NIPER Mock Test! Hardest Trivia Questions
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Questions and Answers
- 1.
Hat is the lmax for the following compound? Use the provided parameters for your calculation.
- A.
234 nm
- B.
244 nm
- C.
273 nm
- D.
283 nm
Correct Answer
D. 283 nmExplanation
The lmax for a compound refers to the wavelength at which it absorbs light most strongly. In this case, the correct answer is 283 nm, indicating that the compound absorbs light most strongly at this wavelength. The provided parameters likely include information about the compound's structure and electronic properties, which determine its absorption characteristics.Rate this question:
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- 2.
What is the name of the following anticancer drug?
- A.
Spongistatin 1
- B.
Cryptophycin 1
- C.
Phyllanthoside
- D.
Maytansine 1
Correct Answer
B. Cryptophycin 1Explanation
Cryptophycin 1 is the name of the anticancer drug mentioned in the question.Rate this question:
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- 3.
The following structure is an inhibitor of the enzyme responsible for adding a farnesyl chain to the Ras protein.Which feature binds to the zinc cofactor of the enzyme?
- A.
The tricyclic system
- B.
The piperazine ring
- C.
The aromatic ring
- D.
The imidazole ring
Correct Answer
D. The imidazole ringExplanation
The imidazole ring binds to the zinc cofactor of the enzyme.Rate this question:
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- 4.
The following structure is a natural product having anticancer properties, which was isolated from a marine worm.What is it called?
- A.
Pancratistatin
- B.
Cephalostatin 1
- C.
Bryostatin 1
- D.
Dolostatin 15
Correct Answer
B. Cephalostatin 1Explanation
Cephalostatin 1 is a natural product isolated from a marine worm that possesses anticancer properties. It is a structure that has been found to have potential in fighting cancer cells.Rate this question:
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- 5.
Which of the following inhibits angiogenesis?
- A.
VEGF
- B.
FGF-2
- C.
Angiostatin
- D.
Interleukin-6
Correct Answer
C. AngiostatinExplanation
Angiostatin is the correct answer because it is known to inhibit angiogenesis. Angiogenesis is the process of forming new blood vessels, which is important for tumor growth and metastasis. Angiostatin works by blocking the growth of endothelial cells, which are the cells that line the inside of blood vessels. This prevents the formation of new blood vessels and limits the blood supply to tumors, thereby inhibiting their growth and spread. VEGF, FGF-2, and Interleukin-6, on the other hand, are all pro-angiogenic factors that promote angiogenesis.Rate this question:
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- 6.
The tetracyclines act as intercalating anticancer agents.Which of the following is a simplified analogue of tetracyclines?
- A.
Mitoxantrone
- B.
Teniposide
- C.
Daunorubicin
- D.
Dactinomycin
Correct Answer
A. MitoxantroneExplanation
Mitoxantrone is a simplified analogue of tetracyclines because it also acts as an intercalating anticancer agent. Intercalating agents are molecules that can insert themselves between the DNA base pairs, disrupting DNA replication and causing cell death. Tetracyclines, a class of antibiotics, have been found to have anticancer properties due to their intercalating ability. Mitoxantrone, a chemotherapeutic agent, has a similar mechanism of action and is used in the treatment of various cancers. Therefore, Mitoxantrone is a simplified analogue of tetracyclines.Rate this question:
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- 7.
Pentostatin is a natural product used for the treatment of leukaemia, What is the role of the region shown in blue?
- A.
Transition-state isostere
- B.
Transition-state analogue
- C.
Suicide substrate
- D.
Transition-state bioisostere
Correct Answer
A. Transition-state isostereExplanation
The role of the region shown in blue is to mimic the transition state of a biochemical reaction. A transition-state isostere is a molecule or a part of a molecule that closely resembles the transition state of a specific reaction. In the case of pentostatin, the blue region likely mimics the transition state of a reaction involved in the treatment of leukemia. By acting as a transition-state isostere, pentostatin can interfere with the biochemical process associated with leukemia, ultimately helping to treat the disease.Rate this question:
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- 8.
What is the name of the following anticancer drug?
- A.
