Alkene Questions in English

(C) The structure of $2-$methyl$-2-$butene is $(CH_3)_2C=CH-CH_3$.
Ozonolysis involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
When $2-$methyl$-2-$butene reacts with $O_3$ followed by $Zn/H_2O$,the double bond breaks to form acetone $(CH_3)_2C=O$ and acetaldehyde $CH_3CHO$.
Since the products are both a ketone (acetone) and an aldehyde (acetaldehyde),the correct option is $C$.

(B) In hydroboration-oxidation reaction:
$1$. The reaction involves the syn-addition of $BH_3$ across the double bond.
$2$. The reaction proceeds through a concerted mechanism involving a four-membered cyclic transition state.
$3$. The product obtained is an alcohol (specifically,an anti-Markovnikov alcohol),not a vicinal diol.
$4$. The reaction follows the Anti-Markovnikov rule,but it is a syn-addition,not anti-addition.
Therefore,statements $I$ and $II$ are correct.

(A) $1$. The first step is oxymercuration-demercuration of $3\text{-methylbut-1-ene}$,which follows Markovnikov's rule to give $3\text{-methylbutan-2-ol}$ as product $A$.
$2$. The second step is acid-catalyzed dehydration of $A$ $(H^\oplus/\Delta)$. The major product $B$ is the more stable alkene,$2\text{-methylbut-2-ene}$,formed via the Saytzeff rule.
$3$. The third step is ozonolysis of $B$ $(2\text{-methylbut-2-ene})$ using $O_3/Zn/H_2O$. Ozonolysis of $CH_3-C(CH_3)=CH-CH_3$ cleaves the double bond to yield acetone $(CH_3COCH_3)$ and acetaldehyde $(CH_3CHO)$.
$4$. Thus,$C$ and $D$ are $CH_3COCH_3$ and $CH_3CHO$.

(B) Ozonolysis of an unbranched alkene involves the cleavage of the $C=C$ double bond.
For an unbranched alkene of the general form $R_1-CH=CH-R_2$,the reaction with $O_3$ followed by reductive workup $(Zn/H_2O)$ results in the formation of two aldehyde molecules.
The general reaction is: $R_1-CH=CH-R_2 \xrightarrow{O_3, Zn/H_2O} R_1-CHO + R_2-CHO$.
Since the carbon atoms involved in the double bond are bonded to at least one hydrogen atom,the products are aldehydes.

(C) Hydroboration-oxidation of alkenes is an anti-Markovnikov addition of water $(H_2O)$ across the double bond.
For but$-1-$ene $(CH_3CH_2CH=CH_2)$,the reaction proceeds as follows:
$CH_3CH_2CH=CH_2 + (BH_3)_2 \rightarrow CH_3CH_2CH_2CH_2BH_2$
Followed by oxidation with $H_2O_2/OH^-$:
$CH_3CH_2CH_2CH_2BH_2 + 3H_2O_2 + OH^- \rightarrow CH_3CH_2CH_2CH_2OH + B(OH)_4^-$
The final product is butan$-1-$ol.

(C) Hydroboration-oxidation of alkenes follows anti-Markovnikov addition of water across the double bond.
For but-$1$-ene $(CH_3CH_2CH=CH_2)$,the hydroxyl group $(-OH)$ attaches to the terminal carbon atom.
Therefore,the product formed is butan-$1$-ol $(CH_3CH_2CH_2CH_2OH)$.

(A) The reaction proceeds via the electrophilic addition of $H_2SO_4$ to propene according to Markovnikov's rule.
$CH_3-CH=CH_2 + HOSO_3H \rightarrow CH_3-CH(OSO_3H)-CH_3$ (Isopropyl hydrogen sulphate).
On heating with water (hydrolysis),the isopropyl hydrogen sulphate yields propan$-2-$ol.
$CH_3-CH(OSO_3H)-CH_3 + H_2O \xrightarrow{\Delta} CH_3-CH(OH)-CH_3 + H_2SO_4$.

