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Kpk Chemistry 2019 — Solved Past Paper with Answers
All 18 MCQs from Kpk Chemistry 2019, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT KPK / ETEA past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all KPK / ETEA papers.
Attempt this paper interactively →Q1. Which of the given techniques is used for determination of functional group in a compound?
- A. IR✓
- B. UV
- C. NMR
- D. Mass spectroscopy
Explanation: Infrared spectroscopy is used to identify functional groups within a compound. It measures the vibrations of molecules when exposed to infrared light. Different functional groups absorb infrared radiation at specific frequencies, producing a spectrum that can be used to identify the presence of various functional groups in the compound.
Why the other options are wrong
- B. Ultraviolet spectroscopy is primarily used to study the electronic transitions in molecules, particularly in conjugated systems and aromatic compounds. It is more focused on the transitions between electronic energy levels rather than identifying specific functional groups.
- C. Nuclear magnetic resonance spectroscopy provides detailed information about the molecular structure, dynamics, and environment of the atoms within a molecule. While NMR is extremely powerful for elucidating the structure of a compound, including identifying functional groups, it is not the primary technique used specifically for identifying functional groups alone.
- D. Mass spectroscopy is used to determine the molecular weight and structure of a compound by ionizing chemical species and measuring the mass-to-charge ratio of the ions. It provides information about the molecular formula and structural features of the compound but is not specifically used to identify functional groups directly.
Q2. Pollutants have adverse effect over......?
- A. Biosphere
- B. Ecosystem
- C. Hydrosphere
- D. All these✓
Explanation: Since pollutants impact the biosphere, ecosystems, and the hydrosphere, the correct answer encompasses all these areas. Therefore, D is correct.
Why the other options are wrong
- A. The biosphere includes all living organisms on Earth. Pollutants can harm plants, animals, and humans by contaminating the air, water, and soil, leading to health problems and disruptions in ecosystems.
- B. An ecosystem is a community of living organisms interacting with their physical environment. Pollutants can disrupt these interactions, harming species, reducing biodiversity, and altering ecosystem functions and services.
- C. The hydrosphere encompasses all water bodies on Earth, including oceans, rivers, lakes, and groundwater. Pollutants can contaminate water, affecting aquatic life, water quality, and availability for human use and consumption.
Q3. Which is trun for DOT? It is...?
- A. Not a pollutant
- B. An antibiotc
- C. An antiseptic
- D. A non-degradable pollutant✓
Explanation: Correct. DDT is a persistent organic pollutant, meaning it does not easily break down in the environment. It accumulates in the fatty tissues of living organisms and biomagnifies up the food chain, leading to adverse effects on wildlife and human health.
Why the other options are wrong
- A. Incorrect. DDT is indeed a pollutant.
- B. Incorrect. DDT is not an antibiotic, which are drugs used to treat bacterial infections.
- C. Incorrect. DDT is not an antiseptic, which are substances used to prevent infection by killing or inhibiting the growth of microorganisms on living tissues.
Q4. Ethanol is produced from strach by the process of.....?
- A. Hydrolysic
- B. Hydrogennation
- C. Decomposition
- D. Fermentation✓
Explanation: Fermentation is the correct process for producing ethanol from starch. During fermentation, enzymes break down starch into simpler sugars, such as glucose, which are then converted into ethanol and carbon dioxide by the action of yeast or bacteria. This biochemical process is widely used in the production of alcoholic beverages and biofuels.
Why the other options are wrong
- A. Hydrolysis is a chemical process in which a compound is broken down by the addition of water. While hydrolysis can break down starch into simpler sugars like glucose, it is not the process by which ethanol is produced from starch. Instead, hydrolysis can be a preliminary step in converting starch into fermentable sugars, which are then fermented to produce ethanol.
- B. Hydrogenation is a chemical reaction between hydrogen and another compound, usually in the presence of a catalyst. This process is commonly used in the industrial production of saturated fats from unsaturated fats. Hydrogenation is not involved in the production of ethanol from starch.
- C. Decomposition refers to the breakdown of organic substances into simpler compounds, usually through the action of microorganisms or chemical processes. While decomposition is a general term for the breakdown of organic matter, it does not specifically describe the biochemical process of converting starch into ethanol.
Q5. Which of the given orders of relative strengths of acids is correct?
- A. CL CH2 COOH>FCH2 COOH>BrCH2 COOH
- B. CL CH2 COOH>BrCH2 COOH>FCH2 COOH
- C. BrCH2 COOH>FCH2 COOH>CICH2 COOH
- D. FCH2 COOH>CI CH2 COOH>BrCH2 COOH✓
Explanation: The correct order based on the electron-withdrawing inductive effects of the halogen substituents is FCH2 COOH > ClCH2 COOH > BrCH2 COOH. Therefore, d is correct.
