
- Long answer type questions
- JEE Mains Solutions 2020, 4th september, chemistry, second shift
- JEE Mains Solutions 2020, 4th september, chemistry, First shift
- 25th June 2022 first shift
Long answer type questions
1. An alkene ‘A’ (Mol. formula C5H10 on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and forms an iodoform on treatment with I2 and NaOH. Compound ‘C’ does not give Fehling’s test but forms an iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C. |
Ans: Fehling's test is positive when compound 'B' is used. It gives an iodoform test and proves that it is an aldehyde. Because it fails Fehling's test, compound 'C' is a ketone.
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2. An aromatic compound ‘A’ (Molecular formula C8H8O]) gives a positive 2, 4-DNP test. It gives a yellow precipitate of compound ‘B’ on treatment with iodine and sodium hydroxide solution. Compound ‘A’ does not give Tollen’s or Fehling’s test. On drastic oxidation with potassium permanganate it forms a carboxylic acid ‘C’ (Molecular formula C7H6O2), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved. |
Ans: The aromatic compound ‘A’ does not give Tollen’s reagent test, it is not an aromatic aldehyde. As it responds to an iodoform test called methyl ketone.
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3. Write down functional isomers of a carbonyl compound with molecular formula C3H6O. Which isomer will react faster with HCN and why? Explain the mechanism of the reaction also. Will the reaction lead to the completion with the conversion of whole reactant into product at reaction conditions? If a strong acid is added to the reaction mixture what will be the effect on concentration of the product and why? |
Ans: Functional isomers
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4. When liquid ‘A’ is treated with a freshly prepared ammoniacal silver nitrate solution, it gives bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogensulphite. Liquid ‘B’ also forms a white crystalline solid with sodium hydrogensulphite but it does not give test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also. |
Ans: Liquid ‘A’ reduces ammoniacal silver nitrate and ‘B’ is a ketone which forms white crystalline solid on treatment with hydrogen sulphite
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Commonly asked questions
An alkene ‘A’ (Mol. formula C5H10 on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and forms an iodoform on treatment with I2 and NaOH. Compound ‘C’ does not give Fehling’s test but forms an iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C.
An aromatic compound ‘A’ (Molecular formula C8H8O]) gives a positive 2, 4-DNP test. It gives a yellow precipitate of compound ‘B’ on treatment with iodine and sodium hydroxide solution. Compound ‘A’ does not give Tollen’s or Fehling’s test. On drastic oxidation with potassium permanganate it forms a carboxylic acid ‘C’ (Molecular formula C7H6O2), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.
Write down functional isomers of a carbonyl compound with molecular formula C3H6O. Which isomer will react faster with HCN and why? Explain the mechanism of the reaction also. Will the reaction lead to the completion with the conversion of whole reactant into product at reaction conditions? If a strong acid is added to the reaction mixture what will be the effect on concentration of the product and why?
JEE Mains Solutions 2020, 4th september, chemistry, second shift
JEE Mains Solutions 2020, 4th september, chemistry, second shift
Commonly asked questions
The Crystal Field Stabilization Energy (CFSE) of is :
The processes of calcination and roasting in metallurgical industries, respectively, can lead to:
of a waste solution obtained from the workshop of a goldsmith contains and . The solution was electrolyzed at by passing a current of for 15 minutes. The metal/metals electrodeposited will be :
JEE Mains Solutions 2020, 4th september, chemistry, First shift
JEE Mains Solutions 2020, 4th september, chemistry, First shift
Commonly asked questions
Among statements (a) - (d), the correct ones are:
(a) Lime stone is decomposed to during the extraction of iron from its oxides.
(b) In the extraction of silver, silver is extracted as an anionic complex.
(c) Nickel is purified by Mond's process.
(d) Zr and Ti are purified by Van Arkel method.
The decreasing order of reactivity of the following organic molecules towards solution is:
The region in the electromagnetic spectrum where the Balmer series lines appear is:
25th June 2022 first shift
25th June 2022 first shift
Commonly asked questions
Bonding in which of the following diatonic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?
(A) NO
(B) N2
(C) O2
(D) C2
(E) B2
Choose the most appropriate answer from the options given below
Incorrect statement for Tyndall effect is:
Leaching of gold with dilute aqueous solution of NaCN in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold an another complex [B], [A] and [B], respectively are
Chemistry NCERT Exemplar Solutions Class 12th Chapter Twelve Exam