Alkene Questions in English

(A) The stability of an alkene is directly proportional to the number of hyperconjugative structures (number of $\alpha$-hydrogens).
$1$. $(II) CH_3CH_2CH_2CH = CH_2$: $2 \ \alpha$-hydrogens (monosubstituted).
$2$. $(IV) CH_3CH_2C(CH_3) = CH_2$: $5 \ \alpha$-hydrogens (disubstituted).
$3$. $(III) (CH_3)_2C = CHCH_3$: $7 \ \alpha$-hydrogens (trisubstituted).
$4$. $(I) (CH_3)_2C = C(CH_3)_2$: $12 \ \alpha$-hydrogens (tetrasubstituted).
Since stability increases with the number of $\alpha$-hydrogens,the order is $(II) < (IV) < (III) < (I)$.

(A) The reaction of but$-1-$ene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition mechanism (Kharasch effect or peroxide effect).
In this mechanism,the bromine atom attaches to the carbon atom with more hydrogen atoms (the terminal carbon).
Thus,$1-$bromobutane $(CH_3CH_2CH_2CH_2Br)$ is formed as the major product,and $2-$bromobutane $(CH_3CH_2CH(Br)CH_3)$ is formed as the minor product.

(A) The molecular formula $C_{5}H_{10}$ corresponds to an alkene. Ozonolysis of an alkene $R_{1}R_{2}C=CR_{3}R_{4}$ yields carbonyl compounds.
$B$ gives a positive Fehling's test,so it must be an aldehyde. Since it also gives the iodoform test,it must be acetaldehyde $(CH_{3}CHO)$.
$C$ does not give Fehling's test,so it is a ketone. Since it gives the iodoform test,it must be a methyl ketone $(CH_{3}COR)$. Given the total carbon count is $5$,$C$ is propanone $(CH_{3}COCH_{3})$.
Thus,the alkene $A$ is $2$-methylbut-$2$-ene $(CH_{3}-C(CH_{3})=CH-CH_{3})$.
Ozonolysis reaction:
$CH_{3}-C(CH_{3})=CH-CH_{3} \xrightarrow{O_{3}, Zn/H_{2}O} CH_{3}COCH_{3} (C) + CH_{3}CHO (B)$
Iodoform test:
$CH_{3}CHO + 3I_{2} + 4NaOH \rightarrow CHI_{3} + HCOONa + 3NaI + 3H_{2}O$
$CH_{3}COCH_{3} + 3I_{2} + 4NaOH \rightarrow CHI_{3} + CH_{3}COONa + 3NaI + 3H_{2}O$

(C) The reaction involves the acid-catalyzed hydration of $1-$methyl$-1-$vinylcyclopentane.
$1$. Protonation of the double bond leads to the formation of a secondary carbocation.
$2$. The five-membered ring undergoes ring expansion to form a more stable six-membered ring carbocation.
$3$. This carbocation then undergoes rearrangement and elimination of a proton $(-H^+)$ to form the most stable alkene,which is $1,2$-dimethylcyclohexene.
Thus,the major product is $1,2$-dimethylcyclohexene.

(A) The dehydration of neopentyl alcohol $(CH_3-C(CH_3)_2-CH_2OH)$ in the presence of an acid proceeds via the formation of a primary carbocation,which undergoes a $1,2$-methyl shift to form a more stable tertiary carbocation $(CH_3-C^+(CH_3)-CH_2-CH_3)$.
From this tertiary carbocation,two alkenes can be formed by the loss of a proton $(H^+)$:
$1$. Loss of a proton from the $CH_2$ group leads to $2$-methyl-$2$-butene $(CH_3-C(CH_3)=CH-CH_3)$,which is the more substituted and thus more stable alkene (major product,$85\%$).
$2$. Loss of a proton from the $CH_3$ group leads to $2$-methyl-$1$-butene $(CH_3-C(CH_3)-CH=CH_2)$,which is less substituted (minor product,$15\%$).
Therefore,the mixture consists of $2$-methyl-$2$-butene and $2$-methyl-$1$-butene.

