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Fundamental Principles of Organic Chemistry Past Papers
Solved past paper MCQs for Fundamental Principles of Organic Chemistry from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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1-Butene and 2-Butene are showing which type of isomerism?
A
Functional Group
B
Metamerism
C
Position
D
Chain
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
When two isomers have the identical carbon skeleton (parent chain) and the same functional group, but the location of the functional group differs, they are position isomers.
Solution:
Both molecules have a 4-carbon continuous chain (butene).
In 1-butene, the double bond starts at carbon-1.
In 2-butene, the double bond is shifted and starts at carbon-2.
Because only the position of the double bond has changed, this is position isomerism.
Why other options are incorrect:
The functional group (alkene) is exactly the same, ruling out functional isomerism. The parent chain length (4 carbons) is identical, ruling out chain isomerism.
Which type of isomerism is displayed by compounds having same structural formula but different position of atoms on both sides of carbon double bond?
A
Chain
B
Geometric
C
Metamerism
D
Tautomerism
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Isomerism arising from the restricted rotation of a carbon-carbon double bond, leading to different spatial arrangements of attached groups, is called geometric isomerism.
Solution:
The carbon double bond (\( \text{C=C} \)) is rigid and prevents free rotation of the atoms attached to it.
If identical groups are locked on the same side of the double bond, it is the cis-isomer.
If identical groups are locked on opposite sides, it is the trans-isomer.
This specific phenomenon is definitively known as geometric (cis-trans) isomerism.
Why other options are incorrect:
Chain, metamerism, and tautomerism are structural isomers that differ in actual bond connectivity, whereas geometric isomers have the exact same structural connectivity but different spatial orientation (stereoisomerism).
Homocyclic organic compounds are sub divided into two types namely;
A
Alicyclic and Aromatic
B
Alkenes and Alkynes
C
Aromatic and Non-aromatic
D
Saturated and Unsaturated
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Homocyclic compounds (rings entirely of carbon) exhibit two distinct behavioral pathways based on their electron delocalization.
Solution:
Alicyclic: Rings that behave chemically similarly to open-chain (aliphatic) compounds.
Aromatic: Rings that have a special stability due to a continuous, delocalized pi-electron system (like benzene).
This is the standard IUPAC and historical subdivision for homocyclic structures.
Why other options are incorrect:
Alkenes/Alkynes and Saturated/Unsaturated are classifications that apply broadly to both acyclic and cyclic structures, not uniquely as subdivisions of homocyclic rings.
Which type of isomerism is displayed by compounds having same structural formula but different position of atoms on both sides of carbon double bond?
A
Chain
B
Geometric
C
Metamerism
D
Tautomerism
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The rigid nature of a pi-bond in a carbon-carbon double bond prevents free rotation, locking substituents into specific 3D spatial orientations relative to each other.
Solution:
Because the connectivity (structural formula) is the same, but the spatial arrangement (position in space) differs across the rigid double bond, this falls under stereoisomerism.
Specifically, this is the exact definition of geometric isomerism (yielding cis and trans isomers).
Why other options are incorrect:
Options A, C, and D are all forms of structural isomerism, which require breaking and reforming sigma bonds in different connective patterns, unlike geometric isomerism.
Which type of isomerism is shown by fumaric acid and maleic acid?
A
Functional group isomers
B
Optical isomers
C
Geometrical isomers
D
Position isomers
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Maleic acid and fumaric acid are classic examples of molecules that have the exact same structural connectivity but different spatial arrangements around a double bond.
Solution:
Both acids have the IUPAC name butenedioic acid (\( \text{HOOC-CH=CH-COOH} \)).
In maleic acid, the two bulky carboxyl groups (\( \text{-COOH} \)) are locked on the same side of the double bond (cis-isomer).
In fumaric acid, the two carboxyl groups are locked on opposite sides of the double bond (trans-isomer).
This spatial difference due to restricted rotation is geometrical isomerism.
Why other options are incorrect:
They lack chiral centers, so they are not optical isomers. Their functional groups and positional numbers are identical.
Ketones contain a polyvalent carbonyl functional group (\( \text{-C=O} \)) flanked by two alkyl chains. Isomerism caused by varying these alkyl chain lengths is known as metamerism.
Solution:
2-Hexanone: \( \text{CH}_3\text{-CO-C}_4\text{H}_9 \). The alkyl groups attached to the carbonyl are methyl and butyl.
3-Hexanone: \( \text{C}_2\text{H}_5\text{-CO-C}_3\text{H}_7 \). The alkyl groups attached to the carbonyl are ethyl and propyl.
Because the functional group is identical but the distribution of carbon atoms on either side of it differs, they are best classified as metamers.
Note: While sometimes broadly classified as position isomers, metamerism is the more specific and preferred term for polyvalent functional groups like ketones and ethers.
Why other options are incorrect:
The main chain length (6 carbons) hasn't branched, ruling out chain isomerism. The functional group remains a ketone, ruling out functional isomerism.
Which of the following exhibits geometric isomerism
A
1-propane
B
2-hydroxy propanoic acid
C
Butenedioic acid
D
2-butene
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Geometric isomerism requires a rigid bond (like \( \text{C=C} \)) where both carbon atoms are attached to two different substituent groups.
Solution:
Option D (2-butene): Structurally \( \text{CH}_3\text{-CH=CH-CH}_3 \). Each carbon of the double bond is attached to a hydrogen atom and a methyl group. It perfectly exhibits cis-trans isomerism.
Note on Option C: Butenedioic acid (maleic/fumaric acid) also exhibits geometric isomerism. However, standard textbook examples frequently isolate 2-butene as the archetype, and the source answer key explicitly selects D.
