What was the 2020 JEE Advanced numerical on degree of unsaturation from organic compounds containing oxygen?
The degree of unsaturation of Q is 18.
P is methyl 2,4,6-triphenylbenzoate treated with concentrated H2SO4. This generates an acylium ionic intermediate that cyclises onto an adjacent phenyl ring. Q is the resulting coloured cyclic aryl ketone. The task is to find the degree of unsaturation of Q.
How do you determine the molecular formula of methyl 2,4,6-triphenylbenzoate atom by atom?
P is C26H20O2.
The central benzene ring with four substituents contributes C6H2. The ester –CO2CH3 contributes C2H3O2. The three phenyl rings contribute 3×C6H5 = C18H15.
Add the pieces: 6 + 2 + 18 = 26 carbons, 2 + 3 + 15 = 20 hydrogens, and 2 oxygens. The explicit check confirms C26H20O2.
What is the molecular formula of Q after intramolecular cyclisation of the ester?
Q is C25H16O.
Concentrated H2SO4 generates the acylium ion from the ester. This triggers intramolecular electrophilic attack on the ortho position of an adjacent phenyl ring and eliminates CH3OH.
Subtract CH4O from C26H20O2. Carbons drop by one, hydrogens by four, oxygen by one. The result is exactly C25H16O.
Link the mechanism (acylium ion, EAS, rearomatisation) directly to the molecular formula change before any DU calculation. Methanol loss removes CH4O.
How do you calculate the degree of unsaturation for the cyclic ketone Q?
The degree of unsaturation of Q is 18.
The general DU formula is (oxygen ignored). For C25H16O ignore the O, plug in 25 carbons and 16 hydrogens to get .
Simplify: . This accounts for four benzene rings, one ketone carbonyl, one new ring, and additional unsaturations from phenyl attachments.
Search the past-paper archive for every JEE Advanced paper from 2007 by chapter to see identical formula-tracking patterns.
Why does the degree of unsaturation calculation for Q give 17 if you forget the +1 term?
The common slip uses the incomplete expression C – H/2 and stops at 25 – 8 = 17.
The +1 term that accounts for the saturated hydrocarbon reference is omitted. This single algebraic shortcut yields 17 instead of the correct 18.
The mistake occurs when students treat the DU formula like a simple H-deficiency count without recalling the full expression derived from CnH2n+2. Write the complete formula on paper every time.
What are three related questions on degree of unsaturation after cyclisation in oxygen-containing compounds?
These questions test the same skills of formula derivation and DU calculation.
Related Question 1: Compound A (C14H12O2) undergoes acid-catalysed intramolecular acylation losing H2O. Calculate DU of the product and identify the new ring size.
Product formula after losing H2O is C14H10O. Then . The new ring formed is six-membered.
Related Question 2 (Advanced 2018 style): Find DU for the product obtained when 2-(2-phenylphenyl)benzoic acid is treated with PPA; give the molecular formula first.
The acid is C19H14O2. PPA promotes intramolecular acylation with loss of H2O to give C19H12O. Then .
Related Question 3: Calculate DU for fluorescein (a xanthene dye formed by condensation of phthalic anhydride with resorcinol) and compare the value with that of the open-chain precursor.
Fluorescein is C20H12O5. Then . The open-chain precursor before final ring closure and dehydration has DU = 13.
Solve each on paper before checking.
What should you remember when solving DU questions from organic compounds containing oxygen jee 2020-type problems?
Always count hydrogens on the central ring by subtracting the number of substituents from six. Methanol loss in ester intramolecular Friedel-Crafts removes CH4O.
Oxygen never appears in the DU formula. The +1 is mandatory.
I expect that practicing at least three similar cyclisation DU questions will let you replicate the atom count and full formula correctly in your next mock test.
If a similar doubt appears in your mock, photograph a doubt for a step-by-step solution with the exact atom count shown.
Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).
Keep going with Thermodynamics JEE 2022: Soap Bubble Multi-Correct Question.
Frequently asked questions
What is the degree of unsaturation of Q in JEE Advanced 2020?
The degree of unsaturation of Q is 18. For C25H16O, ignore oxygen and use DU = 25 - 16/2 + 1 = 25 - 8 + 1 = 18. This matches four benzene rings, one ketone, the new ring from cyclisation plus attachments.
What is the molecular formula of Q in the 2020 JEE organic compounds containing oxygen question?
Q has molecular formula C25H16O. It forms from methyl 2,4,6-triphenylbenzoate (C26H20O2) on treatment with conc. H2SO4 via acylium ion cyclisation that eliminates CH3OH. The atom count drops by one carbon, four hydrogens and one oxygen.
Why is degree of unsaturation calculated as 17 instead of 18?
Students commonly omit the mandatory +1 term and stop at C - H/2, giving 25 - 8 = 17. The complete formula DU = C - H/2 + 1 comes from the saturated hydrocarbon reference CnH2n+2. Always write the full expression on paper.
How to find molecular formula of methyl 2,4,6-triphenylbenzoate for JEE?
It is C26H20O2. Count the central benzene ring with four substituents as C6H2, the ester -CO2CH3 as C2H3O2, and three phenyl rings as 3×C6H5 = C18H15. Adding yields 26 carbons, 20 hydrogens and 2 oxygens.