Spongistatin 1
- B.
Cryptophycin 1
- C.
Phyllanthoside
- D.
Maytansine 1
Correct Answer
C. Phyllanthoside -
- 9.
Imatinib is an important anticancer drug which targets a protein kinase.The piperazine ring has an important interaction with an amino acid in the binding site. Which amino acid is involved?
- A.
Aspartic acid
- B.
Glutamic acid
- C.
Methionine
- D.
Leucine
Correct Answer
B. Glutamic acidExplanation
The piperazine ring of Imatinib has an important interaction with a specific amino acid in the binding site of the protein kinase. This interaction suggests that the amino acid involved must have a complementary structure or charge to interact effectively with the piperazine ring. Among the given options, Glutamic acid is more likely to have this interaction due to its negatively charged side chain, which can form electrostatic interactions with the positively charged piperazine ring. Therefore, Glutamic acid is the most plausible amino acid involved in this interaction.Rate this question:
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- 10.
The following structure is called marimastat and is a matrix metalloproteinase inhibitor that has undergone clinical trials for the treatment of breast and prostate cancers.What role is played by the hydroxamic acid group?
- A.
It acts as a transition-state isostere.
- B.
It binds to the zinc ion cofactor.
- C.
It binds to a binding sub pocket in the active site.
- D.
It acts as a steric shield.
Correct Answer
B. It binds to the zinc ion cofactor.Explanation
The hydroxamic acid group in marimastat binds to the zinc ion cofactor. This is important because matrix metalloproteinases (MMPs), the target enzymes of marimastat, require zinc ions for their catalytic activity. By binding to the zinc ion cofactor, the hydroxamic acid group effectively inhibits the activity of MMPs, preventing them from degrading the extracellular matrix and promoting cancer cell invasion and metastasis.Rate this question:
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- 11.
The following structure is a natural product currently being studied as an anticancer agent.What is the name of the structure?
- A.
Thalidomide
- B.
Fumagillin
- C.
Depsipeptide
- D.
Aclarubicin
Correct Answer
C. DepsipeptideExplanation
Depsipeptide is a natural product that is currently being studied as an anticancer agent. It is a compound that contains both peptide and ester bonds, which gives it unique properties and potential therapeutic effects. Researchers are investigating its ability to inhibit cancer cell growth and promote cell death, making it a promising candidate for cancer treatment.Rate this question:
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- 12.
To what extent are the three nitrogens of histamine ionised at blood pH?
- A.
All three nitrogens are fully ionised
- B.
All three nitrogens are not ionised at all
- C.
The side chain nitrogen is fully ionised and the heterocyclic nitrogens are not ionised
- D.
The side chain nitrogen and one of the heterocyclic nitrogens are fully ionised
Correct Answer
C. The side chain nitrogen is fully ionised and the heterocyclic nitrogens are not ionisedExplanation
At blood pH, the side chain nitrogen of histamine is fully ionized, meaning it has gained a positive charge. On the other hand, the two heterocyclic nitrogens of histamine are not ionized at all, meaning they do not gain or lose any charge. This is because the pKa values of the side chain nitrogen and the heterocyclic nitrogens are different, with the side chain nitrogen having a lower pKa value and thus becoming fully ionized at blood pH.Rate this question:
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- 13.
Three binding regions were proposed to be present in the binding site of the H2 receptor. Which of the following statements is incorrect?
- A.
There is a binding region for the imidazole ring of histamine analogues which is common for agonists and antagonists.
- B.
There is a binding region which interacts ionically with the α-nitrogen of histamine and results in agonist activity.
- C.
There is a binding region further away from the imidazole ring that produces an antagonist effect if occupied.
- D.
The α-nitrogen of histamine can only bind to the agonist binding region while the guanyl group of Nα-guanylhistamine can only bind to the antagonist binding region.
Correct Answer
D. The α-nitrogen of histamine can only bind to the agonist binding region while the guanyl group of Nα-guanylhistamine can only bind to the antagonist binding region.Explanation
The incorrect statement is that the α-nitrogen of histamine can only bind to the agonist binding region while the guanyl group of Nα-guanylhistamine can only bind to the antagonist binding region. This is incorrect because the α-nitrogen of histamine can also bind to the antagonist binding region, while the guanyl group of Nα-guanylhistamine can bind to both the agonist and antagonist binding regions.Rate this question:
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- 14.
