IChO 2026 · Theory problem 9

Cyclodextrin Chemistry

T9. Cyclodextrin Chemistry

7%

9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 \(\sum\)
2 6 8 3 2 4 6 18 4 53

Cyclodextrins (CD) are a family of cyclic oligosaccharides, consisting of glucose subunits joined by α-1,4-glycosidic bonds. The three most common cyclodextrins (α-, β-, γ-cyclodextrin) contain 6, 7, and 8 α-D-glucopyranoside units, respectively.

9.1 Calculate the molar mass of β-CD. Assume \(M_w(\text{glucose})=180.16\ \mathrm{g\,mol^{-1}}\). 2.0 pt

Figure 1

CDs combine macrocyclic properties with glucose chemistry, which was demonstrated in the synthesis of X. Primary and secondary hydroxy groups show different reactivity due to being a part of CDs. Stoddart et al. synthesised compound K, where only one OH group remained free per glucopyranoside unit.

Figure 2

equiv.= equivalent

9.2 Tick the favorable chair conformation and draw the structure of K by completing the CD template. Show stereochemistry unambiguously. 6.0 pt

Figure 3

9.3 Determine the ring size, \(rs\), of macrocycle X. Give the number of stereocentres, \(sc\), in X. 8.0 pt

α-Cycloaltrin, a synthetic analogue of α-CD, was synthesised as follows.

Figure 4

9.4 Draw the structure of Y with stereochemistry by completing the CD template. 3.0 pt

Figure 5

In 2000, Sinay et al. demonstrated that a single protic group (NH and OH) at unit 1 directs the next reductive debenzylation of a primary OH group to unit 4 in the macrocyclic ring (or to unit 3 if the unit 4 position is not available). This allowed the synthesis of the following β-CD dimer as a mixture of constitutional linkage isomers.

Figure 6

9.5 Draw the structure of L on the β-CD template. Fill in all the boxes. 2.0 pt

Figure 7

9.6 Determine the number of isomers of the β-CD dimer that can form during the synthesis by Sinay et. al. 4.0 pt

To confirm the positions of debenzylation, L was subjected to the “hexo-5-enose degradation” reaction, as shown below. The products were then analysed by mass spectrometry.

Figure 8

9.7 Calculate the \(m/z\) value for the two \(\ce{[M+Na]+}\) peaks that were observed for the degradation products from L. Use integer values of atomic mass. 6.0 pt

Later, Sollogoub et al. applied the ability of alkenes to direct the debenzylation to the “ortho”-glucopyranose unit for the synthesis of hexadifferentiated α-CD S:

Figure 9

9.8 Draw the structures of O–S on the α-CD templates. Fill in all the boxes. You may draw structures outside the boxes if appropriate. Assume 1,2-unit modification happens clockwise, while 1,3-unit modification happens counterclockwise. Protic groups have stronger directing effects than alkenes. 18.0 pt

Figure 10
Figure 11

9.9 Determine the number of all possible arrangements of the functional groups on a hexadifferentiated α-CD (assume only the \(\ce{CH2OH}\) groups have been modified). 4.0 pt

Periodic Table of Elements

Figure 12

Official English text of the problem.

Answers

  • 9.1: \(\boxed{M(\beta\text{-CD})=1135.0\ \mathrm{g\,mol^{-1}}}\).
  • 9.2: Tick the right-hand chair, \(^{1}C_{4}\). K is heptakis(3,6-anhydro)-β-cyclodextrin: every residue has a C3–O–C6 bridge and a free C2–OH.
  • 9.3: \(\boxed{rs=35,\quad sc=21}\).
  • 9.4: Y is the per-6-O-TBS-protected, per-2,3-anhydro manno derivative of α-CD. Each epoxide oxygen lies on the same face as the C5–\(\ce{CH2OTBS}\) substituent.
  • 9.5: L: primary boxes 1 and 4 contain \(\ce{CH2OH}\); boxes 2, 3, 5, 6, and 7 contain \(\ce{CH2OBn}\); the external box contains \(\ce{OBn}\), repeated 14 times.
  • 9.6: \(\boxed{3}\) constitutional linkage isomers.
  • 9.7: \(\boxed{m/z=1299\text{ and }1731}\).
  • 9.8: The complete substitution patterns are given below. All twelve secondary oxygen substituents remain \(\ce{OBn}\).
  • 9.9: \(\boxed{120}\) arrangements.

For 9.8, the numbers identify glucose units, clockwise as on the answer sheet. Each entry is the substituent attached to that unit’s C5.

