Practical Organic Chemistry — Page 6 | JEE Chemistry

Q51

In base vs. acid titration, at the end point methyl orange is present as

A quinonoid form

B heterocyclic form

C phenolic form

D benzenoid form

Correct Answer

Option A

Solution

Hence at the end point methyl orange is present as quinonoid form.

Q52

The group of chemicals used as pesticide is :

A Dieldrin, Sodium arsinite, Tetrachloroethene

B Aldrin, Sodium chlorate, Sodium arsinite

C DDT, Aldrin

D Sodium chlorate, DDT, PAN

Correct Answer

Option C

Solution

To find out which group of chemicals is used as a pesticide, let's examine each option : Option A : Dieldrin : a pesticide Sodium Arsinite : a pesticide and herbicide Tetrachloroethene (also known as perchloroethylene) : used mainly as a solvent and not as a pesticide.

Option B : Aldrin : a pesticide Sodium Chlorate : mainly used as a herbicide, not a pesticide.

Sodium Arsinite : a pesticide and herbicide Option C : DDT (Dichloro-Diphenyl-Trichloroethane) : a well-known pesticide Aldrin : a pesticide Option D : Sodium Chlorate : mainly used as a herbicide, not a pesticide.

DDT : a pesticide PAN (Peroxyacetyl Nitrate) : not used as a pesticide, it is a pollutant found in smog.

From the options given, it appears that Option C (DDT and Aldrin) contains chemicals exclusively used as pesticides.

Both chemicals in this option are known pesticides, while the other options contain chemicals that are not used as pesticides or are used for other primary purposes.

Hence, the correct answer would be : Option C : DDT, Aldrin

Q53

Given below are two statements: Statement I: Morphine is a narcotic analgesic. It helps in relieving pain without producing sleep. Statement II: Morphine and its derivatives are obtained from opium poppy. In the light of the above statements, choose the correct answer from the options given below

A Both Statement I and Statement II are true

B Statement I is true but Statement II is false

C Both Statement I and Statement II are false

D Statement I is false but Statement II is true

Correct Answer

Option D

Solution

Let's examine each statement individually: Statement I: Morphine is indeed a narcotic analgesic, which means it helps in relieving pain.

However, the second part of the statement is incorrect because morphine can produce sleep or drowsiness as it is a central nervous system depressant.

Statement II: This statement is true.

Morphine and its derivatives are obtained from the opium poppy (Papaver somniferum).

Therefore, the correct answer is Option D: Statement I is false but Statement II is true.

Q54

Match with . th, (Compound) (Use)

List - I		List - II
(A)	Carbon tetrachloride	(I)	Paint remover
(B)	Methylene chloride	(II)	Refrigerators and air conditioners
(C)	DDT	(III)	Fire extinguisher
(D)	Freons	(IV)	Non Biodegradable insecticide

A (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

B (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

C (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

D (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Correct Answer

Option B

Solution

Each of the compounds listed in List I has distinct uses.

Let's match them correctly with the uses mentioned in List II: Carbon tetrachloride (A) was historically used as a cleaning agent, among other things, but due to its toxicity and role in ozone depletion, it is no longer widely used for this purpose; hence it does not match with any of the uses.

However, it was previously used in fire extinguishers, so one might think of it in association with (III).

Methylene chloride (B), also known as dichloromethane, is commonly used as a solvent and paint remover, so it matches with (I).

DDT (C) is a well-known, though now largely banned, insecticide due to its persistence and bioaccumulation, making it not biodegradable.

It matches with (IV).

Freons (D) is a trade name for a group of halogenated compounds used primarily as refrigerants, so it matches with (II).

Considering these associations, we can match the compounds with their correct uses: (A) Carbon tetrachloride - (III) Fire extinguisher (B) Methylene chloride - (I) Paint remover (C) DDT - (IV) Non-biodegradable insecticide (D) Freons - (II) Refrigerators and air conditioners So, the correct match according to the given options will be: Option B: (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Q55

The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is

A Fe4[Fe(CN)6]3

B Na4[Fe(CN)5NOS]

C Fe(CN)3

D Na3[Fe(CN)6]

Correct Answer

Option A

Solution

If nitrogen is present in organic compound, then sodium extract contains NaCN. Na + C + N

\overset{{Fuse}}\longrightarrow

NaCN FeSO4 + 6NaCN $\to$ Na4[Fe(CN)6] + Na2SO4 Na4[Fe(CN)6] changes to prussian blue Fe4[Fe(CN)6]3 on reaction with FeCl3.