The following structures show some of the important molecules leading to the discovery of burimamide (B).What strategy was used in developing burimamide from SK&F 91581?
- A.
Extension
- B.
Chain extension
- C.
Substituent variation
- D.
Isosteric replacement
Correct Answer
B. Chain extensionExplanation
The strategy used in developing burimamide from SK&F 91581 is chain extension. This means that additional chemical groups or chains were added to SK&F 91581 to create burimamide. This process of extending the chemical structure can lead to the discovery of new compounds with different properties and potential applications.Rate this question:
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- 15.
The following structures show some of the important molecules leading to the discovery of burimamide (B).Which of the following statements concerning burimamide is untrue?
- A.
It established the existence of H2-receptors
- B.
It was a good antagonist at H2 receptors with only weak partial agonist activity
- C.
It inhibited gastric acid release from parietal cells
- D.
It indicated that binding to the antagonist binding region involved hydrogen bonding and not ionic bonding
Correct Answer
B. It was a good antagonist at H2 receptors with only weak partial agonist activityExplanation
Burimamide was not a good antagonist at H2 receptors with weak partial agonist activity.Rate this question:
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- 16.
The following diagram shows development of H2-antagonists from burimamide (structure B).A sulphur atom was inserted into the side chain of structure C. What effect did this change have?
- A.
It introduced an extra binding interaction
- B.
It stabilised the molecule
- C.
It increased the percentage population of the active heterocyclic tautomer
- D.
It prevented ionisation of the terminal functional group
Correct Answer
C. It increased the percentage population of the active heterocyclic tautomerExplanation
The insertion of a sulfur atom into the side chain of structure C increased the percentage population of the active heterocyclic tautomer. This means that the modified molecule is more likely to exist in a form that is biologically active and can interact with its target effectively.Rate this question:
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- 17.
The following diagram shows development of H2-antagonists from burimamide (structure B).What was the rationale for the introduction of the coloured methyl group?
- A.
To block metabolism at that region of the heterocyclic ring
- B.
To introduce a group which would be metabolised in a predictable fashion
- C.
To introduce an electron withdrawing group on the heterocyclic ring to reduce the chance of ionisation
- D.
To introduce an electron donating group on the heterocyclic ring to favour the active tautomer
Correct Answer
D. To introduce an electron donating group on the heterocyclic ring to favour the active tautomerExplanation
The rationale for the introduction of the coloured methyl group is to introduce an electron donating group on the heterocyclic ring to favor the active tautomer. This means that the methyl group helps stabilize the active form of the molecule, making it more likely to be in its active state and therefore more effective as a H2-antagonist.Rate this question:
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- 18.
The following diagram shows development of H2-antagonists from burimamide (structure B). Why was the thiourea functional group in structure D changed to a guanidine group in structure E?What was the rationale for the introduction of the coloured methyl group?
- A.
To introduce a basic group which could ionise and allow ionic interactions with the binding region.
- B.
To replace an unnatural functional group with a naturally occurring group in order to reduce side effects.
- C.
To increase the number of hydrogen bond donors present to acquire extra binding interactions.
- D.
To change the geometry and stereochemistry of the functional group such that it fitted the binding region more closely.
Correct Answer
B. To replace an unnatural functional group with a naturally occurring group in order to reduce side effects.Explanation
The thiourea functional group in structure D was changed to a guanidine group in structure E in order to replace an unnatural functional group with a naturally occurring group. This change was made to reduce side effects that may be caused by the unnatural functional group. By using a naturally occurring group, the likelihood of adverse reactions or unwanted side effects is minimized.Rate this question:
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- 19.
The following diagram shows development of H2-antagonists from burimamide (structure B). Why was the cyanide group introduced into structure F (cimetidine)?
- A.
It is an electron donating group and increases the basicity of the functional group such that it protonates and becomes ionised.
- B.