(A) The dehydration of $2-$Methylhexan$-3-$ol $(CH_3-CH(CH_3)-CH(OH)-CH_2-CH_2-CH_3)$ with concentrated $H_2SO_4$ proceeds via an $E1$ mechanism involving a carbocation intermediate.
$1$. Protonation of the $-OH$ group followed by the loss of water generates a secondary carbocation at the $C3$ position: $CH_3-CH(CH_3)-CH^+-CH_2-CH_2-CH_3$.
$2$. This secondary carbocation can undergo a $1,2-$hydride shift to form a more stable tertiary carbocation at the $C2$ position: $CH_3-C^+(CH_3)-CH_2-CH_2-CH_2-CH_3$.
$3$. Elimination of a proton from the adjacent carbon atoms leads to the formation of alkenes. According to Saytzeff's rule,the most substituted alkene is the major product.
$4$. The tertiary carbocation can lose a proton from $C3$ to form $2-$Methylhex$-2-$ene $(CH_3-C(CH_3)=CH-CH_2-CH_2-CH_3)$,which is a trisubstituted alkene and the most stable product.
Therefore,the major product is $2-$Methylhex$-2-$ene.

(D) When $2-$methylpropan$-2-$ol (tert-butyl alcohol) is passed over heated alumina $(Al_2O_3)$ at $423 \ K$,it undergoes dehydration to form $2-$methylpropene (isobutylene).
The reaction is: $(CH_3)_3C-OH \xrightarrow{Al_2O_3, 423 \ K} CH_3-C(CH_3)=CH_2 + H_2O$.
$2-$methylpropene is commonly known as isobutylene.

(B) The reaction of alkenes with cold and dilute alkaline $KMnO_4$ (Baeyer's reagent) results in the syn-hydroxylation of the double bond to form vicinal diols,commonly known as glycols.
For example,the reaction of ethene with cold and dilute alkaline $KMnO_4$ yields ethane$-1,2-$diol (ethylene glycol):
$CH_2=CH_2 + H_2O + [O] \xrightarrow{\text{alkaline } KMnO_4} HOCH_2-CH_2OH$

(D) Ozonolysis of propene $(CH_3-CH=CH_2)$ involves the reaction with ozone $(O_3)$ followed by reductive cleavage with zinc and water $(Zn/H_2O)$.
The reaction breaks the double bond and forms carbonyl compounds:
$CH_3-CH=CH_2 \xrightarrow[(ii) Zn/H_2O]{(i) O_3} CH_3CHO + HCHO$
The products obtained are acetaldehyde $(CH_3CHO)$ and formaldehyde $(HCHO)$.

(B) Hardening of oil involves the addition of hydrogen $(H_2)$ to unsaturated vegetable oils in the presence of a catalyst such as $Ni$,$Pd$,or $Pt$ to form saturated solid fats.
This process is called hydrogenation.

(A) The oxidation of $prop-1-ene$ $(CH_3-CH=CH_2)$ with acidic $KMnO_4$ leads to the cleavage of the double bond.
The terminal $CH_2$ group is oxidized to $CO_2$ and $H_2O$,while the $CH_3-CH=$ group is oxidized to ethanoic acid $(CH_3COOH)$.
The reaction is: $CH_3-CH=CH_2 + [O] \xrightarrow{acidic \ KMnO_4} CH_3COOH + CO_2 + H_2O$.

(C) When phenylethene $(C_6H_5CH=CH_2)$ is treated with acidic potassium permanganate ($KMnO_4$ in dilute $H_2SO_4$),it undergoes oxidative cleavage.
The double bond is broken,and the terminal $CH_2$ group is oxidized to $CO_2$ and $H_2O$,while the phenyl-substituted carbon is oxidized to a carboxylic acid group.
Thus,phenylethene is oxidized to benzoic acid $(C_6H_5COOH)$.

(D) The reaction of cyclohexene with hot acidic $KMnO_4$ (oxidative cleavage) leads to the breaking of the double bond.
Since the double bond is part of a ring,the ring opens to form a dicarboxylic acid.
The product formed is hexane$-1,6-$dioic acid,commonly known as adipic acid.

(D) Adipic acid is a dicarboxylic acid with the formula $HOOC-(CH_2)_4-COOH$.
Oxidation of cyclic alkenes with strong oxidizing agents like $KMnO_4$ in acidic medium leads to the cleavage of the double bond and the ring,resulting in the formation of dicarboxylic acids.
Cyclohexene,when treated with $KMnO_4$ in dil. $H_2SO_4$,undergoes oxidative cleavage to form adipic acid (hexanedioic acid).
The reaction is: $\text{Cyclohexene} + [O] \xrightarrow{KMnO_4/H^+} HOOC-(CH_2)_4-COOH$.

(A) Alkaline $KMnO_4$ is known as Baeyer's reagent.
It is used as an oxidizing agent in organic chemistry to test for the presence of unsaturation (double or triple bonds) in organic compounds.

