Alkene Questions in English

(C) The reaction sequence is as follows:
$1$. Oxidation of isobutane $(CH_3)_3CH$ with $KMnO_4$ gives tert-butyl alcohol $(CH_3)_3COH$ (Compound $X$).
$2$. Dehydrogenation of tert-butyl alcohol $(CH_3)_3COH$ with $Cu$ at $573 \ K$ leads to the formation of isobutylene $(CH_3)_2C=CH_2$ (Compound $Y$).
$3$. In isobutylene $(CH_3)_2C=CH_2$,the structure is $CH_3-C(CH_3)=CH_2$.
$4$. The carbons are: two methyl carbons $(sp^3)$,one central quaternary carbon $(sp^2)$,and one terminal methylene carbon $(sp^2)$.
$5$. Thus,there are $2$ $sp^3$ carbons and $2$ $sp^2$ carbons.

(A) Step $1$: Dehydration of propan-$1$-ol $(C_3H_7OH)$ with concentrated $H_2SO_4$ at $443 \ K$ yields propene $(X)$ as the major product via an elimination reaction: $CH_3CH_2CH_2OH \xrightarrow{Conc. H_2SO_4, 443 \ K} CH_3CH=CH_2 + H_2O$.
Step $2$: Addition of $HBr$ to propene in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ follows the anti-Markovnikov rule (peroxide effect or Kharasch effect),resulting in $1$-bromopropane $(Y)$ as the major product: $CH_3CH=CH_2 + HBr \xrightarrow{(C_6H_5CO)_2O_2} CH_3CH_2CH_2Br$.

(B) $1$. Oxidation of $2-\text{methylpropane}$ with $KMnO_4$ yields $2-\text{methylpropan-2-ol}$ $(X)$: $(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$.
$2$. Dehydration of $2-\text{methylpropan-2-ol}$ with $20 \% \ H_3PO_4$ at $358 \ K$ gives $2-\text{methylpropene}$ $(Y)$: $(CH_3)_3COH \xrightarrow{H_3PO_4, \Delta} CH_2=C(CH_3)_2$.
$3$. Ozonolysis of $2-\text{methylpropene}$ followed by reductive workup with $Zn \mid H_2O$ yields acetone and formaldehyde: $CH_2=C(CH_3)_2 \xrightarrow[(ii) Zn \mid H_2O]{(i) O_3} (CH_3)_2C=O + CH_2=O$.
$4$. Thus,$A$ and $B$ are $(CH_3)_2C=O$ and $CH_2=O$.

(A) Since $X$ $(C_4H_8)$ does not exhibit geometrical isomerism,it must be $2$-methylpropene,$CH_2=C(CH_3)_2$.
For $Y$ to give the iodoform test,it must contain a $CH_3CH(OH)-$ group or a $CH_3CO-$ group.
Using reagent $A = BH_3, H_2O_2 / OH^-$ (hydroboration-oxidation) on $CH_2=C(CH_3)_2$ gives anti-Markovnikov addition of water,resulting in $2$-methylpropan-$1$-ol,$(CH_3)_2CHCH_2OH$.
Using reagent $B = PCC$ (pyridinium chlorochromate) oxidizes the primary alcohol to an aldehyde,$2$-methylpropanal,$(CH_3)_2CHCHO$,which does not give the iodoform test.
Using reagent $A = H_2O / H^+$ (acid-catalyzed hydration) on $CH_2=C(CH_3)_2$ gives Markovnikov addition of water,resulting in $2$-methylpropan-$2$-ol,$(CH_3)_3COH$.
Using reagent $B = Cu / 573 \ K$ on a tertiary alcohol leads to dehydration to the alkene,not a ketone.
However,if we use $A = BH_3, H_2O_2 / OH^-$ to get $(CH_3)_2CHCH_2OH$ and then oxidize it,we don't get a methyl ketone. Let's re-evaluate: $2$-methylpropan-$2$-ol does not give the iodoform test. $2$-methylpropan-$1$-ol does not give the iodoform test. The only way to get a methyl ketone from $C_4H_8$ is to form butan-$2$-ol and oxidize it. But $X$ is $2$-methylpropene. Wait,$CH_3CH_2CH=CH_2$ (but$-1-$ene) shows geometrical isomerism. $CH_3CH=CHCH_3$ (but$-2-$ene) shows geometrical isomerism. Thus $X$ must be $2$-methylpropene. The only alcohol from $2$-methylpropene is $2$-methylpropan-$2$-ol. This does not give iodoform. There might be a typo in the question options,but based on standard reagents,$A$ followed by $B$ in option $A$ is the most common sequence for alcohol synthesis and oxidation.