It is an electron withdrawing group and increases the basicity of the functional group such that it protonates and becomes ionised.
- C.
It is an electron donating group and decreases the basicity of the functional group such that it does not become protonated and remains un-ionised.
- D.
It is an electron withdrawing group and decreases the basicity of the functional group such that it does not become protonated and remains un-ionised.
Correct Answer
D. It is an electron withdrawing group and decreases the basicity of the functional group such that it does not become protonated and remains un-ionised.Explanation
The cyanide group, being an electron withdrawing group, decreases the basicity of the functional group in structure F (cimetidine). This means that the functional group does not become protonated and remains un-ionized. This is important because H2-antagonists function by blocking the action of histamine on the H2 receptors, which are located on cells in the stomach lining. By remaining un-ionized, the cimetidine molecule can effectively bind to the H2 receptors and inhibit the secretion of gastric acid.Rate this question:
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- 20.
Two regions of cimetidine are susceptible to metabolism. Which regions?Why was the cyanide group introduced into structure F (cimetidine)?
- A.
A and B
- B.
A and C
- C.
B and D
- D.
A and D
Correct Answer
B. A and CExplanation
The two regions of cimetidine that are susceptible to metabolism are regions A and C. The cyanide group was introduced into structure F (cimetidine) because it acts as a potent inhibitor of histamine H2 receptors, which helps to reduce stomach acid production.Rate this question:
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- 21.
The following diagram shows various conformations for the cyanoguanidine group of cimetidine.Two of these conformations were found to be disfavoured. Which ones and what was the implication of this for receptor binding?
- A.
EZ and ZE. It proved the chelation theory of hydrogen bonding.
- B.
EZ and ZZ. It established that there was only one hydrogen bonding interaction with the receptor in this region.
- C.
EE and ZZ. It established that there were two hydrogen bonding interactions to different groups within the same binding region.
- D.
EE and ZE. No conclusions could be drawn.
Correct Answer
C. EE and ZZ. It established that there were two hydrogen bonding interactions to different groups within the same binding region.Explanation
The correct answer is EE and ZZ. This implies that there are two hydrogen bonding interactions to different groups within the same binding region. This suggests that the cyanoguanidine group of cimetidine can form multiple hydrogen bonds with the receptor, potentially increasing the binding affinity and stability of the drug-receptor complex.Rate this question:
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- 22.
Once activated, the proton pump inhibitors bind to exposed amino acids in the proton pump. Which amino acid is involved?
- A.
Serine
- B.
Cysteine
- C.
Lysine
- D.
Histidine
Correct Answer
B. CysteineExplanation
Proton pump inhibitors (PPIs) are medications that work by binding to the proton pump in the stomach, which reduces the production of stomach acid. Once activated, PPIs specifically bind to exposed amino acids in the proton pump. The amino acid involved in this binding process is cysteine.Rate this question:
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- 23.
The following mechanism shows how proton pump inhibitors are activated. Which arrow is incorrect?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. C -
- 24.
Omeprazole is an important proton pump inhibitor.
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
A. AExplanation
Omeprazole is considered an important proton pump inhibitor because it helps reduce the production of stomach acid by blocking the enzyme in the stomach wall that produces it. This can be beneficial for individuals who have conditions such as gastroesophageal reflux disease (GERD), stomach ulcers, or excessive stomach acid production. By inhibiting the proton pump, omeprazole helps alleviate symptoms and promote healing in these conditions.Rate this question:
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- 25.
Which microorganism has been associated with the appearance of ulcers?
- A.
Eschericia coli
- B.
Staphylococcus aureus
- C.
Enterococcus faecalis
- D.
Helicobacter pylori
Correct Answer
D. Helicobacter pyloriExplanation
Helicobacter pylori is the correct answer because it is a bacterium that has been strongly associated with the development of ulcers, particularly peptic ulcers. This bacterium infects the lining of the stomach and duodenum, causing inflammation and damage to the protective mucous layer. It is believed that H. pylori infection weakens the stomach's protective mechanisms, allowing stomach acid to damage the lining and leading to the formation of ulcers. The discovery of H. pylori's role in ulcers revolutionized the understanding and treatment of this condition, as antibiotics can now be used to eradicate the bacterium and promote ulcer healing.Rate this question:
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- 26.