(A) The reaction of an alkene with $HBr$ proceeds via an electrophilic addition mechanism involving a carbocation intermediate.
Step $1$: Protonation of the double bond can form two possible secondary carbocations:
$(I)$ $CH_3-CH_2-CH^{+}-CH(CH_3)_2$
$(II)$ $CH_3-CH^{+}-CH_2-CH(CH_3)_2$
Step $2$: Carbocation $(I)$ undergoes a $1,2-$hydride shift from the adjacent tertiary carbon to form a much more stable tertiary carbocation:
$CH_3-CH_2-CH^{+}-CH(CH_3)_2 \xrightarrow{1,2-H \text{ shift}} CH_3-CH_2-CH_2-C^{+}(CH_3)_2$
Step $3$: The bromide ion $(Br^{-})$ attacks this highly stable tertiary carbocation to form the major product:
$CH_3-CH_2-CH_2-C^{+}(CH_3)_2 + Br^{-} \longrightarrow CH_3-CH_2-CH_2-C(Br)(CH_3)_2$
Thus,the major product is $2-$bromo$-2-$methylpentane.

(D) The reaction proceeds in two steps:
$1$. Dehydrohalogenation (or elimination of the leaving group) using $KO^tBu$ (a bulky base) and $\Delta$ (heat) follows the $E2$ mechanism to form the most stable alkene (Hofmann product is not favored here as the substrate allows for the formation of the more substituted alkene).
Starting material: $CH_3-CH(CH_3)-CH(OSO_2CH_3)-CH_3$.
Elimination of $OSO_2CH_3$ and a proton from the adjacent carbon leads to the formation of $CH_3-CH(CH_3)-CH=CH_2$ ($3$-methylbut$-1-$ene) or $CH_3-CH(CH_3)-CH=CH-CH_3$ ($2$-methylbut$-2-$ene). The more substituted alkene,$CH_3-C(CH_3)=CH-CH_3$ ($2$-methylbut$-2-$ene),is the major product.
$2$. Ozonolysis $(O_3/H_2O_2)$ of $CH_3-C(CH_3)=CH-CH_3$ leads to oxidative cleavage.
$CH_3-C(CH_3)=CH-CH_3 \xrightarrow{O_3/H_2O_2} (CH_3)_2C=O + CH_3COOH$.
Thus,the major products are acetone and acetic acid.

(B) $1$. $B$ gives a yellow precipitate with $I_2 + NaOH$ (iodoform test) and a silver mirror with $Ag_2O$ (Tollens' test). This indicates that $B$ is an aldehyde containing a $CH_3CO-$ group. The simplest such aldehyde is acetaldehyde $(CH_3CHO)$.
$2$. $C$ does not give a yellow precipitate with $I_2 + NaOH$,but upon reduction with $LiAlH_4$,it gives $D$,which forms white turbidity with Lucas reagent $(ZnCl_2 + conc. HCl)$ within $5$ minutes. This indicates $D$ is a secondary alcohol. Thus,$C$ must be a ketone,specifically diethyl ketone $(CH_3CH_2COCH_2CH_3)$.
$3$. The ozonolysis of $A$ $(C_7H_{14})$ yields $B$ $(CH_3CHO)$ and $C$ $(CH_3CH_2COCH_2CH_3)$.
$4$. Reconstructing $A$ from the products: $CH_3CH=O + O=C(CH_2CH_3)_2 \rightarrow CH_3CH=C(CH_2CH_3)_2$.
$5$. The structure $CH_3CH=C(CH_2CH_3)_2$ is $3$-ethylpent-$2$-ene.

(C) The reaction involves the electrophilic addition of $2HBr$ to a conjugated diene system.
In the given molecule,the $NO_2$ group is a strong electron-withdrawing group,which deactivates the double bond adjacent to it.
The other double bond is more electron-rich and undergoes protonation first to form the most stable carbocation.
Following the addition of the first $Br^-$ and then the second $HBr$ molecule,the major product is formed by following Markovnikov's rule and considering the stability of the intermediate carbocation.
The final product is $3,4-dibromo-1-methyl-4-nitrocyclohexane$.

(D) The formation of methanal $(HCHO)$ during ozonolysis indicates the presence of a terminal double bond $(=CH_2)$ in the alkene.
Among the given options,the structure in option $D$ (allylcyclohexane,which is $cyclohexyl-CH_2-CH=CH_2$) contains a terminal double bond.
Upon ozonolysis,it undergoes cleavage at the double bond to produce methanal $(HCHO)$ and cyclohexylacetaldehyde $(C_6H_{11}-CH_2-CHO)$.

(B) The reaction of $CH_3CH_2CH=CH_2$ with $B_2H_6$ followed by oxidation with $H_2O_2$ in the presence of $OH^-$ is known as hydroboration-oxidation.
This reaction follows anti-Markovnikov's rule,where the $OH$ group is added to the less substituted carbon atom of the double bond.
Thus,the product $Z$ formed is $CH_3CH_2CH_2CH_2OH$ (butan$-1-$ol).

