA (aq) + H2O (l) ⇔ H3O+ (aq) + A (aq)
The equilibrium constant for the above reaction is called the acid dissociation constant, Ka. It is a constant to measure the relative strength of an acid. The expression for Ka is:
The larger the Ka value, the stronger the ability of the acid to donate protons, and the stronger the acid is. (Technically, when the Ka value is larger than 10, the acid can be regarded as a strong acid.)
For the conjugate acid-base pair, the stronger the acid, the weaker the conjugate base is, and vice versa.
3.02: Organic Acids and Bases and Organic Reaction Mechanism
3.2.1 Organic Acids
The acids that we talked about in General Chemistry usually refers to inorganic acids, such as HCl, H2SO4, HF etc. If the structure of the acid contains a “carbon” part, then it is an organic acid. Organic acids donate protons in the same way as inorganic acids, however the structure may be more complicated due to the nature of organic structures.
Carboxylic acid, with the general formula of R-COOH, is the most common organic acid that we are familiar with. Acetic acid (CH3COOH), the ingredient of vinegar, is a simple example of a carboxylic acid. The Ka of acetic acid is 1.8×10-5.
Another common organic acid is the organic derivative of sulfuric acid H2SO4.
The replacement of one OH group in H2SO4 with a carbon-containing R (alkyl) or Ar (aromatic) group leads to the organic acid named “sulfonic acid”, with the general formula of RSO3H, or ArSO3H. Sulfonic acid is a strong organic acid with a Ka in the range of 106. The structure of a specific sulfonic acid example called p -toluenesulfonic acid is shown here:
Other than the acids mentioned here, technically any organic compound could be an acid, because organic compounds always have hydrogen atoms that could potentially be donated as H+. Only a few examples are shown here with the hydrogen atoms highlighted in blue:
Therefore, the scope of acids has been extended to be much broader in an organic chemistry context. We will have further discussions on the acidity of organic compounds in section 3.3, and we will see more acid-base reactions applied to organic compounds later in this chapter.
3.2.2 Organic Bases
While it is relatively straightforward to identify an organic acid since hydrogen atoms are always involved, sometimes it is not that easy to identify organic bases. According to the definition, a base is the species that is able to accept the proton. Organic bases may involve a variety of different structures, but they must all share the common feature of having electron pairs that are able to accept protons. The electron pairs could be lone pair electrons on a neutral or negative charged species, or π electron pairs. Organic bases could therefore involve the following types:
• Negatively charged organic bases: RO(alkyloxide), RNH(amide), R(alkide, the conjugate base of alkane). Since the negatively charged bases have a high electron density, they are usually stronger bases than the neutral ones.
Note: Keep in mind that the lone pairs are usually omitted in organic structures as mentioned before. For example, with the formula of CH3NH given, you should understand that the N actually has two pairs of lone pair electrons (as shown in the above structure) and it is a base.
• Neutral organic bases, for example amine, C=O group and C=C group
• Amine: RNH2, R2NH, R3N, ArNH2 etc (section 2.3). As organic derivatives of NH3, which is an inorganic weak base, amines are organic weak bases with lone pair electrons on N that are able to accept the proton.
• Functional groups containing oxygen atoms: carbonyl group C=O, alcohol R-OH, ether R-O-R. The lone pair electrons on O in these groups are able to accept the proton, so functional groups like aldehyde, ketone, alcohol and ether are all organic bases. It may not that easy to accept this concept at the first, because these groups do not really look like bases. However, they are bases according to the definition because they are able to accept the proton with the lone pair on the oxygen atom.
Adjust your thinking here to embrace the broader scope of acids and bases in an organic chemistry context.
Here, we will take the reaction between acetone and H+ as an example, to understand the reaction deeply by exploring the reaction mechanism, and learn how to use the curved arrows to show it.
A reaction mechanism is the step-by-step electron transfer process that converts reactants to products. Curved arrows are used to illustrate the reaction mechanism. Curved arrows should always start at the electrons, and end in the spot that is receiving the electrons. The curved arrows used here are similar to those for resonance structures ( section 1.4 ), but are not exactly same though. Please note that in resonance structures, the curved arrows are used to show how the electrons are transferred within the molecule, leading to another resonance structure. For mechanism purposes, there must be arrows that connect between species.
Notes for the above mechanism:
• For the acid-base reaction between C=O group and the proton, the arrow starts from the electron pair on O, and points to the H+ that is receiving the electron pair. A new O-H bond is formed as a result of this electron pair movement.
• In this acid-base reaction, ketone is protonated by H+, so this reaction can also be called the “protonation of ketone”.
• The product of the protonation is called an “oxonium ion”, which is stabilized with another resonance structure, carbocation.
• Alkene (C=C): Although there are no lone pair electrons in the C=C bond of alkene, the π electrons of the C=C double bond are able to accept proton and act as base. For example:
Example: Organic acid and base reaction
Predict and draw the products of following reaction and use curved arrow to show the mechanism.
Approach: If H+ is the acid as in previous examples, it is rather easy to predict how the reaction will proceed. However, if there is no obvious acid (or base) as in this example, how do you determine which is the acid, and which is the base?
Methanol CH3OH is neutral, and the other reactant, NH2, is a negatively charged amide. The amide with a negative charge has higher electron density than the neutral methanol, therefore amide NH2should act as base, and CH3OH is the acid that donates H+.
Solution:
Exercises 3.1
Predict and draw the products of following reaction and use curved arrow to show the mechanism.
Answers to Practice Questions Chapter 3 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/03%3A_Acids_and_Bases-_Organic_Reaction_Mechanism_Introduction/3.01%3A_Review_of_Acids_and_Bases_and_Ka.txt |
As we mentioned before, all organic compounds could be acids, because they all have hydrogen atoms that could potentially be donated. Most organic acids are weak acids with a small Ka. For example, acetic acid CH3COOH has a Ka of 1.8×10-5. Lots of other organic acids are even weaker than acetic acid, and it is this weak acidity that makes it difficult to realize that some organic compounds are actually acids.
However, this weak acidity is very important in Organic Chemistry. Since it is not that very convenient to say or to remember Ka values like 1.8×10-5, pKa is used more often in Organic Chemistry to refer to the relative acidity of different acids. The definition of pKa is:
pKa = -logKa
The smaller the pKa value, the larger the Ka, and the stronger the acidity is.
The pKa of most organic acids range between 5~60. While it is impossible to know the pKa of every organic compound, it is very useful to understand the pKa (and acidity) based on the functional groups involved, because the same functional groups usually have similar pKas. The approximate ranges of pKa values for seven major functional groups are listed in Table 3.1, which serves as a very valuable starting point for us to predict and understand the acidity of any organic molecule. The strongest organic acid listed here is carboxylic acid, with a pKa of about 5; the weakest organic acids are the alkanes with pKa values of over 50. Since approximate ranges of pKa values are listed in the table, the exact pKa value of a group varies for different compounds because of the structural differences. Fortunately however, it is usually not necessary to know the exact pKa values for most cases in organic chemistry, and the approximate range is good enough.
• Acidity is the ability of a compound to donate H+, so when we talk about the acidity (Ka and pKa) of an organic compound, it must be about a specific H atom (highlighted blue in the table). For different H atoms in the same compound, the acidity and pKa are different. As for the example of methanol:
• It is very useful to memorize the approximate ranges of pKa listed in Table 3.1.
• The acidity of the functional groups in the table decreases from top to the bottom, and the basicity of the conjugate bases in the last column increases from top to bottom, because the stronger the acid, the weaker the conjugate base is.
Predict the Outcome of Organic Acid-Base Reaction — Use pKa as Criterion
With the knowledge of acidity and pKa, we are now ready to see how to apply this information to the understanding of organic reactions from an acid-base perspective.
The following reaction is an example in Section 3.2. If you take a closer look at the reactants and products, you will find that the “product” side also contains an acid (ammonia NH3), and a base (methoxide CH3O). Now the question is, how can we be so sure that the reaction proceeds to the “product” side as written? The question can also be asked in a different way: if equilibrium is established for the reaction mixture, which side will the position of the equilibrium predominantly favour? Left or right?
To answer that question, we will learn about a general rule for acid-base reaction: Acid-base reactions always favour the formation of the weaker acid and the weaker base. This is because the equilibrium always favours the formation of more stable products, and weaker acids and bases are more stable than stronger ones.
With pKa values available at hand, the relative acidity of reactants vs products can be compared by comparing their pKa values, and the reaction will proceed to the side of the acid with a larger pKa (larger pKa means smaller Ka, therefore weaker acid).
So for this reaction, the pKa check indicates that ammonia NH3 is a weaker acid than methanol CH3OH, so the reaction does proceed to the right side with CH3O and NH3 as the major products.
Notes: Only comparing between acids is good enough for this purpose, because if CH3OH is stronger than NH3, then the conjugate base CH3Omust be weaker than the other base NH2.
Examples
Show the products of the following reactions and predict the predominant side of the equilibrium.
Reaction2
Are there any practical applications for such a prediction? Yes! Let’s compare the two reactions in the exercises above. Reaction 1 indicates that if ethyne (HC≡CH) and amide (NH2) are mixed together, the reaction does proceed to the products side, meaning HC≡CH could be deprotonated by amide NH2. However, if HC≡CH and hydroxide OHare mixed together as shown in reaction 2, no reaction occurs, or we can say that HC≡CH can not be deprotonated by OH because OH is not strong enough! So if you are working in the lab and have the option of choosing between NH2or OHto deprotonate HC≡CH, you now know which one to choose.
The idea that OH is not a strong enough base may bother you a lot, since it conflicts with the “common knowledge” that we learned in General Chemistry, where OH is a strong base. Generally speaking, OH is a pretty strong base; however, it is just barely not strong enough to deprotonate HCCH, which is a very weak acid, with a pKa of about ~25. Since HCCH is much weaker than the “weak acids” we learned in General Chemistry, a much stronger base, like NH2, is required to deprotonate it. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/03%3A_Acids_and_Bases-_Organic_Reaction_Mechanism_Introduction/3.03%3A_pKa_of_Organic_Acids_and_Application_of_pKa_to_Predict_Acid-Base_Reaction_Outcome.txt |
We have learned that different functional groups have different strengths in terms of acidity. In this section, we will gain an understanding of the fundamental reasons behind this, which is why one group is more acidic than the other one. Many of the concepts that we will learn here will continue to apply throughout this course as we tackle many other organic topics.
A. Periodic Trend: Electronegativity
The element effect is about the individual atom that connects with the hydrogen (keep in mind that the acidity is about the ability to donate a certain hydrogen). Let’s compare the acidity of hydrogens in ethane, methylamine and ethanol as shown below.
A clear trend in the acidity of these compounds is: the acidity increases for the elements from left to right along the second row of the periodic table, C to N, and then to O. This is consistent with the increasing trend of electronegativity along the period from left to right. The connection between electronegativity and acidity can be explained as the atom with a higher electronegativity being able to better accommodatethe negative charge of the conjugate base, therefore stabilizing the conjugate base in a better way. Therefore, the more stable conjugate base, the weaker the conjugate base is, and the stronger the acid is. For the discussions in this section, the trend in the stability (or basicity) of the conjugate bases often helps to explain the trend of the acidity.
The relative acidity of elements in the same period is:
For elements in the same period, the more electronegative an atom, the stronger the acid is; t he acidity increases from left to right across the period.
B. Group (vertical) Trend: Size of the atom
When moving vertically within a given group on the periodic table, the trend is that acidity increases from top to bottom. This can be illustrated with the haloacids HX and halides as shown below: the acidity of HX increases from top to bottom, and the basicity of the conjugate bases Xdecreases from top to bottom.
The acidity of the H in thiol SH group is also stronger than the corresponding alcohol OH group, following the same trend. For example, the pKa of CH3CH2SH is ~10, which is much more acidic than ethanol CH3CH2OH with a pKa of ~16.
In order to make sense of this trend, we will once again consider the stability of the conjugate bases. When moving vertically in the same group of the periodic table, the size of the atom overrides its electronegativity with regards to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly larger volume, so I is more stable and less basic, making HI more acidic.
The relative acidity of elements in the same group is:
For elements in the same group, the larger the size of the atom, the stronger the acid is; the acidity increases from top to bottom along the group.
3.4.2. Resonance Effect
The resonance effect accounts for the acidity difference between ethanol and acetic acid. For both ethanol and acetic acid, the hydrogen is bonded with the oxygen atom, so there is no element effect that matters. However, the pKa values (and the acidity) of ethanol and acetic acid are very different. What makes a carboxylic acid so much more acidic than an alcohol? As stated before, we begin by considering the stability of the conjugate bases, remembering that a more stable (weaker) conjugate base corresponds to a stronger acid.
For acetate, the conjugate base of acetic acid, two resonance contributors can be drawn and therefore the negative charge can be delocalized (shared) over two oxygen atoms. However, no other resonance contributor is available in the ethoxide ion, the conjugate base of ethanol, so the negative charge is localized on the oxygen atom. As we have learned in section 1.3, the species that has more resonance contributors gains stability, therefore acetate is more stable than ethoxide, and is weaker as the base, so acetic acid is a stronger acid than ethanol.
The charge delocalization by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the big difference of over 10 pKa units between ethanol and acetic acid. Because pKa = –log Ka, that means that there is a factor of about 1010 between the Ka values for the two molecules!
Examples
The pKa of the OH group in alcohol is about 15, however OH in phenol (OH group connected on a benzene ring) has a pKa of about 10, which is much stronger in acidity than other alcohols. Explain the difference.
Solution:
The difference can be explained by the resonance effect. There is no resonance effect on the conjugate base of ethanol, as mentioned before. However, the conjugate base of phenol is stabilized by the resonance effect with four more resonance contributors, and the negative is delocalized on the benzene ring, so the conjugate base of phenol is much more stable and is a weaker base. Therefore phenol is much more acidic than other alcohols.
Exercises 3.2
• Practice drawing the resonance structures of the conjugate base of phenol by yourself!
• It is because of the special acidity of phenol (and other aromatic alcohols), that NaOH can be used to deprotonate phenol effectively, but not to normal alcohols, like ethanol. Show the reaction equations of these reactions and explain the difference by applying the pKa values.
Answers to Practice Questions Chapter 3
3.4.3 Inductive Effect
Let’s compare the pKa values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:
The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, and the argument here apparently does not have to do with the element effect. The resonance effect does not have to do with it either, because no additional resonance contributors can be drawn for the chlorinated molecules. Rather, the explanation for this phenomenon involves something called the inductive effect. A chlorine atom is more electronegative than hydrogen, and is thus able to ‘induce’, or ‘pull’ electron density towards itself via σ bonds in between, and therefore helps to spread out the electron density of the conjugate base, the carboxylate, and stabilize it. The chlorine substituent can be referred to as an electron-withdrawing group because of the inductive effect.
The inductive effect is the charge dispersal effect of electronegative atoms through σ bonds. The inductive effect is addictive; more chlorine atoms have an overall stronger effect, which explains the increasing acidity from mono, to di-, to tri-chlorinated acetic acid. The following diagram shows the inductive effect of trichloro acetate as an example.
Because the inductive effect depends on electronegativity, fluorine substituents have a stronger inductive effect than chlorine substituents, making trifluoroacetic acid (TFA) a very strong organic acid.
In addition, because the inductive effect takes place through covalent bonds, its influence decreases significantly with distance — thus a chlorine that is two carbons away from a carboxylic acid group has a weaker effect compared to a chlorine just one carbon away.
3.4.4 Hybridization Effect
To introduce the hybridization effect, we will take a look at the acidity difference between alkane, alkene and alkyne.
The hydrogen atom is bonded with a carbon atom in all the three functional groups, so the element effect does not invoke. Also considering about the conjugate base of each, there is no extra resonance contributor possible.
The key difference between the conjugate base anions is the hybridization of the carbon atom, that is sp3, sp2 and sp respectively for alkane, alkene and alkyne. Different hybridizations leads to different s character, that is the percent of s orbitals out of the total amount of orbitals. The sp3 hybridization means 25% s character (one s and three p orbitals, so s character is 1/4 = 25%), sp2 hybridization has 33.3% s character, and the number is 50% for sp hybridization. Electrons of 2s orbitals are in the lower energy level than those of 2p orbitals because 2s is much closer to the nucleus. So for the anion with more s character, the electrons are closer to the nucleus and experience stronger attraction, therefore the anion has lower energy and is more stable.
The relative stability of the three anions (conjugate bases) can also be illustrated by the electrostatic potential map, in which the lighter color (less red) indicate less electron-density of the anion, and the higher stability.
This can also be stated in a more general way that more s character in the hybrid orbitals make the atom more electronegative. For the same atom, an sp hybridized atom is more electronegative than sp2 hybridized atom, which is more electronegative than sp3 hybridized atom.
3.05: Lewis Acids and Lewis Bases
The Brønsted-Lowry definition works well for the reactions we learned so far, however it also limits the scope of acid-base reactions in a way where the proton H+ must be involved. Lewis acids and Lewis bases are defined in a more inclusive way that was first introduced by G. N. Lewis in 1923.
Lewis Acid: a species that can accept an electron pair;
Lewis Base: a species that can donate an electron pair.
All Brønsted-Lowry acids and bases fit into the Lewis definition, because the proton transfer process is essentially the reaction where the base uses its electron pair to accept a proton, as indicated by the mechanism arrow that we learned earlier. Therefore in the following reaction, the BL acid, H+, is also the Lewis acid, and BL base, NH3, also fits to the definition of the Lewis base.
However, the Lewis definition is broader and covers more situations. For the following reaction, B(CH3)3 is the Lewis acid because boron has an incomplete octet, and the empty 2p orbital on boron is able to accept electrons. (CH3)3N behaves as the Lewis base with the lone pair electron on N that is able to be donated.
The product between Lewis acids and Lewis bases is usually a species that has the acid and base joined together, and the product is called the “LA-LB adduct”.
Other examples of Lewis acids include electron-deficient species, such as H+, M+, M2+, BH3, BF3, AlCl3 etc. Lewis bases can be: amine, ether or other species that have lone pair electrons to donate.
Exercises 3.3
Show the product of the following LA-LB reaction:
Answers to Practice Questions Chapter 3
3.06: Answers to Practice Questions Chapter 3
3.1 Predict and draw the products of following reaction; use curved arrows to show the mechanism.
3.2
• Practice drawing the resonance structures of the conjugate base of phenol by yourself!
Solutions included in the section.
• It is because of the special acidity of phenol (and other aromatic alcohol) that NaOH can be used to deprotonate phenol effectively, but not to normal alcohols, like ethanol. Show the reaction equations of these reactions and explain the difference by applying the pKa values.
3.3 Show the product of the following LA-LB reaction: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/03%3A_Acids_and_Bases-_Organic_Reaction_Mechanism_Introduction/3.04%3A_Structural_Effects_on_Acidity_and_Basicity.txt |
Thumbnail: The ethane molecule spends 99% of its time in a specific conformation. The so-called staggered conformation is reached if, when the molecule is seen from a position on the C-C axis (as in the second half of the animation), the H atoms of the front C atom are exactly between the H atoms of the other C atom. (CC BY 2.5 Generic; ralk via Wikipedia)
04: Conformations of Alkanes and Cycloalkanes
4.1.1 Conformation
At a molecular level, a property of σ (sigma) bonds in alkane is that the bonds keep on rotating. For the example of ethane (CH3CH3), one methyl (CH3) group is able to rotate around the C-C bond freely without any obstacles.
It is highly recommended that the molecular model is used here to “see” the bond rotation. With a molecular model on hand, you can hold one methyl group steady, and rotate the other methyl group.
The C-C bond is formed by the sp3-sp3 orbitals overlapping and the bond is cylindrically symmetrical, so rotation about the bond can occur easily and the molecule does not seem to change. However, a closer look indicates that the rotation of the C-C bond does result in a different spatial arrangement of hydrogen atoms in the molecule, as shown below:
The different spatial arrangements of the atoms/groups that result from the single bond rotation are called conformations. Molecules with different conformations are called conformational isomers or conformers. The two extreme conformations of ethane coming from the C-C rotation shown above are: the staggered conformation with all of the H atoms spread out, and the eclipsed conformation with all of the H atoms overlapped.
In the study of conformation, it is convenient to use certain types of structural formulas. The formula used in the drawing above is the perspective formula (see section 2.1.1) that shows the side-view of the molecule. In perspective formulas, solid and dashed wedges are used to show the spatial arrangement of atoms (or groups) around the sp3 carbons.
Another structural formula is the sawhorse formula that shows the tilted top-view of the molecule.
The most commonly applied formula in conformation analysis is the Newman projection formula.
How to draw a Newman projection
To draw a Newman projection, we will imagine viewing the molecule from one carbon to the next carbon atom directly along a selected CC bond, as shown below, and follow the rules:
• The front carbon atom is shown as a point with three other bonds:
• The rear carbon atom is shown as a circle with three other bonds:
• Put the two carbons together to get the Newman projection of the staggered conformation:
• From the staggered conformation, fix the front carbon in place, and rotate the rear carbon by 60° to get the eclipsed conformation:
4.1.2 Conformation Analysis of Ethane
Next, we will do a conformation analysis of ethane by using the Newman projections. A conformation analysis is an investigation of energy differences and relative stabilities of the different conformations of a compound.
The two conformations of ethane, staggered and eclipsed, are different and therefore should be in different energy levels. You may also predict intuitively that the staggered conformation is more stable and is lower energy, because the C-H bonds are arranged as far apart as possible in that conformation. That is correct! In eclipsed conformations, the H atoms on the front carbon are overlapping with the H atoms on the rear carbon, and this arrangement causes the repulsion between the electrons of C-H bonds of the two carbons. This type of repulsion is called the torsional strain, also known as the eclipsing strain. Due to the torsional strain, the eclipsed conformer is in the energy level that is 12 kJ/mol (or about 2.9 kcal/mol) higher than the staggered one. This can be represented graphically in a potential energy diagram as shown in Figure 4.1f.
Because of this energy difference, an energy barrier must be overcome when the rotation about the C-C bond occurs. However, this energy difference in ethane is rather small, and the kinetic energy of molecules at room temperature is high enough to cover it. So at room temperature, the changes from staggered to eclipsed conformers occur millions of times per second. Because of these continuous interconversions, these two conformers cannot be separated from each other. However, at any given moment, about 99% of the ethane molecules will be in a staggered conformation because of their higher stability.
4.1.3 Conformation Analysis of Propane
A similar analysis can be applied to propane as well. We will find that there are still two types of conformations, staggered and eclipsed, resulting from the rotation. The difference between propane and ethane is that there is a methyl (CH3) group connected on the rear carbon for propane. However, that does not affect the relative stability, and the staggered conformer is more stable and in lower energy.
4.1.4 Conformation Analysis of Butane
There are three C-C bonds in butane, and rotation can occur about each of them. If we pick up C1-C2 (or C3-C4) for the study, the situation is almost the same as propane, with the ethyl CH2CH3 group replace the CH3 group. However, if we consider the rotation about the C2-C3 bond, the situation will be much more complex.
For both carbon atoms, C2 and C3, there are two hydrogen atoms and one methyl CH3 group bonded with. We can start with the conformer in which the two CH3 groups are opposite to each other, then fix the front carbon and do 60° rotations of the rear carbon to investigate all the possible conformations.
Exercises 4.1: Draw all the possible conformers of butane from viewing along the C2-C3 bond. Finish this practice by yourself before continue reading!
Tips for drawing all the possible conformers about a certain C-C bond:
• View along that C-C bond; circle and decide what atoms/groups are connected on each carbon;
• Start with the staggered conformation in which the largest groups on each carbon are opposite (far away) to each other (this is called the “anti”conformation as we will learn later);
• Keep the groups on one carbon “fixed”, and rotate the groups on the other carbon at 60° angles. Repeat the rotation five times, and you should get total of six conformers.
Answers to Practice Questions Chapter 4
Among all the six conformers obtained, there are three staggered and three eclipsed. Staggered conformations C and E should be in the same energy level because the groups are arranged in equivalent way between these two conformers. Similarly, eclipsed conformations F and B are also in the same energy level. So our studies can be focused on the four conformers, A, B, C and D, that are different in terms of energy and stability.
Between the two staggered conformers A and C, A is more stable than C because the two methyl CH3 groups in A are as far apart as possible. This most stable staggered conformation is called the anti conformation (anti is Greek for “opposite”). In anti conformations, the largest groups on the front and rear carbon are 180° opposite to each other. The other staggered conformation C is called a gauche conformation, in which the two large groups are adjacent and are 60° to each other. With the large groups being close to each other in gauche conformers, the molecule experiences steric strain. Steric strain is the strain that is caused when atoms (or groups) are close enough together that their electron clouds repel each other. Steric strain only matters when the groups are close to each other (less or equal to 60°), so steric strain does not apply in anti conformations. The magnitude of steric strain also depends on the size of group; the larger the size, the higher the steric strain. As a result, there is no steric strain between two small hydrogen atoms, even if they are close to each other.
Between the two eclipsed conformers B and D , D is less stable than B , because the two CH3 groups are eclipsing (overlapping) each other in D , causing both torsional and steric strains.
The energy difference of all the conformers obtained from the rotation about the C2-C3 bond are shown in the potential energy diagram Fig. 4.1l . The curve is more complex than that of ethane since there are four different energy levels corresponding to four conformers with different stabilities. Even the energy barriers for the rotations are larger than that of ethane, but they are still not high enough to stop rotation at room temperature.
Exercises 4.2
Draw all conformers for 3-methylpentane by viewing along the C2-C3 bond, and order them from the most stable to least stable.
Answers to Practice Questions Chapter 4 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/04%3A_Conformations_of_Alkanes_and_Cycloalkanes/4.01%3A_Conformation_Analysis_of_Alkanes.txt |
While the open chain alkanes have conformational isomers because of bond rotation, will this apply to cycloalkanes as well? In this section, we will take a look at properties of cycloalkanes first, and then investigate how the different conformers of cycloalkanes contribute to the different stabilities.
The short line structural formulas of cycloalkenes simply look like shapes such as a triangle, square etc. The internal angles of the shapes can be calculated with geometry, as shown below.
An interesting fact about the cycloalkanes is that they have different relative stabilities, and the stability depends on the size of the ring. It has been observed that cyclic compounds found in nature usually are in 5- or 6-membered rings, and the 3- or 4-membered rings are rather rare.
To explain this stability difference, German chemist Adolf von Baeyer proposed the “Bayer Strain Theory”. By assuming all the rings are in a flat (or planar) shape, Bayer Theory suggests that the difference between the ideal bond angle (which is 109.5° for sp3 carbon) and the angle in the planer cycloalkane causes the strain, which is called angle strain. According to the Bayer Theory, cyclopentane would be the most stable because its bond angles, 108°, are closest to the ideal angle of 109.5°. Cyclopropane would be the least stable one since it has the largest angle deviation of 49.5° (60° vs 109.5°). It was also predicted that cyclohexane would be less stable than cyclopentane because of the larger angle deviation (10.5° deviation for cyclohexane vs 1.5° for cyclopentane), and as the number of sides in the cycloalkanes increases beyond six the stability would decrease.
However, experimental results show a different trend. It turns out that cyclohexane is the most stable ring that is strain-free, and is as stable as a chain alkane. Furthermore, cyclic compounds do not become less and less stable as the number of rings increases.
To measure the relative stability of cycloalkanes, the heat of combustion (ΔHcomb) for each cycloalkane was measured. The heat of combustion is the amount of heat released when the compounds burns completely with oxygen. The cycloalkanes will be in higher energy levels than corresponding chain alkanes because of strain energy. Therefore, when cycloalkane burns, more heat will be released, so the difference of ΔHcomb between cycloalkane vs the “strainless” chain alkane is just the amount of strain energy, as shown below. The larger the difference, the higher the strain energy of the cycloalkane. The strain energy for different cycloalkanes measured by this method are listed in Table 4.1.
combustion reaction: (CH2)n + 3n/2 O2 n CO2 + n H2O + heat
cyclopropane cyclobutane cyclopentane cyclohexane
Strain Energy (KJ/mol) 114 110 25 0
Table 4.1 Strain Energies of Cycloalkanes
The major drawback of the Baeyer Theory was that we must assume that all the rings are flat. The highest stability of cyclohexane from experimental results indicate that the rings may not be in a planar shape. We will have a closer look at the actual shape and conformation of 3-, 4-, 5- and 6-membered cycloalkanes.
Cyclopropane
With three carbons for the ring, cyclopropane must be planar.
The bond angle in cyclopropane is 60°, derived significantly from the optimal angle of 109.5°, so it has very high angle strains. The sp3-sp3 orbitals can only overlap partially because of the angle deviation, so the overlapping is not as effective as it should be, and as a result the C-C bond in cyclopropane is relatively weak.
Because of the poor overlapping of sp3-sp3orbitals, the bonds formed in cyclopropane resemble the shape of a banana, and are sometimes called banana bonds.
Other than the angle strains, all the adjacent C-H bonds are eclipsed in cyclopropane, therefore the torsional strains are applied as well. Such a strain can be “viewed” more clearly from the Newman projection of cyclopropane.
The Newman projection of cyclopropane might seems weird at first glance. For cyclopropane, there are three carbons, so the CH2 group connects with both front and rear carbons of the Newman projection.
Because of the high level of angle strains and torsional strains, 3-membered rings are unstable. They rarely exist in nature and undergo ring-opening reaction easily to release the strains.
Cyclobutane
Cyclobutane is not planar. The ring puckers (or folds) slightly due to the efforts of releasing some torsional strain. Meanwhile, cyclobutane still has a considerable amount of angle strains as the internal angles become about 88° with the folded shape. Overall, cyclobutane is an unstable structure with rather high level of strains.
Cyclopentane
Cyclopentane is not planar as well and the total level of strain is lowered quite a lot. It also puckers and adopts a bent conformation where one carbon atom sticks out of the plane of the others, which helps to release the torsional strain by allowing some hydrogen atoms to become almost staggered.
This bent shape of cyclopentane is also called the “envelope” conformation. The envelope conformation can undergo a process called “ring flipping” as a result of C-C bond rotation. More discussions about ring flipping will be included in the section of cyclohexane. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/04%3A_Conformations_of_Alkanes_and_Cycloalkanes/4.02%3A_Cycloalkanes_and_Their_Relative_Stabilities.txt |
Chair conformation of cyclohexane
Cyclohexane is the most stable cycloalkane. It is strain-free, meaning neither angle strains nor torsional strains apply, and it shows the same stability as chain alkanes. This special stability is due to a unique conformation that it adopts. The most stable conformation of cyclohexane is called “chairconformation, since it somewhat resembles a chair.
In the chair conformation of cyclohexane, all the carbons are at 109.5º bond angles, so no angle strain applies. The hydrogens on adjacent carbons are also arranged in a perfect staggered conformation that makes the ring free of torsional strain as well. This will be illustrated more clearly later when we learn about the Newman projection of the chair conformation.
Properties of the chair conformation
In the chair conformation of cyclohexane, the total twelve C-H bonds can be divided into two categories based on the orientations, which are axial (“a”) and equatorial (“e”). In the structure below, the six red coloured bonds are axial and the six blue coloured bonds are equatorial. Axial bonds are vertical and perpendicular to the average plane of the ring, while the equatorial bonds are more “flat” and extend from the perimeter of the ring. For both “a” and “e”, they can either point up ↑(above the ring), or point down ↓(below the ring). The trending of “a” and “e” bonds in the chair conformation can be summarized as:
• Each carbon has one “a” bond and one “e” bond; if one bond points up ↑(above the ring), the other has to point down ↓(below the ring);
• For the same type of bonds, the orientation up ↑ and down ↓ alternates from one carbons to the adjacent carbon, meaning if a certain carbon has a ↑, then the adjacent carbon must has a ↓;
• For the total twelve C-H bonds: 3a↑ , 3 a↓, 3 e↑ , 3 e↓.
How to draw chair conformation
It is important to be able to understand and recognize all the bonds in the chair conformation, and you are also expected to be able to draw the conformation correctly and quickly. The procedure is:
1. Draw two parallel lines of the same length both point slightly down (if connected, they would form a parallelogram with an internal angle of about 60°/120°).
2. Connect the right ending points of the two lines with a “V” shape, so that the vertex of the V points to the upper right
3. Connect the left starting points of the two lines with another “V” shape, so that the vertex of the V points to the bottom leftAdd up all of the “a” bonds on each carbon as the vertical lines, follow the alternating trend on adjacent carbons
4. Add all of the “e” bonds by following the trend in which on a certain carbon, if an “a” bond points up, then an “e” bond must point down, and vice versa. Also notice that the “e” bond is parallel to the C-C bond which is one bond away, as shown below. The “green e” is parallel to the “green C-C bond”, and the “blue e” is parallel to the “blue C-C bond”. (It is more challenging to draw “e” bonds, and following the above trend makes it easier).
Ring flipping
When a cyclohexane ring undergoes a chair-chair conformation conversion, that is known as ring flipping. Ring flipping comes from C-C bond rotation, but since all of the bonds are limited within the ring, the rotation can only occur partially, which leads to the ring “flipping”. Cyclohexane rapidly interconverts between two stable chair conformations because of the ease of bond rotation. The energy barrier is about 45 kJ/mol, and the thermal energies of the molecules at room temperature are great enough to cause about 1 million interconversions to occur per second.
For cyclohexane, the ring after flipping still appears somewhat identical to the original ring, however there are some changes happening on the C-H bonds. Specifically, all the “a” bonds become “e” bonds and all the “e” bonds become “a” bonds, however their relative positions in terms of the ring, up or down, remain the same. The ring flipping is shown in the equation below. Compare the carbon with the same numbering in the two structures to see what happened to the bonds due to ring flipping.
Taking C #1 as an example, you will notice that the red a converted to a red e↓, and the blue econverted to a blue a after ring flipping.
Summary of ring flipping for chair conformation:
• This is NOT rotation, but ring flipping
• The two structures are conformation isomers (or conformers)
• all “a” bonds become “e” bonds and all “e” bonds become “a” bonds
• These two conformations are equivalent for the cyclohexane ring itself (without any substituents), with the same energy level
A molecular model is very useful in understanding ring flipping.
Newman projection of chair conformation
The chair conformation is strain free, with all the C-H bonds in staggered position. However, it is not that easy to see the staggered conformation in the drawings we have so far, and Newman projection help for this purpose.
To draw Newman projections for the chair conformation of cyclohexane, we also need to pick up the C-C bond to view along, just as we did for alkanes. Since there are a total of six C-C bonds, we will pick two of them, and these two need to be parallel to each other.
For the chair conformation example here, the two blue parallel C-C bonds are picked up for viewing, which are C1-C2 and C5-C4. (There are 3 pairs of parallel bonds in the chair conformation, any pair can be picked with the resulting Newman projection looking the same).
For the C1-C2 bond, C1 is the ‘front” carbon and C2 is the “rear” carbon; for the C5-C4 bond, C5 is the ‘front” carbon and C4 is the “rear” carbon. These two bonds will be represented by two “Newman projections” that we are familiar with (two circle things) and each represents two carbons, as shown below:
Keep in mind that there are a total of six carbons in the ring, and the the drawing above only shows four of them with C3 and C6 being left out. Additionally, the two “separated” Newman projections above are actually connected to both C3 and C6, so the overall Newman projection of the chair conformation of cyclohexane looks like this:
The staggered conformation of hydrogens are shown clearly in the Newman projection here!
Notes for Newman projections of the chair conformation (refer to the drawing below):
• The “a” or “e” bonds on four carbons (C1, C2, C4 and C5) are shown explicitly, while the bonds on C3 and C6 are just shown as CH2.
• The vertical red C-H bonds are the “a” bonds, the “flat” blue C-H bonds are the “e” bonds.
• The dashed line in the drawing below can be regarded as the average plane of the ring. Those above the line are the bonds point up ↑, those below the line are the bonds point down ↓.
Other conformation of cyclohexane
The chair conformation is the most stable one with the lowest energy, but it is not the only conformation for cyclohexane. During the ring flipping from one chair conformation to another chair conformation, the ring goes through several other conformations, and we will only talk briefly about the boat conformation here.
The boat conformation comes from partial C-C bond rotations (only flipping one carbon up to convert the chair to a boat) of the chair conformation, and all the carbons still have 109.5º bond angles, so there are no angle strains. However, the hydrogens on the base of the boat are all in eclipsed positions, so there are torsional strains. This can be illustrated by the Newman projection below. The Newman projection is drawn by viewing along C6-C5 and C2-C3 bonds of the above boat conformation.
Other than that, the two hydrogen atoms on C1 and C4 are very close to each other and causes steric strain. This is also called the “flagpole” interaction of the boat conformation. The two types of strains make the boat conformation have considerably higher energy (about 30 kJ/mol) than the chair conformation. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/04%3A_Conformations_of_Alkanes_and_Cycloalkanes/4.03%3A_Conformation_Analysis_of_Cyclohexane.txt |
Monosubstituted cyclohexane
For the cyclohexane ring itself, the two conformers from the ring flipping are equivalent in terms of energy since there are always six hydrogens in axial position and six hydrogens in equatorial position. For substituted cyclohexane however, the two chair conformations are not equivalent any more. Let’s see the example of methylcyclohexane.
Methylcyclohexane has two chair conformations that are interconvertible through the ring flipping. In conformation I the methyl group occupies an axial position, and in conformation II the methyl group occupies an equatorial position. Studies indicate that the conformer II with the equatorial-methyl is more stable, with the energy of about 7.6 kJ/mol lower than the other conformer.
This difference is due to the “1,3-diaxal interaction”. In axial-methyl conformation, the methyl CH3 group (regarded as #1 position) is very close to the axial hydrogens that is one carbon away (regarded as #3 position), and it causes the repulsion between each other that is called the 1,3-diaxal interaction. This type of repulsion is essentially the same as the gauche steric strain because the CH3 group and the CH are in gauche position. While for equatorial-methyl conformer, no such strains applied because the CH3 group and the CH are in anti-position. This interaction could be illustrated more clearly by Newman projection.
For mono-substituted cyclohexane, the equatorial-conformer is more stable than the axial-conformer because of the 1,3-diaxal interaction.
Since 1,3-diaxal interaction is essentially the steric strain, so the larger the size of the substituent, the greater the interaction is. For t-butylcyclohexane, the conformation with the t-butyl group in the equatorial position is about 21 kJ/mol more stable than the axial conformation.
Because of the stability difference between the two chair conformers, the equatorial-conformation is always the predominant one in the equilibrium mixture. The larger the size of the substituent, the larger the energy difference and the equilibrium constant K, so the equilibrium lies more toward the “equatorial” side. For methylcyclohexane, there is about 95% of equatorial-conformer in the mixture, and the percentage is about 99.9% for t-butylcyclohexane.
Disubstituted cyclohexane
When there are two substituents on different carbons of a cycloalkane, there are two possible relative position between the two groups, they can be either on the same side, or opposite side, of the ring, that are called geometric isomers, a type of stereoisomers (more discussions in Chapter 5). The isomer with two groups on the same side of the ring is the “cis” isomer, and the one with two groups on opposite side is called the “trans” isomer. Because the C-C bond can not rotate freely due to the restriction of the ring, the two geometric isomers can not be interconverted.
So now when considering about the conformational isomer, the stereoisomers should be taken into account as well. The general guideline for determining the relative stability of conformers for a certain isomer is:
• The steric effects of all substituents are cumulative, more substituents in equatorial positions, when possible, the more stable the conformation isomer will be.
• For different substituents, the conformer with larger substituent in equatorial positionis more stable.
Let us start with cis-1,2-dimethylcyclohexane, and compare between the two possible chair conformations:
For both conformations, there is one methyl group in equatorial and the other methyl group in axial, so the two conformers are equivalent, have same energy and stability level.
How to tell a isomer in chair conformation is cis or trans? A general way to recognize is to check that whether a group attached by the bond is above the ring (↑, point up), or below the ring (↓, point down). If both groups point to the same side, the compound is cis isomer; otherwise it is trans isomer.
How about the trans-1,2-dimethylcyclohexane? There are also two possible chair conformations:
In one conformation both methyl groups are axial, in the other conformation both methyl groups are equatorial. These two conformers are not equivalent, and the di-equatorial one is the more stable conformation as we would expect.
cis-1-fluoro-4-isopropylcyclohexane is the structure with two different substituents. Both chair conformations have one axial substituent, and one equatorial substituent. According to the guideline, the conformer with larger substituent in equatorial is more stable because if the large group is axial, stronger steric strain will be generated and it is less stable.
Exercises 4.3
Determine which is the more stable isomer, cis -1-ethyl-2-methylcyclohexane or trans -1-ethyl-2-methylcyclohexane?
Tips: draw all the chair conformers of each isomer, and decide which is the most stable one.
Answers to Practice Questions Chapter 4
4.05: Answers to Practice Questions Chapter 4
4.1 Solutions included in the section.
4.2 Draw all conformers for 3-methylpentane by viewing along C2-C3 bond, and order them from the most stable to least stable.
4.3 Determine which is the more stable isomer, cis-1-ethyl-2-methylcyclohexane or trans-1-ethyl-2-methylcyclohexane?
The trans-isomer has the most stable conformer with both substituents are at equatorial positions, therefore trans-isomer is the more stable one. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/04%3A_Conformations_of_Alkanes_and_Cycloalkanes/4.04%3A_Substituted_Cyclohexanes.txt |
Thumbnail: Two enantiomers of a generic amino acid that are chiral. (Public Domain; unknonw author via Wikipedia)
05: Stereochemistry
In stereochemistry topic, we will learn more different types of isomers. To clarify the concepts, it is a good idea to have a summary about the isomers in organic chemistry. There are two major types of isomers, constitutional isomers and stereoisomers. We have had detailed discussions on constitutional isomers in Chapter 2, and will focus on stereoisomers in this chapter. Stereoisomers are molecules with same bonding, but groups are in different spatial arrangement. At beginning of this chapter, we will learn more about geometric isomers in alkenes and the E/Z naming system. Then we will move on to a brand new category of stereoisomers, the isomers with chirality center. The flowchart here (Fig. 5.1a) show the correlation and difference between different types of isomers (pay attention to the definitions included) and provides useful guideline for the learning.
5.02: Geometric Isomers and E Z Naming System
Geometric Isomers of Alkenes
In the discussions about 1,2-dimethylcyclohexane in Chapter 4, we have learned that there are two geometric isomers possible for that compound, that are cis and trans. The restricted C-C bond rotation of cyclic structure result in the cis or trans isomer of 1,2-dimethylcyclohexane. Restricted rotation also can be caused by a double bond, so geometric isomers apply to some alkenes as well.
For the example of 2-butene, the condensed structural formula CH3-CH=CH-CH3 does not really represent the trigonal planar shape of the sp2 carbons with double bonds. To show the shape explicitly, we need to draw the Kekulé structure that show all the bond angles. Then it will be noticed that there are two different shapes of 2-butene, with the CH3 groups on either the same side or opposite side of the double bond.
They are geometric isomers and can be labelled as cis or trans in a similar way as disubstituted cycloalkane. Cis/trans is the common designation for geometric isomers and might be ambiguous for some structures, here we will learn the IUPAC naming system for geometric isomers of alkene, that is the E/Z naming system.
E/Z Naming System
To do the E/Z designation, at first, the groups connected on each sp2 double bond carbon will be assigned the priority based on the atomic number (see following guidelines for details), then the isomer with same priority group on the same side of double bond is assigned as “Z”, and the isomer with the same priority group on the opposite side of double bond is called “E”. Both E and Z come from German, “Zusammen” means same side and “Entgegen” means opposite.
The guidelines for assigning group priority in E/Z naming system
1. Priority is assigned based on the atomic number of the atoms bonded directly to the sp2 double bond carbon, the larger the atomic number, the higher the priority (isotopes with higher mass number has higher priority). For example: S > O > N > C > H.
For the above structure of 2-penetene: on the left side sp2 carbon, methyl group CH3 is higher than hydrogen atom because C > H; on the right side sp2 carbon, ethyl group CH2CH3, is also higher than hydrogen. With higher priority group on both side of the double bond, this is the Z isomer, the complete name of the compound is (Z)-2-pentene.
The group withhigher priority is labelled as #1, and the group with lower priority is labelled as #2 in this book.
2. If the two groups bonded directly on an sp2 carbon start with the same atom, means there is a tie from step 1, then we move on to the atoms that connected to the “tied” atom, priority increases as the atomic number of the next attached atom increases.
For the above structure, it is obvious that Cl is higher than C (C of CH2CH3 group) on the right side sp2 carbon.
On the left side sp2 carbon, we need to compare between methyl CH3 group and ethyl CH2CH3 group. Both groups has carbon atom attached directly on the sp2 carbon, that is a tie. In CH3 group, the carbon atom is bonded to H, H, H; while in CH2CH3 group, the carbon atom is bonded with H, H, C. So ethyl CH2CH3 is higher than methyl CH3 (see Note below). With higher priority group on opposite side of the double bond, this is the E isomer, the complete name of the compound is: (E)-3-chloro-4-methyl-3-hexene.
Note #1: For this round of comparison between H, H, H and H, H, C, compare the single atom with the greatest number in one group verse the single atom with the greatest number in the other group. So H in one group verse C in the other group, since C > H, therefore CH2CH3 is higher than CH3. Remember do not add the atomic numbers. For example, if one group has C, C, C, and the other group has C, O, H, then the C, O, H side is higher because O is higher than C.
Note #2: The above compound is cis-isomer if using the cis/trans naming system (both ethyl group are on the same side of double bond), but is E-isomer for E/Z system. So the cis/trans and E/Z are two different naming systems, don’t always match.
3. Repeat step 2 if necessary, until the priority is assigned.
Examples: What is the correct structural formula of (E)-2-bromo-3-chloro-2-butene?
The answer is B.
Examples: Draw the structure of (E)-3-methyl-2-pentene
Answer
Examples: Order the following groups based on increasing priority.
Approach:
1st round: C, C, C, C (tie);
2nd round:
A: C bonded to C, C, C; (3rd)
B: C bonded to H, Cl, Cl; (Cl is the 2nd high)
C: C bonded to H, C, C; (4th)
D: C bonded to H, H, Br (Br is the highest)
Solution: C < A < B < D
Exercises 5.1
Order the following groups based on decreasing priority for E/Z naming purpose.
Answers to Practice Questions Chapter 5
4. When multiple bond is part of the group, the multiple bond is treated as if it was singly bonded to multiple of those atoms. Specifically:
For these three groups involve multiple bonds, they all start with the carbon atom (the carbon atom highlighted in blue color), and we should compare the group of atoms that connected on the blue carbon by converting the multiple bond to “multiple single bonds”, as shown above. So, if we compare the order of these three groups, it is:
Examples: Assign E/Z of the circled double bond.
Thinking:
The answer is: Z-isomer. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/05%3A_Stereochemistry/5.01%3A_Summary_of_Isomers.txt |
Other than geometric isomers, there is another type of stereoisomer that is related to a special property called chirality. We will start with the basic concepts of chirality, then expand the topic further from there.
5.3.1 Chiral and Chirality
To talk about chirality, let’s first take a closer look at our hands, left hand and right hand. The left hand can be regarded as the mirror image of the right hand, and vice versa. Now let’s try to superimpose (overlay) the left hand on right hand, can you do that?
No! No matter how hard you try, the left hand can not be superimposed on the right hand. This is because of the special property of hand, that is called chirality. Both left and right hand are chiral (ky-ral), and show chirality. Chiral derived from the Greek word cheir, that means “hand”, and chirality means “handedness”.
The definition of the chirality is the property of any object (molecule) of being non-superimposable on its mirror image. The left and right hand are mirror image to each other, and they are not superimposable, so both left hand and right hand are chiral. Other than that, you can also find lots other objects in daily life that show chirality as well.
If an object is superimposable on its mirror image (for such case the object and its mirror image are exact identical), then this object is not chiral, that can be said as achiral.
In organic chemistry, we are interested in organic molecules that are chiral. Let’s see the following molecular models that represent a molecule and its mirror image.
In the models here, the four balls with different colors represent four different substituents, and the two structures are mirror image to each other. The effort of trying to superimpose one structure to the other does not work. Therefore, according to the definition of chiral/chirality, both molecules are non-superimposable on the mirror image, so they both chiral and show chirality.
The chirality of the molecule results from the structure of the central carbon. When the central carbon is sp3 carbon, and bonded with four different groups (represented by four different colors in the model), the molecule is chiral. The central carbon is called chirality center (or asymmetric center). The molecule with one chirality center must be chiral. Chirality center can also be called asymmetric center. We will use the term chirality center in this book.
It is highly recommended that the molecular model set is used as learning tool in this chapter. Assemble the model as shown above to understand the concept of chiral and chirality. The model is also very useful for the R/S assignment later in this section.
As a summary, a chirality (asymmetric) center should meet two requirements:
• sp3 carbon;
• bonded with four different groups.
Examples
For following compounds, label each of the chirality center with a star.
Approach:
• The carbons in CH3 or CH2 are NEVER chirality centers. The chirality center must be the carbon bonded with a branch (or branches).
• sp2 double bond carbon is NEVER a chirality center.
• Carbon in a ring can also be chirality center as long as it meet the two requirements.
• Not all the above compounds have a chirality center.
Solution:
Exercises 5.2
1. Draw the structure of following compounds, determine which one has an chirality center and label it with a star.
a) 1-bromobutane,
b) 1-pentanol,
c) 2-pentanol,
d) 3-pentanol,
e) 2-bromopropanoic acid
f) 2-methyl cyclohexanone
2. Label all the chirality centers in the following molecules.
Answers to Practice Questions Chapter 5
5.3.2 Stereoisomer with One Chirality Center — Enantiomers
For 2-butanol, we are able to recognize that C2 is the chirality center.
The perspective formula show the 3D structure of 2-butanol in two different ways, and they are non-superimposable mirror images to each other.
The two mirror images are different molecules. They have the same bonding, but differ in the way that the atoms arranged in space. So the two molecules are stereoisomers. This specific type of stereoisomer here is defined as enantiomers. Molecules that are a pair of non-superimposable mirror images of each other are called enantiomers.
Important Properties about Enantiomers:
• Enantiomers are a pair of non-superimposable mirror images.
• Enantiomers are a pair of molecules, they both chiral and show chirality. (Enantiomer must be chiral).
• For any chiral molecule, it must has its enantiomer, that is the mirror image to the molecule.
• Achiral molecule does not have enantiomer. The mirror image of an achiral molecule is the identical molecule to itself.
Examples: Draw the pair of enantiomers of 2-bromopropanoic acid.
Approach:
To draw the 3D structure of any enantiomer, we need to use perspective formula with solid and dashed wedges to show the tetrahedral arrangements of groups around the sp3 carbon (refer to section 2.11). Out of the four bonds on tetrahedral carbon, two bonds lie within the paper plane are shown as ordinary lines, the solid wedge represent a bond that point out of the paper plane, and the dashed wedge represent a bond that point behind the paper plane. For the first enantiomer, you can draw the four groups with any arrangement, then draw the other enantiomer by drawing the mirror image of the first one. Please note, although it seems there are different ways to show the enantiomers, there are only total two enantiomers, we will learn in next section how to identify and designate each of them.
Several possible ways to show the structures are included in the answer here. However, your answer can be different to any of them, as long as a pair of mirror images are shown.
Exercises 5.3
Draw the pair of enantiomers of 2-chloro-1-propanol.
Answers to Practice Questions Chapter 5
5.3.3 R/S Naming System of Chirality Center
The two enantiomers are different compounds, although they are very similar. Therefore we need a nomenclature system to distinguish between them, to give each one a different designation so that we know which one we are talking about. That is the R/S naming system defined in IUPAC. The R/S designation can be determined by following the Cahn-Ingold-Prelog rule, the rule devised by R. S Cahn, C. Ingold and V. Prelog.
For a pair of enantiomers with one chirality center, one enantiomer has the R configuration and the other one has the S configuration.
Cahn-Ingold-Prelog Rule:
1. Assign priorities of the groups (or atoms) bonded to the chirality center by following the same priority rules as for E/Z system (section 5.2). The highest priority group is labelled as #1, and lowest priority group labelled as #4 in this book.
2. Orient the molecule in the way that the lowest priority group (#4) pointing away from you.
• Look at the direction in which the priority decrease for the other three groups, that is 1→23.
• For clockwise direction, designation is R , rectus , means “right” in Latin.
For counterclockwise direction, designation is S, sinister, means “left” in Latin.
Let’s take the following molecule as an example to practice the rule:
Step 1: The priorities are assigned.
Step 2: Re-orient the molecule, so H (#4, lowest priority) is on the position away from us. Then the other three groups will be arranged in this way:
Step 3: Go along the direction from #1→#2→#3, it is in the clockwise direction, so this enantiomer is assigned R configuration, and the complete name of the molecule is (R)-1-chloroethanol.
Now let’s assign the configuration of the other enantiomer:
Following the same steps, put H away from us, and the arrangement of the other three groups is:
The counterclockwise direction gives the S configuration, and the complete name of the molecule is (S)-1-chloroethanol.
Examples: Assign R/S configuration of the chirality center.
1.
Solution:
2.
Solution:
More practical hints about R/S assignment with Cahn-Ingold-Prelog rule:
• Assigning priority is the first possible challenge for applying the C.I.P. rule. Review and practice the guidelines in section 5.2.
• The 2nd challenge is to re-orient the molecule (to arrange the #4 group away from you). The molecule model will be very helpful for this purpose. Assemble a molecular model with four different colors connected on the carbon. Compare your model to the given structure and match the assigned priority to each color, for example, red is #1, blue is #2, etc. Then rotate the model to arrange the lowest (#4) group away from you and see how the other groups locate to get the answer.
For the perspective formula of enantiomers, it is important to know the following properties:
• One (odd number of) switch (interchanging) for a pair of groups invert the configuration of the chirality centre;
• Two (even number of) switches get the original configuration back.
For the structures above:
• One switch of A leads to B, A isR and B isS, so A and B are enantiomers,
• One switch of B leads to C, B isS and C isR, so B and C are enantiomers,
• Two switches of C leads to A, both C and A are R, so C and A are identical.
When you switch between a pair of groups, do it with cautions. Do not switch unless it is really necessary because it is quite easy to get lost. Do R/S assignment is a safer (and easier for most cases) way to compare the relationship between two structures.
Exercises 5.4
Determine the R/S configuration of the chirality center in following compounds.
Answers to Practice Questions Chapter 5
Exercises 5.5
Determine the relationship for each pair of molecules: enantiomers, identical, constitutional isomers, non-isomer: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/05%3A_Stereochemistry/5.03%3A_Chirality_and_R_S_Naming_System.txt |
The two enantiomers are mirror images of each other. They are very alike and share many properties in common, like same b.p., m.p., density, color, solubility etc. In fact, the pair of enantiomers have the same physical properties except the way they interact with plane-polarized light.
In normal light, the electric filed oscillates in all directions. When normal light passed through a polarizing filter, only light oscillating in one single plane can go through, and the resulting light that oscillates in one single direction is called plane-polarized light.
When plan-polarized light interacts with chiral molecules, the plane of polarization will be rotated by the chiral substances. It was first discovered by Jean-Baptiste Biot in 1815 that some naturally occurring organic substances, like camphor, is able to rotate the plane of polarization of plane-polarized light. He also noted that some compounds rotated the plane clockwise and others counterclockwise. Further studies indicate that the rotation is caused by the chirality of substances.
The property of a compound being able to rotate the plane of polarization of plane-polarized light is called the optical activity, and the compound with such activity is labelled as optical active. The stereoisomer that is optical active is also called as optical isomer.
Chiral compound is optical active. Achiral compound is optical inactive.
The sample containing a chiral compound rotates the plane of polarization of plane-polarized light, the direction and angles of the rotation depends on the nature and concentration of the chiral substances. The rotation angles can be measured by using polarimeter (later in this section).
For a pair of enantiomers with same concentration, under the same condition, they rotate the plane of polarization with the same angles but in opposite direction, one is clockwise and the other is counterclockwise.
The enantiomer rotates the plane of polarization clockwise is said to be dextrorotatory (Latin, means to the right), and labelled with the prefix (d) or (+). The enantiomer rotates the plane of polarization counterclockwise is said to be levorotatory (Latin, means to the left), and labelled with the prefix (l) or (). The d/l (or +/-) indicate the direction in which an optical active compound rotates the plane of polarization of plan-polarized light, that has to be determined by experiment to measure the optical rotation. d/l (or +/) symbol has nothing to do with R/S. R/S indicates the arrangement of the groups around the chirality center, that can be determined by knowing the exact spatial arrangement of the groups. That means compound with R configuration can be either d or l, and compound with S configuration can also be either d or l. For the examples below, both compounds are S-isomer, but one is d (+) and the other is l (-).
The only thing we can be sure is that for a pair of enantiomers, if one enantiomer has been determined as d, then the other enantiomer must be l, and vice versa.
Measurement of Optical Rotation
Polarimeter is the instrument that measures the direction and angles of rotation of plane-polarized light. The plane-polarized light pass through the sample tube containing the solution of sample, and the angle of rotations will be received and recorded by the analyzer, as summarized in Fig. 5.4c.
Since the measurement results vary with the wavelength of the light being used, the specific light from a sodium atomic spectrum with the wavelength of 589 nm, which is called the sodium D-line, is used for most polarimeter. The rotation degree measured by the polarimeter is called the observed rotation (α), and the observed rotation depends on the length of the sample tube, concentration of the sample and temperature.
To compare the optical rotation between different compounds under consistent conditions, the specific rotation is used. Specific rotation is the rotation caused by a solution with concentration of 1.0 g/mL in a sample tube of 1.0 dm length. The temperature is usually at 20°C. Based on this definition, the specific rotation can be calculated from the observed rotation by applying the formula:
Please note: In this formula, the unit of concentration (g/mL) and length of the sample tube (dm) are not the units we are familiar with. Also, the unit of the specific rotation is in degree (°), don’t need to worry about the units cancellation in this formula.
Examples: Calculate the specific rotation
The observed rotation of 10.0g of (R)-2-methyl-1-butnaol in 50mL of solution in a 20-cm polarimeter tube is +2.3° at 20 °C, what is the specific rotation of the compound?
Solution
Specific rotation is the characteristic property of an optical active compound. The literature specific rotation values of the authentic compound can be used to confirm the identity of an unknown compound. For the example here, if it has been measured that the specific rotation of ( R )-2-methyl-1-butnaol is +5.75 ° , then we can tell that the other enantiomer ( S )-2-methyl-1-butnaol must have the specific rotation of -5.75 ° , without further measurement necessary.
Optical Activity of Different Samples
When a sample under measurement only contain one enantiomer, this sample is called as enantiomerically pure, means only one enantiomer is present in the sample.
The sample may also consists of a mixture of a pair of enantiomers. For such mixture sample, the observed rotation value of the mixture, together with the information of the specific rotation of one of the enantiomer allow us to calculate the percentage (%) of each enantiomer in the mixture. To do such calculation, the concept of enantiomer excess (ee) will be needed. The enantiomeric excess (ee) tells how much an excess of one enantiomer is in the mixture, and it can be calculated as:
We will use a series of hypothetic examples in next table for detailed explanation.
If the specific rotation of a (+)-enantiomer is +100°, then the observed rotation of the following samples are (assume the sample tube has the length of 1 dm, and the concentration for each sample is 1.0 g/mL):
Sample Number
Sample
Observed rotation (º)
1 pure (+) enantiomer +100
2 Pure (-)-enantiomer -100
3 Racemic mixture of 50% (+)-enantiomer
and 50% (-)-enantiomer
0
4 Mixture of 75% (+)-enantiomer and 25%
(-)-enantiomer
+50
4
Mixture of 20% (+)-enantiomer and 80% (-)-enantiomer
-60
Sample #1 and #2 are straightforward.