FeCl3 + Na4[Fe(CN)6] $\to$ Fe4[Fe(CN)6]3 + 12NaCl

Q56

29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is

A 59.0

B 47.4

C 23.7

D 29.5

Correct Answer

Option C

Solution

Moles of

HCl

taken

= 20 \times 0.1 \times {10^{ - 3}} = 2 \times {10^{ - 3}}

Moles of

HCl

neutralised by

NaOH

solution

= 15 \times 0.1 \times {10^{ - 3}} = 1.5 \times {10^{ - 3}}

Moles of

HCl

neutralised by ammonia

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \times {10^{ - 3}} - 1.5 \times {10^{ - 3}}

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 3}}

\%

of nitrogen

= {{1.4 \times N \times V} \over {w.t.\,\,of\,\,Subs\tan ce}} \times 100

= {{1.4 \times 0.5 \times {{10}^{ - 3}}} \over {29.5 \times {{10}^{ - 3}}}} \times 100 = 23.7\%

Q57

For standardizing NaOH solution, which of the following is used as a primary standard ?

A Ferrous Ammonium Sulfate

B dil. HCl

C Oxalic acid

D Sodium tetraborate

Correct Answer

Option C

Solution

The solution that can be prepared directly by dissolving an accurately weighed amount of the substance in water and making up the volume to a known amount by water is known as primary standard solution.

Oxalic acid is an example of primary standard while NaOH is a secondary standard solution.

Hence, strength of NaOH solution can be determined by titrating against oxalic acid.

This process is known as standardisation.

Q58

Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is :

A Fe4[Fe(CN)6]3

B K5Fe[Fe(CN)6]2

C HFe[Fe(CN)6]

D KFe[Fe(CN)6]

Correct Answer

Option A

Solution

Prussian blue is a complex compound that has the formula $\text{Fe}_{4}[\text{Fe(CN)}_{6}]_{3}$ .

It is formed by the reaction of ferric ions with ferrocyanide ions.

The reaction can be explained as follows: Potassium ferrocyanide, $\text{K}_{4}[\text{Fe(CN)}_{6}]$ , dissociates in water to release ferrocyanide ions, $[\text{Fe(CN)}_{6}]^{4-}$ .

The ferric ions, $\text{Fe}^{3+}$ , present in the acidic ferric chloride solution will react with the ferrocyanide ions.

In an acidic medium, the ferric ions hydrolyze, forming a complex with ferrocyanide ions, leading to the formation of Prussian blue.

The overall reaction can be represented as: $\text{Fe}^{3+} + 4[\text{Fe(CN)}_{6}]^{4-} \rightarrow \text{Fe}_{4}[\text{Fe(CN)}_{6}]_{3}$ The Prussian blue complex is a colloidal species and exhibits an intense blue color.

It has been historically used as a pigment in paints and also has applications in medical treatments as an antidote for certain types of heavy metal poisoning.

Q59

A, B and C are three biomolecules. The results of the tests performed on them are given below:

\begin{array}{|l|l|l|l|} \\ & \begin{array}{l} \text{ Molisch's } \\ \text{ Test } \end{array} & \begin{array}{l} \text{ Barfoed } \\ \text{ Test } \end{array} & \begin{array}{l} \text{ Biuret } \\ \text{ Test } \end{array} \\ \\ \text{ A } & \text{ Positive } & \text{ Negative } & \text{ Negative } \\ \\ \text{ B } & \text{ Positive } & \text{ Positive } & \text{ Negative } \\ \\ \text{ C } & \text{ Negative } & \text{ Negative } & \text{ Positive } \\ \\ \end{array}

A, B and C are respectively :

A A = Lactose, B = Glucose, C = Albumin

B A = Lactose, B = Fructose, C = Alanine

C A = Lactose, B = Glucose, C = Alanine

D A = Glucose, B = Fructose, C = Albumin

Correct Answer

Option A

Solution

The given table provides the results of three tests: Molisch's test, Barfoed's test, and Biuret test for three different biomolecules A, B, and C.

Molisch's Test: Detects carbohydrates.

Positive results for A and B indicate that they are carbohydrates.

Barfoed's Test: Specifically for detecting monosaccharides.

A positive result for B indicates that B is a monosaccharide, while A's negative result indicates it's not a monosaccharide, implying it might be a disaccharide.

Biuret Test: Used to detect the presence of proteins.

Positive result for C indicates that it is a protein.

Given the options, we can analyze them as follows: Option A: Lactose (disaccharide), Glucose (monosaccharide), Albumin (protein) - fits the given results.

Option B: Lactose (disaccharide), Fructose (monosaccharide), Alanine (amino acid) - not a match, as Alanine won't give a positive Biuret test.

Option C: Lactose (disaccharide), Glucose (monosaccharide), Alanine (amino acid) - same issue as Option B.

Option D: Glucose (monosaccharide), Fructose (monosaccharide), Albumin (protein) - not a match, as A must be a disaccharide.

Therefore, the correct option is: Option A: A = Lactose, B = Glucose, C = Albumin.

Q60

If you spill a chemical toilet cleaning liquid on your hand, your first aid would be

A aqueous NaOH

B aqueous NaHCO3

C aqueous NH3

D vinegar

Correct Answer

Option B

Solution

Toilet cleaning liquid has about 10.5% w/v HCl ; to neutralise its affect aqueous NaHCO3 is used while NaOH is avoided for this purpose because of its highly corosive in nature and can burn body.