(C) The reaction follows Markovnikov's addition of $HCl$ to the alkene.
$1$. The $H^+$ ion attacks the terminal carbon of the double bond to form the most stable carbocation.
$2$. The intermediate formed is a $3^{\circ}$ benzylic carbocation $(C_6H_5-CH^+-CH_2-CH_3)$,which is highly stabilized by resonance with the phenyl ring.
$3$. The $Cl^-$ ion then attacks this carbocation to form the final product: $1$-chloro-$1$-phenylpropane $(C_6H_5-CH(Cl)-CH_2-CH_3)$.

(D) The reaction is an electrophilic addition of $HBr$ to a conjugated diene $(2\text{-methylbuta-1,3-diene})$ at low temperature $(-80^{\circ}C)$,which favors the kinetic product $(1,2\text{-addition})$.
$1$. The proton $(H^+)$ from $HBr$ attacks the terminal double bond to form the most stable carbocation.
$2$. Protonation at the $C_1$ position leads to a tertiary allylic carbocation,which is more stable than the secondary allylic carbocation formed by protonation at the $C_4$ position.
$3$. The bromide ion $(Br^-)$ then attacks the tertiary carbocation at the $C_2$ position to form the $1,2\text{-addition}$ product.
$4$. The final product $A$ is $2\text{-bromo-2-methylbut-3-ene}$.

(D) The conversion $CH_2=CH-CH_2-CHO \to CH_3-CH_2-CH_2-CH_2OH$ involves the reduction of both the carbon-carbon double bond $(C=C)$ and the aldehyde group $(-CHO)$.
$LiAlH_4$,$NaBH_4$,and $Na + C_2H_5OH$ are selective reducing agents that reduce the aldehyde group to a primary alcohol but do not reduce isolated $C=C$ double bonds.
Catalytic hydrogenation using $H_2 / Ni$ (or $Pt/Pd$) is the appropriate method to reduce both the alkene and the aldehyde functional groups to an alkane and a primary alcohol,respectively.

(B) The reaction of $1\text{-methyl-1-vinylcyclohexane}$ with $HBr$ proceeds via an electrophilic addition mechanism.
$1$. Protonation of the double bond occurs to form the most stable carbocation. Initially,a secondary carbocation is formed at the terminal carbon of the vinyl group.
$2$. This secondary carbocation undergoes a $1,2\text{-methyl shift}$ from the cyclohexane ring to the adjacent carbon to form a more stable tertiary carbocation at the ring position.
$3$. Finally,the bromide ion $(Br^-)$ attacks this tertiary carbocation to form the major product,which is $1\text{-bromo-1-isopropylcyclohexane}$.

(C) The reaction shows the chlorination of cyclohexene to form a dichlorocyclohexene derivative.
Specifically,the reaction involves the substitution of hydrogen atoms at the allylic positions.
Free radical halogenation using $Cl_2$ in the presence of $UV$ light is the standard method for allylic substitution in alkenes.
Therefore,the correct reagent and condition is $A = Cl_2$ with $UV$ light.

(C) The reaction of $HI$ with $3,3$-dimethyl-$1$-butene $(CH_3-C(CH_3)_2-CH=CH_2)$ proceeds via a carbocation intermediate.
First,protonation of the alkene forms a secondary carbocation: $CH_3-C(CH_3)_2-C^+H-CH_3$.
To increase stability,a $1,2$-methyl shift occurs from the adjacent quaternary carbon,resulting in a more stable tertiary carbocation: $CH_3-C^+(CH_3)-CH(CH_3)-CH_3$.
Finally,the nucleophilic attack of the iodide ion $(I^-)$ on this tertiary carbocation yields the major product,$2$-iodo-$2,3$-dimethylbutane $(CH_3-CI(CH_3)-CH(CH_3)-CH_3)$.

(A) $1$. The reaction of $1-$methylcyclopentene with dilute alkaline $KMnO_4$ at $273 \ K$ (Baeyer's reagent) is a syn-hydroxylation reaction,which adds two hydroxyl groups across the double bond to form $1-$methylcyclopentane$-1,2-$diol $(A)$.
$2$. The product $A$ is a vicinal diol. The secondary alcohol group is more easily oxidized than the tertiary alcohol group. Treatment with $CrO_3$ (Jones reagent or similar chromium-based oxidant) selectively oxidizes the secondary alcohol to a ketone,resulting in $2-$hydroxy$-2-$methylcyclopentanone $(B)$.