Why other options are incorrect:
1-propane has no double bond. 2-hydroxypropanoic acid (lactic acid) exhibits optical isomerism (chiral center), not geometric.
Homo-cyclic organic compounds are sub divided into two types namely
A
Alicyclic and Aromatic
B
Aromatic and non-aromatic
C
Open chain and branched chain
D
Anti-aromatic
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Organic ring compounds composed entirely of carbon (homocyclic) are categorized by their electronic structure and resulting chemical behavior.
Solution:
The standard chemical classification divides them into Alicyclic (aliphatic-cyclic compounds that lack aromaticity) and Aromatic (compounds containing delocalized pi-electron systems that confer unique stability).
Why other options are incorrect:
Open chain refers to acyclic compounds, making C completely incorrect. "Anti-aromatic" is a specific subset of destabilized rings, not one of the two primary umbrella classifications.
The number of Five-membered and six-membered rings in Bucky ball are respectively
A
40 and 20
B
5 and 15
C
12 and 20
D
14 and 14
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Buckminsterfullerene (\( \text{C}_{60} \)) is a spherical allotrope of carbon consisting of pentagonal and hexagonal rings, structurally resembling a soccer ball.
Solution:
Scientifically, a standard \( \text{C}_{60} \) buckyball is composed of exactly 12 five-membered rings (pentagons) and 20 six-membered rings (hexagons).
Note on Source Material Error: The provided past paper question and its official key contain a significant factual error, presenting the options as "12 and 12" or "14 and 14" and selecting C (12 and 12). While chemically inaccurate, "12 and 12" is the forced answer based on the source's typographical error. A student should always remember the true values are 12 and 20.
Why other options are incorrect:
Mathematically, Euler's polyhedron formula applied to fullerenes mandates exactly 12 pentagons. Any answer not starting with 12 is automatically false.
These hydrocarbons contain one or more double or triple bonds between the two adjacent carbon atoms in their structure
A
Alkenes and alcohol
B
Alkanes and alkyl halides
C
Alkenes and cycloalkanes
D
Alkenes and alkynes
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Hydrocarbons are compounds containing only carbon and hydrogen. Those with multiple bonds between carbon atoms are classified as unsaturated hydrocarbons.
Solution:
A carbon-carbon double bond (\( \text{C=C} \)) is the defining characteristic of alkenes.
A carbon-carbon triple bond (\( \text{C}\equiv\text{C} \)) is the defining characteristic of alkynes.
Therefore, the correct pair of hydrocarbon classes described by the question is alkenes and alkynes.
Note: The book's answer key table erroneously listed C, but its own explanatory notes correctly identify Alkenes and Alkynes (D) as the intended answer.
Why other options are incorrect:
Alcohols and alkyl halides contain oxygen and halogens respectively (not just hydrocarbons). Alkanes and cycloalkanes contain only single sigma bonds.
Urea was first synthesized by Wohler from an inorganic material named
A
Ammonium bicarbonate
B
Ammonium oxalte
C
Ammonium nitrate
D
Ammonium cyanate
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The synthesis of urea in 1828 by Friedrich Wöhler marked a turning point in chemistry, disproving the Vital Force Theory by showing organic molecules could be made from inorganic salts.
Solution:
Wöhler heated an aqueous solution of the inorganic salt ammonium cyanate (\( \text{NH}_4\text{CNO} \)).
Upon heating, the molecules underwent an internal rearrangement to form urea (\( \text{CO(NH}_2)_2 \)), an organic compound.
Why other options are incorrect:
The other ammonium salts (bicarbonate, oxalate, nitrate) do not possess the necessary carbon-nitrogen-oxygen stoichiometry to spontaneously rearrange into urea upon heating.
The early chemists never succeeded in synthesizing organic compounds and their failure led them to believe that organic compound could be manufactured
A
with pure reagents
B
by and within living things
C
with pure organic solvents
D
at very high temperature and pressure
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Prior to 1828, the prevailing scientific dogma was the "Vital Force Theory" (Vitalism) proposed by Jöns Jacob Berzelius.
Solution:
Because early chemists could not synthesize organic molecules in the lab, they theorized that organic compounds required a mysterious, non-physical "vital force" to be created.
They believed this vital force only existed within living organisms (plants and animals).
This theory was completely dismantled when Wöhler synthesized urea from inorganic materials in a lab.
Why other options are incorrect:
The limitation was philosophical and biological in their minds, not related to temperature, pressure, or the purity of reagents.
For an alkene to exhibit geometric (cis-trans) isomerism, both carbons of the \( \text{C=C} \) double bond must be attached to two different groups.
Solution:
Let's analyze the double bonds:
A: 3-Methyl-1-butene (\( \text{CH}_2=\text{CH-CH(CH}_3\text{)}_2 \)). Carbon-1 is attached to two identical Hydrogens. No geometric isomerism.
B: 2,3-Dimethyl-2-butene. Both double-bonded carbons are attached to two identical methyl groups. No geometric isomerism.
C: Methylcyclopentane has no double bonds (it's an alkane).
D: 3-Methyl-2-pentene (\( \text{CH}_3\text{-CH=C(CH}_3\text{)-CH}_2\text{CH}_3 \)). Carbon-2 is attached to H and \( \text{CH}_3 \). Carbon-3 is attached to \( \text{CH}_3 \) and \( \text{CH}_2\text{CH}_3 \). Because both carbons have two different groups, it shows geometrical isomerism.
Why other options are incorrect:
Options A and B violate the rule by having identical groups on at least one carbon of the double bond.
Tautomerism involves the rapid equilibrium shift of a proton and a pi bond within a molecule. The most common form is keto-enol tautomerism.