Compound Unit 1 Unit 2 Unit 3 Unit 4 Unit 5 Unit 6
O \(\ce{CH=CH2}\) \(\ce{CH2OBn}\) \(\ce{CH2OBn}\) \(\ce{CH2OBn}\) \(\ce{CH2OBn}\) \(\ce{CH2OBn}\)
P \(\ce{CH=CH2}\) \(\ce{CH2N3}\) \(\ce{CH2OBn}\) \(\ce{CH2OBn}\) \(\ce{CH2OBn}\) \(\ce{CH2OBn}\)
Q \(\ce{CH=CH2}\) \(\ce{CH2NH}\), bridge end \(\ce{CH2OBn}\) \(\ce{CH2OBn}\) \(\ce{CH2O}\), bridge end \(\ce{CH2OBn}\)
R \(\ce{CH=CH2}\) \(\ce{CH2N(Boc)}\), bridge end \(\ce{CH2OBn}\) \(\ce{CH2OBn}\) \(\ce{CH2O}\), bridge end \(\ce{CH2N3}\)
S \(\ce{CH=CH2}\) \(\ce{CH2N(Bn)}\), bridge end \(\ce{CH2OH}\) \(\ce{CH2OBn}\) \(\ce{CH2O}\), bridge end \(\ce{CH2NH2}\)

In Q, R, and S, join the indicated nitrogen at unit 2 to the oxygen at unit 5 through

\[ \ce{N-CH2-C(=CH2)-CH2-O}. \]

Thus the complete bridge runs from unit 2 to unit 5 as

\[ \ce{CH2-N(R)-CH2-C(=CH2)-CH2-O-CH2}, \]

where \(R=\ce{H}\), Boc, and Bn in Q, R, and S, respectively.

Solution

The SMILES below specify complete molecular connectivity and stereochemistry. Chair conformations and the positions on the printed templates are specified separately in words.

The substituent abbreviations are benzyl, Bn, *Cc1ccccc1; acetyl, Ac, *C(=O)C; tert-butyldimethylsilyl, TBS, *[Si](C)(C)C(C)(C)C; and tert-butoxycarbonyl, Boc, *C(=O)OC(C)(C)C. Here * marks the attachment point of a substituent.

9.1

A cyclic assembly of seven glucose molecules contains seven glycosidic bonds. Consequently, formation of β-CD eliminates seven water molecules:

\[ \ce{7C6H12O6 -> C42H70O35 + 7H2O}. \]

Using the supplied atomic masses,

\[ M(\ce{H2O})=2(1.008)+16.00=18.016\ \mathrm{g\,mol^{-1}}, \]

and therefore

\[ \begin{aligned} M(\beta\text{-CD}) &=7(180.16)-7(18.016)\\ &=1135.008\ \mathrm{g\,mol^{-1}}\\ &\approx\boxed{1135.0\ \mathrm{g\,mol^{-1}}}. \end{aligned} \]

Glucose in the α-D-glucopyranose form has SMILES OC[C@H]1O[C@H](O)[C@H](O)[C@@H](O)[C@@H]1O; water has SMILES O.

β-Cyclodextrin, \(\ce{C42H70O35}\), has SMILES:

OC[C@H]1O[C@@H]2O[C@@H]3[C@H](O)[C@@H](O)[C@@H](O[C@@H]4[C@H](O)[C@@H](O)[C@@H](O[C@@H]5[C@H](O)[C@@H](O)[C@@H](O[C@@H]6[C@H](O)[C@@H](O)[C@@H](O[C@@H]7[C@H](O)[C@@H](O)[C@@H](O[C@@H]8[C@H](O)[C@@H](O)[C@@H](O[C@@H]1[C@H](O)[C@H]2O)O[C@@H]8CO)O[C@@H]7CO)O[C@@H]6CO)O[C@@H]5CO)O[C@@H]4CO)O[C@@H]3CO

9.2

The seven equivalents of tosyl chloride preferentially activate the seven primary C6 hydroxyl groups. Under basic conditions, the C3 alkoxide of each glucose residue attacks its own C6 carbon by intramolecular \(S_\mathrm{N}2\) displacement of tosylate. This produces a 3,6-anhydro ether bridge.

The C2 hydroxyl group remains free. Substitution occurs at the non-stereogenic C6 carbon, so the existing configurations of the glucose stereocentres are retained.