(A) The reaction of ethylene glycol $(HOCH_2-CH_2OH)$ with oxalic acid $((COOH)_2)$ at $110^{\circ}C$ forms a cyclic ester (ethylene oxalate).
Upon further heating to $210^{\circ}C$,this cyclic ester undergoes decarboxylation to produce ethene $(CH_2=CH_2)$ and carbon dioxide $(CO_2)$.
Therefore,the major product $X$ formed at $210^{\circ}C$ is ethene $(CH_2=CH_2)$.

(A) The reaction involves the electrophilic addition of $HBr$ to a conjugated diene,$2$-methylbuta-$1,3$-diene (isoprene).
When $HBr$ is added to a conjugated diene,both $1,2$-addition and $1,4$-addition products can be formed.
In the presence of excess $HBr$,the reaction proceeds further. However,the question asks for the major product of the initial addition step or the primary outcome of the reaction.
The $1,4$-addition product is generally more stable due to the formation of a more substituted alkene.
The reaction of $2$-methylbuta-$1,3$-diene with $HBr$ yields $1$-bromo-$3$-methylbut-$2$-ene as the major $1,4$-addition product.

(B) The reaction involves the acid-catalyzed dehydration of $3,3-dimethylbutan-2-ol$.
$1$. Protonation of the $-OH$ group leads to the formation of a secondary carbocation: $(CH_3)_3C-CH^+(CH_3)$.
$2$. This secondary carbocation undergoes a $1,2-methyl$ shift to form a more stable tertiary carbocation: $(CH_3)_2C^+-CH(CH_3)_2$.
$3$. Loss of a proton from the adjacent carbon atom results in the formation of the most stable alkene,which is $2,3-dimethylbut-2-ene$ $(CH_3-C(CH_3)=C(CH_3)_2)$.
Thus,the major product is $2,3-dimethylbut-2-ene$.

(C) The reaction of propene $(CH_3-CH=CH_2)$ with bromine water $(Br_2/H_2O)$ proceeds via the formation of a cyclic bromonium ion intermediate.
The water molecule then attacks the more substituted carbon atom of the bromonium ion,which is consistent with the regioselectivity predicted by the Markovnikov rule (as the more substituted carbon can better stabilize the partial positive charge during the transition state).
This leads to the formation of $1-$bromopropan$-2-$ol $(CH_3-CH(OH)-CH_2Br)$.
Therefore,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(B) The reagent $Na/H_{2}$ is not a standard reducing agent used in organic synthesis.
$Pt/C, H_{2}$ and $Pd/C, H_{2}$ are common catalysts for catalytic hydrogenation.
$Zn/H_{2}O$ is used for specific reductions like the reduction of ozonides.
Therefore,$Na/H_{2}$ is the correct choice.

(A) In the presence of peroxides like benzoyl peroxide $((C_6H_5CO)_2O_2)$,the addition of $HBr$ to unsymmetrical alkenes follows the Anti-Markovnikov rule (also known as the peroxide effect or Kharasch effect).
In this mechanism,the bromine radical attacks the less substituted carbon atom of the double bond to form a more stable radical intermediate.
Reaction: $CH_3-CH(CH_3)-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH(CH_3)-CH_2-CH_2-Br$ ($1$-bromo$-3-$methylbutane).

(A) The reaction of cyclohexene with $KMnO_4 / H_2SO_4 / \Delta$ (hot acidic $KMnO_4$) causes oxidative cleavage of the double bond,resulting in the formation of a dicarboxylic acid (adipic acid) as compound '$A$'.
The reaction of cyclohexene with $KMnO_4 / H_2O / 273 \ K$ (cold alkaline $KMnO_4$,also known as Baeyer's reagent) results in syn-hydroxylation,forming a diol (cyclohexane$-1,2-$diol) as compound '$B$'.
Therefore,compound '$A$' is a dicarboxylic acid and compound '$B$' is a diol.

