IChO 2026 · Theory problem 6

Carbon Nanorings

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T6. Carbon Nanorings

7%

6.1 6.2 6.3 6.4 6.5 6.6 6.7 \(\sum\)
10 11 2 12 24 20 4 83

In 2019, \(\ce{C18}\), the first example of a new class of carbon allotropes called cyclo[n]carbons (\(\ce{C_n}\)) was detected. Cyclocarbons are monocyclic, all-carbon compounds with high strain energy and low stability. They can be generated and analysed by bond-resolved Atomic Force Microscopy (AFM). Cyclocarbons have interesting aromatic properties based on their \(\pi\) systems. \(\ce{C13}\) has triplet (\({}^{3}\ce{C13}\), spin \(S=1\)), and singlet (\({}^{1}\ce{C13}\), spin \(S=0\)) forms.

6.1 Based on Hückel’s rule, fill in the table with the number of distinct aromatic (A) and anti-aromatic (AA) \(\pi\)-systems present in each of \(\ce{C18}\), \(\ce{C16}\), \({}^{3}\ce{C13}\), and \({}^{1}\ce{C13}\). Fill in all blanks. 10.0 pt

Figure 1

Later, cyclo[n]carbons \(\ce{C10}\), \(\ce{C12}\), \(\ce{C14}\), \(\ce{C20}\), and \(\ce{C26}\) were generated by a combination of dehalogenation and retro-Bergman (1) reactions on a surface. AFM images taken during the synthesis of \(\ce{C14}\) are shown below:

Figure 2
Figure 3

6.2 Draw structures B-D. Note they may contain one or more unpaired electrons. The structure of A is given as an example. 11.0 pt

Figure 4

6.3 The voltages applied during the AFM-mediated synthesis of \(\ce{C_n}\) may vary. Using the given C–X bond energies, choose the possible halogen(s) in the halogenated reagent for \(\ce{C18}\) synthesis via a retro-Bergman reaction with a voltage of 2.5 V acting on the electrons. 2.0 pt

Bond (C–X) Bond Dissociation Energy / \(\mathrm{kJ\,mol^{-1}}\)
C–F 467
C–Cl 346
C–Br 290
C–I 228

In 2025, the first relatively stable cyclo[48]carbon was synthesised. The molecule was stabilised by catenation (forming interlocking rings) with macrocycle E and was first characterised by electrospray mass spectrometry in positive mode.

Figure 5

6.4 In the mass spectrum obtained, four intense peaks were observed \(m/z\): 591, 783, 879, and 1174. Suggest the identity of these ions. Use integer atomic masses and assume no fragmentation happened. The ion corresponding to \(m/z=591\) is given as an example. 12.0 pt

Figure 6

Cycloparaphenylene (CPP) is another cyclic, highly-strained molecule commonly called a carbon nanohoop. The synthesis of the smallest known member, [5]CPP, is shown below, where \(\ce{PhI(OAc)2}\) acts as a two electron oxidant.

Figure 7

(equiv. = Equivalent, excess = Excess)

6.5 Draw the structures of F-L. 24.0 pt

Next generation nanobelts were constructed from porphyrin rings and triple bonds using the template synthesis method. Hint: M is the thermodynamic product and has four types of protons.

Figure 8

(equiv. = Equivalent)

6.6 Draw the structures of M-R. 20.0 pt

When aromatic rings are joined to form a macrocycle, oxidation or reduction can produce global aromaticity, in which \(\pi\) electrons are delocalised around the entire macrocycle. Only the \(\pi\) system forming the continuous conjugated pathway around the ring participates in this delocalisation. As an example, [n]CPPs can become globally aromatic upon addition or removal of electrons. The number of \(\pi\) electrons is counted only along the bold pathway shown below.

Figure 9

When P6 is oxidised, it can exhibit global aromaticity and global anti-aromaticity.