Why the other options are wrong
- A. This is incorrect because fluorine, being more electronegative than chlorine, would make fluoroacetic acid stronger than chloroacetic acid.
- B. This is incorrect because it suggests that chloroacetic acid is stronger than both bromoacetic and fluoroacetic acids, and that bromoacetic acid is stronger than fluoroacetic acid. This contradicts the correct order of electronegativity and inductive effects.
- C. This is incorrect because it places bromoacetic acid as the strongest acid, which is not accurate given bromine's weaker -I effect compared to both fluorine and chlorine.
Q6. Clemmensen reduction is carried out with.....?
- A. LiAIH4✓
- B. Zn-Hg and HCI
- C. LiAIH4 and HCI
- D. All these
Explanation: Lithium aluminum hydride is a strong reducing agent commonly used in organic chemistry to reduce various functional groups such as aldehydes, ketones, carboxylic acids, esters, and amides to their respective alcohols or amines. However, it is not used in the Clemmensen reduction, which specifically involves the reduction of ketones and aldehydes to alkanes using zinc amalgam (Zn-Hg) and hydrochloric acid (HCl) under acidic conditions.
Why the other options are wrong
- B. This combination of reagents is used in the Clemmensen reduction. Zinc amalgam (Zn-Hg) serves as the reducing agent, and hydrochloric acid (HCl) provides the acidic conditions necessary for the reaction. Together, they convert ketones and aldehydes into alkanes by removing the carbonyl group and replacing it with hydrogen atoms.
- C. While lithium aluminum hydride (LiAlH4) is a strong reducing agent, it is not used in the Clemmensen reduction. Combining LiAlH4 with HCl would not achieve the Clemmensen reaction but could lead to other reduction reactions of different functional groups.
- D. This option is incorrect because the Clemmensen reduction specifically requires the use of zinc amalgam (Zn-Hg) and hydrochloric acid (HCl). While LiAlH4 is a versatile reducing agent, it is not used in the Clemmensen reduction reaction.
Q7. Which of the given has highest solubility in water?
- A. (CH3)2CHOH
- B. (CH3)3COH
- C. C2H5OH✓
- D. MeOH
Explanation: Ethanol is highly soluble in water due to its ability to form hydrogen bonds with water molecules. This allows ethanol to mix uniformly with water at all proportions.
Why the other options are wrong
- A. Isopropanol is moderately soluble in water. It can form hydrogen bonds with water molecules but not as effectively as ethanol due to its larger hydrophobic alkyl group.
- B. Tert-butanol is relatively insoluble in water compared to ethanol. Its large, bulky tert-butyl group hinders its ability to form hydrogen bonds with water molecules effectively.
- D. Methanol is also highly soluble in water, similar to ethanol, due to its ability to form hydrogen bonds with water molecules. However, ethanol is slightly more soluble in water than methanol due to the longer hydrophobic alkyl chain in ethanol, which enhances its solubility in water.
Q8. The strongest reducing agent in these is.....?
- A. HI✓
- B. HF
- C. HBr
- D. All these
Explanation: HI is a strong reducing agent because iodine (I) is in its highest oxidation state (-1) in this compound. It can easily donate an electron to another species, making it a strong reducing agent.
Why the other options are wrong
- B. HF is not a strong reducing agent. Fluorine (F) is already in its most electronegative form (-1 oxidation state) in HF, making it less likely to donate electrons and act as a reducing agent.
- C. HBr is a stronger acid than HF but weaker than HI. Bromine (Br) is in the -1 oxidation state in HBr, making it a weaker reducing agent compared to HI.
- D. This option is incorrect because not all of these acids are equally strong reducing agents. HI is the strongest reducing agent among the given options due to the ease with which iodine can donate electrons.
Q9. Aqueous KOH cause SN-reaction in alkylhaide.On which of the given alkylhaide KOH would like to attack easily?
- A. CH3-CH2-CI
- B. CH3-CH2-Br
- C. CH3-CH2-F✓
- D. CH3-CH2-I
Explanation: Fluorine is the smallest halogen and the least electronegative among the options, making the carbon atom less hindered and more easily attacked by the nucleophile in an SN2 reaction.
Why the other options are wrong
- A. This alkyl chloride is less reactive towards SN2 reactions compared to the other options because chlorine is a larger atom and more electronegative than fluorine and iodine, making it more difficult for the nucleophile to attack the carbon atom.
