Tandem Reactions and Super Acids

Chemistry Level 3

You are tasked to synthesise the above compound's derivative.

You start with Succinaldehyde , Benzylamine and b-ketoglutaric acid whose formulae are $C_{4}H_{6}O_{2}$ , $C_{7}H_{9}N$ , $C_{5}H_{6}O_{5}$ respectively.

These compounds undergo a tandem reaction to give precursor P . Now,

P is reduced with $NaBH_{4}$ and then reacts with $NaCN$ in presence of $TsOH$ to give A
A reacts with DIBAL-H to form B after hydrolysis.
B reacts with 2-chloro ethanol in presence of piperidine and a catalytic amount of ${H}^{+}$ to give C after hydrolysis.
C is now oxidised in acidic $KMnO_{4}$ to give D
D is decarboxylated to give precursor E whose formula is $C_{16}H_{21}O_{2}N$
E undergoes hydrogenation in presence of palladium to give the precursor F
F forms its silver salt and on reaction with $Br_{2}$ gives G
G undergoes an intramolecular reaction to form the target molecule.

You call the target molecule X and carry out the next set of reactions on it:

X on hofmann exhaustive methylation gives an unstable compound Y
Y immediately rearranges in presence of ${H}^{+}$ and heat to give W
W on mono chlorination followed by reaction with $AgF$ gives a compound T

T reacts with $SbF_{5}$ to give a complex of antimony and a very stable cation V .

However V can be thought of as a tropylium cation derivative and it exceeds the tropylium carbocation's mass by b units.

In B let a denote the sum of double bonds between carbons and the number of oxygen atoms.

Calculate the value of b - a - 1

Inspirations:

Problem 7 in 2016's IChO
Teleanu Florin's problem

The answer is 9.

1 solution

Vitthal Yellambalse
Oct 11, 2016

The structures of the starting components are:

These components, when combined under go a tandem reaction called Robinson's Troponine synthesis . This reaction was involved in Problem 7 of 2016's IChO . These components form P its structure is shown:

P is first reduced and then substituted to give A .

Since A contain's a cyanide group and DIBAL-H is a mild reducing agent, an aldehyde is formed after hydrolysis of the resultant imine. This is B , it contains 3 double bonds and 1 oxygen. Hence $\boxed{a = 4}$

B reacts with the next set of reagents to add a carbon chain at the $\alpha$ position of the aldehyde. Hence C is:

In presence of acidic $KMnO_{4}$ , both alcohol and aldehyde are oxidised to carboxylic acids, to give D :

On decarboxylation, one of the carboxy groups is lost resulting in compound E . The fact that only one group is lost can be seen by observing E 's molecular formula.

During hydrogenation the benzyl group bonded to nitrogen is replaced with a hydrogen. This gives us F

The silver salt and addition of bromine are the reagents used in Hunsdiecker Reaction to give us G :

G contains a nucleophillic nitrogen and primary carbon bonded to a bromine. This results in an intramolecular $S_{N}2$ reaction, giving us the target molecule X :

On hofmann exhaustive methylation we get compound Y . It is unstable since its double bonds aren't in conjugation and can hence rearrange to ensure conjugation is achieved.

Y now rearranges to form a conjugated system in presence of ${H}^{+}$ . On heating, this intermediate can undergo a [1 7] electrocyclic shift that results in compound W . Teleanu Florin's problem looks into the interesting chemistry brought about by one such electrocyclic shift.

W 's major mono chlorination product is at its tertiary $s{p}^{3}$ carbon. This reacts with $AgF$ to precipitate $AgCl$ out to give T :

Super acids played a very large role in isolating carbocations . Tl;dr, we get the superacid $SbF_{6}^{-}$ and carbocation V :

Tropylium has a mass of 91 units and the species V has a mass of 105 units. It exceeds tropylium's mass by 14 units. Hence $\boxed{b = 14}$ .

Therefore the answer is 14 - 4 - 1 = $\boxed{9}$

Tandem Reactions and Super Acids

The answer is 9.

1 solution

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