Alkene Questions in English

(C) The empirical formula of $Y$ is calculated as follows:

Element	$\%$ of element	At mass	Moles	Ratio
$C$	$22.22$	$12$	$1.85$	$2$
$H$	$3.71$	$1$	$3.71$	$4$
$Br$	$74.07$	$80$	$0.92$	$1$

The empirical formula is $C_2H_4Br$. Since the reaction is $X + Br_2 \rightarrow Y$,the molecular formula of $Y$ is $C_4H_8Br_2$.
The alkene $X$ is $C_4H_8$. Ozonolysis of $X$ gives only one product,which implies $X$ is symmetric. Among the options,$2$-butene $(CH_3-CH=CH-CH_3)$ is symmetric and yields two molecules of acetaldehyde upon ozonolysis.

(B) The reaction of $but-2-ene$ $(CH_3-CH=CH-CH_3)$ with $O_3$ followed by reductive workup with $Zn / H_2O$ is known as reductive ozonolysis.
In this reaction,the double bond is cleaved to form two molecules of acetaldehyde $(CH_3CHO)$.
The reaction mechanism is as follows:
$CH_3-CH=CH-CH_3 + O_3 \rightarrow \text{Ozonide intermediate}$
$\text{Ozonide} + Zn / H_2O \rightarrow 2CH_3CHO + ZnO + H_2O$
Therefore,the major product is acetaldehyde $(CH_3CHO)$.

(D) The reaction of but$-2-$ene $(CH_3-CH=CH-CH_3)$ with acidic $KMnO_4$ is an oxidative cleavage reaction.
Acidic $KMnO_4$ is a strong oxidizing agent that breaks the carbon-carbon double bond.
Since but$-2-$ene is a symmetrical alkene,the oxidative cleavage of the double bond results in the formation of two molecules of acetic acid $(CH_3COOH)$.
The reaction is: $CH_3-CH=CH-CH_3 + [O] \xrightarrow{KMnO_4/H^+} 2CH_3COOH$.

(A) Step $1$: Ethanol $(CH_3CH_2OH)$ undergoes dehydration in the presence of concentrated sulphuric acid $(H_2SO_4)$ at $170^{\circ} C$ to produce ethene $(CH_2=CH_2)$ gas.
$CH_3CH_2OH \xrightarrow{Conc. H_2SO_4, 170^{\circ} C} CH_2=CH_2 + H_2O$
Step $2$: The ethene gas is then treated with bromine $(Br_2)$ in carbon tetrachloride $(CCl_4)$,which is an electrophilic addition reaction,resulting in the formation of $1,2$-dibromoethane.
$CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2Br-CH_2Br$
Therefore,the major product is $1,2$-dibromoethane.

(A) The reaction involves the acid-catalyzed hydration of $1$-methylcyclopentene.
In the presence of an acid catalyst $(H^+)$,water adds across the double bond of the alkene.
This reaction proceeds via an electrophilic addition mechanism,where the proton $(H^+)$ first adds to the less substituted carbon of the double bond to form the more stable carbocation intermediate.
For $1$-methylcyclopentene,the proton adds to the $CH$ group,resulting in a tertiary carbocation at the carbon atom already bearing the methyl group.
Subsequently,a water molecule attacks this tertiary carbocation,followed by deprotonation to yield $1$-methylcyclopentan-$1$-ol.
This follows the Markownikoff rule,which states that the nucleophile $(-OH)$ attaches to the more substituted carbon atom.
Therefore,the major product is structure $I$.

(D) The reaction of $2$-butene $(CH_3-CH=CH-CH_3)$ with cold,dilute alkaline $KMnO_4$ (Baeyer's reagent) is a syn-hydroxylation reaction.
This reaction adds two hydroxyl $(-OH)$ groups across the double bond to form a vicinal glycol.
The product formed is butane-$2,3$-diol.

(C) The reaction sequence is as follows:
$X$ $\xrightarrow{\Delta, Zn} Y$ $\xrightarrow{O_3, Zn/H_2O} 2 CH_3CH_2CHO$ (Propionaldehyde)
Since the ozonolysis product is two moles of propionaldehyde $(CH_3CH_2CHO)$,the alkene $Y$ must be hex$-3-$ene $(CH_3CH_2CH=CHCH_2CH_3)$.
Heating a vicinal dibromide with $Zn$ dust causes debromination to form an alkene.
Therefore,$X$ must be $3,4-$dibromohexane $(CH_3CH_2CH(Br)CH(Br)CH_2CH_3)$.
Comparing this with the given options,the structure corresponds to $3,4-$dibromohexane.