(D) The molecular formula for pentene is $C_5H_{10}$. The acyclic structural isomers are:
$1$. $Pent-1-ene$ $(CH_2=CH-CH_2-CH_2-CH_3)$
$2$. $Pent-2-ene$ $(CH_3-CH=CH-CH_2-CH_3)$
$3$. $2-Methylbut-1-ene$ $(CH_2=C(CH_3)-CH_2-CH_3)$
$4$. $3-Methylbut-1-ene$ $(CH_2=CH-CH(CH_3)_2)$
$5$. $2-Methylbut-2-ene$ $(CH_3-C(CH_3)=CH-CH_3)$
Now,considering geometrical isomers:
$Pent-2-ene$ exists as $cis$ and $trans$ isomers.
Therefore,the total number of acyclic isomers is $5$ (structural) $+ 1$ (additional geometrical isomer for $pent-2-ene$) $= 6$ isomers.
These are:
$1$. $Pent-1-ene$
$2$. $cis-Pent-2-ene$
$3$. $trans-Pent-2-ene$
$4$. $2-Methylbut-1-ene$
$5$. $3-Methylbut-1-ene$
$6$. $2-Methylbut-2-ene$
Thus,the total count is $6$.

(B) The reaction of an alkene with $KMnO_4 / H^{+}$ (oxidative cleavage) or ozonolysis ($O_3$ followed by $Zn/H_2O$) breaks the double bond.
Compound $'A'$ $(C_4H_8)$ yields $C_3H_6O$ (acetone) as one of the products.
$2-$Methylpropene $(CH_3-C(CH_3)=CH_2)$ undergoes oxidative cleavage with $KMnO_4 / H^{+}$ to form acetone $(CH_3COCH_3)$ and $CO_2$.
Similarly,ozonolysis of $2-$Methylpropene gives acetone $(CH_3COCH_3)$ and formaldehyde $(HCHO)$.
Since the question specifies that $'A'$ yields $'B'$ $(C_3H_6O)$,$2-$Methylpropene is the correct structure for $'A'$.

(D) The reaction of $cis-but-2-ene$ with $Br_2$ in $CCl_4$ proceeds via an anti-addition mechanism.
This results in the formation of a racemic mixture of $(2R, 3R)-2,3-dibromobutane$ and $(2S, 3S)-2,3-dibromobutane$.
Since these two enantiomers are stereoisomers,the total number of stereoisomers formed as product $'P'$ is $2$.

(D) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the addition of oxygen atoms to each carbon atom of the double bond.
Formaldehyde is $HCHO$ and $2-$methylpropanal is $(CH_{3})_{2}CHCHO$.
By joining the carbonyl carbons of these products by a double bond,we get the structure of the original alkene $X$:
$CH_{3}-CH(CH_{3})-CH=O + O=CH_{2} \xrightarrow{Zn/H_{2}O} CH_{3}-CH(CH_{3})-CH=CH_{2} + H_{2}O$
The structure $CH_{3}-CH(CH_{3})-CH=CH_{2}$ corresponds to $3-$methylbut$-1-$ene.
Therefore,the correct option is $D$.

(D) Statement $I$ is incorrect. Alkenes are generally considered more reactive than alkanes due to the presence of the $\pi$-bond,but the stability of a molecule depends on its total bond energy. Alkenes are not inherently 'less stable' than alkanes in a way that makes them unstable compounds; they are simply more reactive towards electrophilic addition.
Statement $II$ is correct. $A$ carbon-carbon double bond $(C=C)$ consists of one $\sigma$-bond and one $\pi$-bond. The total bond energy of a $C=C$ bond (approx. $610 \ kJ/mol$) is significantly higher than that of a $C-C$ single bond (approx. $348 \ kJ/mol$). Therefore,the strength of the double bond is greater than that of the single bond.

(A) The reaction of $3,4-dihydro-2H-pyran$ with $Br_2$ in the presence of $CH_3OH$ proceeds via the formation of a cyclic bromonium ion intermediate.
$1.$ $Br_2$ adds to the double bond to form a cyclic bromonium ion.
$2.$ The nucleophile $CH_3OH$ attacks the more substituted carbon atom of the bromonium ion due to electronic effects and the stability of the transition state.
$3.$ This results in an anti-addition product where the $-OCH_3$ group and the $-Br$ atom are in a trans-configuration relative to each other.
Thus,the major product is the one where $-Br$ and $-OCH_3$ are trans to each other.