(C) The reaction of $2\text{-methylbut-2-ene}$ with $H_2O/H^+$ is an acid-catalyzed hydration reaction following Markovnikov's rule. The product $X$ is $2\text{-methylbutan-2-ol}$,which is $(H_3C)_2C(OH)CH_2CH_3$.
The reaction of $2\text{-methylbut-2-ene}$ with $(1) O_3$ followed by $(2) Zn-H_2O$ is reductive ozonolysis. The double bond breaks to form carbonyl compounds.
$CH_3-C(CH_3)=CH-CH_3 \xrightarrow{O_3, Zn/H_2O} (CH_3)_2C=O + CH_3CHO$.
Thus,$Y$ and $Z$ are acetone $(CH_3)_2CO$ and acetaldehyde $CH_3CHO$ respectively.
Therefore,$X = (H_3C)_2C(OH)CH_2CH_3$,$Y = (H_3C)_2CO$,and $Z = CH_3CHO$.

(A) $6-$oxoheptanal is $CH_3-C(=O)-CH_2-CH_2-CH_2-CH_2-CHO$.
Reductive ozonolysis of a cyclic alkene involves the cleavage of the double bond to form a dicarbonyl compound.
For $1-$methylcyclohexene,the double bond is between $C_1$ and $C_2$. Cleavage results in a $7-$carbon chain with a ketone at $C_2$ and an aldehyde at $C_7$ (numbering from the ketone end),which corresponds to $6-$oxoheptanal.
Therefore,the correct reactant is $1-$methylcyclohexene.

(B) The reaction is an ozonolysis of an alkene followed by reductive workup with $Zn$ dust and $H_2O$.
Ozonolysis cleaves the $C=C$ double bond to form carbonyl compounds.
$Ph-CH_2-CH=CH-CH_3 \xrightarrow[Zn \text{ dust } + H_2O]{1) O_3} Ph-CH_2-CHO + CH_3-CHO$.
The products formed are $2-$phenyl ethanal $(Ph-CH_2-CHO)$ and ethanal $(CH_3-CHO)$.

(B) The oxidation of $cyclohexene$ with a strong oxidizing agent like acidic $KMnO_4$ results in the cleavage of the double bond and the formation of a dicarboxylic acid.
Specifically,$cyclohexene$ $(C_6H_{10})$ undergoes oxidative cleavage to form $adipic$ $acid$ $(HOOC-(CH_2)_4-COOH)$,which is a $6$-carbon dicarboxylic acid.
The chemical reaction is as follows:
$3C_6H_{10} + 8KMnO_4 + 4H_2O \rightarrow 3HOOC(CH_2)_4COOH + 8KOH + 8MnO_2$
Thus,the product formed is $adipic$ $acid$.

(B) In the $cis-isomer$,the methyl groups are on the same side of the double bond,resulting in a net dipole moment.
In the $trans-isomer$,the methyl groups are on opposite sides,causing the individual bond dipoles to cancel each other out,resulting in a net dipole moment of $0.00 \ D$.
Due to the higher polarity,the $cis-isomer$ has stronger intermolecular dipole-dipole interactions compared to the $trans-isomer$,leading to a higher boiling point.
The boiling point of $cis-but-2-ene$ is $277 \ K$ and the dipole moment of $trans-but-2-ene$ is $0.00 \ D$.

(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ gives propene $(A)$ as the major product.
$CH_3-CH(Br)-CH_3 \xrightarrow{alc. KOH} CH_3-CH=CH_2 (A) + HBr$
Step $2$: Addition of $HBr$ to propene in the presence of peroxide $( (C_6H_5CO)_2O_2 )$ follows the Anti-Markownikoff rule (Kharasch effect).
$CH_3-CH=CH_2 + HBr \xrightarrow{peroxide} CH_3-CH_2-CH_2-Br (B)$
Thus,the final product $B$ is $1$-bromopropane.
Hence,option $C$ is correct.

(B) The de-bromination of vicinal dihalides like $1, 2$-dibromoethane is carried out using zinc $(Zn)$ dust in the presence of alcohol (usually ethanol) upon heating.
This reaction is a type of elimination reaction where the two bromine atoms are removed to form an alkene.
The chemical equation is:
$BrCH_2-CH_2Br + Zn \xrightarrow{\text{alcohol}, \Delta} CH_2=CH_2 + ZnBr_2$
Thus,the correct metal is $Zn$.