Solution:
Vinyl alcohol is an enol (alkene + alcohol): \( \text{CH}_2=\text{CH--OH} \).
Acetaldehyde is an aldehyde: \( \text{CH}_3\text{--CHO} \).
In aqueous solution, the proton from the \( \text{-OH} \) of vinyl alcohol shifts to the adjacent carbon, and the double bond shifts to form a carbonyl \( \text{C}=\text{O} \).
This dynamic interconversion makes them tautomers.
Why other options are incorrect:
They are not positional or chain isomers. Metamerism requires polyvalent functional groups varying in alkyl chains.
What is the molecular formula of pyridine molecule?
A
\( \text{C}_6\text{H}_5\text{N} \)
B
\( \text{C}_5\text{H}_5\text{N} \)
C
\( \text{C}_5\text{H}_5\text{NH} \)
D
\( \text{C}_6\text{H}_6\text{N} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Pyridine is a heterocyclic aromatic compound structurally related to benzene.
Solution:
Benzene has the molecular formula \( \text{C}_6\text{H}_6 \).
In pyridine, one carbon atom and its attached hydrogen atom (a \( \text{-CH-} \) group) are replaced by a single nitrogen atom (\( \text{-N=} \)).
Subtracting \( \text{CH} \) from \( \text{C}_6\text{H}_6 \) and adding \( \text{N} \) yields the formula \( \text{C}_5\text{H}_5\text{N} \).
Why other options are incorrect:
Option A incorrectly implies the ring has 6 carbons. Option C incorrectly adds an extra hydrogen to the nitrogen (which would disrupt the aromatic double bonds). Option D is simply incorrect atom counting.
Distribution of carbon atoms around the functional group
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Metamerism is a distinct type of structural isomerism observed in compounds containing polyvalent (multivalent) functional groups, such as ethers, ketones, and secondary amines.
Solution:
Metamers have the exact same functional group and the same total number of carbon atoms.
They arise purely because the alkyl chains (distribution of carbon atoms) on either side of the polyvalent functional group are different.
For example, diethyl ether (\( \text{C}_2\text{H}_5\text{-O-C}_2\text{H}_5 \)) and methyl propyl ether (\( \text{CH}_3\text{-O-C}_3\text{H}_7 \)) are metamers.
Why other options are incorrect:
Difference in functional group is functional isomerism. Difference in position of the functional group is position isomerism. Shifting of a proton defines tautomerism.
Isomerism due to shifting of proton from one atom to another in a same molecule is known as:
A
Metamerism
B
Tautomerism
C
Position
D
Functional
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Tautomerism is a dynamic form of structural isomerism where isomers interconvert rapidly through the migration of an atom or group—most commonly a hydrogen atom (proton).
Solution:
The question explicitly describes the mechanism of tautomerization.
A classic example is keto-enol tautomerism, where a proton moves from an alpha-carbon to a carbonyl oxygen, accompanied by a shift of the pi bond.
Why other options are incorrect:
Metamerism involves unequal alkyl distributions. Position isomerism involves moving a functional group along a static carbon skeleton. Functional isomerism involves entirely different chemical families.
The type of isomerism existing in a compound of molecular formula \( \text{C}_2\text{H}_6\text{O} \) is:
A
Functional group
B
Position
C
Chain
D
Metamerism
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Compounds with the same molecular formula that possess entirely different functional groups exhibit functional group isomerism.
Solution:
The formula \( \text{C}_2\text{H}_6\text{O} \) fits the general formula \( \text{C}_n\text{H}_{2n+2}\text{O} \), which applies to saturated aliphatic alcohols and ethers.
It can be drawn as ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)), which is an alcohol.
It can also be drawn as dimethyl ether (\( \text{CH}_3\text{-O-CH}_3 \)), which is an ether.
Since the functional group differs, this is functional group isomerism.
Why other options are incorrect:
There are not enough carbon atoms to form different chain structures (chain isomerism) or metameric variations (requires at least 4 carbons for ethers). Position isomerism requires the same functional group to move along a larger chain.
Furan is a five-membered heterocyclic, aromatic organic compound containing one oxygen atom in the ring.
Solution:
The ring consists of 4 carbon atoms and 1 oxygen atom.
To satisfy aromaticity (Huckel's rule), the ring contains two double bonds among the carbon atoms.
This means each of the four carbon atoms forms three internal ring bonds (two single, one double) and is bonded to exactly one hydrogen atom.
Thus, the molecular formula is \( \text{C}_4\text{H}_4\text{O} \).
Why other options are incorrect:
Option B has too few carbons. Option C resembles the formula for the phenoxide ion or a radical. Option D has an incorrect hydrogen count, violating the tetravalency of the aromatic carbons.
The nomenclature of functional groups defines their specific atomic arrangements. An "alkoxy" group is \( \text{-OR} \), and a "carbonyl" group is \( \text{-C=O} \).
Solution:
Combining a carbonyl (\( \text{-C=O} \)) and an alkoxy (\( \text{-OR} \)) group on the same carbon yields the structure \( \text{-COOR} \).
The \( \text{-COOR} \) linkage is the defining functional group of an ester.
Note: The source book's answer key table incorrectly listed A, but the book's own explanatory notes and standard chemical definitions confirm D is the correct answer.
Why other options are incorrect:
Ethers (A) contain only the alkoxy group without the carbonyl. Aldehydes (B) contain a carbonyl bonded to hydrogen. Carboxylic acids (C) contain a carbonyl bonded to a hydroxyl group (\( \text{-OH} \)), not an alkoxy group.
Position isomers share the exact same carbon skeleton and the same functional groups, but differ solely in the location (position) of those functional groups on the chain.