(B) The reaction of $2,3$-dimethylbut-$2$-ene with bromine $(Br_2)$ in $CCl_4$ is an electrophilic addition reaction,which yields $2,3$-dibromo-$2,3$-dimethylbutane.
$CH_3-C(CH_3)=C(CH_3)-CH_3 + Br_2 \rightarrow CH_3-C(CH_3)(Br)-C(CH_3)(Br)-CH_3$
When $2,3$-dibromo-$2,3$-dimethylbutane is heated with alcoholic $KOH$,it undergoes a dehydrohalogenation reaction (elimination of two molecules of $HBr$) to form the conjugated diene,$2,3$-dimethylbuta-$1,3$-diene.
$CH_3-C(CH_3)(Br)-C(CH_3)(Br)-CH_3 \xrightarrow{\text{alc. } KOH, \Delta} CH_2=C(CH_3)-C(CH_3)=CH_2 + 2HBr$
Thus,the major product is $2,3$-dimethylbuta-$1,3$-diene.

(D) The reaction of $1-$ethylcyclopentene with $BH_3 / THF$ followed by $H_2O_2 / NaOH$ is a hydroboration-oxidation reaction.
This reaction proceeds via the syn-addition of $H$ and $OH$ across the double bond.
Since the addition is syn,the $H$ atom and the $OH$ group add to the same face of the double bond.
This results in the formation of a mixture of enantiomers where the $OH$ group and the ethyl group are in a cis-relationship.
Specifically,the product is $2-$ethylcyclopentan-$1-$ol,where the $OH$ and ethyl groups are cis to each other.
Looking at the options,the structure representing the cis-isomer is the correct product.

(C) The hydrocarbon with molecular formula $C_5H_{10}$ is an alkene.
Acid-catalyzed hydration of $2$-methyl$-2$-butene $(CH_3-C(CH_3)=CH-CH_3)$ follows Markovnikov's rule to form $2$-methyl$-2$-butanol,which is a tertiary $(3^{\circ})$ alcohol.
Oxidative cleavage of $2$-methyl$-2$-butene with acidic $KMnO_4$ breaks the double bond to produce acetone $(CH_3COCH_3)$ and acetic acid $(CH_3COOH)$,which are a ketone and a carboxylic acid respectively.
Therefore,the correct hydrocarbon is $2$-methyl$-2$-butene.

(C) The reaction involves the electrophilic addition of $HBr$ to $1$-cyclobutyl$-1-$methylethene.
Step $1$: Protonation of the double bond by $H^+$ from $HBr$ forms a more stable tertiary carbocation.
Step $2$: The cyclobutyl ring undergoes a ring expansion to form a more stable $5$-membered cyclopentyl ring carbocation.
Step $3$: The $Br^-$ ion attacks the carbocation to form the final major product,which is $1$-bromo$-1,2-$dimethylcyclopentane.

(A) The reaction of $2$-methylbut-$1$-ene with $BH_3, THF$ followed by $H_2O_2/OH^-$ is a hydroboration-oxidation reaction,which follows Anti-Markovnikov addition of water to the double bond. This yields $2$-methylbutan-$1$-ol as product $A$.
The reaction of $2$-methylbut-$1$-ene with $Hg(OAc)_2, H_2O$ followed by $NaBH_4$ is an oxymercuration-demercuration reaction,which follows Markovnikov addition of water to the double bond. This yields $2$-methylbutan-$2$-ol as product $B$.

(A) $1$. Calculate the moles of the hydrocarbon: Molar mass of $C_{10}H_{16} = (10 \times 12) + (16 \times 1) = 136 \, g/mol$. Moles of hydrocarbon $= \frac{17 \times 10^{-3} \, g}{136 \, g/mol} = 1.25 \times 10^{-4} \, mol$.
$2$. Calculate the moles of $H_2$ gas: At $STP$ ($0^{\circ}C$,$760 \, mm \, Hg$),$22400 \, mL$ corresponds to $1 \, mol$. Moles of $H_2 = \frac{8.40 \, mL}{22400 \, mL/mol} = 3.75 \times 10^{-4} \, mol$.
$3$. Determine the number of double bonds: Each double bond consumes $1 \, mol$ of $H_2$ for hydrogenation. Number of double bonds $= \frac{\text{moles of } H_2}{\text{moles of hydrocarbon}} = \frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}} = 3$.