(B) The stability of alkenes increases with an increase in the number of alkyl groups attached to the doubly bonded carbon atoms due to the hyperconjugation effect and inductive effect.
The order of stability is:
$R_2C=CR_2 > R_2C=CHR > R_2C=CH_2$
Thus,the correct option is $B$.

(D) The molecular formula $C_5H_{10}$ corresponds to the general formula $C_nH_{2n}$,which represents an alkene.
To find the structural isomers,we consider the position of the double bond and the carbon chain branching:
$1$. $Pent-1-ene$ $(CH_2=CH-CH_2-CH_2-CH_3)$
$2$. $Pent-2-ene$ $(CH_3-CH=CH-CH_2-CH_3)$
$3$. $2-Methylbut-1-ene$ $(CH_2=C(CH_3)-CH_2-CH_3)$
$4$. $3-Methylbut-1-ene$ $(CH_2=CH-CH(CH_3)_2)$
$5$. $2-Methylbut-2-ene$ $(CH_3-C(CH_3)=CH-CH_3)$
Thus,there are $5$ possible structural isomers for $C_5H_{10}$.

(A) For an alkene to exhibit cis-trans isomerism,each carbon atom of the double bond must be attached to two different groups.
In but$-1-$ene $(CH_3-CH_2-CH=CH_2)$,the terminal carbon atom $(C_1)$ is bonded to two identical hydrogen atoms.
Therefore,it cannot exhibit cis-trans isomerism.

(B) The alkenyl groups are derived by removing one hydrogen atom from an alkene. For $C_{4}H_{8}$ (butenes),the possible isomers are $1$-butene,$2$-butene,and $2$-methylpropene.
Removing a hydrogen atom from these structures gives the following $C_{4}H_{7}$ groups:
$(i)$ $CH_{3}CH_{2}CH=CH-$ ($1$-butenyl)
$(ii)$ $CH_{3}CH=CHCH_{2}-$ ($2$-butenyl)
$(iii)$ $CH_{2}=CHCH_{2}CH_{2}-$ ($3$-butenyl)
$(iv)$ $CH_{3}C(CH_{3})=CH-$ ($2$-methyl$-1-$propenyl)
$(v)$ $CH_{2}=C(CH_{3})CH_{2}-$ ($2$-methyl$-2-$propenyl)
Thus,there are $5$ possible alkenyl groups.

(D) The ease of formation of an alkene by dehydrohalogenation follows the stability of the resulting alkene. According to $Saytzeff$ rule,the more substituted alkene is more stable due to hyperconjugation and inductive effects. The stability order of alkenes is: $R_2C=CR_2 > R_2C=CHR > RCH=CHR > R_2C=CH_2$. Therefore,the most substituted alkene,$R_2C=CR_2$,is formed most easily.

(A) The reaction of isobutylene $(CH_3)_2C=CH_2$ with hydrogen bromide $(HBr)$ follows Markovnikov's rule.
According to this rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom having a lesser number of hydrogen atoms.
The reaction proceeds via the formation of a more stable $3^{\circ}$ carbocation intermediate.
$(CH_3)_2C=CH_2 + HBr \rightarrow (CH_3)_3C-Br$
The major product formed is $tert-$butyl bromide.

(C) The peroxide effect (Kharasch effect) is only applicable to the addition of $HBr$ to unsymmetrical alkenes.
It does not apply to $HCl$ or $HI$.
Therefore,the addition of $HCl$ to propene follows Markovnikov's rule $(M.R.)$ even in the presence of peroxide.
According to Markovnikov's rule,the negative part of the addendum $(Cl^-)$ attaches to the carbon atom with fewer hydrogen atoms.
Reaction: $CH_3-CH=CH_2 + HCl \xrightarrow{Na_2O_2} CH_3-CHCl-CH_3$ ($2-$Chloropropane).

(C) The reaction of $HCl$ with an unsymmetrical alkene like propene follows Markovnikov's rule,even in the presence of peroxide.
Peroxide effect (Kharasch effect) is only applicable to $HBr$ and not to $HCl$ or $HI$.
According to Markovnikov's rule,the negative part of the reagent $(Cl^-)$ attaches to the carbon atom with fewer hydrogen atoms.
Thus,$CH_3-CH=CH_2 + HCl \rightarrow CH_3-CHCl-CH_3$ ($2-$chloropropane).