6.7 Based on Hückel’s rule, find the minimum number of electrons, \(n(e)\), from P6 that should be removed to achieve global aromaticity. Write the total number of \(\pi\) electrons, \(n(t)\), in the global aromatic system. 4.0 pt

Physical Constants and Equations

Constants

Constant Value
Avogadro constant \(N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}\)
Universal gas constant \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\)
Faraday constant \(F=96\,485\ \mathrm{C\,mol^{-1}}\)
Planck constant \(h=6.626\times10^{-34}\ \mathrm{J\,s}\)
Speed of light in vacuum \(c=2.998\times10^8\ \mathrm{m\,s^{-1}}\)
Mass of electron \(m_e=9.109\times10^{-31}\ \mathrm{kg}\)
Charge of electron \(e=1.602\times10^{-19}\ \mathrm{C}\)
Standard pressure \(P^\circ=1\ \mathrm{bar}=10^5\ \mathrm{Pa}\)
Ionic product of water at \(25\,^\circ\mathrm C\) \(K_w=1.00\times10^{-14}\)
Zero of the Celsius scale \(0\,^\circ\mathrm C=273.15\ \mathrm K\)
Electronvolt \(1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J\)
Watt \(1\ \mathrm W=1\ \mathrm{J\,s^{-1}}\)
Coulomb \(1\ \mathrm C=1\ \mathrm{A\,s}\)
Nanometre \(1\ \mathrm{nm}=1\times10^{-9}\ \mathrm m\)
Ångstrom \(1\ \text{Å}=1\times10^{-10}\ \mathrm m\)
metric ton \(1\ \mathrm{ton}=1000\ \mathrm{kg}\)
litre \(1\ \mathrm L=1\ \mathrm{dm^3}\)

Official English text of the problem.

Answers

The structures named here are specified explicitly by SMILES in the corresponding Solution entries.

  • 6.1: In the order \(\ce{C18}\), \(\ce{C16}\), \({}^{3}\ce{C13}\), \({}^{1}\ce{C13}\):
    A: \(2,\ 0,\ 0,\ 1\).
    AA: \(0,\ 2,\ 0,\ 1\).

  • 6.2:
    B: A benzene-fused ten-membered ring containing four triple bonds; the retained benzenoid ring has chlorine at C5, C6 and C7, and a radical at C8. Formula \(\ce{C14Cl3}\), one unpaired electron.
    C: A monocyclic \(\ce{C14Cl}\) radical containing six triple bonds and an adjacent \(\ce{C(Cl)=C^{\bullet}}\) pair.
    D: 1,6,7-Trichloro-2,3-bis(chloroethynyl)naphthalene-4,5,8-triyl, \(\ce{C14Cl5}\), three unpaired electrons.

  • 6.3: \(X=\mathrm I\), iodine only.

  • 6.4:
    \(591:\ \ce{[E + H]+}\).
    \(783:\ \ce{[C48 + 3E + 3H]^3+}\).
    \(879:\ \ce{[C48 + 2E + 2H]^2+}\).
    \(1174:\ \ce{[C48 + 3E + 2H]^2+}\).

  • 6.5:
    F: The cis cyclohexadiene diol formed by addition of the TBS-protected biphenylyllithium reagent.
    G: Its bis-TES ether.
    H: The terminal phenol obtained by selective TBS removal.
    I: The terminal para-quinol obtained by oxidative dearomatisation of H.
    J: The second cis diol, formed by addition of 4-bromophenyllithium to I.
    K: The tetra-TES-protected dibromide.
    L: The cyclic product formed by coupling K’s two terminal brominated aryl carbons.

  • 6.6:
    M: 3,5-Di-tert-butyltoluene.
    N: 3,5-Di-tert-butylbenzaldehyde.
    O: Zinc(II) 5,15-bis(3,5-di-tert-butylphenyl)porphyrin.
    Q: Its 10,20-dibromo derivative.
    R: Its 10,20-diethynyl derivative.

  • 6.7: \(\boxed{n(e)=2,\qquad n(t)=82}\).

Solution

6.1 — Separate counts for the two perpendicular π systems

Each cyclocarbon has two perpendicular sets of π orbitals: one perpendicular to the molecular plane and one lying in that plane.

Hückel’s electron-counting criterion is applied separately to each system:

\[ 4k+2\ \text{π electrons: aromatic},\qquad 4k\ \text{π electrons: anti-aromatic}. \]

The electron distributions in the idealised configurations considered here are:

Species π electrons in the two systems A AA
\(\ce{C18}\) \(18,\ 18\) 2 0
\(\ce{C16}\) \(16,\ 16\) 0 2
\({}^{3}\ce{C13}\) \(13,\ 13\) 0 0
\({}^{1}\ce{C13}\) \(14,\ 12\) 1 1

For \(\ce{C18}\), \(18=4(4)+2\), so both systems are aromatic. For \(\ce{C16}\), \(16=4(4)\), so both are anti-aromatic.

