Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: savy2020 on January 20, 2010, 08:48:10 PM

Title: Cycloaddition chemistry question (INChO 2007 Q4)
Post by: savy2020 on January 20, 2010, 08:48:10 PM
The question is from INChO - Indian National Chemistry Olympiad. I'm typing the question as such first.... I'll show my working also
I just want to know if I'm correct....

4.8 Draw the structure of dimethyl 1,2-cyclobutene dicarboxylate (J)

When J is heated with maleic anhydride(butenedioic anhydride), an unusual reaction takes place to form compound K. When K is boiled with aq. NaOH and the solution is acidified, compound L (C10H10O8), which is optically inactive, is obtained. 1.0g of L reacts with 77.5mL of 0.2M NaOH.

4.9 Equivalent weight of compound L is _________

4.10 The no. of -COOH groups present in compound L is ________

4.11 L is expected to contain (Mark X for all the appropriate choices)
(i) Cyclobutane ring   ___
(ii) Cyclohexane ring  ___
(iii) open chain structure ___
(iv) one double bond ___

4.12 Draw the possible structure/s of compound L and compound K.

4.13 Draw the structure of a possible intermediate in the reaction.

L on heating forms M (C10H6O6) which on reaction with N gives a polymer O.

4.14 Draw the structure of a representative segment of polymer O.

4.8 That's pretty easy ;). This is my structure..

4.9 As 1.0g of L reacts with (77.5)(0.2) m.eq. of NaOH, equivalent weight is 1000/(15.5) ~64.5 g/eq

4.10 no. of -COOH groups in L = (Molecular Wt.)/(Eq. Weight) = 4

4.11 Well for this question I've calculated DBE(Double bond equivalent) of L and got '6'. 4 of the 'double bonds' are with 4 -COOH s.
So that leaves L with 2 more unsaturation units .
I expect it may contain a ring (hopefully a cyclohexane ring) and 'may' be a double bond.

Since I've expected L to contain cyclo hexane ring and a double bond, I think the possible reaction in accordance with it is

4.13 No idea how the reaction proceeds.. Is it by free-radical mechanism or ... ???

I  have some vague idea about this but first let's agree upon 'L'.

But there are other approaches also which yield different products..
I'll post them later..