(A) The chemical reaction is: $C_{5}H_{10} + Br_{2} \rightarrow C_{5}H_{10}Br_{2}$
From the stoichiometry,$1 \ mol$ of pent$-1-$ene reacts with $1 \ mol$ of $Br_{2}$.
Molar mass of pent$-1-$ene $(C_{5}H_{10})$ $= 5 \times 12 + 10 \times 1 = 70 \ g/mol$.
Molar mass of $Br_{2} = 2 \times 80 = 160 \ g/mol$.
Moles of pent$-1-$ene $= \frac{5.0 \ g}{70 \ g/mol} = \frac{1}{14} \ mol$.
Since the molar ratio is $1:1$,moles of $Br_{2}$ required $= \frac{1}{14} \ mol$.
Mass of $Br_{2}$ required $= \text{moles} \times \text{molar mass} = \frac{1}{14} \times 160 \ g = 11.4285 \ g$.
Expressing in terms of $10^{-2} \ g$: $11.4285 \times 100 \times 10^{-2} \ g = 1142.85 \times 10^{-2} \ g$.
Rounding to the nearest integer,we get $1143 \times 10^{-2} \ g$.

(A) The reaction is Oxymercuration-Demercuration of $3,3-$dimethylbut$-1-$ene.
This reaction involves the addition of $H_2O$ across the double bond following Markovnikov's rule.
An important feature of this reaction is that it proceeds without carbocation rearrangement.
Therefore,the $OH$ group attaches to the more substituted carbon ($C$-$2$) and the $H$ atom attaches to the less substituted carbon ($C$-$1$),resulting in $3,3-$dimethylbutan$-2-$ol as the major product.

(C) Step $1$: The terminal alkyne $n-Bu-C \equiv CH$ reacts with $n-BuLi$ to form the acetylide anion $n-Bu-C \equiv C^- Li^+$.
Step $2$: The acetylide anion undergoes an $S_N2$ reaction with $n-C_5H_{11}Cl$ to form the internal alkyne $n-Bu-C \equiv C-C_5H_{11}$.
Step $3$: The hydrogenation of the internal alkyne using $Lindlar$ catalyst and $H_2$ results in the formation of a $cis$-alkene.
Therefore,the major product is the $cis$-isomer where the $n-Bu$ and $C_5H_{11}$ groups are on the same side of the double bond,which corresponds to option $C$.

(D) Ozonolysis of $A$ gives $\text{ethane-}1,2\text{-dicarbaldehyde}$ (succinaldehyde) and $\text{glyoxal}$ (ethanedial). This indicates $A$ is $\text{cyclohex-}1,3\text{-diene}$.
Ozonolysis of $B$ gives $5\text{-oxohexanal}$. This indicates $B$ is $1\text{-methylcyclopent-}1\text{-ene}$.
Thus,$A$ is $\text{cyclohex-}1,3\text{-diene}$ and $B$ is $1\text{-methylcyclopent-}1\text{-ene}$.

(C) The reaction is an electrophilic addition of $HBr$ to the alkene $Ph-CH=CH-CH_2Br$.
$1$. The proton $(H^+)$ from $HBr$ attacks the double bond to form the most stable carbocation.
$2$. The carbocation formed at the benzylic position $(Ph-CH^+-CH_2-CH_2Br)$ is resonance-stabilized by the phenyl group.
$3$. The bromide ion $(Br^-)$ then attacks this stable carbocation to form the major product.
$4$. Thus,the major product is $Ph-CH(Br)-CH_2-CH_2Br$.

(D) The molecular formula $C_4H_8$ corresponds to alkenes.
Oxidation of alkenes with acidic $KMnO_4$ leads to cleavage of the double bond.
$2-$methylpropene $(CH_3)_2C=CH_2$ on oxidation with acidic $KMnO_4$ gives acetone $(CH_3)_2C=O$ and $CO_2$ gas,which causes effervescence.
The reaction is: $(CH_3)_2C=CH_2 + [O] \xrightarrow{KMnO_4/H^+} (CH_3)_2C=O + CO_2 + H_2O$.
Thus,compound $A$ is $2-$methylpropene.

(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in the molecule,which is allyl acrylate $(CH_2=CH-COO-CH_2-CH=CH_2)$.
$1$. The first $HBr$ molecule adds to the more reactive double bond. The double bond in the allyl group (alkene part) is more electron-rich than the double bond conjugated with the carbonyl group (acrylate part). Thus,$HBr$ adds to the allyl double bond following Markovnikov's rule to form $CH_2=CH-COO-CH_2-CH(Br)-CH_3$.
$2$. The second $HBr$ molecule adds to the remaining double bond in the acrylate part. This addition follows anti-Markovnikov's rule (if peroxide is present,though usually,in the absence of peroxide,it follows Markovnikov's rule,but here the electron-withdrawing carbonyl group directs the addition to the terminal carbon to form the more stable carbocation or via conjugate addition). The addition to the $CH_2=CH-$ group results in the formation of $Br-CH_2-CH_2-COO-CH_2-CH(Br)-CH_3$.
Therefore,the final product is $1-$bromopropan$-2-$yl $3-$bromopropanoate.