Sample #3 is for a mixture with equal amount of two enantiomers, and such mixture is called racemic mixture or racemate. Racemic mixtures do not rotate the plane of polarization of plane-polarized light, that means racemic mixtures are optical inactive and have the observed rotation of zero! This is because that for every molecule in the mixture that rotate the plane of polarization in one direction, there is an enantiomer molecule that rotate the plan of polarization in the opposite direction with the same angle, and the rotation get cancelled out. As a net result, no rotation is observed for the overall racemic mixture. The symbol (±) sometimes is used to indicate a mixture is racemic mixture.
Sample #4, the (+)-enantiomer is in excess. Since there are 75% (+)-enantiomer and 25%(-)-enantiomer, the enantiomeric excess (ee) value of (+)-enantiomer is 75% – 25% = 50%, this can also be calculated by the formula: ee =
In this sample of mixture, the rotation of the (-)-enantiomer is cancelled by the rotation caused by part of the (+)-enantiomer, so the overall net observed rotation depends on how much “net amount” of (+)-enantiomer present. This can be shown by the diagram below that helps to understand.
Sample #5, the (-)-enantiomer is in excess, and because there is 80% (-)-enantiomer and 20% (+)-enantiomer, the enantiomeric excess (ee) value of (-)-enantiomer is 80% – 20% = 60%, this can also be calculated by the formula: ee =
Please note: to calculate the e.e value, it is not necessary to include the sign of the rotation angle, as long as keep in mind that the sign (+ or ) of the observed rotation indicates that which enantiomer is in excess.
Exercises 5.6
Draw the diagram for Sample #5 by referring to the diagram for Sample #4 .
Answers to Practice Questions Chapter 5
Examples: An advanced level of calculation
The (+)-enantiomer of a compound has specific rotation ([α]20D) of +100°. For a sample (1 g/ml in 1dm cell) that is a mixture of (+) and (-) enantiomers, the observed rotation α is -45°, what is the percentage of (+) enantiomer present in this sample?
Solution
The observed rotation is in “-”, so (-)-enantiomer is in excess.
ee of (-)-enantiomer is:
From here, we will see two ways of solving such type of question:
Method I: solving algebra
% of (-)-enantiomer is set as “x”; % of (+)-enantiomer is set as “y”
x + y = 100%
x – y = 45%
Solve x = 72.5%; y = 27.5%;
So there is 72.5% (-)-enantiomer and 27.5% of (+)-enantiomer in the sample.
Method II: using diagram, the answer is in blue color, there is 27.5% of (+)-enantiomer.
Chirality and Biological Properties
Other than optical activity difference, the different enantiomers of a chiral molecule usually show different properties when interacting with other chiral substances. This can be understood by using the analogue example of fitting a hand into the respective glove: right hand only fits into right glove, and it feels weird and uncomfortable if you wear left glove on the right hand. This is because both right hand and right glove are chiral. A chiral object only fit into a specific chiral environment.
In human body, the biological functions are modulated by a lot of enzymes and receptors. Enzymes and receptors are essentially proteins, and proteins are made up of amino acids. Amino acids are examples of naturally exist chiral substances. With the general formula given below, the carbon with amino (NH2) group is the chirality (asymmetric) center for most amino acids, and only one enantiomer (usually S-enantiomer) exist in nature. A few examples of amino acids are given below with the general formula.
Because amino acids are chiral, proteins are chiral so enzymes and receptors are chiral as well. The enzyme or receptor therefore form the chiral environment in human body that distinguish between R or S enantiomer. Such selectivity can be illustrated by the simple diagram below.
The binding site of enzyme or receptor is chiral, so it only binds with the enantiomer whose groups are in the proper positions to fit into the binding site. As shown in the diagram, only one enantiomer binds with the site, but not the other enantiomer.
A couple of common examples to showcase such binding selectivity of different enantiomer may include limonene and carvone.
Limonene has two enantiomers, and they smell totally different to human being because they interact with different receptors that located on the nerve cells in nose. The (R)-(+)-limonene is responsible for the smell of orange, and the (S)-(-)-limonene gives the bit smell of lemon.
If you like caraway bread, that is due to the (S)-(+)-carvone; and the (R)-(-)-carvone that is found in spearmint oil gives much different odor.
More dramatic examples of how chirality plays important role in biological properties are found in many medicines. For the common over-counter anti-inflammatory drug ibuprofen (Advil), for example, only (S)-enantiomer is the active agent, while the (R)-enantiomer has no any anti-inflammatory action. Fortunately, the (R)-enantiomer does not have any harmful side effect and slowly converts to the (S)-enantiomer in the body. The ibuprofen is marketed usually as a racemate form.
The issue of chiral drugs (the drug contain a single enantiomer, not as a racemate) was not in the attention of drug discovery industry until 1960. Back then, drugs were approved in racemate form if a chirality center involved, and there was no further study about biological difference on different enantiomers. These were all changed by the tragic incident of thalidomide. Thalidomide was a drug that was sold in more than 40 countries, mainly in Europe, in early 1960s as a sleeping aid and to pregnant women as antiemetic (drug that preventing vomiting) to combat morning sickness. It was not recognized at that time that only the R-enantiomer has the property, while the S-enantiomer was a teratogen that causes congenital deformations. The drug was marketed as a racemic mixture and caused about 10,000 children had been damaged until it was withdrawn from the market in Nov. 1961. This drug was not approved in US however, attributed to Dr. Frances O. Kelsey, who was a physician for the FDA (Food and Drug Administration) at that time and had insisted on addition tests on some side effects. Thousands of life were saved by Dr. Kelsey, and she was awarded the President’s medal in 1962 for preventing the sale of thalidomide. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/05%3A_Stereochemistry/5.04%3A_Optical_Activity.txt |
For the discussions so far, the perspective formula with solid and dashed wedges have been used to represent the 3D arrangement of groups bonded to a chirality center. Other than that, there is another broadly applied formula for that purpose, that is the Fisher projection. A Fisher projection is a shortcut for showing the spatial group arrangement of a chirality center, it is more easily to be drawn and recognized, and is particularly useful for showing the structures with more than one chirality centers.
In Fisher projection, the chirality center is shown as the intersection of two perpendicular lines. The horizontal lines represent the bonds point out of the plane, and the vertical lines represent the bonds that point behind the plane.
It is very important to keep in mind that the lines in Fisher projection are not just bonds, they represent the bonds with specific spatial arrangements and stereochemistry.
Assigning R/S Configuration in Fisher projection
Taking the following compound as an example:
1. Assign group priority as we usually do.
1. If the lowest priority group (#4 group) is on a vertical bond, determine the priority decrease direction from #1→#2→#3 as usual to get the configuration, clockwise is R and counterclockwise is S.
So, the example here is a R-isomer, and the complete name of the compound is (R)-2-chlorobutane.
1. If the lowest priority group is on a horizontal bond (as the case in the following structure), determine the priority decrease direction as in step 2, then reverse the answer to opposite way, to get the final configuration.
So, the example here is a S-isomer, and the complete name of the compound is (S)-2-chlorobutane.
Exercises 5.6
Explain that why in step 3 of the above procedure, the answer should be reversed to get the final (actual) configuration?
Answers to Practice Questions Chapter 5
Exercises 5.7: Indicate the configuration of the following structures.
Properties of Fisher projection:
1. One switch (interchange) of two groups in a Fisher projection invert the configuration, two switches bring the original isomer back.
For above structures:
• one switch of A leads to B, A and B are enantiomers;
• one switch of B leads to C, B and C are enantiomers;
• two switches of C leads to A, A and C are identical.
2. Rotate the Fisher projection 180º get same structure, with the configuration retained.
• 180º rotation of A leads to B, A and B are identical.
3. Rotate the Fisher projection 90º get the configuration inverted.
• 90º rotation of A leads to B, A and B are enantiomers.
5.06: Compounds with More Than One Chirality Centers
5.6.1 Diastereomers
It is very common that there are more than one chirality centers in an organic compound. For the example of 2-bromo-3-chlorobutane below, there are 2 chirality centers, C2 and C3. With each chirality center has two possible configurations, R and S, the total number of possible stereoisomers for this compound is four, with configurations on C2 and C3 as RR, SS, RS and SR respectively.
As a general rule, for a compound has n chirality centers, the maximum number of stereoisomers for that compound is 2n.
The four stereoisomers of 2-bromo-3-chlorobutane consist of two pairs of enantiomers. Stereoisomers A and B are a pair of non-superimposable mirror images, so they are enantiomers. So are the isomers C and D. Then what is the relationship between isomer A and C?
A and C are not identical, not enantiomers, and they are stereoisomers (have the same bonding but differ in the spatial arrangement of groups). Such type of stereoisomers are defined as diastereomers. Diastereomers are stereoisomers that are not enantiomers. For the four stereoisomers here, there are four pairs of diastereomers: A and C, A and D, B and C,B and D. The relationship between the four stereoisomers can be summarized as:
With the introduction of diastereomer concept, the way to categorize isomers can be revised, and the summary in Fig. 5.1a can be replaced by the updated version in Fig. 5.6a. The stereoisomer then has two sub-types, enantiomers and diastereomers, because any stereoisomers that are not enantiomers can always be called diastereomers. Based on such definition, the geometric isomers we learned earlier also belong to the diastereomer category.
As mentioned earlier, enantiomers are very alike to each other, and they share same physical properties except optical activity (opposite sign for specific rotation). Enantiomers also generally have same chemical properties, except the reaction with other chiral reagents (not topics in this course).
However, diastereomers are not that closely related. Diastereomers have different physical properties, for example, different b.p, color, density, polarity, solubility etc. They also have different chemical properties.
Next, we will go through the examples of cyclic compounds, to see how the new concept of diastereomer relates to the knowledge about cyclic compounds we learned before.
Examples
Draw the structures of all the stereoisomers for 1-bromo-2-chlorocyclobutane, and indicate the relationship between any two stereoisomers.
Approach:
There are two chirality centers for 1-bromo-2-chlorocyclobutane molecule. So the maximum number of stereoisomer is four. To work on the stereoisomers for cyclic compound, we can start with cis/trans isomer, and then check does the enantiomer apply to each case.
Solution:
There are two cis-isomers, A and B, and they are enantiomers of each other; similarly, there are also two trans-isomers C and D that are enantiomers of each other as well.
The relationship between any of the cis-isomer to any of the trans-isomer is diastereomers (A and C, A and D, B and C, B and D). Since they are geometric isomers, and remember that the geometric isomers can also be called diastereomers.
All geometric isomers are diastereomers (it is always correct to call a pair of geometric isomers as diastereomers), however not all the diastereomers are geometric isomers!
Examples:
What is the relationship between the following pair of compounds, enantiomers, identical, diastereomers, constitutional isomers, non-isomers?
1.
Method I: The basic way is to determine the configuration of each chirality center. As shown below that the configuration for both chirality centres are right opposite between the structure A and B. So they are enantiomers.
Method II: For the cyclic structures, sometimes rotate or flip a given structure in a certain way helps us to tell the relationship (using the molecular model helps the rotate or flip part). For this example, flipping structure B horizontally leads to structure C, B and C are identical. Then it is easy to tell that A and C are just non-superimposable mirror images to each other, so A and C are enantiomers, then A and B are enantiomers as well.
If this method looks confusing to you, then you can stick to Method I.
2.
You can use either of the above methods, the answer is “identical”.
5.6.2 Meso compound
Next, we will see another example of a compound containing two chirality centers, 2,3-dichlorobutane, the compound that has the same substituents on C2 and C3 carbons.
Theoretically, there are maximum four stereoisomers, the structures are shown by Fisher projections here.
Stereoisomer A and B are non-superimposable mirror images, so they are enantiomers.
We will take a detailed look at stereoisomer C and D. Yes, they are mirror images, but are they really non-superimposable? If isomer C is rotated 180° (180° rotation still get the same structure back for Fisher projection), then it could get superimposed on isomer D. So, isomer C and D are superimposable mirror images, that means they are the same, identical!
Then “C” and “D” are just different drawings for the same stereoisomer. The next questions is, is this stereoisomer chiral? We have confirmed that this isomer does get superimposed on its mirror image, that means it is achiral.
This is so weird! How come a compound that contain two chirality centers (C2 and C3) is achiral?
Yes, it does happen! A compound that is achiral but contain chirality centers is called meso compound. A meso compounds is achiral and optical inactive (does NOT rotate the plane of polarization of plan-polarized light), but it does have multiple chirality centers.
Because that one stereoisomer is meso compound, the total number of stereoisomers for 2,3-dichlorobutane is three.
Attention, 2n is the maximum number of stereoisomers. Some compounds may have less than the maximum, because of the existence of meso compounds.
Examples: Draw all the stereoisomers of 1,2-dibromocyclopentane.
Solutions: there are total three stereoisomers.
Exercises 5.8
• Draw all stereoisomers for 1-ethyl-3-methylcyclohexane.
• Draw all stereoisomers for 1-ethyl-4-methylcyclohexane.
• Draw all stereoisomers for 1,2-dimethylcyclohexane.
Answers to Practice Questions Chapter 5
5.6.3 Chiral or achiral by looking for Plane of symmetry
The existence of chirality centers does not guarantee the chirality of a molecule, for example of the meso compound. Following the definition of chirality always involve the comparison between original structure and its mirror image, that needs extra work. Is there any easier way to tell whether a moleculeis chiral or achiral?
We can check the plane of symmetry. Plane of symmetry is a plane that cuts the molecule in half and that one half is the mirror image of the other.
• If a molecule does have a plan of symmetry, then the molecule is achiral.
• The molecule that does not have a plane of symmetry in any conformation is chiral.
For the meso isomer of 2,3-dichlorobutane, the plane of symmetry is the plane that is labelled in the structure below.
Examples:
Determine whether the following molecule is chiral or achiral.
Solution:
Checking the plane of symmetry provides a quick way to determine the chirality of a molecule. But sometimes you may need to look for the proper conformation to get the plane of symmetry. See following example.
Examples: What is the relationship of the following pair of structures?
Approach: Determine the R/S configuration of each carbon.
For both structures, the chirality centres are bonded with the same groups, and structure I has R and S, structure II has S and R. Are they enantiomers?
A bit further investigation is necessary to get the conclusion. Let’s rotate the groups around the 2nd chirality centre of structure I (you can use the molecular model to do the rotation, that is very helpful for visualizing the spatial arrangement of the groups):
Rotation of the groups around the chirality centre does not change the configuration, however it does change the conformation to eclipsed conformation. In the eclipsed conformation, it is easier to tell that the structure has a plane of symmetry, so it is a meso compound that is achiral. Achiral compound does not have enantiomer, so structure II is also meso compound that is identical to structure I.
Solution: Identical
(You can rotate, or do switches to compare between the two structures, but make sure to keep track on any action. If it is easy to get lost by rotating or switch, assign R/S configuration is a safer way.)
Examples
Thinking: Determine the relationship between the molecule in each question with the given one, and apply the knowledge of specific rotation.
5.07: Answers to Practice Questions Chapter 5
5.1 Order the following groups based on decreasing priority for E/Z naming purpose.
Answer: D > C > A > B
5.2
1. Draw the structure of following compounds, determine which one has an chirality center and label it with a star.
2. Label all the chirality centers in the following molecules.
5.3 Draw the pair of enantiomers of 2-chloro-1-propanol.
5.4 Determine the R/S configuration of the chirality center in following compounds.
5.5 Determine the relationship for each pair of molecules: enantiomers, identical, constitutional isomers, non-isomer:
5.6 Draw the diagram for Sample #5 by referring to the diagram for Sample #4.
5.7 Explain that why in step 3 of the above procedure, the answer should be reversed to get the final (actual) configuration?
According to the definition of Fisher projection, the horizontal bond is the bond pointing towards the viewer. Therefore when the lowest priority group is on a horizontal bond, it is on the position just opposite to the way defined by the Cahn-Ingold-Prelog rule, so the actual configuration should be the reversed version of whatever obtained initially.
5.8 Indicate the configuration of the following compounds.
5.9
• Draw all stereoisomers for 1-ethyl-3-methylcyclohexane.
• Draw all stereoisomers for 1-ethyl-4-methylcyclohexane.
• Draw all stereoisomers for 1,2-dimethylcyclohexane. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/05%3A_Stereochemistry/5.05%3A_Fisher_Projection.txt |
Electromagnetic radiation is the radiation composed of oscillating electrical and magnetic fields. The whole electromagnetic spectrum covers the radiation in very broad range from gamma rays (emitted by the nuclei of certain radioactive elements), X-rays (used for medical examination of bones), to ultraviolet (UV) light (is responsible for sunburn, can also be used for dis-infection purpose), microwaves, and radio-frequency waves (used for radio and television communication, and of the cell phone signal). Visible light, the radiation that is visible to our bare eyes and what we commonly refer to as “light”, just accounts for a very narrow band out of the full electromagnetic spectrum.
Electromagnetic radiation exhibits wave-like properties. As a general property of waves, the wavelength (λ, Greek ‘lambda‘) and frequency (ν, Greek ‘nu’, in unit of Hz or s-1, 1Hz = 1s-1) of electromagnetic radiation fits in the formula of:
c =λν Formula 6.1
where c is the speed, usually referred to as the “speed of light”, with the constant value of 2.998×108m/s in vacuum (the speed of light in air is a little bit slower than this constant but is usually regarded as the same). Because electromagnetic radiation travels at a constant speed, wavelength (λ) and frequency (ν) are inversely proportional to each other, the longer waves have lower frequencies, and shorter waves have higher frequencies.
The energy of electromagnetic radiation can be calculated based on formula:
E = = hcFormula 6.2
where E is energy of each photon in unit of Joule (J) and h is the Planck’s constant with value of 6.626×10-34J·s.
So radiations with higher frequencies correspond to higher energy. High energy radiation, such as gamma radiation and X-rays, is composed of very short waves – as short as 10-16m. Longer wavelengths are much less energetic, and thus are less harmful to living things. Visible light waves are in the range of 400 – 700 nm (nanometer, 1nm = 10-9m), while radio waves can be several hundred meters in length.
In a molecular spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a lower energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are not absorbed by the sample molecule, so they pass on through. A detector records which wavelengths were absorbed, and how much were absorbed.
As we will see in this chapter, we can learn a lot about the structure of an organic molecule by quantifying how it absorbs (or does not absorb) different wavelengths in the electromagnetic spectrum. The IR spectroscopy involves absorption of radiation in the infrared region and radio waves are applied in the NMR technique.
6.02: Infrared (IR) Spectroscopy Theory
In IR spectroscopy, how the vibration mode of covalent bonds are affected by absorbing the infrared electromagnetic radiation is studied. Covalent bonds in organic molecules are not rigid sticks, they behave as if they were vibrating springs instead. At room temperature, organic molecules are always in motion that involves several vibration modes, such as stretching, bending, and twisting as illustrated in Fig. 6.2a.
Stretching is the vibration occurring along the line of the bond that changes the bond length. Bending is the vibration that like swing, it does not occur along the line, but change the bond angles. The specific bending mode are often referred to by the descriptive terms like scissoring, twisting etc.
One covalent bond may vibrate in different vibrational modes, for example, the C-H bond can be in stretching and bending mode. Each vibrational mode for a given bond occurs with a characteristic ground state frequency, that corresponds to the frequency of infrared region (1013 to 1014Hz, or 2.5 to 17 μm in wavelength) of the electromagnetic spectrum. If a molecule is exposed to infrared radiation, it will absorb the radiation that matches the frequency of the vibration of one of its bonds. The IR radiation absorbed allows the bond to vibrate a bit more, that is increase the amplitude of vibration, but the vibrational frequency will remain the same.
In an infrared spectrophotometer (Fig. 6.2b) a beam of IR radiation passed through the sample and some radiation is absorbed by the sample, the remaining go through. Another beam of IR radiation pass through the cell with blank (no sample, no absorption) and all light go through. The detector in the instrument record and compare the radiation transmitted through the sample with that transmitted in the absence of the sample. Any frequencies absorbed by the sample will be apparent by the difference. The computer plots the result as a graph showing transmittance vs frequency (in format of wavenumber that will be explained next). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.01%3A_Electromagnetic_Radiation_and_Molecular_Spectroscopy.txt |
Xin Liu
With the basic understanding of the IR theory, we will take a look at the actual output from IR spectroscopy experiments, and learn how to get structural information from IR spectrum. Below is the IR spectrum for 2-hexanone.
Notes for interpreting IR spectra:
• The vertical axis is ‘% transmittance’, which tells how strongly light was absorbed at each frequency. The solid line traces the values of % transmittance for every wavelength passed through the sample. At the high end of the axis, 100% transmittance means no absorption occurred at that frequency. Lower values of % transmittance mean that some of the energy is absorbed by the compound, and gives the downward spikes. The spikes are called absorption bands in an IR spectrum. A molecule have a variety of covalent bonds, and each bond have different vibration modes, so the IR spectrum of a compound usually show multiple absorption bands.
• The horizontal axis indicates the position of an absorption band. But instead of using frequency to show the absorbed radiation, wavenumbers (, in unit of cm-1) are used as a conventional way in IR spectra. The wavenumber is defined as the reciprocal of wavelength (Formula 6.3), and the wavenumbers of infrared radiation are normally in the range of 4000 cm-1 to 600 cm-1 (approximate corresponds the wavelength range of 2.5 μm to 17 μm of IR radiation).
Please note the direction of the horizontal axis (wavenumber) in IR spectra decrease from left to right. The larger wavenumbers (shorter wavelengths) are associated with higher frequencies and higher energy.
The power of infrared spectroscopy arises from the observation that the covalent bonds characterizing different functional groups have different characteristic absorption frequencies (in wavenumber, Table 6.1). The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest.
For example, the most characteristics absorption band in the spectrum of 2-hexanone (Figure 6.3a) is that from the stretching vibration of carbonyl double bond C=O, at 1716 cm-1. It is a very strong band comparing to the others on the spectrum. A strong absorbance band in the 1650-1750 cm-1 region indicate that a carbonyl group (C=O) is present. Within that range, carboxylic acids, esters, ketones and aldehydes tend to absorb in the higher wavenumber/frequency end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the lower wavenumber/frequency end (1650-1700 cm-1).
Stretching Vibrations
Generally, stretching vibrations the stretching vibrations require more energy and show absorption bands in the higher wavenumber/frequency region. The characteristics stretching vibration bands associated with the bonds in some common functional groups are summarized in Table 6.1.
Formula
Bond
Characteristic IR Frequency range (cm-1)
alcohol
O-H stretching
3200 – 3600 (broad)
carbonyl
C=O stretching
1650 – 1750 (strong) aldehyde
C-H stretching
~ 2800 and ~ 2700 (medium) carboxylic acid
C=O stretching
1700 – 1725 (strong)
O-H stretching
2500 – 3300 (broad) alkene
C=C stretching
1620 – 1680 (weak)
vinyl =C-H stretching
3020 – 3080 benzene
C=C stretching
~ 1600 and 1500 – 1430 (strong to weak) alkyne
C≡C stretching
2100 – 2250 (weak)
terminal ≡C-H stretching
3250 – 3350 alkane
C-H stretching
2850-2950 amine
N-H stretching
3300-3500 (medium)
Table 6.1 Characteristic IR Frequencies of Stretching Vibrations
The information in Table 6.1 can be summarized in the diagram that is easier to be identified (Figure 6.3b), in which the IR spectrum is divided in several regions, with the characteristic band of certain groups labelled.
The absorption bands in IR spectra have different intensity, that can usually be referred to as strong (s), medium (m), weak (w), broad and sharp. The intensity of a absorption band depends on the polarity of the bond, the bond with higher polarity will show more intense absorption band. The intensity also depends on the number of bonds responsible for the absorption, the absorption band with more bonds involved has higher intensity.
The characteristic IR frequencies of stretching vibrations in Table 6.1 and Figure 6.3b provide very useful information to identify the presence of certain functional group, that can be generally summarized as:
The polar O-H bond (in alcohol and carboxylic acid) usually shows strong and broad absorption bands that are easy to be identified. The broad shape of the absorption band results from the hydrogen bonding of the OH groups between molecules. The OH bond of alcohol group usually has absorption in the range of 3200-3600 cm-1, while the OH bond of carboxylic acid group occurs at about 2500-3300 cm-1 (Figure 6.4a and Figure 6.4c).
The polarity of N-H bond (in amine and amide) is weaker than OH bond, so the absorption band of N-H is not as intense, nor that broad as O-H, and the position is in 3300-3500 cm-1 region.
The C-H bond stretching of all hydrocarbons occur in the range of 2800-3300 cm-1, and the exact location can be used to distinguish between alkane, alkene and alkyne. Specifically:
• ≡C-H (sp C-H) bond of terminal alkyne give absorption at about 3300 cm-1
• =C-H (sp2 C-H) bond of alkene give absorption at about 3000-3100 cm-1
• -C-H (sp3 C-H) bond of alkane give absorption at about ~2900 cm-1 (see the example of IR spectrum of 2-hexanone in Figure 6.3a, the C-H absorption band at about 2900 cm-1)
A special note should be taken for the C-H bond stretching of an aldehyde group that shows two absorption bands, one at ~2800 cm-1 and the other at ~ 2700 cm-1. It is therefore relative easy to identify the aldehyde group (together with the C=O stretching at about 1700 cm-1) since essentially no other absorptions occur at these wavenumbers (see the example of IR spectrum of butanal in Figure 6.4d ).
The stretching vibration of triple bonds C≡C and C≡N have absorption bands of about 2100~2200 cm-1. The band intensity are in medium to weak level. The alkynes can generally be identified with the characteristic weak but sharp IR absorbance bands in the range of 2100-2250 cm-1 due to stretching of the C≡C triple bond, and terminal alkynes can be identified by their absorbance at about 3300 cm-1, due to stretching of sp C-H.
As mentioned earlier, the C=O stretching has strong absorption band in the 1650-1750 cm-1 region. Other double bonds like C=C and C=N have absorptions in bit lower frequency regions of about 1550-1650 cm-1. The C=C stretching of an alkene only shows one band at ~1600 cm-1 (Figure 6.4b), while a benzene ring is indicated by two sharp absorption bands, one at ~1600 cm-1 and one at 1500-1430 cm-1 (see the example of IR spectrum of ethyl benzene in Figure 6.4e).
You will notice in Figure 6.3a and 6.3b that a region with the lower frequency 400-1400 cm-1 in the IR spectrum is called the fingerprint region. Kind of like a human fingerprint, the pattern of absorbance bands in the fingerprint region is characteristic of the compound as a whole. Even if two different molecules have the same functional groups, their IR spectra will not be identical and such difference will be reflected in the bands in the fingerprint region. Therefore the IR from an unknown sample can be compared to a database of IR spectra of known standards in order to confirm the identification of the unknown sample.
6.04: IR Spectrum Interpretation Practice
Now, let’s take a look at the more IR spectrum for examples. It is very important to keep in mind that generally we do not try to identify all the absorption bands in an IR spectrum. Instead, we will look at the characteristic absorption band to confirm the presence or absence of a functional group. An IR spectrum usually does not provide enough information for us to figure out the complete structure of a molecule, and other instrumental methods have to be applied in conjunction with, such as NMR that we will learn in later sections, that is a more powerful analytical method to give more specific information about molecular structures.
In the IR spectrum of 1-hexanol, there are sp3 C-H stretching bands of alkane at about 2800-3000 cm-1 as expected. Other than that, there is a very broad peak centered at about 3400 cm-1, that is the characteristic band of the O-H stretching mode of alcohols.
The spectrum for 1-octene shows two bands that are characteristic of alkenes: the one at 1642 cm -1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm -1 is due to stretching of the σ bond between the sp 2 -hybridized alkene carbons and their attached hydrogens.
The following IR spectrum are taken from Spectral Database for Organic Compounds, the free organic compounds spectral database. The key bands for each compound are labelled on the spectra. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.03%3A_IR_Spectrum_and_Characteristic_Absorption_Bands.txt |
37
Although the other techniques provide valuable information about a molecule, they do not tell us what about the overall molecular structure, or the framework about C-C and C-H bonds. Nuclear magnetic resonance (NMR) spectroscopy is an immensely powerful analytical technique that provides such information. NMR works by the same principles as an Magnetic Resonance Imaging (MRI) scanner in a hospital. MRI is a scanning technique to detect the hidden medical problems without causing any harm of pain to the patient. While doctors use MRI peer inside the human body, we will see how NMR allows organic chemists to piece together, atom by atom and bond by bond, the structure of an organic molecule.
NMR-active Nuclei
The basis for NMR is the phenomenon that some atomic nuclei spin about their axes and as a result generate their own magnetic field, or magnetic moment, therefore these nuclei are called NMR-active. Not all nuclei have a magnetic moment though, only those nuclei with an odd number of proton and/or neutron have. Fortunately nuclei that are important for organic compounds, such as the 1H isotope of hydrogen, the 13C isotope of carbon, the 14N isotope of nitrogen, 19F and the 31P are all NMR-active and therefore can be observed by NMR. Other nuclei, such as the common 12C isotopes of carbon and 16O isotope of oxygen, do not have magnetic moments, and cannot be directly observed by NMR.
In practice, the 1H and 13C nuclei are most commonly observed by NMR spectroscopy, and we will focus on these techniques in this chapter, beginning with 1H NMR. 1H NMR is usually called proton NMR, because the nucleus of 1H atom is actually a single proton. The name of ‘proton’ and ‘hydrogen’ will be used interchangeably in this chapter for 1H NMR purpose.
Spin State and Magnetic Resonance
We will take proton, the nucleus of 1H atom, as an example for the discussions here.
When a sample of an organic compound is sitting in a flask on a laboratory bench, the magnetic moments of all of its protons are oriented randomly. However, when the same sample is placed within the field of a strong magnet in an NMR instrument (this field is referred to as the applied external magnetic field, B0, in NMR), each proton will assume one of two possible orientations with respect to the external magnetic field. These two orientations corresponds to the two spin states that can be labelled as α and β. In the α spin state, the proton’s magnetic moment is aligned with the direction of external magnetic field B0, while in the β spin state it is aligned opposed to the direction of B0 (Fig. 6.5b).
The α spin state is slightly lower in energy than the β state, and the energy gap between them, ΔE, depends upon the strength of B0: a stronger applied external magnetic field results in a larger ΔE. For a large population of organic molecules in an external magnetic field, slightly more than half of the protons will occupy the lower energy α spin state, while slightly less than half will occupy the higher energy β spin state. It is this population difference between the two spin states that is exploited by NMR, and the difference increases with the strength of the applied magnetic field B0.
Energy is required to excite the proton from the lower energy state (α spin state) to the higher energy state (β spin state). In an NMR spectrometer the energy is supplied by electromagnetic radiation in the radio frequency (RF) region. When a proton in an external magnetic field is exposed to RF radiation with the energy that matches the energy gap ΔE, the energy of the RF is absorbed and the proton will flip its magnetic moment from the lower energy state (α spin state) to the higher energy state (β spin state), the nuclei are said to be in resonance with the electromagnetic radiation.
The frequency of radiation absorbed by a proton (or any other nucleus) during a spin transition in the NMR experiment is called its resonance frequency, ν. As a result, the resonance frequency also depends on B0, the larger B0 the higher resonance frequency, and the relationship fits to the specific formula: (Formula 6.4 is for your information purpose only)
γ is the magnetogyric (or gyromagnetic) ratio, different nucleus has different value of γ. For a proton, the γ value is 26.753 rad · s -1 · tesla -1 .
Calculations indicate that if external magnetic field B0≅1.41 Tesla, the energy difference corresponds to RF with the frequency of 60×106 Hz (60 MHz) for proton; when B0≅7.04 Tesla, the corresponding RF frequency is 300×106 Hz (300 MHz) for proton. This frequency is the most important parameter for a NMR spectrometer (the instrument that run NMR experiments), the higher the frequency, the more sensitive the instrument and the higher resolution the resulting NMR spectrum is.
The NMR Experiment
In this book we will just explain how the NMR experiment and NMR spectrometer work in a simplified way (again with proton as an example), the full version is out of the scope of this course.
When a sample of compound is placed in the strong applied external magnetic field B0 of the instrument, the protons begin to spin with one of the two spin states. Initially, slightly more than half of the protons have the magnetic moments in aspin states (aligned with B0), and slightly less than half are in bspin states (aligned against B0). Then, the sample is exposed to a range of radio frequencies. Out of all of the frequencies which hit the sample, only the frequencies that matches the resonance frequency of the protons are absorbed, causing those protons which are aligned with B0 to ‘spin flip’ so that they align themselves against B0. When the ‘flipped’ protons flip back down to their ground state, they emit energy, again in the form of radio-frequency radiation. The NMR instrument detects and records the frequency and intensity of this radiation by making using of a mathematical technique known as a Fourier transform (FT). Fourier Transform convert the signal from a time versus amplitude signals to a frequency versus amplitude signals, that is what we observe in a NMR spectrum.
Most modern FT-NMR spectrometers use superconducting magnets that have very high magnetic fields, therefore operate with high resonance frequency from 100 MHz to 800 MHz. Superconducting magnets operate in a bath of liquid nitrogen or liquid helium at very low temperature.
Despite the powerfulness and high resolution of the high frequency NMR spectrometers, it is very costly for purchase and maintenance of the instrument. For teaching purpose, the bench top NMR are becoming more and more popular recently. The frequency of bench top NMR are usually in the range of 60-90 MHz, however, they can provide spectra with good resolution for lots basic organic structures that are used for undergraduate organic chemistry class. With the low-cost bench top NMR available, students have chance to gain hand-on NMR experiences in sample preparation, instrument operation and spectrum processing.
Shielding and Deshielding
If all hydrogen atoms (and protons) in organic molecules had the same resonance frequency, then they all show the same signal, NMR spectroscopy would not be that useful for chemists. Fortunately, however, resonance frequencies are different for different protons in a molecule. Specifically, the resonance frequency varies according to the electronic environment that a given proton inhabits.
For hydrogen atoms in any bonds, such as C-H, O-H etc, the external magnetic field B0 causes the s electrons to circulate in a way that generate an induced local magnetic field (Blocal) at the proton, and the direction of the local field Blocal is opposite to the external field B0. The proton thus experiences a net magnetic field, which is called Beff, that is smaller than the applied magnetic field:
Beff= B0 – Blocal
As a result, the proton responses to a lower frequency (resonance frequency is proportional to the magnetic field as mentioned early). This Blocal, to a small but significant degree, shield the proton from experiencing the full force of B0, so this effect is called shielding effect. Different hydrogen atoms in organic structures are in different electronic environment, have different selectron density, therefore have different Blocal and different Beff as well. That is why different hydrogens (and protons) are in different resonance frequency and show different signals in the spectrum.
For hydrogen atoms close to electronegative groups, electronegative groups withdraw electron density from nearby atoms, so diminishing the shielding of the protons by circulating electrons. The hydrogen atoms near an electronegative groups are said to be deshielded from the external magnetic field, and have a higher resonance frequency than those shielded protons. As the electronegativity of the substituent increase, so does the extent of the deshielding effect (and so the chemical shift, see section 6.6.2 for more discussions about chemical shift) as shown in the examples below. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.05%3A_NMR_Theory_and_Experiment.txt |
Understanding the basics of NMR theory gets us ready to move on to the most important and practical part in this section, that is how to understand the 1H NMR spectrum and elucidate the structure of a compound from 1H NMR spectrum information. Let’s first take a look at an actual 1H NMR spectrum.
Generally, the information about the structure of molecule can be obtained from four aspects of a typical 1H NMR spectrum:
• Chemical equivalent and non-equivalent protons (total number of signals)
• Chemical shift
• Integration
• Signal splitting
6.6.1 Chemical Equivalent and Non-Equivalent Protons
In the above 1H NMR spectrum of methyl acetate (Fig. 6.6a), we can see that there are three signals. The peak at the far right is for the standard reference compound tetramethylsilane (TMS, more discussions in chemical shift section 6.6.2), not for the compound. So the compound methyl acetate shows two signals in 1H NMR spectrum. Why only two signals for a compound containing total six hydrogens?
This is because of chemical equivalence. The total six hydrogens can be divided to two groups, the three Ha protons in the methyl group that bonded with C=O are all in the same chemical environment, therefore they are chemical equivalent. All chemical equivalent hydrogens have the same resonance frequency with applied to an external magnetic field, so show only one signal in 1H NMR spectrum. The three Hb protons in the methyl group bonded with O atom are chemical equivalent as well and show the other signal. That is why there are total two signals for compound methyl acetate.
The ability to recognize chemical equivalent and non-equivalent protons in a molecule is very important in understanding NMR spectrum. For the compound with structure given, we should be able to predict how many signals are there in 1H NMR spectrum. On the other side, if the 1H NMR spectrum is available for an unknown compound, counting the number of signals in the spectrum tells us the number of different sets of protons in the molecule, and that is the very important information to determine the structure of the compound.
Here we will go through several examples for the first situation, that is to predict the number of signals in 1H NMR spectrum with the structure of a compound given. To do that, we need to count how many distinct proton sets are included in the molecule.
For each of the following molecule, the chemically equivalent protons are labelled in the same color to facilitate the understanding.
• Benzene: all six protons are chemical equivalent (have the same bonding and in the same chemical environment) to each other and have the same resonance frequency in an 1H NMR experiment, therefore show only one signal.
• Acetone: both methyl groups (two CH3) bonded with C=O bond, so they are in the same chemical environment, and as a result all the six protons are chemical equivalent that show only one signal.
Notes: As you probably already realized, chemical equivalence or non-equivalence in NMR is closely related to symmetry. The protons that are symmetric to each other by a certain plane of symmetry are chemical equivalent.
The molecules in the next figure contains more sets of chemically equivalent protons.
• Acetaldehyde: The three Ha protons in the methyl group are chemical equivalent, and they all bonded to an sp3-hybridized carbon; but they are different to the Hb proton that is bonded to an sp2hybridized carbonyl carbon. Two signals total in 1H NMR spectrum.
• 1,4-dimethylbenzene: all four aromatic protons in are chemically equivalent because of the symmetry. The two methyl groups are equivalent to each other as well. Two signals total in 1H NMR spectrum.
• 1,2-dimethylbenzene: both Ha protons are adjacent to a methyl substituent, while both Hc protons are two carbons away. So the four aromatic protons are divided to two sets. Both methyl groups are in the same bonding and symmetric to each other, they are equivalent. Three signals total in 1H NMR spectrum.
• 1,3-dimethylbenzene: Hb is situated between two methyl groups, the two Hc protons are one carbon away from a methyl group, and Hd is two carbons away from a methyl group. Therefore, the four aromatic protons can be divided to three sets. The two methyl groups are equivalent. Four signals total in 1H NMR spectrum.
Exercises 6.1
How many 1H NMR signals would you predict for each of the following molecules?
Answers to Practice Questions Chapter 6
6.6.2 Chemical Shift
As seen in the 1H NMR spectrum of methyl acetate (Fig. 6.6a), the x-axis units of NMR spectrum are in ppm (not in Hz as we would expect for frequency), and the two signals stand at different position along the x-axis. Let’s explain how that works and what information can be obtained.
The position of a signal along the x-axis of an NMR spectra is called chemical shift, or δ, of the signal. Chemical shift is determined by the structural electronical environment of the nuclei producing that signal. Protons in different chemical environments (non-equivalent) show signals at different chemical shift. The direction of chemical shift scale in x-axis is opposite to what we are familiar with, that is the smaller value is at right-hand side, and the larger value is at the left-hand side (Fig. 6.6b).
• Smaller chemical shift (δ) values correspond with lower resonance frequency;
• Larger chemical shift (δ) values correspond with higher resonance frequency.
By convention, the right-hand side of an NMR spectrum with smaller chemical shift values is called upfield, and the left-hand direction is called downfield (Fig. 6.6b).
For protons that are shielded, because of the Blocal caused by circulating electrons, the magnetic field experienced by the proton, Beff, is smaller than applied external field, Bo, so the protons resonance at lower frequency and have smaller chemical shift values.
• Shielded protons have lower resonance frequency, and smaller chemical shift (d) values;
• Deshielded protons have higher resonance frequency, and larger chemical shift (d) values.
In 1H NMR spectrum, the absorption of the protons of TMS (tetramethylsilane) is defined as “zero” on the chemical shift (δ) scale, and the absorption of other protons are reported as relative shift compared with that of TMS.
TMS was chosen as a reference compound and defined as “zero” for several reasons. Since silicon is less electronegative than carbon, the hydrogens of TMS are in high electron-density environment, therefore are highly shielded with very low resonance frequency and rarely interfere with the signals of other compounds. Also there are twelve equivalent hydrogens in TMS that show a single signal, so the signal is rather strong even with very little amount of TMS. TMS is also quite inert and easy to be removed with the boiling point of 27 ºC. A small amount of TMS was used to be added in the sample as an internal standard for NMR measurement, and removed by evaporation afterwards. However, for contemporary NMR spectrometer (including the bench top NMR), it is no longer necessary to actually add TMS since the computer can calibrate the chemical shift electronically based on resonance frequencies of the solvent used.
The unit of chemical shift (δ) is ppm. The ‘ppm’ label stands for ‘parts per million’. The chemical shift relative to TMS in ppm is defined as the formula below.
The reason for using a relative value of chemical shift in ppm, rather than the actual resonance frequency in Hz is that every NMR instrument will have a different magnetic field strength, so the actual value of resonance frequencies expressed in Hz will be different on different instruments – remember that ΔE for the magnetic transition of a nucleus depends upon the strength of the externally applied magnetic field Bo. However, the chemical shift expressed in ppm will always be the same whether measured with an instrument operating at 400 MHz or 60 MHz. In the 1H NMR of methyl acetate, the two signals are at 2.0 and 3.6 ppm represents the two sets of protons in methyl acetate have resonance frequencies about 2.0 and 3.6 parts per million higher than the resonance frequency of the TMS protons. If, for example, the spectrum is measured by the 400 MHz NMR spectrometer, then the chemical shift in Hz will be 800 Hz and 1440 Hz respectively.
Most protons in organic compounds have chemical shift values between 0 and 12 ppm relative to TMS, although values below 0 ppm and above 12 ppm are occasionally observed. The chemical shift value of hydrogens in certain structural environment, or common organic functional groups, are listed in chart (Fig. 6.6c) and table (Table 6.2) below.
The importance of chemical shift information is that it gives critical clues about molecular structures. Several highlights here:
• Usually the hydrogens in C-H bond, without any other functional groups nearby, are in the range of 1-2 ppm;
• For hydrogen in C-H bond beside double bond, like C=C or C=O bond, the signal goes downfield to 2-2.5 ppm;
• With electronegative atoms connected on the carbon, like O-C-H, the hydrogens get deshielded and chemical shift move further downfield to 3-4 ppm;
• The hydrogens bonded directly to double bond carbon have the chemical shift at around 4.5-6 pm;
• The aromatic hydrogens (H on benzene ring) show chemical shift around 7 ppm;
• The chemical shift of hydrogens in OH (alcohol) or NH (amine) group vary in a rather large range, from 1-5 ppm;
• The hydrogen in aldehyde (-CHO) and carboxylic acid (COOH) group has the chemical shift rather downfield at about 9-10 ppm and 10-12 ppm respectively.
When referring to the chemical shift table (or chart) for a certain compound, it is useful to keep in mind that the exact value may vary a bit to the given range, sometimes the difference up to 0.5 ppm unit may happen depends on the specific structure and the solvent used.
With chemical shift information available, we can now assign the signals in the 1H NMR spectrum of methyl acetate. According to Fig. 6.6c, the protons in CH3 group beside C=O bond are supposed to be in the range of 2-3 ppm, and protons in CH3 group connected with O directly have δ value of about 3-4 ppm. So the 2.0 ppm signal is for the Ha group and 3.6 ppm signal is for Hb group.
Chemical Shift of Protons Near π Electrons — Anisotropy Effect
The chemical shift values of aromatic protons and vinylic protons (those directly bonded to an alkene carbon) resonate much further downfield (higher frequency, higher chemical shift) than can be accounted for simply by the deshielding effect of nearby electronegative atoms. These chemical shifts result from the anisotropy effect.
Let’s investigate the aromatic protons first. In benzene ring (and many other aromatic structures), the total six π electrons form delocalized big π bond around the ring (more discussions in Organic II). When the molecule is exposed to the external magnetic field Bo, these π electrons begin to circulate in a ring current and generating their own induced magnetic field Binduced. Whether shielding or deshielding occurs depends on the location of the protons in the induced magnetic field, and this is called anisotropy (means “non-uniformity”) effect. This can be illustrated specifically in the figure below by comparing between point A and B.
If a proton is at point A, it feels the induced magnetic field pointing to the opposes direction of Bo, sothe proton experiences shielding effect. For the proton at point B, however, it feels the induced magnetic field to the same direction as Bo, so the proton experiences deshielding effect.
The protons on benzene ring are at the position equivalent of ‘point B’, that means that the induced current in this region of space is oriented in the same direction as B0, so it adds to B0 and result in a deshilelding effect and the benzene protons resonance at a higher frequency and have larger chemical shifts.
As a result, due to the anisotropy of the induced field generated by the circulating π electrons, the benzene protons are highly deshielded. Their chemical shift is far downfield, in the range of 6.5–8.5 ppm.
Anisotropy is also responsible for the downfield (high frequency) chemical shifts of vinylic protons (4–6.5 ppm) and aldehyde protons (9.5–11 ppm). The π electrons in these groups also circulate in such a way to generate an induced magnetic field that adds to external field Bo in the spots occupied by the protons. Carboxylic acid protons are even further downfield (9.5–12 ppm) due to the combined influence of the electronegative oxygen atom and the nearby π bond. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.06%3A_H_NMR_Spectra_and_Interpretation_%28Part_I%29.txt |
6.7.1 Integration of Signal Areas
The computer in the NMR instrument can be instructed to mathematically integrate the area under a signal or group of signals. The signal integration process is very useful in 1H NMR spectrum, because the area under a signal is proportional to the number of protons to which the signal corresponds.
The Fig. 6.7a is the 1H NMR spectrum of 1,4-dimethylbenzene with integration line (blue lines). The integration line generated by the computer is always in curve shape that resemble steps. The integration numbers are also generated by the computer together with the curve, that show the relative area of each signal (the integration numbers in the actual spectra are usually with decimals, whole numbers are shown here for simplicity).
As we discussed earlier, the molecule of 1,4-dimethylbenzene has two sets of equivalent protons: the four aromatic (Ha) protons and the six methyl (Hb) protons. The integration of the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm. Please note that the integration number show the relative ratio of the number of protons, not the actual number. The ratio 3 to 2 here matches the ratio of actual number 6 to 4. This integration information, along with the chemical shift knowledge we have learned before allow us to assign the peaks: peak at 7.4 ppm correspond to protons (Ha) on the benzene ring, and the peak at 2.6 ppm correspond to two methyl groups (Hb).
6.7.2 Signal Splitting (Coupling)
In the 1H NMR spectra that we have seen so far, each set of protons generates a single NMR signal. This is not that common for 1HNMR actually. In fact, the 1H NMR spectra of most organic molecules contain signals that are ‘split’ into two or more peaks that is called splitting (or coupling). The spectra with peak splitting may looked more complicated, however, this splitting behavior provides very useful information about the structure of a compound.
Let’s consider the spectrum for 1,1,2-trichloroethane (Fig. 6.7b). In this and in other spectra to follow, the expansions of individual signals are shown so that the signal splitting patterns are recognizable.
The signal at 3.96 ppm, corresponding to the two Ha protons, is split into two peaks of equal height (and area) – this is referred to as a doublet. The Hb signal at 5.76 ppm, on the other hand, is split into three peaks, with the middle peak higher than the two outside peaks and the integration ratio between the three peaks is 1:2:1, such splitting signal is called a triplet.
Signal splitting is caused by spin-spin coupling, a term that describes the magnetic interactions between non-equivalent hydrogen atoms that are with 2 or 3 bonds of the hydrogens producing the signal. The nearby protons have magnetic moment that can be either against or with the external magnetic field, therefore splits the energy levels of the protons whose signal is being observed, and result in the splitting of the signal into multiple peaks (the terms ‘splitting’ and ‘coupling’ are often used interchangeably when discussing NMR).
The most typical coupling we observed in this course is from non-equivalent vicinal hydrogens that are 3 bonds away, that is the hydrogens on adjacent carbons. This is also called vicinal coupling or three-bond coupling.
A simple rule that applies for predicting the number of peaks (or splitting pattern) expected from coupling and the rule in 1H NMRis:
number of peaks = n + 1
(n is the number of vicinal non-equivalent hydrogens)
We will exam the splitting pattern with different number of n:
• When n=0, the signal is a singlet, or has only one peak, as the signals observed in Fig. 6.6d and Fig. 6.7a.
• When n=1, the signal is a doublet with two peaks. The area ratio of the two peaks for a doublet is 1:1. The space between the two peaks is called coupling constant, Jab, measured in Hz.
For the example of compound 1,1,2-trichloromethane, the signal of Ha protons fits into this situation. With only one vicinal proton, Hb, on the adjacent carbon, the signal of Ha show as a doublet.
• When n=2, the signal is a triplet with three peaks. The three peaks of triplet has the ratio of the area as 1:2:1.
In the same compound 1,1,2-trichloromethane, the signal of Hb proton fits into this situation. With two vicinal protons, 2Ha, on the adjacent carbon, the signal of Hb show as a triplet.
• When n=3, the signal is a quartet, that means four peaks. The four peaks of quartet has the area ratio of 1:3:3:1.
For the spectrum of ethyl acetate (Fig. 6.7e), the signal of Hb is a quartet, because there are three vicinal protons 3Hc on the adjacent carbon. Please note that the carbon with Hb connected with oxygen on the other side, and there are no hydrogen atoms on that oxygen atom, so only the coupling with three vicinal protons apply.
• When n≥4, the signal can be called a multiplet. Theoretically, with n increase the signal split into more peaks and the total number of peaks is “n+1”. However, the small peaks on the sides may or may not be able to be observed since they might be merged into noise. The signal with more than four peaks are generally called as a multiplet, and it is not that critical to tell exactly how many peaks involved in a multiplet.
Extra notes about signal splitting:
1. Splitting (coupling) only occurs between nonequivalent protons. For equivalent protons, there is no coupling. In the spectrum of succinic acid (Fig. 6.7f) for example, the protons on the two middle carbons are equivalent (Ha), so there is no coupling between them and they show a singlet.
2. Protons in OH or NH generally do not couple with vicinal hydrogens. OH and NH protons are acidic enough to rapidly exchange between different molecules, so the neighboring protons never actually ‘feels’ their influence. See the specific example of 1-heptanol spectrum in Fig. 6.7g.
Signal assignment based on the given structure
With the structure of a compound given, we can apply all the knowledge about 1H NMR to assign the signals in the spectrum, that is to identify a certain signal comes from which hydrogen(s).
Examples
Match the 1H NMR spectrum below to its corresponding compound, and assign all of the signals.
a) cyclopentanone b) 3-pentanone c) butaldehyde d) 2-pentanone
e) 4-heptanone f) 1-butene
Approach: It is good idea to draw the structure of each compound and try to see which matches to the spectrum.
The spectrum has four signals: triplet (~0.7 ppm), multiplet (~1.4 ppm), singlet ( ~1.9 ppm) and triplet (~2.2 ppm). Based on the structure of each compound, compound c), d) and f) should have four signals in the 1H NMR spectrum.
• There is no signals at about 9 ppm for the aldehyde hydrogens in the spectra, so the spectrum is not for compound c) , butaldehyde.
• There is no signals at about 4~5 ppm for the alkene hydrogens in the spectra, so the spectrum is not for compound f) , 1-butene.
• The signals in the spectrum match with what are expected for compound d), 2-pentanone.
Solution: The spectrum is for 2-pentanone.
Structure Determination based on 1H NMR spectrum
For an advanced level of practice, we are supposed to be able to determine the exact structure of a compound with 1H NMR spectrum given (and other necessary information). As we have learned, there are a lot valuable information about the structure of a compound can be obtained from an 1H NMR spectrum. For a summary, analyzing the four features of the spectrum is critical to elucidate the structure of a compound:
• The number of signals indicate how many different sets of protons there are in the molecule;
• The chemical shift of the signal tells us about the electronic environment of each set of protons;
• The integration under each signal provides information about how many protons there are in the set being measured (keep in mind that the integration values are for the ratio, not actual number of protons);
• The splitting pattern of each signal tells about the number of protons on atoms adjacentto the one whose signal is being measured. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.07%3A_H_NMR_Spectra_and_Interpretation_%28Part_II%29.txt |
For carbon element, the most abundant isotope 12C (with ~99% natural abundance) does not have a nuclear magnetic moment, and thus is NMR-inactive. The C NMR is therefore based on the 13C isotope, that accounts for about 1% of carbon atoms in nature and has a magnetic dipole moment just like a proton. The theories we have learned about 1H NMR spectroscopy also applies to 13C NMR, however with several important differences about the spectrum.
The magnetic moment of a 13C nucleus is much weaker than that of a proton, meaning that 13C NMR signals are inherently much weaker than proton signals. This, combined with the low natural abundance of 13C, means that it is much more difficult to observe carbon signals. Usually, sample with high concentration and large number of scans (thousands or more) are required in order to bring the signal-to-noise ratio down to acceptable levels for 13C NMR spectra.
Chemical Equivalent
For carbons that are chemical equivalent, they only show one signal in 13C NMR as like protons for 1HNMR. So it is very important to be able to identify equivalent carbons in the structure, in order to interpret 13C NMR spectrum correctly. Taking toluene as an example, there are five sets of different carbon atoms (shown in different colors), so there are five signals in the 13C NMR spectrum of toluene.
Chemical Shift
13C nuclei has different value of g (the magnetogyric ratio) comparing to 1H nuclei, so the resonance frequencies of 13C nuclei is different to those of protons in the same applied field (referring to formula. 6.4, in section 6.5). In an instrument with a 7.05 Tesla magnet, protons resonate at about 300 MHz, while carbons resonate at about 75 MHz. This allows us to look at 13C signals using a completely separate ‘window’ of radio frequencies. Just like in 1H NMR, tetramethylsilane (TMS) is also used as the standard compound in 13C NMR experiments to define the 0 ppm, however it is the signal from the four equivalent carbon atoms in TMS that serves as the standard. Chemical shifts for 13C nuclei in organic molecules are spread out over a much wider range of about 220 ppm (see Table 6.3).
The chemical shift of a 13C nucleus is influenced by essentially the same factors that influence the chemical shift a proton: the deshielding effect of electronegative atoms and anisotropy effects tend to shift signals downfield (higher resonance frequency, with higher chemical shifts). In addition, sp2 hybridization results in a large downfield shift. The 13C NMR signals for carbonyl carbons are generally the furthest downfield (170-220 ppm), due to both sp2 hybridization and to the double bond to oxygen.
Integration and Coupling in 13C NMR
Unlike 1H NMR, the area under a 13C NMR signal cannot easily be used to determine the number of carbons to which it corresponds. The signals for some types of carbons are inherently weaker than for other types, for example peaks corresponding to carbonyl carbons are much smaller than those for methyl or methylene (CH2) peaks. For this reason, signal integration is generally not useful in 13C NMR spectroscopy.
Because of the low natural abundance of 13C nuclei, the spin-spin coupling between two nonequivalent 13C atoms is negligible. 13C nuclei are coupled to nearby protons, however, which results in complicated spectra. For clarity, chemists generally use a technique called broadband decoupling, which essentially ‘turns off’ C-H coupling, resulting in a spectrum in which all carbon signals are singlets. Below is the proton-decoupled 13C NMR spectrum of ethyl acetate in CDCl3 (Fig. 6.8a), showing the expected four signals, one for each of the carbons.
For our class purpose, 13C NMR spectra are usually used as supporting information to confirm the structure of a compound.
Exercises 6.2
Below are 13C NMR spectra for methylbenzene (common name toluene) and methyl methacrylate. Refer to Table 6.3 to match the spectra to the correct structure.
Answers to Practice Questions Chapter 6 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.08%3A_C_NMR_Spectroscopy.txt |
1H NMR provides a powerful tool for determining the structure of unknown compound. Other than that 1H NMR, additional information includes molecular formula, IR and 13C NMR spectrum are usually provided as well. Solving the structure of an unknown compound based on all the given information is an important type of question we will work on for this chapter. We will take the C5H12O constitutional isomer example to go through the strategy for solving this type of question.
Example: Constitutional Isomers with Formula C5H12O
The 1H NMR below are all for compounds with molecular formula of C5H12O (the relative integration area for each signal are given as numbers on the spectra). The IR spectra of these compounds do not have any strong band at above 3000 cm-1, nor are there strong bands at 1700 cm-1. Propose a reasonable structure for each compound that is consistent with the data given.
Step 1: Calculate the degree of unsaturation (or IHD, section 2.3) based on the given molecular formula, and get hints about structure/functional group according to the degree of unsaturation. This is usually the first step to solving this type of question.
Degree of unsaturation =
From what we learned about the degree of unsaturation, zero degree means there is no any ring nor double bond in the structure, that means all the compounds in this question have open chain structures with single bonds only. With one oxygen atom involved, the possible functional group therefore will be open chain alcohol, or open chain ether.
Step 2: Narrow down the possible functional groups with IR information.
IR indicate that there is noany strong bands at above 3000 cm-1 for the compound, that exclude the option of alcohol, so the only choice left is the open chain ether.
Step 3: Use available spectroscopy data (mainly 1H NMR, with 13C NMR as supporting if available) to identify discrete parts of the structure.
Step 4 : Try to put the pieces of the puzzle together, and double check if everything fit the available data.
Step 3 and 4 are the most challenging parts since there is no simple rule to follow about how to do that. It takes practices to do the interpretation of 1H NMR signals and translating that into the structure of unknown compound. Checking the four aspects of 1H NMR as we learned in section 6.6.5. The relative integration areas are given for this question to make it bit easier.
Compound 1:
We can start with the simplest spectrum that have least signals:
• There are only two signals (both are singlet) in this spectrum, indicate that there are two sets of non-equivalent hydrogens.
• The integrations of the two signals are 3 and 1, means the ratio of the number of hydrogens in these two sets are 3:1. And since there are total 12 hydrogens, the actual number of hydrogens should be 9 and 3 in each group.
• 3 hydrogens imply a CH3 methyl group, and 9 hydrogens could be three CH3 groups. Also since all the 9 hydrogens are equivalent, that means the three CH3 groups are equivalent. The only way to have three equivalent CH3 groups is that there is a t-butyl group.
• So the structure is the ether with a methyl group and a t-butyl group connected with the oxygen atom.
• The structure of compound 1 is given below, with the chemical shift valued included.
For the remaining compounds, the integration for each signal could be a very good starting point, since generally the integration value indicates the possible structural unit like CH3, CH2 or CH. Then the structural units can be put together in a logical way like putting pieces of a puzzle together.
Compound 2:
• Based on the integration, it is determined that there are:
• one CH3 group show a triplet;
• two equivalent CH3 groups show a doublet;
• one CH group show a multiplet;
• one CH2 group show a quartet.
• The triplet CH3 could connect with quartet CH2 as a CH2CH3 ethyl group, that makes sense based on the splitting pattern.
• Also, the two equivalent CH3 groups with a CH could give an isopropyl group, that is consistent to the splitting pattern.
• So the overall structure of compound 2 is isopropyl ethyl ether.
Compound 3:
• Based on the integration, it is determined that there are:
• one CH3 group show a triplet;
• one CH3 groups show a doublet;
• one CH2 group show a multiplet;
• one CH group show a quartet;
• one CH3 group show a singlet.
