Q.14.

Question

Question: Januvia, the trade name for sitagliptin, was introduced in 2006 for the treatment of type 2 diabetes. (a) Explain why the five-membered ring in sitagliptin is aromatic. (b) Determine the hybridization of each N atom. (c) In what type of orbital does the lone pair on each N atom reside?

Step-by-Step Solution

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Answer

Answer

a. The five-membered ring in sitagliptin is aromatic because it has six pi electrons.

b. Hybridization of each N atom:

$s p^{3}$
$s p^{2}$
$s p^{2}$
$s p^{2}$
$s p^{2}$

c. Orbital in which lone pair resides:

$s p^{3}$
p
p
$s p^{2}$
$s p^{2}$

1Step 1: Conditions for lone pair in hybrid orbital or pure orbital

When the heteroatom is bonded to multiple bonds (double bond), its lone pair resides in a hybrid orbital and is not involved in delocalization.

When the heteroatom is not bonded to multiple bonds, its lone pair resides in a p orbital and is involved in delocalization.

2Step 2: Reason for aromaticity of the five-membered ring

(a)

Sitagliptin - N atoms in the five-membered ring

There are six non-bonding electrons in the five-membered ring: four electrons from two double bonds and one lone pair from $s p^{3}$ hybridized N atom.

The total number of non-bonding electrons follows Huckel’s rule and hence the five-membered is aromatic.

3Step 3: Hybridization of each N atom and location of lone pair

(b) and (c)

Sitagliptin - N atoms in the structure

For 1: N atom is attached to four groups - $s p^{3}$ hybridization.

N atom lone pair is not delocalized and therefore resides in $s p^{3}$ hybrid orbital.

For 2: N atom is attached to three groups - $s p^{2}$ hybridization.

N atom lone pair is delocalized and therefore resides in the p orbital.

For 3: N atom is attached to three groups - $s p^{2}$ hybridization.

N atom lone pair is delocalized and therefore resides in the p orbital.

For 4: N atom is attached to three groups - $s p^{2}$ hybridization.

N atom lone pair is not delocalized and therefore resides in $s p^{2}$ hybrid orbital.