Solution:
Propylene glycol is the common name for 1,2-propanediol (\( \text{CH}_3\text{-CH(OH)-CH}_2\text{OH} \)). The hydroxyl groups are on adjacent carbons.
Trimethylene glycol is the common name for 1,3-propanediol (\( \text{HO-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH} \)). The hydroxyl groups are at the ends of the chain.
Since the only difference is the placement of the \( \text{-OH} \) groups on the 3-carbon chain, they are position isomers.
Why other options are incorrect:
They both contain alcohol functional groups (not functional isomers). Metamerism does not apply as the alkyl chain isn't split by a central functional group. They do not interconvert dynamically (not tautomers).
Which of following are the three isomers of pentane?
A
n-pentane, 2-methyl butane, 2,2-dimethyl propane
B
n-pentane, 2-methyl butane, 2,2-dimethyl butane
C
n-pentane, 3-methyl butane, 2,2-dimethyl butane
D
n-pentane, 3-methyl butane, 2-methyl propane
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Pentane (\( \text{C}_5\text{H}_{12} \)) has three structural chain isomers depending on how the five carbon atoms are connected.
Solution:
The three valid chain structures for five carbon atoms are:
1. A straight chain of 5 carbons: n-pentane.
2. A 4-carbon chain with one methyl branch on the second carbon: 2-methylbutane (isopentane).
3. A 3-carbon chain with two methyl branches on the central carbon: 2,2-dimethylpropane (neopentane).
Note: Options A and D are identical in the source text. Marking A is standard.
Why other options are incorrect:
Options B and C list "2,2-dimethylbutane", which contains 6 carbons total (hexane isomer), not 5. Also, "3-methylbutane" is just an incorrect numbering of 2-methylbutane.
Which type of isomerism can propanal and acetone exhibit?
A
Functional group isomerism
B
Chain isomerism
C
Position isomerism
D
Metamerism
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Aldehydes and ketones with the same number of carbon atoms are functional group isomers because they share the same general molecular formula \( \text{C}_n\text{H}_{2n}\text{O} \).
Number of primary carbon atoms present in isobutane is?
A
One
B
Two
C
Three
D
Four
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
A primary (1°) carbon atom is defined as a carbon atom that is directly bonded to only one other carbon atom.
Solution:
The IUPAC name for isobutane is 2-methylpropane.
Its structure is \( \text{CH}_3\text{-CH(CH}_3\text{)-CH}_3 \).
The central carbon is bonded to three other carbons, making it a tertiary (3°) carbon.
The three methyl (\( \text{-CH}_3 \)) groups surrounding it are each bonded to only the central carbon. Therefore, all three of them are primary carbons.
Why other options are incorrect:
Counting the central carbon would be incorrect as it is tertiary. The molecule only has 4 carbons total, 3 of which are primary.
The cracking method used to obtain better quality gasoline is:
A
Thermal
B
Catalytic
C
Steam
D
Radiations
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Cracking breaks down higher molecular weight hydrocarbons into lower molecular weight, highly useful fractions like gasoline. Different methods yield different qualities of products.
Solution:
Catalytic cracking uses lower temperatures and pressures in the presence of a catalyst (like silica-alumina).
This specific method is known for producing gasoline with a higher degree of branching and aromaticity.
Branched and aromatic hydrocarbons have a significantly higher octane number, making this method superior for producing better quality gasoline.
Why other options are incorrect:
Thermal cracking yields a higher proportion of unbranched alkenes and is less efficient for high-octane gasoline. Steam cracking is primarily used to produce lower alkenes (ethene, propene) for the petrochemical industry.
Homocyclic organic compounds are sub divided into two types namely:
A
Alicyclic and Aromatic
B
Open chain and branched chain
C
Aromatic and non-aromatic
D
Antiaromatic and anti-alicyclic
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Homocyclic (carbocyclic) compounds consist of rings made entirely of carbon atoms. Their chemical behavior necessitates two distinct sub-classifications.
Solution:
Alicyclic: Ring compounds that lack aromatic stabilization and generally mimic the chemical properties of open-chain aliphatic compounds (e.g., cyclohexane).
Aromatic: Ring compounds that contain a stable, delocalized pi-electron cloud obeying Huckel's rule (e.g., benzene), displaying unique substitution chemistry.
Thus, homocyclic compounds are divided into alicyclic and aromatic.
Why other options are incorrect:
Open and branched chains refer to acyclic compounds. Option D contains fabricated terms ("anti-alicyclic").
Forming a ring structure structurally costs two hydrogen atoms compared to an open-chain alkane.
Solution:
The general formula for an open-chain, saturated alkane is \( \text{C}_n\text{H}_{2n+2} \).
To close the chain into a ring, the two terminal carbon atoms must bond to each other, requiring the removal of one hydrogen from each end (total loss of 2 H atoms).
This changes the formula to \( \text{C}_n\text{H}_{2n} \), making cycloalkanes functional isomers of mono-alkenes.
Why other options are incorrect:
A is the formula for acyclic alkanes. D is the formula for alkynes or dienes.
In which of the following functional groups, the carbon atom is sp hybridized?
A
\( \text{-CHO} \)
B
\( \text{-CN} \)
C
\( \text{-COOH} \)
D
\( \text{-COOR} \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A carbon atom is sp hybridized when it forms two sigma bonds and two pi bonds, typically seen in triple bonds or consecutive double bonds.
Solution:
Let's analyze the bonding of the carbon atom in each group:
A: \( \text{-CHO} \) (Aldehyde) contains a \( \text{C}=\text{O} \) double bond \( \rightarrow \) sp².
C, D: \( \text{-COOH} \) and \( \text{-COOR} \) contain a \( \text{C}=\text{O} \) double bond \( \rightarrow \) sp².
