AHSEC HS 2nd Year Chemistry Model Question Paper Answer

Read the given passage and answer the questions number 1 to 5 that follow:

Question 1:

The substitution reaction of alkyl halide mainly occurs by SN1 or SN2 mechanism. Whatever mechanism alkyl halides follow for the substitution reaction to occur, the polarity of the carbon halogen bond is responsible for these substitution reactions. The rate of SN1 reactions are governed by the stability of carbocation whereas for SN2 reactions steric factor is the deciding factor. If the starting material is a chiral compound, we may end up with an inverted product or racemic mixture depending upon the type of mechanism followed by alkyl halide. Cleavage of ethers with HI is also governed by steric factor and stability of carbocation, which indicates that in organic chemistry, these two major factors help us in deciding the kind of product formed.

1. Predict the stereochemistry of the product formed if an optically active alkyl halide undergoes substitution reaction by SN1 mechanism.
2. Name the instrument used for measuring the angle by which the plane polarised light is rotated.
3. Predict the major product formed when 2-Bromopentane reacts with alcoholic KOH.
4. Give one use of CHI3.
5. Write the structures of the products formed when anisole is treated with HI.

Answers

1. SN1 on an optically active alkyl halide → racemic mixture of enantiomers (attack on the planar carbocation from either face). (Note: often there can be slight enantiomeric excess due to ion-pairing or neighbouring-group effects, but the general result is racemization.)
2. Instrument: Polarimeter (measures the angle of rotation of plane-polarized light).
3. 2-Bromopentane + alcoholic KOH → elimination (Zaitsev) major product 2-pentene (a mixture of E (trans-2-pentene) and Z (cis-2-pentene), with the trans isomer predominating). Minor product: 1-pentene.
   (Reason: alcoholic KOH promotes E2; the more substituted alkene is favored.)
4. One use of CHI₃ (iodoform): used in the iodoform test — a qualitative test to detect methyl ketones (-COCH₃) or ethanol/secondary alcohols with a CH₃CH(OH) group (gives a yellow precipitate of iodoform).
5. Anisole + HI → cleavage to give phenol (C₆H₅OH) and methyl iodide (CH₃I).
   Structures:
   • Phenol: C6H5–OH
   • Methyl iodide: CH3–I

Question 2:

Organic compounds containing amine as functional group are present in a vivid variety of compounds, namely amino acids, hormones, neurotransmitters, DNA, alkaloids, dyes, etc. Drugs including nicotine, morphine, codeine and heroin, etc. which have physiological effects on humans also contain amino group in one form or another. Amines are basic because of the presence of lone pair of electrons on nitrogen. Addition of nitrogen into an organic framework leads to the formation of two families of molecules, namely amines and amides. As chemistry students, we must appreciate the versatility of nitrogen.

1. What are amino acids?
2. Why are amino acids amphoteric?
3. Give one point of difference between acidic and basic amino acid.
4. What are essential amino acids?
5. Name the linkage formed when carboxyl end of one amino acid condenses with
   amino end of other amino acid.

Answer:

1. Amino acids are organic compounds that contain both an amino group (–NH₂) and a carboxyl group (–COOH) on the same molecule. They are the building blocks of proteins and play essential roles in biological processes.

2. Amino acids are amphoteric because they contain both acidic (–COOH) and basic (–NH₂) groups.

• The carboxyl group donates a proton (acts as acid).
• The amino group accepts a proton (acts as base).

Thus, amino acids can behave both as acids and as bases.

3.

Acidic Amino Acid	Basic Amino Acid
Contains more than one –COOH group, so it is acidic in nature.	Contains an extra –NH₂ group, so it is basic in nature.
Example: Aspartic acid, Glutamic acid	Example: Lysine, Arginine

4. Essential amino acids are those which cannot be synthesized by the human body and must be obtained from dietary sources such as meat, milk, eggs, etc.

Examples: Valine, Leucine, Isoleucine, Lysine, Methionine, etc.

5. The linkage formed is called a Peptide bond (–CONH–).

Question 3:

The d-block of the periodic table contains the elements of the groups 3-12 and are known as transition elements. In general, the electronic configuration of these elements is (n-1)d¹–¹⁰ ns^1-2. The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e., 3d, 4d and 5d series. However, Zn, Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states, complex formation, formation of coloured ions and alloys, catalytic activity, etc. Transition metals are hard (except Zn, Cd and Hg) and have a high melting point.

1. Why are Zn, Cd and Hg non-transition elements?
2. Which transition metal of 3d series does not show variable oxidation states?
3. Why do transition metals and their compounds show catalytic activity?
4. Why are melting points of transition metals high ?
5. Why is Cu²⁺ ion coloured while Zn²⁺ ion is colourless in aqueous solution?

Answers:

1. Zn, Cd and Hg are not considered transition elements because their atoms and common ions have completely filled d-subshells (d¹⁰ configuration).

• Example: Zn (3d¹⁰ 4s²), Zn²⁺ (3d¹⁰)
  Since they do not have partially filled d-orbitals, they do not show typical transition metal properties.

2. Scandium (Sc) does not show variable oxidation states.
It shows only +3 oxidation state because after losing three electrons, it attains a stable noble-gas configuration.

3. Transition metals show catalytic activity mainly due to:

• Partially filled d-orbitals, which allow formation of intermediate complexes.
• Ability to adopt variable oxidation states, helping in electron transfer processes.
• Large surface area that can adsorb reactant molecules and lower activation energy.

Thus, they act as excellent heterogeneous and homogeneous catalysts.

4. Transition metals have high melting points because:

The presence of a large number of unpaired d-electrons increases the strength of metallic bonds.

They possess strong metallic bonding, due to involvement of both (n−1)d and ns electrons.

5. Cu²⁺ has a 3d⁹ configuration, which contains unpaired electrons.
The presence of partially filled d-orbitals allows d–d transitions, producing colour.

Zn²⁺ has a 3d¹⁰ configuration, which is completely filled, so no d–d transitions can occur.

Hence, Cu²⁺ is coloured while Zn²⁺ is colourless in aqueous solution.