(C) $(i)$ Propylene dibromide $(CH_3-CHBr-CH_2Br)$ $\xrightarrow{Zn, \Delta}$ Propene $(CH_3-CH=CH_2)$ $(A)$.
$(ii)$ Propene $(A)$ $\xrightarrow{HBr}$ $2-$Bromopropane $(CH_3-CHBr-CH_3)$ $(B)$ (following Markovnikov's rule).
$(iii)$ $2-$Bromopropane $(B)$ $\xrightarrow{Na, \text{ether}}$ $2,3-$Dimethylbutane $(CH_3-CH(CH_3)-CH(CH_3)-CH_3)$ $(C)$ (Wurtz reaction).

(C) The synthesis of $but-1-ene$ $(CH_2=CH-CH_2-CH_3)$ from $CH_3MgI$ (a Grignard reagent) involves a nucleophilic substitution reaction with an alkyl halide containing a terminal double bond.
Specifically,$CH_3MgI$ reacts with allyl chloride $(CH_2=CH-CH_2Cl)$ to form $but-1-ene$ and magnesium salts $(Mg(Cl)I)$:
$CH_2=CH-CH_2Cl + CH_3MgI \rightarrow CH_2=CH-CH_2-CH_3 + Mg(Cl)I$
Thus,the correct reagent is allyl chloride.

(A) The reaction of ethylene $(C_2H_4)$ with oxygen $(O_2)$ in the presence of a palladium catalyst supported on alumina $(Pd / Al_2O_3)$ is a catalytic oxidation process.
The chemical equation for this reaction is:
$2C_2H_4(g) + O_2(g) \xrightarrow{Pd / Al_2O_3} 2CH_3CHO(g)$
As shown in the reaction,the product formed is acetaldehyde $(CH_3CHO)$.

(C) The hydrogenation of ethene $(CH_2=CH_2)$ involves the addition of hydrogen in the presence of a metal catalyst such as finely divided $Ni$,$Pd$,or $Pt$.
The reaction is: $CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3$.

(D) When alkenes are heated with $Br_2$ at high temperature,the hydrogen atom of the allylic carbon is substituted by a halogen atom,resulting in the formation of an allyl halide.
$CH_2=CH-CH_3 + Br_2 \xrightarrow{\Delta} CH_2=CH-CH_2Br + HBr$
This reaction is known as allylic substitution.
The product formed is $3-$bromopropene.

(A) The stability of an alkene increases with the number of alkyl groups attached to the doubly bonded carbon atoms due to the hyperconjugation effect and inductive effect.
$R_2C=CR_2$ is a tetra-substituted alkene,which has the maximum number of alkyl groups.
Therefore,$R_2C=CR_2$ is the most stable alkene.

(B) The stability of alkenes is determined by the number of alkyl groups attached to the doubly bonded carbon atoms (hyperconjugation effect).
More alkyl groups lead to greater stability.
$I) (CH_3)_2C=C(CH_3)_2$ has $12$ alpha-hydrogens.
$II) (CH_3)_2C=CH_2$ has $6$ alpha-hydrogens.
$III) (CH_3)_2C=CHCH_3$ has $9$ alpha-hydrogens.
Therefore,the order of stability is $I > III > II$.

(A) An alkadiene is a hydrocarbon containing two double bonds.
$1$. Isoprene ($2$-methylbuta-$1,3$-diene) has two double bonds,making it an alkadiene.
$2$. $\beta$-Phellandrene is an alkatriene as it contains three double bonds.
$3$. But-$2$-ene and pent-$1$-ene are monoalkenes containing only one double bond.
Therefore,the correct option is $A$.

(D) The stability of alkenes is determined by the number of alkyl groups attached to the doubly bonded carbon atoms due to the hyperconjugation effect and inductive effect.
Greater the number of alkyl groups attached to the $C=C$ bond,greater is the stability of the alkene.
The order of stability is: $R_2C=CR_2 > R_2C=CHR > RCH=CHR > R_2C=CH_2 > RCH=CH_2 > CH_2=CH_2$.
Therefore,the alkene with the maximum number of alkyl groups,$R_2C=CR_2$,is the most stable.