(D) The compound $A$ is $2,4,4-trimethylpent-1-ene$.
Reaction with $O_3, H_2O / Zn$ (reductive ozonolysis) gives formaldehyde $(HCHO)$ as by-product $a$ and a ketone.
Reaction with $KMnO_4 / H^{+}$ (oxidative cleavage) gives formic acid $(HCOOH)$ as by-product $b$ and a ketone.
Formaldehyde $(a)$ on oxidation gives formic acid,which is found in ants.
Thus,$X$ is $O_3, H_2O / Zn$ and $Y$ is $KMnO_4 / H^{+}$.

(B) The reaction with $Hg(OAc)_2, H_2O$ followed by $NaBH_4$ is Oxymercuration-Demercuration,which follows Markovnikov's rule to give product $A$ ($3$,$3$-dimethylbutan$-2-$ol).
The reaction with $(BH_3)_2$ followed by $H_2O_2/OH^-$ is Hydroboration-Oxidation,which follows anti-Markovnikov's rule to give product $B$ ($3$,$3$-dimethylbutan$-1-$ol).
Therefore,$A$ is the Markovnikov product and $B$ is the anti-Markovnikov product.

(A) The reaction is an iodolactonization followed by dehydroiodination.
$1$. Treatment of the unsaturated carboxylic acid with $I_2/NaHCO_3$ leads to the formation of an iodolactone intermediate. The carboxylate oxygen attacks the iodonium ion formed across the double bond,resulting in a bicyclic lactone with an iodine atom at the bridgehead position.
$2$. Subsequent treatment with pyridine and heat $(\Delta)$ promotes an $E2$ elimination reaction,removing the iodine atom and a hydrogen atom from the adjacent carbon to form a double bond.
$3$. The final product is a bicyclic lactone containing a double bond in the six-membered ring,which corresponds to option $A$.

(C) $1$. Ethanol $(CH_{3}CH_{2}OH)$ on heating with concentrated $H_{2}SO_{4}$ undergoes dehydration to produce ethene gas $(CH_{2}=CH_{2})$.
$2$. Ethene reacts with cold dilute aqueous solution of alkaline $KMnO_{4}$ (Baeyer's reagent) to undergo hydroxylation,resulting in the formation of ethane$-1,2-$diol,commonly known as glycol $(HOCH_{2}-CH_{2}OH)$.

(A) $1$. The molecular formula of $X$ is $C_{11}H_{14}$.
$2$. Ozonolysis of $X$ gives $P$ and $Q$. $P$ gives a positive iodoform test $(I_2/NaOH)$ and does not reduce Tollen's reagent,identifying $P$ as a methyl ketone (specifically acetophenone,$Ph-CO-CH_3$).
$3$. $Q$ gives a positive Fehling's test but no iodoform test,identifying $Q$ as an aliphatic aldehyde (specifically propanal,$CH_3CH_2CHO$).
$4$. Combining these fragments ($Ph-CO-CH_3$ and $CH_3CH_2CHO$) via ozonolysis reversal,the structure of $X$ is $Ph-C(CH_3)=CH-CH_2-CH_3$.
$5$. Hydrogenation of $X$ $(Ph-C(CH_3)=CH-CH_2-CH_3 + H_2 \rightarrow Ph-CH(CH_3)-CH_2-CH_2-CH_3)$ produces a chiral center at the carbon attached to the phenyl group,making the product optically active.

(C) The reaction is an ozonolysis of an alkene $P$ $(M.wt. = 210)$.
Number of moles of $P = \frac{420 \times 10^{-3} \,g}{210 \,g/mol} = 2 \times 10^{-3} \,mol$.
Since the yield of both $Q$ and $R$ is $40 \,\%$,the number of moles of $Q$ and $R$ produced is $40 \,\%$ of $2 \times 10^{-3} \,mol = 8 \times 10^{-4} \,mol$.
Product $Q$ is benzaldehyde ($C_6H_5CHO$,$M = 106$) and $R$ is $3$-methoxybenzaldehyde ($CH_3OC_6H_4CHO$,$M = 136$).
Mass of $R = \text{moles} \times \text{molar mass} = 8 \times 10^{-4} \,mol \times 136 \,g/mol = 1088 \times 10^{-4} \,g = 108.8 \,mg$.
Mass of $Q = 8 \times 10^{-4} \,mol \times 106 \,g/mol = 848 \times 10^{-4} \,g = 84.8 \,mg$.
Given that $108.8 \,mg$ of $Q$ is produced,$R$ must be the product with mass $84.8 \,mg$.

