0 . 15 g of an organic compound ( A ) gave 0 . 33 g of CO 2 and 0 . 18 g of H 2 O . The molecular weight of ( A ) is 60 . The compound ( A ) on dehydration gave a hydrocarbon ( B ) containing 85 . 7 % of C . ( B ) , on successive treatment with hydriodic acid and silver hydroxide gave a product ( C ) , isomeric with ( A ) . Find structural formulae of ( A ) , ( B ) and ( C ) .

0 . 33 gm of CO 2 will contain 0 . 09 gm carbon and 0 . 24 gm of oxygen. Moles of carbon = 0 . 09 12 = 0 . 0075 Again, 0 . 18 gm of water contains 0 . 02 gm of hydrogen and 0 . 16 gm of oxygen. Moles of hydrogen = 0 . 02 1 = 0 . 02 Total weight of carbon and hydrogen = ( 0 . 09 + 0 . 02 ) = 0 . 11 gm So, the weight of oxygen would be = 0 . 15 - 0 . 11 = 0 . 04 gm Moles of oxygen = 0 . 04 16 = 0 . 0025 Mole ratio of C : H : O is 0 . 0075 : 0 . 02 : 0 . 0025 Simple ratio of C : H : O would be 0 . 03 : 0 . 08 : 0 . 01 Integer ratio would be 3 : 8 : 1 So, the formula of a compound ( A ) would be : C 3 H 8 O which could be CH 3 CH 2 CH 2 OH as compound ( A ) with molecular weight of 60 . Now, dehydration of compound ( A ) will form alkene compound ( B ) as a product. CH 3 CH 2 CH 2 OH → dehydration CH 3 CH = CH 2 Compound ( B ) is CH 3 CH = CH 2 Compound ( B ) reacts with hydroiodic acid to form 2 - Iodo propane which on further reaction with silver hydroxide forms alcohol with respect to Markownikoff's rule. CH 3 CH = CH 2 → silver hydroxide Hydroiodic acid CH 3 CH OH CH 3 compound ( C ) . The alcohol formed is a position isomer of A .

An organic compound ( A ) has 76 . 6 % C and 6 . 38 % H . Its vapour density is 47 . It gives characteristic colour with aqueous FeCl 3 solution. ( A ) when heated with CO 2 and NaOH at 140 ° C under pressure, gives ( B ) which on acidification gives ( C ) . ( C ) reacts with acetyl chloride to give ( D ) which is a well-known painkiller. Identify ( A ) , ( B ) , ( C ) and ( D ) and also explain the reactions.

Element % by wt % by wt At wt = X i ; X i X smallest C 76 . 6 76 . 6 12 = 6 . 38 6 . 38 1 . 06 = 6 H 6 . 38 6 . 38 1 = 6 . 38 6 . 38 1 . 06 = 6 O 17 . 02 17 . 02 16 = 1 . 06 1 . 06 10 . 6 = 1 ∴ Empirical formula = C 6 H 6 O Empirical wt. = 12 × 6 + 6 + 16 = 94 Mol. Wt. = V. d. × 2 = 47 × 2 = 94 g Molecular weight = Empirical weight = 94 g ∴ Molecular formula = Empirical formula = C 6 H 6 O Compound A gives a characteristic colour with neutral FeCl 3 , so it should be phenol. Phenol reacts with CO 2 and NaOH under the given conditions so, as to form Compound B which is Sodium salicylate. Compound B on acidification forms compound C which is Salicylic acid. Compound C undergoes acetylation to form compound D which is Aspirin or Acetyl salicylic acid.

HARD

JEE Main

IMPORTANT

Earn 100

An alkene $(A)$ , on ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidised to an acid $(B)$ . When $(B)$ is treated with bromine in presence of phosphorus, it yields compound $(C)$ which on hydrolysis gives a hydroxy acid $(D)$ . The acid can also be obtained from acetone by the reaction with hydrogen cyanide followed by hydrolysis. Identify the compounds $(A), (B), (C)$ and $(D)$ .