In the triplet \(\ce{C13}\) configuration, each system has an odd electron count and one unpaired electron; neither satisfies the stated closed-shell Hückel counts. In the paired singlet configuration, the 26 π electrons are distributed as 14 and 12, giving one aromatic and one anti-aromatic system.

6.2 — Ring-opened intermediates

Retro-Bergman reactions break C–C bonds of the original fused framework and generate new triple bonds. Chlorine loss and carbon valence must both be respected.

In the following SMILES, bracketed [c] and [C] atoms identify the carbon radicals through their explicitly unsatisfied valence.

B — Trichlorinated benzene-fused cyclodecatetrayne radical

SMILES: Clc1c(Cl)c(Cl)[c]c2c1C#CC#CC#CC#C2

Retain one six-membered benzenoid ring. Its four nonfusion positions, numbered C5–C8 consecutively, carry:

  • C5: chlorine;
  • C6: chlorine;
  • C7: chlorine;
  • C8: one unpaired electron, adjacent to a fusion carbon.

The two fusion carbons are also connected by the eight-carbon chain

\[ \ce{-C#C-C#C-C#C-C#C-}, \]

forming a fused ten-membered ring. Thus B contains four triple bonds, three chlorine atoms and one radical, with formula \(\ce{C14Cl3}\).

C — Monochlorocyclo[14]carbon radical

SMILES: ClC1=[C]C#CC#CC#CC#CC#CC#C1

Draw a single fourteen-membered carbon ring with six alternating triple bonds and one double bond. One carbon of the double bond carries chlorine; the other carries the unpaired electron:

\[ \ce{-C(Cl)=C^{\bullet}-}. \]

The remaining bonds between these unsaturated units are single bonds. C therefore has formula \(\ce{C14Cl}\), with the radical next to the chlorine-bearing carbon.

D — 1,6,7-Trichloro-2,3-bis(chloroethynyl)naphthalene-4,5,8-triyl

SMILES: Clc1c(C#CCl)c(C#CCl)[c]c2[c]c(Cl)c(Cl)[c]c12

Use the standard naphthalene numbering:

  • C2 and C3 each carry a chloroethynyl arm, \(\ce{-C#C-Cl}\).
  • C1, C6 and C7 each carry chlorine.
  • C4, C5 and C8 each carry one unpaired electron.

The naphthalene core remains intact, while the opened third ring supplies the two chloroethynyl arms. The formula is \(\ce{C14Cl5}\), with three radicals.

The chlorine balances for the displayed sequence are:

\[ \begin{aligned} \ce{C14Cl10 &-> C14Cl8 + 2Cl^{\bullet}} &&\text{formation of A},\\ \ce{C14Cl8 &-> C14Cl3 + 5Cl^{\bullet}} &&\text{A to B},\\ \ce{C14Cl3 &-> C14Cl + 2Cl^{\bullet}} &&\text{B to C},\\ \ce{C14Cl8 &-> C14Cl5 + 3Cl^{\bullet}} &&\text{A to D},\\ \ce{C14Cl &-> C14 + Cl^{\bullet}} &&\text{C to cyclo[14]carbon}. \end{aligned} \]

The final cyclo[14]carbon has seven alternating triple bonds.

Name: cyclo[14]carbon.
SMILES: C1#CC#CC#CC#CC#CC#CC#C1

6.3 — Electron energy and bond dissociation

An electron accelerated through a potential difference \(U\) gains energy \(eU\). Per mole of electrons,

\[ E_{\mathrm{mol}}=FU =(96\,485\ \mathrm{C\,mol^{-1}})(2.5\ \mathrm{J\,C^{-1}}) =241\,212.5\ \mathrm{J\,mol^{-1}}. \]

Thus the available energy is approximately

\[ E_{\mathrm{mol}}=241\ \mathrm{kJ\,mol^{-1}} \approx2.4\times10^2\ \mathrm{kJ\,mol^{-1}}. \]

Only the C–I bond energy lies below this value:

\[ 228<241<290<346<467 \qquad(\mathrm{kJ\,mol^{-1}}). \]

Therefore,

\[ \boxed{X=\mathrm I}. \]

6.4 — Masses, stoichiometries and charge states

Using the supplied formula of E,

\[ M(E)=40(12)+34(1)+2(14)+3(16)=590\ \mathrm{Da}. \]