• The singlet means the CH3 has no any other hydrogens bonded on adjacent atoms, so the CH3 group should be bonded with the oxygen atom, and the value of chemical shift (about 3.2 ppm) confirms.
• The triplet CH3 could connect with quartet CH2 as a CH2CH3 ethyl group, that makes sense based on the splitting pattern.
• The doublet CH3 groups should connect with a CH group, that is consistent to the splitting pattern.
• The chemical shift (about 3 ppm) and splitting of the CH group (quartet) indicate it should connect to the oxygen atom.
• Put all the above pieces together, the structure of compound 3 is sec-butyl methyl ether.
Compound 4
• Based on the integration, it is determined that there are:
• one CH3 group show a triplet;
• another CH3 groups show a triplet;
• one CH2 group show a multiplet;
• two CH2 groups with signals overlapping
• The two CH3 groups both as triplet indicate that they both connect with CH2, so there are two ethyl CH2CH3 groups in the structure, and they are not equivalent.
• Therefore there is only one more CH2 group left.
• There is only one possible structure with two CH2CH3 groups, one CH2 group and one oxygen atom, so the structure of compound 4 is ethyl methyl ether.
6.10: Answers to Practice Questions Chapter 6
Xin Liu
6.1 How many 1H NMR signals would you predict for each of the following molecules? 6.2 Below are 13 C NMR spectra for methylbenzene (common name toluene) and methyl methacrylate. Refer to Table 6.3 to match the spectra to the correct structure. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.09%3A_Structure_Determination_Practice.txt |
Let’s start with a simple substitution reaction example:
In this reaction, the Br in the reactant methylbromide (CH3Br) is replaced by the OH group, and the methanol (CH3OH) is produced as the major product, together with bromide Br-, the side product. It is easy to understand that this is a substitution reaction, because Br is substituted by OH.
Further discussions on this simple reaction require the introduction of some key terms that are critical in understanding why and how the reaction proceed in this way. These terms are electrophile, nucleophile and leaving group.
Electrophile
The reactant CH3Br is an alkyl halide. The C-X bond (X: F, Cl and Br) in alkyl halide is polar because halogen is more electronegative than carbon, and as a result carbon has a partial positive charge and halogen has a partial negative charge.
Because of the partial positive charge on carbon, the carbon atom in C-X bond is electron-deficient, and it is going to seek electron-rich reagent to connect with. Such electron-deficient species is called an electrophile (phile is the Greek suffix means “love”), means the species that loves electrons. The electron-deficient species are usually electrophiles. Other electrophile examples include positive charged ions and atom with incomplete octet, for example: H+, CH3+,BH3, BeF2, AlCl3.
For CH3Br in this reaction, it is the carbon atom that act as the electrophile, and the carbon can be called as electrophilic carbon.
The compound CH3Br that undergoes the substitution usually can be called the substrate. Nucleophile
The hydroxide, OH, is another reactant in above reaction. It is shown clearly with the Lewis structure of OH that the oxygen atom has three lone pair electrons and is negatively charged, so it is an electron-rich species with high electron density.
An electron-rich species is called a nucleophile (“nucleo” comes from nucleus, that means positive charge), that is the reagent seeking positively charged or electron-poor species to react with. OH is the nucleophile for above reaction. Generally, any species with the electron pair available for sharing could be nucleophile. Nucleophile can be either negatively charged (Nu:), or neutral (Nu:), for example: OR, H2O, ROH, NH3, RNH2, RCOO are all possible nucleophiles.
Based on the understanding of the concepts of electrophile and nucleophile, you probably realize that a nucleophile could react with an electrophile! Yes, that is the very important and fundamental rule for organic reaction: when electron-rich nucleophile meet with electron-deficient electrophile, organic reaction would occur.
Leaving Group
To ensure the above substitution occurs, another critical factor is that the Br must leave together with the electron pairs in C-Br bond, and the bromide, Br-, is called the leaving group. The leaving group (LG) leaves with the bonding pair of electrons, and is replaced by the nucleophile in the substitution reaction. Without a proper leaving group, even nucleophile is attracted to electrophile, the substitution reaction still cannot move forward. Leaving group can be negatively charged or neutral, as we will see in detailed discussions later.
Applying the three key terms, the above substitution reaction can be summarized as: the nucleophile displaces the leaving group in a substrate, so such reaction is called nucleophilic substitution reaction. Nucleophilic substitution reaction could therefore be shown in a more general way:
Note: the nucleophile and leaving group are not necessary negatively charged, they could be neutral as mentioned earlier.
Kinetics of Nucleophilic Substitution Reaction
Kinetics is the study that concerns the rate of a chemical reaction, or how fast the reaction occurs. The reaction rate data helps to shine a light on the understanding of reaction mechanism, the step-by-step electron transfer process. Kinetic studies on nucleophilic substitution reactions indicate that there are two different rate law expressions for such reactions. For the two reactions below, reaction 1 is in second order while reaction 2 is in first order. The only reason behind the different kinetic rate is that the reactions go through different reaction mechanism.
Reaction 1 is the substitution reaction we are familiar with already. It is a second-order reaction. That means the reaction rate depends on the concentration of both substrate CH3Br and nucleophile OH. If the concentration of CH3Br doubled, the reaction rate get doubled, and if the concentration of OH doubled, the reaction rate doubled as well. When the concentration of both CH3Br and OH doubled, the reaction rate increased by a factor of four.
Reaction 2 is another substitution reaction example. The substrate here is a tertiary bromide and the nucleophile is neutral water molecule. As a first-order reaction, the reaction rate depends only on the concentration of substrate (CH3)3CBr and has nothing to do with nucleophile.
The two types of reactions correspond to two types of reaction mechanism:
• The second-order reaction goes through the bimolecular reaction mechanism that is called SN2 reaction, meaning Substitution, Nucleophilic and Bimolecular.
• The first-order reaction goes through the unimolecular reaction mechanism that is called SN1 reaction, meaning Substitution, Nucleophilic and Unimolecular.
We will have detailed discussions on SN2 and SN1 mechanism respectively, and then compare the similarities and differences in between. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/07%3A_Nucleophilic_Substitution_Reactions/7.01%3A_Nucleophilic_Substitution_Reaction_Overview.txt |
SN2 Reaction Mechanism
Let’s still take the reaction between CH3Br and OH as the example for SN2 mechanism.
SN2 mechanism involves two electron pair transfers that occur at the same time, nucleophile attacking (red arrow) and leave group leaving (blue arrow). The nucleophile OH approaches the electrophilic carbon from the back side, the side that is opposite to the direction that leaving group Br leaves. With the nucleophile OH getting closer, the Br start to leave as well. The new C—OH bond formation and the old C—Br bond breaking occur at the same time. In a very short transient moment, the carbon atom is partially connected with both OH and Br, that gives a highest energy level state of the whole process called transition state. In the transition state of SN2 reaction, there are five groups around the carbon and the carbon can be called “pentacoordinated”. As the OHcontinues to get closer to the carbon, the Br moves further away from it with the bonding electron pair. Eventually, the new bond is completely formed and the old bond is completely broken that gives the product CH3OH.
In the mechanism, the reaction proceeds in a single step that involves both nucleophile and the substrate, so increasing the concentration of either of them makes the possibility of collision increase, that explains the second-order kinetics of SN2 reaction. With both nucleophile attacking and leaving group leaving happen at the same time, SN2 is also said to be a concerted mechanism, concerted means simultaneous.
Notes for drawing SN2 mechanism:
• The two arrows must be shown when drawing the SN2 mechanism. Both have to be shown with the proper direction: nucleophile attack from the direction that is opposite to the leaving group leaves, i.e., backside attack.
• The transition state is optional (depends on the requirement of the question). However it is important to understand that the reaction process goes through the transition state before producing the products.
• Please pay attention that for the product, the positions of the three hydrogens around carbon are all pushed to the other side, and the overall configuration of the carbon get inverted, like an umbrella flipped inside out in a windstorm. It seems does not really matter for product (CH3OH) in this reaction, however it does make a difference if the carbon is a chirality center.
Energy Diagram of SN2 Mechanism
The energy changes for the above reaction can be represented in the energy diagram shown in Fig. 7.1. SN2 is a single-step reaction, so the diagram has only one curve. The products CH3OH and Br are in lower energy than the reactants CH3Br and OH, indicates that the overall reaction is exothermic and the products are more stable.
The top of the curve corresponds to the transition state, which is the highest-energy structure involved in the reaction. Transition state always involves partial bonds, partially formed bond and partially broken bond, and therefore is very unstable with no appreciable lifetime. The transition state therefore can never been isolated. The structure of the transition states is usually shown in a square bracket with a double-dagger superscript.
The Effect of Alkyl Halide Structure on SN2 Reaction Rate
For the discussions on SN2 mechanism so far, we focused on the reaction of methylbromide CH3Br. Other alkyl halides could undergo SN2 reactions as well. The studies on the reaction rate for SN2 indicate that the structure category of electrophilic carbon in alkyl halide affects the reaction rate dramatically.
Type of Alkyl Halide
Alkyl Halide
Structure
Relative Rate
Methyl
CH3X
30 Primary (1°) RCH2X 1 Secondary (2°)
0.03 Tertiary (3°)
(no SN2 reaction)
negligible
Table 7.1 Relative Reaction Rate of SN2 for Different Type of Alkyl Halide
As shown in Table 7.1, methyl and primary halides are the substrates with the highest rate, the rate decreases a lot for secondary halides, and the tertiary halides do not undergo SN2 reaction at all because the rate is too low to be practical.
The relative reactivity of alkyl halides towards SN2 reaction can therefore be summarized as:
Why the trend is like this? This can be explained by the mechanism of SN2 reaction. Actually this is one of the experiment evidences scientists based on for proposing the mechanism. A key feature in SN2 mechanism is that the nucleophile attacks from the back side. When nucleophile approaching to the carbon, it is easiest to getting close to the methyl carbon because the hydrogen atoms connected on carbon are small in size. With the size of the groups connected on the carbon getting larger, it is becoming more difficult to access to the carbon, and such approaching is totally blocked for tertiary carbon with three bulky alkyl groups connected. Therefore, the reactivity difference is essentially caused by the steric effect. Steric effect is the effect that based on the steric size or volume of a group. Because of the steric hinderance of bulky groups on the electrophilic carbon, it is less accessible for nucleophile to do back-side attack, so the SN2 reaction rate of secondary (2°) and tertiary (3°) substrates decreases dramatically. Actually the 3° substrates never go with SN2 reaction mechanism because the reaction rate too slow.
The Stereochemistry of SN2 Reaction
Another feature of SN2 reaction mechanism is that the overall configuration of the carbon in the product get inverted comparing to that of the reactant, like an umbrella flipped inside out. Such inversion of configuration is called Walden inversion. let’s see what is the stereochemistry consequence for such inversion.
Start with the (R)-2-bromobutane, the SN2 reaction produces only one enantiomer of 2-butanol product, and it is predictable that the configuration of the product supposed to be S because of the configuration inversion.
Note: Inversion means the arrangement of the groups get inverted, not necessary means the absolute configuration, R/S, inverted. The product does get inverted R/S configuration comparing to the reactant for lot cases, but not guaranteed. The actual configuration of the product has to be determined accordingly.
Exercises 7.1
Show the product of the following SN2 reaction (CN is the nucleophile):
Answers to Practice Questions Chapter 7 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/07%3A_Nucleophilic_Substitution_Reactions/7.02%3A_SN2_Reaction_Mechanism_Energy_Diagram_and_Stereochemistry.txt |
Leaving Group
When alkyl halides undergo nucleophilic substitution reactions, halogen is the leaving group. Not only halogens can be leaving group, other appropriate groups could be leaving groups as well. Generally speaking, nucleophilic substitution reaction requires good leaving group. How to determine whether a leaving group is good or not then? When leaving group departs, it takes the electron pair from the broken bondtogether it. So the good leaving group should be the one that can accommodate the electron pair very well with it, or it can be said the good leaving group should be stable with the pair of electrons.
The stability of a group with electron pair is related to the basicity of the group, since basicity means the ability of the species to share its electron pair. As a result, strong base has the high reactivity to share the electron pair, so it is not stable, and cannot be good leaving group. On the other side, weak base with low tendency to share the electron pair, is more stable base and therefore is good leaving group. So the general trend is:
The weaker the basicity of a group, the better leaving group it is .
Our knowledge in acid-base topic will be very helpful here to compare the strength between different leaving groups.
For alkyl halides, the relative reactivities as leaving group is:
(best leaving group) I> Br> Cl> F(weakest leaving group)
This order matches with the relative basicity of halide anions, Iis the weakest base and also the best leaving group.
Beside halides, other groups can be leaving groups as well. In acid-base chapter we have learned about some examples of strong organic acids, for example, tosylic acid, TsOH, etc. Since the conjugate base of strong acid is very weak bases, the conjugate bases of those acids are good choice of leaving group as well. Examples include (the leaving group is highlighted in blue color):
Strong bases such as OH, RO, NH2, Rare therefore very poor leaving groups and cannot go with nucleophilic substitution reactions. For OHor RO however, upon protonation they can be converted to neutral H2O or ROH molecules, that are good leaving groups suitable to substitution. This topic will be covered in section 7.6.
Note : with the scope of leaving group expanded, the substitution reaction not only limited to alkyl halide. Any compounds with a good leaving group can undergo nucleophilic substitution.
Nucleophile
For SN2 reaction, nucleophile is one of the rate-determining factors, therefore strong nucleophile helps to speed up SN2 reactions.
The relative strength of a nucleophile is called nucleophilicity. Nucleophilicity of a nucleophile is measured in terms of the relative rate of its SN2 reaction with the same substrate. Generally speaking, the nucleophilicity trend depends on several structural features of the nucleophile.
• A nucleophile with negative charge is always stronger than the corresponding neutral one. For example: OH> H2O; RO> ROH.
• Nucleophilicity decrease across a period. For example: NH3 > H2O; RNH2 > ROH
• Nucleophilicity increase across a group. For example:
RSH > ROH; RS> RO;
I > Br > Cl > F (protic solvent)
• Smaller group is better nucleophile than bulky group.
For example, t-BuO is very poor nucleophile because of its bulky size.
To make it more convenient for studying purpose, the commonly applied strong and weak nucleophiles are listed here:
Strong (good) nucleophile: OH, RO(small alkoxide), RS(thiolate), N3(azide), CN(cyanide), Cl, Br, I(halide), RCO2(carboxylate), RNH2 (amine)
Weak (poor) nucleophile: ROH, H2O, t-BuO
With the structure of nucleophiles being so diverse, SN2 reaction can be used to synthesize the compounds with a variety of functional groups, as shown here.
Examples
Exercises 7.2
Show reaction mechanism of the above reactions.
Answers to Practice Questions Chapter 7 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/07%3A_Nucleophilic_Substitution_Reactions/7.03%3A_Other_Factors_that_Affect_SN2_Reactions.txt |
SN1 Reaction Mechanism
The reaction between tert-butylbromide and water proceeds via the SN1 mechanism. Unlike SN2 that is a single-step reaction, SN1 reaction involves multiple steps. Reaction: (CH3)3CBr + H2O → (CH3)3COH + HBr
In step 1, C—Br bond breaks and Br departs with the bonding electron pair to produce a tertiary carbocation and bromide anion Br. This step only involves a highly endothermic bond-breaking process, and this is the slowest step in the whole mechanism. In multiple-step mechanism, the overall reaction rate is determined by the slowest step, such step is therefore called the rate-determining step. In SN1 reaction, step 1 is the slowest step and therefore the rate-determining step. The rate-determining step only involves the alkyl halide substrate, that is why the overall rate law is in first order, because nucleophile does not participate in the rate-determining step.
The product of step 1, carbocation, will be the reactant of next step and is called the intermediate for SN1 reaction. Intermediate is the unstable, highly-reactive species with very short lifetime. The carbocation intermediate is in trigonal planar shape, with the empty 2p orbital particular to the plane. The central carbon is sp2 hybridized and has the incomplete octet, so carbocation is the highly reactive intermediate, that is also the electrophile.
Step 2 is the nucleophilic attack step, that the nucleophile H2O use its lone pair to react with the carbocation intermediate, and produces the protonated t-butyl alcohol (oxonium ion). Because of the planar shape of carbocation intermediate, there is same possibility for the nucleophile to attack from either side of the plane, so possible products are generated with the same amounts. For this reaction, attacking from either side gives the same product (both are still shown for the purpose to illustrate the concept); however it gives different stereoisomers if the electrophilic carbon is the chirality center. In step 3, a water molecule acting as a Bronsted base to accept the proton from the oxonium ion, and the final neutral product t-butyl alcohol is produced. This deprotonation step is very fast, and sometimes can be combined with step 2 together as one step (i.e. step 3 may not be regarded as an individual step).
Energy diagram of SN1 mechanism
Because SN1 is a multiple-step reaction, so the diagram has multiple curves, with each step can be represented by one curve. Out of the three steps, the activation energy for step 1 is the highest, therefore step 1 is the slowest step, that is the rate-determining step.
The connection between the first two curves represent the carbocation intermediate. Generally, intermediate is the product of one step of a reaction and the reactant for the next step. Intermediate is at a relatively lower energy level comparing to transition state (which is at the peak of a curve), but intermediate is also highly reactive and unstable.
The Effect of Substrate Structure on SN1 Reaction Rate
Different substrates have different reaction rates towards SN1 reaction, and the relative reactivity of substrates towards SN1 reaction can be summarized as:
Comparing this trend to that for SN2 reaction, you probably realize that they are just opposite. Tertiary substrate is most reactive towards SN1, but it does not undergo SN2 at all; primary and methyl substrate are unreactive for SN1, but they are the best substrates for SN2. This comparison is very important and useful for us to choose the proper reaction condition for different substrate as we will see in next section. For now, we will need to understand the reasoning of the trend for SN1.
This is because of the stability of carbocation intermediate. The mechanism shows that a carbocation is formed in the rate-determining step, so the more stable the carbocation, the more easily it is formed, the more it facilitates the rate-determining step and speed up the whole reaction. Therefore the more stable the carbocation intermediate is, the faster the rate of a SN1 reaction.
The relative stability of carbocation is given below, that the tertiary carbocations are the most stable and methyl carbocation is the least stable.
The relative stability of carbocations can be explained by the hyperconjugation effect. Hyperconjugation is the partial orbital overlap between filled bonding orbital to an adjacent unfilled (or half-filled) orbital. Carbocation is the electron-deficient species that has the incomplete octet and empty 2p orbital. If there is an alkyl group connected with carbocation, then there are C-C or C-H sigma bonds beside the carbocation carbon, so the filled orbitals of sigma bonds will be able to partially overlap with the empty 2p orbital, therefore sharing the electron density to carbocation and to get the carbocation stabilized. The more R group involved, the stronger hyperconjugation effect is. So tertiary (3°) carbocation is the most stable one. While there is no any R group in methyl carbocation, CH3+, it is least stable.
Stereochemistry of SN1 mechanism
The stereochemistry feature of the SN1 reaction is very different to that of SN2, and of course can be explained well with the SN1 mechanism.
Starting with (S)-3-bromo-3-methylhexane reactant, the SN1 reaction produces a 50:50 mixture of both R and S enantiomers of 3-methyl-3-hexanol, that is the racemic mixture product. This is because the carbocation formed in the first step of an SN1 reaction has the trigonal planar shape, when it react with nucleophile, it may react from either the front side or the back side, and each side gives one enantiomer. There is equal possibility for reaction to occur from either side, so the two enantiomers are formed with the same amount, and the product is a racemic mixture.
A reaction that coverts an optically active compound into a racemic form is said to proceed with racemization. For SN1 reaction that start with (an optical active)one enantiomer as the reactant, and the chirality center is also the electrophilic carbon (i.e. the reaction occurs on the chirality center), it proceeds with racemization as shown above.
Exercises 7.3
Show the detailed mechanism for above reaction of (S)-3-bromo-3-methylhexane and water.
Answers to Practice Questions Chapter 7
Please note that if the chirality center of the reactant is not the reaction center, or if there are more than one chirality center in the reactant, the SN1 reaction does not produce the racemic mixture as example below.
Examples
Show product(s) of the following SN1 reaction:
L eaving Group Effect on SN1
Same as for SN2 reaction, a good leaving group is also required for SN1 mechanism, and all the discussions we had before in section 7.3 apply.
Nucleophile
Unlike a SN2 reaction, the rate-determining step of SN1 reaction does not include nucleophile, so theoretically the strength of nucleophile has no effect on SN1 reaction. However, a strong nucleophile has high tendency to go with SN2 reaction instead of SN1, so a weaker nucleophile is a better choice for SN1. For the examples we had so far, H2O is the nucleophile.
In practice, neutral substances such as H2O, ROH, RCOOH are usually used as nucleophiles in SN1 reaction. When these substances are applied in the reaction, they serve for another function as solvents. So they are used as both nucleophiles and solvents for SN1 reaction, and such reaction is also called the solvolysis reaction. Solvolysis reaction is a nucleophilic substitution in which the nucleophile is a molecule of solvent as well. The term solvolysis comes from: solvent+lysis, that means cleavage by the solvent. A SN1 reaction is usually a solvolysis reaction.
Examples
Show the structures of the products for the following solvolysis reaction.
Solution: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/07%3A_Nucleophilic_Substitution_Reactions/7.04%3A_SN1_Reaction_Mechanism_Energy_Diagram_and_Stereochemistry.txt |
7.5.1 Comparison Between SN1 and SN2 Reactions
Till now, we have finished the basic concepts about SN1 and SN2 reactions. You probably already noticed that the two type of reactions have some similarities, also quite different though. It will be very helpful to put them together for comparison. To help you get in-depth understanding of the two types of mechanism, it is highly recommended that you have a summary in your own way. The following comparison is provided here for your reference.
SN1
SN2
Rate law
Rate = k[electrophile]
Rate = k[nucleophile]×[electrophile]
Mechanism
multiple steps with carbocation intermediate
one step, concerted
Reaction Diagram
Stereochemistry
racemization on reaction center
inversion on reaction center
Electrophilic Substrate tertiary 3° > secondary 2° > primary 1° and methyl
primary 1° and methyl > secondary 2° > tertiary 3°
Nucleophile
weak nucleophile, solvolysis
strong nucleophile
7.5.2 Solvent Effect on Sn1 and SN2 Reactions
Other than the factors we have talked about so far, solvent is another key factor that affect nucleophilic substitution reactions. Proper solvent is required to facilitate a certain mechanism. For some cases, picking up the appropriate solvent is the effective way to control which pathway the reaction proceed.
To understand the solvent effect, we first of all need to have more detailed discussions about solvents, then learn how to choose good solvent for a specific reaction.
Solvents can be divided into three major categories based on the structures and polarities, that is: non-polar, polar protic and polar aprotic solvents.
Non-polar solvents are non-polar compounds. (hexane, benzene, toluene, etc.)
Polar protic solvents are the compounds containing OH or NH group that is able to form hydrogen bonds. Polar protic solvents are highly polar because of the OH or NH group.
Polar aprotic solvents is a group solvents with medium range of polarity. They are polar because of polar bonds like C=O or S=O, but the polarity is not as high as OH or NH group. Typical examples of polar aprotic solvents include acetone, DMSO, DMF, THF, CH2Cl2.
The general guideline for solvents regarding nucleophilic substitution reaction is:
• SN1 reactions are favored by polar protic solvents (H2O, ROH etc), and usually are solvolysis reactions.
• SN2 reactions are favored by polar aprotic solvents (acetone, DMSO, DMF etc).
Polar Protic Solvents Favor SN1 Reactions
In SN1 reaction, the leaving group leaves and carbocation formed in the first step, that is also the rate-determining step. The polar solvent, such as water, MeOH, is able to form hydrogen bonding with the leaving group in the transition state of the first step, therefore lowering the energy of the transition state that leads to the carbocation, and speed up the rate-determining step. As a result, polar protic solvents facilitate SN1 reactions. It is very common that the polar protic solvents serve as nucleophiles as well for SN1 reactions, so usually SN1 reactions are solvolysis reactions as we learned earlier.
Polar Aprotic Solvents Favor SN2 Reactions
Strong nucleophiles are required in SN2 reactions, and strong nucleophile are usually negatively charged species, such as OH, CH3O, CN etc. These anions must stay with cations in salt format like NaOH, CH3ONa etc. Since salts are insoluble in non-polar solvent, therefore non-polar solvents are not appropriate choices, and we need polar solvents that can dissolve the salts.
The issue for polar protic solvent is that the nucleophile anions will be surrounded by a layer of solvent molecules with hydrogen bonds, and this is called the solvation effect. The solvation effect stabilize (or encumber) the nucleophiles and hinder their reactivities in SN2 reaction. Therefore, polar protic solvents are not suitable for SN2 reactions.
As a result the polar aprotic solvents, such as acetone, DMSO etc are the best choice of SN2 reactions. They are polar enough to dissolve the salt format nucleophiles, and also not interact as strongly with anions to hinder their reactivities. The nucleophile anions still move around freely in polar aprotic solvent to act as nucleophile.
The reaction rate for a SN2 reaction in different solvents are provided in the table below, and the polar aprotic solvent DMF proved to be the best choice that speed up the reaction significantly.
reaction: CH3I + Cl→ CH3Cl + I
solvent
relative rate
CH3OH
1
12.5
1,200,000
7.5.3 The Choice of Reaction Pathway: SN1 or SN2?
With all the knowledge about SN1, SN2 reactions and reaction conditions, we should be able to determine that whether a given reaction go with SN1 or SN2 pathway, or design a proper reaction that will produce the desired product(s). The reaction pathway predominantlydepends on the nature of the substrates (primary, secondary or tertiary), and the choice of proper reaction condition serve as a way to facilitate the process.
• Primary and methyl substrates undergo SN2 reaction predominantly.
• Tertiary substrates go with SN1 process.
• The reaction of secondary substrates mainly rely on the conditions applied. The condition include nucleophile, solvent, etc. See examples for more detailed discussions.
Exercises 7.4
Show the product(s) of the following reactions:
1. (S)-2-iodobutane + CH3ONa+ (DMSO →_
2. (S)-2-iodobutane (CH3OH →_
Answers to Practice Questions Chapter 7
Some practical tips for working on SN1, SN2 reactions:
1. As we understand that strong nucleophiles are required for SN2 reaction, and most of the strong nucleophiles are those with negative charges, for example OH, OR. These nucleophiles can be either shown as anions OH, CH3O, C2H5O, or in salt format like NaOH, KOH, CH3ONa, C2H5ONa in the reaction conditions. You should understand that it is the same thing. The anion format are easy to identify and also highlight the nature of these species, however since anions must stay together with counter cations as salt, the salt format show the actual chemical formula of the compound used in the reaction.
1. Since polar aprotic solvent favors SN2 reactions, so any of above anions or salt can be used together with DMSO, DMF etc, such as OH/DMSO, CH3ONa/DMF etc .
However, sometimes you may see the combination like CH3ONa/CH3OH, that is the combination of CH3O together with its conjugate acid CH3OH. It may seems contradictory, why a strong nucleophile for SN2 combine with solvent for SN1? The reality is that CH3ONa here still act as strong nucleophile and can be used for SN2 reaction and CH3OH is the solvent for CH3ONa. The reason why CH3OH is used together as solvent is that the CH3ONa can be prepared by treating an alcohol with Na. For example:
Other alcohol can also react with Na metal (or potassium metal, K) to generate the corresponding RONa.
The reaction between alcohol and NaH can be used as well.
Since alcohol are in excess in the above reactions, it is also a good solvent for the resulting alkoxide, and RO/ROH combination is used commonly together. The RO in this combination can be used as strong nucleophile for SN2 reaction, or base in elimination reaction (Chapter 8). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/07%3A_Nucleophilic_Substitution_Reactions/7.05%3A_SN1_vs_SN2.txt |
Our discussions so far focus on the fundamental concepts about SN1 and SN2 mechanism, and the reactions we learned about proceed in the regular way. There are some other conditions can be “added” to the basic nucleophilic substitution reactions, to make the reaction look different, or more challenge. However, understanding the basic concepts well is very helpful for us to deal with various situations. The reaction may looks different, but essentially it is still the same.
7.6.1 SN1 Reaction with Carbocation Rearrangement
Let’s take a look at a SN1 reaction.
With the secondary substrate and neutral nucleophile (CH3COOH), this is a SN1 reaction, and solvolysis that CH3COOH acts as both solvent and nucleophile. It is supposed to give the acetate as product, with the acetate replace the Br. However, as shown in the reaction equation that the acetate was not introduced on the carbon with leaving group Br, but was connected on the next carbon instead. What is the reason for the unexpected structure of the product?
For reactions involve carbocation intermediate, it is a common phenomena that the carbocation might rearrange, if such rearrangement leads to a more stable carbocation, and this is called carbocation rearrangement. Because of the carbocation rearrangement, the product of the above reaction is different than expected. This can be explained with the step-by-step mechanism below.
When Br leaves, the initial carbocation formed is a secondary one. The CH3 group on the next carbon then shift with its bonding electrons to the positively charged carbon, and creating a new more stable tertiary carbocation. The tertiary carbocation then reacts with nucleophile CH3COOH to give the final acetate product. The CH3 group shift with the electron pair, and such move is called 1,2-methanideshift. “1,2-“ here refer to the movement occur between two adjacent carbons, not necessarily means C1 and C2.
Other than CH3 group, the H atom in other reactions could shift as well with the electron pair, if such shift can lead to a more stable carbocation. The shift of hydrogen is called 1,2-hydride shift. A couple of notes about the carbocation rearrangement:
• Any reaction that involves carbocation intermediate might have rearrangement.
• Not all carbocations rearrange. Carbocations only rearrange if they become more stable as a result of the rearrangement.
• The shift is usually 1,2-shift, that means it occur between two adjacent carbons.
7.6.2 Intramolecular Nucleophilic Substitution Reaction
For the reactions we learned before, the substrate with leaving group and the nucleophile are always two separate compounds. It is actually possible for one compound containing both leaving group and nucleophile, and the reaction occurs within the same molecule. Such reaction is called the intramolecular (intra, Latin for “within”) reaction. Cyclic product is obtained from intramolecular reaction.
Let’s talk about the reaction mechanism that rationalize the structure and stereochemistry of the product for following reaction.
In the above reaction, the reactant has two functional groups, bromide (Br) and alcohol (OH). A compound with two functional groups is called bifunctional molecule. In this reactant, Br is connected on a tertiary carbon that is a good substrate for SN1 reaction, and OH is a good nucleophile for SN1 as well, so the substitution reaction could occur within the same molecule via SN1 mechanism. So the reaction occurs between one end of the molecule, Br, that acts as the leaving group, and the other part of the molecule, OH, which acts as the nucleophile. As a result, a six-membered cyclic ether is formed as the product.
Since the reaction occurs with SN1 mechanism, the carbocation intermediate is in trigonal planar shape, and the nucleophile can attack from either side of the carbocation to give both enantiomers. Therefore, the product is the racemic mixture that is optical inactive. This is consistent with the stereochemistry feature of SN1 reaction we learned before.
Usually if the intramolecular reaction could produce five- or six-membered ring as the product, the reaction will be highly favored because of the special stability of five- or six-membered ring.
7.6.3 Converting Poor Leaving Group to Good Leaving Group
In early discussions about leaving groups (section 7.3), we have mentioned the importance of a good leaving group for both SN1 and SN2 reactions, that the substitution reaction will not occur is a poor leaving group present. For some situations however, the poor leaving group could be converted to a good leaving group to make the reaction feasible. We will see a couple of strategies for such purpose.
By Acid Catalyst H+
Example: Propose the mechanism to rationalize the reaction.
The last three steps in the above mechanism are the standardsteps of SN1 mechanism. However, the reaction won’t proceed without the first step. In the first step, which is an acid-base reaction, a proton is rapidly transferred to the OH group, and get the alcohol protonated. By protonation, the OH group is converted to H2O, that is a much weaker base therefore a good leaving group. In step 2, water molecule departs with the electron pair and leave behind a carbocation intermediate. The following steps are just SN1, that explains why the product is the racemic mixture. The acid H+ was regenerated in step 4 and can be reused for further reactions, therefore only catalytical amount of H+ is necessary to start the process.
By Sulfonyl Chloride
Another commonly applied method for converting OH group to a better leaving group is by introducing a sulfonate ester. When alcohol reacts with sulfonyl chloride, with the presence of weak base, the sulfonate ester is formed.
As the example shown above, when p-toluenesulfonyl chloride (tosyl chloride, TsCl) is used, the resulting ester is p-toluenesulfonate (tosylate, OTs). Does tosyl group look familiar to you? Yes, we learned about with this species in section 3.2. As the conjugate base of strong acid p-toluenesulfonic acid (TsOH), OTs is the very weak base and therefore an excellent leaving group. Pyridine here acts as the weak base to neutralize the side product HCl and facilitate the reaction to completion. The detailed mechanism for this reaction is not required in this course.
Other than introducing OTs, other commonly applied sulfonyl chlorides include MsCl and TfCl, and the sulfonate ester OMs (mesylate) and OTf (triflate) are formed respectively.
Once the primary alcohol has been converted to OTs (or OMs, OTf), it is then the good substrate for SN2 reaction. With the appropriate nucleophile added in a separate step, for example CH3O, the SN2 reaction takes place readily to give ether as the final product, as shown below.
The overall synthesis of butyl methyl ether from 1-butanol involves two separate steps: the conversion of OH to OTs, and then the replacement of OTs by CH3O through SN2 reaction. The two steps have to be carried out one after the other, however the whole synthesis scheme can also be shown as below:
Note:
• Figure 7.6j represents the common and conventional way to show the multiple-step synthesis in organic chemistry. The reaction conditions (reagent, catalyst, solvent, temperature etc.) for each step are shown on top and bottom of the equation arrow. Only the structures of starting material and final product(s) are shown, and the structures of the intermediate products for each step are not included.
• The individual steps need to be labelled as 1), 2) etc. for the proper order, they can not be mixed together.
7.07: Answers to Practice Questions Chapter 7
7.1 Show the product of the following SN2 reaction (CN- is the nucleophile):
7.2 Show reaction mechanism of the reactions.
7.3 Show the detailed mechanism for above reaction of (S)-3-bromo-3-methylhexane and water.
7.4 Show the product(s) of the following reactions: | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/07%3A_Nucleophilic_Substitution_Reactions/7.06%3A_Extra_Topics_on_Nucleophilic_Substitution_Reaction.txt |
Thumbnail: tert-Butyl cation, demonstrating planar geometry and sp2 hybridization. (CC BY-SA 3.0; via Wikipedia)
08: Elimination Reactions
8.1.1 E2 Mechanism
E2 mechanism is the bimolecular elimination mechanism, that the reaction rate depends on the concentration of both substrate and base. We will take the elimination reaction of 2-bromo-2-methylpropane as an example for discussion.
It was mentioned earlier that HX is the side product of dehydrohalogenation, why there is no HX (HBr for this reaction) in the reaction equation? It could be understood as that with the presence of excess base (OH) in the reaction mixture, HBr reacts with OH to give H2O and Br. The following discussion of the mechanism will help you to understand this better.
E2 mechanism is also a single-step concerted reaction, similar to SN2, with multiple electron pair transfers happen at the same time.
Base, OH, uses its electron pair to attack a β-hydrogen on β-carbon, and starts to form a bond; at the same time, the β C-H sigma bond begins to move in to become the π bond of a double bond, and meanwhile Br begins to depart by taking the bonding electrons with it. A transition state is formed in the reaction process with partially breaking and partially forming bonds. At the completion of the reaction, the C=C double bond and H2O molecule are fully formed, with Br leaves completely.
Since both the substrate (halide) and the base are involved in the single-step mechanism, E2 is the second order reaction.
8.1.2 Regioselectivity of E2 reaction: Zaitsev’s Rule vsHofmann Rule
For the reaction we talked in above section, there are three β-carbons in the substrate 2-bromo-2-methylpropane, however they are all identical, so the reaction gives only one single elimination product 2-methylpropene.
For other alkyl halides, if there are different β-carbons in the substrate, then the elimination reaction may yield more than one products. For example, the dehydrohalogenation of 2-bromo-2-methylbutane can produce two products, 2-methyl-2-butene and 2-methyl-1-butene, by following two different pathways.
Between the two possible products, 2-methyl-2-butene is a trisubstituted alkenes, whereas 2-methyl-1-butene is monosubstituted. For alkenes, the more alkyl groups bonded on the double bond carbons, the more stable the alkene is. Generally, the relative stability of alkenes with different amount of substituents is:
tetrasubstituted > trisubstituted > disubstituted > monosubstituted > ethene
Therefore, 2-methyl-2-butene is more stable than 2-methyl-1-butene. When a small size base is used for the elimination reaction, such as OH, CH3O, EtO, it turned out that the relative stability of the product is the key factor to determine the major product. As a result, 2-methyl-2-butene is the major product for above reaction.
As a general trend, when small base is applied, the elimination products can be predicted by Zaitsev’s rule, that said the more substituted alkene is obtained preferably. So the Zaitsev’s rule essentially can be explained by the higher stability of the more substituted alkenes.
However, if a bulky base is applied in the elimination, such as t-BuOK, the reaction favors the formation of less substituted alkenes.
This is mainly because of steric hinderance. With t-BuO attacking the β-hydrogen, it is difficult for this big bulky base to approach the hydrogens from the β-carbon that is bonded with more substituents (as shown in pathway (a) below), while the hydrogen of the methyl group is much easily to be accessed (in pathway (b) instead. When the elimination yields the less substituted alkene, it is said that it follows the Hofmann rule.
8.1.3 Stereochemistry of E2 Reaction
The E2 mechanism has special stereochemistry requirement to ensure it does proceed. First, the bond connected with the leaving group and the bond connected with the H must be in the same plane, to allow the proper orbital overlapping of the two carbons in the formation of π bond of the alkene product. Second, the leaving group and H must be in anti-position to each other. This is because the anti-position allows the transition state of the reaction is in the more stable staggered conformation, that helps to lower down the energy level of the transition state and speed up the reaction. Overall, E2 reaction proceeds with the leaving group and H are in anti coplanar conformation.
Because of the anti-coplanar conformation requirement for E2 reaction, one stereoisomer will be produced preferably over the other, and this is called stereoselectivity. For the following example, the elimination of (2S,3S)-2-bromo-3-phenylbutane produces the E isomer specifically, not the Z isomer at all. This is because when H is in anti-position to the leaving group Br, the whole compound is in staggered conformation, and the other groups retain their relative position in elimination that leads to the E isomer.
Exercises 8.1
Show elimination product of the following reactions
1.
2.
Answers to Practice Questions Chapter 8
8.1.4 Bases in E2 Reactions (Brief Summary)
The most commonly applied bases in E2 reaction are hydroxide OH, and alkoxide RO. Specifically, the combination of base with corresponding alcohol are used broadly, such as: CH3ONa/CH3OH, C2H5ONa/C2H5OH.
Examples of small bases: OH, CH3O, C2H5O, NH2
Examples of big bulky bases: t-BuO, LDA (lithium diisopropylamide) | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/08%3A_Elimination_Reactions/8.01%3A_E2_Reaction.txt |
E1 Mechanism
Similar to substitutions, some elimination reactions show first-order kinetics. These reactions go through E1 mechanism, that is the multiple-step mechanism includes the carbocation intermediate.
When t-butyl bromide reacts with ethanol, small amount of elimination products obtained via E1 mechanism.
The overall elimination involves two steps:
Step 1: The bromide dissociates and forms a tertiary (3°) carbocation. This is a slow bond-breaking step, it is also the rate-determining step for the whole reaction. Since only the bromide substrate involved in the rate-determining step, the reaction rate law is first order. That is the reaction rate only depends on the concentration of (CH3)3Br, and has nothing to do with the concentration of base, ethanol.
Step 2: The hydrogen on β-carbon (β-carbon is the one beside the positively charged carbon) is acidic because of the adjacent positive charge. The base, EtOH, reacts with the β-H by removing it, and the C-H bond electron pair moves in to form the C-C π bond.
The base ethanol in this reaction is a neutral molecule, and therefore a very weak base. Since strong base favors E2, so weak base is a good choice for E1, by discouraging from E2. Ethanol acts as the solvent as well, so E1 reaction is also the solvolysis reaction.
For E1 reaction, if more than one alkenes could be possibly formed as product, the major product will also be the more substituted alkene, like E2, because of the stability of those alkenes.
Since E1 reaction involves a carbocation intermediate, the carbocation rearrangement might occur if such rearrangement leads to a more stable carbocation. For the following example, the initially formed secondary carbocation undergoes a 1,2-methanide shift to give the more stable tertiary benzylic carbocation, that lead to the final elimination product.
Examples: Show elimination product of the following reaction.
8.03: E1 E2 Summary
The comparison between E1 and E2, in terms of rate law, mechanism, reaction condition etc., are summarized in the following table.
E1
E2
Rate law
Rate = [substrate]
Rate = [substrate]×[base]
Mechanism
multiple steps with carbocation intermediate
one step, concerted
Product More substituted, more stable alkenes
small base: more substituted alkenes (Zaitsev’s rule)
bulky base: less substituted alkenes (Hoffmann rule)
Substrate tertiary 3° > secondary 2° > primary 1° (no E1)
tertiary 3° > secondary 2° > primary 1°
Base weak base, (H2O, ROH) strong base (OH, RO, etc.)
The competition between E1 and E2, or whether a substrate goes through E1 or E2 mainly depends on the nature of the substrate, that is:
• Primary 1º substrates go with E2 only, because primary carbocations are too unstable to be formed.
• Secondary 2º and tertiary 3º substrates can go with either E1 or E2 reaction, and appropriate reaction conditions are necessary to facilitate a specific mechanism. E2 reaction is favored by a high concentration of strong base (OH, RO, or NH2) and a polar aprotic solvent. E1 reaction is favored by a weak base, and polar protic compound, H2O, ROH, can be both base and solvent (solvolysis).
For study purpose, the comparison between E1 and E2 mechanism help us to understand the two process in depth. In practice, however, the competition between E1 and E2 will not be an issue because they require rather different reaction conditions. More important actually , it is the competition between elimination and substitution. We will have detailed discussions next for the comparison and competition between all the four types of reactionsSN1, SN2, E1 and E2. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/08%3A_Elimination_Reactions/8.02%3A_E1_Reaction.txt |
For a certain substrate, it may have chance to go through any of the four reaction pathways. So it seems rather challenging to predict the outcome of a certain reaction. We will talk about the strategies that can be applied in solving such problem, and explain the reasonings behind.
It is very important to understand that the structural nature of a substrate (primary, secondary or tertiary) is the most critical factor to determine which reaction pathway it goes through. For example, primary substrates never go with SN1 or E1 because the primary carbocations are too unstable. If the substrate could go with a couple of different reaction pathways, then the reaction conditions, including the basicity/nucleophilicity of the reagent, temperature, solvent etc., play the important role to determine which pathway is the major one. Our discussions therefore will start from the different type for substrates, then explore the condition effects on that substrate.
Methyl
This is the easiest case. Methyl substrate only go with SN2 reaction, if any reaction occurs. Elimination is not possible for methyl substrates, and no SN1 reaction either because CH3+ is too unreactive to be formed, so the only possible way is SN2.
Primary (1 °)
Primary (1°) substrates cannot go with any unimolecular reaction, that is no SN1/E1, because primary carbocations are too unstable to be formed. Since primary substrates are very good candidates for SN2 reaction, so SN2 is the predominant pathway when good nucleophile is used. The only exception is that when big bulky base/nucleophile is used, E2 becomes the major reaction.
Examples of reactions for primary substrates:
Secondary (2 °)
It is most complicated or challenging to predict the reaction of a secondary substrate (2°), because all the pathways are possible. The reaction conditions then become very key factor. The total four types of reactions can be separated into 3 pathway, that is:
• E2: favored by a strong base
• SN2: favored by a good nucleophile (relatively weaker base)
• SN1/E1: It is hard to separate SN1 and E1 completely apart, because they both go through carbocation intermediates, and are favored by poor nucleophile/weak base, for example, H2O or ROH (solvolysis). Under such neutral condition, SN1 and E1 usually occur together for secondary substrates, and increasing the reaction temperature favors E1 over SN1.
It is relatively easy to separate SN2 and E2 pathways from SN1/E1, since both SN2 and E2 require strong nucleophile or strong base that are usually negatively charged species, while SN1/E1 require neutral conditions.
In order to distinguish SN2 from E2, we need to be able to determine whether a negatively charged anion is a strong nucleophile (for SN2) or a strong base (for E2)? All nucleophiles are potential bases, and all bases are potential nucleophiles, because the reactive part of both nucleophile and base is lone pair electrons. Whether an anion is a better nucleophile or a better base depends on its basicity, size and polarizability. Generally speaking, the relative stronger bases have the stronger tendency to act as base; and relative weaker base, with small size and good polarizability, have the better tendency to act as nucleophile, see the list given below.
Strong bases: OH, RO(R: small size alkyl group), NH2
Good nucleophiles (relatively weaker bases): Cl, Br, I, RS, N3, CN, RCO2, RNH2
Please note that bulky bases, such as t-BuOand LDA, always favor E2 and generate elimination products that follow Hofmann rule, because they are too big to do back-side attack in SN2.
Examples of reactions for secondary substrates:
Tertiary (3 °)
Tertiary (3°) substrates do not go with SN2 reactions because of steric hinderance. So E2 reaction is the choice when strong base applied, or SN1/E1 pathway with neutral condition (poor nucleophile/weak base). Theoretically speaking, E2 and E1 supposed to give the same elimination product. However, in order to synthesize an alkene from a tertiary substrate, it is a better choice to use a strong base that encourage E2 process rather go with E1. This is because that E1 always combine together with SN1, and it is almost impossible to avoid the substitution product.
The above discussions can be briefly summarized in the table below, followed by several examples. To predict the reaction outcome, or to design synthesis route for a certain case, it is highly recommend that you do the analysis by following the logics mentioned above, instead of just refer to the table. Also, practice makes perfect!
Substrate
Preferred Reaction Pathways
Methyl
SN2 reaction
Primary
Predominantly SN2 reaction;
Exception: E2 reaction for bulky base
Secondary
SN2 reaction with good nucleophile (e.g., RS, RCO2, etc)
E2 reaction with strong base (e.g., OH, OR)
SN1/E1 with neutral condition (e.g., H2O, ROH)
Tertiary E2 reaction with strong base (e.g., OH, OR)
SN1/E1 with neutral condition (e.g., H2O, ROH)
Examples: Show major organic product(s) for following reactions.
8.05: Answers to Practice Questions Chapter 8
8.1 Show elimination product of the following reactions.
2.
The anti coplanar conformation of H and leaving group OTs is shown more clearly in the chair conformation of the cyclohexane. Please note that the other β-H can not be anti to the leaving group OTs. Also, in order to fit to the anti coplanar requirement, both H and OTs have to be in axial positions, so this conformation is the one that undergoes the elimination although it is not the most stable one. Since the most stable conformation does not fit the E2 stereochemistry requirement, so the elimination has to go through the less stable conformation. Heat is preferred to facilitate the reaction. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/08%3A_Elimination_Reactions/8.04%3A_Comparison_and_Competition_Between_SN1_SN2_E1_and_E2.txt |
For the reactions we learned so far, bond breaking occurs in the way that one part of the bond takes both electrons (the electron pair) of the bond away. For example of SN1 reaction, the leaving group Br leaves with the electron pair to form Br and carbocation intermediate.
This process is called heterolytic bond cleavage, the σ bond breaks heterolytically. As we have always been doing, an arrow with the double-barbs is used to show heterolytic cleavage, that is the transfer of electron pair specifically:
There is another type of bond breaking process, in which each part of the σ bond takes one electron away, as shown below:
This is called homolytic cleavage , or homolysis . The electron pair separate evenly to each part, and as a result both products contain a single electron. The species that contain one or more single electron is called radical (or free radical). Radicals are produced from homolytic cleavage. The arrow with sing-barb (like the shape of a fishhook) is used to show homolytic cleavage, that is single electron transfer specifically:
Homolysis occurs mainly for non-polar bonds, heat or light (delta is the symbol for heat; hv is used to show light) is needed to provide enough energy for initiating the process. For example:
Radical is another highly reactive reaction intermediate, because of the lack of octet. The substitution reaction we will learn in this chapter involves the radical intermediate.
9.02: Halogenation Reaction of Alkanes
When alkanes react with halogen (Cl2 or Br2), with heat or light, hydrogen atom of the alkane is replaced by halogen atom and alkyl halide is produced as product. This can be generally shown as:
A specific example is:
Such type of reaction can be called as substitution because hydrogen is substituted by halogen; can also be called halogenation because halogen is introduced into the product. For this book, both terms are used in this chapter, interchangeably.
The net reaction for halogenation seems straightforward, the mechanism is more complicated though, it go through multiple steps that include initiation, propagation and termination.
We will take the example of mono-chlorination of methane, for the discussion of reaction mechanism.
CH4 + Cl2 → CH3Cl + HCl
Mechanism for mono-chlorination of methane:
Initiation: Production of radical
With the energy provided from heat or light, chlorine molecule dissociates homolytically, each chlorine atom takes one of the bonding electrons, and two highly reactive chlorine radicals, Cl•, are produced.
Propagation: Formation of product and regeneration of radical
The propagation step involve two sub-steps. In the 1st step, the Cl• takes a hydrogen atom from the methane molecule (this is also called as hydrogen abstraction by Cl•), and C-H single bond breaks homolytically. A new σ bond is formed by Cl and H each donate one electron and HCl is produced as the side product. The CH3 radical, CH3•, the critical intermediate for the formation of product in next step, is formed as well.
In the 2nd step, the CH3• abstracts a chlorine atom to give final CH3Cl product, together with another Cl•. The regenerated Cl• can attack another methane molecule and cause the repetition of step 1, then step 2 is repeated, and so forth. Therefore the regeneration of the Cl• is particularly significant, it makes the propagation step self-repeat hundreds or thousands of time. The propagation step is therefore called the self-sustaining step, only small amount of Cl• is required at the beginning to promote the process.
Initiation and propagation are productive steps for the formation of product. This type of sequential, step-wise mechanism in which the earlier step generate the intermediate that cause the next step of the reaction to occur, is call the chain reaction.
The chain reaction will not continue forever though, because of the termination steps.
Termination: Consumption of radicals
When two radicals in the reaction mixture meet with each other, they combine to form a stable molecule. The combination of radicals lead to the decreasing of the number of radicals available to propagating the reaction, and the reaction slows and stops eventually, so the combination process is called termination step. A few examples of termination are given above, other combinations are possible as well.
The propagation steps are the core steps in halogenation. The energy level diagram helps to provide further understanding of the propagation process.
The 1st step in propagation is endothermic, while the energy absorbed can be offset by the 2nd exothermic step. Therefore the overall propagation is exothermic process and the products are in lower energy level the than reactants.
The reaction heat (enthalpy) for each of the propagation step can also be calculated by referring to the homolytic bond dissociation energies (Table 9.1). For such calculation, energy absorbed for bond-breaking step, so the bond energy was given “+” sign, and energy released for bond-forming step, and the “-” sign applied.
Bond
kJ/mol
Bond
kJ/mol
Bond
kJ/mol
A — B → A • + B •
F — F 159 H —Br 366 CH3 — I 240
Cl — Cl 243 H — I 298 CH3CH2 —H 421
Br — Br 193 CH3 — H 440 CH3CH2 —F 444
I — I 151 CH3 — F 461 CH3CH2 —Cl 353
H — F 570 CH3 — Cl 352 CH3CH2 — Br 295
H — Cl 432 CH3 — Br 293 CH3CH2 — I 233
Table 9.1 Homolytic Bond Dissociation Energies for Some Single Bonds
Examples
Calculation reaction energy for the propagation step of mono-chlorination of methane (referring to the corresponding bond energies in Table 9.1.)
Solution:
Step 1: H — CH3 + •Cl → CH3• + H — Cl
The H — CH3 bond broken, absorb energy, so +440 kJ
The H — Cl bond formed, release energy, so – 432 kJ
ΔH1 = +440 + (-432) = +8 kJ
Step 2: Cl — Cl + CH3• → CH3 — Cl + •Cl
The Cl — Cl bond broken, absorb energy, so +243 kJ
The CH3 — Cl formed, release energy, so -352kJ
ΔH2 = +243 + (-352) = – 109kJ
ΔHpropagation = ΔH1 + ΔH2 = +8 + ( – 109 ) = – 101kJ
The calculated data does match with the data from the energy diagram.
Reactivity Comparison of Halogenation
The energy changes for halogenation (substitution) with the other halogens can be calculated in the similar way, the results are summarized in Table 9.2.
Reaction
F2
Cl2
Br2
I2
Step 1:
H — CH3 + •X → CH3• + H — X
-130 +8 +74 -142
Step 2:
X — X + CH3• → CH3 — X + • X
-322 -109 -100 -89
Overall propagation:
H — CH3 + X — X → CH3 — X + HX
-452 -101 -26 +53
Table 9.2. Enthalpy of the Propagation Steps in Mono-halogenation of Methane (kJ/mol)
The data above indicate that the halogen radicals have different reactivity, fluorine is most reactive and iodine is least reactive. The iodine radical is very unreactive with overall “+” enthalpy, so iodine does not react with alkane at all. On the other side, the extreme high reactivity of fluorine is not a benefit either, the reaction for fluorine radical is so vigorous or even dangerous with lots heat released, and it is not practical to apply this reaction for any application because it is hard to control it. So Cl2 and Br2, with reactivity in the medium range, are used for halogen substitutions of alkanes. Apparently Cl2 is more reactive than Br2, and this leads to the different selectivity and application between the two halogens, more discussions in section 9.4. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/09%3A_Free_Radical_Substitution_Reaction_of_Alkanes/9.01%3A_Homolytic_and_Heterolytic_Cleavage.txt |
Alkyl radical is the key intermediate for halogenation reaction of alkanes, so the relative stability of radical determines the relative reactivity. Based on the energy diagram, the alkane that generate the more stable carbon radical exhibits the higher reactivity.
The alkyl radicals with different structures show different stabilities. Specifically, tertiary radical is most stable and the primary and methyl radicals are least stable, that follow the same trend as the stability of carbocations.
This trend can be explained by two reasonings:
• Hyperconjugation effect of alkyl (R) group: alkyl groups are electron-donating groups through hyperconjugation effect (refer to section 7.4), that is the electron density of C-C or C-H σ bond overlap with the half-filled p orbital of carbon radical. Similar to the carbocation, carbon radical is also the electron deficient species, so the electron-donating effect of alkyl groups help to stabilize it. With more alkyl groups involved, the radical is more stable.
• Homolytic bond dissociation energy comparison: Homolytic cleavage of C–H bond produces carbon radical. The C–H bond in different structure has different bond dissociation energy. Let’s compare two different types below, primary vs secondary:
Since both radicals come from the same compound, propane, so the higher the homolytic bond dissociation energy means the higher the energy level of the resulting carbon radical. The bond energy of the 1° C–H is 10 kJ/mol higher in energy than the bond energy of the 2° C–H, therefore the secondary radical is more stable than the primary one.
Other than the above reasons, there is another effect that affect the stability of radicals. For example, the following radical exhibits special stability, that is even more stable than other regular tertiary radical, although it is a primary radical. Why? This is because of another effect — resonance effect!
The radical here is not a regular primary radical, it is on the position that is beside the benzene ring. The position right next to the benzene ring is call the benzylic position, and this radical is a benzylic radical. Because of the presence of benzene ring, the benzylic radical has total five resonance contributors. According to resonance effect, the more resonance contributors available, the better the electron density dispersed, the more stable the species is.
The resonance effect also helps to stabilize the allylic radical as well. The carbon that is right next to the C=C double bond is the allylic position. The resonance structures of an allylic radical example are shown below. Both benzylic and allylic radicals are more stable than the tertiary alkyl radicals because of resonance effects.
9.04: Chlorination vs Bromination
9.4.1 Monochlorination
First we will focus on monochlorination product, by assuming that chlorination only occur once. Since chlorine is a rather reactive reagent, it shows relative low selectivity, that means Cl2 does not discriminate greatly among the different types of hydrogens atoms (primary, secondary or tertiary) in an alkane. As a result, for the reaction of alkane with different hydrogen atoms, a mixture of isomeric monochlorinated products are obtained.
The experimental results of the monochlorination of propane indicate that 45% primary chloride (1-chloropropane) and 55% secondary chloride (2-chloropropane) are produced. How to explain this result?
To predict the relative amount of different chlorination product, we need to consider two factors at the same time: reactivity and probability.
It has been discussed in section 9.3, that different radicals (primary, secondary or tertiary) have different stability and reactivity. The relative reaction rate of alkyl radicals for chlorination have been measured and has the approximate values of:
Probability simply depends how many hydrogen atoms are there for each type. With more hydrogen atoms available, the chance for that type of hydrogen to react is higher statistically.
So the overall amount of each isomeric product should be estimated by accounting for both reactivity and probability, that is:
the amount of a certain type of product = number of that type of hydrogens × relative reactivity
For the example of monochlorination of propane, the calculation is:
Amount of 1-chloropropane: 6 (number of 1°hydrogens) × 1.0 (relative reactivity) = 6.0
Amount of 2-chloropropane: 2 (number of 2°hydrogens) × 3.8 (relative reactivity) = 7.6
yield % of 1-chloropropane: 6.0/13.6 = 44 %
yield % of 1-chloropropane 7.6/13.6 = 56 %
The calculated values are consistent with the experiment results.
Exercises 9.1
Predict the percentage yield of each product for monochlorination of isobutane by calculation, and compare your calculated numbers to the experiment results. Are they consistent?
Answers to Practice Questions Chapter 9
For the alkane with only one type of hydrogen, the problem of isomeric mixture can be prevented of course since only one product produced. For the following chlorination of cylclopentane, only one monochloride is produced.
9.4.2 Multichlorination
Although we assume that chlorination occurs once in last section discussions, this is not the actual case unfortunately. A common issue with chlorination is that multiple substitution always happen. A simple example is the chlorination of methane, that a mixture of multiple chlorination product were obtained as we learned before.
The mechanism for the formation of multichlorination product is similar to that of monochloride. When chloromethane (or methylchloride) reacts with Cl2, another hydrogen is replaced by chlorine atom to give dichloromethane, dichloromethane reacts with Cl2 again to give trichloromethane, and trichloromethane reacts further to produce tetrachloromethane. All the reactions still go through similar propagation steps with radical mechanism.
Examples
Show the mechanism of propagation steps for the formation of dichloromethane from chloromethane.
Solution:
Practically, to minimize the problem ofmultichlorination products, the reaction conditions can be controlled in certain ways, for example:
• Use high concentration of alkane relative to Cl2, to decrease the possibility of multichlorination;
• Control reaction time: stop reaction after “short” time to favor monochlorination product.
These methods help to reduce the amount of multichlorination products, but the problem still cannot be completely avoided.
9.4.3 Bromination
Because of the two major problems for chlorination, lack of selectivity and multi-substitution, chlorination is not useful as a synthesis method to prepare a specific alkyl halide product. Instead, bromination with Br2 can be applied for that purpose. The relative lower reactivity of bromine makes it exhibits a much greater selectivity. Bromine is less reactive, means it reactive more slowly, therefore it has chance to differentiate between the different types of hydrogens, and selectively reacts with the most reactive one. The relative reaction rate of bromination for different radical is shown here, and you can see the big difference to that of chlorination:
For bromination, the reactivity difference between different types of position is so high that the reactivity factor become predominant for determining the product. Therefore bromination usually occurs selectively on the most reactive position (the position that forms the most stable carbon radical intermediate), and gives one major product exclusively, as the example here for bromination of isobutane.
As a result, bromination has the greatest utility synthesis of alkyl halide.