The 3,6 bridge requires the inverted \(^{1}C_4\) chair, which is the right-hand template. To complete it:

  • Place the C5–C6 substituent above the ring.
  • Connect C6 through oxygen to C3, also on the upper face.
  • Draw the remaining C2–OH axial down.
  • Retain the α-glycosidic connections already shown.

K, heptakis(3,6-anhydro)-β-cyclodextrin, has formula \(\ce{C42H56O28}\) and SMILES:

O[C@H]1[C@H]2O[C@@H]3[C@@H]4OC[C@H]3O[C@H](O[C@@H]3[C@@H]5OC[C@H]3O[C@H](O[C@@H]3[C@@H]6OC[C@H]3O[C@H](O[C@@H]3[C@@H]7OC[C@H]3O[C@H](O[C@@H]3[C@@H]8OC[C@H]3O[C@H](O[C@@H]3[C@@H]9OC[C@H]3O[C@H](O[C@@H]3[C@@H]1OC[C@H]3O2)[C@@H]9O)[C@@H]8O)[C@@H]7O)[C@@H]6O)[C@@H]5O)[C@@H]4O

9.3

Periodate cleaves the C2–C3 bond of every glucose residue because C2 and C3 bear a vicinal diol. The corresponding general atom balance is

\[ \ce{R-CH(OH)-CH(OH)-R' + IO4- -> R-CHO + R'-CHO + IO3- + H2O}. \]

Borohydride reduces both aldehydes to primary alcohols. Acetic anhydride then acetylates the hydroxyl groups at the original C2, C3, and C6 positions.

After C2–C3 cleavage, the surviving macrocyclic path through each residue is

\[ \mathrm{C1-O5-C5-C4-O4}, \]

followed by C1 of the next residue along that path. There are five macrocyclic atoms per glucose-derived unit:

\[ \boxed{rs=7\times5=35}. \]

C2 and C3 become \(\ce{CH2OAc}\) groups and lose their stereochemistry. The configurations at C1, C4, and C5 remain:

\[ \boxed{sc=7\times3=21}. \]

X, the peracetylated, periodate-cleaved and reduced β-cyclodextrin derivative, has formula \(\ce{C84H126O56}\) and SMILES:

CC(=O)OC[C@H]1O[C@H](COC(C)=O)O[C@H](COC(C)=O)[C@@H](COC(C)=O)O[C@H](COC(C)=O)O[C@H](COC(C)=O)[C@@H](COC(C)=O)O[C@H](COC(C)=O)O[C@H](COC(C)=O)[C@@H](COC(C)=O)O[C@H](COC(C)=O)O[C@H](COC(C)=O)[C@@H](COC(C)=O)O[C@H](COC(C)=O)O[C@H](COC(C)=O)[C@@H](COC(C)=O)O[C@H](COC(C)=O)O[C@H](COC(C)=O)[C@@H](COC(C)=O)O[C@H](COC(C)=O)O[C@@H]1COC(C)=O

9.4

TBS chloride protects the six primary C6 hydroxyl groups. Sodium hydride deprotonates the twelve remaining secondary hydroxyl groups.

Tosylation at O2 followed by intramolecular attack of O3 at C2 produces a 2,3-epoxide, with inversion at C2. Consequently, the epoxide has the manno configuration: its oxygen lies above the ring, on the same face as the C5–\(\ce{CH2OTBS}\) group.

To complete the template, draw:

  • \(\ce{CH2OTBS}\) at C5, with the original D configuration;
  • an oxygen bonded to both C2 and C3 on the upper face;
  • no free hydroxyl groups.

The formula provides an independent check. Before epoxide formation, the six TBS-protected residues give \(\ce{C72H144O30Si6}\). Formation of six epoxides corresponds to loss of six water equivalents from the carbohydrate framework:

\[ \ce{C72H144O30Si6 -> C72H132O24Si6 + 6H2O}. \]

Subsequent fluoride treatment removes the TBS groups. Water opens the epoxide at C3 with inversion, giving the C2-up/C3-down hydroxyl arrangement of the drawn α-cycloaltrin.