(A) The reaction of $2$-methylpropene with cold $H_2SO_4$ follows Markovnikov's addition,where the proton $H^+$ adds to the terminal carbon to form a stable tertiary carbocation $(CH_3)_3C^+$. This carbocation then reacts with the hydrogen sulfate ion $HSO_4^-$ to form the alkyl hydrogen sulfate,$2$-methyl$-2-$propyl hydrogen sulfate,which is product '$A$'.
At higher temperatures $(80^{\circ}C)$,the reaction involves the dimerization of the alkene. The tertiary carbocation $(CH_3)_3C^+$ formed initially attacks another molecule of $2$-methylpropene to form a more stable carbocation,which then undergoes deprotonation to yield the dimer,$2,4,4$-trimethylpent-$2$-ene,which is product '$B$'.

(D) $1$. The hydrocarbon '$X$' has the molecular formula $C_6H_8$.
$2$. It consumes two moles of $H_2$ upon catalytic hydrogenation,indicating the presence of two double bonds.
$3$. Ozonolysis of '$X$' yields two moles of methane dicarbaldehyde $(OHC-CH_2-CHO)$.
$4$. Methane dicarbaldehyde is also known as malonaldehyde.
$5$. Cyclohexa$-1, 4-$diene undergoes ozonolysis to break the two double bonds,resulting in two molecules of $OHC-CH_2-CHO$.
$6$. Therefore,the hydrocarbon '$X$' is cyclohexa$-1, 4-$diene.

(C, D) The reaction of $but-2-yne$ with $Na/liq. NH_3$ (Birch reduction) yields $trans-but-2-ene$ $(B)$.
The reaction of $but-2-yne$ with $Pd/C$ (Lindlar's catalyst) yields $cis-but-2-ene$ $(A)$.
$A$ $(cis-but-2-ene)$ has a dipole moment $\mu \neq 0$,while $B$ $(trans-but-2-ene)$ has $\mu = 0$.
Statement $A$: $cis$ isomers are generally more polar and have higher boiling points,but solubility depends on crystal packing; $trans$ isomers $(B)$ pack better,leading to higher melting points.
Statement $B$: $A$ $(cis)$ has a higher boiling point than $B$ $(trans)$,but $B$ has a higher melting point due to better symmetry.
Statement $C$: $A$ is more polar than $B$ because $A$ has a non-zero dipole moment,while $B$ is zero. The statement claims $A$ is more polar because its dipole moment is zero,which is false.
Statement $D$: Both $A$ and $B$ are alkenes and react similarly with $Br_2$; there is no significant difference in ease of addition. Thus,$D$ is also incorrect.
Since $C$ and $D$ are incorrect,and no option matches,the question is flawed.

(C) The synthesis of alkenes involves elimination reactions.
$(A)$ Treatment of alkynes with $Na$ in liquid $NH_3$ (Birch reduction) produces trans-alkenes.
$(B)$ Heating alkyl halides with alcoholic $KOH$ undergoes dehydrohalogenation to form alkenes.
$(C)$ Treating alkyl halides with aqueous $KOH$ results in nucleophilic substitution,producing alcohols $(R-X + KOH_{(aq)} \rightarrow R-OH + KX)$,not alkenes.
$(D)$ Treating vicinal dihalides with $Zn$ metal undergoes debromination/dehalogenation to form alkenes.
Therefore,option $C$ is the incorrect method for alkene synthesis.

(B) The ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
Ethanal is $CH_3CHO$ and propanone is $(CH_3)_2CO$.
By joining these two fragments at the carbonyl carbons,we get the structure of the alkene $(X)$:
$CH_3-CH=C(CH_3)_2$.
The chemical formula of this hydrocarbon is $C_5H_{10}$.
The molar mass of $C_5H_{10} = (5 \times 12) + (10 \times 1) = 60 + 10 = 70 \ g\ mol^{-1}$.