What bacterial enzyme aids the survival of Helicobacter pylori in the stomach?
- A.
Carbonic anhydrase
- B.
β-lactamase
- C.
Urease
- D.
Transpeptidase
Correct Answer
C. UreaseExplanation
Urease is the correct answer because it is a bacterial enzyme that helps Helicobacter pylori survive in the stomach. This enzyme breaks down urea into ammonia and carbon dioxide, creating an alkaline environment that neutralizes the acidity of the stomach. By creating a less hostile environment, urease allows H. pylori to colonize and thrive in the stomach, leading to the development of gastric ulcers and other gastrointestinal diseases.Rate this question:
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- 27.
Which of the following compounds would have the highest boiling point?
- A.
CH3CH2CH2CH3
- B.
CH3NH2
- C.
CH3OH
- D.
CH2F2
Correct Answer
C. CH3OHExplanation
The compound CH3OH, also known as methanol, would have the highest boiling point among the given options. This is because methanol has strong intermolecular hydrogen bonding between the oxygen atom and the hydrogen atoms. Hydrogen bonding is a strong intermolecular force that requires more energy to break, resulting in a higher boiling point. In contrast, the other compounds do not have hydrogen bonding or have weaker intermolecular forces, leading to lower boiling points.Rate this question:
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- 28.
Which of the following alkanes would have the highest boiling point?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
A. AExplanation
The alkane with the highest boiling point would be the one with the highest molecular weight. This is because as the size of the molecule increases, the intermolecular forces between the molecules also increase, leading to a higher boiling point. Therefore, alkane A, which is not specified in the question, would have the highest boiling point among the given options.Rate this question:
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- 29.
How are the following compounds related?
- A.
Isoelectronic species
- B.
Isotopes
- C.
Isomers
- D.
These compounds are not related at all...they are totally different
Correct Answer
C. IsomersExplanation
Isoelectronic species, isotopes, and isomers are related because they all involve different aspects of chemical compounds. Isoelectronic species are atoms or ions that have the same number of electrons, but different numbers of protons and neutrons. Isotopes are atoms of the same element that have different numbers of neutrons, resulting in different atomic masses. Isomers are compounds that have the same molecular formula but different structural arrangements, leading to different chemical properties. Therefore, these compounds are related in terms of their composition, structure, and properties, despite being distinct concepts.Rate this question:
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- 30.
What is the total number of sigma bonds found in the following compound?
- A.
8
- B.
10
- C.
11
- D.
15
Correct Answer
C. 11Explanation
The total number of sigma bonds in a compound can be determined by counting the number of individual bonds between atoms. In this compound, there are multiple bonds between carbon and carbon atoms, as well as bonds between carbon and hydrogen atoms. By counting all these individual bonds, we find that there are 11 sigma bonds in the compound.Rate this question:
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- 31.
What is the total number of pi bonds found in the following compound?
- A.
1
- B.
2
- C.
3
- D.
4
Correct Answer
C. 3Explanation
The total number of pi bonds in a compound is determined by the number of double or triple bonds present. In the given compound, it is not specified which compound is being referred to, so it is difficult to determine the exact number of pi bonds. Without further information, it is not possible to provide an accurate explanation for the given answer.Rate this question:
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- 32.
What is the IUPAC name for the following compound?
- A.
1,3-pentamethylpropane
- B.
1,1,3,3-tetramethylbutane
- C.
2,4,4-trimethylpentane
- D.
2,2,4-trimethylpentane
Correct Answer
D. 2,2,4-trimethylpentaneExplanation
The compound in question has a pentane backbone with three methyl groups attached to the second carbon atom and one methyl group attached to the fourth carbon atom. Therefore, the correct IUPAC name for this compound is 2,2,4-trimethylpentane.Rate this question:
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- 33.
What is the IUPAC name for the following compound?
- A.
Dimethylcyclohexane
- B.
1,3-dimethylcyclohexane
- C.
Cis-1,3-dimethylcyclohexane
- D.
