Exams › JEE Advanced › Chemistry
How many carbon atoms are present in the first (simplest) member of the homologous series of esters?
- 1
- 2
- 3
- 4
Correct answer: 2
Solution
The first member of the ester homologous series is methyl formate (HCOOCH3), which contains 2 carbon atoms — one in the formyl group and one in the methyl group.
Related JEE Advanced Chemistry questions
- Three carbon acids are listed with their pKa values: (I) Triphenylmethane (Ph3CH), pKa = 33.3 (II) Benzene (C6H6), pKa = 43 (III) Methane (CH4), pKa = 50 Arrange them in increasing order of acidity (least acidic first).
- From the following compounds, how many will produce effervescence (release CO2 gas) when treated with aqueous NaHCO3 solution? (i) 2-methylbenzoic acid (o-toluic acid) (ii) cyclohexylsulfonic acid (iii) 4-aminobenzaldehyde (iv) 4-hydroxycyclohexanone (v) cyclobutane-1,2,3,4-tetraol
- Which of the following correctly represents the order of acid strength for the given substituted benzoic acids?
- Match the following chemical tests in List-I with the functional groups or compounds they identify in List-II. List-I: (A) Baeyer test, (B) Ceric ammonium nitrate test, (C) Tollen's dye test (phthalein dye test), (D) Schiff test List-II: (I) Phenol, (II) Aldehyde, (III) Alcoholic -OH group, (IV) Unsaturation Choose the correct matching:
- In the Dumas method for nitrogen estimation, 0.30 g of an organic compound liberated 82.1 mL of nitrogen gas collected over water at 300 K and 775 mm Hg total pressure. Given that the aqueous tension (vapour pressure of water) at 300 K is 15 mm Hg, calculate the percentage of nitrogen in the compound.
- 0.8 g of an organic compound is analysed for nitrogen by Kjeldahl's method. If the nitrogen content is found to be 42%, how many mL of (M/2) H2SO4 are neutralized by the ammonia liberated? (N = 14)
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →