JEE Hydrocarbons - Alkenes Questions

We need to identify the incorrect statements regarding geometrical isomerism.

(A) Propene shows geometrical isomerism.

Propene is $$CH_3-CH=CH_2$$. One carbon of the double bond has two identical groups (two H atoms). For geometrical isomerism, each carbon of the double bond must have two different substituents. Since one carbon has 2 H atoms, propene does NOT show geometrical isomerism. Statement (A) is INCORRECT.

(B) Trans isomer has identical atoms/groups on the opposite sides of the double bond.

By definition, trans isomers have identical groups on opposite sides of the double bond. This is CORRECT.

(C) Cis-but-2-ene has higher dipole moment than trans-but-2-ene.

In cis-but-2-ene, the two methyl groups are on the same side, so the bond dipoles add up, giving a non-zero dipole moment. In trans-but-2-ene, the methyl groups are on opposite sides, and the dipoles cancel out, giving zero (or nearly zero) dipole moment. So cis has a higher dipole moment. This is CORRECT.

(D) 2-methylbut-2-ene shows two geometrical isomers.

2-methylbut-2-ene is $$CH_3-C(CH_3)=CH-CH_3$$. The carbon bearing the double bond at position 2 has two methyl groups (identical substituents). Since one carbon of the double bond has two identical groups, it cannot show geometrical isomerism. Statement (D) is INCORRECT.

(E) Trans-isomer has lower melting point than cis isomer.

Generally, trans isomers have higher melting points than cis isomers because they are more symmetrical and pack better in the crystal lattice. So the statement that trans has lower melting point is INCORRECT.

The incorrect statements are (A), (D), and (E).

The correct answer is Option 2: (A), (D) and (E) Only.

To determine the correct IUPAC name of the given cyclic compound, we apply the standard IUPAC nomenclature rules by identifying the principal functional group and assigning the appropriate numbering.

The molecule contains both a hydroxyl group $$(-OH)$$ and a carbon-carbon double bond $$\left(C=C\right)$$ in a five-membered ring. According to IUPAC priority rules, the hydroxyl group has higher priority than the double bond. Therefore, the suffix of the compound is “-ol”, and the carbon bearing the $$-OH$$ group is assigned position 1.

After fixing the hydroxyl-bearing carbon as carbon 1, the ring is numbered in the direction that gives the double bond the lowest possible locant. With this numbering, the double bond begins at carbon 2, and the ethyl substituent is located at carbon 4.

Thus, the numbering is:

• Hydroxyl group at C1
• Double bond between C2 and C3
• Ethyl substituent at C4

The name is constructed as follows:

• Substituent: 4-ethyl
• Parent ring: cyclopent
• Double bond: 2-en
• Principal suffix: 1-ol

Hence, the correct IUPAC name of the compound is

4-ethylcyclopent-2-en-1-ol.

To determine the number of optically active products, we analyze the fragments produced by the complete ozonolysis of the given compound and examine each for chirality.

Complete ozonolysis cleaves all carbon-carbon double bonds $$\left(C=C\right)$$ and converts them into carbonyl groups $$\left(C=O\right).$$

The given polyene can be represented as

$$
CH_3-CH=CH-C^(CH_3)(H)-CH=CH-C^(H)(CH_3)-CH=CH-CH_3.
$$

Breaking the molecule at each of the three double bonds gives the following fragments:

1. $$CH_3CHO$$ (Ethanal)
2. $$OHC-CH(CH_3)-CHO$$ (2-methylpropanedial)
3. $$OHC-CH(CH_3)-CHO$$ (2-methylpropanedial)
4. $$CH_3CHO$$ (Ethanal)

Now, let us examine the optical activity of each product.

Ethanal $$\left(CH_3CHO\right)$$ does not contain any stereocentre and is therefore achiral.

For 2-methylpropanedial $$\left(OHC-CH(CH_3)-CHO\right),$$ the central carbon is attached to:

• $$-H$$
• $$-CH_3$$
• $$-CHO$$
• $$-CHO$$

Since two of the substituents are identical $$\left(-CHO\right),$$ the central carbon is not a stereogenic centre. The molecule possesses a plane of symmetry and is therefore achiral.

Although the original reactant contains chiral centres, ozonolysis converts the neighbouring alkene fragments into identical aldehyde groups, thereby destroying the chirality of the molecule.

Hence, none of the products formed after complete ozonolysis are optically active.

Therefore, the number of optically active products is

$$
\boxed{0}.
$$

To determine which compound yields 3-methyl-6-oxoheptanal upon ozonolysis, we reverse the ozonolysis process.

The target product is 3-methyl-6-oxoheptanal, whose structure can be written as

$$CHO-CH_2-CH_2-CH(CH_3)-CH_2-C(=O)-CH_3.$$

The name indicates a seven-carbon chain containing an aldehyde group at C1, a ketone group at C6, and a methyl substituent at C3.

Ozonolysis of an alkene followed by reductive workup ($$O_3$$ followed by $$Zn/H_2O$$) cleaves the carbon-carbon double bond and converts the two alkene carbons into carbonyl groups. Therefore, to identify the original alkene, we reverse the process by joining the two carbonyl carbons with a double bond.

In the given product, the two carbonyl carbons are:

• The aldehyde carbon ($$CHO$$) at C1.
• The ketone carbon ($$C=O$$) at C6.

Connecting these two carbons reconstructs the original double bond and produces a six-membered ring. The methyl substituent present at C3 of the product remains attached to the corresponding carbon in the ring, while the methyl group bonded to the ketone carbon remains attached to the alkene carbon.

The resulting structure is 1,4-dimethylcyclohex-1-ene.

Examining the given options, Option B represents 1,4-dimethylcyclohex-1-ene. On ozonolysis, cleavage of its $$C=C$$ bond produces:

• An aldehyde group at one end.
• A ketone group at the other end.

This gives the product

$$CHO-CH_2-CH_2-CH(CH_3)-CH_2-C(=O)-CH_3,$$

which is 3-methyl-6-oxoheptanal.

Therefore, the correct answer is Option B.

The problem can be solved by first identifying the products formed in each reaction step and then carrying out the required stoichiometric calculations.

In the first step, the starting compound is 2-bromopentane. When treated with alcoholic KOH, it undergoes a $\beta$-elimination reaction.

According to Zaitsev's rule, the more substituted alkene is formed as the major product.

The reaction is

$$\text{2-Bromopentane}+\text{alc. KOH}\longrightarrow\text{Pent-2-ene}+KBr+H_2O.$$

Hence, product P is

$$\boxed{\text{Pent-2-ene }(\mathrm{CH_3-CH=CH-CH_2-CH_3}).}$$

In the second step, pent-2-ene reacts with bromine through electrophilic addition across the double bond.

The reaction is

$$\text{Pent-2-ene}+Br_2\longrightarrow\text{2,3-Dibromopentane}.$$

Therefore, product Q is

$$\boxed{\mathrm{CH_3-CH(Br)-CH(Br)-CH_2-CH_3}.}$$

Next, calculate the molar masses.

For 2-bromopentane,

$$\mathrm{C_5H_{11}Br},$$

the molar mass is

$$M=(5\times12)+(11\times1)+80$$

$$=60+11+80$$

$$=151\ \text{g mol}^{-1}.$$

For 2,3-dibromopentane,

$$\mathrm{C_5H_{10}Br_2},$$

the molar mass is

$$M=(5\times12)+(10\times1)+(2\times80)$$

$$=60+10+160$$

$$=230\ \text{g mol}^{-1}.$$

Initially,

$$151\ \text{g}$$

of 2-bromopentane is taken.

Hence,

$$\text{Number of moles}=\frac{151}{151}=1\ \text{mol}.$$

The first reaction proceeds with an

$$80%$$

yield.

Therefore, the actual amount of product P formed is

$$1\times0.80=0.8\ \text{mol}.$$

The second reaction has

$$100%$$

yield and follows a

$$1:1$$

stoichiometric ratio.

Hence,

$$\text{Moles of Q}=0.8\ \text{mol}.$$

Finally, the mass of product Q is

$$\text{Mass}=0.8\times230$$

$$=184\ \text{g}.$$

Therefore, the required mass of product Q is

$$\boxed{184\ \text{g}}.$$

Methods used for purification of organic compounds (such as distillation, crystallization, chromatography, etc.) depend on both the nature of the compound being purified and the type of impurities present.

For example, distillation is used when the compound and impurity have different boiling points, while crystallization depends on solubility differences. The choice of method depends on both factors.

The correct answer is Option 1: nature of compound and presence of impurity.

In chromatography: polar compounds interact more strongly with the polar stationary phase (silica) and move slower, giving smaller Rf values. Non-polar compounds move faster with larger Rf.

The correct answer is Option (3): Rf of a polar compound is smaller than that of a non-polar compound.

We need to evaluate two statements about the Kjeldahl method and its applicability to pyridine.

Analysis of Statement (I): "Kjeldahl method is applicable to estimate nitrogen in pyridine."

This statement is FALSE. The Kjeldahl method is NOT applicable to compounds where nitrogen is present in a ring structure (as in pyridine, quinoline, and similar heterocyclic compounds), or in nitro groups ($$-NO_2$$) and azo groups ($$-N=N-$$). This is because the nitrogen in these structures is particularly resistant to being converted to ammonium sulphate by digestion with concentrated sulphuric acid.

Analysis of Statement (II): "The nitrogen present in pyridine can easily be converted into ammonium sulphate in Kjeldahl method."

This statement is also FALSE. The nitrogen in pyridine's ring structure cannot be easily liberated and converted to ammonium sulphate $$(NH_4)_2SO_4$$ during the digestion step of the Kjeldahl method. In the Kjeldahl method, the organic compound is digested with concentrated $$H_2SO_4$$ in the presence of a catalyst (such as $$K_2SO_4$$ and $$CuSO_4$$), which converts organic nitrogen to $$(NH_4)_2SO_4$$. However, the nitrogen locked in the aromatic ring of pyridine resists this conversion because of the extra stability of the aromatic ring.

Since both statements are false, the correct answer is Option (2): Both Statement I and Statement II are false.

We need to determine which statements about purification techniques are correct.

Statement A: "Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point."

Glycerol has a normal boiling point of about 290 degC, but it begins to decompose around 250 degC. Vacuum distillation lowers the boiling point by reducing the pressure, allowing glycerol to be distilled at a lower temperature without decomposition. This statement is correct. ✓

Statement B: "Aniline can be purified by steam distillation as aniline is miscible in water."

While it is true that aniline can be purified by steam distillation, the reason given is incorrect. Steam distillation works for compounds that are immiscible (or sparingly soluble) in water, not miscible. Aniline is only slightly soluble in water (about 3.6 g/100 mL at 20 degC), which is why it can be steam-distilled - the two immiscible liquids distil at a temperature below the boiling point of either component. The statement incorrectly claims aniline is miscible in water. This statement is incorrect. ✗

Statement C: "Ethanol can be separated from ethanol-water mixture by azeotropic distillation because it forms an azeotrope."

Ethanol and water form a minimum boiling azeotrope at 95.6% ethanol (by weight) at 78.1 degC. Simple distillation cannot separate them beyond this composition. Azeotropic distillation (using a third component like benzene or cyclohexane as an entrainer) can break the azeotrope and separate pure ethanol. This statement is correct. ✓

Statement D: "An organic compound is pure, if mixed M.P. is remained same."

The mixed melting point test is a standard purity test: if a sample is mixed with a pure reference compound and the melting point remains unchanged (no depression), then the sample is the same pure compound. If impurities are present, the melting point would be depressed. This statement is correct. ✓

Conclusion: Statements A, C, and D are correct.

The correct answer is Option 2: A, C, D only.

The reagents $$O_3$$ followed by $$Zn/H_2O$$ indicate reductive ozonolysis. In this reaction, the carbon-carbon double bond is cleaved completely, and the resulting fragments are converted into aldehydes or ketones.

The given alkene is trans-2-butene, $$CH_3-CH=CH-CH_3$$. Cleavage of the $$C=C$$ bond divides the molecule into two identical fragments. Each fragment forms acetaldehyde, $$CH_3CHO$$. Therefore, the product obtained is acetaldehyde.

A single molecule of acetaldehyde contains one oxygen atom.

Hence, the number of oxygen atoms present in one molecule of product $$P$$ is $$1$$

3-Methylhex-2-ene: $$CH_3CH=C(CH_3)CH_2CH_2CH_3$$

In the presence of peroxides, HBr adds via anti-Markovnikov (free radical) mechanism. The Br adds to the less substituted carbon of the double bond.

The product is: $$CH_3CHBr-CH(CH_3)-CH_2CH_2CH_3$$ (Br on C-2).

Let's check for stereocenters:

C-2: $$CHBr$$ bonded to $$CH_3$$, $$H$$, $$Br$$, and $$CH(CH_3)CH_2CH_2CH_3$$ — this is a chiral center.

C-3: $$CH(CH_3)$$ bonded to $$CH_3$$, $$H$$, $$CHBrCH_3$$, and $$CH_2CH_2CH_3$$ — this is a chiral center.

With 2 chiral centers, the maximum number of stereoisomers = $$2^2 = 4$$.

Since the two chiral centers have different substituents (they are not equivalent), all 4 stereoisomers are distinct (no meso form).

The number of possible stereoisomers is $$\boxed{4}$$.

We need to convert trans-PhCH=CHCH$$_3$$ to cis-PhCH=CHCH$$_3$$.

Trans to cis conversion cannot be done directly. The approach is:

1. Convert the double bond to a triple bond (via addition-elimination)

2. Then selectively reduce the triple bond to give the cis alkene

1. Br$$_2$$: Anti addition to the trans alkene gives a vicinal dibromide

2. Alc. KOH: Double dehydrohalogenation (two eliminations) gives the alkyne PhC$$\equiv$$CCH$$_3$$

3. NaNH$$_2$$: Strong base for the second elimination (if needed) to ensure complete conversion to alkyne

4. H$$_2$$/Lindlar catalyst: Selective syn-hydrogenation of the alkyne gives the cis alkene

The key is that Lindlar's catalyst (Pd/CaCO$$_3$$/quinoline) gives cis (syn) addition, while Na in liquid NH$$_3$$ (Birch conditions) would give the trans product.

Option B: Br$$_2$$, alc KOH, NaNH$$_2$$, H$$_2$$ Lindlar Catalyst - this correctly uses Lindlar's catalyst for cis selectivity.

The correct answer is Option B: Br$$_2$$, alc KOH, NaNH$$_2$$, H$$_2$$ Lindlar Catalyst.

Water gas is a mixture of carbon monoxide ($$CO$$) and hydrogen ($$H_2$$), produced by passing steam over hot coke:

$$C + H_2O \rightarrow CO + H_2$$

When water gas is passed over a cobalt ($$Co$$) catalyst under appropriate conditions of temperature and pressure, it produces methanoic acid (formic acid).

The balanced reaction is:

$$CO + H_2O \rightarrow HCOOH$$

Here, carbon monoxide reacts with water (present as steam) in the presence of a cobalt catalyst to form methanoic acid ($$HCOOH$$).

Let us verify why the other options are incorrect:

Option A (Methanal / Formaldehyde, $$HCHO$$): Methanal is obtained by catalytic oxidation of methanol using a silver or iron-molybdenum oxide catalyst. It is not a direct product of water gas with a cobalt catalyst.

Option C (Ethanol, $$C_2H_5OH$$): Ethanol can be produced from syngas but requires different catalysts (such as rhodium-based catalysts) and specific conditions. Cobalt catalyst does not directly produce ethanol from water gas.

Option D (Methanol, $$CH_3OH$$): Methanol is produced from syngas ($$CO + 2H_2$$) using a $$ZnO$$-$$Cr_2O_3$$ catalyst at high pressure. The reaction is $$CO + 2H_2 \rightarrow CH_3OH$$. This requires a different catalyst, not cobalt.

Therefore, the correct answer is Option B: Methanoic acid.

We need to find which compound gives the maximum number of isomeric alkenes on dehydrohalogenation (excluding rearrangement).

1-Bromo-2-methylbutane: CH$$_2$$Br-CH(CH$$_3$$)-CH$$_2$$-CH$$_3$$

The only $$\beta$$-carbon is C2. Elimination gives: 2-methylbut-1-ene.

Number of isomeric alkenes = 1

2-Bromopropane: CH$$_3$$-CHBr-CH$$_3$$

$$\beta$$-hydrogens are on C1 and C3 (equivalent). Elimination gives only: propene.

Number of isomeric alkenes = 1

2-Bromopentane: CH$$_3$$-CHBr-CH$$_2$$-CH$$_2$$-CH$$_3$$

$$\beta$$-hydrogens at C1 and C3:

Elimination from C1: gives pent-1-ene

Elimination from C3: gives pent-2-ene, which exists as both (E)-pent-2-ene and (Z)-pent-2-ene

Number of isomeric alkenes = 3 (maximum)

2-Bromo-3,3-dimethylpentane: CH$$_3$$-CHBr-C(CH$$_3$$)$$_2$$-CH$$_2$$-CH$$_3$$

C3 is quaternary (no $$\beta$$-H). Only $$\beta$$-H at C1.

Elimination from C1: gives 3,3-dimethylpent-1-ene

Number of isomeric alkenes = 1

2-Bromopentane gives the maximum number of isomeric alkenes (3).

The correct answer is Option 3: 2-Bromopentane.

The reaction is an acid-catalyzed dehydration of an alcohol using concentrated (H_2SO_4).

Under acidic conditions, the hydroxyl group is first protonated to form a good leaving group ((-OH_2^+)). Loss of water then generates a secondary carbocation at the carbon originally bearing the hydroxyl group.

Since a more stable carbocation can be formed, the secondary carbocation undergoes a (1,2)-hydride shift from the adjacent tertiary carbon, producing a tertiary carbocation.

From this rearranged carbocation, elimination of a (\beta)-hydrogen takes place to form an alkene. According to Zaitsev’s rule, the major product is the most substituted and therefore the most thermodynamically stable alkene.

Among the given options, Option A corresponds to the tetrasubstituted alkene formed after carbocation rearrangement and subsequent elimination. This is more stable than the other possible alkene products and is therefore the major product.

Hence, the correct answer is

$$\boxed{\text{Option A}}.$$

The answer key indicating the last option is incorrect because it does not account for the carbocation rearrangement that occurs before elimination.

The starting compound is 2-methylbut-1-ene.

$$CH_2=C(CH_3)CH_2CH_3$$

Both reactions convert the alkene into an alcohol by adding $$H$$ and $$OH$$ across the double bond. The difference lies in the position at which the $$OH$$ group is introduced.

For reaction A, the reagents are $$BH_3/THF$$ followed by $$H_2O_2/OH^-$$.

This is the hydroboration-oxidation reaction.

Hydroboration-oxidation follows anti-Markovnikov addition. Therefore, the $$OH$$ group is introduced at the less substituted carbon of the double bond.

The product formed is

$$HOCH_2CH(CH_3)CH_2CH_3$$

which is 2-methylbutan-1-ol.

Thus, product A is a primary alcohol.

For reaction B, the reagents are $$Hg(OAc)_2,\ H_2O$$ followed by $$NaBH_4$$.

This is the oxymercuration-demercuration reaction.

Oxymercuration-demercuration follows Markovnikov addition without carbocation rearrangement. Therefore, the $$OH$$ group is introduced at the more substituted carbon of the double bond.

The product formed is

$$CH_3C(OH)(CH_3)CH_2CH_3$$

which is 2-methylbutan-2-ol.

Thus, product B is a tertiary alcohol.

Therefore,

$$A=\text{2-methylbutan-1-ol}$$

and

$$B=\text{2-methylbutan-2-ol}$$

Hence, the correct answer is

$$\text{Option A}$$

The amount of hydrogen consumed during hydrogenation can be used to determine the number of double bonds present in the hydrocarbon.

At STP, one mole of a gas occupies (22{,}400\ \text{mL}). Therefore, the number of moles of hydrogen used is

$$n_{H_2}=\frac{8.40}{22400}=3.75\times10^{-4}\ \text{mol}.$$

The molar mass of (C_{10}H_{16}) is

$$M=(10\times12)+(16\times1)=136\ \text{g mol}^{-1}.$$

Hence, the number of moles of hydrocarbon present is

$$n_{\text{hydrocarbon}}=\frac{0.017}{136}=1.25\times10^{-4}\ \text{mol}.$$

The number of moles of hydrogen consumed per mole of hydrocarbon is

$$\frac{n_{H_2}}{n_{\text{hydrocarbon}}}=\frac{3.75\times10^{-4}}{1.25\times10^{-4}}=3.$$

Since each mole of hydrogen saturates one double bond, the hydrocarbon contains

$$\boxed{3\ \text{double bonds}}.$$

This result is also consistent with the ozonolysis products obtained—acetone, two molecules of formaldehyde, and succinaldehyde—which collectively indicate the presence of three carbon-carbon double bonds in the original molecule.

Therefore, the hydrocarbon contains

$$\boxed{3\ \text{double bonds}}.$$

Hydrocarbon (X) on ozonolysis consumes one mole of O₃ per mole of X and gives one mole each of ethanal (CH₃CHO) and propanone (CH₃COCH₃).

Since each C=C double bond consumes one mole of O₃ in ozonolysis, the consumption of one mole of O₃ indicates the presence of a single C=C double bond in the molecule.

Ozonolysis cleaves this double bond into two carbonyl fragments:

$$>C=C< \xrightarrow{O_3} >C=O + O=C<$$

One fragment is ethanal (CH₃CHO), implying one side of the original double bond was CH₃-CH=, and the other fragment is propanone ((CH₃)₂CO), implying the other side was =C(CH₃)₂.

Rejoining these fragments across the double bond produces the structure

$$CH_3\text{-}CH=C(CH_3)_2$$

which corresponds to 2-methylbut-2-ene.

The molecular formula of this hydrocarbon is C₅H₁₀, and its molar mass is

$$M = 5 \times 12 + 10 \times 1 = 60 + 10 = 70 \text{ g mol}^{-1}$$

Therefore, the molar mass of hydrocarbon X is 70 g mol⁻¹.

During ozonolysis of a rigid Kekulé benzene structure, the $$\mathrm{C=C}$$ bonds are cleaved while the $$\mathrm{C-C}$$ single bonds remain intact.

The resulting two-carbon fragments depend on the substituents attached to the intact single bonds:

$$\mathrm{-CH-CH- \rightarrow Glyoxal\ (CHO-CHO)}$$

$$\mathrm{-C(CH_3)-CH- \rightarrow Methylglyoxal\ (CH_3-CO-CHO)}$$

$$\mathrm{-C(CH_3)-C(CH_3)- \rightarrow Dimethylglyoxal\ (CH_3-CO-CO-CH_3)}$$

For $$\mathrm{o\text{-}xylene}$$, the two Kekulé structures produce different fragment distributions.

In the first structure:

$$\mathrm{2\ Methylglyoxal + 1\ Glyoxal}$$

are formed.

In the second structure:

$$\mathrm{1\ Dimethylglyoxal + 2\ Glyoxal}$$

are formed.

Thus, shifting the localized double bonds changes the product distribution in $$\mathrm{o\text{-}xylene}$$.

For $$\mathrm{m\text{-}xylene}$$, $$\mathrm{p\text{-}xylene}$$, and toluene, both Kekulé structures give identical fragment distributions.

Correct Option: $$\mathrm{(C)}$$

We need to evaluate both statements about alkenes and alkanes.

Statement I: "The presence of weaker $$\pi$$-bonds make alkenes less stable than alkanes."

The $$\pi$$ bond in alkenes is formed by the lateral overlap of unhybridized p-orbitals. This lateral overlap is less effective than the head-on overlap in $$\sigma$$ bonds, making the $$\pi$$ bond weaker and more reactive. Due to the presence of this weaker $$\pi$$ bond, alkenes are more reactive (less stable) than alkanes, which contain only strong $$\sigma$$ bonds.

Statement I is correct.

Statement II: "The strength of the double bond is greater than that of carbon-carbon single bond."

The bond energy of a C=C double bond is approximately $$614$$ kJ/mol, while the bond energy of a C-C single bond is approximately $$347$$ kJ/mol. Since $$614 > 347$$, the double bond is indeed stronger than the single bond overall (even though the $$\pi$$ component individually is weaker than a $$\sigma$$ bond).

Statement II is correct.

Note: There is no contradiction between the two statements. The double bond as a whole is stronger than a single bond, but the presence of the weaker $$\pi$$ bond component makes the molecule more reactive at that site.

The correct answer is Option A: Both Statement I and Statement II are correct.

The starting compound is allyl acrylate, which contains two different carbon-carbon double bonds.

Since:

$$\mathrm{2\ equivalents\ of\ HBr}$$

are used, both double bonds undergo electrophilic addition.

The double bond on the allyl side:

$$\mathrm{-O-CH_2-CH=CH_2}$$

undergoes normal Markovnikov addition.

The proton adds to the terminal carbon to generate the more stable secondary carbocation.

$$\mathrm{-CH=CH_2 \longrightarrow -C^+H-CH_3}$$

The bromide ion then attacks the carbocation.

Thus:

$$\mathrm{-O-CH_2-CH=CH_2 \longrightarrow -O-CH_2-CH(Br)-CH_3}$$

The double bond on the acrylate side:

$$\mathrm{CH_2=CH-C(=O)-}$$

is conjugated with the carbonyl group.

The electron-withdrawing:

$$\mathrm{C=O}$$

group alters the regiochemistry of addition.

Addition occurs such that bromine attaches to the terminal:

$$\mathrm{\beta}$$

carbon.

Thus:

$$\mathrm{CH_2=CH-C(=O)- \longrightarrow Br-CH_2-CH_2-C(=O)-}$$

Combining both additions, the final major product is:

$$\mathrm{Br-CH_2-CH_2-C(=O)-O-CH_2-CH(Br)-CH_3}$$

The reaction shown is an oxymercuration-demercuration reaction.

This reaction converts alkenes into alcohols according to:

$$\mathrm{Markovnikov's\ Rule}$$

The reaction proceeds through a cyclic mercurinium ion intermediate instead of a free carbocation.

Hence, carbocation rearrangements such as hydride or methyl shifts do not occur.

In the first step, $$\mathrm{Hg(OAc)_2/H_2O}$$ adds across the double bond and water attacks the more substituted carbon.

In the second step, $$\mathrm{NaBH_4}$$ replaces the mercury group with hydrogen.

For $$\mathrm{3,3\text{-}Dimethyl\text{-}1\text{-}butene}$$, the hydroxyl group $$\mathrm{(-OH)}$$ attaches to the secondary carbon, while hydrogen attaches to the terminal carbon.

The adjacent $$\mathrm{tert\text{-}butyl}$$ group remains unchanged because no rearrangement occurs.

The final major product is:

$$\mathrm{3,3\text{-}Dimethyl\text{-}2\text{-}butanol}$$

Its structural formula is:

$$\mathrm{(CH_3)_3C-CH(OH)-CH_3}$$. Thus, the right option is A.

The reaction shown is an acid-catalysed alkylation reaction used industrially for preparation of highly branched alkanes.

The acid catalyst $$\mathrm{(H^+)}$$ protonates the double bond of $$\mathrm{2\text{-}Methylpropene}$$.

According to Markovnikov’s rule, protonation generates a highly stable tertiary carbocation $$\mathrm{(CH_3)_3C^+}$$.

This tert-butyl carbocation attacks another molecule of isobutylene, forming a larger $$\mathrm{C_8}$$ tertiary carbocation intermediate.

A hydride transfer then occurs from $$\mathrm{Isobutane}$$ to the carbocation.

This neutralises the intermediate and regenerates the tert-butyl carbocation.

The final major product formed is $$\mathrm{2,2,4\text{-}Trimethylpentane}$$, commonly called $$\mathrm{Isooctane}$$.

Its condensed structural formula is $$\mathrm{(CH_3)_3C-CH_2-CH(CH_3)_2}$$.

Both reactions involve cleavage of the terminal double bond in compound $$\mathrm{A}$$.

Cleavage produces the major ketone shown and a one-carbon by-product.

The by-product $$\mathrm{a}$$ on oxidation gives the substance produced by ants: $$\mathrm{HCOOH}$$.

Hence, by-product $$\mathrm{a}$$ must be: $$\mathrm{HCHO}$$ since oxidation of formaldehyde gives formic acid.

Therefore, reagent $$\mathrm{X}$$ must be a reagent capable of cleaving the terminal alkene to form formaldehyde.

Thus, the right option is D.

The starting alkene is:

$$\mathrm{3,3\text{-}dimethylbut\text{-}1\text{-}ene}$$

For formation of product $$\mathrm{A}$$, the reagents used are:

$$\mathrm{Hg(OAc)_2 \cdot H_2O \ followed\ by\ NaBH_4}$$

This reaction is:

$$\mathrm{Oxymercuration\text{-}Demercuration}$$

The reaction proceeds through a mercurinium ion intermediate and does not involve carbocation rearrangement.

Water attacks the more substituted carbon of the double bond.

Hence, the hydroxyl group attaches according to:

$$\mathrm{Markovnikov\ addition}$$

Therefore, product $$\mathrm{A}$$ is a Markovnikov product.

For formation of product $$\mathrm{B}$$, the reagents used are:

$$\mathrm{(BH_3)_2 \ followed\ by\ H_2O_2/OH^-}$$

This reaction is:

$$\mathrm{Hydroboration\text{-}Oxidation}$$

Boron attaches to the less substituted carbon due to steric and electronic factors.

During oxidation, boron is replaced by:

$$\mathrm{-OH}$$

Thus, the hydroxyl group finally appears on the less substituted carbon.

Hence, hydration occurs through:

$$\mathrm{Anti\text{-}Markovnikov\ addition}$$

Therefore, product $$\mathrm{B}$$ is an anti-Markovnikov product.

Thus:

$$\mathrm{A = Markovnikov\ Product}$$

$$\mathrm{B = Anti\text{-}Markovnikov\ Product}$$

The conversion requires transformation of a terminal alkene into an aldehyde.

Hydroboration-oxidation using $$\mathrm{BH_3}$$ followed by $$\mathrm{H_2O_2/OH^-}$$ converts the alkene into a primary alcohol through anti-Markovnikov addition.

$$\mathrm{R-CH=CH_2 \xrightarrow[BH_3]{H_2O_2/OH^-} R-CH_2-CH_2OH}$$

The primary alcohol is then oxidised using $$\mathrm{PCC}$$ which selectively converts primary alcohols into aldehydes without further oxidation to carboxylic acids.

$$\mathrm{R-CH_2OH \xrightarrow{PCC} R-CHO}$$

Therefore, the required reagents are:

$$\mathrm{BH_3,\ H_2O_2/OH^- \ followed\ by\ PCC}$$

Thus, the right option is going to be $$\mathrm{C}$$

The alkene reacted with $$\text{HBr}$$ acts as a nucleophile and attacks the proton ($$\text{H}^+$$). The double bond breaks to form the most stable possible carbocation intermediate.

The positive charge on the carbon directly attached to the benzene ring is a benzylic carbocation. It is exceptionally stable because the positive charge can be delocalized throughout the $$\pi$$-system of the aromatic ring via resonance.
Once the stable benzylic carbocation intermediate is formed, the bromide ion ($$\text{Br}^-$$) attacks the electron-deficient benzylic carbon.

Therefore, Option B is correct

We need to find compounds A and B based on the ozonolysis products.

For compound A:

A on ozonolysis gives ethane-1,2-dicarbaldehyde ($$OHC-CH_2-CH_2-CHO$$) and glyoxal ($$OHC-CHO$$).

In ozonolysis of cyclic dienes, each double bond is cleaved. For a six-membered ring with two double bonds:

Cyclohex-1,3-diene has double bonds at C1=C2 and C3=C4 positions. On ozonolysis followed by reductive workup ($$Zn/H_2O$$):

- Cleavage at C1=C2 and C3=C4 gives two fragments.

- The fragment from C2-C3 (single bond between the two double bonds) gives glyoxal: $$OHC-CHO$$

- The fragment from C4-C5-C6-C1 gives $$OHC-CH_2-CH_2-CHO$$ (ethane-1,2-dicarbaldehyde)

So compound A is cyclohex-1,3-diene.

For compound B:

B on ozonolysis gives 5-oxohexanal ($$CH_3-CO-CH_2-CH_2-CH_2-CHO$$).

Since only one product is formed, B must be a cyclic compound with one double bond. On ozonolysis, the ring opens at the double bond to give a single dicarbonyl compound.

5-oxohexanal has the structure: $$CH_3-CO-CH_2-CH_2-CH_2-CHO$$

The two carbonyl carbons (C1 aldehyde and C2 ketone at C5 position counted from the aldehyde end) were originally connected by a double bond in the ring.

Reconnecting these carbons: we get a 5-membered ring with a methyl substituent. The compound is 1-methylcyclopent-1-ene.

Verification: 1-methylcyclopent-1-ene has the double bond between C1 (bearing $$CH_3$$) and C2. Ozonolysis cleaves this to give a ketone at C1 ($$-CO-CH_3$$) and an aldehyde at C2 ($$-CHO$$), with the remaining chain $$-CH_2-CH_2-CH_2-$$ connecting them, giving $$CH_3CO-CH_2CH_2CH_2-CHO$$ = 5-oxohexanal. This is correct.

Therefore, A is cyclohex-1,3-diene and B is 1-methylcyclopent-1-ene.

Hence, the correct answer is Option C.

To determine the mass of bromine that reacts completely with 5.0 g of pent-1-ene, we first consider the addition reaction of pent-1-ene ($$CH_2{=}CHCH_2CH_2CH_3$$) with bromine across the double bond:
$$CH_2{=}CHCH_2CH_2CH_3 + Br_2 \rightarrow CH_2BrCHBrCH_2CH_2CH_3$$ Because one mole of pent-1-ene reacts with one mole of $$Br_2$$, the stoichiometric ratio is 1:1.

Next, the molar mass of pent-1-ene is calculated from its molecular formula $$C_5H_{10}$$:
$$M = 5(12) + 10(1) = 60 + 10 = 70 \text{ g/mol}$$ As a result, the number of moles in 5.0 g of pent-1-ene is:
$$n = \frac{5.0}{70} = \frac{1}{14} \text{ mol}$$.

Since the molar ratio between pent-1-ene and $$Br_2$$ is 1:1, the required amount of $$Br_2$$ is also $$\frac{1}{14}$$ mol. Given that the molar mass of $$Br_2$$ is 2 × 80 = 160 g/mol, the mass of $$Br_2$$ needed becomes:
$$\text{Mass of } Br_2 = \frac{1}{14} \times 160 = \frac{160}{14} = 11.4286 \text{ g}$$.

Finally, expressing this mass in the desired format gives:
$$11.4286 \text{ g} = 1142.86 \times 10^{-2} \text{ g} \approx 1143 \times 10^{-2} \text{ g}$$ therefore the correct answer is 1143.

The reactant is:

$$\mathrm{2,7\text{-}Dimethyl\text{-}2,6\text{-}octadiene}$$

In presence of acid $$\mathrm{(H^+)}$$, protonation of one double bond occurs according to Markovnikov’s rule, generating a tertiary carbocation.

$$\mathrm{(CH_3)_2C^+ - CH_2 - CH_2 - CH_2 - CH=C(CH_3)_2}$$

The remaining double bond attacks this carbocation intramolecularly, causing cyclization.

Formation of the $$\mathrm{5}$$-membered ring is favored because it generates a more stable tertiary carbocation intermediate.

Finally, deprotonation occurs according to Zaitsev’s rule, forming the most stable alkene.

The final major product contains only one double bond.

Each carbon atom involved in a double bond is $$\mathrm{sp^2}$$ hybridised.

Since one double bond contains exactly two carbon atoms, the number of $$\mathrm{sp^2}$$ hybridised carbon atoms is: $$\mathrm{2}$$

Correct Answer: $$\mathrm{2}$$

To determine the major product of this reaction, we need to analyze the electrophilic addition of hydrogen bromide $$\left(HBr\right)$$ to the conjugated diene, isoprene.

The word "excess" under the diene indicates that the diene is present in large excess relative to $$HBr$$. Therefore, only one equivalent of $$HBr$$ reacts with the diene, eliminating the possibility of double addition products. Hence, options A and D can be ruled out immediately.

The first step is protonation of the diene.

The proton adds to the terminal carbon of the methyl-substituted double bond because this generates the most stable intermediate, a tertiary allylic carbocation.

Adding the proton to the opposite end would produce only a secondary allylic carbocation, which is less stable.

The tertiary allylic carbocation formed is resonance stabilized. The positive charge is delocalized over two carbon atoms as shown below.

$$CH_3-\overset{+}{C}(CH_3)-CH=CH_2 \longleftrightarrow CH_3-C(CH_3)=CH-\overset{+}{CH_2}$$

The bromide ion can now attack either of these positively charged centers.

Attack at the tertiary carbon gives the 1,2-addition product, 3-bromo-3-methylbut-1-ene. This product forms rapidly because the tertiary carbon bears a greater share of the positive charge. However, the resulting alkene is only monosubstituted and is therefore less stable.

Attack at the terminal carbon gives the 1,4-addition product, 1-bromo-3-methylbut-2-ene. Although this product forms more slowly, it contains a trisubstituted internal double bond, making it thermodynamically more stable.

Under normal or higher temperature conditions, electrophilic addition to conjugated dienes is thermodynamically controlled. Therefore, the more stable 1,4-addition product is formed in greater amount.

Hence, the major product is 1-bromo-3-methylbut-2-ene, corresponding to Option (C).

We have trans-but-2-ene, that is $$\mathrm{CH_3\!-\!CH=CH\!-\!CH_3}$$ with the two $$\mathrm{CH_3}$$ groups on opposite sides of the double bond. During the electrophilic addition of bromine, the reaction always proceeds through a cyclic bromonium ion. The first step is

$$$\mathrm{CH_3\!-\!CH=CH\!-\!CH_3 + Br_2 \;\longrightarrow\;} \mathrm{CH_3\!-\!CH\!\!\overset{+}{\!\!\!\!Br}\!-\!CH\!-\!CH_3} \;+\;Br^-$$$

Now, bromide ion $$Br^-$$ opens the three-membered ring from the side opposite to the $$\overset{+}{Br}$$ (anti-addition). So the two bromine atoms always enter from opposite faces of the original double bond plane.

To see the stereochemical result clearly we assign the substituents around each sp3 carbon after addition. Let us choose the configuration in which on carbon-2 ($$C_2$$) the incoming nucleophilic $$Br^-$$ attacks from the front while on carbon-3 ($$C_3$$) it attacks from the rear. We can write the skeletal formula obtained:

$$$\mathrm{CH_3\,(front\;Br)\,CH\!-\!CH\,(back\;Br)\,CH_3}$$$

The molecule so obtained possesses two chiral centres, $$C_2$$ and $$C_3$$. We assign priorities by the C.I.P. rules (stated briefly: atom with higher atomic number gets higher priority). On each centre

Priority order: $$$Br(1) > CH_2CH_3(2) > CH_3(3) > H(4).$$$

Because the bromines are on opposite faces, one centre becomes $$R$$ while the other automatically becomes $$S$$. Explicitly, if $$C_2$$ is $$R$$, $$C_3$$ is $$S$$; the mirror image would have $$C_2$$ as $$S$$ and $$C_3$$ as $$R$$. However these two descriptions represent the same molecule: it has an internal mirror plane passing through the middle of the C-C bond and bisecting the molecule. Hence the compound is superimposable on its mirror image and is optically inactive. Such a stereoisomer is called a meso form.

If, in the bromonium opening step, we imagine the attack sequence reversed—i.e. bromide coming from the rear on $$C_2$$ and from the front on $$C_3$$—we draw

$$$\mathrm{CH_3\,(back\;Br)\,CH\!-\!CH\,(front\;Br)\,CH_3}$$$

Again, priority assignment shows $$C_2$$ to be $$S$$ and $$C_3$$ to be $$R$$, producing exactly the same $$RS$$ (meso) configuration as above. Thus the two routes give two structures that are identical, not different stereoisomers. No $$RR$$ or $$SS$$ combination can arise because anti-addition on a trans alkene never places the two bromines on the same face, a prerequisite for those diastereomeric pairs.

Therefore, electrophilic anti-addition of $$Br_2$$ to trans-but-2-ene furnishes only one unique stereoisomer, viz. the meso-2,3-dibromobutane. The phrase “2 identical mesomers” in the option list simply recognises that the reaction pathway can be visualised in two ways but both lead to the same meso product.

Hence, the correct answer is Option A.

We are told that the hydrocarbon ‘A’ has molecular formula $$C_4H_8$$. The general formula $$C_nH_{2n}$$ corresponds either to an alkene (one C=C bond) or to a cycloalkane (all C-C single bonds inside a ring). So ‘A’ must be either an alkene or a four-membered ring.

After hot acidic permanganate oxidation (written as $$KMnO_4/H^+$$) the only organic product obtained is ‘B’ with formula $$C_3H_6O$$. Hot $$KMnO_4$$ completely cleaves a C=C bond and converts each carbon of that double bond as follows:

• “Internal” carbon (no attached H) → Ketone.
• “Secondary” carbon (one attached H) → Carboxylic acid.
• “Terminal” vinyl carbon $$( =CH_2)$$ → $$CO_2$$.

Because only one organic fragment remains (three carbons long) and one carbon has been lost, the cleavage must have destroyed a terminal $$=CH_2$$ group, giving $$CO_2$$, while the other carbon of the double bond (which had no hydrogens) became the three-carbon ketone $$C_3H_6O$$.

Let us write the unknown alkene in the general form $$R_2C=CH_2$$. Applying the above rule:

$$$\displaystyle R_2C=CH_2 \;\xrightarrow[\text{hot}]{KMnO_4/H^+}\; R_2C=O\;+\;CO_2$$$

The ketone obtained has formula $$C_3H_6O$$. For a ketone the carbonyl carbon is already counted, so its alkyl parts must supply three carbons in total. The only way to supply three carbons with the pattern $$R_2C=O$$ is to make both $$R$$ groups methyl, because

$$$ (CH_3)_2C=O\;\text{ is acetone (propan-2-one), formula }C_3H_6O.$$$

Therefore

$$$ R = CH_3 \quad\Longrightarrow\quad A = (CH_3)_2C=CH_2 $$$

This structure is named 2-methylpropene.

Now we verify the second fact, ozonolysis. Ozonolysis also cleaves the C=C bond but under reductive work-up (Zn/H2O) stops at carbonyl compounds:

$$$ (CH_3)_2C=CH_2 \;\xrightarrow[\;Zn/H_2O\;]{O_3}\; (CH_3)_2C=O \;+\; H_2C=O $$$

The products are acetone $$(C_3H_6O)$$ and formaldehyde $$(CH_2O)$$. Thus the same compound ‘B’ (acetone) is again formed, satisfying the condition in the statement.

Any of the other options fails:

• $$\text{But-2-ene}\;(CH_3CH=CHCH_3)$$ would give two molecules of ethanal $$(C_2H_4O).$$
• $$\text{Cyclobutane}$$ and $$\text{1-methylcyclopropane}$$ contain no C=C bond; their oxidative cleavage patterns cannot furnish a single $$C_3H_6O$$ fragment while losing only one carbon.

Hence, the only structure consistent with all the data is 2-methylpropene.

Hence, the correct answer is Option A.

Oxidation: Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent) produce vicinal glycols. Decolorisation of KMnO4 solution is used as a test for unsaturation. 

Acidic potassium permanganate or acidic potassium dichromate oxidises alkenes to ketones and/or acids depending upon the nature of the alkene and the experimental conditions.

$$ \mathrm{CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4/H^+} 2CH_3COOH} $$

The reaction involves electrophilic addition of:

$$\mathrm{HBr/CCl_4}$$

to the alkene:

$$\mathrm{1\text{-}methyl\text{-}1\text{-}vinylcyclohexane}$$

The reaction follows Markovnikov addition and proceeds through a carbocation intermediate.

First, protonation of the double bond occurs.

According to Markovnikov’s rule, the proton adds to the terminal carbon:

$$\mathrm{=CH_2}$$

forming the more stable secondary carbocation.

$$\mathrm{-CH=CH_2 \longrightarrow -C^+H-CH_3}$$

This secondary carbocation lies adjacent to a quaternary carbon of the cyclohexane ring.

To form a more stable carbocation, a:

$$\mathrm{1,2\text{-}methyl\ shift}$$

takes place.

The methyl group migrates from the ring carbon to the carbocation centre.

This produces a highly stable tertiary carbocation on the ring carbon.

Simultaneously, the side chain becomes:

$$\mathrm{-CH(CH_3)_2}$$

(isopropyl group).

Ring expansion is less favoured because converting a six-membered ring into a seven-membered ring increases ring strain.

Finally, the bromide ion attacks the tertiary carbocation.

$$\mathrm{C^+ + Br^- \longrightarrow C-Br}$$

Thus, the major product formed is:

$$\mathrm{1\text{-}bromo\text{-}1\text{-}isopropylcyclohexane}$$

We have the alkene propene, whose condensed formula is $$\mathrm{CH_3-CH=CH_2}$$.

When bromine is added in the presence of excess water, the actual reagent in aqueous medium is written as $$\mathrm{Br_2/H_2O}$$. The overall transformation is the formation of a bromohydrin, that is, simultaneous addition of $$\mathrm{Br}$$ and $$\mathrm{OH}$$ across the C=C double bond.

The first elementary step is electrophilic attack of $$\mathrm{Br_2}$$ on the double bond. The $$\pi$$ electrons of the C=C bond act as a nucleophile and displace $$\mathrm{Br^-}$$ from $$\mathrm{Br_2}$$, producing a cyclic bromonium ion. This step may be written as

$$\mathrm{CH_3-CH=CH_2 \;+\; Br_2 \;\longrightarrow\; [CH_3-CH-CH_2]^+}$$
with $$\mathrm{Br}$$ bridging the two carbons, and a free $$\mathrm{Br^-}$$ in the medium.

Now water, which is abundant and a better nucleophile than $$\mathrm{Br^-}$$ because of its higher concentration, attacks this three-membered bromonium ring. A key point is that the attack is Markovnikov. Before using that rule, we state it explicitly:

Markovnikov rule: In an electrophilic addition to an alkene, the nucleophilic part of the reagent attaches to the more substituted carbon atom of the double bond, while the electrophilic part attaches to the less substituted carbon.

In the bromonium ion derived from propene, carbon 2 (the middle carbon) is secondary, whereas carbon 1 (the terminal carbon) is primary. According to Markovnikov rule, the nucleophile $$\mathrm{H_2O}$$ must therefore attack carbon 2. We write the attack step as

$$\mathrm{[CH_3-CH-CH_2]^+ + H_2O \;\longrightarrow\; CH_3-CH(OH)-CH_2Br + H^+}$$

The proton released is immediately taken up by $$\mathrm{Br^-}$$ or by another water molecule, regenerating neutral medium; the essential organic product remains:

$$\mathrm{CH_3-CH(OH)-CH_2Br}$$

Let us name this compound systematically. We number the longest carbon chain from either end, but we keep the lowest possible locant for the functional group with priority, which here is the hydroxyl group. Thus carbon 2 bears the $$\mathrm{OH}$$ group and carbon 1 bears the $$\mathrm{Br}$$ atom:

$$\mathrm{HO-CH(CH_3)-CH_2Br}$$ = 1-bromopropan-2-ol

So, the experimental observation given in the Assertion is completely correct.

Now we examine the Reason. It says that the attack of water on the bromonium ion follows Markovnikov rule and therefore gives the same product, 1-bromopropan-2-ol. We have just followed the mechanism and indeed found that application of Markovnikov orientation is precisely why $$\mathrm{OH}$$ ends up on C 2 and $$\mathrm{Br}$$ on C 1. Thus the Reason is also true and it is the exact explanation of the Assertion.

In the list of options, the statement “Both (A) and (R) are true and (R) is the correct explanation of (A)” corresponds to Option C (the third option).

Hence, the correct answer is Option C.

The molecule is 3,3-dimethylbut-1-ene. When this alkene undergoes a typical electrophilic addition reaction, such as hydration with an acid catalyst or reaction with a hydrogen halide like $$HBr$$ or $$HI$$, the process is entirely dictated by the formation and rearrangement of a carbocation intermediate to achieve maximum thermodynamic stability.

The reaction begins when the electron-rich pi bond of the alkene attacks an electrophile, typically a proton ($$H^+$$). According to Markovnikov's rule, the proton adds to the terminal carbon atom (the $$CH_2$$ group at the end of the double bond). This specific placement is favored because it leaves the resulting positive charge on the adjacent, more substituted internal carbon atom. This initial step forms a secondary carbocation, where the positively charged carbon is directly bonded to two other carbon groups.

While secondary carbocations have a moderate level of stability, this specific molecule has a structural feature that allows it to reach a much lower, more stable energy state. Immediately adjacent to the positively charged secondary carbon is a quaternary carbon atom, which is densely packed with three attached methyl groups. Because the molecule is dynamic, it will spontaneously rearrange to relieve energy. An entire methyl group, along with its bonding pair of electrons, physically migrates from the quaternary carbon over to the adjacent positively charged carbon. This fundamental organic mechanism is known as a 1,2-methyl shift.

When the methyl group moves, the positive charge shifts back to the carbon that just lost the group. The newly formed intermediate is now a tertiary carbocation, meaning the positively charged carbon is bonded to three other carbon atoms. This tertiary carbocation is significantly more stable than the initial secondary one. This massive boost in stability is primarily due to hyperconjugation, an effect where the electron density from the numerous adjacent carbon-hydrogen bonds partially overlaps with and stabilizes the empty p-orbital of the positive carbon.

Because the tertiary carbocation is so exceptionally stable, the 1,2-methyl shift happens very rapidly. The final step of the reaction occurs when a nucleophile in the solution, such as a bromide ion or a water molecule, attacks this stable tertiary center, resulting in a rearranged major product rather than a simple addition product. Thus, the correct option is B.

For pentene ($$\text{C}_5\text{H}_{10}$$), we need to count all acyclic structural isomers including geometrical (cis-trans) isomers.

The possible structures are: (1) pent-1-ene: $$\text{CH}_2{=}\text{CH-CH}_2\text{-CH}_2\text{-CH}_3$$, (2) trans-pent-2-ene: $$\text{CH}_3\text{-CH}{=}\text{CH-CH}_2\text{-CH}_3$$, (3) cis-pent-2-ene, (4) 2-methylbut-1-ene: $$\text{CH}_2{=}\text{C(CH}_3\text{)-CH}_2\text{-CH}_3$$, (5) 3-methylbut-1-ene: $$\text{CH}_2{=}\text{CH-CH(CH}_3\text{)}_2$$, and (6) 2-methylbut-2-ene: $$\text{CH}_3\text{-C(CH}_3\text{)}{=}\text{CH-CH}_3$$.

Pent-1-ene has no geometrical isomers because one carbon of the double bond bears two hydrogen atoms. Pent-2-ene has both cis and trans forms since each doubly-bonded carbon has two different groups. 2-methylbut-1-ene has no geometrical isomers (terminal $$=\text{CH}_2$$). 3-methylbut-1-ene has no geometrical isomers (terminal $$=\text{CH}_2$$). 2-methylbut-2-ene has no geometrical isomers because one carbon bears two methyl groups.

The total count is $$6$$.

We are told that compound $$A$$ is a chloro-compound and that on ozonolysis followed by hydrolysis it produces only aldehydes. Recall that:

Whenever an alkene $$\,{\displaystyle R_1CH=CHR_2}\,$$ is treated with ozone and then hydrolysed, the double bond is split and each carbon of the double bond becomes the carbon of a carbonyl group. If every carbon of the original double bond possessed at least one hydrogen, both carbonyl products will be aldehydes. Hence we infer that $$A$$ is an alkene that contains one chlorine atom and possesses exactly one C=C double bond.

Let its molecular formula be written as $$\mathrm{C_nH_mCl}$$. We next determine its molar mass from the vapour-density data.

At STP, one mole of any gas occupies $$22.4\ \text{L}$$. Here,

Volume of vapour obtained  =  $$448\ \text{mL}=0.448\ \text{L}$$.

So, number of moles present is

$$n_{\text{mole}}=\dfrac{\text{volume}}{22.4\ \text{L mol}^{-1}} =\dfrac{0.448}{22.4}=0.020\ \text{mol}.$$

The given mass that produced this vapour is $$1.53\ \text{g}$$, hence the molar mass $$M$$ of $$A$$ is

$$M=\dfrac{\text{mass}}{\text{moles}} =\dfrac{1.53\ \text{g}}{0.020\ \text{mol}} =76.5\ \text{g mol}^{-1}.$$

Writing the molar mass in terms of the unknown numbers of atoms, we have

$$M = 12n + m + 35.5,$$

because one carbon contributes $$12\ \text{u}$$, one hydrogen gives $$1\ \text{u}$$, and one chlorine contributes $$35.5\ \text{u}$$ to the molar mass.

Equating this to the experimentally obtained value,

$$12n + m + 35.5 = 76.5 \quad\Longrightarrow\quad 12n + m = 41.$$

Next, we employ the idea of the degree of unsaturation (also called the index of hydrogen deficiency, IHD). For a straight-chain saturated chloro-alkane with $$n$$ carbons we would expect $$2n+2$$ hydrogens. Each ring, double bond or halogen atom reduces this count by the following rule:

IHD formula: $$\text{IHD}=\dfrac{2n+2 - (m + X)}{2},$$ where $$X$$ is the number of halogen atoms.

In our case there is one Cl atom, so $$X=1$$. Because we have already concluded that there is one C=C double bond and no rings, $$\text{IHD}=1$$. Therefore

$$1=\dfrac{2n+2 - (m+1)}{2} \;\Longrightarrow\; 2=2n+2 - m -1 \;\Longrightarrow\; m = 2n -1.$$

We now possess two simultaneous relations, namely

$$12n + m = 41,$$

$$m = 2n - 1.$$

Substituting the second into the first,

$$12n + (2n - 1) = 41 \;\Longrightarrow\; 14n - 1 = 41 \;\Longrightarrow\; 14n = 42 \;\Longrightarrow\; n = 3.$$

Thus the molecule contains exactly $$3$$ carbon atoms. Immediately we can verify the hydrogen count:

$$m = 2n - 1 = 2(3) - 1 = 5,$$

which gives the molecular formula $$\mathrm{C_3H_5Cl}$$ and satisfies the molar mass check:

$$M = 12(3) + 5 + 35.5 = 36 + 5 + 35.5 = 76.5\ \text{g mol}^{-1},$$

in full agreement with the experimental value.

Hence, the correct answer is Option 3.

Step 1: Analyze Hydrogen Absorption

The problem states that X absorbs two moles of hydrogen ($$\text{H}_2$$) during catalytic hydrogenation. This directly tells us that X contains two carbon-carbon double bonds (a diene).

Step 2: Work Backwards from the Final Product B

Product B is given as 3-oxo-hexanedicarboxylic acid (commonly known as 3-oxohexanedioic acid). Its chemical structure is:

$$\text{HOOC--CH}_2\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_2\text{--COOH}$$

Product B is obtained by treating intermediate A with Tollens' reagent ($$[\text{Ag}(\text{NH}_3)_2]^+$$). Tollens' reagent is a mild oxidizing agent that selectively oxidizes aldehyde groups ($$-\text{CHO}$$) into carboxylic acids ($$-\text{COOH}$$) without altering ketones.

Therefore, intermediate A must be a keto-dialdehyde with the following structure:

$$\text{OHC--CH}_2\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_2\text{--CHO}$$

Step 3: Deduce the Structure of X via Ozonolysis Cleavage

Intermediate A is formed via the reductive ozonolysis ($$\text{O}_3\text{, then Zn/H}_2\text{O}$$) of hydrocarbon X. Ozonolysis cleaves double bonds and inserts oxygen atoms to form carbonyl compounds.

To reconstruct X, we connect the carbonyl carbons of A where the cleavage took place. Since one of the carbonyls is a ketone inside a continuous chain flanked by two aldehydes, X must be a cyclic system containing one endocyclic double bond and one exocyclic methylene ($$=\text{CH}_2$$) group:

• Connecting the aldehyde groups at positions 1 and 6 forms a 6-membered ring.
• The ketone at position 3 implies that an exocyclic double bond ($$=\text{CH}_2$$) was attached there before cleavage (yielding a side product of formaldehyde, $$\text{HCHO}$$).

This corresponds perfectly to 4-methylenecyclohex-1-ene.

Ozonolysis Mechanism Verification:

$$\text{4-methylenecyclohex-1-ene} \xrightarrow{\text{(i) } \text{O}_3 \text{ / (ii) } \text{Zn, H}_2\text{O}} \text{OHC--CH}_2\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_2\text{--CHO} + \text{HCHO}$$

$$\text{OHC--CH}_2\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_2\text{--CHO} \xrightarrow{[\text{Ag}(\text{NH}_3)_2]^+} \text{HOOC--CH}_2\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_2\text{--COOH}$$

Conclusion:

The structural features match 4-methylenecyclohex-1-ene, which corresponds to the layout presented in Option C.

Answer: Option C

For structural isomers or compounds sharing the exact same molecular formula, the heat of combustion is inversely proportional to the thermodynamic stability of the molecule:

$$\text{Heat of Combustion} \propto \frac{1}{\text{Stability}}$$

A more stable molecule sits at a lower potential energy level, thereby releasing less energy (heat) when completely burned into $$\text{CO}_2$$ and $$\text{H}_2\text{O}$$.

Factors Influencing Alkadiene Stability:

1. Conjugation: Conjugated dienes (alternating double-single-double bonds) are significantly more stable than isolated dienes due to continuous $$\pi$$-electron delocalization.
2. Steric Geometries (Cis vs. Trans): Trans-alkenes experience less steric hindrance than cis-alkenes, making the trans,trans-isomer more stable than cis,trans or cis,cis arrangements.

Evaluating the Alkadienes:

• Structure (a): Trans,Trans-Conjugated Diene

  This structure has maximum conjugation along with the most sterically favorable trans,trans geometry. Because it has the lowest internal potential energy, it is the most stable and releases the least heat of combustion.




• Structure (b): Cis,Trans-Conjugated Diene

  This structure maintains the stable conjugated network, but contains a less favorable cis-alkene alignment creating minor steric strain. It possesses intermediate stability and intermediate heat of combustion.




• Structure (c): Cis,Cis-Conjugated or Non-Conjugated / Sterically Strained Diene

  This configuration experiences severe steric hindrance due to the crowded cis,cis arrangement of groups. Being the least stable isomer, it sits at the highest potential energy level and releases the highest heat of combustion.

Conclusion:

Reversing the overall stability trend ($$\text{a} > \text{b} > \text{c}$$) gives the correct increasing order of heat of combustion:

$$\mathbf{(a) < (b) < (c)}$$

Answer: Option A — (a) < (b) < (c)

To determine the major product formed in the reaction of $$CH_2=CHCH(CH_3)_2$$ with $$HBr$$, we apply Markovnikov’s rule while considering the possibility of carbocation rearrangement.

The reaction proceeds through an electrophilic addition mechanism. In the first step, the proton ($$H^+$$) from $$HBr$$ adds to the double bond. According to Markovnikov’s rule, the proton attaches to the terminal carbon atom, generating the more stable secondary carbocation.

The protonation step is represented as

$$CH_2=CH-CH(CH_3)_2 \xrightarrow{H^+} CH_3-CH^+-CH(CH_3)_2.$$

The resulting secondary carbocation is adjacent to a carbon atom bearing a hydrogen atom. Consequently, a 1,2-hydride shift occurs, producing a more stable tertiary carbocation.

The rearrangement is represented as

$$CH_3-CH^+-CH(CH_3)_2 \longrightarrow CH_3-CH_2-C^+(CH_3)_2.$$

Finally, the bromide ion ($$Br^-$$) attacks the tertiary carbocation to give the final product,

$$CH_3CH_2C(Br)(CH_3)_2.$$

Thus, owing to the formation of the more stable tertiary carbocation through hydride shift, the major product of the reaction is

$$\boxed{CH_3CH_2C(Br)(CH_3)_2},$$

which corresponds to option D.

• Step 1: Protonation and Formation of the Carbocation

  The starting material is a cyclopentane ring with a tertiary alcohol substituent group (specifically containing two methyl groups at the alpha position, as derived from the structures). Protonation of the hydroxyl group ($$\text{-OH}$$) by the acid turns it into an excellent leaving group ($$\text{-OH}_2^+$$). Water is eliminated, leaving behind a secondary carbocation outside the ring.

• Step 2: Wagner-Meerwein Ring Expansion

  The initial carbocation is adjacent to a strained five-membered cyclopentane ring. To relieve angle strain and form a more stable ring, a 1,2-alkyl shift occurs where one of the C-C bonds of the cyclopentane ring breaks and migrates to the carbocation carbon. This expands the five-membered ring into a much more stable six-membered cyclohexyl carbocation ring system.

• Step 3: Deprotonation to Form the Most Stable Alkene

  Finally, a water molecule acts as a weak base and removes a proton ($$\text{H}^+$$) from the adjacent carbon atom to establish a double bond. Following Zaitsev's Rule, elimination happens to yield the most highly substituted, conjugated, and thermodynamically stable tetrasubstituted alkene possible.

Conclusion:

The ring expansion transforms the cyclopentane framework into a cyclohexane ring containing a double bond between two carbons that both bear a methyl group. This corresponds to 1,2-dimethylcyclohexene, which is illustrated as the correct option in Option B.

Answer: Option B

• Step 1: Free Radical Substitution

  When cyclohexane is treated with bromine in the presence of light ($$\text{Br}_2/h\nu$$), it undergoes photochemical free-radical bromination to form bromocyclohexane:

  $$\text{Cyclohexane} \xrightarrow{\text{Br}_2/h\nu} \text{Bromocyclohexane}$$



• Step 2: Dehydrohalogenation (Formation of A)

  When bromocyclohexane is treated with a strong, hot base like alcoholic potassium hydroxide ($$\text{KOH (alc.)}$$), it undergoes an $$\text{E}2$$ elimination reaction. The base abstracts a adjacent proton along with the loss of the bromide leaving group to introduce a carbon-carbon double bond:

  $$\text{Bromocyclohexane} \xrightarrow{\text{KOH (alc.)}} \text{Cyclohexene (A)}$$

Validation via Subsequent Steps:

We can confirm that A is cyclohexene by tracking the remaining reactions shown in the diagram:

1. Reductive Ozonolysis: Treating cyclohexene with $$\text{O}_3$$ followed by $$\text{Me}_2\text{S}$$ oxidatively cleaves the double bond down into a dicarbonyl compound, hexanedial (adipaldehyde).
2. Intramolecular Aldol Condensation: Heating hexanedial with aqueous base ($$\text{NaOH (aq.), }\Delta$$) triggers a ring-closing aldol pathway, yielding the final 5-membered cyclic conjugated product, cyclopent-1-ene-1-carbaldehyde.

Answer: Cyclohexane

• Step 1: Hydroboration-Oxidation (Formation of Alcohol)

  The starting alkene undergoes hydroboration-oxidation using $$\text{B}_2\text{H}_6$$ followed by alkaline $$\text{H}_2\text{O}_2$$. This reaction results in the net anti-Markovnikov addition of water across the double bond with syn-stereochemistry, yielding the secondary alcohol:

  $$\text{2,4-dimethylhexan-3-ol} \quad \left[\text{H}_3\text{C--CH}\left(\text{CH}(\text{CH}_3)_2\right)\text{--CH(OH)--CH}_2\text{--CH}_3\right]$$



• Step 2: Carbocation Formation via Dehydration

  Treating this alcohol with dilute sulfuric acid and heat ($$\text{dil. H}_2\text{SO}_4/\Delta$$) leads to protonation of the hydroxyl group followed by the loss of a water molecule ($$-\text{H}_2\text{O}$$). This generates a secondary carbocation:

  $$\text{H}_3\text{C--CH}\left(\text{CH}(\text{CH}_3)_2\right)\text{--}\text{CH}^+\text{--CH}_2\text{--CH}_3$$



• Step 3: Carbocation Rearrangement (1,2-Hydride Shift)

  To achieve greater stability, a 1,2-hydride shift ($$\text{H}^-$$ shift) occurs from the adjacent tertiary carbon to the positively charged secondary carbon atom. This rearranges the intermediate into a highly stable tertiary carbocation.




• Step 4: Deprotonation to the Saytzeff Product (Formation of B)

  Finally, elimination of a proton ($$-\text{H}^+$$) occurs from the adjacent carbon following Saytzeff's rule to form the most highly substituted, conjugated, and thermodynamically stable alkene as the major product:

  $$\text{Product B} = \text{2,3-dimethylhex-2-ene} \quad \left[(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3\right]$$

Conclusion:

The reaction sequence goes through an anti-Markovnikov alcohol formation, followed by a carbocation generation that undergoes a thermodynamic hydride shift to yield the most substituted alkene (2,3-dimethylhex-2-ene).

Answer: Option B — 2,3-dimethylhex-2-ene

First we recall the general statement of Markovnikov’s rule: on electrophilic addition of $$\mathrm{HX}$$ to an unsymmetrical alkene, the proton $$\left(\mathrm{H^+}\right)$$ attaches to that double-bonded carbon which already bears the larger number of hydrogen atoms, because this mode of attack generates the more stable carbocation. Put algebraically, if the alkene is $$\mathrm{C_a=C_b}$$, then in the Markovnikov pathway $$\mathrm{H^+}$$ goes to the carbon rich in hydrogens and the intermediate carbocation appears on the other carbon.

The stability of the intermediate carbocation is decided mainly by two electronic effects: the inductive effect $$\left(-I\ \text{or}\ +I\right)$$ and the mesomeric or resonance effect $$\left(-M\ \text{or}\ +M\right)$$ of the substituent already attached to that carbon. A group which donates electron density by $$+I$$ or $$+M$$ stabilises the positive charge, whereas a group which withdraws electron density by a strong $$-I$$ effect destabilises it. Hence, if a substituent has a very powerful $$-I$$ effect and lacks any meaningful $$+M$$ donation, the carbocation situated adjacent to it becomes highly unstable, and the reaction may proceed by the opposite orientation, i.e. the anti-Markovnikov mode.

Now we analyse each given alkene one by one.

Option A : $$\mathrm{CH_3O{-}CH=CH_2}$$

The methoxy group $$\left(\mathrm{-OCH_3}\right)$$ exerts a $$+M$$ resonance donation because of the lone pairs on oxygen. In the Markovnikov pathway the proton adds to the terminal $$\mathrm{CH_2}$$ carbon and the carbocation appears on the carbon bearing $$\mathrm{-OCH_3}$$:

$$\mathrm{CH_3O{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ CH_3O{-}\overset{+}{\mathrm{CH}}{-}CH_3$$

The positive charge is strongly stabilised by resonance with the oxygen lone pair, so this path is preferred. Hence the product is Markovnikov, not anti-Markovnikov.

Option B : $$\mathrm{Cl{-}CH=CH_2}$$

Chlorine possesses a weak $$+M$$ effect as well as a $$-I$$ effect. When the carbocation is placed on the chlorine-bearing carbon (Markovnikov orientation) the lone pairs of chlorine can offer some resonance stabilisation:

$$\mathrm{Cl{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ Cl{-}\overset{+}{\mathrm{CH}}{-}CH_3}$$

The $$+M$$ relief outweighs the inductive withdrawal; consequently the Markovnikov product remains major. No anti-Markovnikov preference is observed.

Option C : $$\mathrm{H_2N{-}CH=CH_2}$$

The amino group $$\left(\mathrm{-NH_2}\right)$$ is a very strong $$+M$$ donor. Exactly as in Option A, the Markovnikov pathway places the positive charge adjacent to the donor nitrogen, which is highly stabilising by resonance:

$$\mathrm{H_2N{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ H_2N{-}\overset{+}{\mathrm{CH}}{-}CH_3}$$

Therefore this alkene also obeys Markovnikov’s rule.

Option D : $$\mathrm{CF_3{-}CH=CH_2}$$

The trifluoromethyl group $$\left(\mathrm{-CF_3}\right)$$ is one of the strongest electron-withdrawing groups known; it shows an intense $$-I$$ effect and has no $$+M$$ resonance donation, because the fluorines are not conjugated with the carbocation centre through a lone-pair bridge as chlorine can be. Examine the two possible orientations:

Markovnikov attack (proton to the terminal $$\mathrm{CH_2}$$ carbon):

$$\mathrm{CF_3{-}CH=CH_2\ \overset{H^+}{\underset{\text{Markovnikov}}{\rightarrow}}\ CF_3{-}\overset{+}{\mathrm{CH}}{-}CH_3}$$

The carbocation $$\mathrm{CF_3{-}\overset{+}{CH}{-}CH_3}$$ is directly attached to the very powerful $$-I$$ group $$\mathrm{CF_3}$$, so the positive charge is pulled even further electron-deficient and becomes extremely unstable.

Anti-Markovnikov attack (proton to the $$\mathrm{CF_3{-}CH}$$ carbon):

$$\mathrm{CF_3{-}CH=CH_2\ \overset{H^+}{\underset{\text{anti-M}}{\rightarrow}}\ CF_3{-}CH_2{-}\overset{+}{\mathrm{CH}}_2$$

Now the positive charge resides on the terminal carbon, which is not adjacent to the $$\mathrm{-CF_3}$$ group, so it does not suffer the strong inductive withdrawal. Even though both carbocations are primary, the one formed in the anti-Markovnikov pathway is appreciably more stable because it is farther from the electron-withdrawing group. Therefore the reaction proceeds predominantly by this orientation, giving the anti-Markovnikov product $$\mathrm{CF_3{-}CH_2{-}CH_2Cl}$$.

Thus, among the four alkenes, only the trifluoromethyl-substituted alkene in Option D furnishes the anti-Markovnikov addition product with $$\mathrm{HCl}$$.

Hence, the correct answer is Option D.

Elimination reaction takes place along with formation of double bond towards the side where conjugation is possible. Next, oxidation takes place and the C-C bonds break at the site of unsaturation getting replaced by Oxygen.

Cl2 in the presence of H2O results in HOCl formation, which disassociates into OH- and Cl+. Hence, after formation of Carbocation at the middle carbon option B would be formed as product.

Mechanism of Allylic Bromination

When an alkene like but-1-ene (compound E) reacts with bromine (Br2) in the presence of light (hv), it undergoes a free radical substitution reaction rather than electrophilic addition.

The ease of replacing a hydrogen atom depends entirely on the stability of the free radical intermediate formed after the hydrogen is abstracted:

• Allylic free radicals are exceptionally stable because the unpaired electron is delocalized via resonance over the adjacent carbon-carbon double bond.
• Saturated alkyl radicals (secondary or primary) that are not adjacent to a double bond lack this resonance stabilization and are much harder to form.

Evaluating the Carbons in Compound (E)

Let's look at the types of hydrogens attached to each labeled position in CH3(delta)-CH2(gamma)-CH(beta)=CH2(alpha):

• alpha Position (CH2=): These are vinylic hydrogens. Breaking a vinylic C-H bond forms an unstable vinylic radical.
• beta Position (=CH-): These are also vinylic hydrogens, which are highly resistant to homolytic cleavage.
• gamma Position (-CH2-): These carbons are directly adjacent to the sp2 carbon of the double bond. Therefore, these are allylic hydrogens. Abstraction of a hydrogen from this position forms a resonance-stabilized allylic radical.
• delta Position (-CH3): These are standard primary aliphatic hydrogens, which form an unstable primary radical and are much less reactive.

We are given the alkene but-2-ene, whose condensed structural formula is $$\mathrm{CH_3-CH=CH-CH_3}$$. In an oxidation reaction with hot, alkaline potassium permanganate ($$\mathrm{KMnO_4 + OH^-}$$), followed by acidification (addition of $$\mathrm{H^+}$$), the double bond undergoes oxidative cleavage.

First, recall the general rule (state the formula):

$$\text{Alkene}\;(\mathrm{R_1-CH=CH-R_2})\; +\;[\mathrm{O}]\;\xrightarrow{\text{hot } \mathrm{KMnO_4}/\mathrm{OH^-}} \; \text{cleaves to}\; \mathrm{R_1-COO^-} + \mathrm{R_2-COO^-}$$

After this oxidative cleavage step, the basic medium gives carboxylate ions. Subsequent acidification converts each carboxylate ion into its corresponding carboxylic acid:

$$\mathrm{R-COO^- + H^+ \;\longrightarrow\; R-COOH}$$

Now we apply the rule to but-2-ene. Identify $$R_1$$ and $$R_2$$ on either side of the double bond:

We have $$\mathrm{CH_3-CH=CH-CH_3}$$, so $$R_1 = \mathrm{CH_3}$$ attached to the left carbon and $$R_2 = \mathrm{CH_3}$$ attached to the right carbon. Both sides are identical.

Performing oxidative cleavage:

$$ \mathrm{CH_3-CH=CH-CH_3} \;+\;[\mathrm{O}] \;\xrightarrow{\text{hot } \mathrm{KMnO_4}/\mathrm{OH^-}} \; \underbrace{\mathrm{CH_3-COO^-}}_{\text{acetate ion}} \;+\; \underbrace{\mathrm{CH_3-COO^-}}_{\text{another acetate ion}} $$

Next we acidify (add $$\mathrm{H^+}$$):

$$ \mathrm{CH_3-COO^- + H^+ \;\longrightarrow\; CH_3-COOH} $$

The same step occurs for the second acetate ion. So we finally obtain two molecules of acetic acid:

$$ \mathrm{CH_3-COOH + CH_3-COOH} $$

No aldehyde or diol is produced because the oxidative cleavage under these strong conditions fully oxidises each alkene carbon to the carboxylic acid level.

Therefore, the product set is “2 molecules of $$\mathrm{CH_3COOH}$$”, which matches Option A.

Hence, the correct answer is Option A.

1. The reaction begins with the generation of an initial carbocation intermediate (typically via the protonation and loss of a leaving group like a hydroxyl group or via electrophilic addition to an alkene). The carbocation forms.




2. Nucleophilic Attack by Chloride Ion ($$\text{Cl}^\ominus$$):

   The chloride ion ($$\text{Cl}^\ominus$$) acts as a nucleophile. It attacks the electron-deficient sp2 hybridized carbocation carbon:

   $$\text{R}^\oplus + \text{Cl}^\ominus \rightarrow \text{R--Cl}$$

Analyzing the Final Product Structure:

Based on the provided mechanism diagram:

• The parent framework consists of a fused bicyclic ring system (a 6-membered ring fused to a 5-membered or 6-membered ring).
• An exocyclic group contains an isopropyl-like unit where the rearrangement takes place.
• The chloride ion attacks the tertiary carbon center bearing the two methyl ($$\text{--CH}_3$$) groups, resulting in a tertiary alkyl chloride derivative as the final stable major product: $$\text{--C(Cl)(CH}_3)_2$$.

Conclusion:

The major product is formed via a thermodynamic pathway favored by the generation of a stable tertiary carbocation intermediate, followed by direct nucleophilic capture by the chloride ion.

Answer: The option corresponding to the structure containing the chlorine atom attached to the tertiary carbon carrying the two methyl groups on the bicyclic framework (Option D).

We have to find out which reagent converts an alkyne ($$R-C \equiv C-R'$$) into a trans (that is, $$E$$-) alkene. In organic chemistry two standard partial‐reduction methods are famous:

1. The Lindlar catalyst, written as $$H_{2} / Pd-CaCO_{3} \ or\ Pd/BaSO_{4} \ (poisoned)$$, gives syn addition of hydrogen. Because both hydrogens add from the same face, the product is the cis (or $$Z$$-) alkene.

2. The dissolving-metal reduction, usually $$Na / liq. NH_{3}$$ (or $$Li / liq. NH_{3}$$), proceeds through radicals and delivers the two hydrogens from opposite faces (anti-addition). Anti-addition produces the trans (or $$E$$-) alkene.

Let us write the key steps for the dissolving-metal (Birch) reduction of an internal alkyne:

$$ R-C \equiv C-R' \;\overset{Na,\ liq.\ NH_{3}}{\underset{\text{-33 °C}}{\rightarrow}} R{-}\underset{\text{trans}}{\text{CH}=\text{CH}}{-}R' $$

Step-wise mechanism (shown only to keep every algebraic/chemical step explicit):

1. Electron addition from $$Na$$ gives a radical anion: $$R-C \equiv C-R' + e^- \to R-C \equiv C-R'^{\bullet-}$$

2. Protonation by liquid ammonia ($$NH_{3}$$ is the proton source) forms a vinyl radical: $$R-C \equiv C-R'^{\bullet- + NH3 \to R-C(H)=C^{\bullet}-R' + NH2^-}$$

3. A second electron addition gives a vinyl anion: $$R-C(H)=C^{\bullet-R' + e^- \to R-CH= C^{-}-R'}$$

4. Final protonation furnishes the trans-alkene: $$R-CH= C^{--R' + NH3 \to R-CH=CH-R' + NH2^-}$$

Because the two protonation steps take place from opposite sides of the molecule, the overall result is anti-addition, giving the trans product.

Now we test each option given in the question:

A. $$Sn/HCl$$ - this reagent is used for reducing nitro groups to amines, not for selective alkyne reduction. It does not give trans-alkenes.

B. $$H_{2}-Pd/C, BaSO_{4}$$ - this is exactly the Lindlar (poisoned) catalyst discussed above; it furnishes cis alkenes, not trans.

C. $$NaBH_{4}$$ - sodium borohydride is a mild hydride donor; it reduces aldehydes and ketones, but it does not reduce carbon-carbon triple bonds at all.

D. $$Na / liq. NH_{3}$$ - as explained, this is the dissolving-metal (Birch) reduction that converts an alkyne specifically into a trans alkene.

So, the only reagent that satisfies the requirement is Option D.

Hence, the correct answer is Option D.

We have the alkene 3-methyl-pent-2-ene, whose condensed form may be written as $$CH_3-CH=C(CH_3)-CH_2-CH_3$$. The double bond is between carbon-2 and carbon-3.

In the presence of peroxide, the addition of $$HBr$$ follows the anti-Markovnikov (peroxide) rule. Stated in words, this rule says that in a radical addition of $$HBr$$, the bromine radical adds first to the double-bond carbon that will give the more stable carbon radical; afterwards that radical abstracts a hydrogen from another $$HBr$$ molecule. Hence the final product places bromine on the less substituted alkene carbon.

Applying the rule, let us compare the two possibilities:

• If $$Br\cdot$$ added to C-3, the radical would reside on C-2, a secondary radical.
• If $$Br\cdot$$ added to C-2, the radical would reside on C-3, a tertiary radical (more stable).

Because the tertiary radical is preferred, $$Br\cdot$$ adds to C-2 and the radical forms on C-3; this radical then captures $$H\cdot$$ from another $$HBr$$. So the net result is

$$CH_3-CH=C(CH_3)-CH_2-CH_3 \;\xrightarrow[\text{peroxide}]{HBr}\; CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3$$

The IUPAC name of the product is 2-bromo-3-methylpentane.

Now we examine stereochemistry. In the product the only π bond has been converted to σ bonds, so any stereoisomerism will arise solely from chiral (asymmetric) carbon atoms.

Checking each carbon:

• Carbon-2 bears $$Br$$, $$H$$, $$CH_3$$ (left side), and $$CH(CH_3)-CH_2-CH_3$$ (right side). The four substituents are all different, therefore carbon-2 is a chiral centre.

• Carbon-3 bears $$H$$, $$CH_3$$ (its methyl substituent), $$CH_2-CH_3$$ (to the right), and $$CH(Br)-CH_3$$ (to the left). Again the four groups are different, so carbon-3 is also a chiral centre.

• All other carbons possess at least two identical substituents and are achiral.

Thus the molecule contains $$n = 2$$ stereogenic centres. The maximum number of stereoisomers obtainable from $$n$$ independent chiral centres is given by the formula $$2^{\,n}$$.

Applying the formula, $$2^{\,2} = 4$$ possible stereoisomers are predicted: the pairs $$(R,R)\;,\;(S,S)\;,\;(R,S)\;,\;(S,R)$$. No internal mirror plane exists in the molecule, so none of these coincide or collapse into a meso form, and all four are distinct.

Hence, the correct answer is Option C.

First we recall the meanings of the two special names given in the statement:

• A vinyl group is the fragment $$CH_2=CH-$$. In other words, the double bond must be at the very end of the molecule so that the terminal carbon is $$CH_2$$ and is directly attached to the rest of the skeleton.

• A neopentyl group is the fragment $$(CH_3)_3CCH_2-$$. The characteristic feature is a $$CH_2-$$ unit that is bonded to a quaternary carbon bearing three methyl groups.

The required hydrocarbon must therefore contain both of these pieces somewhere in its structure and, when all carbon atoms are counted, the total must be seven.

Now we test every option one by one.

Option A : 2,2-dimethyl-4-pentene

Writing the carbon chain explicitly, we have

$$CH_3C(CH_3)_2CH_2CH_2CH=CH_2$$

Counting carbons: the pentane parent contributes $$5$$ and the two methyl groups contribute $$2$$, so the total is indeed $$7$$. However the double bond is shown at position 4 on the parent chain, so the fragment at the end is $$CH_3CH=$$, not $$CH_2=CH-$$. Therefore the terminal vinyl group is missing. Hence this option cannot satisfy the condition.

Option B : 4,4-dimethylpent-1-ene

First we write the parent chain of five carbons and place the double bond at carbon 1 as demanded by “pent-1-ene”:

$$CH_2=CH-CH_2-C-CH_3$$

The carbon labelled $$C$$ (carbon 4 in the name) still needs the two methyl substituents written in the name. Adding them gives

$$CH_2=CH-CH_2-C(CH_3)_2-CH_3$$

Now count the carbons:

• Parent chain = $$5$$.
• Two extra methyl groups = $$2$$.
Total $$= 5 + 2 = 7$$.

Next we look for the required fragments:

1. The left end is $$CH_2=CH-$$, exactly the vinyl group.

2. If we disconnect the bond immediately to the right of the $$CH_2-$$ unit, what remains on the right is $$C(CH_3)_3$$ : a quaternary carbon bonded to three methyl groups. Thus the fragment $$CH_2-C(CH_3)_3$$ is present, which is precisely the neopentyl group.

Hence this single structure simultaneously contains a vinyl group, a neopentyl group, and has seven carbon atoms. So Option B satisfies every requirement.

Option C : isopropyl-2-butene

Convert the name into a structure:

$$CH_3CH=CHCH_3$$ is the but-2-ene parent, and an isopropyl group $$CH_3CH(CH_3)-$$ is attached. Joining them gives a total of $$4 + 3 = 7$$ carbons; however the double bond is internal, so the molecule ends in $$CH_3CH=$$ rather than $$CH_2=CH-$$. Therefore the vinyl criterion fails, and there is no neopentyl fragment either.

Option D : 2,2-dimethyl-3-pentene

Its chain is

$$CH_3C(CH_3)_2CH=CHCH_3$$

Again the double bond lies inside the chain, giving no terminal $$CH_2=CH-$$ unit. Consequently the vinyl group is absent, so this option is also ruled out.

Among the four possibilities only Option B, 4,4-dimethylpent-1-ene, fulfils all the conditions laid down in the question.

Hence, the correct answer is Option B.

We have propene, whose formula is $$CH_3 - CH = CH_2$$. When chlorine is passed through water, the well-known reaction $$Cl_2 + H_2O \rightarrow HOCl + HCl$$ takes place, so the real reagent that finally adds to the alkene is hypochlorous acid $$HOCl$$.

For the addition to begin, we recall the general rule that $$HOCl$$ polarises to give an electrophile $$Cl^+$$ and a nucleophile $$OH^-$$. The electrophile is the species that first attacks the electron-rich carbon-carbon double bond.

So, in the first step, propene donates the $$\pi$$ electrons of its double bond to the electrophile $$Cl^+$$. Because an electrophilic addition always proceeds through the more stable carbocation, we have to decide to which carbon the $$Cl^+$$ should attach.

Let us examine the two possibilities one by one:

1. Suppose $$Cl^+$$ adds to the middle carbon (C-2). The skeleton after this attack would be $$CH_3 - CHCl - CH_2^+.$$ The positive charge now resides on C-3, which is a primary carbocation and therefore relatively unstable.

2. Suppose instead that $$Cl^+$$ adds to the terminal carbon (C-3). The skeleton produced in that case is $$CH_3 - CH^+ - CH_2Cl.$$ Here the positive charge is on C-2. This centre is attached to two carbon atoms (C-1 and C-3), so it is a secondary carbocation. A secondary carbocation is significantly more stable than a primary one.

Because the mechanism always proceeds through the more stable intermediate, the second alternative is preferred. Hence the reaction passes through the carbocation

$$CH_3 - CH^+ - CH_2 - Cl.$$

This structure corresponds exactly to Option D in the list supplied.

Hence, the correct answer is Option 4.

Under $$Br_2/h\nu$$, the reaction proceeds via a free-radical mechanism.

A bromine radical abstracts an allylic hydrogen, forming a resonance-stabilized allylic radical. Bromination therefore occurs at a single allylic position while the double bond remains unchanged.

Therefore, the major product is 3-bromocyclohexene.

First, recall the basic condition for geometrical (cis-trans) isomerism in alkenes. We state the rule:

For an alkene $$R^1R^2C=CR^3R^4$$ to show geometrical isomerism, each of the two doubly‐bonded carbons must be attached to two different groups. Because rotation about the $$C=C$$ bond is restricted, different spatial arrangements (cis and trans or the more general $$E$$/$$Z$$) become possible only when the four substituents $$R^1,\,R^2,\,R^3,\,R^4$$ are not all identical in pairs on either carbon.

Now we examine every option one by one, drawing or visualising the structure of the compound and checking the substituents on each doubly-bonded carbon.

Option A : 1,1-Diphenyl-1-propane

The word “propane’’ tells us there is no double bond at all; the compound is saturated. Without a $$C=C$$ bond (or a suitably constrained ring), geometrical isomerism cannot arise. So Option A fails the very first requirement.

Option B : 1-Phenyl-2-butene

Let us write its carbon skeleton. Numbering starts from the end nearer the double bond:

$$\text{Ph-CH}_2-CH=CH-CH_3$$

• The double bond is between C-2 and C-3.
• On C-2 we have the groups $$\text{Ph-CH}_2-$$ and $$H$$, clearly two different substituents.
• On C-3 we have $$CH_3$$ and $$H$$, again two different substituents.

Since both doubly-bonded carbons possess two different groups, the alkene fulfils the stated rule. Rotation is locked, so we can arrange similar groups on the same side (cis) or on opposite sides (trans). Therefore Option B does exhibit geometrical isomerism.

Option C : 3-Phenyl-1-butene

Its structure is

$$CH_2=CH-CH_2-CH_2-\text{Ph}$$

The double bond is now at the very end, between C-1 and C-2.

• Carbon C-1 carries two identical substituents, namely $$H$$ and $$H$$ (it is $$CH_2=$$).
Because the two groups on C-1 are identical, the required difference is missing; no cis-trans pair can be generated. Hence Option C does not show geometrical isomerism.

Option D : 2-Phenyl-1-butene

The formula becomes

$$CH_2=CH-CH(\text{Ph})-CH_3$$

Again the double bond is terminal (between C-1 and C-2).

• Carbon C-1 still has two identical $$H$$ atoms.
Thus the same objection raised for Option C applies here as well. Option D cannot display geometrical isomerism.

Among the four possibilities, only Option B satisfies the necessary condition on both double-bonded carbons. All the others fail because either there is no double bond or one of the double-bonded carbons bears two identical substituents.

Hence, the correct answer is Option 2.

We first write the name of the required ozonolysis product exactly as given in the question:

$$\text{5-keto-2-methyl hexanal}$$

This name tells us four vital pieces of information.

1. “hex-” shows that the longest carbon skeleton in the required fragment contains $$6$$ carbon atoms.

2. “-anal’’ tells us that carbon-1 of that skeleton is an aldehydic carbon, that is, $$\,-CHO$$.

3. “5-keto’’ tells us that carbon-5 of the same chain is a ketonic carbonyl $$\bigl(>C\! =\! O\bigr)$$.

4. “2-methyl’’ tells us that a $$CH_3$$ group is attached to carbon-2 of the chain.

So in the product we must have

$$\begin{aligned}

\text{C-1}&: && \; -CHO\\

\text{C-2}&: && \; -CH(CH_3)-\\

\text{C-3}&: && \; -CH_2-\\

\text{C-4}&: && \; -CH_2-\\

\text{C-5}&: && \; -CO-\\

\text{C-6}&: && \; -CH_3

\end{aligned}$$

The structure obtained by placing these pieces in order is therefore

$$\boxed{\,OHC-CH(CH_3)-CH_2-CH_2-CO-CH_3\,}$$

Ozonolysis is nothing but cleavage of a $$C=C$$ double bond, each vinylic carbon being converted into a carbonyl centre:

• if the olefinic carbon possessed at least one hydrogen, it ends up as an aldehyde $$\bigl(-CHO\bigr);$$
• if the olefinic carbon possessed no hydrogen, it ends up as a ketone $$\bigl(>C\!=\!O\bigr).$$

Thus our required product contains both kinds of carbonyls. Evidently, during the cleavage

• one alkene carbon must have borne at least one H atom (that carbon is now C-1 of the product and appears as $$-CHO$$);
• the other alkene carbon must have carried no hydrogen (that carbon is now C-5 of the product and appears as $$>C\!=\!O$$).

We therefore back-translate the product into the only double bond that can furnish it:

remove the oxygen of the aldehyde group and remove the oxygen of the ketone group, then join the two carbon centres thus freed by a double bond. This “reverse ozonolysis” step gives

$$

\begin{aligned}

&O\;{\Large\strikethrough{\text{HC}}}-CH(CH_3)-CH_2-CH_2-{\Large\strikethrough{\text{CO}}}-CH_3\\[4pt]

&\;\;\;\;\;\;\;\downarrow\;\text{join the two centres}\\[4pt]

&CH_2\;=\;C(CH_3)-CH_2-CH_2-CO-CH_3

\end{aligned}

$$

The fragment we have reconstructed is

$$\boxed{\,CH_2= C(CH_3)-CH_2-CH_2-CO-CH_3\,}$$

Notice carefully:

• the left-hand alkene carbon ($$CH_2$$) carries two hydrogens and will therefore give an aldehyde (specifically an $$-CHO$$) on ozonolysis;
• the right-hand alkene carbon is tertiary, $$C(CH_3)-$$, and therefore possesses no hydrogen; on ozonolysis it must therefore give a ketone.

Exactly that pattern - a terminal $$CH_2= C<$$ group flanked two carbons away from a pre-existing $$>C\!=\!O$$ - occurs in compound (3) of the options. The other three alternatives all fail on at least one count:

• either both alkene carbons possess hydrogen (so two aldehydes would be produced), or
• neither alkene carbon possesses hydrogen (so two ketones would result), or
• cleavage of their double bond does not leave a six-carbon chain that can still contain the embedded ketone at the right position.

Because only option (3) contains the exact $$CH_2= C(CH_3)-$$ sub-unit and the correctly placed pre-existing keto group, it alone can furnish the target fragment together with formaldehyde as the co-product, exactly as demanded by the rules stated above.

Hence, the correct answer is Option C (3).

The reagent $$Hg(OAc)_2/H_2O$$ followed by $$NaBH_4$$ adds an $$-OH$$ group to the more substituted carbon of the alkene and an $$-H$$ to the less substituted carbon.

Unlike acid-catalyzed hydration, this process occurs via a cyclic mercurinium ion intermediate, which prevents carbocation rearrangement.

For allylbenzene, the more substituted carbon in the vinyl group is the secondary carbon ($$C_2$$).

The major product is therefore 1-phenylpropan-2-ol.

Anti-Markownikoff addition is possible only in case of HBr and not in HCl and HI. In HBr, both the chain initiation and propagation steps are exothermic, while in HCl, first step is exothermic, and second step is endothermic. In HI, the first step is heavily endothermic, and the second step is exothermic. Hence HCl and HI do not undergo anti-Markownikoff’s addition.

The hydroboration-oxidation reaction of propene involves two main steps: hydroboration with diborane (B2H6) followed by oxidation with hydrogen peroxide (H2O2) and sodium hydroxide (NaOH). Propene has the structure CH3CH=CH2, which is an unsymmetrical alkene. The double bond is between the second carbon (CH group) and the third carbon (CH2 group), with the third carbon being less substituted.

In the hydroboration step, diborane dissociates into BH3, which acts as an electrophile. Boron attaches to the less substituted carbon (the terminal CH2 group) due to its higher electron density, while hydrogen attaches to the more substituted carbon (the CH group). This addition follows anti-Markovnikov orientation. The reaction proceeds as follows:

CH3CH=CH2 + BH3 $$\rightarrow$$ CH3CH2CH2BH2

However, diborane typically forms trialkylboranes. With excess propene, the reaction yields tripropylborane:

3 CH3CH=CH2 + (1/2) B2H6 $$\rightarrow$$ (CH3CH2CH2)3B

In the oxidation step, the alkylborane is treated with H2O2 and NaOH. This replaces the boron group with a hydroxyl group (OH), attaching it to the same carbon that was bonded to boron. The reaction for the trialkylborane is:

(CH3CH2CH2)3B + 3 H2O2 + NaOH $$\rightarrow$$ 3 CH3CH2CH2OH + sodium borate

The organic product is propan-1-ol, CH3CH2CH2OH, where the OH group is attached to the terminal carbon.

Now, comparing with the options:

• A. CH3CH2CH2OH (propan-1-ol)
• B. CH3CH2OH (ethanol)
• C. CH3CHOHCH3 (propan-2-ol)
• D. (CH3)3COH (tert-butanol)

The product matches option A. This is consistent with anti-Markovnikov addition, where the OH group attaches to the less substituted carbon.

Hence, the correct answer is Option A.

The balanced chemical equation for the complete combustion of an alkene is:

$$C_nH_{2n} + \frac{3n}{2}O_2 \rightarrow nCO_2 + nH_2O$$

$$\frac{\text{Volume of } O_2}{\text{Volume of alkene}} = \frac{3n/2}{1} = \frac{3n}{2}$$

$$\frac{27}{6} = \frac{3n}{2}$$

$$n = \frac{4.5}{1.5} = 3$$

The value $$n = 3$$ corresponds to the molecular formula $$C_3H_6$$, which is Propene.