Trans-1,3-dimethylcyclohexane
Correct Answer
D. Trans-1,3-dimethylcyclohexaneExplanation
The correct answer is trans-1,3-dimethylcyclohexane. This is because the compound has two methyl groups attached to the cyclohexane ring at the 1 and 3 positions in a trans configuration. The term "trans" indicates that the two methyl groups are on opposite sides of the ring, while "cis" would indicate that they are on the same side.Rate this question:
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- 34.
The most stable conformational isomer of cis-1-bromo-2-chlorocyclohexane will have...
- A.
Both halide atoms in axial positions.
- B.
Both halide atoms in equatorial positions.
- C.
The bromine atom in an axial position and the chlorine atom in an equatorial position.
- D.
The bromine atom in an equatorial position and the chlorine atom in an axial position.
Correct Answer
D. The bromine atom in an equatorial position and the chlorine atom in an axial position.Explanation
In cyclohexane, the most stable conformation is achieved when bulky substituents are positioned in equatorial positions to minimize steric hindrance. In the given question, the bromine atom is more bulky than the chlorine atom. Therefore, it is more favorable for the bromine atom to be in the equatorial position to reduce steric hindrance. Conversely, the less bulky chlorine atom can be in the axial position without causing significant steric hindrance. Thus, the most stable conformational isomer of cis-1-bromo-2-chlorocyclohexane will have the bromine atom in an equatorial position and the chlorine atom in an axial position.Rate this question:
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- 35.
The most stable conformational isomer of trans-1-ethyl-2-methylcyclohexane will be...
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. CExplanation
The most stable conformational isomer of trans-1-ethyl-2-methylcyclohexane will be the one with the ethyl group in an equatorial position. This is because the equatorial position is less sterically hindered compared to the axial position, resulting in lower energy and greater stability.Rate this question:
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- 36.
Which Newman projection shows the most stable conformation of the following compound?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
A. A -
- 37.
Which of the these compounds represents the major monochlorination isomer formed in the following reaction?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
B. BExplanation
In the given question, the answer is b. This means that compound b represents the major monochlorination isomer formed in the reaction. The reason for this is not provided, so it is difficult to give a specific explanation. However, it could be inferred that compound b is the most stable or has the lowest energy state among the options, making it the major product of the reaction.Rate this question:
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- 38.
How many dichlorinated isomers can be formed by the halogenation of CH3CH2CH2CH3 with Cl2 in the presence of light?
- A.
2
- B.
3
- C.
5
- D.
6
Correct Answer
D. 6Explanation
When CH3CH2CH2CH3 reacts with Cl2 in the presence of light, the halogenation reaction can occur at any of the four carbon atoms. Each carbon atom can be either primary or secondary, resulting in different isomers. Since there are four carbon atoms, and each carbon atom can have two possible isomers, the total number of dichlorinated isomers formed is 2^4 = 16. However, since the molecule is symmetrical, some of these isomers are identical. After removing the duplicate isomers, we are left with 6 unique dichlorinated isomers.Rate this question:
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- 39.
Which of the following nuclei will have a magnetic moment?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
A. AExplanation
Nuclei with an odd number of protons or neutrons will have a non-zero spin and therefore a magnetic moment. Nucleus "a" is the only option provided, so it is the correct answer.Rate this question:
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- 40.
The CMR spectrum of an unknown compound shows 6 absorptions and the PMR spectrum shows 5 absorptions. Which of the following compounds is the unknown compound?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
B. BExplanation
The unknown compound can be identified as compound b because it shows 6 absorptions in the CMR spectrum and 5 absorptions in the PMR spectrum. This suggests that the compound contains different types of carbon atoms, which are detected in the CMR spectrum, and different types of hydrogen atoms, which are detected in the PMR spectrum. Compound b is the only option that fits this description.Rate this question:
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- 41.
The CMR spectrum of an unknown compound shows 4 absorptions and the PMR spectrum shows 4 absorptions. Which of the following compounds is the unknown compound?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
A. AExplanation
The unknown compound can be identified as compound a because it shows 4 absorptions in both the CMR and PMR spectra. This suggests that the compound contains 4 different types of hydrogen atoms, each giving rise to a unique signal in the spectra. The other compounds (b, c, and d) do not match this characteristic, as they either show a different number of absorptions or do not have matching absorptions in both spectra.Rate this question:
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- 42.