For cyclo[48]carbon,

\[ M(\ce{C48})=48(12)=576\ \mathrm{Da}. \]

For a complex containing one carbon ring, \(a\) macrocycles E and \(z\) added protons,

\[ \frac mz=\frac{576+590a+z}{z}. \]

The assignments are:

\(m/z\) Assignment Calculation
591 \(\ce{[E + H]+}\) \(590+1=591\)
783 \(\ce{[C48 + 3E + 3H]^3+}\) \((576+3\times590+3)/3=783\)
879 \(\ce{[C48 + 2E + 2H]^2+}\) \((576+2\times590+2)/2=879\)
1174 \(\ce{[C48 + 3E + 2H]^2+}\) \((576+3\times590+2)/2=1174\)

The following are explicit component SMILES for these ions. A dot separates covalently disconnected rings; SMILES does not encode their mechanical interlocking. The protonated pyridyl nitrogens shown are representative protonation sites.

\(m/z=591\): monoprotonated bipyridine-containing macrocycle E

SMILES: [nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1

\(m/z=783\): triprotonated [4]catenane containing cyclo[48]carbon and three E rings

SMILES: C1#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#C1.[nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1.[nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1.[nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1

\(m/z=879\): diprotonated [3]catenane containing cyclo[48]carbon and two E rings

SMILES: C1#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#C1.[nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1.[nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1

\(m/z=1174\): diprotonated [4]catenane containing cyclo[48]carbon and three E rings

SMILES: C1#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#CC#C1.[nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1.[nH+]1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1.n1c(-c2c(C)cc(O3)cc2C)cccc1-c1cccc(-c2c(C)cc(OCc4ccc(Oc5ccc(C3)cc5)cc4)cc2C)n1

Their respective formulas are \(\ce{C40H35N2O3+}\), \(\ce{C168H105N6O9^3+}\), \(\ce{C128H70N4O6^2+}\), and \(\ce{C168H104N6O9^2+}\).

6.5 — Construction of [5]CPP

The useful cyclohexadiene building blocks have their two aryl substituents cis to one another; the two oxygen substituents are correspondingly cis on the other face. These bent units allow macrocyclisation before the highly strained aromatic nanohoop is generated.

The paired @ symbols below specify this relative stereochemistry. They do not specify a unique overall macrocycle conformation.

F — cis-1-(4-Bromophenyl)-4-[4′-(tert-butyldimethylsiloxy)biphenyl-4-yl]cyclohexa-2,5-diene-1,4-diol

SMILES: Brc1ccc([C@]2(O)C=C[C@](O)(c3ccc(-c4ccc(O[Si](C)(C)C(C)(C)C)cc4)cc3)C=C2)cc1

First, sodium hydride deprotonates the existing alcohol:

\[ \ce{ROH + NaH -> RONa + H2}. \]

The aryllithium reagent then adds to the ketone carbonyl. After work-up, that carbonyl becomes a tertiary alcohol:

\[ \ce{R2C=O + ArLi + H2O -> R2C(OH)Ar + LiOH}. \]

The original alkoxide is also protonated during work-up. F therefore has two free hydroxyl groups and retains the terminal TBS-protected phenol.

G — Bis(triethylsilyl) ether of F

SMILES: Brc1ccc([C@]2(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c3ccc(-c4ccc(O[Si](C)(C)C(C)(C)C)cc4)cc3)C=C2)cc1

Two equivalents of TESCl protect both tertiary alcohols. With Im denoting imidazole, each protection has the balance

\[ \ce{ROH + Et3SiCl + Im -> ROSiEt3 + [ImH]Cl}. \]

The terminal phenolic TBS ether remains present.

H — Terminal phenol obtained by selective TBS deprotection of G

SMILES: Brc1ccc([C@]2(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c3ccc(-c4ccc(O)cc4)cc3)C=C2)cc1

LiOH hydrolyses the phenolic TBS ether:

\[ \ce{ArOSiMe2(t-Bu) + H2O ->[LiOH] ArOH + HOSiMe2(t-Bu)}. \]

The two tertiary TES ethers remain. This gives exactly the supplied formula:

\[ \boxed{\ce{C36H47BrO3Si2}}. \]

I — Terminal para-quinol formed by oxidative dearomatisation of H

SMILES: Brc1ccc([C@]2(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c3ccc(C4(O)C=CC(=O)C=C4)cc3)C=C2)cc1

The terminal phenol becomes a cyclohexa-2,5-dien-1-one. Its para carbon, already carrying the rest of the molecule, acquires an OH group.