(B) The reaction involves the electrophilic addition of $Br_2$ to the alkene group of pent$-4-$enoic acid.
$1$. $Br_2$ reacts with the double bond to form a cyclic bromonium ion intermediate.
$2$. The base $NaHCO_3$ deprotonates the carboxylic acid group to form a carboxylate anion $(R-COO^-)$.
$3$. This carboxylate anion acts as an internal nucleophile and attacks the more substituted carbon of the bromonium ion,leading to an intramolecular cyclization (halolactonization).
$4$. This results in the formation of a five-membered lactone ring with a bromomethyl group attached,which is the major product $P$.

(A) The reaction of $2-$hexene $(CH_3-CH=CH-CH_2-CH_2-CH_3)$ with $O_3$ followed by $H_2O$ is an oxidative ozonolysis.
In oxidative ozonolysis,the double bond is cleaved,and the resulting aldehydes are further oxidized to carboxylic acids.
$CH_3-CH=CH-CH_2-CH_2-CH_3 \xrightarrow[(i) O_3]{(ii) H_2O} CH_3COOH + CH_3CH_2CH_2COOH$
The products formed are acetic acid $(CH_3COOH)$ and butanoic acid $(CH_3CH_2CH_2COOH)$.

(A) The reaction of $3-$methylhex$-2-$ene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition mechanism.
The product $(A)$ formed is $2-$bromo$-3-$methylhexane.
The structure of $2-$bromo$-3-$methylhexane is $CH_3-CH(Br)-CH(CH_3)-CH_2-CH_2-CH_3$.
This molecule contains $2$ chiral centers at $C-2$ and $C-3$.
Since the two chiral centers are not equivalent (the groups attached to them are different),the number of stereoisomers is given by $2^n$,where $n$ is the number of chiral centers.
Number of stereoisomers $= 2^2 = 4$.

(D) $1$. In the presence of $h\nu$ (ultraviolet light),$Cl_2$ undergoes homolytic cleavage to form chlorine free radicals. These radicals perform allylic substitution on cyclohexene to form $1,4-\text{dichlorocyclohex-2-ene}$ as product $A$.
$2$. In the presence of $CCl_4$ (an inert solvent),$Cl_2$ undergoes electrophilic addition across the double bond of cyclohexene to form $1,2-\text{dichlorocyclohexane}$ as product $B$.

(B) The reaction is the ozonolysis of but$-2-$ene.
$CH_3-CH=CH-CH_3 \xrightarrow[(ii) Zn/H_2O]{(i) O_3} 2CH_3CHO$
Here,the product $(P)$ is acetaldehyde $(CH_3CHO)$.
In one molecule of acetaldehyde $(CH_3CHO)$,there is $1$ oxygen atom.
Therefore,the total number of oxygen atoms present per molecule of the product is $1$.

(A) Option $A$ is correct. The reaction of $1-$methylcyclohexene with $HI$ follows Markovnikov's rule,where the nucleophile $(I^-)$ attaches to the more substituted carbon atom of the double bond,resulting in $1-$iodo$-1-$methylcyclohexane.
Option $B$ is incorrect because the reaction of cyclohexene with $Br_2$ in the presence of $UV$ light/heat typically leads to allylic bromination,not addition.
Option $C$ represents the Hofmann bromamide degradation reaction,which is correctly balanced as: $CH_3CH_2CONH_2 + Br_2 + 4NaOH \rightarrow CH_3CH_2NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.
Option $D$ represents the carbylamine reaction,which is also correctly written.
Given the provided images,the question likely intends to identify the correct chemical transformation shown in the diagrams. Since both $A$ and $C$ are chemically correct,$A$ is the primary reaction depicted in the solution image.

(C) The reaction of $1\text{-methylcyclohexene}$ with $H^+/H_2O$ is an acid-catalyzed hydration reaction,which follows Markovnikov's rule. The carbocation formed is more stable at the tertiary position,leading to the formation of $1\text{-methylcyclohexanol}$ as product $A$.
The reaction of $1\text{-methylcyclohexene}$ with $B_2H_6$ followed by $H_2O_2/NaOH$ is a hydroboration-oxidation reaction,which is an anti-Markovnikov addition of water across the double bond. This leads to the formation of $2\text{-methylcyclohexanol}$ as product $B$.