Important Questions on Empirical, Molecular and Structural Formulae

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 22

An aromatic hydrocarbon

(A)

, containing

91.3 %

C

and

8.7 %

H

on treatment with chlorine gave

3

isomeric monochloro compounds

(B), (C)

and

(D)

, each having

28 %

of chlorine. On oxidation with permanganate, all the three gave a monobasic acid. The acid from

(B)

on distillation with soda lime gave benzene while those from

(C)

and

(D)

gave monochlorobenzene on the same treatment. Assign formulae to

(A), (B), (C)

and

(D)

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 23

0.15 g

of an organic compound

(A)

gave

0.33 g

{CO}_{2}

and

0.18 g

H_{2} O

. The molecular weight of

(A)

60

. The compound

(A)

on dehydration gave a hydrocarbon

(B)

containing

85.7 %

C

(B)

, on successive treatment with hydriodic acid and silver hydroxide gave a product

(C)

, isomeric with

(A)

. Find structural formulae of

(A), (B)

and

(C)

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 24

Compound

(A)

with molecular weight

108,

contained

88.89 %

C

and

11.11 %

H

. It gave a white precipitate with ammoniacal silver nitrate, Complete hydrogenation atom of

(A)

gave another compound

(B)

with molecular weight

112

. Oxidation of

(A)

gave an acid with equivalent weight

128

. Decarboxylation of this acid gave cyclohexane. Give structures of

(A)

and

(B)

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 25

A compound

(X)

containing only

C, H

and

O

is unreactive towards sodium. It does not add bromine. It does not react with Schiff reagent. On refluxing with an excess of hydroiodic acid,

(X)

Yields only one organic product

(Y)

. On hydrolysis,

(Y)

yields a new compound

(Z)

which can be converted to

(Y)

by reaction with red phosphorous and iodine. The compound

(Z)

, on oxidation with

{KMnO}_{4}

gives a carboxylic acid. The equivalent weight of this acid is

60

. What are the compounds

(X), (Y)

and

(Z)

. Write chemical equation leading to the conversion of

(X)

(Y)

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 27

0.369 g

of a bromo derivative of a hydrocarbon

(A)

when vaporised occupied

67.2 mL

NTP

(A)

, on reaction with aqueous

NaOH,

gives

(B)

(B)

, when passed over alumina at

250 ° C

, gives a neutral compound

(C)

while at

300 ° C

, it gives a hydrocarbon

(D)

(D)

, when heated with

HBr

gives an isomer of

(A)

. When

(D)

is treated with concentrated

H_{2} {SO}_{4}

and the product is diluted with water and distilled,

(E)

is obtained. Identify

(A)

(E)

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 28

When

0.90 g

of an organic compound

C_{4} H_{10} O_{2}

(A)

was treated with

Na

224 mL

H_{2}

were evolved at

NTP

. Compound

(A)

could be separated into two fractions

(B)

and

(C)

by crystallisation, of which fraction

(B)

could be resolved into optical isomers,

(D)

and

(E)

. Write down the structural formulae for

(C)

(D)

and

(E)

with proper reasoning.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 29

An organic compound

(A)

has

76.6 % C

and

6.38 %

H

. Its vapour density is

47

. It gives characteristic colour with aqueous

{FeCl}_{3}

solution.

(A)

when heated with

{CO}_{2}

and

NaOH

140 ° C

under pressure, gives

(B)

which on acidification gives

(C)

(C)

reacts with acetyl chloride to give

(D)

which is a well-known painkiller. Identify

(A), (B), (C)

and

(D)

and also explain the reactions.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 10 - Empirical, Molecular and Structural Formulae>PROBLEMS>Q 30

An unknown compound of

C, H

and

O

contains

69.77 %

C

and

11.63 % H

and has a molecular weight of

86

. It does not reduce Fehling's solution but forms a disulphide addition compound and gives a positive iodoform test. What are the possible structures?

Important Questions on Empirical, Molecular and Structural Formulae

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