(A) Step $(i)$ and $(ii)$ represent the hydroboration-oxidation of the alkene,which follows anti-Markownikoff's addition of water across the double bond. This results in the formation of an alcohol where the $-OH$ group attaches to the less substituted carbon.
Step $(iii)$ involves the acid-catalyzed dehydration of the alcohol using $conc. \ H_2SO_4$. This proceeds via an $E1$ mechanism involving a carbocation intermediate,followed by the elimination of a proton to form the most stable alkene according to Zaitsev's rule.
The major product formed is $1-methyl-3-phenylcyclohex-1-ene$ (or a related isomer depending on the specific structure,but based on the provided reaction sequence,the product is the more substituted alkene).

(C) An optically active compound must contain at least one chiral center (a carbon atom bonded to four different groups).
$I$: $2$-ethyl-pent-$1$-ene on hydrogenation gives $3$-methylhexane,which has a chiral center at $C3$.
$II$: $3$-ethyl-hex-$3$-ene on hydrogenation gives $3$-ethylhexane,which is achiral.
$III$: $5$-methyl-hept-$1$-ene on hydrogenation gives $3$-methylheptane,which has a chiral center at $C3$.
$IV$: $4$-methyl-hept-$2$-ene on hydrogenation gives $3$-methylheptane,which has a chiral center at $C3$.
Wait,let us re-evaluate the structures:
$I$: $2$-ethyl-pent-$1$-ene $\xrightarrow{H_2/Ni}$ $3$-methylhexane (Chiral).
$II$: $3$-ethyl-hex-$3$-ene $\xrightarrow{H_2/Ni}$ $3$-ethylhexane (Achiral).
$III$: $5$-methyl-hept-$1$-ene $\xrightarrow{H_2/Ni}$ $3$-methylheptane (Chiral).
$IV$: $4$-methyl-hept-$2$-ene $\xrightarrow{H_2/Ni}$ $3$-methylheptane (Chiral).
Based on the provided image and standard analysis,$I$ and $III$ are the intended answers as they form chiral centers upon hydrogenation of the double bond.

(B) The reaction of cyclic alkenes with strong oxidizing agents like acidic $KMnO_4$ leads to the oxidative cleavage of the double bond.
For cyclohexene,the oxidative cleavage of the $C=C$ double bond results in the formation of a dicarboxylic acid.
The reaction proceeds as follows:
Cyclohexene $\xrightarrow{Acidic \ KMnO_4, \Delta}$ Hexan$-1,6-$dioic acid (Adipic acid).
Thus,the starting compound is cyclohexene.

(A) The reaction is an acid-catalyzed hydration of an alkene.
$1$. The proton $(H^+)$ attacks the double bond to form the most stable carbocation intermediate.
$2$. The protonation of the alkene leads to a tertiary carbocation at the carbon attached to the cyclohexyl group.
$3$. Water $(H_2O)$ then acts as a nucleophile and attacks this tertiary carbocation.
$4$. Subsequent deprotonation yields the tertiary alcohol as the major product,which corresponds to structure $I$.

(C) The reaction of allyl alcohol $(CH_2=CH-CH_2OH)$ with concentrated $HBr$ is a nucleophilic substitution reaction.
First,the hydroxyl group is protonated to form a good leaving group $(-OH_2^+)$.
Then,water is eliminated to form the resonance-stabilized allylic carbocation ($CH_2=CH-CH_2^+$ $\leftrightarrow$ $^+CH_2-CH=CH_2$).
Finally,the bromide ion $(Br^-)$ attacks the carbocation to form allyl bromide $(CH_2=CH-CH_2Br)$.
Thus,the major product is allyl bromide.

(A) The reaction of cyclohexene with bromine in $CCl_4$ in the dark is an electrophilic addition reaction.
Bromine adds across the double bond in an anti-addition manner,resulting in the formation of $trans-1,2-dibromocyclohexane$.
The reaction proceeds through a cyclic bromonium ion intermediate,which is then attacked by the bromide ion from the opposite side,leading to the trans-product.