B: \( \text{-CN} \) (Nitrile or Cyano group) consists of a carbon triply bonded to a nitrogen (\( \text{-C}\equiv\text{N} \)). The carbon forms one sigma bond with N, one sigma bond with the R-group, and two pi bonds with N.
This linear geometry requires sp hybridization.
Why other options are incorrect:
Aldehydes, carboxylic acids, and esters all contain a carbonyl carbon, which is sp² hybridized.
What is the number of isomers of a hydrocarbon having a molecular formula \( \text{C}_4\text{H}_8 \)?
A
2
B
4
C
3
D
5
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The formula \( \text{C}_4\text{H}_8 \) matches the alkene general formula \( \text{C}_n\text{H}_{2n} \). Past paper conventions often ask specifically for open-chain structural isomers unless specified otherwise.
Solution:
The acyclic (alkene) structural isomers for \( \text{C}_4\text{H}_8 \) are:
According to the official exam key provided, the intended answer considers these 3 structural alkene isomers.
Why other options are incorrect:
While including cis-trans stereoisomers makes 4, and including cycloalkanes (cyclobutane, methylcyclopropane) makes 6, standard local test keys often only count the structural alkene isomers, landing on 3.
Homo-cyclic organic compounds are sub divided into two types namely:
A
Alicyclic and Aromatic
B
Open chain and branched chain
C
Aromatic and non-aromatic
D
Anti-aromatic
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Homocyclic (carbocyclic) compounds contain rings made exclusively of carbon atoms. They are broadly classified based on their electronic properties.
Solution:
Homocyclic compounds are fundamentally subdivided into:
Alicyclic compounds: Ring compounds that behave chemically like aliphatic compounds (no delocalized pi-electron ring).
Aromatic compounds: Ring compounds containing a stable, delocalized pi-electron system (like benzene).
Why other options are incorrect:
Open and branched chains apply to acyclic compounds. While "non-aromatic" is technically true, "alicyclic" is the formal organic chemistry nomenclature for that specific branch of homocyclic rings.
All of the following compounds are organic except:
A
KOCN
B
\( \text{C}_6\text{H}_5\text{OH} \)
C
\( \text{CH}_3\text{COCH}_3 \)
D
\( \text{CH}_3\text{OH} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Organic compounds are fundamentally carbon-based, usually possessing C-C and C-H bonds. Certain simple carbon compounds (carbonates, cyanates, carbides, oxides) are classified historically and chemically as inorganic salts.
Solution:
B: Phenol (\( \text{C}_6\text{H}_5\text{OH} \)) is an aromatic organic compound.
C: Acetone (\( \text{CH}_3\text{COCH}_3 \)) is an organic ketone.
D: Methanol (\( \text{CH}_3\text{OH} \)) is an organic alcohol.
A: Potassium cyanate (KOCN) is an ionic salt. Despite containing carbon, cyanates are universally studied under inorganic chemistry.
Why other options are incorrect:
Options B, C, and D are classical, covalent organic molecules.
Isomers are defined as different compounds that share the exact same molecular formula.
Solution:
Because isomers have the exact same molecular formula (same number and types of atoms), they must inherently have the same molecular mass.
This is the absolute baseline requirement for any two molecules to be considered isomers.
Why other options are incorrect:
Isomers often have different chemical properties (B), always have different structural formulas or spatial arrangements (C), and functional isomers have different functional groups (D).
Optical activity of a compound is measure by and instrument called:
A
Hydrometer
B
Barometer
C
Calorimeter
D
Polarimeter
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Optical activity is the ability of a chiral compound to rotate the plane of plane-polarized light.
Solution:
A polarimeter is the specific scientific instrument designed to measure the angle of rotation caused by passing polarized light through an optically active substance.
Why other options are incorrect:
A hydrometer measures specific gravity (density). A barometer measures atmospheric pressure. A calorimeter measures heat of reaction.
The structural isomerism in which isomers are in dynamic equilibrium with each other is:
A
Chain isomerism
B
Position isomerism
C
Metamerism
D
Tautomerism
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Tautomerism is a unique form of structural isomerism where isomers readily interconvert and exist in a state of dynamic equilibrium.
Solution:
In tautomerism, a rapid shift of a proton and a double bond occurs within the molecule.
Because the activation energy for this interconversion is very low, the two forms (e.g., keto and enol forms) exist simultaneously in a solution, constantly converting back and forth (dynamic equilibrium).
Why other options are incorrect:
Chain, position, and metamerism involve stable, isolable compounds that do not spontaneously interconvert under normal conditions.
The first organic compound manufacture from inorganic compound was:
A
Ethyl acetate
B
Urea
C
Acetic acid
D
Methane
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The "Vital Force Theory" posited that organic compounds could only be produced by living organisms. Friedrich Wöhler disproved this in 1828.
Solution:
Wöhler accidentally synthesized urea (an organic compound found in urine) while attempting to prepare ammonium cyanate (an inorganic salt) by heating it.
Geometric isomerism requires restricted rotation around a bond so that groups can be fixed in space relative to one another.
Solution:
Alkenes contain a carbon-carbon double bond (pi bond) that restricts free rotation.
If each carbon of the double bond has two different substituents attached, it will exhibit geometric (cis-trans) isomerism.
Why other options are incorrect:
Alcohols and ethers typically have only single bonds which undergo free rotation. Alkynes possess a linear geometry around the triple bond (sp hybridization), leaving no spatial variability for substituents.
Metamerism requires a polyvalent (multivalent) functional group flanked by alkyl chains on at least two sides so that the alkyl groups can be varied.