The two-electron oxidation is balanced by

\[ \ce{C36H47BrO3Si2 + PhI(OAc)2 + H2O -> C36H47BrO4Si2 + PhI + 2CH3COOH}. \]

Thus I contains the original protected cyclohexadiene unit and a new hydroxycyclohexadienone unit.

J — Second cis-diol addition product

SMILES: Brc1ccc([C@]2(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c3ccc([C@]4(O)C=C[C@](O)(c5ccc(Br)cc5)C=C4)cc3)C=C2)cc1

NaH deprotonates I’s newly formed OH group. Then 4-bromophenyllithium adds to the remaining ketone, as in the formation of F.

The five-ring chain now has the sequence

\[ \text{bromophenyl–protected diene–phenylene–diol diene–bromophenyl}. \]

J has formula \(\ce{C42H52Br2O4Si2}\).

K — Tetra-TES-protected dibromide precursor

SMILES: Brc1ccc([C@]2(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c3ccc([C@]4(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c5ccc(Br)cc5)C=C4)cc3)C=C2)cc1

Two more equivalents of TESCl protect J’s two free hydroxyl groups. K therefore contains four TES ethers and two terminal aryl bromides:

\[ \ce{C54H80Br2O4Si4}. \]

L — Tetra-TES-protected, doubly dearomatised [5]cycloparaphenylene

SMILES: c1%10ccc([C@]2(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c3ccc([C@]4(O[Si](CC)(CC)CC)C=C[C@](O[Si](CC)(CC)CC)(c5ccc-%10cc5)C=C4)cc3)C=C2)cc1

Nickel-mediated intramolecular coupling joins the two aromatic carbons that originally carried bromine. This closes the five-unit hoop.

The net coupling balance is

\[ \ce{C54H80Br2O4Si4 + Ni -> C54H80O4Si4 + NiBr2}. \]

The resulting formula agrees with the problem:

\[ \boxed{\ce{C54H80O4Si4}}. \]

The two equivalents of nickel complex shown are the experimental reagent loading; the net redox balance consumes one equivalent of \(\ce{Ni(0)}\).

Finally, fluoride removes all four TES groups. For each group, a hydrolytic work-up can be represented by

\[ \ce{ROSiEt3 + F- + H2O -> ROH + Et3SiF + OH-}. \]

The resulting tetrol has formula \(\ce{C30H24O4}\). Reductive aromatisation then gives [5]CPP. Its balanced reduction half-equation is

\[ \ce{C30H24O4 + 4H+ + 4e- -> C30H20 + 4H2O}. \]

Tin(II) supplies the electrons:

\[ \ce{2Sn^2+ -> 2Sn^4+ + 4e-}. \]

Name: [5]cycloparaphenylene.
SMILES: c1%10ccc(-c2ccc(-c3ccc(-c4ccc(-c5ccc-%10cc5)cc4)cc3)cc2)cc1

6.6 — Porphyrin building block

Define \(\mathrm{Ar}\) here as 3,5-di-tert-butylphenyl.

M — 3,5-Di-tert-butyltoluene

SMILES: Cc1cc(C(C)(C)C)cc(C(C)(C)C)c1

Under the thermodynamic alkylation conditions, the bulky substituents adopt the alternating 1,3,5 arrangement, avoiding adjacent methyl/tert-butyl substitution.

The net alkylation is

\[ \ce{C6H5CH3 + 2(CH3)3CCl ->[AlCl3] C6H3(CH3)[C(CH3)3]2 + 2HCl}. \]

M has the required four proton environments:

  1. the methyl group, 3 H;
  2. the two equivalent tert-butyl groups, 18 H;
  3. aromatic H2 and H6, 2 H;
  4. aromatic H4, 1 H.

N — 3,5-Di-tert-butylbenzaldehyde

SMILES: O=Cc1cc(C(C)(C)C)cc(C(C)(C)C)c1

NBS with peroxide initiator brominates the benzylic methyl group:

\[ \ce{ArCH3 + NBS -> ArCH2Br + succinimide}. \]

The intermediate is 3,5-di-tert-butylbenzyl bromide.