Exercises 9.2
Show the major bromination product of following reactions.
Answers to Practice Questions Chapter 9 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/09%3A_Free_Radical_Substitution_Reaction_of_Alkanes/9.03%3A_Stability_of_Alkyl_Radicals.txt |
For substitution reaction in which a stereocenter is generated, the stereochemistry can be explained by the structure feature of radical intermediate.
In the structure of carbon radical, carbon has three bonds and one single electron. Based on VSEPR, there are total four electron groups, radical should be in tetrahedral shape. However, experiment evidence indicate that the geometric shape of most alkyl radical is trigonal planar shape, with the carbon in sp2 hybridization, and there is one single unpaired electron in the unhybridized 2p orbital.
We will take the bromination reaction of (±)-3-methylhexane to explain the stereochemistry. The experiment results indicate that the racemic mixture of R and S 3-bromo-3-methylhexane were obtained with the bromination.
This can be explained by the stereochemistry of the propagation steps in the mechanism. The carbon radical generated in step 1 is in trigonal planar shape as mentioned earlier. When the radical reacts with bromine in step 2, the reaction can occur at either side of the plane. Because both sides are identical, the probability of the reaction by either side is the same, therefore equal amount of the R– and S– enantiomer are obtained as a racemic mixture.
The stereochemistry of the radical substitution is similar to that of SN1 reaction, because both carbon radical and carbocation are in trigonal planar shape.
Examples
Show the bromination product(s) with stereoisomers when applied.
Solutions:
The racemic mixture is obtained.
Exercises 9.3
Show all the mono-chlorination products of butane with any stereoisomers when applied.
Answers to Practice Questions Chapter 9
9.06: Synthesis of Target Molecules- Introduction of Retrosynthetic Analysis
We have learned three major types of reactions so far, nucleophilic substitution, elimination and halogenation of alkane (radical substitution), now we will see how to put the knowledge of these reactions together for application, that is to design synthesis route for a target (desired) compound from available starting materials.
Building larger, complex organic molecules from smaller, simple molecules is the goal of organic synthesis. Organic synthesis have great importance for many reasons, from testing the newly developed reaction mechanism or method, to replicate the molecules of living nature, and to produce new molecules that have potential applications in energy, material or medicinal fields.
It usually take multiple steps, from several to 20 or more, to synthesize a desired compounds, and therefore it would be challenging to visualize from the start all the steps necessary. The common strategy to design a synthesis is to work backward, that is instead of looking at the starting material and deciding how to do the first step, we look at the product and decide how to do the last step. This process is called retrosynthetic analysis, the technique applied frequently in organic synthesis. We will introduce the basic ideas of retrosynthetic analysis here, and for practice purpose the starting material is always defined for our examples.
Retrosynthetic analysis can usually be shown in the above way, with the open arrows indicate that the analysis is backward. We first identify the precursor 1 that could react in one step to make the target compound, then identify the next precursor that could react to give precursor 1, and repeat the process until we reach the starting material. Please note that the analysis is the way to show the “thinking or ideas” for solving the problem, so typically the reagents/conditions required for each step are not specified until the synthesis route is written in the forward direction. Also it is very possible you may come up with multiple routes, with different precursors, then the most efficient synthesis route can be determined by evaluating the possible benefits and disadvantages of each path.
Examples
Design the synthesis route of methoxybenzene starting from toluene.
Approach: The target compound is an ether. We have learned that SN2 reaction is a reasonable way to introduce different functional groups by applying different nucleophiles (section 7.3), that said the reaction between CH3O (nucleophile) and halide gives the desired ether, and the halide can be the “precursor 1”. The halide precursor can then be directly connected with the starting material, toluene, through the halogenation that we just learned in this chapter. This is an easy example that only involve two steps.
Solutions:
The analysis can then be transferred to the solution of the question by showing the reactions in forward direction and include the reagents/condition for each step.
Synthesis route design is a rather challenge topic that need lots practices. In order to do that well, you should be very familiar with all types of reactions in terms of how the functional groups transformed, and what reagents and conditions involved. Sometimes some reaction features, like stereochemistry will be useful as well.
Exercises 9.4 Design the synthesis route.
Answers to Practice Questions Chapter 9
9.07: Answers to Practice Questions Chapter 9
9.1 Predict the percentage yield of each product for monochlorination of isobutane by calculation, and compare your calculated numbers to the experiment results. Are they consistent?
Calculation:
Amount of 1°-chloride: 9 (number of 1°hydrogens) × 1.0 (relative reactivity) = 9.0
Amount of 3°-chloride: 1 (number of 3°hydrogens) × 3.8 (relative reactivity) = 3.8
yield% of 1°-chloride: 8.0/12.8 = 70.3 %
yield% of 3°-chloride 3.8/12.8 = 29.7 %
The calculated values are consistent to the experiment results, not exactly same though.
9.2 Show the major bromination product of following reactions.
9.3 Show all the mono-chlorination products of butane with any stereoisomers when applied.
9.4 Design the synthesis route. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/09%3A_Free_Radical_Substitution_Reaction_of_Alkanes/9.05%3A_Stereochemistry_for_Halogenation_of_Alkanes.txt |
Thumbnail: Ball-and-stick model of the ethylene (ethene) molecule, \(\ce{C2H4}\). (Public Domain; Ben Mills via Wikipedia)
10: Alkenes and Alkynes
10.1.1 Dehydrohalogenation of Alkyl Halide
The E2 elimination reaction of alkyl halide is one of the most useful method for synthesizing alkene.
Lots discussions have been given about the mechanism and stereochemistry of E2 reaction in Chapter 8. Here are a few practical hints about making use of E2 reaction to prepare alkene as the desired product:
• Choose a secondary or tertiary substrate if possible, since they prefer E2.
• If primary substrate is necessary, choose a bulky base such as t-BuO, to avoid the competition of substitution reaction. As mentioned early (section 8.4) that primary substrate undergoes SN2 reaction with small species like OH, that is also a good nucleophile.
• High concentration of a strong base at elevated temperature favor E2 reaction.
• Keep in mind that small base produces Zaitsev’s product (more substituted alkene), while bulky base produces Hofmann product (less substituted alkene).
10.1.2 Dehydration of Alcohol
Other than alkyl halides, alcohols can also be the substrates for elimination to produce alkenes. Most alcohols undergo elimination by losing the OH group and an H atom from an adjacent carbon. Since a water molecule is eliminated for the overall reaction, the reaction is also called dehydration. Two dehydration reactions are shown below for synthesizing alkene from alcohol. Dehydration of an alcohol requires a strong acid with heat. Concentrated sulfuric acid (H2SO4) or phosphoric acid (H3PO4) are the most commonly used acids in the lab.
Understanding the dehydration reaction mechanism would be helpful for us to apply the method effectively. Let’s take the dehydration of tert-butyl alcohol as an example.
The elimination mechanism involves the carbocation intermediate, so it is essentially an E1 mechanism. However, not a typical E1, since it start with the protonation step. We have learned in substitution reaction chapter (section 7.6) that OH group is a poor leaving group, so it never leaves. However, with the presence of strong acid (H3O+, H2SO4,etc), OH group is protonated by acidand therefore converted to the good leaving group H2O. The same concepts apply here in elimination as well. Step 1 in the mechanism is the acid-base reaction for the purpose to convert poor leaving group OH to good leaving group H2O. Step 2 and 3 are typical steps for an E1 mechanism. The overall dehydration reaction can be regarded as the E1 reaction of a protonated alcohol.
For E1 mechanism, the rate-determining step is the formation of carbocation, so the relativity stability of carbocation defines the relative reactivity of alcohol towards E1 dehydration. As you can predict that the trend is:
3° alcohol > 2° alcohol > 1° alcohol (not undergoes E1 dehydration)
Another observation in dehydration reaction is that rearrangement occurs. This make sense because the mechanism involves the formation of carbocation. We have learned the concept in section 7.6, that a carbocation will rearrange if the rearrangement produces a more stable carbocation. An example of dehydration with rearrangement is given below:
For the dehydration of 3,3-dimethyl-2-butanol, two alkenes are obtained with 2,3-dimethyl-2-butene as the major product. However, both products have the different carbon skeleton comparing to that of the reactant. This is due to the rearrangement of carbocation intermediate, that is shown explicitly in the mechanism below.
In step 2 of the mechanism, the initially formed secondary carbocation undergoes rearrangement, 1,2-methanide shift, to produce the more stable tertiary carbocation.
In step 3, there are two β-hydrogens available in the tertiary carbocation for removal. The more substituted alkene, which is more stable, is the major product.
Primary Alcohol Elimination
The primary alcohol can also undergo dehydration, however through an E2 mechanism because the primary carbocations are too unstable to be formed. The first step of the mechanism still involves the protonation of OH group though, to convert the poor leaving group to a good leaving group. The second step is the actual E2 of the protonated primary alcohol. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/10%3A_Alkenes_and_Alkynes/10.01%3A_Synthesis_of_Alkenes.txt |
Alkenes undergo a large variety of reactions. At first glance, these reactions appear to be quite different, however detailed studies indicated that the different mechanism all share some common features. The double bond is the reactivity center of alkene, this is mainly because of the relatively loosely held π electrons of the double bond. The π bond is formed by side-by-side overlapping, the relative weak overlapping mode, so π bond is weak and exhibits high reactivity. The π electrons also make the double bond carbons electron-rich, and have the tendency to be attracted to an electrophile. The high reactivity make alkenes an important type of organic compounds, and they can be used to the synthesis a wide variety of other compounds, such as halide, alcohol, ethers, alkanes.
The most common type of reaction for alkene is the addition reaction to C=C double bond. In addition reaction, a small molecule is added to multiple bond and one π bond is converted to two σ bonds (unsaturation degree decreases) as a result of addition. Addition reaction is the opposite process to elimination.
The addition reactions can generally be categorized depends on what small molecule added, our following discussions will also be based that.
10.2.1 Addition of Hydrogen Halide to Alkenes
The addition reaction of a hydrogen halide to an alkene produces an alkyl halide as product. For examples:
In above reactions, the alkenes are in symmetric structures, that means it does not matter which carbon boned with hydrogen and which carbon bonded the halogen, the same product will be obtained in either way.
For the alkene that does not have the symmetric structure, the double bond carbons have different substituents, then the question of which carbon get the hydrogen is very critical. For the example of following reaction, two possible products could be produced, 2-bromo-2-methylpropane and 1-bromo-2-methylpropane, which one is actually formed? Or are both formed?
It turns out that 2-bromo-2-methylpropane is the main product for the reaction. To explain and understand the outcome of the reaction, we need to look at the mechanism of the reaction as we always do.
The mechanism of the addition reaction involves two steps (shown below). In first step, the π electrons of the alkene act as nucleophile and are attracted to the partially positively charged hydrogen (electrophile) of HBr. As the π electrons of the alkenes moving toward the hydrogen, the H-Br bond breaks, with Br moves away with the bonding electrons, and a new σ bond formed between one double bond carbon and hydrogen. A carbocation and a bromide, Br, are formed this step.
In the second step, the bromide, Br, reacts with the positively charged carbocation to give the final product. This step is sort of similar to the second step of SN1 reaction, in which a nucleophile reacts with electrophile (carbocation).
When the new sbond formed between double bond carbon and hydrogen in first step, the hydrogen could possibly be bonded with either carbon, as shown in path (a) and (b), and the carbocations with different structure will be produced. It is obvious to tell that the tertiary carbocation formed in path (a) is much more stable than the primary carbocation in path (b), and will be produced preferably. The tertiary carbocation is then attacked by the Br in the second step, that produces the product 2-bromo-2-methylpropane. It is the stability difference between two carbocations in the first step that accounts the selective formation of 2-bromo-2-methylpropane of the overall reaction.
Because the first step of the above reaction is the addition of an electrophile (H+) to the alkene, the reaction is called an electrophilic addition reaction. Electrophilic addition reaction is a characteristic type of reaction of alkenes, several other addition reactions we will see later also belong to this category.
The two possible products of this reaction are constitutional isomers to each other. For the reaction in which two or more constitutional isomers could be obtained as products, but one of them predominates, the reaction is said to be a regioselective reaction. Regio comes from Latin word regionem that means direction. The regioselectivity trend of the electrophilic addition of HX to alkenes had been summarized as Markovnikov’s rule by Russian chemist Vladimir Markovnikov. One way to state Markovnikov’s rule is that in the addition of HX to an alkene, the hydrogen atom adds to the double bond carbon that has greater number of hydrogen atoms.
The underlying reasoning for Markovnikov’s rule is the stability of carbocation intermediate that involved in reaction mechanism. It seems easy for you to just memorize the rule or just memorize the fact that 2-bromo-2-methylpropane is the product for above reaction, without understanding why. However, you will notice soon that your memorization will be overwhelmed and mixed up with many more reactions coming up. The proper way to study organic reactions is to learn and understand the mechanism, unify the principles of reactions based on mechanism. The mastery of the contents will much easier and a lot more fun in this way, rather than trying to memorize tons of reactions.
Exercises 10.1
Show structure of the major product for following addition reactions.
Answers to Practice Questions Chapter 10
Exercises 10.2
For the addition of HBr to 3-methyl-1-butene, two products were observed. Show the reaction mechanism to explain the formation of both products.
10.2.2 Radical Addition of HBr to Alkenes
In last section we learned that the electrophilic addition of HX to alkene gives addition products that follow Markovnikov’s rule. Here we will learn that the hydrobromide, HBr, can also add to alkene in a way that gives anti-Markovnikov product.
The anti-Markovnikov product are obtained through different mechanism, that is the radical mechanism. To initiate radical mechanism, peroxide must be involved in order to generate the radical in the initiation step of the mechanism. The O-O bond of peroxide is weak (with bond energy of about 150 kJ/mol), and it undergoes the homolytic cleavage readily with heat to produce alkoxyl radicals. The peroxide therefore acts as radical initiator by generating radicals, and the addition is called radical addition. The detailed radical addition mechanism of the above addition of HBr to 2-methylpropene is given here.
Radical Addition Mechanism:
The initiation involves two steps for the radical addition mechanism. The alkoxyl radical generated in step 1 reacts with H-Br to generate bromine radical, Br·, that reacts with alkene to initiate the chain reaction in propagation steps. It shown clearly in the propagation steps that the order of the addition is reversed in radical addition comparing to that of electrophilic addition. Specifically, the bromine radical (Br) is added to the double bond first followed by the abstraction of hydrogen atom (H), therefore the anti-Markovnikov product is produced as a result.
One more note is that only HBr proceed with radical addition in the presence of peroxide, not HCl or HI. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/10%3A_Alkenes_and_Alkynes/10.02%3A_Reactions_of_Alkenes-_Addition_of_Hydrogen_Halide_to_Alkenes.txt |
Addition of Water to Alkenes (Hydration of Alkenes)
An alkene does not react with pure water, since water is not acidic enough to allow the hydrogen to act as an electrophile to start a reaction. However, with the presence of small amount of an acid, the reaction does occur with a water molecule added to the double bond of alkene, and the product is an alcohol. This is the acid-catalyzed addition reaction of water to alkene (also called hydration), and this reaction has great utility in large-scale industrial production of certain low-molecular-weight alcohols.
The acid most commonly applied to catalyze this reaction is dilute aqueous solution of sulfuric acid (H2SO4). Sulfuric acid dissociates completely in aqueous solution and the hydronium ion (H3O+) generated participates in the reaction. Strong organic acid, tosyl acid (TsOH), is used sometimes as well.
The mechanism for acid-catalyzed hydration of alkene is essentially the same as the mechanism for the addition of hydrogen halide, HX, to alkenes, and the reaction therefore follows Markovnikov’s rule as well in terms of regioselectivity. The hydration of 1-methylcyclohexene and the reaction mechanism are shown below.
Since water molecule can be regarded as H—OH, so the regioselectivity of alcohol product that follows Markovnikov’s rule means the hydrogen atom connects to the double bond carbon that has more hydrogen atoms, and OH group adds to the carbon that has less hydrogen atoms. This can be explained again by the formation of more stable carbocation in the first step of the mechanism. The acidic hydronium ion (H3O+) is regenerated in the last deprotonation step, so only a small amount of acid is required to initiate the reaction, the acid therefore is a catalyst.
Comparing the hydration reaction of alkene to the dehydration reaction of alcohol in section 10.1.2, you would recognize that they are reverse reactions, one is addition and the other is elimination. To produce alcohol from alkene via hydration, water should be in excess to ensure the reaction goes to completion. While to prepare alkene from alcohol through dehydration, high concentration of acid with elevated temperature favor the elimination process and the product can be removed by distillation as they formed to push the equilibrium to alkene side.
Addition of Alcohol to Alkenes
With the presence of acid, an alcohol can be added to the alkene in the same way that water does, and ether formed as product. For example:
Examples:
Show the mechanism for above addition reaction of methanol to 2-methyl-1-butene.
Refer to the hydration mechanism.
Solutions:
Mechanism: addition of methanol to 2-methyl-1-butene
Step 1: Electrophilic attack of H3O+ to the alkene, carbocation intermediate formed
Step 2: Methanol reacts with the carbocation
Step 3: Deprotonation to get neutral product
Note: Please keep in mind that for the reaction that involves carbocation intermediate, the rearrangement of carbocation is always an option. Therefore the addition of water/alcohol to alkenes may involve carbocation rearrangement if possible.
Exercises 10.3
Show major product(s) for the following reactions.
Answers to Practice Questions Chapter 10
10.04: Reactions of Alkenes- Addition of Bromine and Chlorine to Alkenes
Addition reaction also occur easily between halogens (Br2 and Cl2) and alkenes. In the presence of aprotic solvent, the product is a vicinal dihalide, as shown here for the addition of chlorine to propene.
The reaction between C=C double bond and bromine (Br2) can be used as a test for the presence of alkene in an unknown sample. The bromine reagent is in reddish color, and the product vicinal dibromide is colorless. When bromine is added to the sample, if the reddish color disappear, that means the sample does contain an alkene. The addition reaction occurs to get reddish bromine consumed and colorless product formed, so color fades off.
Mechanism for the Addition of Halogen to Alkenes
The products for addition of halogen to alkenes seems straightforward, with each halogen added to each double bond carbon. However, the addition proceeds with unique stereochemistry feature that need special attention. It turns out that the halogen atoms are added via anti addition to the double bond, as examples shown here:
The mechanism that accounts for the anti addition of halogen involves the electron pairs transferred in a way that is different to what we are familiar with, and the formation of the cyclic halonium ion intermediate. We will take the addition of bromine to (E)-2-butene as example to explain the mechanism.
When Br2 molecule approaching alkene in the first step, the electron density of the π bond in alkene repels electron density in the bromine, polarizing the bromine molecule and make the bromine atom that is closer to the double bond electrophilic. The alkene donate a pair of π electrons to the closer bromine, causing the displacement of the bromine atom that is further away. The lone pair on the closer bromine atom then acts as nucleophile to attack the other sp2 carbon. Thus, the same bromine atom is both electrophile and the nucleophile, and two single bonds are formed between the two sp2 carbons and the closer bromine that gives the cyclic bromonium ion intermediate.
In the second step, the nucleophilic bromide, Br (generated in step 1), attacks the carbon of the cyclic intermediate. Since the bottom side of the intermediate is blocked by the ring, the Br can only attack from the top side, that results in the anti position of the two Br in the product. The attack is similar to SN2 reaction and cause the ring to open and the formation of vicinal dibromide. For the above example, the two carbons in the bromonium ion intermediate are in same chemical environment, so they both have the same chance to be attacked by Br, as shown in blue and red arrows. The two attacks result in the same product, the meso compound (2R,3S)-2,3-dibromobutane, in this reaction.
Next, let’s exam the addition of bromine to (Z)-2-butene. As you may expect, the reaction goes through the same mechanism that involves the cyclic bromonium ion intermediate, however the products have different stereochemistry features.
In the addition of Br2 to (Z)-2-butene, the attack of Brto either carbon in bromonium ion by following blue or red arrow results in different enantiomer (step 2 in above mechanism). Since both carbons have the same chance to be attacked, so the product is the 50:50 racemic mixture of the two enantiomer.
Starting from the two different diastereomers, (E)-2-butene and (Z)-2-butene,the addition reaction produces different stereoisomers. The addition of (E)-2-butene gives one product, the meso compound (2R,3S)-2,3-dibromobutane, while the addition of (Z)-2-butene produces the racemic mixture of two enantiomers, (2S,3S)-2,3-dibromobutane and (2R,3R)-2,3-dibromobutane. Such reaction, the one where a particular stereoisomer of the starting material yields a specific stereoisomer of the product is called stereospecific reaction. The anti addition of a halogen to an alkene is an example of a stereospecific reaction.
Examples
Show the product of following addition.
Solutions:
The formation of the racemic mixture product can be explained by the mechanism:
Formation of Halohydrin
If water is used as a solvent in the reaction, rather than CH2Cl2, then water takes in part of the reaction and acts as nucleophile to attack the cyclic halonium intermediate in the second step. The major product of the addition will be a vicinal halohydrin as a result. A vicinal halohydrin is the compound that contains a halogen and an OH group on two adjacent carbons.
In second step of the mechanism, both H2O (solvent) and Br (produced in the first step) are nucleophiles and have chance to react with the cyclic bromonium ion. However since H2O is the solvent, its concentration is much higher than that of Br, so the major products come from the attack of H2O.
This reaction is still the stereospecific reaction in which the anti addition occurs, that the halogen and OH group are in anti position. For above example, the addition of bromine water to cyclohexene, the racemic mixture with both enantiomers are obtained.
If the alkene is not in symmetric structure, it is observed that the addition shows the regioselectivity as well, specifically the halogen adds on the carbon atom with greater number of hydrogen atoms, and OH group ends up on the double bond carbon with less amount of hydrogen atoms. How to explain this?
This is due to the difference between the two double bond carbons in the cyclic intermediate. When nucleophile water attacks, the C-Br bond start to breaking and the carbon atom has partial positive charges. The carbon atom with two substituents bears more positive charges and it resembles the more stable tertiary carbocation, and the other carbon atom with one substituent shows secondary carbocation character. As a result, the attack to the carbon with more tertiary carbocation character it is more preferably.
Exercises 10.4
Show major product(s) of the following reactions:
Answers to Practice Questions Chapter 10 | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/10%3A_Alkenes_and_Alkynes/10.03%3A_Reactions_of_Alkenes-_Addition_of_Water_%28or_Alcohol%29_to_Alkenes.txt |
When alkenes react with hydrogen gas in the presence of a variety of metal catalysts, a hydrogen molecule will be added to the double bond in the way that each carbon atom bonded with one hydrogen atom, such addition reaction is called hydrogenation.
Catalysts are must-have for hydrogenation, so the reaction can also be called catalytic hydrogenation. The commonly applied metal catalysts involve palladium and platinum. Palladium, which is used as a powder absorbed on charcoal to maximize the surface area, is the most common catalyst that is referred to as palladium on charcoal (Pd/carbon). Platinum, which is used usually as oxide PtO2, is also employed frequently and referred to as Adams catalyst. These metal catalysts are not soluble in the reaction mixture and therefore are described as heterogeneous catalysts. The heterogeneous catalyst can be easily filtrated out of the reaction mixture after reaction, and then be recycled and reused.
The hydrogenation reaction does not take place without catalyst because of the enormous activation energy. The catalysts lower down the activation energy by weakening the H-H bond, and make the reaction feasible at room temperature. The details of the mechanism of catalytical hydrogenation are not completely clear. What was understood was that hydrogen gas is adsorbed on the surface of the metal, and the alkene also complexes with the metal by overlapping its π orbitals with vacant orbitals of the metal. The reaction occur on the surface of the metal catalyst, with both hydrogen atoms added from the same side of the alkene, to give alkane as the product that diffuses away from the metal surface. This mode of addition that the atoms added from the same side of the alkene is called the syn addition.
Example:
10.06: Two Other Hydration Reactions of Alkenes
As we learned in section 10.2.2, the acid-catalyzed hydration (addition of water) to alkene produces alcohol that follow Markovnikov’s regioselectivity. Here we will investigate two other methods for hydration of alkene, via different reaction conditions and mechanism, and produce either Markovnikov or anti- Markovnikov alcohol product respectively.
10.6.1 OxymercurationDemercuration of Alkenes The oxymercuration-demercuration of alkenes provides an alternative way to synthesize Markovnikov’s alcohol from alkene. It is a fast reaction with lots application in laboratories, and the yield is usually greater than 90%.Comparing to acid-catalyzed hydration, the benefits of oxymercuration-demercuration are: no strong acids required and no carbocation rearrangements involved. The only reason that limits the wide application of this method is the environment concern since mercury (Hg) waste produced.
Oxymercuration-demercuration is a two-step procedure, as shown explicitly below:
The mechanism in the oxymercuration step involves a mercury acting as a reagent attacking the alkene double bond to form a cyclic mercurinium ion intermediate. Because no carbocation intermediate involved, rearrangements are not observed in such reaction. Then a water molecule attacks the most substitutedcarbon to open the mercurium ion bridge, followed by proton transfer to solvent water molecule. For the same reasoning that water molecule attacks the more substituted carbon of the cyclic halonium ion in halohydrin formation (section 10.2.4), the water molecule in this mechanism also attacks the more substituted carbon preferentially, as the partial positive charge is better accommodated on a tertiary carbon than on a primary carbon (if attack occurs on the other carbon).
The organomercury intermediate is then reduced by sodium borohydride, the mechanism for this final step is beyond the scope of our discussions here. Notice that the overall oxymercuration-demercuration mechanism follows Markovnikov’s rule with the OH group is attached to the most substituted carbon and the hydrogen atom adds to the less substituted carbon.
10.6.2 Hydroboration –Oxidation of Alkenes
Hydroboration-oxidation is another method to convert alkene to alcohol, however, in anti-Markovnikov regioselectivity, that is OH is bonded to the carbon with greater number of hydrogens and hydrogen atom bonded to the carbon with less hydrogens.
The overall reaction is also a two-step process:
• First step is hydroboration, that is the addition of boron atom and hydrogen atom to the alkene.
• Second step is oxidation and hydrolysis of the alkylborane formed in step 1, to produce alcohol.
The borane reagent used in the first step is usually available as the solution containing BH3·THF complex. Borane, BH3, is an electron-deficient species because the boron atom has incomplete octet with only six electrons. When BH3 is introduced to THF, they react to form a Lewis acid-Lewis base adduct (Chapter 3.??), which is more stable and relatively easy to be handled and stored. The solution containing BH3·THF is still rather sensitive and must be used in an inert atmosphere (nitrogen or argon) and with care.
Because of the incomplete octet of the boron atom in BH3, it is a good electrophile that reacts with alkene. The mechanism of the hydroboration step is illustrated below with propene as the example.
Mechanism of Hydroboration
When a terminal alkene, for example propene, is treated with BH3·THF, the BH3 molecule adds successively to the C=C double bond of three alkene molecules to form an trialkylborane. In each addition step, the boron atom becomes attached to the less substituted double bond carbon, and a hydrogen atom transferred from the BH3 to the more substituted carbon. In the second step (oxidation and hydrolysis) of the whole process, the borane is oxidized and hydrolyzed to OH group. So the regioselectivity of the hydroboration step defines the anti-Markovnikov regioselectivity of the overall reaction.
Such regioselectivity of the hydroboration step can be explained by both electronic and steric effects. In terms of steric factor, the boron-containing group is more bulky than hydrogen atom, so they can approach the less substituted carbon more easily. The electronic effect lies in the transition state structure for the formation of alkylborane. As shown above, the π electrons from the double bond is donated to the π orbital of boron and a four-atom ringcyclic transition state is approached. In the transition state, electrons shift in the direction of the boron atom and away from the carbon that is not connected to the boron. This make the carbon not connected to the boron bears a partial positive charge, that is better accommodated on the more substituted carbon. As a result the electronic effect also favors the addition of boron on the less substituted carbon.
Stereochemistry of Hydroboration
Hydroboration-oxidation takes place with syn stereochemistry, that the OH group and the hydrogen atom add to the same side of the double bond, as shown in the following example.
This can be explained by the mechanism of the hydroboration step. The four-membered ring transition state requires that the boron atom and the hydrogen atom approach to the same surface of the alkene double bond, so they are added in the syn position to the double bond. Since the boron part is converted to OH group in the second step, that results in the syn addition of OH and H in the product.
Oxidation and Hydrolysis of trialkylboranes
With the hydroboration reaction is over, the trialkylboranes are usually notisolated, they are oxidized and hydrolyzed with the addition of hydrogen peroxide (H2O2) in basic aqueous solution. The mechanism for the oxidation and hydrolysis of trialkylboranes is rather complicated and could be an optional topic, the net result is the boron that initially bonded on the carbon is replaced by the hydroxy OH group.
Summary: Hydration Methods of Alkene
Overall there are three methods for converting alkene to alcohol via addition, they are acid-catalyzed hydration, oxymercuration-demercuration and hydroboration-oxidation. Each method has its own character with benefit and disadvantage. The proper method could be picked up based on the need.
Acid-catalyzed hydration Oxymercuration-demercuration Hydroboration-oxidation
Reaction Conditions cat. H+/H2O 1)Hg(OAc)2/THF·H2O 2)NaBH4 1) BH3·THF 2) NaBH4
Regioselectivity Markovnikov Markovnikov Anti-Markovnikov
Stereochemistry Not controlled Not controlled syn-addition
Rearrangement Yes No No | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/10%3A_Alkenes_and_Alkynes/10.05%3A_Reaction_of_Alkenes-_Hydrogenation.txt |
Alkenes undergo a number of reactions in which the C=C double bond is oxidized. For organic compounds, a conventional way to tell whether the oxidation or reduction occur is to check the number of C–O bonds or the C–H bonds. An oxidation reaction increase the number of C–O bonds or decrease the number of C–H bonds. On the other side a reduction reaction increase the number of C–H bonds or decrease the number of C–O bonds. The relative oxidation state of some common organic functional groups are listed here based on the trend.
10.7.1 Syn 1,2-Dihydroxylation
1,2-Dihydroxylation, the conversion of the C=C double bond to 1,2-diol, is an oxidative addition reaction of alkene. Osmium tetroxide (OsO4) is a widely used oxidizing agent for such purpose. Potassium permanganate can be used as well, although further oxidation is prone to occur to cleave the diol because it is a stronger oxidizing agent (10.7.2).
The traditional method of 1,2-dihydroxylation with osmium tetroxide is a two-step procedure. Osmium tetroxide first reacts with alkene to from a cyclic osmate ester intermediate and this cyclic intermediate involves the syn addition of OsO4 to the double bond. The cleavage of the O—Os bond of the intermediate then take places in the second step with reducing agent NaHSO3, without modifying the stereochemistry of the C—O bond. The diol formed therefore has the syn stereochemistry property.
Catalytic OsO4 1,2-Dihydroxylation
The 1,2-dihydroxylation with osmium tetroxide an effective reaction that used very often in the labs for the preparing diol from alkene. However, this method has major drawbacks because osmium tetroxide is a highly toxic, volatile and expensive reagent. Improved methods have been developed that allow only catalytical amount of OsO4 being used in conjunction with a co-oxidant in stoichiometric amount. N-methylmorpholine N-oxide (NMO) is one of the most commonly employed co-oxidants. In such condition, osmium compounds are re-oxidized by NMO and can be reused to react with more alkenes, so only small molar percentage of OsO4 is necessary in the reaction mixture. The reaction proceeds smoothly with syn diols produced in good yield.
In terms of the stereochemistry of the product, although the syn addition could occur on either side of the alkene plane, that gives the same product which is the meso compound. This can be identified by either looking for the plane of symmetry of the product, or by assigning the absolute configuration on the chirality centers. Review the stereochemistry knowledge.
Examples
Show product of following reaction:
Solution:
The syn addition occurs on either side of the alkene plane, so both enantiomers are obtained with same amount as racemic mixture.
Cleavage with Ozone
With stronger oxidizing agent being applied, the C=C double bond of alkenes can be oxidatively cleaved, and the alkene molecule is cleaved to smaller molecules.
The most effective way for cleaving alkene is to use ozone, O3, by a two-step process. Alkene is first reacted with ozone at very low temperature (-78 °C) and then treated with dimethyl sulfide, (CH3)2S, (or Zn/CH3COOH) to give the cleavage products. The whole process is called ozonolysis.
Ozonolysis results in the cleavage of the double bond, and each double bond carbon get bonded to an oxygen atom with a new double bond. The products of ozonolysis are aldehyde(s) and/or ketone(s), and the exact structures of the products depends on the structure of the initial alkene:
• Disubstituted alkene carbons are oxidatively cleaved to ketone;
• Monosubstituted alkene carbons are oxidatively cleaved to aldehyde;
• Unsubstituted alkene carbons are oxidatively cleaved to formaldehyde (HCHO).
Examples
Show ozonolysis products of following reactions:
Hint: To figure out the structure of ozonolysis product(s), “cut” the double bond, then “add” a “=O” (double bonded oxygen) to each carbon.
As shown with above examples, ozonolysis reaction is useful as a synthetic tool for certain aldehyde and ketone. Meanwhile, it is also a method for determining the position of double bonds in an alkene by working backward from the structure of the products.
Examples
Determine the structure of the alkene:
Approach: To determine the structure of initial alkene, we can work backwards by connecting two C=O bonds in the products together. The two C=O bonds are “connected” to give a C=C bond with all oxygen atoms “removed”. In this example, the two blue C=O bonds gives the blue C=C bond, and the two red C=O bonds gives the red C=C bond.
Mechanism for Ozonolysis
The hints mentioned earlier is to help us solving the problems with ozonolysis reaction, not the reaction mechanism. The mechanism of ozonolysis reaction is rather complicate that involves the formation of initial cyclic ozonide that decompose to fragments, and the fragment recombine to form a new cyclic ozonide, which is reduced to give products.
Cleavage with Potassium Permanganate KMnO4
Potassium permanganate, KMnO4, is another oxidizing agent that cleaves the C=C double bond of an alkene. Under hot basic condition, the oxidative cleavage products of alkenes could involve ketone, salt of carboxylic acid or carbon dioxide depends on the different substituent patterns on the alkene:
• Disubstituted alkene carbons are oxidatively cleaved to ketone;
• Monosubstituted alkene carbons are oxidatively cleaved to the carboxylic acid (in salt format);
• Unsubstituted alkene carbons are oxidatively cleaved to CO2 and H2O.
KMnO4 is a stronger oxidizing agent that further oxidize the initial cleavage products, therefore aldehyde is further oxidized to carboxylic acid (in salt format under basic condition). For terminal unsubstituted alkene carbons, the initial product is HCHO, which is then further oxidized to carboxylate CO32-in basic condition. Acidification of CO32-produces H2CO3 that decomposes to CO2 and H2O. Because of over oxidation, KMnO4 is not the useful reagent for the synthesis of aldehyde/ketone from alkenes. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/10%3A_Alkenes_and_Alkynes/10.07%3A_Oxidation_Reactions_of_Alkenes.txt |
Alkyne is the hydrocarbon that contain C≡C triple bond. In this section, we will explore the methods for the synthesis of alkyne and the chemical reactions of alkynes.
10.8.1 Acidity of Terminal Alkynes and Related Reactions
In the discussions of acids and bases (Chapter 3), we have learned that the hydrogen atom bonded to the terminal alkyne carbon shows higher acidity than the hydrogen atoms bonded to the carbons of an alkene or alkane, and the pKa value of the terminal alkyne hydrogen is about 25.
Because of the relative high acidity, the terminal alkynes can be deprotonated by appropriate strong bases, such as NaH, NaNH2.
The product of the above deprotonation, alkynide anion, is a good nucleophile that can be used in SN2 reaction with primary substrates (since primary substrates work best for such SN2 reaction as we have learned):
New carbon portion is introduced in the product with new carbon-carbon bond formed in the SN2 reaction, and this is a common method to synthesize internal alkynes with longer carbon chain. A specific example for the synthesis of 2-methyl-3-hexyne from 3-methyl-1-butyne is given here:
10.8.2 Synthesis of Alkynes by Elimination
The method in 10.4.1 applies to the synthesis of alkyne with certain structure. The more general way to synthesize alkyne is via the elimination reaction of vicinal dihalides. Recall that vicinal dihalides are the halogenation products of alkenes (section 10.4). The vicinal dihalide can then be subjected to a double dehydrohalogenation reaction with a strong base to produce an alkyne.
The dehydrohalogenation occurs twice, in two steps, the first product is a haloalkene, and the second product is the alkyne. Amide, usually NaNH2, is a base that is strong enough to cause both reactions carried out consecutively in the same mixture. Two molar equivalents of sodium amide per mole of the dihalide are required to ensure the elimination occur two times.
If a terminal alkyne is the desired product, then three molar equivalents of base are required. The terminal alkyne produced after double dehydrohalogenation is deprotonated by sodium amide, the third mole of base is to ensure the deprotonation occurs completely and all the terminal alkyne converted to the salt format. The salt of alkynide was then treated with ammonium chloride (or water, as source of proton) to produce terminal alkyne as the final desired product.
Examples
Design the synthesis route of 1-butyne from 1-butene.
Approach:
Use retro-synthetic analysis:
Hydrogenation of Alkynes
The catalytic hydrogenation applied to the π bonds of C≡C triple bonds as well. Depending on the conditions and catalysts employed, one or two molar equivalents of hydrogen will be added to a triple bond and alkene or alkane produced as the product respectively.
When platinum or palladium catalysts applied, the final product of the hydrogenation is an alkane with sufficient hydrogen provided. The initial product is an alkene, that undergoes the reaction successively to give alkane as the final product.
With certain catalyst used, the hydrogenation of alkyne can be stopped at the alkene stage. The most commonly employed catalyst is the Lindlar catalyst. Lindlar catalyst is prepared by precipitating palladium on calcium carbonate and then treating it with lead (II) acetate and quinoline. The special treatment modifies the surface of the palladium metal by partially deactivating it, and making it more effective at catalyzing the hydrogenation to a triple bond rather than to a double bond.
The mechanism for the catalytic hydrogenation of alkyne is almost the same as that of alkene (10.5). Since both hydrogen atoms are delivered from the surface of the catalyst, they are delivered to the same side of the triple bond, therefore the syn addition occurs. So the hydrogenation of an internal alkyne produces cis-alkene with the Lindlar catalyst.
Internal alkyne can be converted into trans-alkene using sodium (or lithium) in liquid ammonia. The mechanism for this reaction involves successive single electron transfers from the metal (sodium or lithium) and proton transfers from ammonia, with radical intermediates. The sodium metal (or lithium) reacts more rapidly with triple bond than double bond, so the reaction stops at the alkene stage. Low temperature (-78 °C) is necessary to keep ammonia at the liquid state.
The trans-vinylic anion is formed preferentially because of the higher stability with two R groups farther apart. Protonation of the trans-vinylic anion leads to the trans-alkene.
Hydrohalogenation of Alkynes
An alkyne is an electron-rich molecule with high density of pi electrons, therefore it is a good nucleophile that reacts readily with electrophiles. Thus alkynes, like alkenes, also undergo electrophilic addition with hydrogen halide.
• Alkyne reacts with one mole of HX to form haloalkene, and with two moles of HX to form geminal dihalides, the dihalide with both halogen attached to the same carbon. “Geminal” comes from geminusin Latin, that means “twin”.
• Both addition follow Markovnikov’s rule in terms of regioselectivity.
If one molar equivalent of HX available, the addition can be stopped at the first addition to haloalkene. The halo-substituted alkene is less reactive than alkyne for electrophilic addition because a halogen substituent withdraws electrons inductively, therefore decreasing the nucleophilicity of the double bond.
The mechanism for the electrophilic addition to alkyne is rather similar to the addition of alkene, with protonation as the first step. For terminal alkyne, if the protonation occurs on different triple bond carbon, the primary or secondary vinylic cation intermediate will formed. The higher stability of the secondary vinylic cation leading to the Markovnikov’s regioselectivity, that the hydrogen atom attached to the carbon that has the greater number of hydrogen atoms.
If excess hydro halide is present, the addition to alkyne occur twice to give geminal halide that follow the Markovnikov’s regioselectivity.
Hydration of Alkynes
Alkynes also undergo the acid-catalyzed addition of water (hydration), similar to alkenes. As a result, the H added to one triple bond carbon and OH added to the other triple bond carbon, and the product formed is called an enol (“en” comes from “ene” that means double bond, “ol” means OH group). An enol is a compound with a carbon-carbon double bond and an OH group connected on one of the double bond carbon.
Enol is a very unstable compound, it immediately undergoes rearrangement to give more stable carbonyl compound, aldehyde or ketone. The structure of a carbonyl compound and an enol differ in the location of the double bond and a hydrogen atom, and they are called tautomers. The interconversion between the tautomers is called tautomerization. The mechanism is not covered. Enol always undergoes tautomerization rapidly because of the high stability of carbonyl compound, as shown in the general way below.
For symmetrical internal alkyne that has the same group attached to each of the triple bond (sp) carbon, the addition of water forms a single ketone as a product. As in the early example that 2-butanone is produced from the hydration of 2-butyne.
For unsymmetrical internal alkyne with different groups on each of the triple bond carbon, the mixture of two ketones are formed because the initial addition of the proton can occur on either of the sp carbons. The hydration of 2-pentyne is shown here that produce the mixture of 2-pemtanone and 3-pentanone as product.
Terminal alkynes are not as reactive as internal alkynes towards the hydration. The addition of water to a terminal alkyne will occurs if mercuric ion (Hg2+) present as a catalyst. The enol formed from the addition follows the Markovnikov’s rule with the hydrogen atom attached to the terminal carbon, and a methyl ketone (the ketone with a methyl group connected on one side of the C=O bond) is the final product after tautomerization.
Hydroboration-Oxidation of Alkynes:
Hydroboration-oxidation also applies to alkyne in the similar way as to alkene. The two-step process results in the enol, that goes through tautomerization to give carbonyl compound.
Meanwhile, the addition of borane to a terminal alkyne shows the same regioselectivity as observed in borane addition to an alkene. That is boron adds preferentially to the terminal triple bond (sp) carbon (the carbon with more hydrogen atom), or the terminal carbon with less substituents. After oxidation, the boron-containing group is converted to the OH group, so the enol is produced in the anti-Markovnikov way, with OH connected on the terminal carbon. The tautomerization of such enol generates aldehyde as the final product.
Comparing the two hydration methods of alkyne, hydroboration-oxidation produces aldehyde from terminal alkyne, while acid-catalyzed hydration converts terminal alkyne to methyl ketone.
10.09: Answers to Practice Questions Chapter 10
10.1 Show structure of the major product for following addition reactions.
10.2 For the addition of HBr to 3-methyl-1-butene, two products were observed. Show the reaction mechanism to explain the formation of both products.
Mechanism:
10.3 Show major product(s) for the following reactions.
10.4 Show major product(s) of the following reactions. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_I_(Liu)/10%3A_Alkenes_and_Alkynes/10.08%3A_Alkynes.txt |
• 1.1: Prelude to General Techniques
• 1.2: Glassware and Equipment
• 1.3: Transferring Methods
It is often needed to transfer chemicals from one container to another in organic chemistry laboratories. The proper approach to take depends on the nature of what needs to be transferred.
• 1.4: Heating and Cooling Methods
As reaction rate increases with temperature, it is very common to heat solutions in the organic laboratory. There are many heat sources available to the organic chemist, although some methods are better than others depending on the situation. Described in this section are some concepts that are common to all heating methods (e.g. how to boil controllably), followed by specifics on each heat source. Also included are methods used to cool solutions.
• 1.5: Filtering Methods
There are many methods used to separate a mixture containing a solid and liquid. If the solid settles well, the liquid can sometimes be poured off (decanted). If the solid has very small sized particles or forms a cloudy mixture, the mixture can sometimes be centrifuged or passed through a filter pipette (on the microscale, < 5 mL).
01: General Techniques
A Grignard reaction is run under a balloon containing an inert gas.
1.2.01: Storing Samples (Parafilm Teflon Tape)
• 1.2.1: Storing Samples (Parafilm/Teflon Tape)
When samples must be stored for a period of time, they are best stored upright in screw-capped vials. Samples may evaporate through the joint over time, and if a highly volatile sample is used, the joint should be wrapped in Teflon tape (semi-stretchy white film or Parafilm (stretchy plastic film) to create a better seal. Teflon tape is less permeable to solvents than Parafilm, and volatile samples wrapped in Parafilm may still evaporate over a period of weeks.
• 1.2A: Pictures of Glassware and Equipment
• 1.2B: Ground Glass Joints
Most organic glassware uses "ground glass joints," which have a frosted appearance. They are precisely ground to a certain size (which makes them expensive) and have outer (female) and inner (male) joints so that pieces can be connected together with a tight fit. Common joint sizes are 14/20, 19/22, and 24/40. The first number refers to the inner diameter (in millimeters) of a female joint or outer diameter of a male joint. The second number refers to the length of the joint.
• 1.2C: Clamping
Metal clamps are used to connect glassware to ring stands or the metal lattice work. Two common type of clamps are "extension clamps" and "three-fingered clamps". Although in many situations the clamps can be used interchangeably, an extension clamp must be used when clamping to a round bottomed flask as 3-fingered clamps do not hold well.
• 1.2D: Greasing Joints
Ground glass joints are manufactured to fit quite well with one another, and yet they are not perfectly airtight. In some situations (e.g. when using reduced pressure inside an apparatus), grease must be applied to each joint to ensure a good seal. Grease is also used whenever the joint may be in contact with a highly basic solution, as basic solutions can form sodium silicates and etch glass.
• 1.2E: Cleaning Glassware
Glassware should be dismantled and cleaned as soon as possible. Experience with home dishwashing can tell you that dishes are more difficult to clean when allowed to dry. If there is a time constraint, it's best to leave glassware in a tub of soapy water.
• 1.2F: Drying Glassware
Glassware that appears "dry" actually contains a thin film of water condensation on its surface. When using reagents that react with water (sometimes violently!), this water layer needs to be removed.
1.02: Glassware and Equipment
When samples must be stored for a period of time, they are best stored upright in screw-capped vials. Samples may evaporate through the joint over time, and if a highly volatile sample is used, the joint should be wrapped in Teflon tape (semi-stretchy white film, Figures 1.14 a+b) or Parafilm, (stretchy plastic film, Figures 1.14 c-e) to create a better seal. Teflon tape is less permeable to solvents than Parafilm, and volatile samples wrapped in Parafilm may still evaporate over a period of weeks. If a sample is to be stored in a round bottomed flask for some time, it should be stoppered with a cork stopper or plastic cap. A rubber stopper should not be used as it will tend to swell when exposed to organic vapors, and a glass stopper may freeze. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.01%3A_Prelude_to_General_Techniques.txt |
Glassware used in all chemistry courses:
Item Name:
1. Graduated cylinders
2. Wash bottle
3. Side arm
4. Erlenmeyer flask
5. Beaker
6. Funnels
7. Test tubes
8. Watch glasses
Glassware commonly used in organic chemistry for conducting reactions and purifications:
Item Name:
1. Cork ring
2. Fractionating column
3. West condenser
4. Separatory funnel
5. Stopper
6. Round bottomed flask
7. Three-way adapter (distilling head)
8. Claisen adapter
9. Vacuum adapter
10. Thermometer adapter
11. Plastic clip (Keck clip)
12. Rubber fitting for
thermometer adapter
Tools used in all chemistry courses:
Item Name:
1. Crucible tongs
2. Test tube holder
3. Striker
4. Forceps
5. Spatulas
6. Scoopula
7. Glass stirring rod
8. Pasteur pipette
9. Dropper bulb
Various equipment in the organic chemistry lab:
Item Name:
1. Evaporating dish
2. TLC chamber
3. Buchner funnel
4. Hirsch funnel
5. Rubber sleeves for funnels (filter adapters)
6. Clay tile
7. Vials for storage
8. Stir bars and spin vane
9. Drying tube
Clamps for securing apparatuses:
Item Name:
1. Extension clamps (with and without vinyl sleeves)
2. Three-fingered clamps (vinyl and fire-resistant sleeves)
3. Ring clamp (iron rings)
4. Wire mesh
Burners and tubing:
Item Name:
1. Thick-walled vacuum tubing
2. Bunsen burner
3. Wood blocks
4. Rubber tubing
Glassware used in microscale work:
Item Name:
1. Drying tube
2. Hickman head
3. Water condenser
4. Air condenser
5. Conical vial
6. Claisen adapter
7. Spin vane
1.2B: Ground Glass Joints
Most organic glassware uses "ground glass joints," which have a frosted appearance. They are precisely ground to a certain size (which makes them expensive) and have outer (female) and inner (male) joints so that pieces can be connected together with a tight fit (Figure 1.1a). Common joint sizes are 14/20, 19/22, and 24/40. The first number refers to the inner diameter (in millimeters) of a female joint or outer diameter of a male joint. The second number refers to the length of the joint (Figure 1.1b).
It is best if ground glass joints are free of chemicals when pieces are connected, or else the compounds may undergo reactions that cause the joints to "freeze" together, or become inseparable. Solid in the joint can also compromise the seal between the pieces. If chemical residue were to get on the joint during transfer (Figure 1.1c), the joint should be wiped clean with a KimWipe (lint-free tissue, Figure 1.2a) before connecting with another piece. Spillage on the joint can be minimized by using a funnel.
Figures 1.2 b+c shows a "frozen" joint (notice the residue on the frosted joint), where benzaldehyde crept into the joint during storage and probably oxidized to seal the round bottomed flask and stopper together. To separate a frozen joint, first try to gently twist the two pieces apart from one another. If that fails, gently tap on the joint with a spatula or other piece of equipment (Figure 1.2c). If that fails, next try heating the joint in a hot water bath (heat may cause expansion of the outer joint), or sonicating the flask if a sonicator is available. As a last resort, see your instructor, and they may heat the joint briefly with a heat gun. The frozen joint in Figure 1.2 had to be heated to separate the pieces. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.02%3A_Glassware_and_Equipment/1.2A%3A_Pictures_of_Glassware_and_Equipment.txt |
Organic chemistry glassware is often segmented so that pieces can be arranged in a variety of ways to create setups that achieve different goals. It is important that the pieces are securely fastened in an apparatus so that flammable vapors don't escape and pieces don't fall (whereupon the glassware may break or contents may be spilled). Some chemistry labs have a lattice-work of metal bars (Figure 1.3c) secured to the benchtop that can be used for clamping apparatuses, and other labs rely on ring stands (Figure 1.3a). Ring stands should be positioned so that the apparatus is clamped directly over the heavy metal base, not in the opposite direction as the base (Figure 1.3a, not 1.3b)
Metal clamps are used to connect glassware to ring stands or the metal lattice work. Two common type of clamps are "extension clamps" and "three-fingered clamps" (Figure 1.4a). Although in many situations the clamps can be used interchangeably, an extension clamp must be used when clamping to a round bottomed flask (Figure 1.4b), as 3-fingered clamps do not hold well. The extension clamp should securely grasp the neck of a round bottomed flask below the glass protrusion (Figure 1.4b, not Figure 1.4c). Three-fingered clamps are generally used to hold condensers (Figure 1.3b), suction flasks, and chromatography columns (Figure 1.5).
Both types of clamps often come with vinyl sleeves that may be removed if desired. The vinyl sleeves provide a gentle grasp for glassware, but should not be used with hot pieces as they may melt (or in the author's experience catch on fire!). Sometimes fire resistant sleeves are also provided with clamps as an alternative (right-most clamp in Figure 1.4a).
Ring clamps (or iron rings) are also commonly used in the organic lab. They are used to hold separatory funnels (Figure 1.6a), and can be used to secure funnels when filtering or pouring liquids into narrow joints (Figure 1.6b). Furthermore, they can be used along with a wire mesh to serve as a platform for supporting flasks (Figure 1.6c).
Plastic clips (sometimes called "Keck clips" or "Keck clamps") are also commonly used to secure the connections between joints (Figure 1.7). The clips are directional, and if they don't easily snap on, they are probably upside down. Plastic clips should not be used on any part of an apparatus that will get hot, as they may melt at temperatures above 140 ˚C (Figure 1.7b). Metal versions of these clips can be used alternatively in hot areas. Clips should not be relied upon to hold any substantial weight, as they can easily fail (especially if they have been warmed). Therefore, reaction flasks should not be held with just clips, but always supported in some more significant way (e.g. with an extension clamp attached to a ring stand).
1.2D: Greasing Joints
Ground glass joints are manufactured to fit quite well with one another, and yet they are not perfectly airtight. In some situations (e.g. when using reduced pressure inside an apparatus), grease must be applied to each joint to ensure a good seal. Grease is also used whenever the joint may be in contact with a highly basic solution, as basic solutions can form sodium silicates and etch glass.
Grease can be applied with a syringe full of grease (Figure 1.8a), wood splint, or toothpick. Grease should be lightly applied in portions around the male joint, closer to the glass end than the end which will be in contact with reagents (Figure 1.8a). If grease is allowed near the end which will contact the reagents, there is a possibility the reagent will dissolve the grease and become contaminated. The female joint should then be connected, and the joints twisted to spread the grease in a thin layer. The joint should become transparent all the way around the joint, but to a depth of only one-third to one-half of the joint (Figure 1.8b). If the entire joint becomes transparent or if grease is seen spilling out of the joint, too much grease has been used (Figure 1.8c). Excess grease should be wiped off with a KimWipe (one is used in Figure 1.9).
To clean grease from a joint after a process is complete, wipe of the majority of the grease using a paper towel or KimWipe. Then wet a KimWipe with some hydrocarbon solvent and rub the moistened KimWipe onto the joint to dissolve the grease (Figure 1.9). Hydrocarbon solvents (e.g. hexanes) work much better than acetone to dissolve residual grease. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.02%3A_Glassware_and_Equipment/1.2C%3A_Clamping.txt |
Glassware should be dismantled and cleaned as soon as possible. Experience with home dishwashing can tell you that dishes are more difficult to clean when allowed to dry. If there is a time constraint, it's best to leave glassware in a tub of soapy water.
To clean glassware, use the following procedures:
• Use 2-3 mL solvent to rinse residual organic compounds from the glassware into a waste beaker. The compounds should be highly soluble in the solvent. The default solvent is often acetone as it is inexpensive, relatively nontoxic, and dissolves most organic compounds. Some institutions reuse their acetone ("wash acetone") as the solvation ability is not spent after a few uses.
As it will soon become second-nature for most students to use acetone as part of their cleaning ritual, it is worth reminding that the purpose of acetone rinse is to dissolve organic residue in a flask. Not everything dissolves in acetone, for example ionic salts are insoluble in acetone and are more successfully rinsed out with water.
• After a preliminary rinse, glassware should then be washed with soap and water at the bench.
Residual acetone will likely evaporate from the flask, but it is acceptable for small quantities of residual acetone to be washed down the drain. Acetone is a normal biological byproduct of some metabolic processes,$^1$ and has low toxicity as it can be easily excreted by most organisms.
If using undiluted detergent from the store, it is best to use small amounts during washing as they tend to form thick foams that need lots of rinsing (Figure 1.10). Some institutions instead use dilute soap solutions at their cleaning stations for this reason. For cleaning of glassware, the biodegradable detergent "Alconox" is the industry standard.
• Rinse all glassware with a few mL of distilled water, then store wet glassware in a locker atop paper towels to evaporate by the next lab period.
$^1$R. Boyer, Concepts in Biochemistry, 2$^\text{nd}$ edition, 2002, Brooks-Cole, p. 565.
1.2F: Drying Glassware
Quick Drying
If dry glassware is not needed right away, it should be rinsed with distilled water and allowed to dry overnight (in a locker). If dry glassware is promptly needed, glassware can be rinsed with acetone and the residual acetone allowed to evaporate. Rinsing with acetone works well because water is miscible with acetone, so much of the water is removed in the rinse waste. Evaporation of small amounts of residual acetone can be expedited by placing the rinsed glassware in a warm oven for a short amount of time or by using suction from a tube connected to the water aspirator. Residual acetone should not be evaporated inside a hot oven (>$100^\text{o} \text{C}$) as acetone may polymerize and/or ignite under these conditions. It should also not be evaporated using the house compressed air lines, as this is likely to contaminate the glassware with dirt, oil, and moisture from the air compressor.
Oven and Flame Drying
Glassware that appears "dry" actually contains a thin film of water condensation on its surface. When using reagents that react with water (sometimes violently!), this water layer needs to be removed. To evaporate the water film, glassware can be placed in a $110^\text{o} \text{C}$ oven overnight, or at the least for several hours. The water film can also be manually evaporated using a burner or heat gun, a process called "flame drying". Both methods result in extremely hot glassware that must be handled carefully with tongs or thick gloves.
To flame dry glassware, first remove any vinyl sleeves on an extension clamp (Figure 1.11a), as these can melt or catch on fire. Clamp the flask to be dried, including a stir bar if using (Figure 1.11b). Apply the burner or heat gun to the glass, and initially fog will be seen as water vaporizing from one part of the glassware condenses elsewhere (Figure 1.11c). Continue waving the heat source all over the glassware for several minutes until the fog is completely removed and glassware is scorching hot (Figure 1.11d). If the glass is only moderately hot, water will condense from the air before you are able to fully exclude it.
Safety Note
Glassware will be extremely hot after flame drying.
Regardless of the manner in which glassware is heated (oven or flame drying), allow the glassware to cool in a water-free environment (in a desiccator, under a stream of inert gas, or with a drying tube, Figure 1.12) before obtaining a mass or adding reagents.
Drying Tubes
A drying tube is used when moderately but not meticulously dry conditions are desired in an apparatus. If meticulously dry conditions are necessary, glassware should be oven or flame dried, then the air displaced with a dry, inert gas.
Drying tubes are pieces of glassware that can be filled with a drying agent (often anhydrous $\ce{CaCl_2}$ or $\ce{CaSO_4}$ in the pellet form) and connected to an apparatus either through a thermometer adapter (Figures 1.13 b+c) or rubber tubing (Figure 1.13d). Air passing through the tube is removed of water when it comes in contact with the drying agent. Since it is important that air can flow through the drying tube, especially so the apparatus is not a closed system, the drying agent should be fresh as used drying agents can sometimes harden into a plug that restricts air flow. Drying tubes can also be filled with basic solids such as $\ce{Na_2CO_3}$ to neutralize acidic gases. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.02%3A_Glassware_and_Equipment/1.2E%3A_Cleaning_Glassware.txt |
It is often needed to transfer chemicals from one container to another in organic chemistry laboratories. The proper approach to take depends on the nature of what needs to be transferred.
• 1.3A: Transferring Methods - Solids
A solid can be dispensed from its reagent jar directly into a vessel or onto a weighing boat or creased piece of paper. If a solid is to be transferred into a vessel containing a narrow mouth (such as a round bottomed flask), a "powder funnel" or wide-mouth funnel can be used. Alternatively, the solid can be nudged off a creased piece of paper in portions using a spatula.
• 1.3B: Transferring Methods - Liquids
When transferring liquids with volumes greater than 5 mL, they can be poured directly into vessels. Graduated cylinders and beakers have an indentation in their mouth, so they can be poured controllably as long as the two pieces of glass touch one another.
• 1.3C: Transferring Methods - Inert Atmospheric Methods
Meticulously dry or oxygen-free conditions are sometimes necessary when using reagents that react with water or oxygen in the air. To safely and effectively use these reagents, glassware should be oven or flame dried, then the air displaced with a dry, inert gas (often nitrogen or argon). This creates an "inert atmosphere" inside an apparatus, one that will not react with the reagents. Inert gases can be delivered to a flask through gas lines and a gas manifold.