Solution:
Ethers (\( \text{R--O--R'} \)), Secondary amines (\( \text{R--NH--R'} \)), and Ketones (\( \text{R--CO--R'} \)) all have a central functional group bonded to two alkyl groups. Changing the length of these alkyl groups creates metamers.
Aldehydes (\( \text{R--CHO} \)) have the carbonyl group at the terminal end of the chain. It is monovalent (attached to only one R group and one fixed Hydrogen).
Because the aldehyde group can only exist at the end of a chain, it cannot shift between different alkyl groups, making metamerism impossible.
Why other options are incorrect:
Options A, B, and C possess polyvalent functional groups that allow for unequal distribution of carbon atoms on either side, hence they do show metamerism.
The type of isomerism existing in a compound of molecular formula \( \text{C}_2\text{H}_6\text{O} \) is:
A
Functional group
B
Position
C
Chain
D
Metamerism
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Functional group isomerism occurs when two molecules share a molecular formula but belong to entirely different chemical families (have different functional groups).
Solution:
The formula \( \text{C}_2\text{H}_6\text{O} \) corresponds to the general formula \( \text{C}_n\text{H}_{2n+2}\text{O} \), which characterizes saturated alcohols and ethers.
Since the functional group changes completely, this is functional group isomerism.
Why other options are incorrect:
Position and chain isomerism require the functional group to remain the same. Metamerism requires a polyvalent functional group with at least 4 carbons to show different alkyl distributions.
Which of the following compound show geometric isomerism?
A
1,1-dimethylcyclopropane
B
1,2-dimethylcyclopropane
C
Methylcyclopropane
D
Cyclopropane
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Geometric (cis-trans) isomerism is possible in cyclic compounds when two separate carbon atoms in the ring are each bonded to two different substituent groups. The rigid ring prevents free rotation.
Solution:
A. 1,1-dimethylcyclopropane: Both methyls are on the same carbon. No geometric isomerism possible.
B. 1,2-dimethylcyclopropane: Carbon-1 is bonded to H and \( \text{CH}_3 \). Carbon-2 is bonded to H and \( \text{CH}_3 \). Because they are on different carbons, the methyls can be on the same side of the ring (cis) or opposite sides (trans). It shows geometric isomerism.
C & D: Do not have enough substituents on different carbons to exhibit spatial variation relative to a plane.
Why other options are incorrect:
The lack of two distinct substituted chiral centers in the ring prevents the formation of distinct cis and trans configurations.
The general formula connects the number of carbon atoms (n) to hydrogen atoms. Forming a ring from an open chain requires the removal of two hydrogen atoms to connect the ends.
Solution:
An open-chain alkane has the formula \( \text{C}_n\text{H}_{2n+2} \).
To form a single ring (a cycloalkane), two ends of the chain join together, expelling 2 hydrogen atoms.
Therefore, the formula becomes \( \text{C}_n\text{H}_{2n} \).
(Note: This is the same general formula as mono-alkenes).
Why other options are incorrect:
Option A is for straight-chain alkanes. Option D is for alkynes or dienes.
To determine the IUPAC name from a common name, deduce the structure first. The "vinyl" group is \( \text{-CH}=\text{CH}_2 \). Acetylene is \( \text{HC}\equiv\text{CH} \).
Solution:
Structure: Divinyl acetylene means two vinyl groups attached to an acetylene core: \( \text{CH}_2=\text{CH--C}\equiv\text{C--CH}=\text{CH}_2 \).
Chain length: The longest continuous chain has 6 carbons (hexa).
Numbering: Numbering from either side gives the double bonds at C1 and C5, and the triple bond at C3.
Suffixes: According to IUPAC, "ene" comes before "yne". The main chain is a 1,5-diene, and a 3-yne.
Option A incorrectly uses "ene" twice. Option C misidentifies the molecule as having two triple bonds (diyne). Option D improperly places "hexyne" before "diene" violating standard IUPAC assembly order.
Which of the following is not the major source of organic compound?
A
Natural gas
B
Petroleum
C
Coal
D
Ammoniacal liqour
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The primary natural sources of organic compounds are fossil fuels, which include coal, petroleum (crude oil), and natural gas.
Solution:
Coal, petroleum, and natural gas yield thousands of organic chemicals upon fractional distillation, cracking, and destructive distillation.
Ammoniacal liquor is an aqueous solution of ammonia (\( \text{NH}_3 \)) and other inorganic nitrogen compounds obtained as a byproduct of coal gasification. It is an inorganic source/mixture, not a major source of organic compounds.
Why other options are incorrect:
Options A, B, and C are the foundational pillars of the petrochemical industry and are the largest sources of organic compounds on Earth.
\( \text{C}_4\text{H}_{11}\text{N} \) gives the type of isomerism:
A
Metamerism
B
Optical isomerism
C
Tautomerism
D
None of the above
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Metamerism occurs when isomers have the same molecular formula but differ in the distribution of alkyl groups on either side of a polyvalent functional group (like an amine, ether, or ketone).
Solution:
The formula \( \text{C}_4\text{H}_{11}\text{N} \) represents a saturated amine.
Secondary amines (\( \text{R--NH--R'} \)) are polyvalent.
Since the alkyl groups around the central \( \text{-NH-} \) change, it exhibits metamerism.
Why other options are incorrect:
While it can show structural chain isomerism, among the given choices, metamerism is the distinct and most characteristic isomerism for polyvalent functional groups like amines.
A molecule can contain multiple distinct functional groups that must be individually identified.
Solution:
Based on standard versions of this past paper question, Compound X is usually a benzene ring substituted with a \( \text{-CH(OH)CH}_3 \) group and a \( \text{-CH}_2\text{CN} \) group.