(D) The reaction is an electrophilic addition of $D-Cl$ to $1-methylcyclopentene$.
First,the electrophile $D^+$ attacks the double bond to form the more stable tertiary carbocation.
The $D^+$ adds to the $CH$ group (the carbon with more hydrogens) to form a tertiary carbocation at the carbon bearing the $CH_3$ group.
Then,the nucleophile $Cl^-$ attacks the planar carbocation from either the top or bottom face,resulting in a racemic mixture of $1-chloro-1-methyl-2-deuteriocyclopentane$ (where $D$ and $CH_3$ are trans or cis).
Based on the options provided,the structure representing the addition of $D$ and $Cl$ across the double bond is the correct product.

(A) The reaction involves the nucleophilic substitution of bromine atoms by dimethylamine $(Me_2NH)$.
First,one $Me_2NH$ molecule attacks the allylic position,displacing one $Br^-$ ion.
Then,the lone pair on the nitrogen atom of the newly attached dimethylamino group participates in the formation of an aziridinium ion intermediate by displacing the second $Br^-$ ion.
Finally,a second molecule of $Me_2NH$ attacks the aziridinium ring,opening it to form the final product,which is $3,4$-bis(dimethylamino)cyclopent-$1$-ene.

(C) Dipole moment is a vector quantity. In the $cis$-isomer,the bond dipoles of the two $C-CH_3$ bonds are in the same direction,leading to a non-zero net dipole moment.
In the $trans$-isomer,the bond dipoles of the two $C-CH_3$ bonds are equal in magnitude and opposite in direction,which cancel each other out,resulting in a net dipole moment of zero.
Since the $cis$-isomer has a non-zero dipole moment and the $trans$-isomer has a zero dipole moment,the $cis$-isomer is more polar.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(D) The reaction of cyclohexene with hot acidic potassium permanganate $(KMnO_4, H_2SO_4)$ is an oxidative cleavage reaction.
The double bond in the cyclic alkene is broken,and the carbon atoms at the double bond are oxidized to carboxylic acid groups.
Cyclohexene $(C_6H_{10})$ undergoes oxidative cleavage to form hexanedioic acid,which is commonly known as adipic acid $(HOOC-(CH_2)_4-COOH)$.

(C) $(1)$ Acid-catalyzed hydration of propene follows Markovnikov's rule,where the electrophile $H^+$ adds to the terminal carbon to form the more stable secondary carbocation $(CH_3-CH^+-CH_3)$,which is then attacked by $H_2O$ to form propan-$2$-ol as the major product $A$.
$(2)$ Hydroboration-oxidation of propene follows anti-Markovnikov's rule,where the boron atom adds to the terminal carbon,and subsequent oxidation with $H_2O_2/OH^-$ replaces boron with an $-OH$ group,resulting in propan-$1$-ol as the major product $B$.

(A) The reaction of an alkene with hot acidic $KMnO_4$ (oxidative cleavage) results in the cleavage of the $C=C$ double bond.
For the given reactant,$1,2$-dicyclohexylethane,the oxidative cleavage of the double bond yields two molecules of cyclohexanecarboxylic acid as the major product.
The reaction is:
$(C_6H_{11})CH=CH(C_6H_{11}) \xrightarrow{KMnO_4/H^+} 2 C_6H_{11}COOH$
Thus,the major product '$P$' is cyclohexanecarboxylic acid.

(A) The reaction of $NOCl$ with an alkene follows Markovnikov's rule.
$NOCl$ dissociates as $NO^{+}$ and $Cl^{-}$.
The electrophile $NO^{+}$ attacks the carbon atom with more hydrogen atoms (the terminal $CH_2$ group) to form a more stable carbocation intermediate.
The nucleophile $Cl^{-}$ then attacks the more substituted carbon atom ($CH$ group).
Thus,the product $P$ is $CH_3-CH(Cl)-CH_2(NO)$.

(B) $1$. Moles of $P$ ($4$-methyloct$-1-$ene) = $\frac{2.52 \ g}{126 \ g \ mol^{-1}} = 0.02 \ mol$.
$2$. The reaction with $HBr$ in the presence of peroxide follows anti-Markovnikov addition. The combined yield is $50 \%$,so the total moles of bromides formed = $0.02 \times 0.50 = 0.01 \ mol$.
$3$. The bromides are formed in a $9:1$ ratio. The primary alkyl bromide is the major product ($9$ parts out of $10$).
$4$. Moles of primary alkyl bromide = $\frac{9}{10} \times 0.01 \ mol = 0.009 \ mol$.
$5$. The primary alkyl bromide reacts with diethylamine to form a quaternary ammonium salt,which upon treatment with aqueous $K_2CO_3$ yields the non-ionic amine product $S$.
$6$. The molar mass of $S$ ($N,N$-diethyl$-4$-methyloctan$-1-$amine) is $199 \ g \ mol^{-1}$.
$7$. Mass of $S$ = $\text{moles} \times \text{molar mass} = 0.009 \ mol \times 199 \ g \ mol^{-1} = 1.791 \ g = 1791 \ mg$.

(C) Bromination of alkenes proceeds via anti-addition (trans-addition) of $Br_2$.
$(i)$ $trans-but-2-ene$ gives a racemic mixture of $(2R, 3R)$-$2,3-dibromobutane$ $(M)$ and $(2S, 3S)$-$2,3-dibromobutane$ $(N)$.
(ii) $cis-but-2-ene$ gives the meso compound $(2R, 3S)$-$2,3-dibromobutane$. Since the meso compound is achiral,$O$ and $P$ represent the same molecule (meso form).
Statement [$B$] is correct because bromination is a stereospecific anti-addition.
Statement [$C$] is correct because $O$ and $P$ are the same meso compound.
Therefore,the correct statements are [$B$] and [$C$].

(A) $3$-Bromo-$3$-cyclopentylhexane has three different types of $\beta$-hydrogens (not $\alpha$-hydrogens as $\alpha$ is the carbon attached to $Br$). These are marked as $(a)$,$(b)$,and $(c)$ in the structure.
Dehydrobromination leads to the formation of three structurally isomeric alkenes:
$1$. From $\beta$-hydrogen $(a)$: $CH_3CH_2CH=C(cyclopentyl)CH_2CH_3$ (exists as $E$ and $Z$ isomers).
$2$. From $\beta$-hydrogen $(b)$: $CH_3CH_2CH_2C(cyclopentyl)=CHCH_3$ (exists as $E$ and $Z$ isomers).
$3$. From $\beta$-hydrogen $(c)$: $CH_3CH_2CH_2C(cyclopentyl)=CHCH_2CH_3$ (this is the same as the product from $(b)$).
Wait,re-evaluating the structure: The $\beta$-carbons are the two $CH_2$ groups of the hexane chain and the $CH$ group of the cyclopentyl ring.
- Removal of $H$ from $CH_2$ (position $a$): $CH_3CH_2CH=C(cyclopentyl)CH_2CH_3$ ($E/Z$ isomers).
- Removal of $H$ from $CH_2$ (position $b$): $CH_3CH_2CH_2C(cyclopentyl)=CHCH_3$ ($E/Z$ isomers).
- Removal of $H$ from $CH$ (position $c$): $CH_3CH_2CH_2C(cyclopentyl)=C(CH_2CH_3)_2$ (this is a different alkene,no $E/Z$ isomerism).
Total alkenes = $2 (E/Z) + 2 (E/Z) + 1 = 5$.

(D) Hydrogenation of the given compounds results in the following products:
$(A)$ $CH_3-CH=CH-CH(Br)CH_3 \xrightarrow{H_2/Pt} CH_3-CH_2-CH_2-CH(Br)CH_3$. The product has a chiral center and is optically active.
$(B)$ $H_2C=CH-CH(Br)CH_2CH_3 \xrightarrow{H_2/Pt} CH_3-CH_2-CH(Br)CH_2CH_3$. The product is $3$-bromopentane,which has a plane of symmetry and is optically inactive.
$(C)$ $H_2C=C(CH_3)-CH(Br)CH_3 \xrightarrow{H_2/Pt} CH_3-CH(CH_3)-CH(Br)CH_3$. The product has two chiral centers and is optically active.
$(D)$ $H_2C=CH-C(Br)(H)CH_2CH_3 \xrightarrow{H_2/Pt} CH_3-CH_2-CH(Br)CH_2CH_3$. The product is $3$-bromopentane,which is optically inactive.
Thus,compounds $(B)$ and $(D)$ produce optically inactive compounds upon hydrogenation.

