Chemistry Test 296

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Chemistry Test 296

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following is the correct IUPAC name for the complex [Co(NH₃)₃(σ − C₃H₅) (en)]SO₄?

(i) η¹-Allyltriamineethylenediaminecobalt(III) sulfate

(ii) η¹-Allyltriammineethylenediaminecobalt(III) sulfate

(iii) Triamminecyclopropylethylenediaminecobalt(III) sulfate

(iv) Triamminecyclopropylethylenediaminecobaltate(III) sulfate

A
ii only

B
i and ii only

C
ii and iii only

D
ii,iii and iv only

Solution
Question 2
4 / -1

Which one of the following octahedral complex will not show geometric isomerism (A and B are monodentate ligands) :-

A
MA₅B

B
MA₂B₄

C
MA₃B₃

D
MA₄B₂

Solution

The correct option is [MA₅ B]

[MA₃B₃] - 2 geometrical isomers

[MA₂B₄] - 2 geometrical isomers

[MA₄B₂] - 2 geometrical isomers

The complexes of general formula [MA₆] and [MA₅ B] octahedral geometry do not show geometrical isomerism.
Question 3
4 / -1

When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p−aminophenol produced is :-

(Given: Molar mass of p−aminophenol = 109 g/mol)

A
10.9g

B
98.1g

C
109.0g

D
9.81g

Solution
Question 4
4 / -1

Two identical rigid bulbs of volume V each are connected by a thin tube of negligible volume and contain an ideal gas at initial pressure p_i and temperature T₁. One bulb is heated to temperature T₂ while the other remains at T₁. The final equilibrium pressure p_f in the system is

A

B

C

D

Solution
Question 5
4 / -1

For a fixed mass of an ideal gas, which of the following plotted relationships will NOT yield a straight-line graph?

A
PV vs P(at constant temperature)

B
log P vs log V (at constant temperature)

C

D
Density vs T (at constant pressure)

Solution
Question 6
4 / -1

n-butane can be formed by reduction of methylethyl ketone by

A
The Meerwein-Ponndorf-Verley reduction

B
The Wolf-Kishner reduction

C
Mg − Hg,H₂O

D
All of the above

Solution

Step 1: Identify the carbon skeleton of methylethyl ketone (MEK).

Methylethyl ketone = butan-2-one, CH₃–CH₂–CO–CH₃ (4-carbon chain).

Step 2: Recognise the outcome of the Wolff–Kishner reduction.

Wolff–Kishner (NH₂NH₂, KOH, high-boiling solvent, ∆) converts C=O → CH₂, preserving the carbon skeleton.

Step 3: Apply the reduction to MEK.

CH₃CH₂COCH₃ → CH₃CH₂CH₂CH₃ = n-butane.

Step 4: Check the other options.

A. Meerwein–Ponndorf–Verley gives sec-butanol (alcohol, not alkane).

C. Mg–Hg, H₂O (pinacol coupling) yields a 1,2-diol, not an alkane.

Only Wolff–Kishner actually furnishes n-butane.
Question 7
4 / -1

A doctor by mistake administered a Ba(NO₃)₂ solution to a patient for radiography investigation.

Which of the following should be given as best to prevent the absorption of soluble barium?

A
NaCl

B
Na₂SO₄

C
NaClO₄

D
NH₄Cl

Solution
Question 8
4 / -1

For a first order reaction A → Products, the concentration of [A] is reduced from 1M to 0.25M in one hour.

The t_1/2of this reaction (in sec) is :-

A
600

B
300

C
1800

D
0.693/1200

Solution
Question 9
4 / -1

In the formation of Grignard reagent, what is the order of reactivities of methyl halides?

A

B

C

D

Solution
Question 10
4 / -1

Which one is the most reactive towards alcoholic KOH among the following compounds:

A

B

C

D

Solution

Step 1: Alcoholic KOH promotes an E2 elimination:

R–Br + KOH (alc) → Alkene + KBr + H₂O.

The rate depends on how easily the β-hydrogen is removed and on the stability of the incipient double bond (transition-state stability).

Step 3: Compare elimination pathways.

C & D are 1° alkyl halides; E2 is slow and gives only a terminal alkene.

B has an electron-withdrawing carbonyl, so its β-hydrogens are more acidic; still, it is only a 1° bromide and the double bond formed is monosubstituted.

Step 4: Special reactivity of vinyl halide A.

Although CH₂=CH–Br has no β-hydrogen on the sp² carbon, the C–Br bond is extremely weak because the carbon is sp²-hybridised (bond energy ≈ 78 kcal mol⁻¹ vs ≈ 96 kcal mol⁻¹ for saturated C–Br).

Alcoholic KOH acts as a strong base/nucleophile and rapidly cleaves this weak bond, forming acetylene (HC≡CH) via:

CH₂=CH–Br + OH⁻ → HC≡CH + Br⁻ + H₂O.

This elimination is much faster than any E2 on the saturated bromides.

Conclusion: The weakest C–Br bond and the conjugate-stabilised alkyne product make CH₂=CH–Br the most reactive toward alcoholic KOH.

Answer: A. CH₂=CH–Br