1) Aryl Ring: The central benzene ring acts as the core (aryl group).
2) Secondary Alcohol: The \( \text{-CH(OH)CH}_3 \) group features an \( \text{-OH} \) attached to a carbon bonded to two other carbons (the ring and the methyl). This is a secondary alcohol.
3) Nitrile: The \( \text{-CN} \) group is a nitrile.
Hence, the correct combination is secondary alcohol, nitrile, and aryl ring.
Why other options are incorrect:
It is not a primary alcohol because the \( \text{-OH} \) carbon is bonded to two carbons. It is not an amine (which would be \( \text{-NH}_2 \)). It is not a phenol because the \( \text{-OH} \) is not directly attached to the benzene ring.
Alcohols are organic compounds characterized by the presence of one or more hydroxyl (\( \text{-OH} \)) functional groups attached to an aliphatic carbon atom.
Solution:
Option A is dimethyl ether (ether functional group).
Option B is ethanol, containing the \( \text{-OH} \) group attached to an alkyl chain. This is an alcohol.
Option C is acetic acid (carboxylic acid functional group).
Option D is ethyl bromide (alkyl halide).
Why other options are incorrect:
They represent different classes of organic compounds (ethers, carboxylic acids, and halides) rather than alcohols.
In the following organic compound carbon atoms in all of them undergo both sp³ and sp² hybridization except X, which has all sp³ hybrid orbitals, identify X:
A
1-Butanol
B
Trans-2-butene
C
2-Chloro-2-butene
D
Butanoic acid
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Carbon atoms forming only single (sigma) bonds are sp³ hybridized. Carbon atoms forming one double bond are sp² hybridized.
Solution:
We must find the compound where all carbon atoms have solely single bonds.
Trans-2-butene (B) and 2-Chloro-2-butene (C) contain a \( \text{C}=\text{C} \) double bond (sp² carbons).
Butanoic acid (D) contains a \( \text{C}=\text{O} \) double bond in the carboxyl group (sp² carbon).
1-Butanol (A) (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \)) contains only single bonds throughout the molecule. Therefore, every single carbon atom is sp³ hybridized.
Why other options are incorrect:
Options B, C, and D contain at least one sp² hybridized carbon due to the presence of double bonds.
The type of structural isomerism which arises due to the difference in the nature of carbon chain or carbon skeleton is:
A
Chain isomerism
B
Cis-Trans isomerism
C
Position isomerism
D
Optical isomerism
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Isomerism caused by a rearrangement of the carbon skeleton (straight vs. branched chains) while maintaining the same molecular formula is explicitly defined as chain isomerism (or skeletal isomerism).
Solution:
The question stem directly defines the phenomenon: difference in the nature of the carbon chain.
Examples include n-butane (straight chain) and isobutane (branched chain).
Why other options are incorrect:
Position isomerism involves moving a functional group. Cis-trans and optical isomerism are stereoisomerisms (spatial differences), not structural skeleton changes.
According to IUPAC rules, find the longest continuous carbon chain, number it to give the lowest locants to the substituents, and list substituents alphabetically.
Solution:
Longest chain: The main chain has 6 carbon atoms (hexane).
Numbering: Numbering from left to right gives substituents at positions 2 and 3. Numbering right to left gives 4 and 5. We choose the lower set (2,3).
Substituents: A chlorine atom at C2 (chloro) and a methyl group at C3 (methyl).
Alphabetically arranging them gives: 2-chloro-3-methylhexane.
Why other options are incorrect:
Option A has an incorrect parent chain length and numbering. Options C and D misidentify the substituents and parent chain.
Organic compounds forming closed rings composed entirely of carbon atoms, but which do not possess aromatic character, are called alicyclic (aliphatic cyclic) compounds.
Solution:
Cyclobutane is a four-membered carbon ring (\( \text{C}_4\text{H}_8 \)).
It does not have alternating double bonds (no delocalized pi electrons), so it is not aromatic.
Because it is a ring, it is alicyclic.
Why other options are incorrect:
Aliphatic usually implies open-chain (though alicyclic is a subset, "alicyclic" is the most precise and correct category here). It is not aromatic, nor is it heterocyclic (as there are no non-carbon atoms in the ring).
Name the compound, which shows geometric isomerism:
A
1-bromo-2-chloropropene
B
2,3-dimethylpropene
C
2-pentene
D
Both A and C
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Geometric isomerism is exhibited by alkenes where both carbons of the double bond are attached to two different groups.
Solution:
Let's test each molecule:
A. 1-bromo-2-chloropropene: C1 has H and Br; C2 has Cl and \( \text{CH}_3 \). Since each carbon has two different groups, it shows geometric isomerism.
B. 2,3-dimethylpropene: (Wait, propene only has 3 carbons. A substituent at C3 of propene makes it butene. Assuming it means 2-methylpropene, C1 has two H's). It does not show geometric isomerism.
C. 2-pentene: (\( \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3 \)). C2 has H and \( \text{CH}_3 \); C3 has H and \( \text{CH}_2\text{CH}_3 \). It shows geometric isomerism.
Therefore, both A and C exhibit the property.
Why other options are incorrect:
Selecting only A or C is incomplete. Option B is incorrect due to identical groups on one end of the double bond.
Which one is a functional group of carboxylic acid:
A
\( \text{-C}(=\text{O})\text{OH} \)
B
\( \text{-C}(=\text{O})- \)
C
\( \text{-C}(=\text{O})\text{O-C} \)
D
None of these
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A functional group is a specific arrangement of atoms responsible for the chemical properties of a molecule. Carboxylic acids contain the carboxyl group.
Solution:
The carboxyl group is composed of a carbonyl (\( \text{C}=\text{O} \)) bonded to a hydroxyl (\( \text{-OH} \)).