1.03: Transferring Methods
A solid can be dispensed from its reagent jar directly into a vessel or onto a weighing boat or creased piece of paper. If a solid is to be transferred into a vessel containing a narrow mouth (such as a round bottomed flask), a "powder funnel" or wide-mouth funnel can be used (Figure 1.15a). Alternatively, the solid can be nudged off a creased piece of paper in portions using a spatula (Figures 1.15 b+c).
If the solid is the limiting reagent in a chemical reaction, it should ideally be dispensed from the reagent jar directly into the vessel (Figure 1.16a). However, if using a weighing boat, residue should be rinsed off with the solvent that will be used in the reaction (only if the boat is unreactive to the solvent) in order to transfer the reagent in its entirety.
Residue clinging to ground glass joints should also be dislodged with a KimWipe or rinsed into the flask with solvent to prevent joints from sticking, and to make sure the entire reagent makes it to the reaction vessel.
Certain solid compounds (e.g. \(\ce{KOH}\), \(\ce{K2CO3}\), \(\ce{CaCl2}\)) are sticky or hygroscopic (readily absorb water from the air), and these reagents should be dispensed onto glossy weighing paper (used in Figure 1.15b). This weighing paper has a wax coating so that sticky reagents more easily slide off its surface.
For transfer into vessels with very narrow mouths (e.g. NMR tubes), it is sometimes easier to dissolve solids in their eventual solvent and transfer a solution via pipette (Figures 1.16 b+c).
1.3B: Transferring Methods - Liquids
Pouring Liquids
When transferring liquids with volumes greater than $5 \: \text{mL}$, they can be poured directly into vessels. Graduated cylinders and beakers have an indentation in their mouth, so they can be poured controllably as long as the two pieces of glass touch one another (Figure 1.17a). If pouring from an Erlenmeyer flask, or transferring a liquid into a vessel containing a narrow mouth (e.g. a round bottomed flask), a funnel should be used. Funnels can be securely held with a ring clamp (Figure 1.17b), or held with one hand while pouring with the other (Figure 1.17c).
Comments Regarding Measurements
In order to determine a meaningful yield for a chemical reaction, it is important to have precise measurements on the limiting reactant. It is less important to be precise when manipulating a reagent that is in excess, especially if the reagent is in several times excess.
A portion of the liquid measured by a graduated cylinder always clings to the glassware after pouring, meaning that the true volume dispensed is never equivalent to the markings on the cylinder. Therefore, graduated cylinders can be used for dispensing solvents or liquids that are in excess, while more accurate methods (e.g. mass, calibrated pipettes or syringes) should be used when dispensing or measuring the limiting reactant. A graduated cylinder may be used to dispense a limiting reactant if a subsequent mass will be determined to find the precise quantity actually dispensed.
When determining the mass of a vessel on a balance, it's best to not include the mass of a cork ring (Figure 1.18a) or other support (e.g. the beaker in Figure 1.18b). A cork ring might get wet, have reagents spilled on it, or have pieces of cork fall out, leading to changes in mass that cannot be accounted for. Beakers used to support flasks can get mixed up, and every $100$-$\text{mL}$ beaker does not have the same mass. It is also best to transport vessels containing chemicals to the balance in sealed containers, so as to minimize vapors and prevent possible spillage during transport.
Using Pasteur Pipettes
Pasteur pipettes (or pipets) are the most commonly used tool for transferring small volumes of liquids (< $5 \: \text{mL}$) from one container to another. They are considered disposable, although some institutions may clean and reuse them if they have a method for preventing the fragile tips from breaking.
Pasteur pipettes come in two sizes (Figure 1.19a): short (5.75") and long (9"). Each can hold about $1.5 \: \text{mL}$ of liquid, although the volume delivered is dependent on the size of the dropper bulb. The general guideline that "$1 \: \text{mL}$ is equivalent to 20 drops" does not always hold for Pasteur pipettes, and may be inconsistent between different pipettes. The drop ratio for a certain pipette and solution can be determined by counting drops until $1 \: \text{mL}$ is accumulated in a graduated cylinder. Alternatively, a pipette can be roughly calibrated by withdrawing $1 \: \text{mL}$ of liquid from a graduated cylinder and marking the volume line with a permanent marker (Figure 1.19b).
To use a pipette, attach a dropper bulb and place the pipette tip into a liquid. Squeeze then release the bulb to create suction, which will cause liquid to withdraw into the pipette (Figures 1.20 a+b). Keeping the pipette vertical, bring it to the flask where it is to be transferred, and position the pipette tip below the joint of the flask but not touching the sides before depressing the bulb to deliver the material to the flask (Figure 1.20c). The bulb can be squeezed a few times afterward to "blow out" residual liquid from the pipette.
If the receiving flask has a ground glass joint, the pipette tip should be below the joint while delivering so that liquid does not splash onto the joint, which sometimes causes pieces to freeze together when connected. If the pipette is to be reused (for example is the designated pipette for a reagent bottle), the pipette should be held so it does not touch the glassware, where it may become contaminated by other reagents in the flask (Figure 1.20d).
Using Calibrated Pipettes
Calibrated Plastic Pipettes
When some precision is needed in dispensing small volumes of liquid ($1$-$2 \: \text{mL}$), a graduated cylinder is not ideal as the pouring action results in a significant loss of material. Calibrated plastic pipettes have markings at $0.25 \: \text{mL}$ increments for a $1 \: \text{mL}$ pipette, and are economical ways to dispense relatively accurate volumes.
To use a calibrated plastic pipette, withdraw some of the liquid to be transferred into the bulb as usual (Figure 1.21b). Then squeeze the bulb just enough so that the liquid drains to the desired volume (Figure 1.21c), and maintain your position. While keeping the bulb depressed so the liquid still reads to the desired volume, quickly move the pipette to the transfer flask (Figure 1.21d), and depress the bulb further to deliver liquid to the flask (Figure 1.21e).
Calibrated Glass Pipettes
When a high level of precision is needed while dispensing liquids, calibrated glass pipettes (volumetric or graduated) can be used. Volumetric pipettes have a glass bulb at the top of their neck, and are capable of dispensing only one certain volume (for example, the top pipette in Figure 1.22 is a $10.00 \: \text{mL}$ pipette). Graduated pipettes (Mohr pipettes) have markings that allow them to deliver many volumes. Both pipettes need to be connected to a pipette bulb to provide suction.
The volume markings on a graduated pipette indicate the delivered volume, which may seem a bit "backward" at first. For example, when a graduated pipette is held vertically, the highest marking is $0.0 \: \text{mL}$, which indicates that no volume has been delivered when the pipette is still full. As liquid is drained into a vessel, the volume markings increase on the pipette, with the lowest marking often being the total capacity of the pipette (e.g. $1.0 \: \text{mL}$ for a $1.0 \: \text{mL}$ pipette).
Graduated pipettes can deliver any volume of liquid made possible by differences in the volume markings. For example, a $1.0 \: \text{mL}$ pipette could be used to deliver $0.4 \: \text{mL}$ of liquid by: a) Withdrawing liquid to the $0.0 \: \text{mL}$ mark, then draining and delivering liquid to the $0.4 \: \text{mL}$ mark, or b) Withdrawing liquid to the $0.2 \: \text{mL}$ mark and draining and delivering liquid to the $0.6 \: \text{mL}$ mark (or any combination where the difference in volumes is $0.4 \: \text{mL}$).
It is important to look carefully at the markings on a graduated pipette. Three different $1 \: \text{mL}$ pipettes are shown in Figure 1.23a. The left-most pipette has markings every $0.1 \: \text{mL}$, but no intermediary markings so is less precise than the other two pipettes in Figure 1.23a. The other two pipettes differ in the markings on the bottom. The lowest mark on the middle pipette is $1 \: \text{mL}$, while the lowest mark on the right-most pipette is $0.9 \: \text{mL}$. To deliver $1.00 \: \text{mL}$ with the middle pipette, the liquid must be drained from the $0.00 \: \text{mL}$ to the $1.00 \: \text{mL}$ mark, and the final inch of liquid should be retained. To deliver $1.00 \: \text{mL}$ with the right-most pipette, liquid must be drained from the $0.00 \: \text{mL}$ mark completely out the tip, with the intent to deliver its total capacity.
Pipettes are calibrated "to-deliver" (TD) or "to-contain" (TC) the marked volume. Pipettes are marked with T.C. or T.D. to differentiate between these two kinds, and to-deliver pipettes are also marked with a double ring near the top (Figure 1.23b). After draining a "to-deliver" pipette, the tip should be touched to the side of the flask to withdraw any clinging drops, and a small amount of residual liquid will remain in the tip. A "to-deliver" pipette is calibrated to deliver only the liquid that freely drains from the tip. However, after draining a "to-contain" pipette, the residual liquid in the tip should be "blown out" with pressure from a pipette bulb. "To-contain" pipettes may be useful for dispensing viscous liquids, where solvent can be used to wash out the entire contents.
In this section are described methods on how to use a calibrated glass pipette. These methods are for use with a clean and dry pipette. If residual liquid is in the tip of the pipette from water or from previous use with a different solution, a fresh pipette should be used.
Alternatively, if the reagent is not particularly expensive or reactive, the pipette can be "conditioned" with the reagent to remove residual liquid. To condition a pipette, rinse the pipette twice with a full volume of the reagent and collect the rinsing in a waste container. After two rinses, any residual liquid in the pipette will have been replaced by the reagent. When the reagent is then withdrawn into the pipette it will not be diluted or altered in any way.
To use a calibrated glass pipette:
1. Place the pipette tip in the reagent, squeeze the bulb and connect it to the top of the pipette (Figures 1.24 a+b).
2. Partially release pressure on the bulb to create suction, but do not fully release your hand or you may create too great a vacuum, causing liquid to be violently withdrawn into the pipette bulb. Suction should be applied until the liquid rises to just past the desired mark (Figure 1.24c).
3. Break the seal and remove the pipette bulb, then quickly place your finger atop the pipette to prevent the liquid from draining (Figure 1.24d).
4. With a slight wiggling motion or slight release of pressure from your finger, allow tiny amounts of air to be let into the top of the pipette in order to slowly and controllably drain the liquid until the meniscus is at the desired volume (Figure 1.25a shows a volume of $0.00 \: \text{mL}$).
5. Holding the top of the pipette tightly with your finger, bring the pipette to the flask where the liquid is to be delivered and again allow tiny amounts of air into the top of the pipette in order to slowly drain the liquid to the desired mark (Figure 1.25b; Figure 1.25c shows the delivered volume to be slightly below $0.20 \: \text{mL}$).
6. Touch the pipette tip to the side of the container to dislodge any hanging drops and remove the pipette.
7. If liquid was drained to the bottom of the pipette with a T.C. pipette, use pressure from a pipette bulb to blow out the residual drop. Do not blow out the residual drop when using a T.D. pipette.
8. If a volumetric pipette is used, the liquid should be withdrawn with suction to the marked line above the glass bulb (indicated in Figure 1.25d). The liquid can be drained into the new container with your finger fully released from the top. When the liquid stops draining, the tip should be touched to the side of the flask to withdraw any clinging drops, but the residual drop should not be forced out (similar to a T.D. pipette).
Calibrated Pipettes Summary
Place pipette tip in reagent bottle, squeeze pipette bulb, and connect to the pipette.
Partially release your hand to create suction. Do not let go completely or liquid will withdraw forcibly and possibly into the bulb
Apply suction until liquid is withdrawn to just past the desired mark.
Remove the pipette bulb and place your finger atop the pipette.
Allow tiny amounts of air to be let into the top of the pipette by wiggling your finger or a slight release of pressure.
Drain the liquid to the desired mark.
Tightly hold the pipette with your finger, bring it to the transfer flask and deliver the reagent to the desired mark.
Touch the pipette to the side of the container to dislodge the drip at the end of the pipette.
If a pipette is drained to the tip,
• To-deliver (T.D.) pipettes and volumetric pipettes should not be blown out.
• To-contain (T.C.) pipettes should be "blown out".
Note: If pipette is wet with a different solution before use, obtain a fresh one or "condition" the pipette with two rinses of the reagent.
Table 1.4: Procedural summary for using calibrated pipettes.
Dispensing Highly Volatile Liquids
When attempting to dispense highly volatile liquids (e.g. diethyl ether) via pipette, it is very common that liquid drips out of the pipette even without pressure from the dropper bulb! This occurs as the liquid evaporates into the pipette's headspace, and the additional vapor causes the headspace pressure to exceed the atmospheric pressure.
To prevent a pipette from dripping, withdraw and expunge the liquid into the pipette several times. Once the headspace is saturated with solvent vapors, the pipette will no longer drip.
Pouring Hot Liquids
It may be difficult to manipulate a vessel of hot liquid with your bare hands. If pouring a hot liquid from a beaker, a silicone hot hand protector can be used (Figure 1.26a) or beaker tongs (Figures 1.26b+c).
When pouring a hot liquid from an Erlenmeyer flask, hot hand protectors can also be used, but do not hold the awkward shape of the flask very securely. Pouring from hot Erlenmeyer flasks can be accomplished using a makeshift "paper towel holder". A long section of paper towel is folded several times in one direction to the thickness of approximately one inch (and secured with lab tape if desired, Figure 1.27a). This folded paper towel can be wrapped around the top of a beaker or Erlenmeyer flask and pinched to hold the flask (Figures 1.26d + 1.27b).
When pouring hot liquid from an Erlenmeyer flask, the paper towel holder should be narrow enough that the towel does not reach the top of the flask. If it does, liquid will wick toward the paper as it is poured, thus weakening the holder and also removing possibly valuable solution (Figure 1.27c). When the paper towel is a distance away from the top of the flask, liquid can be poured from the flask without absorbing the liquid (Figure 1.27d). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.03%3A_Transferring_Methods/1.3A%3A_Transferring_Methods_-_Solids.txt |
Meticulously dry or oxygen-free conditions are sometimes necessary when using reagents that react with water or oxygen in the air. To safely and effectively use these reagents, glassware should be oven or flame dried, then the air displaced with a dry, inert gas (often nitrogen or argon). This creates an "inert atmosphere" inside an apparatus, one that will not react with the reagents.
Inert gases can be delivered to a flask through gas lines and a gas manifold (in a research setting, Figure 1.28), or through a balloon of inert gas (more common in teaching labs, Figure 1.29).
Step-By-Step Procedures
The techniques shown in this section use balloons of nitrogen gas to create inert atmospheric conditions in a round bottomed flask, and syringes to transfer liquids from dry reagent bottles. These techniques can be easily adapted to use with a gas manifold if available.
Prepare a balloon of inert gas
1. Prepare a needle attachment for a balloon: Cut the end off a plastic $1 \: \text{mL}$ syringe and fit the barrel into a piece of thick rubber tubing. Attach a helium-quality balloon to the rubber tubing, and seal all joints with Parafilm. Alternatively, attach a balloon directly to a $2$-$3 \: \text{mL}$ plastic syringe.
2. Fill the balloon by connecting to a hose on the regulator of a tank of inert gas (nitrogen or argon, Figure 1.30a). Open the gas regulator to fill the balloon to between 7" – 8" in diameter (Figure 1.30b).
[For use with very sensitive reagents, the gas should first be passed through a column of drying agent.]
3. While holding the balloon close to your body, twist the balloon to prevent gas from escaping. Then attach a green needle (#21 gauge, $0.8 \: \text{mm}$ × $25 \: \text{mm}$, safety note: very sharp!) securely to the end of the syringe (Figure 1.30c).
4. To prevent gas from escaping when the balloon is untwisted, insert the needle into a rubber stopper (Figure 1.30d). The balloon can now be set aside while other parts of the setup are prepared.
Prepare the reagent flask
5. Remove the surface water from a reagent flask (with stir bar if applicable), by either flame drying the flask or placing it in a hot oven for several hours. Safety note: flask will be very hot! Use thick gloves to handle the hot glass.
6. Immediately insert a rubber septum (Figure 1.31a) into the ground glass joint. Fold one side of the septum over the lip of the flask and hold it in place while folding the opposite sides over as well (Figures 1.31b–d). This can be difficult to do with thick gloves. An alternative is to hold the flask against your body with the thick gloves, and fold the septum flaps over while using your bare hands (or thinner gloves, Figures 1.32 a+b).
7. Immediately secure the reaction flask to a ring stand or latticework using an extension clamp and insert the needle of the inert gas balloon into the inner circle on the septum (Figure 1.32c, see Figure 1.31d for the circle on the septum).
8. Insert a single needle into the circle on the septum (called an "exit needle") to "flush" the air from the reaction flask (Figure 1.32d). The goal is to use the pressure from the balloon to force inert gas into the reaction flask and displace the air in the flask out the exit needle.
9. Allow the system to flush for at least 5 minutes if using nitrogen gas and perhaps 1-2 minutes if using argon gas (argon is denser than air so will displace the air more easily than nitrogen). Then remove the exit needle and allow the flask to fully cool under the balloon of inert gas.
10. If a mass is required of the empty flask, remove the inert gas balloon (insert the needle into a rubber stopper) and obtain the mass of the cool, empty flask with septum.
Prepare the syringe for reagent transfer
11. Remove a long, flexible needle from a hot oven and immediately screw it into the barrel of a plastic syringe, freshly opened from its packaging (Figure 1.33a).
The syringe needs to be able to hold a volume larger than the volume of reagent intended to deliver in order to have enough flexibility to properly manipulate the reagent. For example, a $10$-$\text{mL}$ syringe is too small to deliver $10 \: \text{mL}$ of reagent, but could be used to deliver $7 \: \text{mL}$ of reagent.
Hold the syringe such that the volume markings are visible, and connect the bent needle pointed upwards, so that when screwed on (which normally requires roughly a half turn) the bent needle points downwards with the numbers visible. With this approach, the volume markings can be seen while withdrawing liquid, instead of being inconveniently on the back face of the syringe (as in Figure 1.33d).
Glass syringes are often used with air-sensitive reagents dissolved in nonpolar solvents (e.g. hexanes), and require some further considerations that are not described in this section. Consult with your instructor for further instructions if you are to use a glass syringe.
12. Wrap the joint between the needle and syringe with Teflon tape or Parafilm (Figure 1.33b).
13. Flush the needle with inert gas: Insert the needle into the septum of an empty, dry flask attached to a balloon of inert gas (Figure 1.33c), withdraw a full volume of inert gas (Figure 1.33d), then expunge it into the air.
14. Immediately insert the flushed syringe into the reagent flask septum if nearby , or into a rubber stopper until the syringe is to be used.
Withdraw the reagent
15. A balloon of inert gas must be inserted into the reagent bottle in order to equalize pressures during withdrawal of liquid. A platform (e.g. ring clamp/wire mesh) should also be used beneath the reagent bottle if positioned above the bench, to provide support in case the bottle slips from the grasp of the clamp.
16. Insert the needle of the flushed syringe into the septum of the air-sensitive reagent, and into the liquid (Figure 1.34a).
17. Slowly withdraw some liquid into the syringe. If the plunger is pulled back too quickly, the low pressure inside the syringe may cause air to seep through the joint between the needle and syringe (through or around the Teflon tape or Parafilm).
18. Inevitably a bubble will form in the syringe. Keeping the syringe upside down and vertical (Figure 1.34b), push on the plunger to force the gas pocket back into the bottle.
19. Slowly withdraw liquid to $1$-$2 \: \text{mL}$ greater than the desired volume (Figure 1.34c), then keeping the syringe vertical, expunge liquid back to the desired volume (Figure 1.34d shows $2.0 \: \text{mL}$ of liquid).
Withdrawing greater than the desired volume at first allows you to be confident that no gas bubbles are in the needle, and that you have measured an accurate volume.
20. The needle should be full of the air-sensitive reagent at this point, and if it were removed from the bottle the reagent would come into contact with the atmosphere at the needle tip. This can have disastrous consequences if the reagent is quite reactive (smoking or potentially fire). Safety note: It is therefore essential that a "buffer" of inert gas (Figure 1.36) is placed between the air-sensitive reagent and the atmosphere before removing the needle.
21. To create the "inert gas buffer":
a. Place the needle into the headspace of the reagent bottle (Figures 1.35 a+b).
b. Keeping the syringe upside down and vertical, gently pull back on the plunger until a bubble is seen in the barrel (approximately $20\%$ of the syringe capacity, Figure 1.35c).
Immediately insert the syringe into the reaction flask septum if nearby, or into a rubber stopper if the flask is a distance away (Figure 1.35d).
Deliver the reagent
22. With an inert gas balloon inserted in the reaction flask, place the syringe with reagent into the reaction flask septum. Keeping the syringe vertical, push on the plunger to first deliver the inert gas buffer (Figure 1.37a), then slowly deliver reagent to the flask.
23. Stop delivering reagent when the rubber plunger of the syringe meets the end of the barrel (Figure 1.37b). Do not invert the syringe and push out the residual liquid: this would result in delivering a larger volume of reagent than measured by the syringe.
24. The needle will still be full of the air-sensitive reagent, so with the needle tip still in the headspace of the reaction flask, withdraw an inert gas buffer into the syringe. Insert the needle tip into a rubber stopper if the cleaning station is not nearby.
Clean the needle and syringe
25. The syringe and needle should be cleaned as soon as possible, as over time deposits may form in the needle creating a plug. To clean the syringe and needle:
a. Withdraw into the syringe a few $\text{mL}$ of clean solvent similar to the solvent used in the air-sensitive solution (Figure 1.37c). For example, the pictures in this section show transfer of a $\ce{BH_3}$ reagent dissolved in THF. An ideal rinse solvent would then be THF. As THF was not available, diethyl ether was a good substituted as the two solvents are structurally similar (they are both ethers).
b. Expunge the solvent into a waste beaker. Repeat with another solvent rinse, being sure to rinse the entire area in the syringe where the reagent touched.
c. Rinse the syringe once with water to dissolve and remove any inorganic salts.
d. Further rinse the syringe and needle twice with a few $\text{mL}$ of acetone.
e. Remove the needle from the syringe and retain for future use. The plastic syringe should not be reused, but instead thrown away: solvent present in many air-sensitive solutions degrade the rubber plunger on the syringe, causing them to swell and be ineffective after one use.
Inert Atmospheric Methods Summary
Table 1.5: Procedural summary for inert atmospheric methods.
Prepare the balloon
Fill a balloon with an inert gas (nitrogen or argon) to 7-8 inches in diameter.
Twist the balloon to prevent gas from escaping, then attach a needle and insert into a rubber stopper to plug.
Prepare the reaction flask
Flame or oven dry a reaction flask (with stir bar, and fold a rubber septum over the joint while wearing thick gloves.
Clamp the hot flask to the ring stand or latticework and insert the balloon of inert gas.
Insert an "exit needle" and allow the flask to flush for ~5 minutes (to displace the air).
Remove the exit needle and allow the flask to cool.
Prepare the syringe
Obtain a needle from the hot oven and screw it into the tip of a freshly opened plastic syringe.
Wrap the needle/syringe joint with Teflon tape or Parafilm.
Flush the syringe with inert gas by using an empty flask attached to a balloon of inert gas. Withdraw a volume of gas and expunge into the air.
Insert the needle tip into a rubber stopper to plug.
Withdraw the reagent
Use the prepared syringe to slowly withdraw some reagent (or air may enter the syringe).
Push out the gas bubble.
Withdraw reagent to a slightly greater volume than needed, then expunge liquid to the exact volume.
Place the needle in the headspace of the bottle and withdraw an "inert gas buffer", ~$20\%$ of the volume of the syringe.
Insert the needle into a rubber stopper for transport.
Deliver the reagent
Into the reaction flask (with inert gas balloon), deliver first the inert gas buffer, then the air-sensitive reagent.
Don't push out the residual liquid.
Withdraw an inert gas buffer into the mostly empty syringe.
Clean by rinsing with two portions (few $\text{mL}$ each) of solvent (similar to reagent solvent), then one portion of water and two portions of acetone. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.03%3A_Transferring_Methods/1.3C%3A_Transferring_Methods_-_Inert_Atmospheric_Methods.txt |
• 1.4A: Methods and Flammability
As safety is an important factor in making laboratory choices, it's important to consider the flammability of the liquid to be heated. Almost all organic liquids are considered "flammable," meaning they are capable of catching on fire and sustaining combustion (an important exception is that halogenated solvents tend to be non-flammable). However, this doesn't mean that all organic liquids will immediately ignite if placed near a heat source.
• 1.4B: Controlled Boiling
Boiling solutions always have the potential to "bump", where bubbles vigorously erupt from superheated areas of the solution: areas where the temperature is above the boiling point of the solvent, but gas bubbles have not yet formed due to lack of a nucleation site. Bumping can splash hot material out of a flask: onto your hand or onto a hotplate surface where it might start a fire. Bumping is hazardous, not to mention frightening when a bubble unexpectedly erupts.
• 1.4C: Adjustable Platforms
Adjustable platforms come in many forms. A lab jack is the easiest to manipulate, and can be adjusted up or down by turning the knob. Unfortunately, lab jacks are expensive so are likely to be used in research settings but not in teaching labs. A simple platform can be made from anything stackable, such as wood blocks or KimWipe boxes, although at some height they can be easily tipped. A more secure platform can be created by placing a wire mesh atop a ring clamp.
• 1.4D: Bunsen Burners
Bunsen burners are generally used to rapidly heat high-boiling liquids with low flammability (such as water). Burners do have their place in the organic lab. Burners are often used in steam distillation.
• 1.4E: Hotplates
Hotplates are perhaps the most versatile heat source in the laboratory and can be used to heat beakers, Erlenmeyer flasks, and various hot baths (water, sand, and oil baths). They can also be used to develop stained TLC plates.
• 1.4F: Steam Baths
A steam bath is a relatively safe way to heat flammable organic liquids. They are designed to heat beakers, Erlenmeyer flasks, and round-bottomed flasks, and have a series of concentric rings that can be removed to adjust to the size of the flask. Many science buildings have in house steam lines in their labs, allowing for this convenient and safe method to heat various solvents.
• 1.4G: Heating Mantles
Heating mantles are a relatively safe way to heat flammable organic liquids in a round bottomed flask. The mantles are cup-shaped and designed for different sizes of round bottomed flask. If a mantle does not fit a round bottomed flask perfectly, sand can be added to ensure good thermal contact.
• 1.4H: Water, Sand, and Oil Baths
Water, sand, and oil baths are related heat sources as they envelop a flask in a warm material (liquid or sand). A thermometer is often used to monitor the temperature of the bath, and is used to approximate the internal temperature of liquid in a flask (the bath is often slightly hotter than the liquid in the flask).
• 1.4I: Heat Guns
Heat guns are inexpensive tools for delivering strong heat in a more flexible manner than other heating methods. Heat can be directed from every direction, and the gun can be manually waved about in order to dissipate the heating intensity.
• 1.4J: Cooling Baths
On occasion a solution may need to be cooled: to minimize evaporation of volatile liquids, induce crystallization, or to favor a certain reaction mechanism. Several cold baths are used for certain applications, with the simplest being an ice bath. When preparing an ice bath, it is important to use a mixture of ice and water, as an ice-water slurry has better surface contact with a flask than ice alone.
• 1.4K: Reflux
A reflux setup allows for liquid to boil and condense, with the condensed liquid returning to the original flask. A reflux setup is analogous to a distillation, with the main difference being the vertical placement of the condenser. The liquid remains at the boiling point of the solvent (or solution) during active reflux.
1.04: Heating and Cooling Methods
In some contexts, the choice of what heat source to use is critical while in other contexts several could work equally well. The choice of which heat source to use depends on several factors:
• Availability (does your institution own the equipment?)
• Rate of heating (do you want to heat gradually or quickly?)
• Flexibility of heating (does the heat need to be waved around an apparatus?)
• Final temperature required (low boiling liquids require a different approach than high boiling liquids)
• Flammability of the content
Table 1.6: Summary of heating methods.
As safety is an important factor in making laboratory choices, it's important to consider the flammability of the liquid to be heated. Almost all organic liquids are considered "flammable," meaning they are capable of catching on fire and sustaining combustion (an important exception is that halogenated solvents tend to be non-flammable). However, this doesn't mean that all organic liquids will immediately ignite if placed near a heat source. Many liquids require an ignition source (a spark, match, or flame) in order for their vapors to catch on fire, a property often described by the liquid's flash point. The flash point is the temperature where the vapors can be ignited with an ignition source. For example, the flash point of $70\%$ ethanol is $16.6^\text{o} \text{C}$,$^2$ meaning it can catch on fire at room temperature using a match (Figure 1.38). A Bunsen burner is an excellent ignition source (and can reach temperatures of approximately $1500^\text{o} \text{C}$),$^3$ making burners a serious fire hazard with organic liquids, and a heat source that should often be avoided.
Another important property in discussing flammability is a liquid's autoignition temperature: the temperature where the substance spontaneously ignites under normal pressure and without the presence of an ignition source. This property is particularly insightful because it does not require a flame (which is often avoided in the organic lab), but only a hot area. A hotplate surface turned up to "high" can reach temperatures up to $350^\text{o} \text{C}$.$^3$ Safety note: as diethyl ether, pentane, hexane, and low-boiling petroleum ether have autoignition temperatures below this value (Table 1.7), it would be dangerous to boil these solvents on a hotplate as vapors could spill out of the container and ignite upon contact with the surface of the hotplate. In general, caution should be used when using a hotplate for heating any volatile, flammable liquid in an open vessel as it's possible that vapors can overrun the hotplate's ceramic covering and contact the heating element beneath, which may be hotter than $350^\text{o} \text{C}$. It is for this reason that hotplates are not the optimal choice when heating open vessels of volatile organic liquids, although in some cases they may be used cautiously when set to "low" and used in a well-ventilated fume hood.
Table 1.7: Autoignition temperature data for selected solvents.$^4$
Compound Boiling Point $\left( ^\text{o} \text{C} \right)$ Autoignition Temperature $\left( ^\text{o} \text{C} \right)$
Diethyl Ether 34 180
Pentane 36 260
Low boiling Petroleum Ether 30-40 246
Acetone 56 465
Methanol 65 464
Hexane 69 225
Ethyl Acetate 77 426
Carbon Tetrachloride 77 n/a
Ethanol 78 363
Benzene 80 498
As combustion is a reaction in the vapor phase, liquids with low boiling points (< $40^\text{o} \text{C}$) tend to have low flash points and autoignition temperatures as they have significant vapor pressures (Table 1.7). All low boiling liquids should be treated more cautiously than liquids with moderate boiling points (> $60^\text{o} \text{C}$).
$^2$From the SDS (Safety Data Sheets) of $70\%$ denatured ethanol.
$^3$ As reported in the Fischer Scientific catalog.
$^4$ Data from Handbook of Chemistry and Physics, 84$^\text{th}$ ed., CRC Press, 2003-2004, 16-16 to 16-31. Petroleum ether autoignition temperature is from the SDS. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4A%3A_Methods_and_Flammability.txt |
Boiling solutions always have the potential to "bump", where bubbles vigorously erupt from superheated areas of the solution: areas where the temperature is above the boiling point of the solvent, but gas bubbles have not yet formed due to lack of a nucleation site. Bumping can splash hot material out of a flask: onto your hand or onto a hotplate surface where it might start a fire. Bumping is hazardous, not to mention frightening when a bubble unexpectedly erupts. Several methods can be used to prevent bumping and ensure smooth boiling.
Boiling Stones (Boiling Chips)
Boiling stones (or boiling chips) are small pieces of black porous rock (often silicon carbide) that are added to a solvent or solution. They contain trapped air that bubbles out as a liquid is heated, and have high surface area that can act as nucleation sites for formation of solvent bubbles. They should be added to a cool liquid, not one that is near its boiling point, or a vigorous eruption of bubbles may ensue. When a liquid is brought to a boil using boiling stones, the bubbles tend to originate primarily from the stones (Figure 1.39b). Boiling stones cannot be reused, as after one use, their crevices fill with solvent and they can no longer create bubbles.
Boiling stones should not be used when heating concentrated solutions of sulfuric or phosphoric acid, as they may degrade and contaminate the solution. For example, Figure 1.40 shows a Fischer esterification reaction that uses concentrated sulfuric acid. When a stir bar is used for bump prevention, the solution remains colorless (Figure 1.40a). When the same reaction is conducted using a boiling stone, the solution darkens during heating (Figure 1.40b) and eventually turns the entire solution a deep purple-brown color (Figure 1.40c). Besides contaminating the solution, the dark color makes manipulation of the material with a separatory funnel difficult: two layers are present in Figure 1.40d, although it is very difficult to see.
Boiling Sticks (Wood Splints)
"Boiling sticks" (wood splints) are also used to encourage smooth boiling. They are plunged directly into a solvent or solution, and act much the same as boiling stones: they too are highly porous and contain nucleation sites. When a liquid is brought to a boil using a boiling stick, the bubbles tend to originate primarily from the surface of the stick (Figure 1.41b).
When choosing between boiling stones and boiling sticks, the main advantage of boiling stones is that they are small and so will fit in any flask. They also absorb very little compound, unlike boiling sticks. The main advantage of boiling sticks is that they can be easily removed from a solution. This is useful in crystallization, as the stick can be easily removed before crystals form (Figure 1.41 c+d).
Stir Bars and Spin Vanes
Stir bars (stirring bars, or spin vanes in microscale work, Figure 1.42a) are Teflon-coated magnets that can be made to spin with a magnetic stir plate (Figure 1.42b). Stirring is often used with heating as stirring encourages homogeneity, allowing for liquids to more quickly heat or cool, and disrupts superheated areas. In the context of chemical reactions, stirring also increases the rate of mixing (especially for heterogeneous mixtures) and thus increases reaction rates.
A stir bar can be removed from a flask with a magnet, called a "stir bar retriever" (Figure 1.42d). If a solution is to be subsequently poured through a funnel and into a separatory funnel, they are also easily removed by the funnel. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4B%3A_Controlled_Boiling.txt |
It is quite useful for an apparatus to rest on a platform that can be adjusted up or down. For example, when heating a round bottomed flask in a distillation or reflux, the heat source should be held in such a way that it can be easily lowered and removed from the flask. This is an important safety measure, as it allows for adjustment if the system heats too rapidly (as is evidenced by bumping and foaming), or if anything unexpected occurs (smoking or charring). Removal of the heat source is also of course necessary at the end of a process, and it is best if the heat can be removed while leaving the apparatus intact to cool.
Adjustable platforms come in many forms. A lab jack (Figure 1.43a) is the easiest to manipulate, and can be adjusted up or down by turning the knob. Unfortunately, lab jacks are expensive so are likely to be used in research settings but not in teaching labs. A simple platform can be made from anything stackable, such as wood blocks or KimWipe boxes (Figure 1.43b), although at some height they can be easily tipped. A more secure platform can be created by placing a wire mesh atop a ring clamp (Figure 1.43c).
Adjustable platforms should be used underneath any flask that is clamped in an apparatus well above the benchtop and contains chemicals, especially if they are to be heated or are extremely reactive. If a clamp were to fail for some reason, a platform is a fail-safe and prevents hot or reactive chemicals from falling onto a heat source or splashing on the benchtop where they may become a hazard. Figure 1.43d shows withdrawal of an air-sensitive reagent by syringe, and the reagent bottle is secured by a clamp and supported with a ring clamp/wire mesh platform. If the reagent were to slip from the grip of the clamp, the platform ensures that the hazardous material will not fall.
1.4D: Bunsen Burners
Bunsen burners are generally used to rapidly heat high-boiling liquids with low flammability (such as water). Safety note: It is important to know that they can reach temperatures of approximately $1500^\text{o} \text{C}$,$^5$ and can easily ignite most organic compounds. If an apparatus is improperly set up, or if there is a small gap that allows organic vapors to escape from an apparatus, these vapors can ignite with a burner. Therefore, it is generally recommended to use other heat sources to warm flammable organic liquids (for example in distillation or reflux). Bunsen burners should never be used with highly flammable solvents such as diethyl ether.
However, burners do have their place in the organic lab. Burners are often used in steam distillation (Figure 1.44a) as the vapors are generally not flammable. In this context, a wire mesh set atop a ring clamp is often used under the flask to dissipate the heat and avoid overheating one area. Burners are also used in the Beilsten test for halogens (Figure 1.44b), with Thiele tubes in melting and boiling point determinations (Figure 1.44c), and for softening pipettes to create capillary TLC spotters (Figure 1.44d). They may also be used in sublimations.
Burners come in several different forms. The common Bunsen burner is six inches tall and has two models differing in how the gas and air are adjusted (a Bunsen burner is in Figure 1.45a, and a Tirrill burner is in Figure 1.45b). Small burners (microburners, Figure 1.45c) and large burners (Meker burners, Figure 1.45d) are also sometimes used.
To light a burner
1. Connect the rubber tubing on the burner to the gas line on the benchtop.
2. Open the gas valve on the burner one "turn" from closed, by either turning the gold arm on a Tirrill burner (Figure 1.46a) or notched dial near the bottom of a Bunsen burner (Figure 1.46c).
3. Open the air valve slightly so that a small opening is observed in the slats or on the screw portion of the burner (Figure 1.46 b+d).
4. Open the gas valve on the benchtop until a faint hiss of gas can be heard, then use a striker to create a spark and light the burner. If matches are instead used, first light the match and then turn on the gas. If the burner fails to light, there is either too much or too little of either gas or air. Try adjusting both and observe the effect.
5. Once the burner is lit, adjust the gas and air until a blue triangular flame appears (a "blue cone", Figure 1.47d). The flame should be 1-2 inches high and accompanied with an audible hissing of the flame. An orange flame (Figure 1.47b) forms when there is incomplete combustion of the fuel, is cooler than a blue flame, and if used to heat glassware will deposit black charcoal onto the glass. To convert an orange flame into a blue conic flame, allow more air into the burner. The tip of the blue cone is the hottest part of the flame.
$^5$As reported in the Fischer Scientific catalog. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4C%3A_Adjustable_Platforms.txt |
Hotplates are perhaps the most versatile heat source in the laboratory (Figure 1.48) and can be used to heat beakers, Erlenmeyer flasks, and various hot baths (water, sand, and oil baths). They can also be used to develop stained TLC plates.
Hotplates work by passing electricity through a heating element covered by a ceramic top. Safety note: the hotplate surface can reach temperatures up to $350^\text{o} \text{C}$,$^6$ which is hot enough to ignite many low-boiling solvents. Diethyl ether, pentane, hexane, low-boiling petroleum ether, and acetone should therefore never be heated in an open vessel with a hotplate.
Caution should be used when heating any flammable organic liquid in an open vessel on a hotplate, as organic vapors may spill out of containers and ignite upon contact with the heating element, which may be hotter than the ceramic surface. This can be especially true if the hotplate is turned to "high". Therefore, a low setting must be used when cautiously heating certain flammable organic liquids (e.g. ethanol) with a hotplate.
$^6$As reported in the Fischer Scientific catalog.
1.4F: Steam Baths
A steam bath (Figure 1.49) is a relatively safe way to heat flammable organic liquids. They are designed to heat beakers, Erlenmeyer flasks, and round-bottomed flasks, and have a series of concentric rings that can be removed to adjust to the size of the flask. Many science buildings have in house steam lines in their labs, allowing for this convenient and safe method to heat various solvents.
The steam line should be connected to the upper arm of the steam bath, and condensation should be allowed to drain from the lower arm of the bath to the sink (Figure 1.50a). The steam tap should be adjusted so that a moderate amount of steam can be seen coming only from the central opening in the bath, then a flask should be set atop the opening to warm the flask (Figure 1.50b). Steam should only warm the bottom of the flask, and steam should not be visible coming out anywhere else in the bath. When warming highly volatile solvents, the flask may need to be held above the opening of the bath to control the rate of heating (Figure 1.50c), or the steam rate lowered.
1.4G: Heating Mantles
Heating mantles are a relatively safe way to heat flammable organic liquids in a round bottomed flask (Figure 1.52). The mantles are cup-shaped and designed for different sizes of round bottomed flask (Figure 1.51a). If a mantle does not fit a round bottomed flask perfectly, sand can be added to ensure good thermal contact (Figure 1.51c).
The mantles should never be connected directly to the outlet, but first to a "Variac" (blue piece of equipment in Figure 1.51b) which then connects to the outlet and delivers variable voltage to the mantle. A Variac set to "100" would be equivalent to plugging the mantle directly into the wall $\left( 100\% \right)$, while a setting of "50" means the delivered voltage is halved $\left( 50\% \right)$. By controlling the delivered voltage, Variacs are used to regulate the temperature of a heating mantle. There is variation between devices, and settings must be experimented with to determine appropriate heating rates.
Heating mantles take some time to warm up (so may be pre-heated during setup of an apparatus), and also take some time to cool down. The mantle will remain warm even after turning off the Variac, and therefore flasks have to be removed from the mantle in order to cool (Figure 1.52c).
Safety note: the main hazard with heating mantles is that flammable organic liquids spilled on the surface of a hot mantle do have the possibility of ignition. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4E%3A_Hotplates.txt |
Water, sand, and oil baths are related heat sources as they envelop a flask in a warm material (liquid or sand). A thermometer is often used to monitor the temperature of the bath, and is used to approximate the internal temperature of liquid in a flask (the bath is often slightly hotter than the liquid in the flask).
Water baths, heated on a hotplate, are most commonly used to heat solutions to $100^\text{o} \text{C}$ (boiling baths, Figures 1.53 + 1.54a). They may also be used to heat to lower temperatures, although it can be difficult to maintain a constant temperature. Water baths can be covered with aluminum foil to prevent excessive evaporation, or to prevent excess moisture from entering open vessels. Cold water baths can also be used to cool apparatuses in a quick manner (Figure 1.54b).
Sand baths, also heated on a hotplate, can be used to heat solutions to a wide variety of temperatures, including very high temperatures (> $250^\text{o} \text{C}$). Sand can be placed inside Pyrex crystallizing dishes (as is used with the water bath in Figure 1.54b), although Pyrex may crack if heated at too fast a rate. Sand can also be heated to moderate temperatures inside thin aluminum foil pie dishes, but should not be used at high temperatures as aluminum will then oxidize, causing the thin foil to disintegrate. Thick metal pie tins (Figure 1.54c) are indestructible alternatives when high temperatures are required (but may interfere with the stirring mechanism if used).
A vessel should be buried in a sand bath as much as possible as the surface is often much cooler than the sand below. A scoopula or metal spatula can be used to pile the sand up to at least the height of liquid inside the flask. Sand takes a long time to heat up, and a long time to cool down. To save time, a sand bath may be preheated while an apparatus is assembled as long as it is preheated a distance away from volatile organic liquids.
If the sand overheats and causes a liquid to boil uncontrollably, the flask can be partially lifted out of the sand, or the sand moved with a metal spatula away from contact with the liquid. Sand will remain warm even after turning off the hotplate, and therefore flasks have to be lifted out of the sand bath in order to cool (Figure 1.54d). Hot sand baths can be cooled atop a ceramic tile.
Oil baths are much like water baths, but use silicone or mineral oils in order to enable temperatures hotter than the boiling point of water (> $100^\text{o} \text{C}$). Silicone oil baths can be heated to $250^\text{o} \text{C}$, while mineral oil baths can be heated to $300^\text{o} \text{C}$.$^7$ Mineral oil is composed of mixtures of long-chain alkanes, and so is combustible. Direct contact with open flames should therefore be avoided.
Oil baths can be heated in a Pyrex crystallizing dish atop a hotplate. It is also quite common for the oil to be electrically heated, through immersion of a coiled wire connected to a "Variac" (light blue piece of equipment in Figure 1.55a). A Variac connects to the outlet and can deliver variable voltage through the wire. A Variac set to "100" would be equivalent to plugging the system directly into the wall $\left( 100\% \right)$, while a setting of "50" means the delivered voltage is halved $\left( 50\% \right)$. By controlling the delivered voltage, Variacs are used to regulate the temperature of the oil. If you have previous experience using a Variac with heating mantles, the settings will not translate to the oil bath as oil bath wires heat more rapidly than a heating mantle's wires. A paper clip can also be used in an oil bath (Figure 1.55c) and stirred with a stirring plate in order to dissipate heat. This allows for the temperature of an oil bath to quickly respond to adjustments up or down.
$^7$J. A. Dean, Lange's Handbook of Chemistry, 15$^\text{th}$ ed., McGraw-Hill, 1999, Sect. 11.3. Additionally, the boiling point of mineral oil is $310^\text{o} \text{C}$. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4H%3A_Water_Sand_and_Oil_Baths.txt |
Heat guns are inexpensive tools for delivering strong heat in a more flexible manner than other heating methods. Heat can be directed from every direction, and the gun can be manually waved about in order to dissipate the heating intensity. Heat guns are commonly used to quickly develop stained TLC plates (Figures 1.56a+b), and result in more even heating and less charring than when using a hotplate. They are also ideal for sublimations (Figure 1.56c), as the heat can be directed to the sides of the flask to coax off crystals deposited on the sides. A disadvantage of using heat guns is that they must be continually held, which makes them most ideal for short processes.
Safety note: A heat gun is not simply a hair dryer, and the nozzle gets quite hot (temperatures can be between $150$-$450^\text{o} \text{C}$)!$^8$ Care should be taken to not touch the nozzle after use, and the gun should be set down carefully, as it may mark the benchtop or cord.
$^8$As reported in the Fischer Scientific catalog.
1.4J: Cooling Baths
On occasion a solution may need to be cooled: to minimize evaporation of volatile liquids, induce crystallization, or to favor a certain reaction mechanism. Several cold baths are used for certain applications, with the simplest being an ice bath. When preparing an ice bath, it is important to use a mixture of ice and water, as an ice-water slurry has better surface contact with a flask than ice alone.
Sometimes salts are added to ice in order to create baths colder than $0^\text{o} \text{C}$ (freezing point depression). It is also quite common to cool a solvent with dry ice (solid $\ce{CO_2}$) in order to achieve dramatically colder baths. In dry ice baths, dry ice is added to the solvent until a portion of dry ice remains. The most common dry ice bath is made with acetone and dry ice, and achieves a cold bath of $-78^\text{o} \text{C}$. Other cold baths are shown in Table 1.8.
Table 1.8: Commonly used cold baths.$^9$
Cold Bath Composition T $\left( ^\text{o} \text{C} \right)$
$36 \: \text{g}$ $\ce{NaCl}$ + $100 \: \text{g}$ ice -10.0
$75 \: \text{g}$ $\ce{NaNO3}$ + $100 \: \text{g}$ ice -5.3
$66 \: \text{g}$ $\ce{NaBr}$ + $100 \: \text{g}$ ice -28
Acetonitrile + dry ice -40
Acetone + dry ice -78
Ethyl acetate + dry ice -84
Hexane + dry ice -95
Methanol + dry ice -98
Ice baths can be made in Tupperware containers, beakers, or almost any container (Figure 1.57a). Baths colder than $-10^\text{o} \text{C}$ should be made in an insulating container or else they cannot be easily handled and will lose heat too quickly to the room. The most common cold bath container is a wide-mouthed Dewar (Figure 1.57b), which has a vacuum jacket between the bath and the exterior. An inexpensive homemade insulating bath can be made by nesting two crystallizing dishes, filling the gap with vermiculite (a packing material) and sealing the gap with silicone caulking (Figure 1.57c).
Safety note: Dry ice should not be handled with your bare hands or else you may get frostbite. Also avoid direct contact with baths colder than $-10^\text{o} \text{C}$.
$^9$J. A. Dean, Lange's Handbook of Chemistry, 15$^\text{th}$ ed., McGraw-Hill, 1999, Sect 11.1. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4I%3A_Heat_Guns.txt |
Overview of Reflux
A reflux setup (Figure 1.58) allows for liquid to boil and condense, with the condensed liquid returning to the original flask. A reflux setup is analogous to a distillation, with the main difference being the vertical placement of the condenser. The liquid remains at the boiling point of the solvent (or solution) during active reflux.
A reflux apparatus allows for facile heating of a solution, but without the loss of solvent that would result from heating in an open vessel. In a reflux setup, solvent vapors are trapped by the condenser, and the concentration of reactants remains constant throughout the process.
The main purpose of refluxing a solution is to heat a solution in a controlled manner at a constant temperature. For example, imagine that you want to heat a solution to $60^\text{o} \text{C}$ for one hour in order to conduct a chemical reaction. It would be difficult to maintain a warm water bath at $60^\text{o} \text{C}$ without special equipment, and it would require regular monitoring. However, if methanol was the solvent, the solution could be heated to reflux, and it would maintain its temperature without regular maintenance at the boiling point of methanol $\left( 65^\text{o} \text{C} \right)$. True, $65^\text{o} \text{C}$ is not $60^\text{o} \text{C}$ and if the specific temperature were crucial to the reaction, then specialized heating equipment would be necessary. But often the boiling point of the solvent is chosen as the reaction temperature because of its practicality.
Step-by-Step Procedures
1. Pour the solution to be refluxed into a round bottomed flask, and clamp it to the ring stand or latticework with an extension clamp (Figure 1.59a). The flask should be no more than half full.
2. Add a stir bar or few boiling stones for bump prevention. Boiling stones should not be used when refluxing concentrated solutions of sulfuric or phosphoric acid, as they will colorize the solution. For example, when a stir bar is used for bump prevention with concentrated sulfuric acid, the solution remains colorless (Figure 1.59b). When the same reaction is conducted using a boiling stone, the solution darkens during heating (Figure 1.59c) and eventually turns the entire solution a deep purple-brown color (Figure 1.59d).
1. Place rubber hoses on a condenser (wet the ends first to allow them to slide on), then attach the condenser vertically to the round bottomed flask. If using a tall condenser, clamp the condenser to the ring stand or latticework (Figure 1.60a). Be sure the condenser fits snugly into the flask. Safety note: if the pieces are not properly connected and flammable vapors escape, they may be ignited by the heat source. Do not connect the round bottomed flask and condenser with a plastic clip, as shown in Figure 1.60c. Plastic clips can sometimes fail (especially when they are heated), and this setup does not allow for the flask to be reliably removed from the heat source at the end of the reflux.
2. Connect the hose on the lower arm of the condenser to the water faucet and allow the hose on the upper arm to drain to the sink (Figure 1.60b). It is important that water goes in the bottom of the condenser and out the top (so water flows against gravity) or else the condenser will be ineffective as it will not fill completely.
1. If multiple solutions will be refluxed at the same time (e.g. if many students are performing a reflux side by side), the hoses from each reflux setup can be connected in series (Figure 1.62). To accomplish this, the upper arm of "Setup A" which normally drains to the sink is instead connected to the lower arm of "Setup B." The upper arm of Setup B then drains to the sink. Connecting apparatuses in series minimizes the use of water, as water exiting one condenser enters the next. Several reflux setups can be connected in series, and the water flow should be monitored to ensure that all setups are adequately cooled.
Figure 1.62: Connecting reflux condensers in series.
1. Begin circulating a steady stream of water through the hoses (not so strong that the hose flops around from the high water pressure). Check again that the pieces of glassware securely fit together, then position the heat source under the flask. Turn on the stirring plate if using a stir bar.
1. If using a heating mantle, hold it in place with an adjustable platform (e.g. a wire mesh / ring clamp). Allow a few inches below the mantle so when the reaction is complete, the mantle can be lowered and the flask cooled. If the heating mantle is not a perfect fit for the size of the round bottomed flask, surround the flask with sand to create a better contact (Figure 1.63a).
2. If using a sand bath, bury the flask in the sand so that sand is at least as high as the level of liquid in the flask (Figure 1.63b).
3. If the setup will eventually be left unattended for a long period of time (e.g. overnight), tighten copper wire over the hose attachments to the condenser to prevent changes in water pressure from causing them to pop off.
1. If the heat source was preheated (optional), the solution should begin to boil within five minutes. If it does not, increase the rate of heating. The appropriate heating rate occurs when the solution is vigorously boiling and a "reflux ring" is seen roughly one-third of the way up the condenser. A "reflux ring" is the upper limit of where hot vapors are actively condensing. With some solutions (e.g. aqueous solution), the reflux ring is obvious with easily visible droplets in the condenser (Figures 1.64 a+b). With other solutions (e.g. many organic solvents) the reflux ring is subtler, but can be seen with close observation (Figure 1.64c). Subtle movement may be seen in the condenser as liquid drips down the sides of the condenser, or background objects may appear distorted from refraction of light through the condensing liquid (in Figure 1.64d the ring stand pole is distorted).
2. If following a procedure in which you are to reflux for a certain time period (e.g. "reflux for one hour"), the time period should begin when the solution is not just boiling but actively refluxing in the bottom third of the condenser.
3. The heat should be turned down if the reflux ring climbs to half-way up the condenser or higher, or else vapors could escape the flask.
4. After the reflux is complete, turn off the heat source and remove the flask from the heat by either lifting the reflux apparatus up, or dropping the heat source down (Figure 1.65a).
Do not turn off the water flowing through the condenser until the solution is only warm to the touch. After a few minutes of air cooling, the round bottomed flask can be immersed in a tap water bath to accelerate the cooling process (Figure 1.65b).
Summary
Pour liquid into the flask along with a stir bar or boiling stones.
Use an extension clamp on the round bottomed flask to connect to the ring stand or latticework.
Attach the condenser, and connect the hoses so that water travels against gravity (cooling water comes into the bottom and drains out the top).
Be sure there is a secure connection between the round bottomed flask and condenser, as vapors escaping this joint have the potential to catch on fire.
Circulate water through the condenser, then begin heating the flask (by using a heating mantle, sand, water, or oil bath).
Use an adjustable platform so the heat can be lowered and removed at the end of the reflux, or if something unexpected occurs.
Heat so that the "reflux ring" is seen in the lower third of the condenser.
Turn down the heat if the refluxing vapors reach higher than halfway up the condenser.
At the end of the reflux period, lower the heat source from the flask or raise the apparatus.
Keep circulating water in the condenser until the flask is just warm to the touch.
After air cooling somewhat, the flask can be quickly cooled by immersing in a container of tap water.
Table 1.9: Procedural summary for reflux. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4K%3A_Reflux.txt |
There are many methods used to separate a mixture containing a solid and liquid. If the solid settles well, the liquid can sometimes be poured off (decanted). If the solid has very small sized particles or forms a cloudy mixture, the mixture can sometimes be centrifuged or passed through a filter pipette (on the microscale, < 5 mL).
• 1.5A: Overview of Methods
The most common methods of solid-liquid separation in the organic lab are gravity and suction filtration. Gravity filtration refers to pouring a solid-liquid mixture through a funnel containing a filter paper, allowing the liquid to seep through while trapping the solid on the paper. Suction filtration is a similar process with the difference being the application of a vacuum beneath the funnel in order to pull liquid through the filter paper with suction.
• 1.5B: Decanting
When there is a need to separate a solid-liquid mixture, on occasion it is possible to pour off the liquid while leaving the solid behind. This process is called decanting, and is the simplest separation method.
• 1.5C: Gravity Filtration
Gravity filtration is generally used when the filtrate (liquid that has passed through the filter paper) will be retained, while the solid on the filter paper will be discarded.
• 1.5D: Suction Filtration
Suction filtration (vacuum filtration) is the standard technique used for separating a solid-liquid mixture when the goal is to retain the solid (for example in crystallization). Similar to gravity filtration, a solid-liquid mixture is poured onto a filter paper, with the main difference being that the process is aided by suction beneath the funnel.
• 1.5E: Hot Filtration
A hot filtration is generally used in some crystallization, when a solid contains impurities that are insoluble in the crystallization solvent. It is also necessary in crystallization when charcoal is used to remove highly colored impurities from a solid, as charcoal is so fine that it cannot be removed by decanting.
• 1.5F: Pipette Filtration (Microscale)
For the separation of small volumes (< 10mL ) of solid-liquid mixtures, pipette filters are ideal as filter papers absorb a significant amount of material. Pipette filtration may also be used if small amounts of solid are noticed in NMR or GC samples, as both instruments require analysis of liquids without suspended solids.
• 1.5G: Centrifugation
Centrifugation is used for the separation of solid-liquid mixtures that are stubborn to settle or difficult to otherwise filter. It uses centrifugal force by rapidly spinning samples so that the solid is forced to the bottom of the tube. In this section is shown centrifugation of a suspension of yellow lead(II) iodide in water (Figure 1.90b).
1.05: Filtering Methods
Overview of Filtering Methods
There are many methods used to separate a mixture containing a solid and liquid. If the solid settles well, the liquid can sometimes be poured off (decanted). If the solid has very small sized particles or forms a cloudy mixture, the mixture can sometimes be centrifuged or passed through a filter pipette (on the microscale, < $5 \: \text{mL}$).
The most common methods of solid-liquid separation in the organic lab are gravity and suction filtration. Gravity filtration refers to pouring a solid-liquid mixture through a funnel containing a filter paper, allowing the liquid to seep through while trapping the solid on the paper (Figure 1.66a). Suction filtration is a similar process with the difference being the application of a vacuum beneath the funnel in order to pull liquid through the filter paper with suction (Figure 1.66b).
Gravity and suction filtration have pros and cons, but what helps decide which method to use is generally whether the solid or filtrate is to be retained. The "filtrate" refers to the liquid that has passed through a filter paper (as indicated in Figure 1.66a). Gravity filtration is typically used when the filtrate is retained, while suction filtration is used when the solid is retained.
Gravity filtration is preferred when the filtrate is retained as suction has the potential of pulling small solid particles through the filter paper pores, potentially producing a filtrate contaminated with the solid compound. Suction filtration is preferred when the solid is retained as gravity filtration is much less efficient at removing residual liquid from the solid on the filter paper.
1.5B: Decanting
When there is a need to separate a solid-liquid mixture, on occasion it is possible to pour off the liquid while leaving the solid behind. This process is called decanting, and is the simplest separation method. Decanting is often used to remove hydrated sodium sulfate (\(\ce{Na2SO4}\)) from an organic solution. The sodium sulfate often clings to the glassware (Figure 1.67a), enabling the liquid to be poured off (Figure 1.67b). If liquid is to be poured into a small vessel, a funnel could be used or liquid poured down a glass stirring rod to direct the flow (Figure 1.67c). Unfortunately, there are many mixtures that do not decant well.
1.5C: Gravity Filtration
When there is a need to separate a solid-liquid mixture, it is common that the particles are so fine that they swirl and disperse when the flask is tilted. These mixtures cannot be decanted, and an alternative method is gravity filtration. Gravity filtration is generally used when the filtrate (liquid that has passed through the filter paper) will be retained, while the solid on the filter paper will be discarded.
A common use for gravity filtration is for separating anhydrous magnesium sulfate (\(\ce{MgSO4}\)) from an organic solution that it has dried (Figure 1.68b). Anhydrous magnesium sulfate is powdery, and with swirling in an organic solvent creates a fine dispersal of particles like a snow globe.
To gravity filter a mixture, pour the mixture through a quadrant-folded filter paper (Figure 1.69) or fluted filter paper in a funnel and allow the liquid to filter using only the force of gravity (Figure 1.68c). It is best to pour as if attempting to decant, meaning to keep the solid settled in the flask for as long as possible. When solid begins to pour onto the filter paper, it has the possibility of clogging the filter paper pores or slowing filtration. After finished pouring, rinse the solid on the filter paper (and in the flask) with a few portions of fresh solvent to remove residual compound adhering to the solid. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods/1.5A%3A_Overview_of_Methods.txt |
Suction Filtration Overview
Suction filtration (vacuum filtration) is the standard technique used for separating a solid-liquid mixture when the goal is to retain the solid (for example in crystallization). Similar to gravity filtration, a solid-liquid mixture is poured onto a filter paper, with the main difference being that the process is aided by suction beneath the funnel (Figures 1.70 + 1.71).
The process has advantages and disadvantages in comparison to gravity filtration.
Advantages: 1) Suction filtration is much faster than gravity filtration, often taking less than one minute with good seals and a good vacuum source. 2) Suction filtration is more efficient at removing residual liquid, leading to a purer solid. This is especially important in crystallization, as the liquid may contain soluble impurities which could adsorb back onto the solid surface when the solvent evaporates.