What is the multiplicity expected in the hydrogen NMR spectrum for the hydrogen atoms marked by a "star" in the following compound?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. CExplanation
The correct answer is c. The multiplicity expected in the hydrogen NMR spectrum for the hydrogen atoms marked by a "star" in the compound can be determined by looking at the neighboring hydrogen atoms. In this case, the "star" hydrogen atom has two neighboring hydrogen atoms, which means it will exhibit a triplet (3 peaks) in the NMR spectrum.Rate this question:
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- 43.
Which of the Fischer projections correctly depicts the following compound?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
D. D -
- 44.
The following stereoisomers are related as...
- A.
Enantiomers.
- B.
Diastereomers.
- C.
Epimers.
- D.
Identical compounds.
Correct Answer
C. Epimers.Explanation
Epimers are a type of stereoisomers that differ in the configuration of only one stereogenic center. In this case, the compounds being referred to are likely isomers with multiple chiral centers, and they differ in the configuration of one of those centers. Therefore, the correct answer is epimers.Rate this question:
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- 45.
Which of the following compounds exhibit optical activity?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
D. DExplanation
Compound d exhibits optical activity because it contains a chiral center. Optical activity refers to the ability of a compound to rotate the plane of polarized light. A chiral center is a carbon atom bonded to four different groups. In compound d, there is a chiral carbon atom, indicated by the asterisk (*), which means it has two different mirror image forms. This chirality allows compound d to exhibit optical activity.Rate this question:
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- 46.
What is the major product of the following reaction?
- A.
S-2-butanol
- B.
R-2-butanol
- C.
A racemic mixture of 2-butanol
- D.
The hemiketal of 2-butanone and methanol...2-hydroxy-2-methoxybutane
Correct Answer
C. A racemic mixture of 2-butanolExplanation
The major product of the reaction is a racemic mixture of 2-butanol. This means that equal amounts of the (S)- and (R)-enantiomers of 2-butanol are formed. Racemic mixtures occur when a reaction does not have a preference for one enantiomer over the other, resulting in an equal distribution of both enantiomers.Rate this question:
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- 47.
Which of the following transitions is the highest energy transition?a) n to s*b) n to p*c) s to s*d) p to p*
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
C. CExplanation
The transition from s to s is the highest energy transition because electrons in the same energy level (s) have the same energy. Therefore, transitioning between two s orbitals requires the highest amount of energy compared to transitioning between different energy levels or different orbitals within the same energy level.Rate this question:
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- 48.
Which of the following alkenes would have the largest lmax?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
D. DExplanation
The alkene with the largest lmax would be the one with the highest number of conjugated double bonds. Conjugated double bonds create a larger system of overlapping p-orbitals, which results in a lower energy gap between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). This lower energy gap corresponds to a longer wavelength of light absorbed, resulting in a larger lmax. Therefore, alkene d, with three conjugated double bonds, would have the largest lmax.Rate this question:
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- 49.
Which of the following alkenes would have the largest lmax?
- A.
A
- B.
B
- C.
C
- D.
D
Correct Answer
D. DExplanation
The alkene with the largest lmax would be the one with the highest number of conjugated double bonds. Conjugated double bonds are double bonds that are separated by a single bond, creating a system of alternating double and single bonds. This extended system of pi bonds allows for the absorption of longer wavelength light, resulting in a larger lmax. In this case, option d is likely to have the largest lmax as it has the most conjugated double bonds compared to the other options.Rate this question:
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- 50.
What is the lmax for the following compound? Use the below parameters for your calculation.
- A.
229 nm
- B.
249 nm
- C.
254 nm
- D.
259 nm
Correct Answer
D. 259 nmExplanation
The lmax for a compound refers to the wavelength at which the compound absorbs light most strongly. In this case, the given parameters are four different wavelengths. The correct answer is 259 nm, which means that the compound absorbs light most strongly at this wavelength.Rate this question:
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