(B) The reaction of $3-$bromo$-2-$methylpentane with alcoholic $KOH$ is a dehydrohalogenation reaction ($\beta-$elimination).
In this reaction,a hydrogen atom is removed from the $\beta-$carbon and a bromine atom is removed from the $\alpha-$carbon to form an alkene.
According to Saytzeff's rule,the major product is the more substituted alkene.
For $3-$bromo$-2-$methylpentane,elimination can occur at $C-2$ or $C-4$.
Elimination at $C-2$ gives $2-$methylpent$-2-$ene (a trisubstituted alkene),while elimination at $C-4$ gives $3-$methylpent$-2-$ene (a disubstituted alkene).
Thus,$2-$methylpent$-2-$ene is the major product.

(D) The oxidation of cyclohexene with $KMnO_4$ in the presence of dilute $H_2SO_4$ is a vigorous reaction that causes the cleavage of the double bond.
Initially,the double bond is broken to form an intermediate,which is further oxidized to a dicarboxylic acid.
Specifically,cyclohexene undergoes oxidative cleavage to yield $Hexanedioic \ acid$,commonly known as $Adipic \ acid$ $(HOOC-(CH_2)_4-COOH)$.

(C) Hydroboration-oxidation of propene $(CH_3-CH=CH_2)$ involves the addition of borane $(BH_3)$ to the double bond,followed by oxidation with alkaline hydrogen peroxide $(H_2O_2/OH^-)$.
This reaction follows anti-Markovnikov addition,where the hydroxyl group $(-OH)$ attaches to the less substituted carbon atom.
Therefore,propene yields propan-$1$-ol $(CH_3-CH_2-CH_2OH)$ as the final product.

(D) When alkenes are oxidized with strong oxidizing agents like $KMnO_4$ in the presence of dilute $H_2SO_4$ (acidic medium),the double bond undergoes oxidative cleavage.
Depending on the structure of the alkene,the products formed are ketones $(Alkanones)$,carboxylic acids $(Alkanoic \text{ } acids)$,or carbon dioxide and water.
Specifically,if the alkene has a $R_2C=CR_2$ structure,it yields ketones $(Alkanones)$.
If it has $RCH=CR_2$ or $RCH=CHR$ structures,it yields carboxylic acids $(Alkanoic \text{ } acids)$.
However,in the context of general oxidative cleavage of alkenes by $KMnO_4/H^+$,the formation of $Alkanoic \text{ } acids$ is the most common outcome for terminal and internal alkenes that are not fully substituted.
Given the options provided,$Alkanoic \text{ } acid$ is the most appropriate product for general oxidative cleavage.

(C) The reaction of $2-\text{methylbut-}2-\text{ene}$ with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition mechanism (Kharasch effect).
In this mechanism,the $Br^\bullet$ radical attacks the double bond to form the more stable free radical intermediate.
For $CH_3-C(CH_3)=CH-CH_3$,the $Br$ radical adds to the $CH$ carbon to form a tertiary radical intermediate $(CH_3-C^\bullet(CH_3)-CH(Br)-CH_3)$,which is more stable than the secondary radical.
However,in the presence of peroxide,the addition of $HBr$ to alkenes typically follows anti-Markovnikov's rule,where the $Br$ atom attaches to the carbon with more hydrogen atoms.
For $2-\text{methylbut-}2-\text{ene}$,the structure is $CH_3-C(CH_3)=CH-CH_3$. The carbon atoms of the double bond are $C_2$ (which has no $H$) and $C_3$ (which has one $H$).
According to the anti-Markovnikov rule,the $Br$ atom adds to the $C_3$ position,resulting in $2-\text{bromo-}3-\text{methylbutane}$ as the major product.

(A) The reaction of alkenes with $HBr$ follows Markovnikov's rule,where the electrophile $(H^+)$ adds to the carbon with more hydrogen atoms,and the nucleophile $(Br^-)$ adds to the more substituted carbon to form the most stable carbocation intermediate.
For $1-$methylcyclohexene,the double bond is between $C1$ and $C2$. Adding $H^+$ to $C2$ creates a stable tertiary carbocation at $C1$. Then,$Br^-$ attacks the $C1$ carbocation to form $1-$bromo$-1-$methylcyclohexane.
Therefore,$1-$methylcyclohexene is the correct reactant.

(D) Adipic acid is a dicarboxylic acid with the formula $HOOC-(CH_2)_4-COOH$.
Oxidation of cyclic alkenes with strong oxidizing agents like $KMnO_4$ in acidic medium leads to the cleavage of the double bond and the formation of dicarboxylic acids.
Cyclohexene $(C_6H_{10})$ undergoes oxidative cleavage of the $C=C$ bond to form hexanedioic acid,which is commonly known as adipic acid.
The reaction is: $Cyclohexene + [O] \xrightarrow{KMnO_4/H^+} HOOC-(CH_2)_4-COOH$.