Disadvantages: The force of suction may draw fine crystals through the filter paper pores, leading to a quantity of material that cannot be recovered from the filter paper, and possibly an additional quantity that is lost in the filtrate. This method therefore works best with large crystals. On small scales, the loss of material to the filter paper and filtrate is significant, and so other methods are recommended for microscale work.
Rinsing
As the goal of suction filtration is to fully separate a solid from its surrounding liquid, rinsing the solid is necessary if the liquid cannot easily evaporate. In the case of crystallization, the liquid may contain impurities that can reincorporate into the solid if not removed.
To rinse a suction-filtered solid, the vacuum is removed and a small portion of cold solvent is poured over the solid (the "filter cake"). In the case of crystallization, the same solvent from the crystallization is used. The solid is then delicately slushed around in the solvent with a glass rod, and the vacuum is reapplied to remove the rinse solvent.
To demonstrate the importance of a rinse, Figure 1.72 shows the recovery of a white solid from a yellow liquid using suction filtration. The yellow liquid seemed to be somewhat retained by the solid, as the first crystals collected had a yellow tint (Figure 1.72b). However, rinsing with a few portions of cold solvent were effective at removing the yellow liquid (Figure 1.72d), which could have been reincorporated into the solid without the rinse.
Water Aspirator
A vacuum source is necessary for suction filtration (and vacuum distillation). Although many science buildings come equipped with a house vacuum system (Figure 1.73a), solvents evaporating from a suction filter flask over time can degrade the oil pumps used in a house vacuum. Therefore, it is recommended to instead connect a suction flask to a water aspirator.
A water aspirator is an inexpensive attachment to a water spigot, and the nub on the aspirator connects with tubing to the vessel to be evacuated (Figure 1.73b). As water flows through the faucet and the aspirator, suction is created in the flask.
A water aspirator creates suction through the Bernoulli Principle (technically, the Venturi Effect, for liquids). Water coming from the faucet is constricted inside the aspirator (Figure 1.73c). As the water flow must be the same going into the aspirator as it is going out, the water speed must increase in the constricted area in the direction of flow. A similar phenomenon can be seen in creeks and rivers where the water flows the fastest at the narrowest portions of streams. When the water increases its velocity in the direction of the water flow, conservation of energy dictates that its velocity in perpendicular directions must decrease. The result is a lowered pressure adjacent to the fast-moving liquid. In other words, the gain in velocity of the constricted liquid is balanced by a reduction in pressure on the surrounding material (the gas).
For this reason, the speed at which the water flows through the faucet is correlated with the amount of suction experienced in the connected flask. A strong flow of water will have the fastest speeds through the aspirator and the greatest reduction in pressure.
Step-by-Step Procedures
Assemble the suction filtration flask
1. Clamp a side-arm Erlenmeyer flask to a ring stand or latticework and attach a thick-walled rubber hose to its side arm. Connect this thick tubing to a "vacuum trap" (Figure 1.74) and then to the water aspirator. It is best to not bend or strain the tubing as much as is practical, as this may cause poor suction.
A vacuum trap is necessary when connecting apparatuses to a vacuum source as changes in pressure can cause back-suction. When using a water aspirator, back-suction might cause water from the sink to be pulled into the vacuum line and flask (ruining the filtrate), or the filtrate to be pulled into the water stream (contaminating the water supply).
1. Place a rubber sleeve (or filter adapter) and Buchner funnel atop the side-arm Erlenmeyer flask (Figure 1.75a). Alternatively use a Hirsch funnel for small scales (Figure 1.75d).
2. Obtain a filter paper that will fit perfectly into the Buchner or Hirsch funnel. Filter papers are not completely flat and have a subtle arc to their shape (Figure 1.75b). Place the filter paper inside the funnel concave side down (Figure 1.75b+c). The paper should cover all the holes in the funnel, and with the paper arching downward (Figure 1.76a), solid will be less likely to creep around the edges.
1. Turn on the faucet connected to the water aspirator to create a strong flow of water (the degree of suction is related to the water flow). Wet the filter paper with cold solvent (using the same solvent used in crystallization, if applicable, Figure 1.76b).
2. Suction should drain the liquid and hold the moist filter paper snugly over the holes in the filter. If the solvent does not drain or suction is not occurring, you may need to press down on the funnel (Figure 1.76c) to create a good seal between the glass and rubber sleeve.
Lack of suction may also be from a faulty aspirator or a leak in the system: to test for suction remove the tubing from the suction flask and place your finger over the end (Figure 1.76d).
Filter and Rinse the Mixture
1. Swirl the mixture to be filtered in order to dislodge solid from the sides of the flask. If the solid is very thick, use a spatula or stirring rod to free it from the glass (Figure 1.77a).
In the context of crystallization, the flask will have previously been in an ice bath. Use a paper towel to dry water residue from the outside of the flask so water does not accidentally pour onto the solid.
2. With a quick motion, swirl and dump the solid into the funnel in portions (Figure 1.77b). If the solid is very thick, scoop it out of the flask onto the filter paper (Figure 1.77c).
It's best if the solid can be directed toward the middle of the filter paper, as solid near the edges may creep around the filter paper.
3. A small amount of chilled solvent ($1$-$2 \: \text{mL}$ for macroscale work) can be used to help rinse any residual solid from the flask into the funnel (Figure 1.77d). In crystallization, it is not wise to use an excessive amount of solvent as it will decrease the yield by dissolving small amounts of crystals. Again, press on the funnel to create a good seal and efficient drainage if necessary.
1. Rinse the solid on the filter paper to remove contaminants that may remain in the residual liquid.
1. Break the vacuum on the flask by opening the pinch clamp at the vacuum trap (Figure 1.78a) or by removing the rubber tubing on the filter flask. If adjusting the pinch clamp, you will know the system is open when there is an increase in water flow by the faucet. Then turn off the water on the aspirator. It is always important to open the system to the atmosphere before turning off the aspirator in order to prevent back-suction.
2. Add $1$-$2 \: \text{mL}$ of cold solvent (Figure 1.78b). Use a glass stirring rod to break up any solid chunks and distributed the solvent to all portions of the solid (Figure 1.78c), taking care to not rip or dislodge the filter paper.
3. Reapply the vacuum to the flask, and dry the solid with suction for a few minutes.
2. After filtration is complete, again open the flask to the atmosphere by releasing the pinch clamp or opening it elsewhere, and turn off the water connected to the aspirator.
1. Transfer the solid, filter paper and all, to a pre-weighed watch glass using a spatula (Figures 1.79a+b). The filter cake shouldn't be mushy, and if it is, the liquid was not adequately removed (try a different aspirator and repeat suction filtration).
2. Allow the solid to dry overnight if possible before recording a final mass or melting point. The solid will flake off the filter paper more easily when completely dry (Figure 1.79c).
3. If pressed for time, a solid can be quickly dried in the following ways:
• If the solid is wet with water, it can be placed in a $110^\text{o} \text{C}$ oven (if the melting point is not below this temperature). If the solid is wet with organic solvent, it should never be placed in an oven as it may ignite.
• If the solid is wet with organic solvent, it can be pressed between fresh pieces of filter paper (multiple times if needed) to quickly dry them. Inevitably some solid will be lost on the filter paper.
Suction Filtration Summary
Clamp a side-armed Erlenmeyer flask.
Connect thick-walled hosing from the side arm to a vacuum trap and the water aspirator.
Place a vacuum sleeve on the Buchner (or Hirsch) funnel, then filter paper on the funnel so it arches downward.
Turn on the aspirator.
Add a few $\text{mL}$ of the same solvent used in the flask to wet the filter. The solvent should drain with suction.
Swirl the mixture to be filtered to dislodge the solid from the sides of the flask.
With a quick motion, pour the slurry into the funnel in portions.
In some applications, (e.g. crystallization), rinse with solvent:
• Open the apparatus to the atmosphere, then turn off the aspirator.
• Add a few $\text{mL}$ of cold solvent to the filter paper.
• Delicately swirl the solid in the solvent with a glass rod.
Apply suction again for a few minutes (repeat the rinse step if necessary).
Dry the solid on a watch glass along with the filter paper, overnight if possible. The solid will flake off the paper when dried.
Table 1.10: Procedural summary for suction filtration. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods/1.5D%3A_Suction_Filtration.txt |
Hot Filtration Overview
A hot filtration is generally used in some crystallization, when a solid contains impurities that are insoluble in the crystallization solvent. It is also necessary in crystallization when charcoal is used to remove highly colored impurities from a solid, as charcoal is so fine that it cannot be removed by decanting.
A hot filtration is performed by first pouring a few $\text{mL}$ of solvent through a funnel containing a "fluted filter paper". A fluted filter paper has many indentations and high surface area, which allows for a fast filtration. The funnel is allowed to get hot, while the mixture to be filtered is brought to a boil. The boiling mixture is then poured through the filter paper in portions (Figures 1.81b+d).
It is best to use a ring clamp to secure the filtration funnel, although the funnel could also be simply placed atop the flask. If not using a ring clamp, it is recommended to place a bent paper clip between the flask and funnel to allow for displaced air to escape the bottom flask as liquid drains (Figure 1.81c+d). Without a ring clamp, the setup is more prone to tipping and so using a ring clamp is considerably safer.
A hot filtration is used for filtering solutions that will crystallize when allowed to cool. It is therefore important that the funnel is kept hot during filtration through contact with hot solvent vapors, or crystals may prematurely form on the filter paper or in the stem of the funnel (Figure 1.82).
Crystallization on the filter paper can clog the setup and cause a loss of yield (as the filter paper will be later thrown away). Crystallization in the stem hinders filtration, and can act as a plug on the bottom of the funnel. An advantage of hot filtration is that the boiling solvent in the filter flask helps to dissolve crystals that prematurely form in the stem of the funnel. With hot filtration, it is advised to use a short-stemmed funnel (Figure 1.83a) or stemless funnel if available, instead of a long-stemmed funnel (Figure 1.83b), as material is less likely to crystallize in a short or absent stem.
As it is essential that a solution filters quickly before it has a chance to cool off in the funnel, a "fluted filter paper" (Figure 1.84b+c) is commonly used instead of the quadrant-folded filter paper sometimes used with gravity filtration (Figure 1.84a). The greater number of bends on the fluted filter paper translate into increased surface area and quicker filtration. The folds also create space between the filter paper and glass funnel, allowing for displaced air to more easily exit the flask as liquid drains.
Step-by-Step Procedures
Hot filtration is often used with crystallization, and this procedure should be inserted after the dissolution step, but before setting aside the solution to slowly cool.
Prepare the Filtration Setup
1. Obtain a stemless or short-stemmed funnel (Figure 1.85a), and insert it into a ring clamp, attached to a ring stand or latticework (or alternatively, obtain a bent paper clip for the purpose shown in Figure 1.85b).
2. Flute a filter paper of the correct size for your funnel into an accordion shape (instructions are in Figure 1.86 and the resulting accordion is in Figure 1.85c). When placed in the funnel, the paper should not be shorter than the top of the funnel, or the solution might slip past the filter paper when poured.
1. With a clean Erlenmeyer flask of the correct size for the crystallization beneath the funnel and on the heat source, pour a few $\text{mL}$ of hot solvent into the funnel (Figure 1.85d).
1. If using a ring clamp, adjust the clamp so that there is a small gap between the mouth of the Erlenmeyer and bottom of the funnel: this allows for air to be displaced when liquid flows into the flask. If the gap is too large, hot vapors will escape without heating the funnel.
2. If not using a ring clamp, place a bent paper clip between the flask and funnel (Figure 1.85b).
2. Allow the solvent to boil and get the entire setup hot. If using charcoal, insert that procedure now.
Filter the Solution in Portions
1. When the filter flask is quite hot, and the solution to be filtered is boiling, pour the boiling mixture into the filter funnel in portions. Touch the flask to the filter paper in the funnel as you pour (Figure 1.87a).
2. Safety note: the flask may be quite hot, and hot vapors may scald your hand as you pour (pour sideways so your hand is not above the funnel). If the flask is too hot to hold with your hands, use a "paper towel holder" to hold the flask (Figure 1.87b):
1. Fold a section of paper towel over several times such that the resulting strip is roughly one inch wide. If desired, secure the strip together using a few pieces of tape.
2. When holding a flask, the paper towel holder should be below the lip of the flask. In this way, liquid will not wick toward the paper towel when pouring (towel remains dry in Figure 1.87a), but wet with the too wide towel in Figure 1.87c).
1. When not pouring the mixture to be filtered, return the flask to the heat source (Figure 1.88a).
2. When the mixture is completely filtered, set the empty flask on the benchtop (safety note: do not heat an empty flask, or it may crack). Inspect the funnel: if crystals are seen on the filter paper (as in Figure 1.88b), rinse with a few $\text{mL}$ of boiling solvent to dissolve them. A rinse is not needed in Figure 1.88c.
3. Inspect the filtrate (the liquid that has gone through the filter paper). If charcoal was used and the filtrate is grey, or you can see fine black particles, then charcoal passed through the filter paper either through a hole or by using the wrong filter mesh size. If classmates do not have grey in their solutions, it was likely a hole. Repeat the hot filtration step with a new filter paper and flask.
Hot Filtration Summary
Prepare a fluted filter paper and stemless or short-stemmed funnel clamped above the filter flask. Pour a few $\text{mL}$ of solvent through the funnel, and allow the solvent to boil and get the funnel hot. When the mixture to be filtered is boiling, pour the mixture into the funnel in portions, returning it to the heat source in between additions. Rinse the filter paper with hot solvent if crystals are seen on the paper.
Table 1.11: Procedural summary for hot filtration. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods/1.5E%3A_Hot_Filtration.txt |
For the separation of small volumes (< $10 \: \text{mL}$) of solid-liquid mixtures, pipette filters are ideal as filter papers absorb a significant amount of material. Pipette filtration may also be used if small amounts of solid are noticed in NMR or GC samples, as both instruments require analysis of liquids without suspended solids.
To create a pipette filter, use a long rod to wedge a small piece of cotton into the bottom of a Pasteur pipette (Figure 1.89a-c). For very small volumes (< $2 \: \text{mL}$), a GC vial makes a nice receiving flask. These vials are very easy to tip over, but can be held in place by an upside-down large septum (Figure 1.89e).
Pipette the solution to be filtered through the top of the filter pipette (Figure 1.89d). It's best to allow the liquid to trickle through the filter on its own, and to at first not use pressure from a dropper bulb, or else solid may be forced through. After the majority of the liquid has filtered, the residual liquid that stays in the pipette (Figure 1.89e), can then be gently pushed through with pressure from a dropper bulb.
Pipette filters can be filled with a centimeter or two of solid material on top of the cotton plug to achieve more than just a solid-liquid separation. They can be filled with drying agents (e.g. $\ce{Na_2SO_4}$) to remove small amounts of water from a solution, or basic solids (e.g. $\ce{Na_2CO_3}$) to neutralize a slightly acidic solution. With the addition of solid to the pipette filter, the liquid must then be gently pushed through with pressure from a dropper bulb as it will not drip through on its own.
1.5G: Centrifugation
Centrifugation is used for the separation of solid-liquid mixtures that are stubborn to settle or difficult to otherwise filter. It uses centrifugal force by rapidly spinning samples so that the solid is forced to the bottom of the tube. In this section is shown centrifugation of a suspension of yellow lead(II) iodide in water (Figure 1.90b).
As a centrifuge (Figure 1.90a) can spin up to 10,000 rotations per minute, an unbalanced load will cause the centrifuge to knock and wobble. If severely unbalanced, the centrifuge can even wobble off the benchtop, causing harm to anything in its way (they are heavy!). To prevent wobbling, every sample in the centrifuge needs to be balanced by an equal volume of liquid in the opposite chamber in the centrifuge.
Special test tubes or centrifuge tubes must be used that exactly fit the width of the chambers in the centrifuge. Each tube should be filled to no greater than three-quarters full as the samples will be tilted in the centrifuge and could spill out (Figure 1.91b). If only one sample is to be centrifuged, a tube of water that contains an equal height of liquid should be placed into an opposing slot in the centrifuge (Figure 1.91a-c). More than one sample can be centrifuged at a time, with the only requirement being that each opposing tube must have nearly the same volume. It is acceptable for one pair of tubes to have different volumes as another pair, as long as the entire centrifuge is symmetrically balanced.
To operate the centrifuge, close the lid and turn on the centrifuge. Set the rotation speed if your instrument allows for it (a good general speed is 8,000 rotations per minute) and turn the dial to the recommended amount of time (set by your instructor or experimentation). Allow the system to spin for the designated time and turn it off. Although a centrifuge has a braking mechanism, it is not recommended to use it as the jostling can stir up sediments. After the set amount of time has expired, simply let a centrifuge slow and come to a stop on its own. The solid can then be separated from the liquid by decantation or by pipette. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods/1.5F%3A_Pipette_Filtration_%28Microscale%29.txt |
Chromatography is a technique used to separate the components of a mixture. It can be used as an analytical technique to gain information about what is present in a mixture, or as a purification technique to separate and collect the components of a mixture.
• 2.1: Prelude to Chromatography
A mixture of ferrocene and acetylferrocene is purified with column chromatography.
• 2.2: Chromatography Generalities
Chromatography in the organic chemistry laboratory can be classified into several broad categories: Thin-lay chromatography, Column chromatography and gas chromatography. These techniques follow the same general principles in terms of how they are able to separate mixtures, and so will be discussed collectively in this section. Details specific to each technique will then be further discussed in their own sections.
• 2.3: Thin Layer Chromatography (TLC)
Many students enter organic chemistry having already done a form of chromatography in their academic career, either in grade school or in general chemistry. Most likely, this will have been paper chromatography (Figure 2.4), as the materials are cheap but still effective. Paper chromatography can be used to separate the dyes in pens, markers, and food coloring. Thin layer chromatography (TLC) is an extension of paper chromatography and uses a different stationary phase.
• 2.4: Column Chromatography
Column chromatography is an extension of thin layer chromatography (TLC). Instead of applying a sample on a thin layer of silica or alumina, a sample is deposited on a cylinder of adsorbent and solvent is continually applied with pressure until the components completely drain from the cylinder. With this modification, components can be not only separated but collected into different containers, allowing for purification of mixtures.
• 2.5: Gas Chromatography (GC)
Gas chromatography (GC) is a powerful instrumental technique used to separate and analyze mixtures. A gas chromatograph is a standard piece of equipment in forensics, medical, and environmental testing laboratories.
02: Chromatography
A mixture of ferrocene and acetylferrocene is purified with column chromatography.
2.2A: Overview of Chromatography
Chromatography is a technique used to separate the components of a mixture. It can be used as an analytical technique to gain information about what is present in a mixture, or as a purification technique to separate and collect the components of a mixture. Chromatography in the organic chemistry laboratory can be classified into several broad categories:
Table 2.1: Variations of chromatography.
These techniques follow the same general principles in terms of how they are able to separate mixtures, and so will be discussed collectively in this section. Details specific to each technique will then be further discussed in their own sections.
• 2.2A: Overview of Chromatography
The first uses of chromatography involved separating the colored components of plants in the early 1900's. The pigments in a plant can be separated into yellow, orange, and green colors (xanthophylls, carotenes and chlorophylls respectively) through this method. The Greek name for color is chroma, and graphein is 'to write,' so chromatography can be thought of as "color writing."
• 2.2B: General Separation Theory
In all chromatographic methods, a sample is first applied onto a stationary material that either absorbs or adsorbs the sample: adsorption is when molecules or ions in a sample adhere to a surface, while absorption is when the sample particles penetrate into the interior of another material.
2.02: Chromatography Generalities
The first uses of chromatography involved separating the colored components of plants in the early 1900's. The pigments in a plant can be separated into yellow, orange and green colors (xanthophylls, carotenes and chlorophylls respectively) through this method. The Greek name for color is chroma, and graphein is 'to write,' so chromatography can be thought of as "color writing."
The general idea of chromatography can be demonstrated with food dyes in your kitchen. Commercial green food dye does not contain any green colored components at all, and chromatography can show that green food dye is actually a mixture of blue and yellow dyes. If a drop of green food dye is placed in the middle of a paper towel followed by a few drops of water, the water will creep outwards as it wets the paper (Figure 2.1). As the water expands, the dye will travel with it. If you let the dye expand long enough, you'll see that the edges will be tinted with blue (Figure 2.1d). This is the beginning of the separation of the blue and yellow components in the green dye by the paper and water.
A complete separation of the green food dye can be accomplished using paper chromatography. A dilute sample is deposited on the bottom edge of a piece of paper, the paper is rolled in a cylinder, stapled, and placed vertically in a closed container containing a small amount of solvent$^1$ (Figure 2.2a). The solvent is allowed to wick up the paper through capillary action (called "elution," Figure 2.2b), and through this method complete separation of the blue and yellow components can be achieved (Figure 2.2d).
$^1$The solvent used in this separation is a solution made from a 1:3:1 volume ratio of $6 \: \text{M} \: \ce{NH_4OH}$:1-pentanol:ethanol. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.01%3A_Prelude_to_Chromatography.txt |
The main chromatographic techniques (thin layer chromatography, column chromatography, and gas chromatography) follow the same general principles in terms of how they are able to separate mixtures.
In all chromatographic methods, a sample is first applied onto a stationary material that either absorbs or adsorbs the sample: adsorption is when molecules or ions in a sample adhere to a surface, while absorption is when the sample particles penetrate into the interior of another material. A paper towel absorbs water because the water molecules form intermolecular forces (in this case hydrogen bonds) with the cellulose in the paper towel. In chromatography, a sample is typically adsorbed onto a surface, and can form a variety of intermolecular forces with this surface.
After adsorption, the sample is then exposed to a liquid or gas traveling in one direction. The sample may overcome its intermolecular forces with the stationary surface and transfer into the moving material, due to some attraction or sufficient thermal energy. The sample will later readsorb to the stationary material, and transition between the two materials in a constant equilibrium (Equation \ref{1}). If there is to be any separation between components in a mixture, it is crucial that there are many equilibrium "steps" in the process (summarized in Figure 2.3).
$\ce{X}_\text{(stationary)} \leftrightharpoons \ce{X}_\text{(mobile)} \label{1}$
The material the sample adsorbs onto is referred to as the "stationary phase" because it retains the sample's position. The moving material is called the "mobile phase" because it can cause the sample to move from its original position.
The main principle that allows chromatography to separate components of a mixture is that components will spend different amounts of time interacting with the stationary and mobile phases. A compound that spends a large amount of time mobile will move quickly away from its original location, and will separate from a compound that spends a larger amount of time stationary. The main principle that determines the amount of time spent in the phases is the strength of intermolecular forces experienced in each phase. If a compound has strong intermolecular forces with the stationary phase it will remain adsorbed for a longer amount of time than a compound that has weaker intermolecular forces. This causes compounds with different strengths of intermolecular forces to move at different rates.
How these general ideas apply to each chromatographic technique (thin layer chromatography, column chromatography, and gas chromatography) will be explained in greater detail in each section.
2.3A: Overview of TLC
Many students enter organic chemistry having already done a form of chromatography in their academic career, either in grade school or in general chemistry. Most likely, this will have been paper chromatography (Figure 2.4), as the materials are cheap but still effective. Paper chromatography can be used to separate the dyes in pens, markers, and food coloring.
When components separate in paper chromatography it is because the components of the mixtures have different attractions to the stationary phase (the cellulose in the paper) and the mobile phase (the solvent wicking up the paper), and spend different amounts of time in each. Thin layer chromatography (TLC) is an extension of paper chromatography and uses a different stationary phase.
• 2.3A: Overview of TLC
Using thin layers of stationary phase for separations is called "thin layer chromatography" (TLC), and is procedurally performed much the same way as paper chromatography
• 2.3B: Uses of TLC
TLC is a common technique in the organic chemistry laboratory because it can give quick and useful information about the purity of a sample and whether or not a reaction in progress is complete. When low polarity solvents are used, a TLC plate can be complete in less than 5 minutes.
• 2.3C: The Retention Factor
A convenient way for chemists to report the results of a TLC plate in lab notebooks is through a "retention factor", 2 or Rf value, which quantitates a compound's movement
• 2.3D: Separation Theory
TLC is an excellent analytical tool for separating mixtures in a sample. In this section are discussed the details of the separation.
• 2.3E: Step-by-Step Procedures for Thin Layer Chromatography
A step-by-step procedures for performing Thin Layer Chromatography in the laboratory is shown. Basic troubleshooting including streaky or "blobby" spots or uneven spotsd.
• 2.3F: Visualizing TLC Plates
Organic compounds most commonly appear colorless on the white background of a TLC plate, which means that after running a TLC, chemists often cannot simply see where compounds are located. Visualization methods can be either non-destructive (compound is unchanged after the process) or destructive (compound is converted into something new after the process. Viewing a TLC plate under ultraviolet light is non-destructive, while using a chemical stain is destructive.
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online.
2.03: Thin Layer Chromatography (TLC)
Two commonly used stationary phases in chromatography are silica ($\ce{SiO2} \cdot x \ce{H2O}$) and alumina ($\ce{Al2O3} \cdot x \ce{H2O}$). Both of these materials are fine white powders that, unlike paper, do not stand up by themselves, so they are deposited in a thin layer on some sort of backing (either glass, aluminum, or plastic). Using these thin layers of stationary phase for separations is called "thin layer chromatography" (TLC), and is procedurally performed much the same way as paper chromatography (Figure 2.5).
It is very common for organic compounds to appear colorless on the white adsorbent background, which poses the challenge of seeing the separation that occurs in a TLC. TLC plates often have to be "visualized," meaning something has to be done to the plate in order to temporarily see the compounds. Common methods of visualization are to use UV light (Figures 2.6 a+b) or a chemical stain (Figure 2.6c).
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.02%3A_Chromatography_Generalities/2.2B%3A_General_Separation_Theory.txt |
TLC is a common technique in the organic chemistry laboratory because it can give quick and useful information about the purity of a sample and whether or not a reaction in progress is complete. When low polarity solvents are used, a TLC plate can be complete in less than 5 minutes.
Assessing Purity
One of the uses of TLC is to assess the purity of a sample. In Figure 2.7 are TLC plates of acetophenone and cinnamaldehyde: samples that were diluted from their reagent bottle, run, and visualized with UV light. Acetophenone appeared as only one spot on the TLC plate, indicating the reagent is likely pure. Conversely, cinnamaldehyde is unquestionably impure as its TLC showed two large spots, and had a few fainter spots as well. Aldehydes are prone to air oxidation, and it is common for aldehydes to be found in their reagent bottles alongside with their corresponding carboxylic acids. TLC is one method that can be used to determine how much an aldehyde has degraded.
Assessing Reaction Progress
Use of a Co-Spot
TLC can be used to analyze a chemical reaction, for example to determine if the reactants have been consumed and a new product has formed. A pure sample of the reactant can be spotted in one lane of a TLC, and the product mixture in another lane. Often the central lane is used for reference, where both reactant and product mixture are spotted over top of one another, in what is called the "co-spot."
For example, the reaction shown in Figure 2.8a is analyzed by TLC in Figure 2.8b. In the first lane of Figure 2.8b (labeled F) is spotted a pure sample of the reactant ferrocene. In the last lane (labeled AF) is spotted the product mixture, which is assumed to be acetylferrocene. In the central lane (labeled co) is spotted both pure ferrocene and the product mixture. The right-most "AF" lane shows that the reaction appears to be a success: the higher spot of ferrocene is absent (meaning it has been consumed), and a new product spot is present. More tests would have to be done to confirm that the lower spot is the expected product of acetylferrocene, but the TLC results look promising.
The co-spot is used for reference, and can be useful in interpreting certain situations. For example, at times the solvent may run with a slight diagonal, causing identical components to elute to slightly different heights. The co-spot can be useful in tracking these variations. It can also be helpful in identifying spots as different if they have similar \(R_f\) values (a feature difficult to be certain of in Figure 2.9a). Two compounds with slightly different \(R_f\) values eluted over top of one another in the co-spot lane may produce a spot with an elongated shape (Figure 2.9b), making it more obvious they are different compounds. Furthermore, a cospot can be helpful in deciphering a TLC plate if the reaction solvent persists in the product mixture lane (for example if the solvent has a relatively high boiling point like DMF or DMSO), such that the residual solvent has an effect on the component's \(R_f\) values.
Monitoring a Reaction by TLC
TLC can be used to monitor the progress of a reaction. This method is often used in research, and one such journal article reporting its usage is shown in Figure 2.10.
To use TLC in this manner, three lanes are spotted on a TLC plate: one for the limiting reactant, one for the co-spot, and one for the reaction mixture. The goal is to note the disappearance of the limiting reactant in the reaction mixture lane and the appearance of a new product spot. When the limiting reactant has completely disappeared, the chemist deduces that the reaction is complete, and can then be "worked up".
To demonstrate how TLC can be used to monitor a reaction, the transesterification reaction in Figure 2.11 was studied over time (Figure 2.12). In the first lane of each TLC plate (marked "BA") was spotted a dilute sample of the reactant benzyl acetate, while in the third lane of each was spotted the reaction mixture (marked "Pr") at different times. In the central lane (marked "Co" for the co-spot), both benzyl acetate and the reaction mixture were delivered over top of one another.
Figure 2.12 shows the progress of the reaction over time. In Figure 2.12a, with the third lane representing the reaction mixture after one minute of mixing, a faint spot near the middle of the plate corresponds to unreacted benzyl acetate. A new spot appears below it, representing the benzyl alcohol product. Over time (Figure 2.12b-e), the top benzyl acetate spot disappears in the reaction mixture lane, and the lower benzyl alcohol spot intensifies. It is apparent from the TLC plates that the reaction was nearing completion at 10 minutes, and was complete at 20 minutes. The TLC demonstrates that the reaction mixture could be worked up after 20 minutes of mixing.
Obtaining an Aliquot of a Reaction in Progress
To monitor a reaction's progress by TLC, an "aliquot" (or tiny sample) of the reaction mixture is necessary.
If the reaction is run at room temperature or with only mild heating, and the concentration of reactants is conducive to TLC, a capillary spotter can be directly inserted into the flask where the reaction is taking place (Figure 2.13a). A long spotter is ideal if one is available. The aliquot can then be directly spotted on the TLC plate.
If the sample is expected to be UV active, it is a good idea to view the TLC plate under UV light before eluting the plate. If the sample spot is not visible before elution it will not be visible afterwards, as compounds diffuse during elution.
If the sample is expected to be UV active, and only a faint hint of material is seen on the baseline, the material can be deposited multiple times before elution (Figure 2.13b): deliver a small spot of sample on the baseline, and let it fully dry before delivering another spot over top of the first. If the spots are not allowed to dry in between applications, the spot will be too large. Check the plate under UV light again, and if necessary spot more times.
It is important to fully allow a spot to dry on the TLC plate before placement in the TLC chamber. In Figure 2.13c the spot for the reaction mixture (labeled Pr) was not fully dry before elution, and the ethanol solvent likely affected the appearance of the lower \(R_f\) spot. In Figure 2.13d the sample was allowed to fully dry before elution, and the lower \(R_f\) spot was more distinct. Furthermore, residual ethanol was likely the reason why the compound spotted in the reagent lane (labeled "BA") had a different \(R_f\) than it did in the central co-spot lane ("Co").
To obtain an aliquot of a refluxing solution, briefly remove the condenser and insert a spotter into the reaction mixture (Figure 2.13e). Immediately re-connect the condenser and adjust the clamps while holding the aliquot. Alternatively, lift the flask from the heat source to temporarily cease the reflux before inserting the spotter. The sample may be able to be spotted directly on the TLC plate, but if too concentrated it can be first diluted by running an appropriate solvent (e.g. acetone) through the pipette and into a small vial.
2.3C: The Retention Factor
A convenient way for chemists to report the results of a TLC plate in lab notebooks is through a "retention factor",$^2$ or $R_f$ value, which quantitates a compound's movement (Equation \ref{2}).
$R_f = \dfrac{\text{distance traveled by the compound}}{\text{distance traveled by the solvent front}} \label{2}$
To measure how far a compound traveled, the distance is measured from the compound's original location (the baseline marked with pencil) to the compound's location after elution (the approximate middle of the spot, Figure 2.14a). Due to the approximate nature of this measurement, ruler values should be recorded only to the nearest millimeter. To measure how far the solvent traveled, the distance is measured from the baseline to the solvent front.
The solvent front (Figure 2.14b) is essential to this $R_f$ calculation. When removing a TLC plate from its chamber, the solvent front needs to be marked immediately with pencil, as the solvent will often evaporate rapidly.
The $R_f$ value is a ratio, and it represents the relative distance the spot traveled compared to the distance it could have traveled if it moved with the solvent front. An $R_f$ of 0.55 means the spot moved $55\%$ as far as the solvent front, or a little more than halfway.
Since an $R_f$ is essentially a percentage, it is not particularly important to let a TLC run to any particular height on the TLC plate. In Figure 2.15, a sample of acetophenone was eluted to different heights, and the $R_f$ was calculated in each case to be similar, although not identical. Slight variations in $R_f$ arise from error associated with ruler measurements, but also different quantities of adsorbed water on the TLC plates that alter the properties of the adsorbent. $R_f$ values should always be regarded as approximate.
Although in theory a TLC can be run to any height, it's customary to let the solvent run approximately $0.5 \: \text{cm}$ from the top of the plate to minimize error in the $R_f$ calculations, and to achieve the best separation of mixtures. A TLC plate should not be allowed to run completely to the top of the plate as it may affect the results. However, if using a saturated, sealed TLC chamber, the $R_f$ can still be calculated.
$^2$Sometimes the $R_f$ is called the retardation factor, as it is a measurement of how the movement of the spots is slowed, or retarded. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.03%3A_Thin_Layer_Chromatography_%28TLC%29/2.3B%3A_Uses_of_TLC.txt |
General Theory
TLC is an excellent analytical tool for separating mixtures in a sample. In this section are discussed the details of the separation, and expand upon the general discussion of Section 2.1.B.
In all forms of chromatography, samples equilibrate between stationary and mobile phases. In almost all applications of TLC, the stationary phase is a silica or alumina adsorbent and the mobile phase is an organic solvent or solvent mixture (the "eluent") that rises up the plate (equation 3).
$\ce{X}_\text{(silica/alumina)} \rightleftharpoons \ce{X}_\text{(solvent)} \label{3}$
Silica gel (shown in Figure 2.16) is composed of a network of silicon-oxygen bonds, with $\ce{O-H}$ bonds on its surface, as well as a layer of water molecules. Silica gel $\left( \ce{SiO_2} \cdot x \ce{H_2O} \right)$ is used in this discussion, but is structurally analogous to alumina $\left( \ce{Al_2O_3} \cdot x \ce{H_2O} \right)$. This very polar stationary phase is paired with a relatively nonpolar mobile phase (an organic solvent or solution), in what is referred to as "normal phase" TLC. Although this is the most common form of TLC (and what will be focused on in this section), "reverse phase" TLC (with a nonpolar stationary phase and a polar mobile phase) is sometimes used.
Figure 2.16 shows how acetophenone would cling to the surface of silica gel through intermolecular forces (IMF's). In this case, acetophenone can hydrogen bond (the IMF indicated in Figure 2.16a) to the silica surface through its oxygen atom. As eluent flows over the sample (Figure 2.16b), an equilibrium is established between the sample being adsorbed on the stationary phase and dissolved in the mobile phase. When in the mobile phase, the compound moves up the plate with the flow of liquid (Figure 2.16c) to later readsorb on the stationary phase further up the plate. The resulting $R_f$ of the compound is dependent on the amount of time spent in the stationary and mobile phases.
The equilibrium distribution between the two phases depends on several factors:
1. It depends on the strength of intermolecular forces between the sample and the stationary phase.
A compound that forms strong IMF's with the silica or alumina will often favor the stationary phase, and will spend much of the elution time adhered to the plate. This means it will spend less time in the mobile phase (which is the only means for it to travel up the plate), causing it to end up low on the TLC plate, and have a low $R_f$.
Compounds that have oxygen or nitrogen atoms should be able to hydrogen bond with the stationary phase (have strong IMF's with the stationary phase), and thus will have lower $R_f$ values than compounds of similar size that can only interact through London dispersion forces (LDF's).
2. It depends on the strength of interaction between the sample and the mobile phase.
As the mobile phase is always less polar than the stationary phase in normal phase TLC, polar compounds will tend to have a lesser affinity for the mobile phase than nonpolar compounds (based on the "like dissolves like" principle). Therefore, polar compounds tend to spend less of the elution time mobile than a nonpolar compound, so will travel "slower" up the plate, and have a low $R_f$.
The degree of attraction by a compound to the stationary and mobile phases lead to the same conclusion:
• The stronger IMF's possible with the stationary phase (often the more polar functional groups on a compound), the more time the compound will be stationary $\rightarrow$ lower $R_f$.
• The more polar functional groups present on a compound, the less it tends to be attracted to the less polar eluent, and the less time the compound will be mobile $\rightarrow$ lower $R_f$.
Thus, a compound with a lower $R_f$ tends to have more polar functional groups than a compound with a higher $R_f$ (summarized in Figure 2.17).
Structural Considerations
To demonstrate the effect of structural features on $R_f$, an eluted TLC plate of benzyl alcohol, benzaldehyde, and ethylbenzene is shown in Figure 2.18. The relative order of $R_f$ reflects the polarity trend in the series.
Benzyl alcohol and benzaldehyde have polar functional groups so thus had lower $R_f$ values than ethylbenzene, which is completely nonpolar. Both compounds are able to hydrogen bond to the polar stationary phase (Figure 2.19a+b), so are more strongly attracted to the stationary phase than ethylbenzene, which interacts only through weak London dispersion forces (Figure 2.19c). As the least "polar" of the series, ethylbenzene is also the best dissolved by the weakly polar eluent. For these reasons, ethylbenzene spent the least time in the stationary phase and the most time in the mobile phase, which is why it traveled the furthest up the plate and had the highest $R_f$ of the series.
Both benzaldehyde and benzyl alcohol are capable of hydrogen bonding with the stationary phase, but benzyl alcohol had the lower $R_f$ because it can form more hydrogen bonds (through both the oxygen and hydrogen atoms of the $\ce{OH}$ group, Figure 2.19a). This caused benzyl alcohol to be more strongly adhered to the silica/alumina than benzaldehyde, causing it to spend more time in the stationary phase.
To demonstrate a different structural effect on $R_f$, an eluted TLC plate of acetophenone and benzophenone is shown in Figure 2.20. Both compounds are similar in that they can hydrogen bond to the stationary phase through their oxygen atom. However, the larger size of benzophenone causes it to have a slightly higher $R_f$ than acetophenone.
This result can be explained in multiple ways:
• The oxygen atom in benzophenone is more crowded by the aromatic rings than the oxygen atom of acetophenone, which may impede its ability to strongly hydrogen bond with the silica gel. This can lead to a smaller amount of time adsorbed by the stationary phase.
• The additional nonpolar bulk of benzophenone makes it dissolve better in the weakly polar eluent, causing it to spend more time in the mobile phase.
Mobile Phase Polarity
The ability of chromatography to separate components in a mixture depends on equilibration of a compound between the stationary and mobile phases. Since the mobile phase is an important factor, it is possible to change the $R_f$ of a compound by changing the polarity of the mobile phase.
When a mobile phase is made more polar than originally, all compounds travel further and have a higher $R_f$.
This general trend is demonstrated in Figure 2.21b+c, where the TLC of three UV-active compounds (lanes 2-4) was run using two different mixed solvents. The first plate was run using a 6:1 hexane:ethyl acetate mixture, which means the solvent was created by using 6 volumes of hexane for every 1 volume of ethyl acetate. This mixed solvent is mostly nonpolar due to the high percentage of hexane, but is more polar than straight hexane, due to the presence of some ethyl acetate (which has polar bonds, Figure 2.21a). The second plate was run using a 3:2 hexane:ethyl acetate mixture, which is more polar than the 6:1 mixture because there is a higher percentage of ethyl acetate present.
Table 2.2: Summary of Rf values for Figure 2.21
Lane $R_f$ in 6:1 (less polar eluent) $R_f$ in 3:2 (more polar eluent)
2 0.33 0.54
3 (bottom spot) 0.02 0.17
3 (top spot) 0.28 0.52
4 0.49 0.65
Note that in Figure 2.21c all spots maintained their relative order but traveled to a greater height on the plate and increased their $R_f$ values (Table 2.2) in the more polar eluent.
An increase in solvent polarity increases $R_f$ values for two reasons:
1. Moderately polar compounds have a greater attraction to the mobile phase.
When equilibrating between a polar stationary phase and nonpolar eluent, a polar compound tends to favor the polar stationary phase and have a low $R_f$. If the eluent is made to be moderately polar, polar compounds are then more attracted to the mobile phase, causing the equilibrium to change such that the compound spends more time in the mobile phase, resulting in a higher $R_f$.
2. The polar solvent may occupy binding sites on the silica or alumina surface, such that they displace the sample from the stationary phase.
If a polar solvent is able to hydrogen bond and therefore strongly associated with the stationary phase, it may "lock up" binding sites, and force less polar compounds to spend more time in the mobile phase. The result is an increased $R_f$ for polar and nonpolar compounds alike. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.03%3A_Thin_Layer_Chromatography_%28TLC%29/2.3D%3A_Separation_Theory.txt |
General TLC Procedure
The TLC pictured in this section shows elution of a TLC plate containing several samples of red food dye at different aqueous dilutions (0 = undiluted, 3 = 1 drop dye + 3 drops water, etc.).
Dissolve the Samples
1. Dissolve all the samples you want to run (solids and liquids) in small vials using a volatile solvent in which they are soluble (e.g. acetone, diethyl ether, or dichloromethane). Ideally the vials will have a lid to minimize vapors and preserve the samples if tipped over (Figure 2.23a).
The optimal concentration for TLC is typically determined empirically, but a good place to start is to use 50-100 times as much solvent as sample (i.e. 1 drop sample for $\sim 1 \: \text{mL}$ solvent). If you have already prepared an NMR sample ($\sim 5 \: \text{mg} / 0.75 \: \text{mL}$ for high field NMR), you can use that sample directly.
Label the vials. Also prepare a vial of clean acetone (or other solvent, e.g. $\ce{CH_2Cl_2}$) to be used to rinse the spotters in a later step. (A rinse vial of water is pictured in Figure 2.23a, as the dye solutions are water soluble.)
Prepare the TLC Chamber and Plate
1. Obtain a TLC chamber with lid. An inexpensive chamber can be made using a beaker and watch glass (Figure 2.23b). Cut a piece of filter paper (or two) so that when placed in the chamber, the filter paper fits inside the chamber and is flat on the bottom but not obscuring your view of the inside (Figure 2.23c).
The filter paper keeps the chamber saturated with vapors so when the eluent rises on the plate it doesn't easily evaporate, but continues to climb and undergo the chromatography. If the eluent evaporated, movement would stop, but could also change the local composition of a mixed eluent and affect the results.
2. Add a portion of a prepared solvent for chromatography (Figure 2.23c, $5$-$10 \: \text{mL}$ for this type of TLC chamber). Close the lid (or place the watch glass) and tilt the chamber to wet the filter paper.
3. Obtain a TLC plate, touching the plate only on the back or edges, but not on the white surface. Use a rule to lightly draw a straight line with a pencil$^3$ roughly $1 \: \text{cm}$ from the bottom. Do not gouge the silica or alumina.
1. Label the areas with pencil where you plan to place the samples (Figure 2.23d). A one-inch wide TLC plate can comfortably accommodate three samples (have three lanes), and if the spot size is kept small it can fit at maximum five spots. If more than five spots are necessary on one plate, TLC plates can be purchased in sheets and cut wider than one inch.
The lanes should not be placed too close to the edge (keep at least $5 \: \text{mm}$ away from each edge), as it is not unusual for solvent to travel slightly "fast" at the edge where capillary action of the solvent is greater (Figure 2.24).
The lanes should also not be placed too close to one another, or the spots may overlap after elution. A spot is always larger after elution compared to its original size (Figure 2.24b+c) because diffusion occurs in all directions (contact with a liquid spreads the material in both the horizontal and vertical directions). Broadening also occurs as solutes in the mobile phase move at different rates as the flow of eluent is weaker near the adsorbent surface. This can even sometimes cause compounds to get stuck in adsorbent pores where the flow of eluent is especially weak.$^4$. Spot broadening means that samples deposited right next to each other on the baseline of a TLC plate will probably bleed together during elution.
Spot the TLC plate with sample
1. Obtain a capillary spotter (a very thin hollow piece of glass open at both ends). In some institutions, you may need to make your own spotter by stretching a softened pipette. Place your spotter into the diluted sample you want to analyze to withdraw liquid into the spotter through capillary action. If the liquid level is low in your vial and your spotter short, you may have to tip the vial to withdraw liquid (Figure 2.25a).
2. Keeping the spotter mostly vertical, make a practice "spot" on a paper towel or scrap piece of silica or alumina to familiarize yourself with how the liquid delivers from the spotter. The spots should be very small, around $2 \: \text{mm}$ in diameter. Deliver a very small spot of material on the pencil line of the appropriate lane (Figure 2.25b) using a quick "up and down" motion with your hand. Don't gouge the silica or alumina with the spotter.
3. Still keeping the spotter vertical, immediately touch the spotter to a paper towel to expunge all of the liquid remaining in the spotter (Figure 2.25c). If the spotter is set down without immediate draining the liquid, air may get into the spotter making it impossible to dispense and use again.
4. Rinse the spotter with acetone (or another volatile solvent with which your compounds are soluble, Figure 2.25d). Place the empty spotter in your rinse vial to withdraw liquid and drain the solvent onto a paper towel. Rinse once or twice before reusing the spotter for other samples.
Place the TLC plate in the chamber to "elute"
1. Use forceps to delicately place the TLC plate into the chamber (Figure 2.26a). Don't allow the liquid to splash onto the plate.
The liquid level must be below the pencil line where the samples are spotted or the compounds will dissolve in the pool of eluent instead of traveling up the plate. Cap the chamber delicately while keeping it vertical, and don't touch it again until the TLC is complete.
2. Allow the TLC to develop (Figure 2.26b-d). As liquid moves up the TLC plate it will appear transparent and wet. A dark background will allow the solvent front to be more easily seen. If the eluent is very polar (e.g. contains large amounts of ethanol or water), elution will take a relatively long time (can be 30-40 minutes). If the eluent is very nonpolar (e.g. contains large amounts of hexane or petroleum ether), elution will be relatively quick (can be 2-5 minutes for a $10 \: \text{cm}$ tall plate).
3. Depending on the goals of the TLC experiment, the chromatography can be stopped when the solvent level is anywhere between halfway to roughly $0.5 \: \text{cm}$ from the top of the plate. It is best to let the TLC run to around $0.5 \: \text{cm}$ from the top of the plate to get the best separation of spots and to minimize error in $R_f$ calculations.
Remove the TLC plate from the chamber
1. Open the TLC chamber, and remove the TLC plate with forceps. Immediately mark the solvent line with a pencil (Figure 2.26e) to enable an $R_f$ calculation, as the solvent often readily evaporates. Alternatively, the silica can be lightly gouged with the forceps to mark the solvent front.
2. If the compounds on the TLC plate are colored, the process is complete. If the compounds are colorless, they need to next be visualized.
Thin Layer Chromatography Summary
Place a small portion of solvent ($5$-$10 \: \text{mL}$ for this chamber) into a TLC chamber with lid, along with a cut piece of filter paper. Dissolve liquid or solid samples (1 drop per $\sim 1 \: \text{mL}$ solvent) using a low boiling solvent (e.g. acetone or dichloromethane).
Draw a pencil line on a TLC plate $\sim 1 \: \text{cm}$ from the bottom with a ruler, and mark the lanes.
Don't put lanes too close to the edge or to each other.
Spot a dilute sample on the pencil line of the correct lane, making very small spots ($2 \: \text{mm}$ in diameter).
Rinse the spotter with a solvent (e.g. acetone) if going to use it for another sample.
Place the sample in the TLC chamber with forceps, cap it, and leave it alone.
Remove the plate when the solvent line is $\sim 0.5 \: \text{cm}$ from the top.
Immediately mark the solvent line with a pencil.
Visualize if necessary.
Table 2.3: Procedural summary for TLC.
TLC Troubleshooting
The spots are streaky or "blobby"
The components of a sample can appear as long streaks or "blobby" spots on a TLC plate if the samples are run at too high a concentration.
For example, Figure 2.27b shows an eluted TLC plate containing five red food dye samples of different concentrations (lane 0 contains the dye in the concentration found at the grocery store; lane 3 contains 1 drop of dye diluted with 3 drops of water, etc.). After elution, the red and pink components from the undiluted dye (lane 0) streaked severely, as the TLC plate was "overloaded". When this happens proper equilibration between stationary and mobile phases does not occur. With further dilution (lane 12), the streaking disappeared and the spot shapes sharpened.
If streaking is seen on a TLC plate, the sample should be diluted and run again.
Figure 2.27c also demonstrates how dilution can improve the shape of a spot after elution. In this TLC, alkene and alkyne samples were spotted at somewhat high concentrations, while an improved dilution was used in Figure 2.27d. Note how the $R_f$ appears to change in the two TLC plates. The more accurate $R_f$ is of the diluted sample. Running TLC on concentrated samples gives inaccurate $R_f$ values and may hide multiple spots.
The spots ran unevenly
At times the solvent front may run unevenly on a TLC plate. This may occur if the plate was placed in the chamber at a slight tilt, if eluent splashed onto the plate during placement in the chamber, or if the chamber was jostled during elution. In cases where the front is dramatically different from one position to the other, the front should be measured for each lane of the plate (if calculating an $R_f$) instead of only once.
Making Capillary TLC Spotters
Although capillary spotters for TLC can be purchased, some chemists prefer to create their own by stretching Pasteur pipettes (Figure 2.28). To make a spotter, hold a pipette by the edges while wearing thick gloves and position the middle of the pipette into the flame of a large Meker burner (Figure 2.28a). Warm the pipette until the glass becomes quite pliable (borosilicate glass softens at $820^\text{o} \text{C}$,$^5$ so this will take some time). Only rotate, but do not stretch the pipette at all while the pipette is in the flame. Then remove the pipette from the flame and immediately and rapidly pull the pipette to an arm's length. The thin sections can be broken into 6-12 inch segments and used for TLC.
Notebook Record of TLC's
Some chemists scan, photograph, or electronically record their developed TLC plates, but it is much more common to copy a likeness of a TLC plate by hand into a laboratory notebook. It is important to copy a TLC plate "to scale", meaning the dimensions should be the same in the notebook as they are in actuality.
To accomplish this, the TLC plate can be placed atop a notebook page and a rendering created beside it (Figure 2.29c). The back of the TLC plate should previously be wiped clean or else staining reagents may degrade the paper. It is important to copy the TLC plate as accurately as possible, drawing the spots exactly as they appear, even if they are streaky or blobby. All spots seen in a lane should be recorded, even if they are faint. Good record keeping means to record all observations, even if the importance is unknown; a faint, unexpected spot may become relevant at a later time.
Several other notations should be made along with the sketch of the TLC plate (Figure 2.29d). The solvent system and identity of what was spotted in each lane must be recorded. For each spot, an $R_f$ should be calculated (some chemists like to write the measurements on the TLC plate; see the plate in Figure 2.29d) along with notation of UV activity and stain color. If a spot changes appearance over time, as the orange spot in Figure 2.29b faded to light green over time (Figure 2.29d), the initial appearance should be recorded.
$^3$The graphite in pencil will not travel with the eluent, but pen ink would.
$^4$J. Sherman and B. Fried, Handbook of Thin Layer Chromatography, 1991.
$^5$G. L. Weissler, Vacuum Physics and Technology, 2$^\text{nd}$ edition, 1979, p. 315. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.03%3A_Thin_Layer_Chromatography_%28TLC%29/2.3E%3A_Step-by-Step_Procedures_for_Thin_Layer_Chromatography.txt |
Organic compounds most commonly appear colorless on the white background of a TLC plate, which means that after running a TLC, chemists often cannot simply see where compounds are located. The compounds have to be "visualized" after elution, which means to temporarily convert them into something visible.
Visualization methods can be either non-destructive (compound is unchanged after the process) or destructive (compound is converted into something new after the process. Viewing a TLC plate under ultraviolet light is non-destructive, while using a chemical stain is destructive.
Visualization Summary
Below is a summary of various visualization techniques, and the functional groups that generally react with each. A more detailed discussion of each technique is provided later in this section.
UV Light: For aromatics + conjugated systems Iodine: Visualizes ~half the time. Strongly reacts with aromatics p-Anisaldehyde: For many aldehydes, ketones, and alcohols Vanillin: For many aldehydes, ketones, and alcohols
Permanganate: For alkenes, alkynes, or oxidizable groups (aldehydes, alcohols) Phosphomolybdic Acid (PMA): For alcohols, phenols, alkenes, and many carbonyl compounds Iron(III) Chloride: For phenols Bromocresol Green: For acidic compounds
Table 2.4: Summary table for TLC visualization methods.
Ultraviolet Absorption
The most common non-destructive visualization method for TLC plates is ultraviolet (UV) light. A UV lamp can be used to shine either short-waved $\left( 254 \: \text{nm} \right)$ or long-waved $\left( 365 \: \text{nm} \right)$ ultraviolet light on a TLC plate with the touch of a button. Most commercially bought TLC plates contain a fluorescent material (e.g. zinc sulfide) in the silica or alumina, so the background of the plate will appear green when viewing with short-waved UV light. If a compound absorbs $254 \: \text{nm}$ UV light, it will appear dark, as the compound prevents the fluorescent material from receiving the UV light.
This method is so quick and easy that it is often the first visualization method tried. It is most useful for visualizing aromatic compounds and highly conjugated systems, as these strongly absorb UV. Most other functional groups do not absorb UV light at the wavelengths used and will not appear dark under the UV lamp even though they are still there. It doesn't hurt to try UV after performing TLC with all compounds just in case. Since the compounds remain unchanged after viewing with UV light, a further visualization technique can be used afterwards on the same plate.
Procedure for UV visualization of TLC plate:
1. Use a UV lamp to look at your developed TLC plate by pressing the short-waved button. Depending on what is available at your institution, you might tilt the UV lamp at an angle in a darkened area of the room (Figure 2.31a), or you might look at the plate inside a box designed to protect eyes from UV damage (Figure 2.31b).
Safety note: Care should be taken to never look directly at the UV source, and to minimize exposure to eyes.
2. The plate background will appear green under short-waved UV light, and UV-active compounds will appear dark (Figure 2.31c). Use a pencil to lightly circle spots, as they will disappear when the UV light is removed.
3. Some compounds themselves fluoresce (Figure 2.32), appearing a variety of colors when exposed to either short- or long-waved UV light (bright purple or blue is the most common). Record these types of observations in your notebook if you see them, as they are rare, and are therefore an excellent identification tool. They are most common with highly conjugated compounds.
1. The TLC plate can be further visualized with some other method if desired (iodine or a chemical stain).
Iodine
A commonly used semi-destructive visualization method is to expose a developed TLC plate to iodine $\left( \ce{I_2} \right)$ vapor. An "iodine chamber" can be created by adding a few iodine crystals to a TLC chamber, or by adding a few iodine crystals to a chamber containing a portion of powdered silica or alumina (Figure 2.33a). When a developed TLC plate is placed in the chamber and capped, the iodine sublimes and reacts with the compounds on the plate, forming yellow-brown spots (Figure 2.33d). The coloration occurs because iodine forms colored complexes with many organic compounds. This stain will work with approximately half the compounds you may encounter.
This method is considered "semi-destructive" because complexation is reversible, and the iodine will eventually evaporate from the TLC plate, leaving the original compound behind. When the coloration fades, it is theoretically possible to use another visualization technique on the TLC plate, although it's possible the compound may have also evaporated by that time.
Procedure for visualization of TLC plate with iodine:
1. If not already prepared, make an "iodine chamber" (Figure 2.33a): in a fume hood, place a few centimeters of powdered silica or alumina in a screw-capped TLC chamber and add a few crystals of solid iodine (safety note: silica and alumina are lung irritants, and iodine vapor is considered an irritant and toxic). A beaker and watch glass will not work in this context as the iodine vapors will escape. Let the silica or alumina and iodine sit together for a while with periodic swirling, and eventually the powder will become orange from adsorbing the iodine vapor.
2. In a fume hood, place the developed TLC plate in the iodine chamber with forceps (Figure 2.33b) and close the lid. Gently shake the chamber to bury the TLC plate in the iodine-stained silica or alumina (Figure 2.33c) until the spots become colored (Figure 2.33d). Many spots will appear yellow-brown almost immediately, and may darken with extended time. For many compounds, it takes less than 10 seconds to develop a plate, but some compounds require 10 minutes or longer. (Note: alcohols, carboxylic acids, and alkyl halides often do not stain.)
3. Promptly record appropriate observations of the TLC in your notebook, or circle the spots with a pencil, as the colors will soon fade as the iodine evaporates from the plate. Further visualization may be attempted after the color fades.
Chemical Stains
There are a variety of destructive visualization methods that can turn colorless compounds on a TLC plate into colored spots. A plate is either sprayed with or dipped in a reagent that undergoes a chemical reaction with a compound on the TLC plate to convert it into a colored compound, enabling the spot to be seen with the naked eye. Since a chemical reaction is occurring in the process, it is common to gently heat a plate after exposure to the reagent to speed up the reaction, although this may be unnecessary with some stains. Not every compound can be visualized with every reagent if they do not react together, and stains are often designed to work with only certain functional groups. The specific stain should be chosen based on the presumed structure of the compounds you want to visualize.
General Staining Procedure
1. While wearing gloves and holding the TLC plate with forceps in a fume hood, quickly dip the plate into and out of the chemical dip jar so that the stain covers the area where the solvent traveled on the plate (Figure 2.34b).
2. Let excess liquid drip off the plate for a few seconds, then wipe the back of the plate with a paper towel (Figure 2.34c).
3. Gently heat the plate to develop the spots. Preferably use a heat gun (Figure 2.34d), but a hotplate can also be used (Figure 2.35, charring is common).
1. If using a heat gun, hold the TLC plate with forceps and wave the heat gun back and forth onto the front of the plate. The "high" setting can be used at first, with the setting turned to "low" if the plate begins to char. Safety note: Heat guns are not simple hair dryers, and can get quite hot. Be careful to not touch the nozzle, and remember that it remains hot for a long time after heating has ceased. The hot nozzles can even mar benchtops, so be cautious when setting the gun down.
2. If using a hot plate, place the TLC plate on the warm surface (set between low and medium-low, and covered in foil to prevent dip residue from staining the ceramic surface). Periodically move the plate around to distribute the heat.
1. A TLC plate cannot be further visualized after using a stain.
p-Anisaldehyde/Vanillin Stains
Generalities
The p-anisaldehyde and vanillin stains are general purpose, and work for many strong and weak nucleophiles (alcohols, amines), and for many aldehydes and ketones. They do not work on alkenes, aromatics, esters, or carboxylic acids. The TLC plates need to be mildly heated, and will develop a light pink to dark pink background.
A TLC of four samples visualized with three different techniques is shown in Figure 2.36. The plate is visualized with UV light (Figure 2.36b), p-anisaldehyde stain (Figure 2.36c), and vanillin stain (Figure 2.36d). 4-heptanone (lane #1) and acetophenone (lane #2) showed similar colorations using the two stains. Ethyl benzoate (lane #4) was unreactive to both. Cinnamaldehyde (lane #3) was reactive to p-anisaldehyde but not vanillin, while its impurity (cinnamic acid, on the baseline of lane #3) showed the opposite behavior.
