Question

An optically pure aldehyde with molecular formula C5H10O react with excess of formaldehyde gives product P and sodium formate in presence of conc. NaOH. Product P further react with cyclo pentanone in mild acidic medium to gives product Q. Calculate the value of x – y. Whereas x is number of methylene ch2 groups in product Q and Y is number of oxygen atoms in product Q.

Answer 1

4

Answer 2

Step 1: Identify the optically pure aldehyde (C$_5$H$_{10}$O)

The molecular formula C$_5$H$_{10}$O corresponds to saturated aldehydes or ketones. For an aldehyde to be optically pure, it must have a chiral carbon atom. Let's examine the possible aldehyde structures:

1. Pentanal (CH$_3$CH$_2$CH$_2$CH$_2$CHO): No chiral carbon.
2. 3-Methylbutanal ((CH$_3$)$_2$CHCH$_2$CHO): No chiral carbon.
3. 2-Methylbutanal (CH$_3$CH$_2$CH(CH$_3$)CHO): The carbon at position 2 is bonded to -H, -CH$_3$, -CH$_2$CH$_3$, and -CHO groups. This carbon is chiral.
4. 2,2-Dimethylpropanal ((CH$_3$)$_3$CCHO): No alpha hydrogen, no chiral carbon.

Therefore, the optically pure aldehyde is 2-Methylbutanal.

Step 2: Determine product P from the reaction of 2-Methylbutanal with excess formaldehyde in conc. NaOH

2-Methylbutanal has one alpha hydrogen. In the presence of strong base (conc. NaOH) and excess formaldehyde (which has no alpha hydrogens and is very reactive), a crossed aldol condensation occurs first, followed by a crossed Cannizzaro reaction.

The alpha hydrogen of 2-Methylbutanal is removed by the base to form an enolate: CH$_3$CH$_2$CH(CH$_3$)CHO $\xrightarrow{OH^-}$ CH$_3$CH$_2$C$^-$(CH$_3$)CHO

The enolate attacks the carbonyl carbon of formaldehyde: CH$_3$CH$_2$C$^-$(CH$_3$)CHO + HCHO $\rightarrow$ CH$_3$CH$_2$C(CH$_3$)(CH$_2$O$^-$)CHO $\xrightarrow{H_2O}$ CH$_3$CH$_2$C(CH$_3$)(CH$_2$OH)CHO This product is 2-(hydroxymethyl)-2-methylbutanal. It has no alpha hydrogens.

In the presence of excess formaldehyde and concentrated base, this aldehyde undergoes a crossed Cannizzaro reaction. Formaldehyde is oxidized to formate, and the 2-(hydroxymethyl)-2-methylbutanal is reduced to the corresponding alcohol. CH$_3$CH$_2$C(CH$_3$)(CH$_2$OH)CHO $\xrightarrow{HCHO, conc. NaOH}$ CH$_3$CH$_2$C(CH$_3$)(CH$_2$OH)CH$_2$OH + HCOONa The product P is 2-(hydroxymethyl)-2-methylbutan-1-ol. Its molecular formula is C$_6$H$_{14}$O$_2$.

Step 3: Determine product Q from the reaction of P with cyclopentanone in mild acidic medium

Product P is a diol: CH$_3$CH$_2$-C(CH$_3$)-(CH$_2$OH)$_2$. Diols react with ketones in mild acidic conditions to form cyclic ketals. The two hydroxyl groups of the diol react with the carbonyl group of the ketone, eliminating water.

The two -CH$_2$OH groups in P react with the carbonyl carbon of cyclopentanone to form a cyclic ketal. The structure of Q is a spiro compound where the spiro carbon is the ketal carbon originating from cyclopentanone. One ring is the cyclopentane ring (5 carbons). The other ring is formed by the two oxygen atoms from P, the two CH$_2$ groups attached to these oxygens, and the carbon atom (C*) from P to which these CH$_2$ groups are attached. This forms a 6-membered ring (-O-CH$_2$-C*-CH$_2$-O-C$_{ketal}$-). The carbon C* is bonded to a CH$_3$ group and a CH$_2$CH$_3$ group.

Let's visualize the structure of Q: The central carbon (C*) from P is bonded to CH$_3$, CH$_2$CH$_3$, and two -CH$_2$- groups. These two -CH$_2$- groups are part of a 6-membered ring formed by linkage through oxygen atoms to the ketal carbon (C$_{ketal}$). C$_{ketal}$ is also part of the 5-membered cyclopentane ring.

Structure of Q:

      O----CH2
     /      |
C_k--         C*(CH3)(CH2CH3)
     \      |
      O----CH2
      |
(Cyclopentane ring)

C$_k$ is the spiro carbon. It is bonded to two oxygen atoms and two carbons of the cyclopentane ring. C* is bonded to two CH$_2$ groups (in the ring), a CH$_3$ group, and a CH$_2$CH$_3$ group.

Step 4: Calculate x (number of methylene CH$_2$ groups in Q)

Methylene groups (CH$_2$) in Q are located:

1. Within the 6-membered ring: There are two CH$_2$ groups (-O-CH$_2$-C*-CH$_2$-O-).
2. Within the 5-membered cyclopentane ring: The cyclopentane ring has 5 carbons. One carbon is the spiro carbon (C$_k$), which is quaternary. The other four carbons are methylene groups (-CH$_2$-)$_4$. Total number of methylene groups (x) = 2 (from 6-membered ring) + 4 (from cyclopentane ring) = 6.

Step 5: Calculate y (number of oxygen atoms in Q)

The ketal structure contains two oxygen atoms, originating from the diol P. Total number of oxygen atoms (y) = 2.

Step 6: Calculate x - y

x - y = 6 - 2 = 4.