(C) The reaction is an electrophilic addition of $HBr$ to a conjugated diene,$2-\text{methylbuta}-1,3-\text{diene}$ (isoprene).
$1$. Protonation of the terminal double bond $(C_1)$ occurs to form the most stable carbocation.
$2$. Protonation at $C_1$ gives a tertiary allylic carbocation: $H_3C-C^+(CH_3)-CH=CH_2$,which is resonance stabilized.
$3$. The resonance structure is $H_3C-C(CH_3)=CH-CH_2^+$.
$4$. The bromide ion $(Br^-)$ attacks the primary carbocation site $(C_4)$ to give the thermodynamically more stable product (more substituted alkene).
$5$. The major product is $1-\text{bromo}-3-\text{methylbut}-2-\text{ene}$,which corresponds to option $C$.

(D) Step $1$: Acid-catalyzed dehydration of the starting alcohol leads to the formation of a carbocation,which then undergoes rearrangement and elimination to form the diene $P$ (a decalin derivative with two double bonds).
Step $2$: The diene $P$ reacts with aqueous dilute $KMnO_4$ (Baeyer's reagent) at $0^{\circ}C$. This reagent performs syn-dihydroxylation of double bonds.
Step $3$: Since $P$ has two double bonds,both are converted into diols. The resulting product $Q$ contains four hydroxyl groups (two from each double bond).
Thus,the number of hydroxyl groups in $Q$ is $4$.

(B) Ozonolysis involves the cleavage of double bonds to form carbonyl compounds. By analyzing the structure of each isomer in List-$I$ and performing the oxidative cleavage of the double bonds,we can determine the corresponding products in List-$II$.
$(A)$ The isomer has two double bonds,which upon ozonolysis yield a dialdehyde with a methyl substituent. This matches product $(III)$.
$(B)$ The isomer has two double bonds,which upon ozonolysis yield a keto-aldehyde. This matches product $(IV)$.
$(C)$ The isomer has two double bonds,which upon ozonolysis yield a dialdehyde. This matches product $(I)$.
$(D)$ The isomer has two double bonds,which upon ozonolysis yield a keto-aldehyde. This matches product $(II)$.
Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

(A) The oxidation products are $CH_3-CO-CH_3$,$CH_3-COOH$,and $CH_3-CO-CH_2-CH_2-COOH$.
By analyzing the cleavage products,the structure of compound $X$ is determined to be $CH_3-C(CH_3)=CH-CH_2-CH_2-CH=C(CH_3)_2$.
This structure contains $2$ double bonds,which is consistent with the absorption of $2$ moles of hydrogen.
To count the $\sigma$ bonds:
- The carbon skeleton has $10$ carbon atoms,which implies $9$ $C$-$C$ $\sigma$ bonds.
- The number of hydrogen atoms is $3+3+1+2+2+1+3+3 = 18$.
- Thus,there are $18$ $C$-$H$ $\sigma$ bonds.
- Total $\sigma$ bonds = $9$ ($C$-$C$) + $18$ ($C$-$H$) = $27$.

(B) The reaction of alkenes with $HBr_{(aq)}$ proceeds via an electrophilic addition mechanism,where the rate-determining step involves the formation of a carbocation intermediate.
Greater stability of the carbocation intermediate leads to a faster reaction rate.
In molecule $Q$,the protonation of the double bond leads to a carbocation that is stabilized by resonance from the lone pair of the adjacent oxygen atom (forming an oxocarbenium ion).
This resonance stabilization makes the carbocation derived from $Q$ significantly more stable than the alkyl carbocations formed from $P$,$R$,or $S$.
Therefore,molecule $Q$ reacts at the fastest rate.

(B) Step $(i)$: Hydration of propyne $(CH_3-C \equiv CH)$ in the presence of $Hg^{2+}$ and $H_2SO_4$ follows Markovnikov's rule to form acetone $(CH_3-CO-CH_3)$.
Step $(ii)$: Nucleophilic addition of $HCN$ to acetone forms a cyanohydrin $(CH_3-C(OH)(CN)-CH_3)$.
Step $(iii)$: Catalytic hydrogenation of the cyanohydrin using $H_2/Ni$ reduces the $-CN$ group to a primary amine group $(-CH_2NH_2)$.
The final product $(A)$ is $2-hydroxy-2-methylpropan-1-amine$,which corresponds to the structure shown in option $(B)$.

