When aniline is reacted with benzoyl chloride, the correct product is

Which of the following effectively inhibits the growth of tumours?

When phenol is treated with bromine water, the product formed is :

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As aldehydes ketones possess the carbonyl functional group, they show similar reactions. They exhibit nucleophilic addition reactions, reduction, oxidations, reactions due to alpha-Hydrogen atom, and many other reactions, like Cannizzaro reaction to electrophilic substitution reactions (in case of aromatic aldehydes ketones).
These reactions help in the preparation of a number of important compounds, like acetic acid, vinyl acetate, polymers and in perfume and dye industries.
Nucleophilic addition reactions form cyanohydrins, acetals, alcohols and N -substituted derivatives like imine, oxime, semicarbazone, phenylhydrazone Schiff's base. Reactions due to alpha-hydrogen atoms help in formation of aldols and alpha, beta-unsaturated aldehydes ketones.
Which of the following combinations of Grignard's reagent and carbonyl compound can give \(C _6 H _5- C ( OH )\left( CH _3\right)- CH \left( CH _3\right)_2\)
A. \(C _6 H _5 CH _2 MgBr +\left( CH _3\right)_2 CH - CHO\)
B. \(C _6 H _5 MgBr + CH _3- CO - CH \left( CH _3\right)_2\)
C. \(CH _3 MgBr + C _6 H _5- CO - CH \left( CH _3\right)_2\)
D. \(C _6 H _5- CO - CH _3+\left( CH _3\right)_2 CHMgBr\)
E. \(C _6 H _5 CH ( MgBr ) CH _3+\left( CH _3\right)_2 CO\)
Choose the correct answer from the option given below :

The amines which are soluble in water:
(A) Methanamine
(B) Ethanamine
(C) Propanamine
(D) Benzylamine
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Match List-I with List-II.

List-I	List-II
A. Reimer - Tiemann reaction	I. Benzene
B. Kolbe's reaction	II. Picric acid
C. Nitration of phenol with conc. \(HNO _3\)	III. Salicylic acid
D. Reaction of phenol with zinc dust	IV. Salicyaldehyde

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Aldehydes, carboxylic acids and ketones are widespread in the animal and plant kingdoms. They play a pivotal role in various biochemical processes of life. Their presence in nature adds fragrance and flavor. These compounds are widely used in food products, pharmaceuticals, paints, resins and other important product industries. These compounds are prepared by various laboratory methods, which mainly include oxidation, formylation, acylation and reduction. Due to the polar nature of the carbonyl group in aldehydes and ketones, they can exhibit different reactions like nucleophilic addition. They do exhibit redox and various condensation reactions which lead to the formation of various important compounds. On the other hand, the carboxylic acids are mainly prepared by oxidation and hydrolysis of different compounds. The carboxylic acid consists of a carbonyl group and the hydroxy group (attached to the carbonyl carbon atom). This makes it possible for the carboxylic acid to participate in various chemical reactions which involve cleavage of the C-OH bond and the O-H bond along with the reactions involving the complete - COOH group.
The reagent which doesn't react with both acetone and benzaldehyde is :

The function \(f(x)=x^3+1\) is:

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\(KMnO _4\) is prepared by the fusion of \(MnO _2\) with an alkali metal hydroxide and an oxidizing agent like \(KNO _3\) to give a dark-green manganate ion which disproportionate to give permanganate as follows.
\(\begin{array}{l}2 MnO _2+4 KOH + O _2 \rightarrow 2 K_2 MnO _4+2 H _2 O \\ 3 K_2 MnO _4+4 H ^{+} \rightarrow 2 KMnO _4+ MnO _2+2 H _2 O \end{array}\)
On heating \(K _2 MnO _4\) decomposes at 513 K to give \(K _2 MnO _4\). Permanganate ion is tetrahedral and diamagnetic
Acidified \(KMnO _4\) acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
Equivalent weight of \(KMnO _4\) in its oxidising reactions in acidic medium is taken as

Secretions of which one of the following structures does not help in maturation and motility of sperms?

The second order derivative of \(x^3\) log x is:

\(\tan \left[\frac{1}{2} \cos ^{-1}\left(\frac{4}{5}\right)\right]=\)

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The degeneracy of the \(d\) orbitals has been removed due to ligand electron-metal electron repulsions in the octahedral complex 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 \(\Delta_{\circ}\) 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 ( \(\Delta_{\circ}>P\) ) or high spin ( \(\Delta_{\circ}<P\) ) 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.
Which of the following complex is not expected to be coloured?