Answer the question on basis of passage given below :
A ketone \([A]\) which undergoes haloform reaction gives compound \([B]\) on reduction. Compound
\([B]\) on heating with sulphuric acid gives compound \([C]\) which forms monozonide \([D]\). The
compound \([D]\) on hydrolysis in the presence of zinc dust gives compound \([E]\). 
Compound \([E]\) on oxidation gives \(C _2 H _4 O _2\) [F]
Compound \([ F ]\left( C _2 H _4 O _2\right)\) on treatment with chlorine, in the presence of small amount of red phosphorus, gives :

Arrange the following in decreasing order of the \(\text{pK}_\text{b}\) value :
(A) \(\text{C}_2\text{H}_5\text{NH}_2\)
(B) \(\text{C}_6\text{H}_5\text{NHCH}_3\)
(C) \((\text{C}_2\text{H}_5)_2\text{NH}\)
(D) \(\text{C}_6\text{H}_5\text{NH}_2\)
(E) \((\text{CH}_3)_3\text{N}\)
Choose the correct answer from the options given below :

The rate constant for a first order reaction becomes 2 times when the temperature is raised from 330 K to 380 K . Then the energy of activation for the reaction will be:\[[\log 2=0.3010]\]

Choose the correct statement(s) for intermolecular dehydration of alcohols to form ethers
(A) The formation of ether may involve a nucleophilic bimolecular \(S_N2\) reaction or \(S_N1\) reaction
(B) The method is successful with secondary and tertiary alcohols.
(C) The reaction requires protic acids
(D) The reaction is carried out at low temperature.
Choose the correct answer from the options given below:

Read the passage carefully and answer the questions:
The degeneracy of the \(d\) orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral comples to yield three orbitals of lower energy, \(t_{2 g}\) set and two orbitals of higher energy, \(e_g\) set. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation is denoted by \(\Delta_0\). Thus energy of the two \(e_g\) orbitals will increase by \((3 / 5) \Delta_0\) and that of three \(t_{2 g}\) will decrease by \((2 / 5) \Delta_0\). The crystal field splitting \(\Delta_0\) depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields, in which case the splitting will be large, whereas others produce weak fields and consequently result in small splitting of \(d\) orbitals. Relative magnitude of crystal field splitting energy \(\Delta_{\circ}\) and pairing energy, \(P\) (energy required for electron pairing in a single orbital) determine the formation of low spin ( \(\Delta_0>P\) ) or high spin ( \(\Delta_0<P\) ) complex. In tetrahedral coordination entity formation, the \(d\)-orbital splitting is inverted and is smaller as compared to the octahedral field splitting. The crystal field theory attributes the colour of the complex to \(d-d\) transition of the electron. The colour of the coordination compounds depends on the crystal field splitting. In the absence of ligand, crystal field splitting does not occur and hence the substance is colourless.
Choose the correct order of the ligands in their increasing crystal field splitting:

Read the passage carefully and answer the Questions
Molar conductivity \(\left(\Lambda_m\right)\) of a solution at a given concentration (c) is the conductance of volume, V of solution, containing one mole of electrolyte kept between the two electrodes with area of cross-section A and at a distance of unit length. It increases
with the decrease in concentration and when the concentration approaches zero, the molar conductivity is called limiting molar conductivity \(\left(\Lambda_m^0\right)\). For a strong electrolyte, \(\Lambda_m\) increases linearly with dilution and is given by \(\Lambda_m=\Lambda_m^0-A c^{1 / 2}\). The value of the constant A for a given solvent depends on the type of electrolyte along with temperature. According to Kohlrausch law, the value of \(\left(\Lambda_m^0\right)\) for an electrolyte is \(\Lambda_m^0=\nu_{+} \lambda_{+}^0+\nu_{-} \lambda_{-}^0\), where \(\nu_{+}\)and \(\nu_{-}\) are the number of cations and anions, respectively, per
molecule of the electrolyte and \(\lambda_{+}^0\) and \(\lambda_{-}^0\) are limiting molar conductivities of cation and anion, respectively. Kohlrausch law
finds many applications, like determining the solubility of a sparingly soluble salt, determining the degree of dissociation \(\left(\Lambda_m / \Lambda_m^0\right)\), and the dissociation constant of a weak electrolyte.
The unit of constant A is :

The presence or absence of hydroxyl group on which carbon atom of sugar differentiate RNA and DNA :

The matrix A \(=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 5\end{array}\right]\) is :

Match List-I with List-II for convex lens (symbols have their usual meanings)

List-I (Position of Object)	List-II (Details of image)
(A) at 2F	(l) \(m>1\), virtual
(B) at F	(II) \(m <1\), real
(C) between lens and F	(III) m = 1 , real
(D) between \(\infty\) and 2F	(IV) \(m \gg 1\), real

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Read the passage carefully and answer the questions.
Amines are considered to be derivatives of ammonia obtained by replacement of hydrogen atoms with alkyl or aryl groups. They are usually prepared from nitro compounds, halides, amides, imides etc. The presence of hydrogen bonding influences its physical properties. In alkylamines, a combination of electron releasing steric and hydrogen bonding factors influences the stability of substituted ammonium cations in protic polar solvents and thus affects the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia which is more basic than water. Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding. This intermolecular association is more in primary amines than in secondary amines. Tertiary amines do not have intermoleulcar association due to the absenceof hydrogen atom available for hydrogen bonding. Thus, the boiling point of amines depends on hydrogen bonding. In aromatic amines, electron releasing and withdrawing groups increase and decrease their basic character. Aryl diazonium salts usually obtained from arylamine undergo replacement of the diazonium group with a variety of nucleophile. Their coupling reaction of aryl diazonium salts with phenol or arylamine results in formation of dyes.
Carbylamine reaction is used as a test for

Match List - I with List - II

List - I	List - II
(A) Corner point of a feasible region	(I) The line segment joining any two arbitrary points of the region always lies entirely within the region
(B) Bounded feasible region	(II) can not be enclosed within a circle
(C) Unbounded feasible region	(III) can be enclosed within a circle
(D) Convex region	(IV) is a point of intersection of two boundary lines in the feasible region

Choose the correct answer from the options given below:

Which of the following component was not the part of S.L. Miller's experiment?

\[\int e^x\left(1+x^2\right) d x=\]