(B) The reaction of propene with $HBr$ in the presence of peroxide (Kharasch effect) proceeds via a free radical mechanism.
In this mechanism,the bromine radical $(\dot{Br})$ attacks the double bond to form a more stable secondary free radical intermediate $(CH_3-\dot{C}H-CH_2Br)$.
This secondary free radical then abstracts a hydrogen atom from $HBr$ to form the final product,$1$-bromopropane $(CH_3-CH_2-CH_2Br)$.

(C) Lindlar catalyst is a partially deactivated catalyst composed of $BaSO_4$ coated with $Pd$ poisoned with quinoline.
It reduces $(C \equiv C)$ to $(C=C)$ bond,occurs via syn-addition,and gives a $cis$-alkene.
For $1$-phenyl-but-$1$-yne $(C_2H_5-C \equiv C-C_6H_5)$,the reduction yields $cis$-$1$-phenyl-but-$1$-ene,which corresponds to option $C$.

(C) The simplest ketone is acetone,which is $CH_3COCH_3$ (propan$-2-$one).
$3-$Pentanone is $CH_3CH_2COCH_2CH_3$.
Ozonolysis of an alkene involves breaking the $C=C$ double bond and adding oxygen atoms to each carbon.
To find the alkene $X$,we combine the two ketones by removing the oxygen atoms and joining the carbonyl carbons with a double bond:
$CH_3-C(CH_3)=C(CH_2CH_3)-CH_2CH_3$.
The structure is $CH_3-C(CH_3)=C(CH_2CH_3)-CH_2CH_3$.
Numbering the chain to give the lowest locants: the longest chain has $6$ carbons (hexane).
The double bond is at position $2$.
There is a methyl group at position $2$ and an ethyl group at position $3$.
Thus,the $IUPAC$ name is $3-$ethyl$-2-$methylhex$-2-$ene. However,checking the options,the structure $CH_3-C(CH_3)=C(CH_2CH_3)-CH_2CH_3$ corresponds to $3-$ethyl$-2-$methylhex$-2-$ene. Re-evaluating the options provided,option $C$ is $3-$ethyl$-2-$methylpent$-2-$ene,which would yield $3-$pentanone and acetone if the structure was $CH_3-C(CH_3)=C(CH_3)-CH_2CH_3$. Given the standard nature of this problem,the correct structure is $3-$ethyl$-2-$methylhex$-2-$ene,but based on the provided options,$C$ is the closest match if we assume a typo in the chain length.

(D) $1$. The alkene $X$ is $but-2-ene$ $(CH_3CH=CHCH_3)$,which exhibits geometrical isomerism.
$2$. Oxidation of $but-2-ene$ with acidic $KMnO_4$ yields acetic acid $(CH_3COOH)$ as $Y$.
$3$. The sodium salt of $Y$ is sodium acetate $(CH_3COONa)$.
$4$. Heating sodium acetate with soda lime $(NaOH + CaO)$ undergoes decarboxylation to produce methane $(CH_4)$ as $Z$.

(B) $1$. The reaction of $C_3H_4$ (propyne) with $H_2O$ in the presence of $Hg^{2+}/H^+$ at $333 \ K$ is a hydration reaction that follows Markovnikov's rule to form an enol intermediate,which then tautomerizes to form $X$ (acetone,$CH_3COCH_3$).
$2$. The second part of the reaction shows that $Y$ undergoes ozonolysis $((i) O_3, (ii) Zn/H_2O)$ to produce $X$ (acetone).
$3$. Ozonolysis of an alkene $R_2C=CR_2$ produces two carbonyl compounds. To obtain acetone $(CH_3COCH_3)$ as the product,the alkene $Y$ must be $2,3-$dimethyl$-2-$butene $((CH_3)_2C=C(CH_3)_2)$.
$4$. Looking at the options provided,the structure in option $B$ represents $2,3-$dimethyl$-2-$butene.

(A) Step $1$: The hydrogenation of propyne $(CH_3-C \equiv CH)$ with $H_2 / Pd-C$ (Lindlar's catalyst or similar) reduces the alkyne to an alkene,specifically propene $(CH_3-CH=CH_2)$,which is product $A$.
Step $2$: The reaction of propene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ follows the anti-Markovnikov addition (peroxide effect).
Step $3$: According to the anti-Markovnikov rule,the bromine atom attaches to the carbon atom with more hydrogen atoms,resulting in $1$-bromopropane $(CH_3-CH_2-CH_2Br)$ as the major product $B$.

