English
Karnataka Board PUCPUC Science 2nd PUC Class 12

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 - Chemistry

Advertisements
Advertisements

Question

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.

Chemical Equations/Structures
Long Answer
Advertisements

Solution

(a) Finding the molecular formula of a compound:

Percentage of carbon = 69.77%

Percentage of hydrogen = 11.63%

∴ Percentage of oxygen = 100 – (69.77 + 11.63)

= 18.6%

C : H : O = `69.77/12 : 11.63/1 : 18.6/16`

= 5.81 : 11.63 : 1.16

∴ Simple ratio = 5 : 10 : 1

Empirical formula of the given compound = C5H10O

Empirical formula mass = 5 × 12 + 10 × 1 + 1 × 16 = 86

Molecular mass = 86  ...(Given)

Molecular formula = C5H10O × `86/86` = C5H10O

Thus, the molecular formula of the given compound is C5H10O.

(b) Determining the structure of a compound:

  1. Since the given compound forms a divalent compound with sodium hydrogensulphite, it must be an aldehyde or a ketone.
  2. Since the compound does not reduce Tollen's reagent, it cannot be an aldehyde. Hence, it must be a ketone.
  3. Since the compound gives an iodoform test, the given compound is methyl ketone.
  4. Since the given compound gives a mixture of ethanoic acid and propanoic acid on vigorous oxidation, methyl ketone is pentane-2-one. Its structure is as follows:

\[\begin{array}{cc}
\ce{O}\phantom{........}\\
||\phantom{........}\\
\ce{\underset{Pentan-2-one}{CH3 - C - CH2CH2CH3}}
\end{array}\]

(c) Details of the reactions involved:

\[\begin{array}{cc}
\phantom{.................}\ce{O}\phantom{...........................}\ce{OH}\phantom{.....}\ce{S\overset{-}{O}_3N\overset{+}{a}}\\
\phantom{..........}||\phantom{...............................}\backslash\phantom{...}/\\
\ce{CH3 - \underset{Pentan-2-one}{C - CH2}CH2CH3 + NaHSO3 -> C}\\
\phantom{..........................................}/\phantom{...}\backslash\\
\phantom{..................................}\ce{\underset{Sodium Hydrogen Sulphite Additive Products}{CH3CH2H2C \phantom{.....}CH3}}
\end{array}\]

\[\begin{array}{cc}
\ce{O}\phantom{....................................................................................}\\
||\phantom{....................................................................................}\\
\ce{CH3 - \underset{Pentan-2-one}{C - CH2}CH2CH3 + 3I2 + 4NaOH ->[Iodoform][reaction]\underset{(Yellow ppt)}{\underset{Iodoform}{CH3\downarrow}} + CH3CH2CH2COONa + 3NaI + 3H2O}
\end{array}\]

\[\begin{array}{cc}
\ce{O}\phantom{...............................................}\\
||\phantom{...............................................}\\
\ce{CH3 - \underset{Pentan-2-one}{C - CH2}CH2CH3->[K2Cr2O7][H2SO4] \underset{Ethanoic acid}{CH3COOH} + \underset{Propanoic acid}{CH3CH2COOH}}
\end{array}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Aldehydes, Ketones and Carboxylic Acids - Exercises [Page 257]

APPEARS IN

NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 8 Aldehydes, Ketones and Carboxylic Acids
Exercises | Q 8.19 | Page 257

RELATED QUESTIONS

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

Ethanal and Propanal


Distinguish between:

C6H5-COCH3 and C6H5-CHO


A and B are two functional isomers of compound C3H6O.On heating with NaOH and I2, isomer B forms yellow precipitate of iodoform whereas isomer A does not form any precipitate. Write the formulae of A and B.


How will you convert ethanal into the following compound?

But-2-enoic acid


Complete the synthesis by giving missing starting material, reagent or product.


Which of the following compounds will give butanone on oxidation with alkaline \[\ce{KMnO4}\] solution?


Solvent used for dewaxing of petroleum products are


Acetone and acetaldehyde are differentiated by


Fehilng's test is positive for


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. \[\ce{RNH2 ->[HNO2]}\]

In Tollen's test for aldehyde, the overall number of electrons(s) transferred to the Tollen's reagent formula \[\ce{[Ag(NH3)2]+}\] per aldehyde group to form silver mirror is ______. (Round off to the nearest integer)


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.

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.


Benzaldehyde is obtained from Rosenmund’s reduction of:


Fehling’s solution ‘A’ is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×