- B. Bromine is a better leaving group than chlorine but not as good as iodine. However, in this case, the size of the bromine atom makes it more favorable for SN2 reactions compared to chlorine.
- D. Iodine is the most reactive towards SN2 reactions among the halogens because it is the largest and the least electronegative, making it easier for the nucleophile to attack the carbon atom. However, in this case, fluorine is a better choice for an SN2 reaction due to its smaller size and lower electronegativity.
Q10. Which of the given is more stable?
- A. Cyclopropane
- B. Cyclobutane
- C. Cyclopentane
- D. Cyclohexan✓
Explanation: Cyclohexane is the most stable among the given options. It adopts a chair conformation, which allows all carbon-carbon bonds to be staggered, minimizing steric hindrance and ring strain. Additionally, the bond angles in cyclohexane are close to the ideal tetrahedral angle, further contributing to its stability.
Why the other options are wrong
- A. Cyclopropane is the least stable among the given options due to its high ring strain. The bond angles in cyclopropane are approximately 60 degrees, significantly deviating from the ideal tetrahedral angle of 109.5 degrees, leading to high ring strain.
- B. Cyclobutane is more stable than cyclopropane but less stable than cyclopentane and cyclohexane. It experiences some ring strain due to its bond angles being approximately 90 degrees, which is still far from the ideal tetrahedral angle.
- C. Cyclopentane is more stable than cyclobutane and cyclopropane but less stable than cyclohexane. Its bond angles are closer to the ideal tetrahedral angle, resulting in lower ring strain compared to cyclobutane and cyclopropane.
Q11. The order in the ease of dehydration of alcohol is......?
- A. Tertiary>Secondary>Primary
- B. Primary>Secondary>Tertiary
- C. Secondary>Primary>Tertiary✓
- D. Primary>Tertiary>Secondary
Explanation: Secondary alcohols are more easily dehydrated than primary alcohols because the secondary carbocation formed is stabilized by two alkyl groups. Primary alcohols are more easily dehydrated than tertiary alcohols for the reasons mentioned earlier.
Why the other options are wrong
- A. This order is incorrect. Tertiary alcohols are the least easily dehydrated because the tertiary carbocation formed during dehydration is highly unstable due to the lack of alkyl groups to stabilize it. Primary alcohols are more easily dehydrated than tertiary alcohols because the primary carbocation formed is stabilized by two alkyl groups.
- B. This order is also incorrect. Tertiary alcohols are generally the least easily dehydrated, as explained above. Primary alcohols are more easily dehydrated than secondary alcohols because the primary carbocation formed is stabilized by two alkyl groups, while the secondary carbocation formed from a secondary alcohol is only stabilized by one alkyl group.
- D. This order is incorrect. Tertiary alcohols are generally the least easily dehydrated due to the instability of the tertiary carbocation formed. Primary alcohols are more easily dehydrated than tertiary alcohols because the primary carbocation formed is stabilized by two alkyl groups. Secondary alcohols are more easily dehydrated than tertiary alcohols, but less easily dehydrated than primary alcohols, due to the stabilization of the secondary carbocation by two alkyl groups.
Q12. Which of the given is an organic compound?
- A. KCN
- B. NH4OCN
- C. CS2
- D. None of these✓
Explanation: Option D is correct because none of the given compounds (KCN, NH4OCN, CS2) contain carbon-hydrogen (C-H) bonds, which are characteristic of organic compounds. Organic compounds are compounds that contain carbon atoms bonded to hydrogen atoms, often along with other elements such as oxygen, nitrogen, sulfur, and halogens. Since none of the compounds listed contain C-H bonds, they are considered inorganic compounds.
Why the other options are wrong
- A. KCN is an inorganic compound. It is a salt of the cyanide ion and potassium ion and is commonly used in organic synthesis as a source of the CN⁻ nucleophile in reactions such as the Strecker synthesis.
- B. NH4OCN is also an inorganic compound. It is the salt of ammonium ion and cyanate ion and is known for its role in the synthesis of urea.
- C. CS2 is an inorganic compound composed of carbon and sulfur. It is a colorless liquid with a strong odor and is used in the manufacture of rayon and cellophane.
Q13. Furan is......organic compound?
- A. Homocyclic
- B. Aromatic
- C. Heterocyclic✓
- D. Both B and C
Explanation: Heterocyclic compounds are compounds that contain a ring structure with at least one atom (other than carbon) in the ring. Furan contains an oxygen atom in its ring, so it is classified as a heterocyclic compound.
