Question 1

How many 2∘ Hydrogen atoms are present in the given following compound?

Accepted Answer

2∘H are seven.

Question 2

Correct energy profile for amine inversion and hybridization of nitrogen in the transition state is:

Accepted Answer

(d) Potential energy of (A) and (B) is same and transition state has the characteristics of sp2.

Question 3

Correct order of the heats of combustion of above compounds is:

Accepted Answer

(a) More the number of carbons more will be the heat of combustion.

Question 4

Which of the following is not a resonance structure of the others?

Accepted Answer

(d) is not the resonating structure of others.

Question 5

Rank the hydrogen atoms (Ha​,Hb​,Hc​) present in the following molecule in decreasing order of their acidic strength.

Accepted Answer

The correct order is (d) c>b>a.c-H atom is highly polar while a-H is least polar.

Question 6

The correct relation between the bond lengths a and b is:

Accepted Answer

(b) due to resonance, this bond has partial double bond characters

Question 7

The number of sp2−sp2 sigma bonds in the compound given below is:

Accepted Answer

(c)

Question 8

The total number of lone pair of electrons in the given molecule is :

Accepted Answer

(b) Total ' 3 ' lone pairs in the molecule (two lone pairs O and one lone pair N).

Question 9

Which of the following rings is highly strained ?

Accepted Answer

(c) Ring strain ∝ size of ring 1​

Question 10

The functional groups present in Cortisone are :

Accepted Answer

(b) Alcohol, ketone, alkene and ether.

Question 11

Select the acid with the highest Ka​ (i.e., lowest pKa​ ).

Accepted Answer

(b) Acidic strength ∝−I;−H;−M

Question 12

Most stable carbocation among the following is :

Accepted Answer

Tertiary carbocation is more stable than secondary and primary carbocation

Question 13

Arrange the following in increasing order of their pKa​ values.

Accepted Answer

(b) Stability of resonance hybrid ∝ no. of equivalent R.S. So stability order is ; So Acidic strength order is x>y>z so pKa​ is z>y>x.

Question 14

In the given pair identify the most acidic compound in (A) and (B). Most basic in (C) and (D).

Accepted Answer

(b) The ortho-effect : This is a special effect that is shown by o-substituents of benzene and its derivatives, but is not necessarily just a steric effect, e.g., the basicities of some o-substituted anilines were explained in terms of steric effects and differences in crowding round the nitrogen atom. This ortho-effect also operates with the benzoic acids. Irrespective of the polar type, nearly all o-substituted benzoic acids are stronger than benzoic acid. As we have seen, benzoic acid is a resonance hybrid, and so the carboxyl groups is coplanar with the ring. An o-substituent tends to prevent this coplanarity. Thus resonance is diminished (or prevented), and so the O-atom of the OH groups has a greater positive charge, resulting increased acid strength. It follows from this that the greater the steric inhibition to resonance, the stronger is the acid. Support for this is the following order of strengths of substituted benzoic acids.2,6-di-Me (pK . ) a321 > 2-t-Bu (pK . ) a3 46 > 2-Me (pK . ) a391 .Here again, if we consider the stability of the anion, steric inhibition of resonance prevents the +R effect of the ring coming into operation (see above), and since this weakens acid strength, its absence results in increased acid strength.Ka​ order: (due to ortho effect)Kb​ order:For options B and D, no ortho effect is valid and order of acidity and basicity is calculated by nearly examining the inductive effect.

Question 15

Several factors (steric, electronic, orbital interactions etc.) can affect the inversion barrier of an amine. In the given pair which data is correctly placed ?(d)

Accepted Answer

(a) The barrier to inversion for i-Pr 2NMe is less than for Me3​ N because going from sp3 (tetrahedral) to sp2 (trigonal planar) spreads the bulky isopropyl groups further apart and relieves steric crowding. In other words the isopropyl groups destabilize the pyramidal amine more than the planar transition structure.(b) The sp2 transition state has ideal bond angle of ∼120∘. The three member ring T.S. is highly strained because it cannot achieve 120∘ due to the small ring locking the angles at ∼60∘. The 5-membered ring can easily adopt 120∘ angle.(c) As the nitrogen goes from sp3→sp2, it becomes effectively more electronegative. (Remember : more s-character means more electronegative) The electronegative chlorine atoms pull electron density away from N more than the methyl groups, creating an electron - deficient N, and destabilising the T.S.

Question 16

Circle represents the most basic atoms in these molecules. Which of the following is the correct representation?

Accepted Answer

(d) All ' N ' atoms of circle have localised l.p. which do not take part in resonance.

Filters

Filters

book

topic

difficulty

Solutions for Advanced Problems in Organic Chemistry for JEE by M.S. Chouhan

Found the solution, but did not understand the concept?

Ask my first question