Recipe (p-Anisaldehyde): $135 \: \text{mL}$ absolute ethanol, $5 \: \text{mL}$ concentrated $\ce{H_2SO_4}$, $1.5 \: \text{mL}$ glacial acetic acid, and $3.7 \: \text{mL}$ p-anisaldehyde. This stain is susceptible to degradation by light, so store wrapped in aluminum foil (Figure 2.37e), ideally in the refrigerator when not in use. Compared to other stains, this stain has a somewhat short shelf life (approximately half a year). The stain will at first be colorless (Figure 2.37a), but over time will turn to a light then dark pink (Figure 2.37b-d). The stain is less potent when it darkens, but is often still usable. Safety note: wear gloves while using this highly acidic stain.
Recipe (Vanillin): $250 \: \text{mL}$ ethanol, $15 \: \text{g}$ vanillin, and $2.5 \: \text{mL}$ concentrated $\ce{H_2SO_4}$. This stain is light sensitive and should be stored wrapped in aluminum foil in the refrigerator. It is originally light yellow, but darkens over time (Figure 2.37f+g). It should be discarded if it acquires a blue color. Safety note: wear gloves while using this highly acidic stain.
Reaction Pathways
The p-anisaldehyde and vanillin stains react in a similar manner, and commonly undergo Aldol and acetalization reactions to produce highly conjugated (and thus colored) compounds on TLC plates.
Aldol Reactions
Under the acidic conditions of the stain, some aldehydes or ketones can undergo a keto-enol tautomerism, and the enol can undergo acid-catalyzed nucleophilic addition to p-anisaldehyde or vanillin through an aldol mechanism. Dehydration of the aldol product (encouraged by heating the TLC plate), results in a highly-conjugated compound (Figure 2.38d), which is why spots become colored.
For example, a TLC plate containing acetophenone and benzophenone (as seen with UV, Figure 2.38a), are stained with p-anisaldehyde and vanillin stains. Acetophenone produced a colored spot with these stains (Figures 2.38b+c) while benzophenone did not. The main difference is that benzophenone cannot form an enol, or be a nucleophile to p-anisaldehyde, so the stain is unreactive.
Acetalization Reactions
Some alcohols react with p-anisaldehyde and vanillin stains through acetalization reactions. A proposed reaction of p-cresol with p-anisaldehyde is shown in Figure 2.39b to produce a highly-conjugated cation, a possible structure of the pink spot on the TLC plate in lane #2 of Figure 2.39a. This cationic structure may look unusual, but is a feasible structure in the highly acidic conditions of the stain.
Permanganate Stain
The permanganate ion $\left( \ce{MnO_4^-} \right)$ is a deep purple color, and when it reacts with compounds on a TLC plate (and is consumed), the plate is left with a yellow color (Figure 2.40a). The stain easily visualizes alkenes and alkynes by undergoing addition reactions (Figure 2.40d), and the color change is often immediate with these functional groups.
Permanganate is also capable of oxidizing many functional groups (e.g. aldehydes, lane 1 in Figure 2.40c), and so is considered by some to be a universal stain. Heat may be required to visualize some functional groups, and often improves the contrast between spots and the background. Heating may be done (if needed) until the background color just begins to yellow, but a brown background means the plate was overheated.
Recipe: $1.5 \: \text{g}$ $\ce{KMnO_4}$, $10 \: \text{g} \: \ce{K_2CO_3}$, $1.25 \: \text{mL} \: 10\% \: \ce{NaOH} \left( aq \right)$, and $200 \: \text{mL}$ water. Safety note: wear gloves while using this stain, as permanganate is corrosive and will stain skin brown.
Phosphomolybdic Acid Stain
The phosphomolybdic acid stain (PMA) is considered by some a universal stain, able to visualize a wide variety of compounds (alcohols, alkenes, alkyl iodides, and many carbonyl compounds). The yellow-green PMA reagent $\left( \ce{Mo^{6+}} \right)$ oxidizes the compound on the plate while itself being reduced to molybdenum blue ($\ce{Mo^{5+}}$ or $\ce{Mo^{4+}}$). Vigorous heating is required to develop the spots, but the plate is overheated when the background begins to darken. There is typically no color differentiation between spots, as most compounds visualize as green or blue spots (Figure 2.41c).
Recipe: $5 \: \text{g}$ phosphomolybdic acid in $500 \: \text{mL}$ ethanol. The stain is light sensitive and so should be stored in a jar under aluminum foil. The reagent is expensive, but the stain has a very long shelf life (5+ years).
Iron(III) Chloride Stain
The ferric chloride $\left( \ce{FeCl_3} \right)$ stain is highly specific, and is used mainly to visualize phenols $\left( \ce{ArOH} \right)$. Some carbonyl compounds with high enol content may also be visualized. $\ce{Fe^{3+}}$ forms colored complexes with phenols (often faint blue), in the general sense of what is shown in Figure 2.42c. The actual structure of these complexes is debated$^6$. The coloration fades rather quickly with this stain, so observations should be recorded immediately.
Recipe: $1\%$ $\ce{FeCl_3}$ in water and $\ce{CH_3OH}$ ($50\%$ each). This stain has a high shelf life (5+ years).
Bromocresol Green Stain
The bromocresol green stain is specific for acidic compounds, and should be able to visualize compounds that produce a solution lower than pH 5. Experience has shown that carboxylic acids work moderately well (first lane in Figure 2.43d) but phenols are only barely visible (indicated with an arrow in Figure 2.43d). In theory, the plate does not need to be heated after exposure to this stain, but in practice it often improves the contrast between the spots and the background.
Recipe: $100 \: \text{mL}$ absolute ethanol, $0.04 \: \text{g}$ bromocresol green, and $0.10 \: \text{M} \: \ce{NaOH} \left( aq \right)$ drop wise until solution turns from yellow to blue (green works as well, as in Figure 2.43b).
This stain uses an acid-base indicator, which works in a similar manner to phenolphthalein. Bromocresol green is yellow below pH 3.8 and blue above pH 5.4 (Figure 2.44a). When an acidic compound is spotted on the plate, the acid lowers the pH and causes the indicator to shift to the lower pH yellow form (Figure 2.44b).
Visualization Troubleshooting
The Compound Didn't Show Up
Even when a compound has certainly been applied on the baseline of a TLC plate, it is possible that the compound is not seen on the plate after elution. There are several possible reasons for this:
1. The compound may be too dilute.
A simple solution to a dilution problem is to add more compound to the original sample and run the TLC again using a new plate. If the compound is expected to be UV active (i.e. if it contains an aromatic ring), it is a good idea to view the TLC plate under UV light before eluting the plate (Figure 2.45a). If the sample spot is not visible before elution it will not be visible afterwards, as compounds diffuse during elution.
If the sample is determined to be only slightly too dilute, the material can be deposited multiple times before elution (Figure 2.45b). To do this, deliver a small spot of sample on the baseline, and let it fully dry (it helps to blow on it) before delivering another spot over top of the first. If the spots are not allowed to dry in between applications, the spot will be too large. If the compound is expected to be UV active, check the plate under UV light, and if necessary spot more times before elution.
2. The compound may have evaporated.
A TLC plate should be visualized immediately after elution, so if a moderate amount of time was left between running the TLC and visualizing it, evaporation may be the cause of the problem. A solution to this problem is to run the TLC again and visualize it immediately.
If the compound has a low boiling point, it probably evaporated during elution. For example, 2-pentene (boiling point $36^\text{o} \text{C}$) was spotted in lane #1 of Figure 1.45c. It did not stain with permanganate after elution even though the compound is reactive to the stain (an undiluted, uneluted sample of 2-pentene did stain somewhat on a scrap TLC plate, Figure 2.45d). Compounds with boiling points lower than approximately $120^\text{o} \text{C}$ are difficult to analyze through TLC.
3. The compound may be unreactive to the visualization technique.
Visualization techniques are often tailored toward certain functional groups. For example, ultraviolet light is generally good at visualizing aromatic compounds but poor at other functional groups. If UV, iodine, or a stain fails to visualize a compound, it could mean the compound is simply not reactive to the technique, and another method should be tried.
For example, Figure 2.46 shows for different compounds visualized with UV (Figure 2.46a), p-anisaldehyde stain (Figure 2.46b) and iodine (Figure 2.46c). The compound in lane #1 of all the plates (4-heptanone) was only visible with anisaldehyde stain (blue spot), and not with UV or $\ce{I_2}$. The compound in lane #4 of all the plates (ethyl benzoate) was unreactive to anisaldehyde stain, but could be visualized with UV and $\ce{I_2}$. The impurity present on the baseline of lane #3 (the impurity cinnamic acid) was strongly UV active, but could hardly be seen with the other stains.
The Pencil Marking from UV is Different Than the Stain
Ultraviolet light is often the first visualization technique attempted on an eluted TLC plate because it is nondestructive and rather simple to carry out. If a dark spot is seen with a UV lamp, it is customary to circle the spot with pencil (as in Figure 2.46b), as the spot will be invisible when the lamp is removed. Another visualization technique is often carried out after viewing the plate under UV, and it is not uncommon that the subsequent stain extends to a smaller or larger region than the pencil marking.
For example, the compound in lane 2 of Figure 2.46 (acetophenone) can be easily seen with ultraviolet light (Figure 2.46a), but on the plate visualized with iodine (Figure 2.46c), the pencil markings encapsulate a larger region than is seen darkened by the iodine. This is because acetophenone is very strongly UV active, but only mildly complexes with iodine. It is not uncommon for one technique to visualize a compound more effectively than another technique.
It is therefore important to be cautious in using TLC to interpret the quantity of material present in a sample, for example when assessing the quantity of an impurity (such as in lane #3 of Figure 2.46, which contains cinnamaldehyde and its impurity cinnamic acid). It is likely that a large spot is present in a greater quantity than a small spot, but it could also be that the large spot is more responsive to the visualization technique.
A Stain's Color Faded or Changed With Time
It is very common for the coloration produced by a stain to fade with time, as the compounds eventually evaporate from the plate or other slower reactions take place. For this reason, it may be a good idea to circle the spots with pencil immediately after a plate is visualized, although as spots are generally circled after viewing with UV, additional markings may cause confusion as to which compounds are UV-active. Another alternative is to place clear tape across the plate to prevent the spots from evaporating.
It is possible that the coloration produced by a stain will change with extended heating, or with time. For example, the plate in Figure 2.47 was visualized with p-anisaldehyde stain, and Figure 2.47a shows how the plate appeared immediately after heating. Figure 2.47b shows how the same plate appeared after sitting at room temperature for 30 minutes. The compound in lane #2 (acetophenone) had the most dramatic change in color during that time, changing from a bright orange to a green color. Observations recorded into a lab notebook should be of the original color of a spot.
The Staining Doesn't Make Sense
Certain visualization methods work best for certain functional groups, so a positive result with a stain can give clues about the identity of an unknown spot. However, sometimes a compound stains when it isn't "supposed to", and this can be confusing.
For example, a TLC of benzaldehyde visualized with UV light (Figure 2.48a) shows two spots, and based on relative $R_f$ values, it would make sense that the dark spot is benzaldehyde and the fainter spot near the baseline is benzoic acid (caused by an oxidation of benzaldehyde). Staining of the plate with bromocresol green (a stain for acidic compounds), supports this hypothesis as the lower acidic spot is visualized with this method (Figure 2.48b). This is an example of when the staining results "make structural sense", and can even support the identification of unknown spots.
However, in a similar experiment with cinnamaldehyde, both aldehyde and carboxylic acid spots were strongly visualized with bromocresol green (Figure 2.48d), even though only one is an acidic compound. This result does not at first "make sense", and theories can only be postulated for why the aldehyde reacted with the stain.
The significance of Figure 2.48 is that interpretation of staining or lack of staining can sometimes be used to infer structure, but there may be exceptions that are difficult to explain.
$^6$See Nature 165, 1012 (24 June 1950); DOI: 10.1038/1651012b0
$^7$The TLC plate was left in the $\ce{I_2}$ chamber for only about 2 minutes, and the spots may have developed further with additional time. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.03%3A_Thin_Layer_Chromatography_%28TLC%29/2.3F%3A_Visualizing_TLC_Plates.txt |
Column chromatography is an extension of thin layer chromatography (TLC). Instead of applying a sample on a thin layer of silica or alumina, a sample is deposited on a cylinder of adsorbent and solvent is continually applied with pressure until the components completely drain from the cylinder. With this modification, components can be not only separated but collected into different containers, allowing for purification of mixtures. Column chromatography (also known as "flash chromatography"), is frequently used in research settings, as is evidenced by its common occurrence in the procedural sections of journal articles.
• 2.4A: Macroscale Columns
The same underlying principles of thin layer chromatography (TLC) apply to column chromatography. In fact, a TLC is always run before performing a column to assess the situation and determine the proper solvent ratio. There are a few variables which aren’t applicable to TLC, but which affect the separation of components in column chromatography. These include the column diameter, quantity of adsorbent used, and solvent flow rate.
• 2.4B: Microscale (Pipette) Columns
A macroscale column is much too big for very small quantities of material (< 20 mg). Instead, a column can be constructed using a disposable Pasteur pipette.
2.04: Column Chromatography
Procedural Generalities
The same underlying principles of thin layer chromatography (TLC) apply to column chromatography. In fact, a TLC is always run before performing a column to assess the situation and determine the proper solvent ratio. In order to get good separation, it is ideal if the desired component has an $R_f$ around 0.35 and is separated from other components by at least 0.2 $R_f$ units.$^8$ If the spots to be separated are very close (if the difference in $R_f$ is < 0.2), it's best if the middle of the spots has an $R_f$ of 0.35. An $R_f$ near 0.35 is ideal because it is slow enough that stationary-mobile phase equilibration can occur, but fast enough to minimize band widening from diffusion.
There are a few variables which aren't applicable to TLC, but which affect the separation of components in column chromatography. These include the column diameter, quantity of adsorbent used, and solvent flow rate. Table 2.5 summarizes variable recommendations based on the sample size and degree of separation between components. In all scenarios, the columns are to be prepped between 5-6 inches high.
Table 2.5: Summary of recommended values for column chromatography.$^9$
Column diameter $\left( \text{mm} \right)$ Volume of eluent $\left( \text{mL} \right)$ Sample size $\left( \text{mg} \right)$ If $\Delta R_f$ > 0.2 Sample size $\left( \text{mg} \right)$ If $\Delta R_f$ > 0.1 Typical fraction size $\left( \text{mL} \right)$
10 100 0.2" style="vertical-align: middle;">100 0.1" style="vertical-align: middle;">40 5
20 200 0.2" style="vertical-align: middle;">400 0.1" style="vertical-align: middle;">160 10
30 400 0.2" style="vertical-align: middle;">900 0.1" style="vertical-align: middle;">360 20
40 600 0.2" style="vertical-align: middle;">1600 0.1" style="vertical-align: middle;">600 30
For example, a column of one-inch diameter ($1 \: \text{in}$ is $25.4 \: \text{mm}$) should be able to purify around $400 \: \text{mg}$ of material if the separation is good ($\Delta R_f$ > 0.2, third column in Table 2.5), or around $160 \: \text{mg}$ if the separation is difficult ($\Delta R_f$ > 0.1). The column should be able to be prepared and eluted using around $200 \: \text{mL}$ of solvent, and the fractions can be collected with approximately $10 \: \text{mL}$ of solution each.$^9$
There are multiple variations on how to physically run a column, and your instructor may prefer a certain method. One large difference in methods is how the column is prepared. In the "dry packing" method, dry silica or alumina is added directly to a column, and solvent is allowed to trickle through in portions, then with pressure. In the "wet packing" method, the column is filled with solvent first, then dry silica or alumina is lightly shaken in, then packed with pressure. In the "slurry" method, solvent is added to the silica or alumina in an Erlenmeyer flask, poured onto the column as a sludgy material, then packed with pressure.
It is important to know that heat is liberated when solvent is added to silica or alumina (they have an exothermic heat of solvation). The slurry method is presented in this section, with the main reason being that it allows this exothermic step to happen in an Erlenmeyer flask instead of on the column. If heat is liberated during the packing of the column, it may generate bubbles from the boiling of solvent. These can interfere with the separation of the column if they are not adequately removed, and can crack the adsorbent material in the column.
Step-by-Step Procedures
The column pictured in this section shows purification of a $0.20 \: \text{g}$ sample containing a mixture of ferrocene and acetylferrocene (crude TLC is in Figure 2.51a). Roughly $8 \: \text{mL}$ fractions were collected into small test tubes, and roughly $400 \: \text{mL}$ of eluent was used.
Run a TLC
1. Run a TLC of the sample to be purified (Figures 2.52 a+b) to determine the appropriate solvent for chromatography. The desired component should have an $R_f$ around 0.35 and should ideally be separated from all other spots by at least 0.2 $R_f$ units.
2. Prepare a batch of eluent that gives the proper $R_f$ value. The quantity prepared depends on the quantity of sample, the size of the column, and whether or not the solvent composition is planned to be changed midway. (See Table 2.5 for guidelines and the eluotropic series for trends in "solvent power.")
Prepare the packed column
1. Obtain an appropriate column (see Table 2.5), and be sure there is something near the stopcock that will allow liquid to pass through, but not solid. Columns may have a sintered disk (also known as a "frit," Figure 2.52c), or a plug of cotton or glass wool remaining from the previous user (Figure 2.52d). If no disk or plug is present (Figure 2.52e), wedge a small wad of cotton or glass wool into the bottom of the column using a long rod.
1. Secure your column perfectly vertically to a ring stand or latticework, clamping it with three-fingered clamps in two locations. In the fume hood, pour silica gel or alumina adsorbent into the column to between 5-6 inches high (Figure 2.53a).
Safety note: Powdered silica and alumina are lung irritants, and should always be handled carefully in a fume hood. Spilled dust should be disposed of by mopping it with a wet paper towel (if wet, the fine particles are less dispersive).
2. In the fume hood, pour the adsorbent measured in the column into an Erlenmeyer flask (Figure 2.53b), then add some eluent (Figure 2.53c). Make a loose slurry by swirling and stirring with a glass stirring rod (Figure 2.53d) until all of the adsorbent is completely wet, gas bubbles are released, and the consistency is somewhat thick but pourable.
1. Put a beaker or Erlenmeyer flask beneath the clamped column and open the stopcock. In one quick motion, swirl and pour the silica or alumina slurry into the column using a large-mouthed funnel (Figure 2.54a). Immediately use more eluent to rinse residual slurry out of the Erlenmeyer flask (Figure 2.54b) and onto the column.
2. Immediately rinse any silica or alumina off the sides of the column reservoir using eluent and a swirling motion from a Pasteur pipette (Figures 2.54 c+d). If allowed to dry, the adsorbent will cling to the glass and won't easily be rinsed down.
1. Jostle the column firmly using a cork ring or your knuckles (Figure 2.55a) to dislodge any air bubbles in the column (which could cause poor separation or cracking of the adsorbent in the column), and to promote an even deposition of adsorbent.
2. Apply gentle air pressure to the top of the column (Figure 2.55b) to compress it, stopping when the eluent level is $1 \: \text{cm}$ from the top of the column. If a T-adapter is used with the air line as in Figure 2.55b, fine control of the airflow can be accomplished by adjustment of the pinch clamp on the rubber tubing.
Throughout the entire elution process, keep the white column of adsorbent wet, with the eluent level above the top of the silica or alumina.
Gently break the seal to cease application of pressure, and close the stopcock to prevent liquid from dripping out further.
3. Add a thin layer of sand (Figure 2.55c), approximately $0.5 \: \text{cm}$ high. Rinse the sides of the column with eluent using a swirling motion to dislodge the sand off the sides of the glass (Figure 2.55d). Open the stopcock and allow liquid to drip out until the liquid is just above the sand layer. Apply air pressure if the dripping is too slow.
Add the sample
Once the sample is applied to the column, there is a race against time, as diffusion will start to broaden the material. A sample should not be applied until you are ready to complete the column immediately and in its entirety. This process may take between 15 and 90 minutes! If using test tubes to collect fractions, the test tubes should be arranged in a rack prior to adding the sample, and the column height should be adjusted so that the test tube rack can slide underneath.
1. If the crude sample is a liquid, use it directly (go on to step 13).
2. If the crude sample is a solid, do one of the following things:
1. Ideal situation: dissolve the solid in the minimum amount of the eluent (a few $\text{mL}$ at most).
2. If the solid is not particularly soluble, or will not dissolve in a few $\text{mL}$ of the eluent, dissolve it in the minimum amount of dichloromethane (few $\text{mL}$ at most, Figure 2.56a).
3. If the solid is insoluble in the eluent, an alternate procedure is also possible. Dissolve the solid in a round-bottomed flask using a few $\text{mL}$ of a low-boiling solvent (e.g. dichloromethane or acetone). Add to the flask approximately $1 \: \text{g}$ of silica or alumina, then remove the solvent on the rotary evaporator in order to leave a solid that contains the sample deposited on the adsorbent. With an inch of eluent resting atop the packed column (skip adding the sand layer if this method is used), pour the silica-adsorbed sample onto the column using a wide mouthed funnel. If any dust clings to the glass, rinse it down with more eluent (go on to step 15.)
3. Delicately add the sample to the column via pipette, dripping the liquid or solution directly onto the sand with the pipette tip as close as you can manage, not down the sides (Figure 2.56b). Take care to not squirt liquid in forcibly such that indentations would be caused in the sand or silica /alumina column.
4. Rinse the sample container with a little solvent (or dichloromethane if used, Figure 2.56c) and add the rinsing to the column using the same pipette (in order to rinse the pipette as well).
5. Open the stopcock and allow liquid to drip out until the sample is just past the sand layer (Figure 2.56d) and into the white area of the column (apply air pressure if this takes more than 20 seconds).
6. Gently rinse the sides of the column with a swirling motion using 1-2 pipettes-full of eluent to rinse any splashed sample. Again, allow liquid to drip out (or apply air pressure) until the sample is pushed into the white adsorbent.
Repeat the rinsing step until you feel confident that the entire sample is deposited on the adsorbent. If some of the sample is still located in the sand layer, it may dissolve in the eluent when more solvent is added, leading to a loss of yield. If the compound is colored, the rinsing should be completely clear.
Fill with Eluent and Elute the Column
1. Delicately add more eluent via pipette (Figure 2.57a), swirling down the sides, and then when the sand layer would no longer be disturbed with the additions, carefully pour in larger quantities (Figure 2.57b) of the prepared eluent to fill the reservoir (or fill with as much as may be needed). Clean eluent collected during the packing of the column can be reused.
2. Use air pressure to gently and steadily elute the sample through the column (Figures 2.57 c+d). The more times the pressure is started and stopped, the more likely the column may crack. It's best if the pressure can be kept gentle and steady the entire time.
The optimal drip rate during elution depends on the size of the column. The ideal flow of eluent is when the solvent in the cylindrical section of the column above the adsorbent drops at a rate of 2.0 inches per minute.$^{10}$ Therefore, the drip rate should be slower with a narrow column compared to a wider column.
The drip rate for a one-inch column should be where individual drops can be barely distinguished. A stream of liquid pouring out the stopcock with this size of column is slightly too fast.
Collect fractions
1. Immediately start collecting the eluting liquid into test tubes on a rack (Figure 2.58a). See Table 2.5 for recommendations on volumes to collect in each test tube.
2. When the first test tube fills, or if a certain height of liquid has been collected as recommended by your instructor or Table 2.5, move the rack over to start collecting into a different tube (Figures 2.58 b+c). Fill and keep the tubes in order on the rack.
These different tubes are called "fractions". The goal of a column is to collect small enough fractions that most (or some) fractions contain pure material. If the separation of the mixture is difficult (if the $\Delta R_f$ of the components is low), it may be best to collect small fractions (e.g. half-filled tubes).
1. As liquid drains off the column, it often splashes onto the outsides of the tip of the column, and when the solvent evaporates you may see a ring of material on the tip (you will see a ring of solid if the component is a solid as in Figure 2.59b, or oily droplets if the component is a liquid). If the components are colored, the column tip should be rinsed (Figure 2.59c) when it appears as if one component has completely eluted and before the other component approaches.
2. Periodically keep an eye on the eluent level, and refill before it drops below the sand layer.
Possibly Increase the Solvent Polarity
1. One eluent may be used throughout the entire column, especially if the components to be separated have similar $R_f$ values. However, if the components have very different $R_f$ values, the solvent polarity may be increased after one component has eluted from the column (Figure 2.60a).
Increasing the solvent polarity will make components travel "faster." There are several reasons a faster elution is desired. First, if one component has already exited the column, the column has already done its job with separation, so speeding up the process will not affect the purity of the collected fractions. Second, the longer it takes to run a column, the wider will be the component bands (due to diffusion), and collecting a broad band of material will use (and waste) a lot of solvent.
Table 2.6 contains a partial list of the eluotropic series, a list of common solvents ranked according to their "solvent power" in normal phase chromatography. The more polar solvent causes the most dramatic increase in $R_f$.
Table 2.6: Eluotropic series.
Eluotropic Series (from least to most polar)
Petroleum ether / hexanes
Dichloromethane
Diethyl ether
Ethyl acetate
Acetone
Methanol
1. To increase the solvent polarity, the polar solvent can be dripped directly into the eluent on the column reservoir (Figure 2.60b). For example, if using a hexanes:ethyl acetate mixture, addition of pure ethyl acetate to the eluent currently in the reservoir would increase its polarity. If the eluent level is running low, a solution could be prepared that contains a higher percentage of the more polar component. For example, if the column first used a 4:1 hexanes:ethyl acetate mixture, using a 1:1 mixture would be a more polar solvent.
2. Elute the column with the more polar solvent as before, and always remember to watch the eluent level, and refill (Figure 2.60d) before it drops below the sand layer.
Find and Concentrate the Desired Component
1. In finding the desired component in the test tube fractions, it is helpful to understand the relationship between $R_f$ and elution order in column chromatography.
In column chromatography, the sample is deposited on the top of the column and eluted down, while in thin layer chromatography the sample is spotted on the bottom of the plate and eluted up. Therefore, a column can be thought of like an upside-down TLC plate. A compound with a higher $R_f$ runs "faster," meaning it will end up higher on a TLC plate, and will be collected first with a column.
In the column pictured in this section, the component with the lower $R_f$ (orange on the TLC plate in Figure 2.61a), is the component collected second from the column.
2. First determine which test tubes contain dissolved compound.
1. Spot a sample of each fraction on a TLC plate labeled with fraction numbers corresponding to the order in which they were collected (Figure 2.61c). It may be best to spot each sample 2-3 times over top of one another in case the fractions are dilute.
2. If many fractions have been collected, making you hesitant to sample every fraction, one method to identify colorless fractions that may contain compound is to look for a hint of residue on the top of the test tubes. After evaporation, a solid residue (Figure 2.62) or oily droplets are sometimes left on the top of the test tube, making obvious that those fractions contain more than just solvent. Sample all fractions near the tubes containing visible residue.
1. Visualize the spotted TLC plate using UV light and/or a stain to determine which fractions contain compound (Figure 2.61d).
1. Run a TLC of all fractions that contain compound, spotting up to five samples per 1-inch wide TLC plate. Wider TLC plates may be used for this purpose if available.
2. Identify the compound with the desired $R_f$ by comparison with the original crude TLC plate. Choose to retain the fractions that have the desired compound in pure form, as evidenced by the eluted TLC plate. For example, if the compound with the higher $R_f$ is desired in Figure 2.63a, fractions 6-10 should be kept.
3. Combine the pure fractions into an appropriately sized round-bottomed flask (no more than half-full, Figure 2.63b). Rinse each test tube with a small amount of eluent (or other solvent if solubility is an issue), and add the rinsing to the round-bottomed flask (Figure 2.63c).
4. Evaporate the solvent on the rotary evaporator to leave the purified compound in the flask.
Clean Up the Column
1. To dry the column, apply air pressure to drain the majority of eluent from the column into a waste container. Then further dry the column using one of these methods:
1. Leave connected a gentle stream of air running through the column to further dry it as you clean up other things (Figure 2.64a).
2. Clamp the column upside down over a large waste beaker in the fume hood, so that the adsorbent will fall when it has dried (Figure 2.64b). This will take a long time (until the next class period), but is an option.
2. When dry, the adsorbent can be poured out of the column into a waste container in the fume hood (Figure 2.64c).
Safety note: Powdered adsorbents are a lung irritant, and their hazard is exacerbated if the column contains residual compounds that can now make their way into your lungs. Pouring silica or alumina powders should always be done in the fume hood.
3. When the majority of the adsorbent has been collected in a waste container, use water to rinse any residual solid into the sink, and then rinse the column with acetone into a waste beaker. Further clean the column with soap and water and dry with the stopcock parts separated.
Macroscale Column Summary
Table 2.7: Procedural summary for macroscale column chromatography.
Make sure there is a frit or cotton plug in the bottom of the column.
Fill the column with silica or alumina to 5-6 inches in the fume hood.
Pour the adsorbent into an Erlenmeyer flask and add eluent to make a pourable slurry.
The eluent should give the desired component an $R_f$ of 0.35 by TLC.
Pour in the slurry and immediately rinse the sides of the column with eluent and a pipette.
Jostle the column to remove air bubbles.
Use air pressure to pack the column. Keep it perfectly vertical.
Add $0.5 \: \text{cm}$ of sand.
Rinse the sand off the sides carefully with a swirling motion.
Don't disrupt the top surface of the silica or alumina with rinsing.
Adjust the eluent level to the sand layer, and then add the sample (pure liquid, dissolved in $\ce{CH_2Cl_2}$, or solid adsorbed onto a portion of silica).
Rinse the sides and use air pressure to force the eluent down onto the silica/alumina layer.
Fill the reservoir with eluent (carefully to not disrupt the top surface).
Use steady air pressure to elute the column.
Collect fractions in test tubes in a rack (keep in order).
Rinse the column tip if a component has finished coming off the column.
Possibly increase the eluent polarity to make components elute faster.
Never allow the eluent to drop below the top of the adsorbent column.
Use TLC to determine the purity of the fractions, and combine appropriate fractions.
Remove the solvent with the rotary evaporator.
Troubleshooting
The Pipette Broke in the Column
It is quite common to break the tip of a Pasteur pipette while rinsing the column, which often falls and wedges itself into the delicate column. Unfortunately, a broken pipette in a column can cause problems with the separation of components.
For example, Figure 2.65 shows the effect of a broken pipette on the separation of two components. A broken pipette is embedded in the column and is the nearly vertical line of orange seen between the two bands on the column. Since stationary-mobile phase equilibration does not occur on the glass surface of the pipette, compounds stream down quickly and move faster than they should based on their $R_f$. The orange vertical line is the top component draining into the band of the bottom component, contaminating it.
If a broken pipette pierces the column, and the sand or sample has not yet been applied, attempt to remove the pipette with long forceps. After removal, vigorously jostle the column to pack it again and continue on with the column. If the pipette cannot be removed with forceps, you may consider redoing the column.
If the sand has already been applied, removal of the pipette and jostling the column will often ruin the horizontal surface of the adsorbent, so it will be necessary to re-pack the column.
If the sample has already been applied, there isn't anything to do but continue on and hope for the best. If the column results in an unsuccessful separation, all fractions containing the compound of interest can be combined, solvent evaporated by the rotary evaporator, and another purification method (or a second column) can be attempted.
Air Bubbles are Seen in the Column
Like a broken pipette, an air bubble is an empty pocket where stationary-mobile phase equilibration does not happen, so components move quicker around an air bubble than they should. This may lead to uneven eluting bands, which can cause overlap if the separation of the mixture is difficult (if the components have very close $R_f$ values, as in Figure 2.66).
If air bubbles are seen in the column and the sand or sample has not yet been applied, give the column a good jostling during packing to remove all air bubbles. See your instructor if the bubbles are not budging as you may be approaching the task too delicately. If the sand or sample has already been applied, it's best to leave the column as is and hope the air bubbles do not affect the separation.
The Bands are Eluting Unevenly
If the components of a mixture are colored, it may be obvious when the bands elute in a crooked manner. This is most likely due to the column being clamped at a slight diagonal.
If the column is clamped in a slanted manner, components will travel in a slanted manner (Figure 2.67). This may cause separation problems if the components have a similar $R_f$.
There is no way to fix this problem midway through a column, but if the components have very different $R_f$ values, the slanted bands may have no effect on the separation. In the future, be sure to check that the column is perfectly vertical in both the side-to-side and front-back directions.
$^8$W.C. Still, M. Kahn, A. Mitra, J. Org. Chem., Vol. 43, No. 14, 1978.
$^9$A typical small test tube $\left( 13 x 100 \: \text{mm} \right)$ has a capacity of $9 \: \text{mL}$, and a typical medium test tube $\left( 18 x 150 \: \text{mm} \right)$ has a capacity of $27 \: \text{mL}$.
$^{10}$W.C. Still, M. Kahn, A. Mitra, J. Org. Chem., Vol. 43, No. 14, 1978. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.04%3A_Column_Chromatography/2.4A%3A_Macroscale_Columns.txt |
A macroscale column is much too big for very small quantities of material $\left( < 20 \: \text{mg} \right)$. Instead, a column can be constructed using a disposable Pasteur pipette.
In order to get good separation, it is ideal if the desired component has an $R_f$ around 0.35 and is separated from other components by at least 0.2 $R_f$ units.$^{11}$ If the spots to be separated are very close (< 0.2 $\Delta R_f$), it's best if the middle of the spots has an $R_f$ of 0.35. An $R_f$ near 0.35 is ideal because it is slow enough that stationary-mobile phase equilibration can occur, but fast enough to minimize band widening from diffusion.
Step-by-Step Procedures
The column pictured in this section shows purification of a drop of dilute purple food dye (made from 1 drop red dye, 1 drop blue dye and 15 drops water). The dye is separated as best as possible into its three components: blue, red and pink (as seen in the TLC plate of Figure 2.68a). A 2.5" column of silica gel is used and eluted with a solution made from a 1:3:1 volume ratio of $6 \: \text{M} \: \ce{NH_4OH}$:1-pentanol:ethanol.
1. Run a TLC of the sample to be purified (Figures 2.69 a+b) to determine the appropriate solvent for chromatography. The desired component should have an $R_f$ around 0.35.
Prepare the dry column
1. Use a metal rod or hanger (Figure 2.69d) to wedge a bit of cotton or glass wool into the narrow end of a short-stemmed Pasteur pipette. The cotton should be moderately tight so liquid can trickle through, but not solid.
1. Scoop silica or alumina into the wide end of the pipette column (Figure 2.70a), then invert and raise the column so the powder falls to the bottom. Continue to use this scooping method to fill the pipette column to 2 - 2.5 inches high with silica or alumina (this quantity can be altered depending on the amount of sample).
Alternatively, scoop adsorbent into the wide end of a fresh pipette and use it as a funnel to deliver adsorbent through the narrow tip and into the pipette column secured to a ring stand with a three-fingered clamp (Figures 2.70 a–c).
Safety note: As silica and alumina are fine powders and lung irritants, be sure to work in a fume hood when handling silica or alumina. Also tap the pipette after scooping to dislodge adsorbent on the outside of the pipette (so it doesn't spill when vertical).
2. Gently clamp the pipette column to a ring stand or latticework using a three-fingered clamp (note: they are fragile!) and tap it to make sure the silica / alumina is settled and the top edge is horizontal.
3. Add approximately $0.5 \: \text{cm}$ of sand atop the silica / alumina layer. If using very fine sand, use another pipette to act as a funnel, as described in step 3. For coarse sand, use a small scooper or the wide end of another pipette to aid in its delivery (Figure 2.70d). A complete pipette column is in Figure 2.70e.
Wet the column
1. Position a test tube supported in a small beaker beneath the column. Add a squirt-full of the appropriate eluent (previously determined by TLC in Step 1), gently above the sand layer of the pipette column (Figure 2.71a).
2. Use a pipette bulb (or dropper bulb) to apply gentle air pressure and push the eluent through the column (Figure 2.71b), stopping when the liquid level is in the sand layer. Throughout the entire elution process, keep this white column section wet with eluent.
To apply air pressure using a pipette bulb, create a strong connection between the column and the bulb, and then squeeze the bulb. It is important when releasing the pressure to first break the seal while still keeping your hand clenched (Figure 2.71c) and THEN release your hand (Figure 2.71d). If you release your hand while still connected to the column, the bulb will create suction that can violently pull liquid into the bulb and disrupt the column.
3. Add more eluent if needed, and use bulb pressure until the entire column is saturated with eluent (Figure 2.71e), and the eluent level is in the sand layer.
Add the sample
1. Use a pipette to add the entire sample to the sand layer. If the sample is a liquid, add it directly. If it is a solid, dissolve it in the smallest amount of solvent possible, preferably the eluent. If the solid is not soluble in the eluent, use the minimum amount of dichloromethane. Position the pipette tip near the sand layer and add the sample carefully, trying not to splash compound onto the sides (Figure 2.72a).
2. Rinse the original container with a little solvent and add the rinsing to the column using the same pipette (in order to rinse the pipette as well).
3. Apply pressure with the bulb to force the sample just past the sand layer (Figures 2.72 b+c).
4. Add more eluent (approximately $0.5 \: \text{cm}$ high) to rinse the sides of the column (Figure 2.72d). Again use bulb pressure to force the dye onto the adsorbent (Figure 2.72e), and then fill the pipette above the sand layer as high as possible with eluent.
Elute the Column and Collect Fractions
1. Apply gentle bulb pressure to begin eluting the sample through the column (Figure 2.73a), refilling the pipette whenever the solvent level nears the sand layer.
2. Immediately begin collecting the liquid eluting beneath the column into an empty test tube. Change the test tube for a fresh one periodically (Figure 2.73c), based on your judgment or your instructor's guidance (perhaps when a small test tube fills to about $1 \: \text{cm}$ high).
These different tubes are called "fractions". The goal of a column is to collect small enough fractions that most (or some) fractions contain pure material. If the separation of the mixture is difficult (if the $\Delta R_f$ of the components is low), it may be best to collect even smaller fractions (e.g. $0.5 \: \text{cm}$ high).
3. Keep the test tubes in order on a test tube rack (Figure 2.73d).
Find and Concentrate the Desired Component
1. In finding the desired component in the test tube fractions, it is helpful to understand the relationship between $R_f$ and elution order in column chromatography.
In column chromatography, the sample is deposited on the top of the column and eluted down, while in thin layer chromatography the sample is spotted on the bottom of the plate and eluted up. Therefore, a column can be thought of like an upside-down TLC plate. A compound with a higher $R_f$ runs "faster," meaning it will end up higher on a TLC plate, and will be collected first with a column.
In the pipette column pictured in this section, the pink component had the highest $R_f$ on the TLC plate (Figure 2.74a), and was collected first from the column (Figure 2.74b). A purple fraction was also collected (Figure 2.74c), due to incomplete separation of the red and blue components.
2. Use TLC as described in the section on macroscale columns to determine which tubes contain the desired component.
3. Combine the pure fractions into an appropriately sized round-bottomed flask using a funnel, rinse each tube with a small amount of eluent (or other solvent if solubility is an issue), and add the rinsing to the flask. Remove the solvent on the rotary evaporator.
4. To clean up the pipette column, use bulb pressure to force the excess liquid out of the pipette column, and dispose of the semi-dry column (silica gel and all) in the broken glass container.
Microscale (Pipette) Column Summary
Table 2.8: Procedural summary for microscale (pipette) column chromatography.
Wedge a bit of cotton into the bottom of a pipette.
Use a scooping method to fill silica or alumina to 2-2.5 inches high.
Add a $0.5 \: \text{cm}$ layer of sand. Add eluent to the column and apply pressure with a pipette bulb to force eluent through the column to completely wet it.
Remember to break the seal before letting go of the pipette bulb, or suction will ruin the column.
Refill the column with eluent as necessary.
Adjust the eluent level to the sand layer, and then delicately add the sample.
Use pressure to push the eluent down onto the silica/alumina layer.
Rinse with one portion of eluent and push the solvent onto the column.
Fill the pipette with eluent and apply pressure to elute the column.
Collect liquid into test tubes.
Always keep the white column section wet (refill whenever the eluent level nears the sand layer).
Switch test tubes periodically (perhaps when they are $1 \: \text{cm}$ high in small test tubes) to collect different fractions.
Keep fractions organized in a test tube rack in the order they are eluted.
Use TLC to determine the purity of the fractions, and combine appropriate fractions.
Remove the solvent with the rotary evaporator.
$^{11}$W.C. Still, M. Kahn, A. Mitra, J. Org. Chem., Vol. 43, No. 14, 1978. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.04%3A_Column_Chromatography/2.4B%3A_Microscale_%28Pipette%29_Columns.txt |
Gas chromatography (GC) is a powerful instrumental technique used to separate and analyze mixtures. A gas chromatograph is a standard piece of equipment in forensics, medical, and environmental testing laboratories.
• 2.5A: Overview of GC
Gas chromatography (GC) is a powerful instrumental technique used to separate and analyze mixtures. A gas chromatograph is a standard piece of equipment in forensics, medical, and environmental testing laboratories .
• 2.5B: Uses of Gas Chromatography
A Gas Chromatography (GC) instrument is very good at verifying (or disproving) the purity of samples, and it can often spot trace quantities of impurity. Due to the precision of retention times, GC can be used to identify components of a mixture if pure samples are acquired and if the components separate on the column.
• 2.5C: Separation Theory
Gas Chromatography is an excellent analytical tool for separating mixtures in a sample. In this section are discussed the details of the separation.
• 2.5D: Quantitating with GC
Peak integrations are useful because it is possible to correlate the area under a peak to the quantity of material present in a sample. Note it is the area of a peak that is relevant, not the height. A tall, narrow peak may have a smaller area than a short, wide one. It would be very useful if peak areas would directly reflect mixture composition, but that is unfortunately not always the case. The reason for this has to do with how compounds are detected upon exit from a column.
• 2.5E: GC Parameters
There are many factors that affect the separation (and/or retention times) in gas chromatography, including the type of column, sample concentration, oven temperature, and flow rate of carrier gas. The factors controllable by a student are described in this section, and are related to concentration and temperature.
• 2.5F: Sample Preparation for Gas Chromatography
2.05: Gas Chromatography (GC)
Gas chromatography (GC) is a powerful instrumental technique used to separate and analyze mixtures. A gas chromatograph is a standard piece of equipment in forensics, medical, and environmental testing laboratories (Figure 2.75).
To run a GC, a sample is diluted then injected into the instrument where it is vaporized. The gaseous sample is pushed through a long, thin capillary column (typically 30 meters or approximately 100 feet long) by an inert carrier gas. The column separates components of a mixture, detects them, and a spectrum is generated displaying peaks that correspond to material that has exited the column at certain times. A GC spectrum of 87 octane gasoline is shown in Figure 2.76. Every peak represents one or more compounds, and since there are at least 50 peaks, this GC spectrum demonstrates that gasoline contains at least 50 different compounds!
The quantity of time a compound spends inside a GC column before it is detected is called its "retention time," which represents the time a compound is "retained" on the column. This value is the x-axis of the GC spectrum in minutes. In many instruments, the retention time of each peak can be labeled on the spectra with the click of a button, and the value is displayed above each peak maxima. A compound's retention time is similar to \(R_f\) in thin layer chromatography, and should be reproducible between identical runs. Unlike \(R_f\), however, retention times are very precise, and retention times typically vary by no more than 0.01 minute between identical runs (when the spectra have good peak shapes).
2.5B: Uses of Gas Chromatography
Assessing Purity
A GC instrument is very good at verifying (or disproving) the purity of samples, and it can often spot trace quantities of impurity. In Figure 2.77, the GC spectra of "n-hexane" and "hexanes" are shown, two reagents that can often be found in a chemical stockroom. The GC spectrum of n-hexane (Figure 2.77b) shows one prominent peak, although there are hints of three other peaks on the baseline. The bottle of n-hexane claims it to be $95\%$ pure, which is consistent with what is seen in its GC spectrum. "Hexanes" however, is a true mixture, as there are multiple significant peaks in its GC spectrum (Figure 2.77d). Hexanes contains n-hexane as the major component, but also contains closely related isomers (2-methylpentane, 3-methylpentane, and methylcyclopentane).$^{12}$ Hexanes is a commonly used solvent, where its lack of purity is not crucial, and it is considerably cheaper than pure n-hexane.
A sample that produces a GC spectrum with only one peak is not always pure. In Figure 2.78 is the GC spectrum of a sample containing both 2-pentanol and 3-pentanol, even though only a single peak is observed. The GC column used for this run separates based on differences in boiling point, and these isomers differ in boiling point by only $4^\text{o} \text{C}$, causing them to co-elute.
An impure sample can also produce a GC spectrum with only one peak if some components elute outside of the window of time detected by the instrument. For example, the GC spectrum in Figure 2.78 begins collecting data at $1.50 \: \text{min}$ (the value on the x-axis), and any compounds that elute before that time will not be detected. Components of a mixture may also not exit the column before the run is ceased, in which case they remain on the column and sometimes elute overtop of a subsequent run!
Identifying Components
Due to the precision of retention times, GC can be used to identify components of a mixture if pure samples are acquired and if the components separate on the column.
For example, Figure 2.79a shows a GC spectrum of a sample containing both 3-pentanol and 1-pentanol. Boiling points can be used to predict which peak represents which isomer, but another definitive method is to run a GC of one of the pure components. Figure 2.79b shows a GC spectrum of 1-pentanol, and its retention time closely matches one of the two peaks in the original mixture, thus enabling its identification.
Injection of known standards can be helpful in deciphering confusing mixtures, and for identifying closely eluting components, such as cis-trans isomers, where the difference in boiling points may be minimal. If using a mass spectrometer detector, a mass spectrum computer database is also helpful in identifying components.
$^{12}$Methylcyclopentane $\left( \ce{C_6H_{12}} \right)$ is not a true isomer of n-hexane $\left( \ce{C_6H_{14}} \right)$.
$^{13}$Complete removal of hexanes by rotary evaporation is sometimes difficult, and evidence of its persistence can be seen in $\ce{^1H}$ NMR samples by a doublet present around $0.9 \: \text{ppm}$ from the methyl of the methylcyclopentane impurity. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.05%3A_Gas_Chromatography_%28GC%29/2.5A%3A_Overview_of_GC.txt |
General Theory
GC is an excellent analytical tool for separating mixtures in a sample. In this section are discussed the details of the separation, and expand upon the general discussion of Section 2.1B.
The GC column is what allows for the separation of a mixture's components by the instrument. It is composed of a thin, hollow tube approximately $0.5 \: \text{mm}$ thick, and is lined with a very thin polymeric coating $0.1$-$5 \: \mu \text{m}$ thick$^{14}$ (Figure 2.80a). Columns can be 15-100 meters long$^{14}$ (although commonly are 30 meters in length) and can be swapped in and out of an instrument depending on an experimenter's goals. There are columns specifically designed for analyzing poisons, biodiesel samples, or chiral compounds. An example of a typical academic capillary column is shown in Figure 2.80b, which is lined with a "dimethyl polysiloxane" polymer.
Like all forms of chromatography, compounds equilibrate between a stationary phase and mobile phase. In GC, the stationary phase is the polymeric inner coating of the column, and compounds can interact with this high boiling liquid through various types of intermolecular forces (IMF's). The mobile phase is the carrier gas (usually helium), which is continually pushed through the column. Samples enter the column in the gas phase, subsequently adhere to the column coating, and establish an equilibrium between the stationary an mobile phases (equation 4).
$X_\text{(polymeric coating)} \rightleftharpoons X_\text{(gas)} \tag{4}$
Compounds move into the mobile phase if thermal energy from the oven provides sufficient energy to overcome the intermolecular forces between the compound and the column coating. When in the mobile phase, compounds are swept with the flow of the carrier gas, and re-adhere to the column coating further down the column. Compounds can be thought of as bouncing from one position to the next until they exit the column.
As all compounds have the same length to travel before they exit the column, they all spend the same amount of time in the mobile phase. Thus, separation is due to the varying amount of time spent in the stationary phase, or how long compounds adhere to the column coating. Compounds that have weak IMF's with the column coating spend little time "hung up" in the stationary phase, as heat from the oven allows for their IMF's to be broken. They quickly emerge from the column and thus have a short retention time. Conversely compounds that can form strong IMF's with the column coating will favor the stationary phase, and take a long time to emerge from the column (they will have a long retention time).
The type of IMF's experienced with the column coating often parallel the type of IMF's experienced in the pure liquid phase, meaning a compound that can hydrogen-bond in its liquid phase (Figure 2.81a) can also hydrogen bond with the column coating (Figure 2.81b). Therefore, the strength of interaction with the column coating correlates closely with a compound's boiling point, and boiling points can be used to predict the order of elution in GC.
In summary:
• Compounds that have weak IMF's with the column coating (low boiling points), spend little time in the stationary phase, exit the column early,and have shorter retention times.
• Compounds that have strong IMF's with the column coating (high boiling points), have longer retention times.
Structural Considerations
To demonstrate the effect of structural considerations, Figure 2.82 shows the GC spectrum of a mixture containing heptane, octane, nonane, and decane. The order of elution closely follows the trend in increasing boiling point. Heptane has the lowest boiling point of the series, and exits the column first (has the shortest retention time). As the chain length increases, so does the boiling point, and the retention times increase well (Table 2.9).
Table 2.9: Boiling point and retention time data for the GC spectrum in Figure 2.82.
Compound Boiling Point (°C) Retention Time (min)
Heptane ($C_{7}H_{16}$) 98 1.858
Octane ($C_{8}H_{18}$) 125 2.584
Nonane ($C_{9}H_{20}$) 150 3.526
Decane ($C_{10}H_{22}$) 174 4.446
In another example, Figure 2.83 shows the GC spectrum of a mixture containing heptane, hexanal, and 1-hexanol. The order of elution again follows the trend in boiling point. 1-hexanol has the strongest IMF's, as it can interact with the column through both London dispersion forces (LDF's) and hydrogen bonds. This causes it to adhere to the stationary phase most strongly, resulting in the longest retention time. Heptane interacts with the column coating through weaker LDF's, so spends the least amount of time in the stationary phase and elutes first (Table 2.10).
Table 2.10: Boiling point and retention time data for the GC spectrum in Figure 2.83.
Compound Boiling Point (°C) Retention Time (min)
Heptane 98 2.474
Hexanal 130 3.544
1-hexanol 155 4.288
$^{14}$D.C. Harris, Quantitative Chemical Analysis, 5$^\text{th}$ edition, 1982, p 676. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.05%3A_Gas_Chromatography_%28GC%29/2.5C%3A_Separation_Theory.txt |
Quantitative Challenges
Most modern GC instruments allow for integration of the peaks in a GC spectrum with the click of a button. If an older instrument is used, triangulation may be used to calculate the area under a peak, or the peaks may be cut out from a paper spectrum and weighed on an analytical balance to integrate. Peak integrations are useful because it is possible to correlate the area under a peak to the quantity of material present in a sample. Note it is the area of a peak that is relevant, not the height. A tall, narrow peak may have a smaller area than a short, wide one. It would be very useful if peak areas would directly reflect mixture composition, but that is unfortunately not always the case. The reason for this has to do with how compounds are detected upon exit from a column.
Common detectors at undergraduate institutions are flame-ionization detectors (FID) and mass spectrometry detectors (MS). With an FID, a sample is ignited upon exit from the column, and the ions produced in the flame are "counted" by the amount of electric current. Unfortunately, different compounds produce different amounts of ions when burned, meaning the ion count will not be directly equivalent to the molar quantity. Similarly, a mass spectrometer counts the number of ions produced from all the fragments detected in the instrument (called the "Total Ion Count", TIC, the y-axis on Figure 2.84), which does not always directly represent the quantity of the original material. Some compounds fragment more than others in a mass spectrometer, causing them to be detected in a manner that does not represent their true abundance.
To demonstrate how peak integration does not always correlate with actual abundance, GC spectra of two samples containing equimolar quantities of two compounds is shown in Figure 2.84. Figure 2.84a contains equal molar amounts of 1-butanol and 2-butanol, while Figure 2.84b contains equal molar amounts of 2-butanol and 1-heptanol.
The area under the peaks was tabulated by the mass spectrometer (represented by the column indicated by an arrow in each Figure), and was nearly $50\%$ for each component in the butanol system (Figure 2.84a). In this case, the percentages generated by the instrument were nearly identical to the true composition (although not likely to five significant figures!).
However, the sample containing equimolar quantities of 2-butanol and 1-heptanol was reported as a $\sim 20\%$-$80\%$ composition (Figure 2.84b). The true composition of the mixture is $50\%$ of each component, but the compounds are "counted" disproportionately. In this situation, the integrations do not represent the true abundance as 1-heptanol forms more fragments in its mass spectrum than 2-butanol, so is "counted" more frequently.
It is a general rule that the integrations in a GC spectrum (the percentages) can be used directly as a first approximation of the true composition when comparing isomers (e.g. 1-butanol and 2-butanol). When compounds are compared that are of different sizes (e.g. 2-butanol and 1-heptanol), the reported percentages may need to be translated into actual percentages using either a calibration curve or response factors.
Using a Calibration Curve
One method to translate the integration values given by the GC instrument into meaningful percentages that reflect the mixture's composition is to use a calibration curve. To generate a calibration curve, standard samples of known molar or mass ratios are injected into the GC and the percentages reported by the instrument recorded. The graph in Figure 2.85 shows such a calibration curve for a 2-butanol/1-heptanol mixture, as detected by a mass spectrometer.
For example, if a mixture known to be $60 \: \text{mol}\%$ 2-butanol and $40 \: \text{mol}\%$ 1-heptanol was injected into the instrument, and the resulting GC spectrum reported the mixture to be $42\%$ 2-butanol and $58\%$ 1-heptanol, this data point would be recorded on the graph as (60,42), using only the percentage of 2-butanol to simplify the situation. More data points could be analyzed using several known molar ratios of 2-butanol and 1-heptanol.
The calibration curve can then be used to correlate values reported by the GC instrument into true abundances. Imagine that a sample containing only 2-butanol and 1-heptanol is injected into the GC and reported by the instrument to be $70\%$ 2-butanol. Using the calibration curve in Figure 2.85, a measured value of $70\%$ 2-butanol (y-axis) correlates with an actual value of approximately $87$ $\text{mol}\%$ 2-butanol (x-axis). Therefore, a sample reported by the GC instrument to be $70/30\%$ 2-butanol/1-heptanol, is actually closer to $87/13\%$ 2-butanol/1-heptanol.
Due to the approximate nature of this method, percentage values should never be reported with more precision than the one's place, and probably the true error is on the order of $\pm 5\%$.
Using Response Factors
Another method that can be used to translate the integration values given by the GC instrument into meaningful percentages is to use a response factor. A sample containing equimolar quantities of the compounds to be analyzed is injected into the GC and the peak areas are numerically translated into compound sensitivities.
This process will be demonstrated with a mixture containing only 2-butanol and 1-heptanol. Figure 2.86 shows the GC spectrum of a sample that contains equimolar quantities of 2-butanol and 1-heptanol.
The peaks should have equal integrations because the compounds are present in equal molar amounts, but the peak at $2.38 \: \text{min}$ (1-heptanol) is "counting" more than it should. In the quantitative report, the column with the header "corr.area" can be used, as this shows the calculated area of each peak. The ratio of these area counts (see calculation below) gives a value of 3.86, which means that 1-heptanol is detected 3.86 times "better" than 2-butanol in these mixtures.
$\text{Ratio of corrected areas} = \frac{34558086 \: \text{area counts for 1-heptanol}}{8955039 \: \text{area counts for 2-butanol}} = 3.86$
To use this response factor in correlating values reported by the GC instrument into true abundances, imagine that a sample containing only 2-butanol and 1-heptanol is reported by the GC to be $30\%$ 1-heptanol (30 area counts of 1-heptanol for every 70 area counts of 2-butanol). As 1-heptanol is 3.86 times more responsive than 2-butanol, the true area counts would be more accurately represented as $30/3.86 = 7.77$ area counts of 1-heptanol for every 70 area counts of 2-butanol.$^{15}$
$\frac{30 \: \text{area counts for 1-heptanol}}{70 \: \text{area counts for 2-butanol}} \rightarrow \frac{30/3.86 \: \text{(1-heptanol)}}{70 \: \text{(2-butanol)}} = \frac{7.77 \: \text{(1-heptanol)}}{70 \: \text{(2-butanol)}}$
These hypothetical (and yet more accurate) area counts can then be turned into percentages, as follows:
$\% \: \text{1-heptanol} = \frac{7.77 \: \text{(1-heptanol)} \times 100\%}{70 + 7.77 \: \text{(total)}} = 10\% \: \text{1-heptanol, (meaning} \: 90\% \: \text{2-butanol)}$
Therefore, a sample reported by the GC instrument to be $70/30\%$ 2-butanol/1-heptanol, is actually closer to $90/10\%$ 2-butanol/1-heptanol, which is a similar value to what could be found using a calibration curve.
$^{15}$These calculations could be done more succinctly and with more precision using the actual area counts for each peak, found in the "corr.area" column of a quantitative GC report. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.05%3A_Gas_Chromatography_%28GC%29/2.5D%3A_Quantitating_with_GC.txt |
There are many factors that affect the separation (and/or retention times) in gas chromatography, including the type of column, sample concentration, oven temperature, and flow rate of carrier gas. The factors controllable by a student are described in this section, and are related to concentration and temperature.
Dilution
GC runs must be conducted on very dilute samples. A dilute sample enables adequate stationary-mobile phase equilibrium inside the column, resulting in narrow Gaussian-shaped peaks. If samples are too concentrated, the peak shapes will often be broad and flattened on top (Figure 2.87a), indicating the column (and/or detector) has been overwhelmed. Appropriate abundances (the y-axis on GC spectra) using a mass-spectrometer detector should be in the low millions (e.g. 1,500,000 TIC's or so, see Figure 2.87b).
A GC should never be run on undiluted liquid. Overly concentrated samples do not only cause peaks to broaden and possibly overlap, they can also be harmful to detectors. Mass spectrometers require very small quantities of liquid, or over time, their filaments degrade.
To prepare a routine GC sample, $1.5 \: \text{mL}$ of a low-boiling solvent (e.g. methanol, clean acetone, diethyl ether, or dichloromethane) is added to a GC vial (Figure 2.88) along with one drop of compound. the instrument then injects roughly $1 \: \mu \text{L}$ of this dilute solution into the instrument, which is further diverted internally, resulting in very small quantities entering the narrow capillary column and detector. The normal function of a GC instrument requires tiny quantities of material, which is a major advantage to this technique.
Solvent Delay (With Mass Spectrometers)
A properly prepared GC sample contains a small amount of compound dissolved in a solvent, and yet GC spectra (when paired with a mass spectrometer detector) often do not show any solvent peaks. The reason is that GC methods often include a "solvent delay", where the detector is turned off until a certain amount of time has passed after injection. Even with the dilution inside the GC-MS instrument, the quantity of solvent reaching the detector would overwhelm and eventually degrade it. Therefore, the detector is left inactive until the solvent has passed through the column, and activated afterwards.
If you look at any GC-MS spectrum, the retention time (x-axis) never starts at zero minutes, which represents the time the sample is injected on the column. In the GC spectrum of hexanes, the spectrum starts at 1.40 minutes, (as is indicated with an arrow in Figure 2.89). The solvent would have eluted from the column before this time.