Y, hexakis(2,3-anhydro-6-O-tert-butyldimethylsilyl)-α-cyclomannin, has SMILES:

CC(C)(C)[Si](C)(C)OC[C@H]1O[C@@H]2O[C@@H]3[C@@H]4O[C@@H]4[C@@H](O[C@@H]4[C@@H]5O[C@@H]5[C@@H](O[C@@H]5[C@@H]6O[C@@H]6[C@@H](O[C@@H]6[C@@H]7O[C@@H]7[C@@H](O[C@@H]7[C@@H]8O[C@@H]8[C@@H](O[C@@H]1[C@@H]1O[C@H]21)O[C@@H]7CO[Si](C)(C)C(C)(C)C)O[C@@H]6CO[Si](C)(C)C(C)(C)C)O[C@@H]5CO[Si](C)(C)C(C)(C)C)O[C@@H]4CO[Si](C)(C)C(C)(C)C)O[C@@H]3CO[Si](C)(C)C(C)(C)C

9.5

The first reaction benzylates all 21 hydroxyl groups of β-CD.

The first reductive debenzylation can be designated as occurring at unit 1, since all units were initially equivalent. Its newly formed primary OH directs the second debenzylation to unit 4.

Thus L contains two primary hydroxyl groups and nineteen benzyl ethers:

\[ \begin{array}{c|ccccccc} \text{Unit}&1&2&3&4&5&6&7\\ \hline \text{Primary substituent} &\ce{CH2OH}&\ce{CH2OBn}&\ce{CH2OBn}&\ce{CH2OH} &\ce{CH2OBn}&\ce{CH2OBn}&\ce{CH2OBn} \end{array} \]

The external box is \(\ce{OBn}\), with the printed multiplier 14.

L, the 1,4-primary-diol derivative of per-O-benzyl-β-cyclodextrin, has formula \(\ce{C175H184O35}\) and SMILES:

OC[C@H]1O[C@@H]2O[C@@H]3[C@H](OCc4ccccc4)[C@@H](OCc4ccccc4)[C@@H](O[C@@H]4[C@H](OCc5ccccc5)[C@@H](OCc5ccccc5)[C@@H](O[C@@H]5[C@H](OCc6ccccc6)[C@@H](OCc6ccccc6)[C@@H](O[C@@H]6[C@H](OCc7ccccc7)[C@@H](OCc7ccccc7)[C@@H](O[C@@H]7[C@H](OCc8ccccc8)[C@@H](OCc8ccccc8)[C@@H](O[C@@H]8[C@H](OCc9ccccc9)[C@@H](OCc9ccccc9)[C@@H](O[C@@H]1[C@H](OCc1ccccc1)[C@H]2OCc1ccccc1)O[C@@H]8COCc1ccccc1)O[C@@H]7COCc1ccccc1)O[C@@H]6CO)O[C@@H]5COCc1ccccc1)O[C@@H]4COCc1ccccc1)O[C@@H]3COCc1ccccc1

9.6

The two free primary hydroxyl groups in L are constitutionally distinguishable on the directed cyclodextrin scaffold. Linking two molecules therefore permits the combinations

\[ 1-1',\qquad 1-4',\qquad 4-4'. \]

The combination \(4-1'\) is identical to \(1-4'\) after exchanging the two identical cyclodextrin components.

Any alkene \(E/Z\) distinction generated during metathesis disappears upon hydrogenation. Therefore,

\[ \boxed{N_{\text{isomers}}=3}. \]

9.7

Degradation at units 1 and 4 divides the seven-unit ring into fragments containing three and four original glucose units.

For a fragment containing \(n\) original units, the depicted product contains:

  • \(n-1\) intact protected residue blocks, each \(\ce{C27H28O5}\);
  • one degraded terminal block, \(\ce{C22H25O4}\);
  • one additional acetoxy end group, \(\ce{C2H3O2}\), at the other end.

Their integer masses are

\[ \begin{aligned} M(\ce{C27H28O5})&=27(12)+28+5(16)=432,\\ M(\ce{C22H25O4})&=22(12)+25+4(16)=353,\\ M(\ce{C2H3O2})&=2(12)+3+2(16)=59. \end{aligned} \]

Thus

\[ M_n=432(n-1)+353+59. \]

For the three-unit fragment,

\[ M_3=2(432)+353+59=1276, \]

and for the four-unit fragment,

\[ M_4=3(432)+353+59=1708. \]

Adding one sodium ion of integer mass 23 and charge \(+1\),

\[ \boxed{\left(\frac mz\right)_3=1276+23=1299}, \qquad \boxed{\left(\frac mz\right)_4=1708+23=1731}. \]

The neutral fragments have formulas \(\ce{C78H84O16}\) and \(\ce{C105H112O21}\), respectively.