(A) Step $1$: The partial hydrogenation of propyne $(CH_3-C \equiv CH)$ using $H_2$ over $Pd/C$ (Lindlar's catalyst or similar controlled hydrogenation) yields propene $(CH_3-CH=CH_2)$ as product $[A]$.
Step $2$: Ozonolysis of propene $(CH_3-CH=CH_2)$ followed by reductive workup with $Zn/H_2O$ cleaves the double bond to form acetaldehyde $(CH_3CHO)$ as product $[B]$ and formaldehyde $(HCHO)$ as product $[C]$.
Therefore,the correct sequence is $[A]: CH_3-CH=CH_2, [B]: CH_3CHO, [C]: HCHO$.

(D) The reaction involves the dehydrohalogenation of a vicinal dibromide using an excess of alcoholic $KOH$ (a strong base) under heating conditions. This is an $E2$ elimination reaction that proceeds twice to form a conjugated diene. The elimination occurs to form the most stable,highly substituted,and conjugated product. The structure formed is $2-$phenylhepta$-2,4-$diene,which is stabilized by conjugation with the phenyl ring and the double bonds.

(A) Ozonolysis of a cyclic alkene involves the cleavage of the double bond to form a dicarbonyl compound.
To determine the starting material for $3-$methyl$-6-$oxoheptanal,we can reverse the process by connecting the carbonyl carbons with a double bond.
The product $3-$methyl$-6-$oxoheptanal has a chain of $7$ carbons.
Connecting the aldehyde carbon $(C_1)$ and the ketone carbon $(C_6)$ with a double bond forms a $6-$membered ring with a methyl group at the $4-$position relative to the double bond (or $1,4-$dimethylcyclohex$-1-$ene).
Thus,$1,4-$dimethylcyclohex$-1-$ene undergoes ozonolysis to yield $3-$methyl$-6-$oxoheptanal.

(C) Statement $I :$ Ozonolysis of $cis-2-butene$ $(CH_3-CH=CH-CH_3)$ followed by reductive workup $(Zn, H_2O)$ yields two molecules of ethanal $(CH_3CHO)$. This statement is true.
Statement $II :$ Ozonolysis of $3,6-dimethyloct-4-ene$ $(CH_3-CH_2-CH(CH_3)-CH=CH-CH(CH_3)-CH_2-CH_3)$ yields two molecules of $2-methylbutanal$ $(CH_3-CH_2-CH(CH_3)-CHO)$. The carbon atom at the $2-position$ of $2-methylbutanal$ is bonded to four different groups $(-H, -CH_3, -CH_2CH_3, -CHO)$,making it a chiral carbon atom. Thus,the product contains a chiral center. This statement is false.

(B) Hydroboration-oxidation follows anti-Markownikoff's rule (syn-addition of $H$ and $OH$),while acid-catalysed hydration follows Markownikoff's rule and involves a carbocation intermediate,which may undergo rearrangement.
For symmetric alkenes like $CH_2=CH_2$ or $CH_3-CH=CH-CH_3$,both processes yield the same product.
For cyclic alkenes like cyclohexene,the products are the same (cyclohexanol).
However,for unsymmetrical alkenes like $1$-methylcyclohexene,acid-catalysed hydration leads to a more stable tertiary carbocation,resulting in $1$-methylcyclohexanol,whereas hydroboration-oxidation adds $OH$ to the less substituted carbon,resulting in $2$-methylcyclohexanol. Thus,they give different products.

(C) Statement-$I$ is correct because ozonolysis involves the cleavage of a $C=C$ double bond to form carbonyl compounds,which helps in identifying the position of the double bond.
Statement-$II$ is incorrect because $but-1-ene$ $(CH_3CH_2CH=CH_2)$ on ozonolysis yields propanal $(CH_3CH_2CHO)$ and methanal $(HCHO)$,whereas $but-2-ene$ $(CH_3CH=CHCH_3)$ on ozonolysis yields two moles of ethanal $(CH_3CHO)$.
Therefore,Statement-$I$ is correct but Statement-$II$ is incorrect.

(B) The ozonolysis of $CH_3-CH=CH-CH=CH_2$ involves the cleavage of both double bonds.
Step $1$: The first double bond $(CH_3-CH=CH-)$ cleaves to form $CH_3-CHO$ and $OHC-$.
Step $2$: The second double bond $(-CH=CH_2)$ cleaves to form $OHC-$ and $HCHO$.
Step $3$: Combining the fragments,the products are $CH_3-CHO$,$OHC-CHO$ (glyoxal),and $HCHO$.
Therefore,$CH_3-CH_2-CHO$ is not formed.