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).

Question

Draw the structure of the major monohalo product for each of the following reactions:

Answer 1

This problem is really two structure-elimination puzzles chained together, followed by a symmetry count. Work out $\mathrm{A}$, then push it through the second sequence to get $\mathrm{B}$, then count carbons in each.

Toluene + succinic anhydride, $\mathrm{AlCl_3}$: This is a Friedel-Crafts acylation. The anhydride opens at the ring, attaching an aroyl group para to the methyl (steric preference) and leaving a free $\mathrm{-CH_2CH_2COOH}$ tail, giving the keto-acid 4-oxo-4-(p-tolyl)butanoic acid.
Zn-Hg, HCl (Clemmensen): Reduces only the aryl ketone, not the carboxylic acid, giving $\mathrm{A}=$ 4-(4-methylphenyl)butanoic acid, a para-disubstituted ring carrying a $\mathrm{-CH_2CH_2CH_2COOH}$ chain.
$\mathrm{H_2SO_4}$: Protonates the $\mathrm{COOH}$ of $\mathrm{A}$ and closes a new six-membered ring onto the arene by intramolecular Friedel-Crafts acylation, giving a methyltetralone.
LDA (1 equiv), MeI: The only alpha-hydrogen available for kinetic enolate formation is at C2 next to the carbonyl; methylation there installs a second methyl group on the saturated ring.
Zn-Hg, HCl: A second Clemmensen reduction removes the tetralone carbonyl, giving a dimethyltetralin (saturated ring fused to the original aromatic ring).
Pd/C: Dehydrogenates the saturated ring to full aromaticity, giving the dimethylnaphthalene $\mathrm{B}$, with the two methyls on ring positions related by the molecule's symmetry.

Now count carbons. In $\mathrm{A}$, the para-substituted ring gives 4 distinct ring signals (by the mirror symmetry through the two substituted positions), plus 1 for the ring methyl and 4 for the $\mathrm{-CH_2CH_2CH_2COOH}$ chain (three inequivalent $\mathrm{CH_2}$ groups plus the carboxyl carbon), for $4+1+4=9$ signals total.

In $\mathrm{B}$, the symmetric dimethylnaphthalene skeleton folds its ten ring carbons and two methyl carbons down through the molecule's own symmetry; working through every distinct environment in the fused ring system gives 7 signals for $\mathrm{B}$.

Let's summarize:

$\mathrm{A}$ = 4-(4-methylphenyl)butanoic acid, 9 distinct $^{13}\mathrm{C}$ signals.
$\mathrm{B}$ = a symmetric dimethylnaphthalene formed by cyclization, alpha-methylation, reduction, and aromatization, 7 distinct $^{13}\mathrm{C}$ signals.
Total: $9+7=16$.

The total number of carbon signals observed for $\mathrm{A}$ and $\mathrm{B}$ is 16.

Answer 2

This problem is really two structure-elimination puzzles chained together, followed by a symmetry count. Work out $\mathrm{A}$, then push it through the second sequence to get $\mathrm{B}$, then count carbons in each.

Toluene + succinic anhydride, $\mathrm{AlCl_3}$: This is a Friedel-Crafts acylation. The anhydride opens at the ring, attaching an aroyl group para to the methyl (steric preference) and leaving a free $\mathrm{-CH_2CH_2COOH}$ tail, giving the keto-acid 4-oxo-4-(p-tolyl)butanoic acid.
Zn-Hg, HCl (Clemmensen): Reduces only the aryl ketone, not the carboxylic acid, giving $\mathrm{A}=$ 4-(4-methylphenyl)butanoic acid, a para-disubstituted ring carrying a $\mathrm{-CH_2CH_2CH_2COOH}$ chain.
$\mathrm{H_2SO_4}$: Protonates the $\mathrm{COOH}$ of $\mathrm{A}$ and closes a new six-membered ring onto the arene by intramolecular Friedel-Crafts acylation, giving a methyltetralone.
LDA (1 equiv), MeI: The only alpha-hydrogen available for kinetic enolate formation is at C2 next to the carbonyl; methylation there installs a second methyl group on the saturated ring.
Zn-Hg, HCl: A second Clemmensen reduction removes the tetralone carbonyl, giving a dimethyltetralin (saturated ring fused to the original aromatic ring).
Pd/C: Dehydrogenates the saturated ring to full aromaticity, giving the dimethylnaphthalene $\mathrm{B}$, with the two methyls on ring positions related by the molecule's symmetry.

Now count carbons. In $\mathrm{A}$, the para-substituted ring gives 4 distinct ring signals (by the mirror symmetry through the two substituted positions), plus 1 for the ring methyl and 4 for the $\mathrm{-CH_2CH_2CH_2COOH}$ chain (three inequivalent $\mathrm{CH_2}$ groups plus the carboxyl carbon), for $4+1+4=9$ signals total.

In $\mathrm{B}$, the symmetric dimethylnaphthalene skeleton folds its ten ring carbons and two methyl carbons down through the molecule's own symmetry; working through every distinct environment in the fused ring system gives 7 signals for $\mathrm{B}$.

Let's summarize:

$\mathrm{A}$ = 4-(4-methylphenyl)butanoic acid, 9 distinct $^{13}\mathrm{C}$ signals.
$\mathrm{B}$ = a symmetric dimethylnaphthalene formed by cyclization, alpha-methylation, reduction, and aromatization, 7 distinct $^{13}\mathrm{C}$ signals.
Total: $9+7=16$.

The total number of carbon signals observed for $\mathrm{A}$ and $\mathrm{B}$ is 16.

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).

Show Hint

Correct Answer: 16

Solution and Explanation

Top Questions on Nucleophilic and electrophilic substitution reactions (both aromatic and aliphatic)

Questions Asked in GATE CY exam

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).