Direction (Q. Nos. 1 - 10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c), and (d), out of which ONLY ONE option is correct.
Arrange the following compounds in increasing order of their basic strength.
Aniline (pKa value = 4.87)
Pyridine (pKa value = 5.2)
Piperidine (pKa value = 11.22)
Pyrrole ( pKa value =16.5)
Higher the pKa value (lower acidity), higher will be the basicity.
So the order of basic strength will be
I < II < III < IV
In the compound below, which nitrogen is protonated first when treated with HCI?
Which of the following compounds is more easily oxidised to a carbonyl when treated with MnO2?
MnO2 is used for the selective oxidation of allylic and benzylic carbon and in these compounds these carbons are not very easily oxidisable.
Which among these is the simplest example for polycyclic arenes?
Naphthalene has fused ring of aromaticity and has the simplest structure when compared with other polycyclic aromatic hydrocarbons.
The compound shown below evolve hydrogen gas when refluxed with potassium metal, why?
Which compound below has maximum tendency to form a salt when treated with HBr?
What is the correct order of increasing acidic strength of the following?
How many Kekule structures exists for naphthalene?
We'll use the three double bond symbol simply because it is also routinely used in the text. Keep in mind that if the hexagon contains neither the three double bonds nor the circle, the compound is not aromatic. It is simply cyclohexane and there are two hydrogens on each carbon atom.
How many monobromo derivatives exists for anthracene?
The correct answer is Option B.
There are 3 monobromo derivatives exists for anthracene:
7-bromo-1,3,5-cycloheptatriene exists as ionic species in aqueous solution while 5-bromo-1,3-cyclopentadiene does not ionise even in presence of AgNO3(aq) because
The correct answer is Option C.
The C-Br bond in the case of 7-bromo-1, 3, 5-cycloheptatriene is broken easily because the intermediate carbocation formed is very stable (aromatic as it contains (4n + 2)π e- ie, follows Huckle rule) while it does not break easily in the case of S-bromo-1, 3-cyclopentadiene because carbocation formed here is highly unstable as it is anti aromatic ie, does not follow Huckel rule. (It contains 4π electrons).
Direction (Q. Nos. 11-15) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.
Q. Consider the following compounds.
The correct statement regarding properties of above mentioned compounds is/are
Which of the following systems are aromatic?
The correct answers are Options C and D.
Aromatic compounds are those which follow Huckel's rule i.e, they have (4n+2) π electrons, n must be an integer.
In option C, there are 5 π bonds which means 10 π electrons; so 4n+2 = 10 i.e, n= 2 which is an integer.
In option D, Nitrogen has a lone pair which contains 2 electrons therefore this compound also have 10 π electrons; so n= 2
What is true about the 1,3,5,7-cyclooctatetraene?
1-3-5-7-cyclooctatetraene it has 8 pi electrons, and like stated above, fits the criteria of 4n, to be antiaromatic. to avoid this state of anti-aromaticity (less stable then expected), it becomes non-planar, so it can be more stable then it would be in the antiaromatic state. cyclooctatetraene can do this because it can fold, however other 6 carbon compounds that have 4n electrons and are planar can not and result in an antiaromatic compound.
Potassium cyclooctatetraene is formed by the reaction of cyclooctatetraene with potassium metal:
2 K + C8H8 → K2C8H8
The reaction entails 2-electron reduction of the polyene and is accompanied by a color change from colorless to brown.
What is true regarding the following compound?
The given compound will turn itself to
To gain aromaticity, it will transfer the electrons as follow:-
It is clear that the compound is heterocyclic(the ring constitutes other than C and H). Due to -ve charge on outer O atom, it has high affinity for BF3. However, NaBH4 has no reaction with this. As the cound will turn itself to latter, there is no aldehyde or ketone group present in the compound.
Organic compounds can be classified even based upon the function groups. Identify the one which is not a functional group
Isocyanide is a compound and it is not a functional group.