The question asks why the addition of HI to an alkene in the presence of peroxide does not follow anti-Markovnikov's rule. First, recall that anti-Markovnikov addition occurs for HBr with peroxides due to a free radical mechanism. Peroxides generate radicals that initiate a chain reaction: the bromine radical adds to the less substituted carbon of the alkene, forming a more stable carbon radical, which then abstracts a hydrogen from HBr, giving the anti-Markovnikov product.

Now, consider HI. The options must be evaluated:

Option A suggests HI is a strong reducing agent. While HI can act as a reducing agent, this property does not directly prevent anti-Markovnikov addition. The reducing ability might affect other reactions but is not the primary reason here.

Option B claims the H-I bond is too strong to break homolytically. However, the bond dissociation energy (BDE) of H-I is approximately 299 kJ/mol, which is weaker than H-Br (366 kJ/mol) and H-Cl (431 kJ/mol). Thus, the H-I bond breaks more easily, not less, so this option is incorrect.

Option C states that the iodine atom combines with a hydrogen atom to reform HI. In the free radical mechanism, the iodine radical (I•) adds to the alkene to form a carbon radical, which then abstracts a hydrogen from HI to produce the alkyl iodide and regenerate I•. This propagation step is essential for the chain reaction and does not inherently prevent anti-Markovnikov addition. The reversibility of the addition step is key, not reformation of HI.

Option D states that the iodine atom is not reactive enough to add across a double bond. This is correct. The iodine radical has low reactivity due to its large size and low electronegativity. The carbon-iodine bond formed is weak (BDE ≈ 234 kJ/mol), making the addition to the alkene reversible. For anti-Markovnikov addition to occur, the radical addition must be irreversible and fast enough to sustain the chain reaction. However, the iodine radical's addition is slow and reversible, allowing the ionic mechanism (which follows Markovnikov's rule) to dominate. The ionic mechanism is faster for HI because I⁻ is a good nucleophile, and the reaction proceeds via carbocation intermediates.

Hence, the correct answer is Option D.

Initial protonation of the alkene double bond creates a secondary carbocation: $$\text{Ph}-\text{CH}_2-\text{CH}^+-\text{CH}_3$$.

This secondary carbocation undergoes a 1,2-hydride shift to form a more stable benzylic carbocation: $$\text{Ph}-\text{CH}^+-\text{CH}_2-\text{CH}_3$$. The benzylic carbocation is highly stable due to resonance with the benzene ring.

Nucleophilic attack by $$\text{Cl}^-$$ on this benzylic position yields 1-chloro-1-phenylpropane.

The starting material is propene, $$CH_3CH=CH_2$$.

Step (i) : Hydroboration-oxidation
Reagents $$BH_3$$ / $$H_2O_2$$ / $$OH^-$$ add $$H$$ and $$OH$$ across the double bond in an anti-Markovnikov manner with syn stereochemistry.
Hence the -OH group attaches to the terminal carbon:$$CH_3CH=CH_2 \;\xrightarrow[\;H_2O_2/OH^-\;]{\;BH_3\;} CH_3CH_2CH_2OH$$
Thus $$A$$ is 1-propanol.

Step (ii) : PCC oxidation
Pyridinium chlorochromate (PCC) oxidises a primary alcohol only up to the aldehyde stage:$$CH_3CH_2CH_2OH \;\xrightarrow{\;PCC\;} CH_3CH_2CHO$$
Hence $$B$$ is propanal.

Step (iii) : Grignard addition followed by acidic work-up
A Grignard reagent adds the alkyl group to the carbonyl carbon and converts the $$C=O$$ to $$C-OH$$ after hydrolysis:$$CH_3CH_2CHO + CH_3MgBr \;\xrightarrow[\;H_2O\;]{} CH_3CH_2CH(OH)CH_3$$
The carbonyl carbon of propanal gains a $$CH_3$$ group, giving a secondary alcohol, 2-butanol.

Therefore compound $$C$$ is $$CH_3CH_2CH(OH)CH_3$$, i.e. 2-butanol.

Option A which is: $$CH_3CH_2CH(OH)CH_3$$

The degree of unsaturation (also called index of hydrogen deficiency, IHD) tells us how many $$\pi$$-bonds and/or rings are present in a hydrocarbon.

For a general hydrocarbon $$C_nH_m$$, $$\text{IHD}= \dfrac{2n+2-m}{2}$$.

For $$C_4H_6$$, $$\text{IHD}= \dfrac{2(4)+2-6}{2}= \dfrac{10-6}{2}=2$$.
Thus every valid structure must contain a total of two units of unsaturation. Each ring counts as 1 unit and each double bond also counts as 1 unit.

Because the question asks for cyclic structures, at least one ring must be present. Therefore we have two possible patterns of unsaturation:
Case 1: one ring + one double bond.
Case 2: two rings and no double bond (bicyclic system).

Case 1: One ring + one double bond (four-carbon skeleton)

1. Cyclobutene : $$\mathrm{C_4H_6}$$ with the double bond inside the 4-membered ring.
    Only one constitutional arrangement is possible because all C atoms of cyclobutene are equivalent.

2. Methyl-substituted cyclopropenes (3-membered ring contains the double bond).     a) 1-Methylcyclopropene (methyl on a vinylic carbon, unique because the two vinylic positions are equivalent).     b) 3-Methylcyclopropene (methyl on the saturated ring carbon).     These two are not identical, so they give two distinct structures.

3. Methylenecyclopropane : a 3-membered ring where the double bond is exocyclic, i.e. $$\mathrm{CH_2{=}C_3}$$ (ring carbon 3 is doubly bonded to an external $$\mathrm{CH_2}$$ group).     This satisfies one ring + one double bond and is different from the cyclopropenes above.

Hence, under Case 1 we have 1 (cyclobutene) + 2 (methylcyclopropenes) + 1 (methylenecyclopropane) = 4 distinct cyclic structures.

Case 2: Two rings, no double bond

With four carbons and two rings we need five single C-C bonds (because $$\text{rings} = \text{bonds} - n + 1$$). The unique way to connect four carbon atoms with five single bonds is the “edge-fused” bicyclic skeleton $$\mathrm{bicyclo[1.1.0]butane}$$ (two cyclopropane rings sharing one edge). This gives one additional structure.

Total count

Adding Case 1 (4 structures) and Case 2 (1 structure) gives $$4+1 = 5$$ cyclic structural isomers for $$\mathrm{C_4H_6}$$.

Option B which is: 5

Ozonolysis is the reaction in which an alkene (or alkyne) is cleaved by ozone $$\left(O_3\right)$$ to give carbonyl compounds after reductive work-up.

For an alkene segment $$C_1=C_2$$, the products are:

$$C_1=C_2 \xrightarrow[Zn/H_2O]{O_3} \; C_1=O + C_2=O$$

Therefore, by inspecting the carbonyl products we can identify the carbon skeleton that was present before cleavage.

Formaldehyde $$(H-CHO)$$ contains only one carbon. It can arise in ozonolysis only when one of the double-bonded carbons was a terminal $$=CH2$$ group because:

$$CH_{2} = CR_{2} + {O_3} \rightarrow  {HCHO} + {O=CR_{2}}$$

The fragment $$CH2 =$$ directly attached to the rest of the molecule is called a vinyl group (also written as $$-CH=CH_2$$).

Checking the options:

Option A — a vinyl group: Contains $$\ce{-CH=CH2}$$; its ozonolysis indeed yields formaldehyde, so this option is correct.

Option B — an isopropyl group: $$\ce{-(CH3)2CH-}$$ has no double bond; if part of a double bond $$\ce{(CH3)2C=CH2}$$, the products would be acetone $$\left(\ce{(CH3)2C=O}\right)$$ and formaldehyde. Presence of acetone would also be detected, so isopropyl alone is not the confirming feature.

Option C — an acetylenic triple bond: Alkyne ozonolysis gives carboxylic acids (or ketones/CO2) but not formaldehyde from a $$\ce{-C#C-}$$ fragment, so this is wrong.

Option D — two ethylenic double bonds: Merely having two double bonds does not guarantee formaldehyde; only a terminal $$\ce{=CH2}$$ double bond does.

Hence, formaldehyde formation specifically confirms the presence of a vinyl group.

Final answer: Option A which is: a vinyl group

To recognise optical (enantiomeric) isomerism in an alkene, look for at least one $$sp^3$$‐hybridised carbon atom that is attached to four different groups. Such a carbon is called a chiral (asymmetric) centre. The $$sp^2$$ carbons in the $$C=C$$ bond can never be chiral, so the requisite chiral centre must lie elsewhere in the molecule.

Option A : 3-methyl-2-pentene
Structure: $$CH_3CH=C(CH_3)CH_2CH_3$$.
Every $$sp^3$$ carbon here has at least two identical groups attached (either two hydrogens or two alkyl fragments that are identical when traced out), so no carbon is chiral. Hence no optical isomerism.

Option B : 4-methyl-1-pentene
Structure: $$CH_2=CHCH_2CH(CH_3)CH_3$$.
The only $$sp^3$$ carbon with four substituents is the fourth carbon, $$CH(CH_3)$$. Its four attachments are H, $$CH_3$$, $$CH_3$$ (through the straight chain) and $$CH_2CH=CH_2$$. Two of them are identical (both $$CH_3$$ fragments), so the carbon is achiral. No optical activity.

Option C : 3-methyl-1-pentene
Structure: $$CH_2=CHCH(CH_3)CH_2CH_3$$.
Examine the third carbon (marked with *):
$$CH_2=CH\text{-}C^*(H)(CH_3)\text{-}CH_2CH_3$$.
Its four different groups are:
1. H
2. $$CH_3$$ (methyl side chain)
3. $$CH_2CH_2CH_3$$ (propyl fragment towards the chain end)
4. $$CH_2=CH$$ (allyl fragment towards the double bond)
Because all four groups differ, this carbon is chiral, giving rise to a pair of non-superimposable mirror images. Therefore 3-methyl-1-pentene exhibits optical isomerism.

Option D : 2-methyl-2-pentene
Structure: $$CH_3C(CH_3)=C(CH_3)CH_3$$.
Every $$sp^3$$ carbon again has at least two identical substituents (two hydrogens or two methyl/ethyl fragments). No chiral centre, hence no optical isomerism.

Thus, the only alkene among the four that contains a chiral carbon is 3-methyl-1-pentene.

Final answer: Option C which is: 3-methyl-1-pentene.

In ozonolysis, the $$\mathrm{C=C}$$ double bond is cleaved and each $$\mathrm{sp^2}$$ carbon becomes the carbonyl carbon of either an aldehyde or a ketone.
Thus, for an alkene $$R_1\,\!$$ $$\mathrm{CH=CH}$$ $$\,R_2$$, reductive ozonolysis (or oxidative followed by hydrolysis) gives $$R_1\!-\!CHO$$ and $$R_2\!-\!CHO$$.

Given data:
• The alkene is symmetrical ⇒ $$R_1 = R_2$$.
• Two moles of an aldehyde are obtained, each of molar mass $$44\ \text{u}$$.

Step 1: Identify the aldehyde of molar mass $$44\ \text{u}$$.
Molecular mass calculation for acetaldehyde $$\mathrm{CH_3CHO}$$ (formula $$\mathrm{C_2H_4O}$$):
$$2 \times 12\ (\text{C}) + 4 \times 1\ (\text{H}) + 16\ (\text{O}) = 24 + 4 + 16 = 44$$ u.

Hence the aldehyde produced is acetaldehyde $$\mathrm{CH_3CHO}$$.

Step 2: Work backwards to the alkene.
To give $$2$$ identical molecules of $$\mathrm{CH_3CHO}$$, each carbon of the $$\mathrm{C=C}$$ bond must carry a $$\mathrm{CH_3}$$ group, because cutting the double bond inserts an $$\mathrm{O}$$ and yields a $$\mathrm{CHO}$$ group:
$$\mathrm{CH_3{-}CH=CH{-}CH_3 \xrightarrow[H_2O]{O_3} 2\ CH_3CHO}$$

Thus the required alkene is $$\mathrm{CH_3{-}CH=CH{-}CH_3}$$, i.e. 2-butene (whether cis or trans, both are symmetrical about the double bond).

Verification of other options:
• Propene would yield $$\mathrm{CH_3CHO}$$ + $$\mathrm{HCHO}$$ (not symmetrical).
• 1-Butene would yield $$\mathrm{CH_3CH_2CHO}$$ + $$\mathrm{HCHO}$$.
• Ethene would yield only $$\mathrm{HCHO}$$.

Therefore, the only option matching all conditions is:

Option C which is: 2-butene

Geometrical (cis-trans) isomerism in alkenes arises when each carbon of the $$C = C$$ double bond carries two different groups. This creates restricted rotation about the double bond and the possibility of arranging the substituents either on the same side (cis) or on opposite sides (trans).

Let us analyse every option.

Case A:

Propene: $$CH_2 = CH-CH_3$$
Left (first) carbon of the double bond has two identical groups (two hydrogens). Therefore, no geometrical isomerism is possible.

Case B:

2-Methylpropene (isobutene): $$(CH_3)_2C = CH_2$$
Right carbon possesses two identical hydrogens. Hence geometrical isomerism is not feasible.

Case C:

2-Butene: $$CH_3-CH = CH-CH_3$$
Both carbons of the double bond have two different groups (one $$CH_3$$ and one $$H$$ each). Thus cis ($$CH_3$$ on same side) and trans ($$CH_3$$ on opposite sides) forms are possible. Geometrical isomerism exists.

Case D:

2-Methyl-2-butene: $$(CH_3)_2C = C(CH_3)H$$
Left carbon bears two identical $$CH_3$$ groups, so the necessary condition is not met; geometrical isomerism is absent.

Only 2-butene satisfies the required condition.

Hence, the correct choice is:
Option C which is: 2-butene