(B) The reaction of $2-$methylbut$-2-$ene $(CH_3-C(CH_3)=CH-CH_3)$ with $HCl$ follows Markovnikov's rule.
The electrophilic addition of $H^+$ to the double bond forms the most stable carbocation.
The protonation of the double bond leads to the formation of a $3^{\circ}$ carbocation,$(CH_3)_2C^+-CH_2-CH_3$,which is more stable than a $2^{\circ}$ carbocation.
The chloride ion $(Cl^-)$ then attacks the $3^{\circ}$ carbocation to form $2-$chloro$-2-$methylbutane as the major product.

(A) The reaction is the electrophilic addition of bromine $(Br_2)$ to an alkene,specifically $2-$methylbut$-2-$ene.
In this reaction,the double bond breaks,and one bromine atom attaches to each of the two carbon atoms that were previously part of the double bond.
The starting material is $CH_3-C(CH_3)=CH-CH_3$.
Upon addition of $Br_2$,the product formed is $CH_3-C(Br)(CH_3)-CH(Br)-CH_3$.
This compound is named $2,3-$dibromo$-2-$methylbutane.

(B) The reaction of an alkene with cold and dilute alkaline $KMnO_4$ (also known as Baeyer's reagent) is an oxidation reaction.
In this reaction,the double bond of the alkene is cleaved,and two hydroxyl $(-OH)$ groups are added to the adjacent carbon atoms,resulting in the formation of a vicinal diol,commonly known as a glycol.
For example,ethene $(CH_2=CH_2)$ reacts with cold dilute alkaline $KMnO_4$ to form ethane$-1,2-$diol $(HO-CH_2-CH_2-OH)$.

(C) The ozonolysis of propene $(CH_3-CH=CH_2)$ involves the reaction with ozone $(O_3)$ to form a cyclic ozonide intermediate.
This intermediate is then cleaved using zinc dust and water $(Zn/H_2O)$ to produce carbonyl compounds.
The reaction is as follows:
$CH_3-CH=CH_2 + O_3 \rightarrow \text{Propene ozonide}$
$\text{Propene ozonide} + H_2O/Zn \rightarrow CH_3CHO + HCHO + ZnO$
The products formed are $CH_3CHO$ (Ethanal) and $HCHO$ (Methanal).

(D) The ozonolysis of $but-2-ene$ $(CH_3-CH=CH-CH_3)$ involves the addition of ozone $(O_3)$ to form an ozonide intermediate.
This ozonide is then cleaved by $Zn$ dust and $H_2O$ (reductive cleavage).
The reaction is: $CH_3-CH=CH-CH_3 + O_3$ $\rightarrow \text{ozonide}$ $\xrightarrow{Zn/H_2O} 2CH_3CHO$.
The product formed is $2$ moles of $acetaldehyde$ $(CH_3CHO)$.

(C) The reaction of $1$-methylcyclohexene with $HBr$ follows Markovnikov's rule.
In this reaction,the electrophilic $H^+$ adds to the carbon atom of the double bond that has more hydrogen atoms,and the nucleophilic $Br^-$ adds to the more substituted carbon atom,resulting in the formation of $1$-bromo-$1$-methylcyclohexane.

(A) The reaction of $Pent-2-ene$ $(CH_3-CH_2-CH=CH-CH_3)$ with ozone $(O_3)$ followed by reductive cleavage with $Zn/H_2O$ is known as ozonolysis.
In this reaction,the double bond is cleaved to form two carbonyl compounds.
$CH_3-CH_2-CH=CH-CH_3 \xrightarrow[(ii) Zn / H_2O]{(i) O_3} CH_3-CH_2-CHO + CH_3-CHO$
Here,$CH_3-CH_2-CHO$ is $Propanal$ and $CH_3-CHO$ is $Ethanal$.
Thus,$X$ and $Y$ are $CH_3CHO$ and $CH_3CH_2CHO$ respectively.

(D) When an alkene is heated with $Cl_2$ at high temperature,the hydrogen atom of the allylic carbon is substituted by a halogen atom,resulting in the formation of an allyl halide.
$CH_2=CH-CH_3 + Cl_2 \xrightarrow{\Delta} CH_2=CH-CH_2Cl + HCl$
This product is $3-$Chloropropene (also known as allyl chloride).