Sodium adduct of the three-unit degradation fragment, \(\ce{[C78H84O16 + Na]+}\), has SMILES:

C=C[C@H](O[C@H]1O[C@H](COCc2ccccc2)[C@H](O[C@H]2O[C@H](COCc3ccccc3)[C@H](OC(C)=O)[C@H](OCc3ccccc3)[C@H]2OCc2ccccc2)[C@H](OCc2ccccc2)[C@H]1OCc1ccccc1)[C@H](OCc1ccccc1)[C@H](COC(C)=O)OCc1ccccc1.[Na+]

Sodium adduct of the four-unit degradation fragment, \(\ce{[C105H112O21 + Na]+}\), has SMILES:

C=C[C@H](O[C@H]1O[C@H](COCc2ccccc2)[C@H](O[C@H]2O[C@H](COCc3ccccc3)[C@H](O[C@H]3O[C@H](COCc4ccccc4)[C@H](OC(C)=O)[C@H](OCc4ccccc4)[C@H]3OCc3ccccc3)[C@H](OCc3ccccc3)[C@H]2OCc2ccccc2)[C@H](OCc2ccccc2)[C@H]1OCc1ccccc1)[C@H](OCc1ccccc1)[C@H](COC(C)=O)OCc1ccccc1.[Na+]

The disconnected sodium component in these SMILES specifies the adduct composition; the problem does not establish a particular sodium coordination geometry.

9.8

Throughout this sequence, the twelve secondary benzyl ethers remain intact.

Formation of O. Swern oxidation converts the free \(\ce{CH2OH}\) group at unit 1 into an aldehyde. Wittig methylenation then converts that side chain into \(\ce{CH=CH2}\). Units 2–6 retain \(\ce{CH2OBn}\).

O, the unit-1-vinyl α-cyclodextrin derivative, has SMILES:

C=C[C@H]1O[C@@H]2O[C@@H]3[C@H](OCc4ccccc4)[C@@H](OCc4ccccc4)[C@@H](O[C@@H]4[C@H](OCc5ccccc5)[C@@H](OCc5ccccc5)[C@@H](O[C@@H]5[C@H](OCc6ccccc6)[C@@H](OCc6ccccc6)[C@@H](O[C@@H]6[C@H](OCc7ccccc7)[C@@H](OCc7ccccc7)[C@@H](O[C@@H]7[C@H](OCc8ccccc8)[C@@H](OCc8ccccc8)[C@@H](O[C@@H]1[C@H](OCc1ccccc1)[C@H]2OCc1ccccc1)O[C@@H]7COCc1ccccc1)O[C@@H]6COCc1ccccc1)O[C@@H]5COCc1ccccc1)O[C@@H]4COCc1ccccc1)O[C@@H]3COCc1ccccc1

Formation of P. The alkene at unit 1 directs the first debenzylation to its clockwise neighbour, unit 2. Mesylation followed by azide displacement converts the resulting \(\ce{CH2OH}\) into \(\ce{CH2N3}\).

P, the unit-2-azidomethyl derivative of O, has SMILES:

C=C[C@H]1O[C@@H]2O[C@@H]3[C@H](OCc4ccccc4)[C@@H](OCc4ccccc4)[C@@H](O[C@@H]4[C@H](OCc5ccccc5)[C@@H](OCc5ccccc5)[C@@H](O[C@@H]5[C@H](OCc6ccccc6)[C@@H](OCc6ccccc6)[C@@H](O[C@@H]6[C@H](OCc7ccccc7)[C@@H](OCc7ccccc7)[C@@H](O[C@@H]7[C@H](OCc8ccccc8)[C@@H](OCc8ccccc8)[C@@H](O[C@@H]1[C@H](OCc1ccccc1)[C@H]2OCc1ccccc1)O[C@@H]7COCc1ccccc1)O[C@@H]6COCc1ccccc1)O[C@@H]5COCc1ccccc1)O[C@@H]4COCc1ccccc1)O[C@@H]3CN=[N+]=[N-]

Formation of Q. DIBAL-H first reduces the azide at unit 2 to a primary amine. Its protic NH group then directs debenzylation to the opposite unit, unit 5.

The two available groups are consequently \(\ce{CH2NH2}\) at unit 2 and \(\ce{CH2OH}\) at unit 5. Reaction with 1,3-dichloro-2-methylenepropane, C=C(CCl)CCl, joins them through an N,O bridge:

\[ \ce{CH2-NH-CH2-C(=CH2)-CH2-O-CH2}. \]

One N–H bond remains, so the nitrogen can direct the next debenzylation.