Why the other options are wrong
- A. Homocyclic compounds are compounds in which all the ring atoms are of the same element. Furan contains one oxygen atom and four carbon atoms in its ring, so it is not homocyclic.
- B. Aromatic compounds are compounds that contain a ring system with alternating single and double bonds and exhibit special stability. Furan is aromatic because it has a ring system with 6 π-electrons, fulfilling the criteria for aromaticity according to the Hückel's rule.
- D. This option is incorrect because while furan is aromatic and heterocyclic, it is not homocyclic.
Q14. Which of the given is diamagnetic ion?
- A. Cu++
- B. Mn++
- C. Sc+-+✓
- D. Co-++
Explanation: Scandium(III) ion has no unpaired electrons in its electronic configuration, as it has the electron configuration [Ar]3d0 4s0. This configuration indicates that all electrons are paired, making Sc3+ diamagnetic.
Why the other options are wrong
- A. Copper(II) ion has an incomplete d orbital (d9), which makes it paramagnetic due to the presence of unpaired electrons.
- B. Manganese(II) ion has five unpaired electrons in its d orbitals (d5), making it paramagnetic.
- D. Cobalt(II) ion has three unpaired electrons in its d orbitals (d7), making it paramagnetic.
Q15. Which of the given ions exhibits colour in aqueous solution?
- A. Ni2+✓
- B. Ti4+
- C. Zn2+
- D. Sc3+
Explanation: Nickel(II) ion often exhibits a green color in aqueous solution due to the presence of unpaired electrons in its d orbitals, which can undergo electronic transitions that absorb certain wavelengths of light, resulting in color.
Why the other options are wrong
- B. Titanium(IV) ion is typically colorless in aqueous solution because it has a d0 electron configuration, meaning there are no unpaired electrons to undergo electronic transitions that would result in color.
- C. Zinc(II) ion is colorless in aqueous solution because it has a full d10 electron configuration, meaning there are no unpaired electrons to absorb light and give rise to color.
- D. Scandium(III) ion is also typically colorless in aqueous solution because it has a d0 electron configuration, similar to titanium(IV) ion, and therefore does not exhibit any unpaired electrons that can absorb light and produce color.
Q16. Alkali metals have high oxidation potential and hence they behave as.....?
- A. Electrolyles
- B. Lewis bases
- C. Oxidizing agents
- D. Reducing agents✓
Explanation: Alkali metals have a strong tendency to donate their outermost electron (valence electron), which makes them excellent reducing agents. This behavior is a result of their low ionization energy and high reactivity, which allow them to easily lose electrons and reduce other substances by causing them to gain electrons.
Why the other options are wrong
- A. While alkali metals do form electrolytes when dissolved in water or other solvents, this property is not directly related to their high oxidation potential. Electrolytes are substances that conduct electricity when dissolved or molten due to the presence of ions.
- B. Alkali metals can behave as Lewis bases by donating their valence electron to form a bond with a Lewis acid (electron pair acceptor). However, their high oxidation potential is more directly related to their behavior as reducing agents.
- C. Alkali metals themselves are not oxidizing agents. They are more likely to be oxidized (lose electrons) rather than to cause other substances to be oxidized.
Q17. Which of the given has the lowest melting points?
- A. Cs
- B. K
- C. Na
- D. Li✓
Explanation: Lithium has the lowest melting point among alkali metals and is the lightest and least dense solid element.
Why the other options are wrong
- A. Cesium has the lowest melting point of all alkali metals and is a very soft, low-melting metal.
- B. Potassium has a higher melting point than cesium but lower than sodium and lithium.
- C. Sodium has a higher melting point than potassium and cesium but lower than lithium.
Q18. Lithium is the strongest reducing agent among alkali metals due to which of the given factors?
- A. Ionization energy✓
- B. Electron affinty
- C. Lattice energy
- D. Hydration energy
Explanation: Lithium has the lowest ionization energy among alkali metals, meaning it requires the least amount of energy to remove its outermost electron. This makes lithium the strongest reducing agent among the alkali metals because it can easily donate its outermost electron to other substances, reducing them.
Why the other options are wrong
- B. Electron affinity is the energy released when an atom gains an electron. It is not directly related to the reducing ability of an element.
- C. Lattice energy is the energy required to separate one mole of a solid ionic compound into its gaseous ions. It is not directly related to the reducing ability of an element.
- D. Hydration energy is the energy released when ions in the gas phase combine with water molecules to form hydrated ions in solution. While hydration energy can influence the solubility and reactivity of ions in solution, it is not directly related to the reducing ability of an element.
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