SMILES: BrCc1cc(C(C)(C)C)cc(C(C)(C)C)c1

HMTA followed by acidic hydrolysis effects the Sommelet conversion to the aldehyde. Writing the amine hydrolysis products in their neutral forms gives the balanced net equation

\[ \ce{ArCH2Br + C6H12N4 + 6H2O -> ArCHO + 5HCHO + CH3NH2 + 3NH3 + HBr}. \]

In the acidic reaction medium, the amines are protonated.

O — Zinc(II) 5,15-bis(3,5-di-tert-butylphenyl)porphyrin

SMILES: CC(C)(C)c1cc(-c2c3[n]4->[Zn+2]56<-[n]7c(cc8ccc2[n-]->58)C=Cc7c(-c2cc(C(C)(C)C)cc(C(C)(C)C)c2)c2ccc(cc4C=C3)[n-]->62)cc(C(C)(C)C)c1

Two aldehydes and two dipyrromethanes furnish the porphyrin framework. The aldehyde-derived carbons become two opposite meso carbons bearing Ar. The other two meso positions retain hydrogen.

Condensation and oxidation by DDQ have the overall balance

\[ \ce{2C15H22O + 2C9H10N2 + 3DDQ -> C48H54N4 + 2H2O + 3DDQH2}. \]

Zinc insertion then replaces the two inner N–H protons:

\[ \ce{C48H54N4 + Zn(OAc)2 -> C48H52N4Zn + 2CH3COOH}. \]

The SMILES explicitly includes N-to-Zn dative bonds. The four porphyrin nitrogens coordinate zinc in an approximately square-planar arrangement; that geometry is not specified by the SMILES itself.

Q — Zinc(II) 10,20-dibromo-5,15-bis(3,5-di-tert-butylphenyl)porphyrin

SMILES: CC(C)(C)c1cc(-c2c3[n]4->[Zn+2]56<-[n]7c(c(Br)c8ccc2[n-]->58)C=Cc7c(-c2cc(C(C)(C)C)cc(C(C)(C)C)c2)c2ccc(c(Br)c4C=C3)[n-]->62)cc(C(C)(C)C)c1

NBS replaces the two remaining meso hydrogens by bromine. The net balance is

\[ \ce{C48H52N4Zn + 2NBS -> C48H50Br2N4Zn + 2succinimide}. \]

R — Zinc(II) 10,20-diethynyl-5,15-bis(3,5-di-tert-butylphenyl)porphyrin

SMILES: C#Cc1c2[n]3->[Zn+2]45<-[n]6c(c(C#C)c7ccc(c(-c8cc(C(C)(C)C)cc(C(C)(C)C)c8)c3C=C2)[n-]->47)C=Cc6c(-c2cc(C(C)(C)C)cc(C(C)(C)C)c2)c2ccc1[n-]->52

Each meso bromide undergoes Sonogashira coupling with trihexylsilyl-protected acetylene:

\[ \ce{P-Br + HC#C-Si(C6H13)3 -> P-C#C-Si(C6H13)3 + HBr}, \]

where P denotes the remaining porphyrin framework.

Fluoride then removes both silyl protecting groups. Per ethynyl group,

\[ \ce{P-C#C-Si(C6H13)3 + F- + H2O -> P-C#CH + (C6H13)3SiF + OH-}. \]

R therefore contains two opposite terminal ethynyl groups and has formula \(\ce{C52H52N4Zn}\). Oxidative coupling of these terminal groups subsequently generates the butadiyne connections between porphyrins in P6.

6.7 — Global π-electron count

Count only the orbitals belonging to one continuous conjugated route around the entire nanoring.

Each porphyrin contributes 10 π electrons along the route between its two opposite linker-bearing meso positions. Each butadiyne bridge,

\[ \ce{-C#C-C#C-}, \]

contributes 4 π electrons: only one of the two perpendicular π systems of each triple bond belongs to the global conjugation pathway.

With six porphyrins and six butadiyne bridges,

\[ N_\pi=6(10)+6(4)=84. \]

After removing \(n(e)\) electrons, the Hückel condition is

\[ 84-n(e)=4k+2. \]

Removing one electron gives 83, which does not satisfy the condition. Removing two gives

\[ 82=4(20)+2. \]

Therefore, under the Hückel-counting criterion requested,

\[ \boxed{n(e)=2,\qquad n(t)=82}. \]

The solution was checked against the official solution.