AIIMS Chemistry Class 11 Questions

In the given compound, chlorine is present on the first position, methyl group is present on the second position and nitro group is present on the fourth position as shown below. Therefore, IUPAC name of the compound is:

Which has least covalent radius.

The complete reaction is shown below.

$\text{A} \xrightarrow{\text{Ph}-\text{SO}_2\text{Cl}} \text{B} \xrightarrow{\text{KOH}} \text{C} \xrightarrow{\text{C}_2\text{H}_5\text{I}} \text{D}$'C' is water solubleCorrect structure of A and D are

The stability of carbocation depends upon +M effect, +I effect and +H effect. Carbocation (i) shows eight hyperconjugation structures. Carbocation (ii) shows six hyperconjugation structures. Carbocation (iii) shows nine hyperconjugation structures. Carbocation (iv) shows +M effect. Therefore, the correct stability order is:

Structure of m-methoxy phenol

The final product formed is shown below: Ph−CH₂−CH≡CH−CH₃ (i) Br₂ → Ph−CH₂−CH−CH−CH₃ | Br (ii) Alc.KOH −2HBr → Ph−CH₂−C≡C−CH₃

The given free radicals are shown below: (I) CH₃–ĊH–CH₃, (II) (III) ĊH₂–CH(CH₃)₂ (IV) ĊH₂–CH₃ The stability of free radicals is mostly affected by the factors like resonance, hyperconjugation and inductive effect. In compound (II), there is high resonance effect due to which it is highly stable than other three compounds. The effect of hyperconjugation is directly proportional to the number alpha hydrogen. So, compound I has six alpha hydrogen atoms, compound III has one alpha hydrogen atom and compound IV has three alpha hydrogen atoms. Therefore, the order is

The final major product formed is shown below: [Diagram showing reaction sequence with carbocation rearrangement]

Assertion: Propene reacts with HI in presence of peroxide give¹⁻iodopropane. Reason: 1° free radical is less stable than 2° free radical

The given compound is shown below: The IUPAC name of the given compound is 5–Methoxy–3–Nitro cyclohexene.

Identify structure of ‘P’ and ‘Q’ Correct order of pKa value is:

(i) $\text{F}_3\text{C}-\text{COOH}$ , (ii) $\text{CH}_3\text{COOH}$ , (iii) $\text{C}_6\text{H}_5\text{COOH}$ ,(iv) $\text{CH}_3\text{CH}_2\text{COOH}$ Correct order of $\text{pK}_\text{a}$ value is:

In case of paper chromatography, there are both stationary and dynamic phases present. So, the statement that it is a stationary phase is incorrect.

Oleum is manufactured by adsorption of sulphur trioxide in concentrated sulphuric acid. The reaction is as follows: H₂SO₄ (l) + SO₃ (g) → H₂S₂O₇ (oleum) The structure of oleum is shown below:

The structure of the given compound

For balmer series, n¹ = 2 and n² = 3, 4, 5, 6,.........

Suitable reagent for following conversion

The given compounds are shown below. In compound (a), NO₂ possesses the -M effect at para position due to which it is highly acidic than other three compounds. The compound (b) has intermolecular hydrogen bonding and it is less acidic than (a). In compound (d), there is absence of an electron donating group, due to which it is more acidic than compound (c), which has electron donating methyl group. Therefore, the correct order of acidic nature is a > b > d > c.

The complete reaction is shown below.

Assertion: Hydroquinone is more acidic than resorcinol. Reason: OH shows I− effect

C + O₂(g) ⟶ CO₂ ...... (i); ΔH = -393 kJ mol⁻¹ H₂ + 1/2 Ω₂ ⟶ H₂O ...... (ii); ΔH = -287.3 kJ mol⁻¹ 2CO₂ + 3H₂O ⟶ C₂H₅OH + 3O₂ ...... (iii); ΔH = 1366.8 kJ mol⁻¹ Find the standard enthalpy of formation of C₂H₅OH(l)

The complete reaction is shown below.

If HBr attacks on left side double bond of the conjugated diene, then the formation of tertiary carbocation occur which is highly stable and if HBr attacks on right side double bond, the formation of secondary carbocation occurs which is less stable.

In isolated system, find the condition for spontaneous reaction:

Which is most stable conformer of ethan-1,2-diol

The structures of BO₃³⁻ and SO₃²⁻ is shown below.

The entropy for a system can be calculated as: ΔS = nCᵥ ln(T₂/T₁) + nR ln(V₂/V₁)……(I)

Which of the following number of lone pair at central atom zero XeO₃, XeO₂F₂, XeO₄, XeO₃F₂, Ba₂XeF₄

The formula is A₀.₈O. This means 80 Å atoms are related to 100 Ω atoms. Let’s assume that 'x' atoms of A are available as A²⁺ ion and (80 − x) are available as A³⁺ ion. The value of x can be calculated as: x ײ⁺ (80 − x) − 3 = 100 × 2 2x + 240 − 3x = 200 x = 40 Therefore, the fraction of A atoms can be calculated as:

In adiabatic expansion, work is done at the cost of internal energy. Hence, the temperature of system is decrease.

The F–C bond has σ and π characteristics because of some back-bonding between d-orbital of metal and p-orbital of carbon.

NF₃ = 3bp + 1 lp = 4 e⁻ pair And H₂O = 2bp + 2 lp = 4 e⁻ pair. Thus, both NF₃ and H₂O are sp³ hybridized.

Assertion: For liquid dishwashing non-ionic type of detergent are used: Reason: Remove grease and oil by micelle formation.

Number starts from terminal carbon which has nearest branching.

When 0.05 M dimethyl amine is dissolved in 0.1 M NaOH solution then the percentage dissociation of dimethyl amine is: (K_b (CH₃)₂NH = 5 x 10⁻⁴)

As n=2 and Z=2 for He⁺ ion, so the wave length is calculated below.

The chemical reaction is shown below:

The state function is dependent only on the initial and final state of the system. Intensive properties are those which are independent of the amount of the substance. For example temperature, melting point, boiling point etc. The internal energy (U) is a state function.

Assertion: $\text{S}_2\text{O}_7^{2-}$ & $\text{Cr}_2\text{O}_7^{2-}$ both exist. Reason: Both have same valence electrons.

According to the 1ˢᵗ law of thermodynamics ΔU = q + w If, w = 0 ΔU = q q depends on the reaction path travelled and thus not a state function.

The chemical reaction is shown below: H₂C=CH₂ + HOCl → H₂C(OH)−CH₂(Cl) (X) → NaHCO₃ → CH₂OH−CH₂OH (Y)

Which of the following can react with K₂Cr₂O₇

The colors of the flame of the given metals are

The equation for standard Gibbs free energy given below. ΔG = ΔG° + RTlnQ…(1) If ΔG = 0 and Q = K then the above equation at equilibrium becomes, ΔG° = −RTlnK …(2) Substituting equation (2) in equation (1), the new relation forms as, ΔG = −RTlnK + RTlnK ΔG = −RTln(K/Q)

Assertion: $\text{ZrI}_4$ is useful in purification of Zirconium (Zr) Reason: $\text{ZrI}_4$ sublimes at room temperature.

According to L.P−L.P > L.P−B.P > B.P−B.P repulsion, the correct order of S−S bond length will be: I > II > III

Which pair of elements has maximum electronegativity difference?

MnO is:

The compound, Sr(OH)₂ has max. solubility in acidic solution that is at low pH as shown below. Sr(OH)₂ → Sr²⁺ + 2OH⁻ It gives high amount of hydroxide ions which are neutralized by the high amount of H⁺ ions.