The GC solvent must be chosen such that the solvent will elute before the compound of interest (the solvent must have a significantly lower boiling point than the compound), so that the solvent delay may be set between elution of the solvent and sample. For example, methanol (b.p. $65^\text{o} \text{C}$) would have been an unsuitable solvent for the sample of hexanes (b.p. $68^\text{o} \text{C}$). If the solvent delay was set such that the hexane signals could be detected, the GC spectrum would have been overwhelmed by signals from the methanol, as the two compounds have very similar boiling points and would elute rather closely. If the detector was activated after the methanol had eluted, the instrument would likely miss detection of the closely eluting hexanes. Therefore, a significantly lower boiling solvent was used (diethyl ether, b.p. $35^\text{o} \text{C}$) to produce the GC spectrum of hexanes in Figure 2.89, and this solvent would have eluted before the solvent delay of 1.40 minutes.
Oven Temperature
The temperature of the GC oven is an easily adjusted variable (although it should be adjusted by an instructor, not student). The oven temperature has a dramatic effect on retention times, much in a similar way that changing the mobile phase in TLC has on $R_f$.
To demonstrate, a sample containing heptane, octane, nonane, and decane was run using three GC methods, differing only in their starting temperature ($65^\text{o} \text{C}$, $45^\text{o} \text{C}$, or $35^\text{o} \text{C}$). The order of elution of the compounds was the same in all methods (Figure 2.90), but the components eluted faster with a hotter oven temperature (had lower retention times). This result can be explained in that the compounds were more able to overcome their intermolecular attraction with the stationary phase (the coating of the GC column) when there was more available thermal energy. This resulted in compounds spending less time adhering to the stationary phase, and less time retained by the column.
Using a Temperature Ramp
A GC can be run at one oven temperature the entire time (an "isothermal" run), or the method may include temperature programming. In a programmed method, the temperature may remain fixed for a certain length of time, and then may include a "ramp", where the temperature is increased at a steady rate during the run. A graphical display of a ramp generated by the GC instrument is shown in Figure 2.91.
Temperature programming is useful because the longer a compound stays in the column, the wider it elutes as diffusion broadens the sample (analogously to how spots broaden during elution in TLC). Figure 2.92a shows an isothermal GC run of four compounds, and the peak shape significantly broadens as the retention time increases. Figure 2.92b shows a GC run of the same four compounds eluted using a temperature ramp. The ramp allows the compounds to elute faster, and for peak shape to sharpen and improve. The optimal ramp for each situation often takes experimentation to achieve the best peak shape and separation of components.
2.5F: Sample Preparation for Gas Chromatography
Liquid GC Samples
1. Fill a GC vial to the $1.5 \: \text{mL}$ mark (Figure 2.93a) with a low boiling solvent (e.g. methanol, clean acetone, diethyl ether, or dichloromethane). Add one drop of the sample to be analyzed. If you think you possibly only added a half-drop, it's probably enough. Two drops are really too much.
2. Cap the vial and invert once or twice to dissolve the sample. If it appears like the sample drop did not dissolve fully, prepare another sample using a different solvent.
3. Alternatively, if you have already prepared a high-field NMR sample, use one-third of this sample and dilute it further with a low-boiling solvent.
4. Run the GC, as demonstrated by your instructor. Procedures vary at each institution.
Solid GC Samples
1. Some relatively volatile solids (never ionic solids!) can be analyzed by GC. Add one or two "specks" of solid$^{16}$ (a pile approximately $2 \: \text{mm}$ in diameter), or a very small spatula-tip of solid to a GC vial (Figure 2.93c). Then add a low boiling solid (e.g. methanol, clean acetone, diethyl ether, or dichloromethane) to the $1.5 \: \text{mL}$ mark.
2. Cap the vial and invert several times to fully dissolve the solid. If the solid dissolves, run the GC.
3. If the solid does not dissolve, do NOT run the GC anyway! Solids can plug the very small microliter syringes used by the instrument.
1. If the quantity of solid does not appear to have changed at all after adding the solvent, try making another sample with a different low boiling solvent.
2. If it appears the solid has decreased in quantity but is not fully dissolved, it is likely enough of it has dissolved to analyze. Use a pipette filter to filter the solution, and run the GC on the filtered liquid.
$^{16}$The quantity of solid may need to be adjusted if the solid is "fluffy". A GC is at the ideal dilution when it produces abundances around one million using a mass spectrometer detector. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.05%3A_Gas_Chromatography_%28GC%29/2.5E%3A_GC_Parameters.txt |
Crystallization is used in the chemistry laboratory as a purification technique for solids. An impure solid is completely dissolved in a minimal amount of hot, boiling solvent, and the hot solution is allowed to slowly cool. The developing crystals ideally form with high purity, while impurities remain in the saturated solution surrounding the solid (called the "mother liquor"). The crystallized solid is then filtered away from the impurities.
03: Crystallization
In many people's opinion, crystals are beautiful specimens of nature (Figure 3.1). Many crystals have long lines and edges, form geometric patterns, and have sheer surfaces that reflect light and cause them to sparkle. The macroscopic appearance of a crystal is a direct result of its microscopic structure, where a regular, repeating unit forms a three-dimensional crystal lattice. The uniformity and structural repetition differentiate a crystal from an amorphous solid. For this reason, there is a difference between precipitation (the rapid formation of a solid), and crystallization (the slow growth of a solid with regular microscopic structure). Precipitated solids tend to have lower purity than crystals, which is why crystallization is of interest to organic chemists (and geologists!).
Crystallization is used in the chemistry laboratory as a purification technique for solids. An impure solid is completely dissolved in a minimal amount of hot, boiling solvent, and the hot solution is allowed to slowly cool. The developing crystals ideally form with high purity, while impurities remain in the saturated solution surrounding the solid (called the "mother liquor"). The crystallized solid is then filtered away from the impurities. A diagram of the crystallization procedure is shown in Figure 3.2.
As an impure solid may have originally formed through partial crystallization, some call this procedure a "recrystallization", as the solid is allowed to crystallize a second time under more careful conditions. In this chapter the technique will be referred to as crystallization, which does not imply any previous process.
3.2A: Reagent Purification
The main use of crystallization in the organic chemistry laboratory is for purification of impure solids: either reagents that have degraded over time, or impure solid products from a chemical reaction.
• 3.2A: Reagent Purification
If NBS were used without previous crystallization, yields would undoubtedly suffer as the quantity of active reagent would be unknown, and side produces might form due to the presence of excess bromine.
• 3.2B: Purification of Product Mixtures
If the crude product of a chemical reaction is a solid, it may be crystallized in order to remove impurities.
3.02: Uses of Crystallization
\(\ce{N}\)-bromosuccinimide (NBS) is a reagent used for benzylic and allylic radical bromination, among other things. It is commonly found in its reagent jar as a yellow or orange solid (Figure 3.3a), as over time the solid degrades and generates \(\ce{Br_2}\), the source of the color. Before use, NBS should be crystallized from hot water, and the resulting crystals are pure white (Figure 3.3c). If NBS were used without previous crystallization, yields would undoubtedly suffer as the quantity of active reagent would be unknown, and side produces might form due to the presence of excess bromine.
3.2B: Purification of Product Mixtures
If the crude product of a chemical reaction is a solid, it may be crystallized in order to remove impurities. For example, benzoic acid can be brominated to produce m-bromobenzoic acid (Figure 3.4). The crude solid product could very likely contain unreactive benzoic acid, and this impurity could be removed through crystallization.
To demonstrate, a mixture containing roughly $85 \: \text{mol}\%$ p-bromobenzoic acid$^1$ (a solid) contaminated with $15 \: \text{mol}\%$ benzoic acid (another solid) had a yellow tint (Figure 3.5a), and after crystallization the resulting solid was pure white (Figure 3.5c). The crystallization appeared to purify the mixture based on the slight improvement in color.
More reliable than appearance, the crystallization was proven to have indeed purified the mixture through melting point and $\ce{^1H}$ NMR analysis of the crude and crystallized solids. The melting point of the crude solid was $221$-$250^\text{o} \text{C}$ while the melting point of the purified solid was $248$-$259^\text{o} \text{C}$ (literature melting point of p-bromobenzoic acid from Aldrich is $252$-$254^\text{o} \text{C}$). The melting point of the crystallized solid sharpened significantly, indicating greater purity. Additionally, in the crude solid's $\ce{^1H}$ NMR spectrum (Figure 3.6a), there are signals from both p-bromobenzoic acid and benzoic acid, and the integrations correlate reasonably well with the original composition of the mixture. In the crystallized solid's $\ce{^1H}$ NMR spectrum (Figure 3.6b), the benzoic acid signals are entirely gone, demonstrating that the crystallization successfully purified the p-bromobenzoic acid. An analogous purification could be done if a bromination reaction produced a mixture of m-bromobenzoic acid and benzoic acid.
$^1$p-Bromobenzoic acid was used instead of m-bromobenzoic acid for analysis purposes. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.01%3A_Overview_of_Crystallization.txt |
The crystallization procedure is possible as most solids tend to become more soluble in solvents as their temperature is increased.
• 3.3A: Ideal Temperature Profile
For the best crystallization, the compound should be very soluble in the hot solvent and minimally soluble (or insoluble) in the cold solvent.
• 3.3B: General Procedures for Removing Impurities
rystallization works well as a purification technique if impurities are present in very small quantities (less than 5 mol% of the solid), or if the impurities have a very different solubility profile from the desired compound. Impurities can be easily removed if they are either much more soluble or much less soluble in the solvent than the compound of interest.
• 3.3C: Determining Which Solvent to Use
The most important factor in the success of crystallization is probably the chosen solvent. Besides having the crucial solubility properties for crystallization (the compound should be soluble in the hot solvent and as insoluble as possible in the cold solvent), there are other factors that determine an appropriate solvent.
• 3.3D: Using Solubility Data
If you are not following a procedure where a crystallization solvent has been specified, it may be helpful to consult solubility data for the desired compound. Qualitative (and sometimes quantitative) solubility data can be found for many compounds in the CRC and Merck Index.
• 3.3E: Experimentally Testing Solvents
To experimentally determine a single solvent for crystallization, use the following procedure.
• 3.3F: Mixed Solvents
When no single solvent can be found that meets all of the criteria for crystallization, it may be possible to use a mixed solvent. A pair of solvents is chosen: one in which the compound is soluble (called the "soluble solvent"), and one in which the compound is insoluble (called the "insoluble solvent"). The two solvents must be miscible in one another so that their solubility with one another does not limit the proportions used.
3.03: Choice of Solvent
The crystallization procedure is possible as most solids tend to become more soluble in solvents as their temperature is increased (Figure 3.7 shows a yellow solid dissolving in ethanol as the temperature increases on a hotplate).
Caffeine is a white solid that follows this usual patter, as 1 gram of caffeine dissolves in $46 \: \text{mL}$ of room temperature water, $5.5 \: \text{mL}$ of $80^\text{o} \text{C}$ water, and $1.5 \: \text{mL}$ of boiling water.$^2$ If you have ever had a difficult time cleaning solidified starch residue from a kitchen colander (perhaps after you have strained pasta water), you may have learned that the colander is more easily cleaned in hot water than cold. Starch has an increased solubility in hot water, and this trend in solubility is partly what makes a dishwasher so effective.
The solubility profile of compounds is an empirical determination. Some solids increase solubility regularly with temperature (Figure 3.8a), and others increase in an exponential fashion (Figure 3.8b). In unusual cases, a solid's solubility can decrease with temperature. For the best crystallization, the compound should be very soluble in the hot solvent and minimally soluble (or insoluble) in the cold solvent.
$^2$From: The Merck Index, 12$^\text{th}$ edition, Merck Research Laboratories, 1996.
$^3$Data from: A. Seidell, Solubilities of Inorganic and Organic Substances, D. Van Nostrand Company, 1907. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.03%3A_Choice_of_Solvent/3.3A%3A_Ideal_Temperature_Profile.txt |
Crystallization works well as a purification technique if impurities are present in very small quantities (less than $5 \: \text{mol}\%$ of the solid), or if the impurities have a very different solubility profile from the desired compound. Impurities can be easily removed if they are either much more soluble or much less soluble in the solvent than the compound of interest.
Figure 3.9 shows the procedural sequence used to remove a "soluble impurity" from an impure soled, meaning an impurity that is embedded in the crystalline matrix, but would be completely soluble in the crystallization solvent. The impure solid is first fully dissolved in the minimum amount of hot solvent to liberate impurities trapped in the solid's interior. Upon cooling, a completely soluble impurity will remain dissolved in the mother liquor while the desired compound crystallizes. The crystallized solid can then be collected by suction filtration.
Figure 3.10 shows the procedural sequence used to remove an "insoluble impurity" from an impure solid, meaning an impurity that is embedded in the crystalline matrix, but would be insoluble in the crystallization solvent. The impure solid is heated in the minimum amount of hot solvent needed to dissolve the desired compound. The insoluble material is then filtered while the solution is kept hot (called "hot filtration"), and then the desired compound is crystallized and collected by suction filtration.
3.3C: Determining Which Solvent to Use
The most important factor in the success of crystallization is probably the chosen solvent. Besides having the crucial solubility properties for crystallization (the compound should be soluble in the hot solvent and as insoluble as possible in the cold solvent), there are other factors that determine an appropriate solvent.
An ideal crystallization solvent should be unreactive, inexpensive, and have low toxicity. It is also important that the solvent have a relatively low boiling point (b.p. often $< 100^\text{o} \text{C}$ as it's best if the solvent readily evaporates from the solid once recovered. Table 3.1 shows a list of common solvents used with crystallization. Toluene has the highest boiling point $\left( 111^\text{o} \text{C} \right)$ of the list, and should be avoided if alternatives exist for this reason (as well as its toxicity and smell). Along with fast evaporation, a relatively low boiling solvent is also ideal for crystallization as it minimizes the probability of a compound "oiling out", where material comes out of solution above its melting point and forms a liquid instead of a solid. When a compound liquefies first, it rarely crystallizes well.
Table 3.1: Boiling point of common solvents in crystallization. Note: petroleum ether is a mixture of hydrocarbons, and is very nonpolar. The term "ether" comes from its volatility, not the functional groups present.
Solvent Boiling Point (°C)
Diethyl ether 35
Acetone 56
Petroleum Ether (low boiling) 30-60
Ligroin (high boiling petroleum ether) 60-90
Methanol 65
Hexanes 69
Athyl acetate 77
Ethanol 78
Water 100
Toluene 111
Solvents with very low boiling points (e.g. diethyl ether, acetone, and low-boiling petroleum ether) are highly flammable and can be difficult to work with as they readily evaporate. They can still be used with care, but if alternatives exist, they are often preferable.
There are some general trends in predicting the appropriate solvent for a particular compound. As the compound needs to be soluble in the boiling solvent, it helps if the compound and solvent have similar intermolecular forces. For example, if a compound can hydrogen bond (alcohols, carboxylic acids, and amines), it sometimes can be crystallized from water. If a compound has moderate polarity it sometimes is crystallized from ethanol. If a compound is mostly nonpolar, it sometimes is crystallized from petroleum ether or hexanes, or may require a mixed solvent.
As it can be difficult to predict the ideal solvent for crystallization, often a published procedure from a journal article will indicate the appropriate solvent. If no solvent is listed, an appropriate solvent can be predicted using solubility data (next section), determined experimentally, or chosen from Perrin's Purification of Organic Chemicals,$^4$ a reference book which lists purification procedures for roughly 5700 known compounds.
$^4$D.D. Perrin, W.L.F. Armarego, Purification of Organic Chemicals, Pergamon Press, 3$^\text{rd}$ edition, 1988. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.03%3A_Choice_of_Solvent/3.3B%3A_General_Procedures_for_Removing_Impurities.txt |
If you are not following a procedure where a crystallization solvent has been specified, it may be helpful to consult solubility data for the desired compound. Qualitative (and sometimes quantitative) solubility data can be found for many compounds in the CRC$^5$ and Merck Index.$^6$
If the goal is to use just one solvent for crystallization, it's best to look for a solvent in which the desired compound is slightly soluble. Although the percent recovery will be less than $100\%$ (a portion will remain in the mother liquid), there is a good chance that a slightly soluble solid will dissolve when heated. It is possible that solvents listed as insoluble will not dissolve the compound at any temperature.
For example, below are entries in the CRC and Merck Index for the compound 1,4-dinitrobenzene:
• CRC Handbook: "i $\ce{H_2O}$; sl $\ce{EtOH}$, chl; s ace, bz, tol" Translation: The compound is insoluble in water; slightly soluble in ethanol and chloroform; and soluble in acetone, benzene, and toluene.
• Merck Index: "White crystals. . . one gram dissolves in $12,500 \: \text{mL}$ cold water, $555 \: \text{mL}$ boiling water, $300 \: \text{mL}$ alcohol; sparingly soluble in benzene, chloroform, ethyl acetate." Translation: The compound is very insoluble in cold water, but much more soluble in hot water. It is slightly soluble in ethanol, and basically insoluble in benzene, chloroform, and ethyl acetate.
1,4-dinitrobenzene is slightly soluble in ethanol, making it a good "first guess" as a crystallization solvent. Indeed Perrin's Purification of Organic Chemicals$^7$ recommends crystallization of 1,4-dinitrobenzene using ethanol or ethyl acetate.
$^5$Handbook of Chemistry and Physics, CRC Press, 84^\text{th}\) edition, 2003-2004.
$^6$The Merck Index, Merck Research Laboratories, 12$^\text{th}$ edition, 1996.
$^7$D.D. Perrin, W.L.F. Armarego, Purification of Organic Chemicals, Pergamon Press, 3$^\text{rd}$ edition, 1988.
3.3E: Experimentally Testing Solvents
To experimentally determine a single solvent for crystallization, use the following procedure. A flow chart describing this process is shown in Figure 3.11.
• Place $100 \: \text{mg}$ of the solid to be crystallized in a test tube and add $3 \: \text{mL}$ of solvent. (These quantities correspond with the usual guideline of solubility, where a compound is deemed "soluble" if $3 \: \text{g}$ of compound dissolves in $100 \: \text{mL}$ solvent.) Note: quantities may be scaled back depending on compound availability.
• While holding the test tube by the top, flick the tube to vigorously mix the contents.
• If the solid completely dissolves at room temperature, the solvent will not work for crystallization since it needs to be insoluble when cold. If the solid appears to remain insoluble, the solvent may work.
• Bring the solid suspended in the solvent to a boil using a steam bath or hot water bath. If the solid dissolves when the solvent is boiling, it may work for crystallization since it needs to be soluble when hot. If it never dissolves in the hot solvent, the solvent won't work.
• If the solid dissolves in the hot solvent, allow it to cool to room temperature, and then submerge it in an ice bath for 10-20 minutes. If most of the crystals return, the solvent should work for crystallization. If few or no crystals return, try to scratch the flask with a glass stirring rod to initiate crystallization. If crystals still don't return, the solvent won't work for crystallization.
This testing procedure is shown for $\ce{N}$-bromosuccinimide in Figure 3.12, using water as the solvent.
3.3F: Mixed Solvents
When no single solvent can be found that meets all of the criteria for crystallization, it may be possible to use a mixed solvent. A pair of solvents is chosen: one in which the compound is soluble (called the "soluble solvent"), and one in which the compound is insoluble (called the "insoluble solvent"). The two solvents must be miscible in one another so that their solubility with one another does not limit the proportions used. Table 3.2 shows a list of common mixed solvents used in crystallization.
Table 3.2: Common mixed solvent pairs in crystallization.
Solvent Pair
Methanol / Water
Ethanol / Water
Acetone / Water
Methanol / Diethyl ether
Petroleum Ether (or hexanes) / Diethyl ether
Hexanes / Ethyl acetate
To perform a crystallization using a mixed solvent, the solid to be crystallized is first dissolved in the minimum amount of hot "soluble solvent", then hot "insoluble solvent" is added dropwise until the solution becomes slightly cloudy. An additional small portion of hot soluble solvent is then added to clarify the solution, and the solution is set aside to slowly cool and crystallize. A diagram describing this process is shown in Figure 3.13.
Although this procedure can produce good results, when possible it is often best to use a single solvent for crystallization. As a mixed solvent is heated, the composition can change as the solvents evaporate at different rates, which can affect the solubility of the compound in the mixed solvent. Also, crystallizations from mixed solvents sometimes "oil out", where the dissolved compound comes out of solution above its melting point and forms a liquid instead of a solid.
To test a mixed solvent for crystallization, use the procedure that follows. This process is demonstrated by crystallizing trans-cinnamic acid from a mixed solvent of water and methanol (Figures 3.14 + 3.15).
1. Use the previously described method ($100 \: \text{mg}$ compound in $3 \: \text{mL}$ solvent), to find a pair of solvents: one solvent in which the compound is soluble and one solvent in which the compound is insoluble (Figure 3.14). The solvents must be miscible with one another.
2. Place a fresh $100 \: \text{mg}$ of the solid to be crystallized in a test tube and add the "soluble solvent" dropwise while heating (with a steam bath or hot water bath, Figure 3.15a) until the solid just dissolves. The suspension should be immersed in the heat source after each drop, and some time should be allowed in between additions for the sometimes slow dissolving process.
3. Add the "insoluble solvent" dropwise while heating until the solution becomes faintly cloudy (Figure 3.15c).
4. Add the "soluble solvent" dropwise while heating until the solution is clarified (clear).
5. Allow the system to cool to room temperature (Figure 3.15e), and then submerge in an ice bath for 10-20 minutes. If crystals return, the mixed solvent may work for the crystallization. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.03%3A_Choice_of_Solvent/3.3D%3A_Using_Solubility_Data.txt |
• 3.4A: Purification
Crystallization is an excellent purification technique for solids because a crystal slowly forming from a saturated solution tends to selectively incorporate particles of the same type into its crystal structure. A pure crystal is often slightly lower in energy than an impure crystal (or has a higher lattice energy), as packing identical particles into a lattice allows for maximized intermolecular forces.
• 3.4B: Cooling Slowly
The difference in crystal lattice energy between pure and impure solids is marginal, so a solution must be cooled SLOWLY to allow for differentiation. If a hot solution is plunged immediately into an ice bath, the system will favor the formation of a solid (any solid!) so strongly that there may be little preference for purity. Impurities can become engulfed in the developing solid and trapped as solutes are deposited unselectively onto the growing solid.
• 3.4C: Using the Minimum Amount of Hot Solvent
• 3.4D: The Unavoidable Loss of Recovery
• 3.4E: Quantitating Crystallization
• 3.4F: Second Crop Crystallization
As previously discussed, a portion of the compound of interest always remains dissolved in the mother liquor and is filtered away. This is not to say that this portion is lost, as it is possible to recover additional compound from the mother liquor. The solvent can be concentrated in the original vessel or on a rotary evaporator, and a second crystallization can be attempted. A second crystallization from the mother liquor of the first crystallization is called a "second crop crystallization".
3.04: Crystallization Theory
Crystallization is an excellent purification technique for solids because a crystal slowly forming from a saturated solution tends to selectively incorporate particles of the same type into its crystal structure (a model of a crystal lattice is in Figure 3.16a). A pure crystal is often slightly lower in energy than an impure crystal (or has a higher lattice energy), as packing identical particles into a lattice allows for maximized intermolecular forces.
If an impurity is smaller than the majority of particles in a crystal lattice, there may be a "dead space" around the impurity, resulting in a region with unrealized intermolecular forces. If the impurity is larger than the other particles, it may instead disrupt intermolecular forces by forcing other particles in the lattice out of alignment.
A developing solid will tend to incorporate particles of the same type in order to create the lowest energy solid, and exclude impurities that disrupt the idealized packing of the solid. This process works best if the impurity is present as a minor component of the crude solid. When higher quantities of impurities are present, the resulting crystal tends to be heterogeneous, with pure regions of compound and pure regions of "impurity" intermixed with impure regions. This sort of heterogeneity is often seen in the crystallization of minerals (Figure 3.16b).
3.4B: Cooling Slowly
The difference in crystal lattice energy between pure and impure solids is marginal, so a solution must be cooled SLOWLY to allow for differentiation. If a hot solution is plunged immediately into an ice bath, the system will favor the formation of a solid (any solid!) so strongly that there may be little preference for purity. Impurities can become engulfed in the developing solid and trapped as solutes are deposited unselectively onto the growing solid. It is only when enough time is allowed for equilibration between the developing solid and the solution that the lowest energy pure solid is formed.
As a metaphor for the slow crystallization process, imagine that you have won a prize and are allowed to go into a tube, where money in the form of \$1 and \$100 bills will be blown around for you to grab. The prize rules state that you're only allowed to keep a total of twenty bills. Imagine you are given just 30 seconds in the money tube to grab twenty bills: you would likely grab whatever bills you can get your hands on, and hope that most of them are \$100 bills. If instead given more time in the money tube (perhaps 10 minutes), you could be less frenzied and more selective about choosing \$100 bills while excluding \$1 bills.
In a similar way, the crystallization process can be though of as the crystal lattice "grabbing" solutes from solution. If the process is hurried, solutes may be "grabbed" indiscriminately, and once embedded in the interior of the solid they become trapped as the equilibrium between solid and solution happens only on the surface.
Along with increased purity, a slow crystallization process also encourages the growth of larger crystals. Figure 3.17 shows crystallization of acetanilide from water with two different rates. The crystals grown in Figure 3.17a were formed much more quickly, and are smaller than the slower grown, larger crystals in Figure 3.17b.
There are several benefits to larger crystals. Most importantly they tend to be purer than small crystals. They also have a lower surface area exposed to the mother liquid, which makes washing of the crystals more effective. Finally, they are more easily collected by suction filtration (Figure 3.18), as very small crystals may pass through or wedge themselves into the pores of the filter paper, preventing solvent from easily trickling through and leading to crystals that are irrecoverable. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.04%3A_Crystallization_Theory/3.4A%3A_Purification.txt |
The quantity of solvent used in crystallization is usually kept to a minimum, to support the goal of recovering the maximum amount of crystals.
Every solid has partial solubility in the solvents used, even at cold temperatures. In Figure 3.19 the cold solvent surrounding a yellow solid is tinted yellow as some compound dissolves. Solids that appear insoluble in a solvent do in fact have a (normally small) portion of material that dissolves. This is analogous to how "water-insoluble" ionic compounds (such as $\ce{AgCl}$ have a non-zero solubility product constant $\left( K_\text{sp} \right)$.
Crystallization is most common with solids that have moderate solubility at low temperatures, so that heat can "tip them over the edge" to completely dissolve. This means that in practice there will be a quantity of compound that dissolves in the mother liquor at low temperatures, which can be significant depending on the compound's solubility profile. Use of the minimal amount of hot solvent lessens the quantity of compound that is lost to the mother liquor.
To demonstrate the importance of using the minimum amount of hot solvent during a crystallization, imagine you are to crystallize $5.0 \: \text{g}$ 2-furoic acid using hot water.$^8$ The calculations from this example are described below, and summarized in Figure 3.21.
2-furoic acid has an increased solubility in boiling water compared to cold water, as shown by the solubility data in Figure 3.20. The quantity of hot solvent needed to dissolve this sample can be calculated using the compound's solubility in hot water, as shown below. This represents the "minimum amount of hot solvent" needed for the crystallization.
$5.0 \: \text{g 2-furoic acid (FA)} \times \frac{4 \: \text{mL hot water}}{1 \: \text{g FA}} = \textbf{20. mL hot water}$
When the 2-furoic acid is dissolved in $20 \: \text{mL}$ of hot water and is allowed to cool, the compound will crystallize as it has a lower solubility in the cold solvent. Ideally the solution would be placed in an ice bath $\left( 0^\text{o} \text{C} \right)$, but as $0^\text{o} \text{C}$ solubility data is not provided, let's imagine the solution is cooled to $15^\text{o} \text{C}$. The solubility of 2-furoic acid in this cold water is as follows.
$20. \: \text{mL cold water} \times \frac{1 \: \text{g FA}}{26 \: \text{mL cold water}} = \textbf{0.77 g FA} \: \text{dissolves in } 15^\text{o} \text{C} \: \text{water}$
This calculation shows that a portion of the 2-furoic acid will remain dissolved in the mother liquor even when placed in the cold bath. Any compound present that exceeds this quantity will crystallize, in this case $4.2 \: \text{g}$ of 2-furoic acid $\left( 5.0 \: \text{g} - 0.77 \: \text{g} \right)$. Note that because of the loss of material to the mother liquor, the maximum theoretical recovery from this process is only $84\%$ $\left( 100 \times 4.2 \: \text{g}/5.0 \: \text{g} \right)$.
Next imagine that instead of using the minimum amount of hot solvent, we use double the volume of hot solvent ($40 \: \text{mL}$ in this example). The calculations from this example are described below, and summarized in Figure 3.22.
When the solvent volume is doubled, the quantity of material dissolved in the mother liquor also doubles, as shown in the calculation below.
$40. \: \text{mL cold water} \times \frac{1 \: \text{g FA}}{26 \: \text{mL cold water}} = \textbf{1.5 g FA} \: \text{dissolves in } 15^\text{o} \text{C} \: \text{water}$
If a $5.0 \: \text{g}$ sample of 2-furoic acid were instead dissolved in $40 \: \text{mL}$ (double the minimal amount of hot solvent necessary), only $3.5 \: \text{g}$ would crystallize $\left( 5.0 \: \text{g} - 1.5 \: \text{g} \right)$. The maximum theoretical recovery for this process is only $70\%$.
Notice when additional solvent is used than the minimum, more compound is soluble in the solvent and lost to the mother liquor, leading to a lower theoretical recovery.
As you might imagine, there is also a quantity of solvent in which the entire $5.0 \: \text{g}$ sample of 2-furoic acid dissolves (in this case $130 \: \text{mL}$, see calculation below). If $130 \: \text{mL}$ were used for the crystallization instead of the minimal amount of hot solvent, no crystals would form at all when cooled to $15^\text{o} \text{C}$ ($0\%$ recovery). Thus the quantity of solvent used in crystallization should be monitored and restricted when possible.
$5.0 \: \text{g FA (2-furoic acid)} \times \frac{26 \: \text{mL cold water}}{1 \: \text{g FA}} = \textbf{130 mL} \: \text{of } 15^\text{o} \text{C} \: \text{water dissolves the whole sample}$
To demonstrate the importance of using the minimum amount of hot solvent, two samples of benzil (each roughly $1.8 \: \text{g}$), were crystallized from hot ethanol (Figure 3.23). The first was dissolved in the minimum amount of hot solvent (approximately $8 \: \text{mL}$), and resulted in a $92\%$ recovery after crystallization. The second was dissolved in roughly double the quantity of hot solvent (approximately $15 \: \text{mL}$) and resulted in an $87\%$ recovery. The solubility profile of benzil led to an excellent recovery in both trials, but the recovery was compromised somewhat when an excess of solvent was used.
$^8$As recommended in D.D. Perrin, W.L.F. Armarego, Purification of Organic Chemicals, Pergamon Press, 3$^\text{rd}$ edition, 1988. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.04%3A_Crystallization_Theory/3.4C%3A_Using_the_Minimum_Amount_of_Hot_Solvent.txt |
A loss of recovery should be expected when performing a crystallization. Although there are ways to maximize the return of crystals, a portion of the desired compound will always be lost. The reasons for this are both inherent to the design of the process and mechanical.
As previously discussed, a portion of the compound of interest will remain dissolved in the mother liquor and be filtered away. Figure 3.24 shows suction filtration in order to recover benzil (a yellow solid) that had been crystallized from ethanol. In Figure 3.24c it is apparent that the filtrate is also yellow (the liquid that has passed through the filter paper), making it obvious that some benzil remained dissolved in the solvent. Further evidence that some compound is always lost to the mother liquor can be seen when solvent evaporates from drips on glassware or the benchtop, revealing residual solid (Figure 3.25a+b).
The loss of solid material can also be witnessed with every manipulation of the solid. Only the majority of the crystals can be delicately scraped off the glassware, Buchner funnel, and filter paper, and there will always be a residue that is left behind (Figure 3.25c).
When there is such an obvious loss of yield from solid clinging to glassware, it may seem wise to use solvent to rinse additional solid out of the flasks. A few rinses with cold solvent are indeed recommended, but it is not recommended to use solvent excessively in an attempt to recover every granule of solid. The more solvent that is used, the more compound will dissolve in the cold mother liquor, decreasing the yield. The loss of material due to residue on the glassware is an unfortunate, but accepted aspect of this technique.
To demonstrate the loss of yield with crystallization, several pure samples of acetanilide and benzil were crystallized. Using pure samples allowed for any loss of material to be only due to the solubility in the mother liquor and adhesion to the glassware, not by the exclusion of impurities, which also have mass. From many trials of crystallizing acetanilide from hot water using different scales (between $0.5 \: \text{g}$-$1.5 \: \text{g}$ each time), the recoveries were surprisingly consistent, between $60$-$65\%$ (Figure 3-26). From many trials of crystallizing benzil from hot ethanol using different scales (between $0.5 \: \text{g}$-$4.5 \: \text{g}$ each time), the recoveries were also quite consistent, between $87$-$92\%$ (benzil is the yellow solid in Figure 3.24). In all situations, the recovery of solid was never $100\%$. Also, the typical recovery for the two systems was different, with the most logical explanation being that there was a larger percentage of compound lost to the mother liquor with acetanilide than with benzil.
It is common for new organic chemistry students to be disappointed by a low yield (anything less than $95\%$!), and to worry that it is somehow their fault. Students are often inclined to cite "user error" as a main cause for a loss of yield in any process. Although spilling solid on the benchtop or addition of too much solvent will of course compromise the yield, the modest recovery of acetanilide in this section should demonstrate that sometimes low yields are inherent to the process and the chosen solvent. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.04%3A_Crystallization_Theory/3.4D%3A_The_Unavoidable_Loss_of_Recovery.txt |
In this section, real solubility data for compounds is used to quantitatively describe the purification of impure solids that contain either a "soluble impurity" or an impurity with similar solubility to the compound of interest.
With a "Soluble Impurity"
The simplest crystallization in terms of purification is when an impurity is very soluble in the cold solvent while the compound of interest is not (see procedural sequence in Figure 3.27). In this situation, the impurity is trapped in the crystal matrix of the impure solid, and needs only be liberated by dissolution. After adding hot solvent to dissolve the impure solid then cooling, the soluble impurity remains dissolved in the mother liquor while the compound of interest crystallizes and can be collected by suction filtration.
An example of this type of system is a sample containing $4.00 \: \text{g}$ acenapthene and $0.50 \: \text{g}$ acetanilide. According to Perrin's Purification of Organic Compounds,$^9$ acenapthene can be purified by crystallization using ethanol, and even though this sample contains a considerable quantity of acetanilide impurity ($13 \: \text{mol}\%$), the calculations in this section will show that this process should work well.
The solubility values of acenapthene and acetanilide in ethanol are shown in Figure 3.28.
$4.00 \: \text{g}$ of pure acenapthene (AN) should completely dissolve in $12.5 \: \text{mL}$ (see calculation below) of nearly boiling ethanol (the boiling point of ethanol is $78^\text{o} \text{C}$).$^{11}$
$4.00 \: \text{g Acenapthene (AN)} \times \frac{100 \: \text{mL hot ethanol}}{32.0 \: \text{g AN}} = \textbf{12.5 mL hot ethanol} \: \left( 70^\text{o} \text{C} \right) \: \text{to dissolve}$
Since acetanilide (AT) is more soluble than acenapthene at this temperature (see Figure 3.28), it should also completely dissolve in $12.5 \: \text{mL}$ of hot ethanol.
If the hot ethanolic solution containing these two components is then cooled in an ice bath, acenapthene should crystallize as it has low solubility in cold ethanol. The calculation below shows how much acenapthene would remain dissolved in the cold ethanol $\left( 0.19 \: \text{g} \right)$. Any quantity of acenapthene present greater than $0.19 \: \text{g}$ would form a solid, in this case $3.81 \: \text{g}$ $\left( 4.00 \: \text{g} - 0.19 \: \text{g} \right)$.
$12.5 \: \text{mL cold ethanol} \times \frac{1.5 \: \text{g acenapthene (AN)}}{100 \: \text{mL cold ethanol}} = \textbf{0.19 g AN} \: \text{remains dissolved in} \: 0^\text{o} \text{C} \: \text{ethanol}$
The acetanilide impurity, however, should not crystallize. The calculation below shows that $1.45 \: \text{g}$ of acetanilide can dissolve in the cold ethanol, and since only $0.50 \: \text{g}$ was originally present, the entire portion would remain dissolved.
$12.5 \: \text{mL cold ethanol} \times \frac{11.6 \: \text{g AT}}{100 \: \text{mL cold ethanol}} = \textbf{1.45 g AT} \: \text{can dissolve in} \: 0^\text{o} \text{C} \: \text{ethanol}$
This process is summarized in Figure 3.29. As only the acenapthene crystallizes, it could be collected through filtration and separated from the acetanilide impurity in the mother liquor.
Purification of this mixture works well in theory because the acetanilide is so much more soluble (approximately ten times as much) in the cold ethanol than the acenapthene, and so can be removed. In practice if the crystallization is done too quickly, there is still the possibility of acetanilide being incorporated into the developing solid.
With an Impurity of Similar Solubility
If an impure solid contains an impurity that has similar solubility properties as the desired compound, it is still possible to purify the mixture through crystallization if the impurity is present in a small amount. An example is a sample containing $4.00 \: \text{g}$ acenapthene and $0.30 \: \text{g}$ phenanthrene (phenanthrene is $6 \: \text{mol}\%$ of the sample). The solubility of these compounds in ethanol is shown in Figure 3.30.
$4.00 \: \text{g}$ of pure acenapthene (AN) should completely dissolve in $12.5 \: \text{mL}$ (see calculation below) of nearly boiling ethanol (the boiling point of ethanol is $78^\text{o} \text{C}$. $0.30 \: \text{g}$ of pure phenanthrene (PH) would dissolve in $1.2 \: \text{mL}$ of hot ethanol (see calculation below), and so would also dissolve in $12.5 \: \text{mL}$ were used.
$4.00 \: \text{g Acenapthene (AN)} \times \frac{100 \: \text{mL hot ethanol}}{32.0 \: \text{g AN}} = \textbf{12.5 mL hot ethanol} \: \left( 70^\text{o} \text{C} \right) \: \text{to dissolve AN}$
$0.30 \: \text{g Phenanthrene (PH)} \times \frac{100 \: \text{mL hot ethanol}}{26.0 \: \text{g PH}} = \textbf{1.2 mL hot ethanol} \: \left( 70^\text{o} \text{C} \right) \: \text{to dissolve PH}$
If this hot solution is then cooled in an ice bath, the calculations below show the quantity of each compound that would remain dissolved in the cold ethanol.
$12.5 \: \text{mL cold ethanol} \times \frac{1.5 \: \text{g acenapthene (AN)}}{100 \: \text{mL cold ethanol}} = \textbf{0.19 g AN} \: \text{remains dissolved in} \: 0^\text{o} \text{C} \: \text{ethanol}$
$12.5 \: \text{mL cold ethanol} \times \frac{2.88 \: \text{g phenanthrene (PH)}}{100 \: \text{mL cold ethanol}} = \textbf{0.360 g PH} \: \text{remains dissolved in} \: 0^\text{o} \text{C} \: \text{ethanol}$
Since greater than this quantity of acenapthene is present, most of it will crystallize $\left( 3.81 \: \text{g} \right)$. However, since only $0.30 \: \text{g}$ of phenanthrene is present, the entire portion should remain in the mother liquor. The two components can then be separated through filtration.
In essence, impurities that have a similar solubility to the compound of interest can be removed like soluble impurities as long as they are present in small enough quantities. It is a general guide that solids containing less than $5 \: \text{mol}\%$ of impurity can be purified in this manner. The crystallization process essentially "sacrifices" a portion of each component to the mother liquor in order to produce a pure crystal.
$^9$D.D. Perrin, W.L.F. Armarego, Purification of Organic Chemicals, Pergamon Press, 3$^\text{rd}$ edition, 1988.
$^{10}$Adapted from data found in: A. Seidell, Solubilities of Inorganic and Organic Substances, D. Van Nostrand Company, 1907.
$^{11}$It is very possible that the acenapthene/acetanilide mixture might require a slightly different quantity of hot ethanol to dissolve than pure samples, but the calculations in this section provide a rough estimate of the solvent volumes needed.
$^{12}$Adapted from data found in: A. Seidell, Solubilities of Inorganic and Organic Substances, D. Van Nostrand Company, 1907. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.04%3A_Crystallization_Theory/3.4E%3A_Quantitating_Crystallization.txt |
As previously discussed, a portion of the compound of interest always remains dissolved in the mother liquor and is filtered away. This is not to say that this portion is lost, as it is possible to recover additional compound from the mother liquor. The solvent can be concentrated in the original vessel or on a rotary evaporator, and a second crystallization can be attempted. A second crystallization from the mother liquor of the first crystallization is called a "second crop crystallization".
To demonstrate this process, a $1.16 \: \text{g}$ sample of trans-cinnamic acid was crystallized from a methanol/water mixed solvent system (Figure 3.31). $0.95 \: \text{g}$ of a "first crop" of crystals was collected, representing an $82\%$ recovery.
To collect a second crop of crystals, the filtrate was pipetted into an Erlenmeyer flask (Figure 3.32a) and boiled to reduce the volume by half (Figures 3.32b+c). When this solution was cooled, another $0.08 \: \text{g}$ of compound crystallized out (Figure 3.32d), making the combined recovery of trans-cinnamic acid $1.03 \: \text{g}$, or $89\%$.
A "second crop" solid should always be kept separate from a "first crop" solid until its purity can be verified. A second crop crystal is usually more impure than a first crop crystal, as it crystallizes from a solution that contains a higher percentage of impurities (the first crop removed more compound, leaving more impurities behind).
To demonstrate the relative purity of first crop and second crop crystals, a $1.5 \: \text{g}$ mixture of 4-bromoacetophenone contaminated with a small amount of acetophenone was crystallized using ethanol. A GC spectrum of the original mixture (Figure 3.33b) shows the sample to have approximately $7 \: \text{mol}\%$ acetophenone impurity.
Quantitation by a GC-MS instrument can sometimes be inaccurate, so the composition was also calculated through $\ce{^1H}$ NMR integrations. The partial $\ce{^1H}$ NMR spectrum of this sample is in Figure 3.34, and integrations indicate the sample is $5.6 \: \text{mol}\%$ acetophenone. There is good agreement between the two quantitative methods.
After the mixture was crystallized from ethanol, a GC spectrum of the resulting solid ($0.93 \: \text{g}$, $63\%$ recovery) showed a dramatic reduction in the acetophenone impurity, to almost negligible amounts (Figure 3.35b). The crystallization purified the 4-bromoacetophenone very well.
The filtrate from the first crystallization was boiled to reduce the solvent level, and a second crop of crystals was obtained ($0.18 \: \text{g}$, combined recovery of $77\%$). The second crop crystals had a noticeable yellow tint while the first crop crystals were pure white. A GC spectrum of the second crop crystals (Figure 3.36b) showed the acetophenone had incorporated somewhat into the crystalline solid, demonstrating that second crop crystals are less pure than first crop crystals.
Recovery of a second crop of crystals is not commonly done in undergraduate settings as the additional recovery from a second crop is usually modest (Figure 3.37). However, the process may be done if the material is quite expensive, or if a large amount remains dissolved in the mother liquor. For the latter reason, it is a good idea to save the mother liquor until a yield can be calculated. If the recovery is quite low, a second crystallization might be attempted.
$^{13}$GC method: $100^\text{o} \text{C}$ hold for 2 min, $25^\text{o} \text{C}$ ramp per min to $210^\text{o} \text{C}$ (6.4 min), solvent delay 2.00 min. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.04%3A_Crystallization_Theory/3.4F%3A_Second_Crop_Crystallization.txt |
• 3.5A: General Procedure
You will often be asked to follow a crystallization procedure that may be written in one of two varieties: Solid A was crystallized using Y solvent, or Solid A was crystallized using X mL of Y solvent. When no volume of solvent is specified, the process should be conducted using the minimum volume of boiling solvent to dissolve the solid. If a volume is specified, it probably means that previous trials have shown that a greater than minimal solvent volume produce higher quality crystals.
• 3.5B: Heat Source and Bump Prevention
Crystallizations are generally conducted using a hotplate or steam bath, and the crystallization solvent may affect which heat source is appropriate. Boiling stones, boiling sticks, or stir bars must be used for "bump" protection as crystallization involves heating solutions to a boil. Boiling stones are often used for solvents, and may be used with solutions if a hot filtration step is planned.
• 3.5C: Charcoal
Activated charcoal is sometimes used to remove small amounts of colored impurities from solution. Activated charcoal has a high affinity for conjugated compounds, whose flat structures wedge themselves well between the graphene sheets. The quantity used should be limited, as charcoal adsorbs all compounds to some extent and could lead to a lower recovery of the desired compound. Charcoal should of course not be used if the product itself is colored.
• 3.5D: Cooling Slowly
After a solution is dissolved in the minimum amount of hot solvent and filtered (if applicable), the solution should be cooled as slowly as possible (keeping in mind time limitations in a lab).
• 3.5E: Initiating Crystallization
At times, crystals will not form even when a solution is supersaturated, as there is a kinetic barrier to crystal formation. At times crystallization may need to be initiated, for example if the solution becomes faintly cloudy as it cools, or if the solution fails to produce crystals even when it is noticeably cooler than originally. The methods described in this section preferably should be used on solutions that are still warm to avoid a too rapid crystallization.
3.05: Procedural Generalities
Most often you will be asked to follow a published crystallization procedure that may be written in one of two varieties:
• Solid A was crystallized using Y solvent, or
• Solid A was crystallized using X mL of Y solvent
When no volume of solvent is specified, the process should be conducted using the minimum volume of boiling solvent to dissolve the solid. If a volume is specified, it probably means the crystallization is tricky and previous trials have shown that a greater than minimal solvent volume produce higher quality crystals, or attempting to dissolve common impurities leads to using too great a quantity of solvent.
The standard crystallization process (Figure 3.38a) is summarized as follows:
1. Dissolve the impure solid in the minimum amount of hot solvent.
2. Slowly cool the solution to induce crystallization.
3. Further cool the solution in an ice bath.
4. Collect the solid by suction filtration.
An additional step may be inserted into this general process (between steps 1 and 2) if insoluble impurities are present, or if impurities will be removed with charcoal (Figure 3.38b). After the compound is dissolved in the minimal amount of hot solvent, charcoal can be used, or insoluble impurities removed with hot filtration. The general process can then be continued on to next cool the solution and induce crystallization. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.05%3A_Procedural_Generalities/3.5A%3A_General_Procedure.txt |
Crystallizations are generally conducted using a hotplate or steam bath, and the crystallization solvent may affect which heat source is appropriate (Table 3.3). If the crystallization solvent is water, a hotplate can be used as there is no issue with flammability. If the crystallization solvent is highly flammable or volatile (e.g., diethyl ether, low-boiling petroleum ether, or acetone), the vapors produced during boiling could ignite on the hot surface of a hotplate, and it is therefore essential to use a steam bath. Solvents with moderate volatility (e.g., ethanol or ethyl acetate), should be preferably heated with a steam bath, but can be safely heated with a hotplate if extreme care and caution is utilized.
Table 3.3: Solvent and preferred heat source.
Solvent Water Diethyl ether, Acetone, Petroleum Ether (low-boiling) Ethanol, Methanol, Ethyl Acetate, Hexanes
Photo of Substance
Heat source Use a hotplate. Use steam bath, never a hotplate Use steam bath preferably. Can use hotplate with caution.
Boiling stones, boiling sticks, or stir bars must be used for "bump" protection as crystallization involves heating solutions to a boil. Boiling stones are often used for solvents, and may be used with solutions if a hot filtration step is planned (or if the plan is to pick the stones out of the crystallized solid). If no filtration step is planned, it may be easier to use a boiling stick or stir bar for bump protection, as these can be easily removed from the hot solution before crystallization.
3.5C: Charcoal
Activated charcoal is sometimes used to remove small amounts of colored impurities from solution. Activated charcoal has a high affinity for conjugated compounds, whose flat structures wedge themselves well between the graphene sheets. The quantity used should be limited, as charcoal adsorbs all compounds to some extent and could lead to a lower recovery of the desired compound. Charcoal should of course not be used if the product itself is colored.
Decolorizing charcoal (Norit) is added after a solid has been dissolved in the minimum amount of hot solvent. A small portion should be used at first; just as much that can fit on the tip of a spatula (the size of half a pea, see Figure 3.39b). When adding charcoal, it's recommended to temporarily remove the solution from the heat source or the charcoal's high surface area may cause superheated areas to immediately boil. Superheated areas are regions where the temperature is above the boiling point of the solvent, but lack a nucleation site to create a bubble and initiate the phase change. If charcoal is added directly to a near boiling solution, the solution may boil over.
For fairly colored solutions, the first portion of charcoal is likely all that is necessary, and the resulting solution should be a faint grey (Figure 3.39c). Additional charcoal can be added if the color remains (Figure 3.40c). Charcoal particles are so fine that hot filtration is necessary to remove them before crystallization.
Charcoal does not need to be used with every solution that's colored, even if the desired compound is white. Colored impurities may remain in the mother liquor after filtration. For example, the sequence in Figure 3.41 shows the purification of acetanilide (a white solid) that has been contaminated with several drops of methyl red solution to produce an orange solid (Figure 3.41a). The solid was crystallized from hot water without the use of charcoal, and even though the mother liquor was yellow (Figure 3.41b), the crystallized solid was still a pure white color (Figure 3.41c). The methyl red "impurity" remained in the mother liquor and was not obviously incorporated into the crystal lattice of acetanilide.
As a comparison, the same solid was decolorized with charcoal, filtered and crystallized, as shown in Figure 3.42. The charcoal removed the methyl red color, and the mother liquor was colorless (as indicated with an arrow in Figure 3.42b), but there was no obvious improvement of color in the resulting crystallized solid (Figure 3.42c). Additionally, there was a loss of yield when using charcoal. When charcoal was not used, the recovery was \(63\%\), which was consistent with various other trials of acetanilide. However, when charcoal was used, the recovery was \(53\%\). The loss of yield may be attributed to charcoal's absorption of the target compound along with the colored impurity, as well as loss of compound on the filter paper during the hot filtration. For this reason, charcoal should only be used if specified in a procedure, or if previous crystallization trials failed to remove colored impurities. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.05%3A_Procedural_Generalities/3.5B%3A_Heat_Source_and_Bump_Prevention.txt |
After a solution is dissolved in the minimum amount of hot solvent and filtered (if applicable), the solution should be cooled as slowly as possible (keeping in mind time limitations in a lab). To achieve this goal:
The mouth of the Erlenmeyer flask should be covered with a watch glass (Figure 3.43a). This retains heat just like placing a lid on a boiling pot. The watch glass also prevents excess evaporation of low-boiling solvents. Condensation can often be seen on the watch glass as the solution cools, which is evidence that warm vapors are trapped with this method.
The flask should be placed on a non-conductive surface so heat is not wicked away from the bottom of the flask. The flask can be placed atop a paper towel folded several times, a wood block, or on an inverted cork ring (Figure 3.43a-c). For solutions that cool very quickly (e.g., when using solvents with low boiling points like diethyl ether and acetone), the flask may also be covered by an inverted beaker (Figure 3.43d) to create an insulating atmosphere around the flask. This is not generally necessary for solvents with relatively high boiling points like water and ethanol.
The correct flask size should be used so that the quantity of hot solvent used reaches a height of more than $1 \: \text{cm}$ in the flask. If the solvent level is too shallow during crystallization, the high surface area will cause the solution to cool and evaporate too quickly (Figure 3.44b). It will also be difficult to filter a shallow volume. Ideally the solution should be to a height of at least $2 \: \text{cm}$ in the flask, which allows for maintenance of heat by the solution's interior (Figure 3.44d). It is common to use between 10-50 times as much solvent as sample, and a rough guide is to use a flask where the sample just covers the bottom in a thin layer.
3.5E: Initiating Crystallization
At times, crystals will not form even when a solution is supersaturated, as there is a kinetic barrier to crystal formation. At times crystallization may need to be initiated, for example if the solution becomes faintly cloudy as it cools, or if the solution fails to produce crystals even when it is noticeably cooler than originally. The methods described in this section preferably should be used on solutions that are still warm, as initiating crystallization on already cool (or cold) solutions will cause too rapid a crystallization.
The easiest method to initiate crystallization is to scratch the bottom or side of the flask with a glass stirring rod (Figure 3.45a), with enough force that the scratching is audible (but of course not so much that you break the glass!). Crystallization often begins immediately after scratching, and lines may be visible showing crystal growth in the areas of the glass that were scratched (Figure 3.46).
Although there is no doubt that this method works, there are differences in opinion as to the mechanism of action. One theory is that scratching initiates crystallization by providing energy from the high-frequency vibrations. Another theory is that tiny fragments of glass are dislodged during scratching that provide nucleation sites for crystal formation. Another hypothesis is that the solvent may evaporate from the glass stirring rod after it is removed, and a speck of solid may fall into solution and act as a "seed crystal" (see next paragraph). It is not well understood how scratching initiates crystallization, but it is in every chemist's arsenal of "tricks".
There are a few other methods that can be used to initiate crystallization when scratching fails:
• Add a "seed crystal": a small speck of crude solid saved from before the crystallization was begun, or a bit of pure solid from a reagent jar. Seed crystals create a nucleation site where crystals can begin growth.
• Dip a glass stirring rod into the supersaturated solution, remove it, and allow the solvent to evaporate to produce a thin residue of crystals on the rod (Figure 3.47a). Then touch the rod to the solution's surface, or stir the solution with the rod to dislodge small seed crystals.
• If scratching and seed crystals do not initiate crystallization, it's possible there is too much solvent present that the compound remains completely soluble (Figure 3.48). To test if this is the case, return the solution to a boil and reduce the volume of solvent, perhaps by half (Figure 3.48c). Allow the reduced solution to cool and see if solid forms. If it does (Figure 3.48d), the quantity of solvent was definitely the problem, and solvent volumes can be experimented with to achieve the best growth of crystals.
• Use a lower temperature bath to try and encourage crystal formation. A salt water-ice bath $\left( -10^\text{o} \text{C} \right)$ or chemical freezer are some options. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.05%3A_Procedural_Generalities/3.5D%3A_Cooling_Slowly.txt |
If a crystallization is to be performed using flammable organic solvents, a steam bath is recommended and in some situations necessary (when using diethyl ether, acetone, or low-boiling petroleum ether).
3.06: Step-by-Step Procedures
The crystallization pictured in this section shows purification of a roughly $1 \: \text{g}$ sample of old $\ce{N}$-bromosuccinimide (NBS), which was found in its reagent bottle as an orange powder. The crystallization uses water as the solvent, which has no flammability issues, and so a hotplate is used.
If a crystallization is to be performed using flammable organic solvents, a steam bath is recommended and in some situations necessary (when using diethyl ether, acetone, or low-boiling petroleum ether). The following procedure should be used as a guideline for the process, and some key differences between using water and organic solvents are discussed in a future section.
Prepare the Setup
1. Transfer the impure solid to be crystallized into an appropriately sized Erlenmeyer flask (Figure 3.50a). If the solid is granular, first pulverize with a glass stirring rod.
It is not recommended to perform crystallizations in a beaker. The narrow mouth of an Erlenmeyer flask allows for easier swirling and minimized evaporation during the process as solvent vapors instead condense on the walls of the flask (they "reflux" on the sides of the flask). The narrow mouth of an Erlenmeyer also allows for a flask to be more easily covered during the cooling stage, or even potentially stoppered for long crystallizations. A round-bottomed flask is also not ideal for crystallization as the shape of the flask makes it difficult to recover solid at the end of the process.
It is important that the flask be not too full or too empty during the crystallization. If the flask will be greater than half-full with hot solvent, it will be difficult to prevent the flask from boiling over. If the flask will contain solvent to a height less than $1 \: \text{cm}$, the solution will cool too quickly. It is common to use between 10-50 times as much solvent as sample, and a rough guide is to use a flask where the sample just covers the bottom in a thin layer.
2. Place some solvent in a beaker or Erlenmeyer flask along with a few boiling stones on the heat source, and bring to a gentle boil. Use a beaker if the solvent will be poured and an Erlenmeyer flask if the solvent will be pipetted. If a hot filtration step is anticipated for later in the procedure, also prepare a ring clamp containing a funnel with fluted filter paper (Figure 3.50b).
Add the Minimum Amount of Hot Solvent
1. When the solvent is boiling, grasp the beaker with a hot hand protector (Figure 3.50d), cotton gloves, or a paper towel holder made by rolling a sheet of paper towel into a long rectangle (Figure 3.50c). To the side of the heat source, pour a small portion of boiling solvent into the flask containing the impure solid, to coat the bottom of the flask. If the crystallization is being performed on a small scale (using a $50 \: \text{mL}$ Erlenmeyer flask or smaller), it may be easier to use a pipette to transfer portions of the solvent to the flask.
It's customary to not place the dry solid atop the heat source before adding solvent or the solid may decompose. When the solid is dispersed in a small amount of solvent, it can then be placed on the heat source.
1. Place the flask containing the impure solid and solvent atop the heat source. Use some method to prevent bumping (boiling stones if you plan to "hot filter", a boiling stick or stir bar if you don't), and bring the solution to a gentle boil (Figure 3.51a).
2. Add solvent in portions (Figure 3.51b), swirling to aid in dissolution, until the solid just dissolves (Figure 3.51d). For $100 \: \text{mg}$-$1 \text{g}$ of compound, add $0.5$-$2 \: \text{mL}$ portions at a time. Note that it may take time for a solid to completely dissolve as there is a kinetic aspect to dissolution. Each addition should be allowed to come completely to a boil before adding more solvent, and some time should be allowed between additions. Not allowing time for dissolution and consequently adding too much solvent is a main source of error in crystallization.
It is not uncommon for droplets of liquid to be seen during the heating process (Figure 3.52). This is when the material "oils out", or melts before it dissolves. If this happens, the liquid droplets are now the compound you are crystallizing, so continue adding solvent in portions until the liquid droplets fully dissolve as well.
Watch the solution carefully to judge whether the size of the solid pieces (or liquid droplets) change with additional solvent: if they don't they may be an insoluble impurity. Addition of excess solvent in an attempt to dissolve insoluble impurities will negatively affect the recovery. If insoluble solid impurities are present, the solution should be filtered (insert a hot filtration step at this point). Colored impurities can also be removed at this point with charcoal.
Allow the Solution to Slowly Cool
1. When the solid is just dissolved, remove the flask from the heat source using a hot hand protector, paper towel holder, or glove, and set it aside to cool. Remove the boiling stick or stir bar if used for bump protection (boiling stones can be picked out of the solid at a later point if used).
To encourage slow cooling, set the flask atop a surface that does not conduct heat well, such as a folded paper towel. Cover the mouth of the Erlenmeyer flask with a watch glass to retain heat and solvent (Figure 3.53a). Allow the solution to slowly come to room temperature.
2. As the solution cools, eventually solid crystals should form (Figure 3.53b). If the solution is only warm to the touch or cloudy and no crystals have formed, use a glass stirring rod to scratch the glass and initiate crystallization.
After crystallization has begun, the crystals should slowly grow as the temperature decreases. An ideal crystallization takes between 5-20 minutes to fully crystallize, depending on the scale. Complete crystallization in less than 5 minutes is too quick (see Troubleshooting section for advice on how to slow it down).
3. When the solution is at room temperature, place the flask into an ice bath (ice-water slurry) for 10-20 minutes to lower the compound's solubility even more and maximize crystal formation (Figure 3.53d). Also place a portion of solvent in the ice bath, to be used later for rinsing during suction filtration.
4. Use suction filtration to recover the solid from the mixture. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.06%3A_Step-by-Step_Procedures/3.6A%3A_Single_Solvent_Crystallization.txt |
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