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Chemistry Test 206

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Chemistry Test 206
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  • Question 1
    4 / -1

    The compound 'A' when treated with methyl alcohol and few drops of H₂SO₄ give wintergreen smell. The compound 'A' is

    Solution

    The answer is b. Salicylic acid.

    Here's why:

    • Wintergreen smell: This is characteristic of methyl salicylate, an ester formed by the reaction of salicylic acid with methanol (methyl alcohol) in the presence of sulfuric acid (H₂SO₄) as a catalyst.
    • Reaction: The specific reaction is known as esterification, where the hydroxyl group (-OH) from salicylic acid reacts with the alcohol group (-OH) from methanol, forming water and methyl salicylate (oil of wintergreen):

    C₇H₆O₃ (salicylic acid) + CH₃OH (methanol) --> C₈H₈O₃ (methyl salicylate) + H₂O

    • Other options:
      • Succinic acid, tartaric acid, and oxalic acid lack the functional groups necessary for esterification and won't produce the wintergreen smell with methanol and sulfuric acid.

    Therefore, based on the characteristic wintergreen smell produced and the chemical reaction involved, compound A is most likely salicylic acid.

     

  • Question 2
    4 / -1

    1-Phenylethanol can be prepared by the reaction of benzaldehyde with

    Solution

    The answer is c. Methyl iodide and magnesium.

    Here's why:

    • Preparation of 1-Phenylethanol: Several methods exist for synthesizing 1-phenylethanol, but the option involving reaction with benzaldehyde is a Grignard reaction.
    • Grignard Reaction: This reaction involves the formation of a Grignard reagent (alkylmagnesium halide) by reacting magnesium metal with an alkyl halide in the presence of an ether solvent. The Grignard reagent then reacts with a carbonyl compound like benzaldehyde to form an alcohol, in this case, 1-phenylethanol.
    • Analysis of options:
      • a. Methyl bromide: While both methyl bromide and methyl iodide are alkyl halides, only methyl iodide can react effectively with magnesium to form a Grignard reagent due to its greater reactivity.
      • b. Ethyl iodide and magnesium: This option mentions both ethyl iodide and magnesium, but the Grignard reagent formed would be ethylmagnesium iodide, not the required methylmagnesium iodide needed for the synthesis of 1-phenylethanol.
      • d. Methyl bromide and aluminium bromide: Aluminium bromide can be used as a Lewis acid catalyst in some Grignard reactions, but the reaction still requires methyl iodide, not methyl bromide, to form the desired Grignard reagent.

    Therefore, the correct option for synthesizing 1-phenylethanol from benzaldehyde is c. Methyl iodide and magnesium.

     

  • Question 3
    4 / -1

    The spectrum produced due to transition of an electron from M to L shell is

    Solution

    The answer is a. Emission.

    When an electron jumps from a higher energy level (M shell) to a lower energy level (L shell), it releases energy in the form of a photon. This emission of energy results in an emission spectrum, which consists of discrete lines of specific wavelengths corresponding to the energy difference between the two levels.

    Here's a diagram to illustrate:

    As the electron falls from M to L, it can emit photons of different energies, depending on the specific sublevels it lands in within the L shell. This gives rise to the discrete lines in the emission spectrum.

    Absorption spectra, on the other hand, occur when a photon with the right energy is absorbed by an atom, causing an electron to jump from a lower to a higher energy level. Continuous spectra, like those emitted by hot objects, involve a range of wavelengths being emitted, not just discrete lines. X-rays are a type of high-energy electromagnetic radiation typically emitted by transitions involving inner shells of atoms, not the M and L shells.

    Therefore, the transition of an electron from M to L shell results in an emission spectrum.

     

  • Question 4
    4 / -1

    The correct order of increasing energy of atomic orbitals is

    Solution

    The correct option is B

    5< 6< 4f < 5d

    According to Aufbau principle electron are filled in various atomic orbital in the increasing order of energy

    1s< 2s< 2p< 3s< 3p< 4s< 3d< 4p< 5s< 4d< 5p< 6s< 4f< 5d<6p< 7s.

     

     

  • Question 5
    4 / -1

    Combination of two AO's lead to the formation of

    Solution

    According to Linear Combination Of Atomic Orbitals(LCAO), the combination of 2 AO', give 2 MO's one of which is bonding MO while the other is antibonding MO.

     

  • Question 6
    4 / -1

    Elimination of bromine from 2- bromobutane results in the formation of

  • Question 7
    4 / -1

    IUPAC name of CH₃-O-C₂H₅ is

  • Question 8
    4 / -1

  • Question 9
    4 / -1

    The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species?

    Solution

    Higher the bond order, shorter will be the bond length. Thus, NO+ is having higher bond order than that of NO so NO+ has shorter bond length.

     

  • Question 10
    4 / -1

    The segment of DNA which acts as the instructional manual for the synthesis of the protein is

    Solution

    A gene is the molecular unit of heredity of a living organism. It is widely accepted by the scientific community as a name given to some stretches ofdeoxyribonucleic acids (DNA) and ribonucleic acids (RNA) that code for apolypeptide or for an RNA chain that has a function in the organism, though there still are controversies about what plays the role of the genetic material

     

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