हिंदी

An organic compound 'A' with the molecular formula C4H8O2 undergoes acid hydrolysis to form two compounds 'B' and 'C'.

Advertisements
Advertisements

प्रश्न

An organic compound 'A' with the molecular formula C4H8O2 undergoes acid hydrolysis to form two compounds 'B' and 'C'. Oxidation of 'C' with acidified potassium permanganate also produces 'B'. Sodium salt of 'B' on heating with soda lime gives methane.

  1. Identify 'A', 'B' and 'C'.
  2. Out of 'B' and 'C', which will have higher boiling point? Give reason.
संक्षेप में उत्तर
Advertisements

उत्तर

  1. A = C4H8O2 [CH3COOC2H5] ester Reactions involved are
    \[\ce{\underset{A}{CH3-COOC2H5} + H2O ->[dil H2SO4] \underset{B}{CH3COOH} + \underset{C}{CH3CH2OH}}\]
    Then
    \[\ce{\underset{C}{CH3CH2OH} ->[KMnO4][{[O]}] \underset{B}{CH3COOH}}\]
    So A = CH3COOC2H5, B = CH3COOH, C = CH3CH2OH
  2. B has a higher boiling point than C. Because of their propensity to generate intermolecular H-bonds, carboxylic acids have higher boiling temperatures than alcohols.
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2022-2023 (March) Delhi Set 1

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Give a simple chemical test to distinguish between the following pair of compounds:

Ethanal and Propanal


An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.


\[C_6 H_5 - OH \to^{{Br}_2 \left( aq \right)} ?\]

Fehilng's test is positive for


Ammonical silver nitrate solution is called


A hydrocarbon (A) with molecular formula C5H10 on ozonolysis gives two products (B) and (C). Both (B) and (C) give a yellow precipitate when heated with iodine in presence of NaOH while only (B) give a silver mirror on reaction with Tollen’s reagent.

  1. Identify (A), (B) and (C).
  2. Write the reaction of B with Tollen’s reagent.
  3. Write the equation for iodoform test for C.
  4. Write down the equation for aldol condensation reaction of B and C.

An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and gives positive iodoform test. The compound is:


The correct set of products obtained in the following reactions:

  1. \[\ce{RCN ->[reduction]}\]
  2. \[\ce{RCN ->[(i) CH3MgBr][(ii) H2O]}\]
  3. \[\ce{RNC ->[hydrolysis]}\]
  4. \[\ce{RNH2 ->[HNO2]}\]

You are given four organic compounds “A”, “B” , “C” and “D”. The compounds “A”, “B” and “C” form an orange-red precipitate with 2, 4 DNP reagent. Compounds “A” and “B” reduce Tollen’s reagent while compounds “C” and “D” do not. Both “B” and “C” give a yellow precipitate when heated with iodine in the presence of NaOH. Compound “D” gives brisk effervescence with sodium bicarbonate solution. Identify “A”, “B”, “C” and “D” given the number of carbon atoms in three of these carbon compounds is three while one has two carbon atoms. Give an explanation for our answer.


Fehling’s solution ‘A’ is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×