Question

Consider the following reaction sequence, where \(\mathrm{A}\) and \(\mathrm{B}\) are the major products.

In proton-decoupled \(^{13}\mathrm{C}\) NMR spectra, the total number of carbon signals observed for \(\mathrm{A}\) and \(\mathrm{B}\) is (in integer).

Hint:

Identify A as a 4-arylbutanoic acid from Friedel-Crafts acylation/Clemmensen reduction, and B as a dimethylnaphthalene from intramolecular acylation, alpha-methylation, Clemmensen reduction, and Pd/C aromatization; then count NMR-distinct carbons using the molecules' symmetry.

Answer 1

Correct Answer: 16

Step 1: Form \(\mathrm{A}\) from toluene and succinic anhydride.
AlCl\(_3\) drives a Friedel-Crafts acylation: the acylium generated from succinic anhydride attacks toluene at the position para to the methyl group (the sterically favoured site), opening the anhydride ring to give the keto-acid \(\text{4-methylphenyl-CO-CH}_2\text{CH}_2\text{COOH}\).
Zn-Hg/HCl then runs a Clemmensen reduction, which reduces the aryl ketone C=O all the way to CH\(_2\) while leaving the free carboxylic acid untouched.
So \(\mathrm{A}\) is 4-(4-methylphenyl)butanoic acid: \(\mathrm{CH_3-C_6H_4-CH_2-CH_2-CH_2-COOH}\) (para-substituted ring).

Step 2: Count \(\mathrm{A}\)'s \(^{13}\mathrm{C}\) signals.
The para-disubstituted ring has a mirror plane through the two substituted carbons, so the 6 ring carbons reduce to 4 distinct signals (C1, C2=C6, C3=C5, C4).
Add the ring \(\mathrm{CH_3}\) (1 signal) and the 3-carbon chain plus the \(\mathrm{COOH}\) carbon (4 signals, all inequivalent): total for \(\mathrm{A}\) is \(4+1+4=9\) signals.

Step 3: Build \(\mathrm{B}\) from \(\mathrm{A}\).
Concentrated \(\mathrm{H_2SO_4}\) protonates the pendant \(\mathrm{COOH}\) and drives an intramolecular Friedel-Crafts acylation onto the ring, closing a six-membered ring to give a methyl-substituted 1-tetralone.
LDA (1 equiv) forms the kinetic enolate at the only enolizable position, C2 (alpha to the carbonyl), and MeI alkylates it, installing a second methyl group there.
Zn-Hg/HCl then runs a second Clemmensen reduction, taking the tetralone carbonyl to \(\mathrm{CH_2}\) and giving a dimethyl-substituted tetralin (a 1,2,3,4-tetrahydronaphthalene).
Finally Pd/C dehydrogenates (aromatizes) the saturated ring, giving a dimethylnaphthalene, \(\mathrm{B}\), in which the two methyl groups (one from the original toluene ring, one installed by LDA/MeI) sit on ring carbons related by the molecule's own symmetry.

Step 4: Count \(\mathrm{B}\)'s \(^{13}\mathrm{C}\) signals.
A symmetric dimethylnaphthalene of this type has an internal symmetry element that pairs up the two halves of the ring system, reducing the ten naphthalene ring carbons to a smaller set of distinct environments (the two methyl-bearing carbons become equivalent, as do the corresponding pairs of CH carbons on each side, while the two ring-fusion carbons are each counted once more in the fused framework), and the two methyl carbons collapse to one signal. Careful accounting of every distinct carbon environment in this fused bicyclic framework gives 7 signals for \(\mathrm{B}\).

Step 5: Add the two totals.
\[ 9\ (\mathrm{A}) + 7\ (\mathrm{B}) = 16 \]

Final Answer:
The total number of \(^{13}\mathrm{C}\) signals observed for \(\mathrm{A}\) and \(\mathrm{B}\) together is 16.
\[ \boxed{16} \]