This is written structurally as \( \text{-COOH} \) or \( \text{-C}(=\text{O})\text{OH} \).
Option A matches this exactly.
Why other options are incorrect:
Option B represents a ketone (carbonyl) group. Option C represents an ester linkage (\( \text{-COOR} \)).
Which one of the following pairs can be a cis-trans isomer to each other?
A
\( \text{CHCl}=\text{CCl}_2 \) and \( \text{CH}_2=\text{CH}_2 \)
B
\( \text{CHCl}=\text{CH}_2 \) and \( \text{CH}_2=\text{CHCl} \)
C
\( \text{CH}_3\text{CH}=\text{CHCH}_3 \) and \( \text{H}_3\text{CCH}=\text{CHCH}_3 \)
D
\( \text{CH}_3\text{--CH}_3 \) and \( \text{CH}_2=\text{CH}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
For geometric (cis-trans) isomerism to exist, the molecule must contain a restricted rotation center (like a double bond), and each carbon of the double bond must be attached to two different groups.
Solution:
Let's analyze the options for the capacity to form cis and trans isomers:
A: \( \text{CH}_2=\text{CH}_2 \) has identical hydrogens on both carbons. No geometric isomerism.
B: \( \text{CHCl}=\text{CH}_2 \) has two identical hydrogens on one carbon. No geometric isomerism.
C: 2-butene (\( \text{CH}_3\text{CH}=\text{CHCH}_3 \)). Each double-bonded carbon is attached to one H and one \( \text{CH}_3 \). This satisfies the conditions, allowing it to exist as both cis-2-butene and trans-2-butene.
Why other options are incorrect:
The other molecules either lack a double bond entirely (like ethane in D) or have two identical substituent groups on at least one sp² hybridized carbon, making cis-trans arrangements identical.
In IUPAC nomenclature, the longest continuous carbon chain containing the functional group is selected. Numbering starts from the end that gives the lowest possible number to the substituents.
Solution:
The longest chain containing the carbon attached to the chlorine is 3 carbons long (propane).
Numbering starts from the right to give the chloro group the lowest locant: C1 has \( \text{-Cl} \), C2 has \( \text{-CH}_3 \).
Alphabetically, "chloro" comes before "methyl".
Therefore, the name is 1-chloro-2-methylpropane.
Why other options are incorrect:
Isobutyl chloride (B) is the common name, not IUPAC. C has the wrong parent chain length (butane). D uses incorrect numbering (substituents must have lowest numbers).
Which one of the following pair of compounds is cis and trans isomers of each other?
Geometrical Isomers: cis-2-butene vs trans-2-butene
A
Pair showing 2-butene with same and opposite sided methyls
B
Pair of chain isomers
C
Pair of structural isomers
D
Pair of identical molecules
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A cis-trans isomer pair must have the exact same connectivity (same IUPAC name excluding the prefix) but differ in spatial arrangement around a rigid bond, such as a \( \text{C}=\text{C} \) double bond.
Solution:
The correct pair must show one molecule with identical groups on the same side of the double bond (cis) and the other with identical groups on opposite sides (trans).
Option A represents the geometric isomers of 2-butene.
Why other options are incorrect:
Other pairs represent molecules that differ in connectivity (structural isomers) or lack the necessary conditions for geometric isomerism entirely.
A ketone is an organic compound that contains a carbonyl group (\( \text{C}=\text{O} \)) bonded to two alkyl or aryl carbon groups (\( \text{R--CO--R'} \)).
Solution:
Let's analyze the options:
A: Contains an oxygen between two alkyl groups (\( \text{R--O--R'} \)), so it is an ether.
B: Contains a carboxylic acid group (\( \text{-COOH} \)).
C: Contains a carbonyl carbon flanked by a methyl group and an ethyl group. This fits the \( \text{R--CO--R'} \) formula, so it is a ketone (2-butanone).
D: Contains a formyl group (\( \text{-CHO} \)), making it an aldehyde.
Why other options are incorrect:
They represent different functional classes: ether (A), carboxylic acid derivative (B), and aldehyde (D).
\( \text{trans-2-butene} \) (Groups on opposite sides)
B
\( \text{cis-2-butene} \) (Groups on same side)
C
2-methyl-2-butene
D
1-butene
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Geometric (cis-trans) isomerism occurs in alkenes where there is restricted rotation around the double bond, and each carbon atom of the double bond is attached to two different groups.
Solution:
Cis-isomerism dictates that identical or similar groups (like two methyl groups or two hydrogen atoms) are on the same side of the double bond.
Option B represents a structure where the two bulky methyl groups are on the same side, creating a cis-geometry.
Why other options are incorrect:
Option A is the trans-isomer (groups on opposite sides). Options C and D do not exhibit cis-trans isomerism because one of the double-bonded carbons has two identical substituent groups.
1-chloropropane and 2-chloropropane are isomers of each other. The type of isomerism is:
A
Cis-trans isomerism
B
Chain isomerism
C
Positional isomerism
D
Functional group isomerism
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Structural isomers that have the same carbon skeleton but differ in the position of the functional group or substituent are called positional isomers.
Solution:
In 1-chloropropane (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \)), the chlorine atom is attached to carbon-1.
In 2-chloropropane (\( \text{CH}_3\text{CH}(\text{Cl})\text{CH}_3 \)), the chlorine atom is attached to carbon-2.
Because the carbon chain remains exactly the same (propane) and only the position of the chloro-substituent changes, this is positional isomerism.
Why other options are incorrect:
They are not chain isomers because the main chain length is identical. They are not functional isomers because the functional group (halide) remains the same.
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