Q, the unit-2/unit-5 N,O-bridged α-cyclodextrin derivative, has SMILES:

C=C[C@H]1O[C@@H]2O[C@@H]3[C@H](OCc4ccccc4)[C@@H](OCc4ccccc4)[C@H]4O[C@@H]5[C@H](OCc6ccccc6)[C@@H](OCc6ccccc6)[C@@H](O[C@@H]6[C@H](OCc7ccccc7)[C@@H](OCc7ccccc7)[C@@H](O[C@@H]7[C@H](OCc8ccccc8)[C@@H](OCc8ccccc8)[C@@H](O[C@@H]8[C@H](OCc9ccccc9)[C@@H](OCc9ccccc9)[C@@H](O[C@@H]1[C@H](OCc1ccccc1)[C@H]2OCc1ccccc1)O[C@@H]8COCc1ccccc1)O[C@@H]7COCC(=C)CNC[C@H]3O4)O[C@@H]6COCc1ccccc1)O[C@@H]5COCc1ccccc1

Formation of R. The secondary NH at unit 2 is the strongest directing group. Its usual 1,4 target, unit 5, is already incorporated into the bridge and has no benzyl ether to remove.

The fallback is therefore a 1,3 modification counterclockwise from unit 2:

\[ 2\longrightarrow1\longrightarrow6. \]

DIBAL-H unveils the primary OH at unit 6. Boc protection blocks the bridge nitrogen, while mesylation and azide displacement convert the newly exposed unit-6 side chain into \(\ce{CH2N3}\).

R, the N-Boc-protected, unit-6-azidomethyl derivative of Q, has SMILES:

C=C[C@H]1O[C@@H]2O[C@@H]3[C@H](OCc4ccccc4)[C@@H](OCc4ccccc4)[C@H]4O[C@@H]5[C@H](OCc6ccccc6)[C@@H](OCc6ccccc6)[C@@H](O[C@@H]6[C@H](OCc7ccccc7)[C@@H](OCc7ccccc7)[C@@H](O[C@@H]7[C@H](OCc8ccccc8)[C@@H](OCc8ccccc8)[C@@H](O[C@@H]8[C@H](OCc9ccccc9)[C@@H](OCc9ccccc9)[C@@H](O[C@@H]1[C@H](OCc1ccccc1)[C@H]2OCc1ccccc1)O[C@@H]8CN=[N+]=[N-])O[C@@H]7COCC(=C)CN(C(=O)OC(C)(C)C)C[C@H]3O4)O[C@@H]6COCc1ccccc1)O[C@@H]5COCc1ccccc1

Formation of S. Trifluoroacetic acid removes Boc. Benzylation then converts the bridge nitrogen into a tertiary N-benzyl amine, eliminating its N–H directing group.

DIBAL-H reduces the azide at unit 6 to \(\ce{CH2NH2}\). This newly formed protic group directs debenzylation to the opposite unit, unit 3, giving \(\ce{CH2OH}\) there. Unit 4 retains its primary benzyl ether.

The final six side chains are therefore all differentiated: vinyl, bridged N-benzylamino, hydroxymethyl, benzyloxymethyl, bridged oxymethyl, and aminomethyl, in that clockwise order.

S, the hexadifferentiated α-cyclodextrin derivative, has SMILES:

C=C[C@H]1O[C@@H]2O[C@@H]3[C@H](OCc4ccccc4)[C@@H](OCc4ccccc4)[C@H]4O[C@@H]5[C@H](OCc6ccccc6)[C@@H](OCc6ccccc6)[C@@H](O[C@@H]6[C@H](OCc7ccccc7)[C@@H](OCc7ccccc7)[C@@H](O[C@@H]7[C@H](OCc8ccccc8)[C@@H](OCc8ccccc8)[C@@H](O[C@@H]8[C@H](OCc9ccccc9)[C@@H](OCc9ccccc9)[C@@H](O[C@@H]1[C@H](OCc1ccccc1)[C@H]2OCc1ccccc1)O[C@@H]8CN)O[C@@H]7COCC(=C)CN(Cc1ccccc1)C[C@H]3O4)O[C@@H]6COCc1ccccc1)O[C@@H]5CO

9.9

Six different substituents can be assigned to six labelled sites in \(6!\) ways. Rotating the whole α-CD ring does not produce a new arrangement, so each arrangement has been counted six times:

\[ N=\frac{6!}{6}=5!=\boxed{120}. \]

There is no further division by two. Reversing the cyclic order is not a symmetry operation of the directed, chiral α-D-glucopyranoside scaffold.

The solution was checked against the official solution.