(A) The oxidative cleavage of alkenes using hot acidic $KMnO_4$ results in the breaking of the $C=C$ double bond.
For reaction $I$: $(CH_3)_2 C=CH_2$ undergoes cleavage to form $(CH_3)_2 C=O$ (acetone,a ketone) and $CO_2 + H_2 O$. Thus,$X$ is a ketone.
For reaction $II$: $CH_3-CH=CH-CH_3$ undergoes cleavage to form $2CH_3COOH$ (acetic acid,a carboxylic acid). However,looking at the provided options and the structure of $Y$ in the image,the product is $CH_3COOH$. Given the options,the correct functional groups are Ketone $(X)$ and Carboxylic acid $(Y)$.

(A) Ozonolysis of an alkene involves the cleavage of the double bond to form two carbonyl compounds.
Given products are $CH_3-CH_2-CHO$ (propanal) and $CH_3-CHO$ (ethanal).
By removing the oxygen atoms from the carbonyl groups and joining the carbon atoms with a double bond,we get:
$CH_3-CH_2-CH=CH-CH_3$.
The structure corresponds to $Pent-2-ene$.
Therefore,the correct option is $A$.

(A) $1$. Ozonolysis of alkene $X$ $(C_6H_{12})$ yields acetaldehyde $(CH_3CHO)$ and ethyl methyl ketone $(CH_3COCH_2CH_3)$.
$2$. By reversing the ozonolysis process,we can determine the structure of $X$ by joining the carbonyl carbons with a double bond: $CH_3CH=C(CH_3)CH_2CH_3$.
$3$. The structure of $X$ is $3-methylpent-2-ene$.
$4$. When $X$ reacts with $HBr$,the reaction follows Markovnikov's rule,where the $H^+$ adds to the carbon with more hydrogen atoms and $Br^-$ adds to the more substituted carbon.
$5$. The reaction is: $CH_3CH=C(CH_3)CH_2CH_3 + HBr \rightarrow CH_3CH_2-C(Br)(CH_3)CH_2CH_3$.
$6$. The product formed is $3-bromo-3-methylpentane$.

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
To find the original alkene,we join the two carbonyl products by removing the oxygen atoms and forming a double bond between the carbon atoms.
The products are $Propan-2-one$ $(CH_3-CO-CH_3)$ and $methanal$ $(H-CHO)$.
Joining these: $(CH_3)_2C=O + O=CH_2 \rightarrow (CH_3)_2C=CH_2$.
The resulting alkene is $2-Methylpropene$.

(B) The addition of $HBr$ to propene in the presence of a peroxide occurs contrary to Markovnikov's rule,which is known as the peroxide effect or Kharasch effect.
This reaction proceeds via a free radical chain mechanism.
The mechanism is initiated by the homolysis of benzoyl peroxide as shown below:
$(i) \ C_6H_5-CO-O-O-CO-C_6H_5$ $\xrightarrow{\text{Homolysis}} 2C_6H_5-CO-O^{\bullet}$ $\rightarrow 2C_6H_5^{\bullet} + 2CO_2$

(A) The addition of $Br_2/\text{water}$ to an alkene is an electrophilic addition reaction. The rate-determining step involves the formation of a carbocation intermediate.
Electron-donating groups (like alkyl groups) increase the electron density of the double bond and stabilize the carbocation intermediate,thereby increasing the rate of reaction.
Electron-withdrawing groups (like $-NO_2$) decrease the electron density of the double bond and destabilize the carbocation intermediate,thereby decreasing the rate of reaction.
Comparing the substituents:
$1$. $CH_3-CH=CH_2$ $(D)$: Has one methyl group (+$I$ effect),stabilizing the carbocation.
$2$. $CH_3-CH_2-CH=CH_2$ $(C)$: Has an ethyl group (+$I$ effect),which is slightly more effective than methyl at stabilizing,but steric factors and hyperconjugation make $D$ and $C$ very reactive.
$3$. $CH_2=CH_2$ $(A)$: No substituents.
$4$. $CH_2=CH-NO_2$ $(B)$: The $-NO_2$ group is a strong electron-withdrawing group (-$I$ and -$M$ effect),making it the least reactive.
Thus,the correct order of reactivity is $C > D > A > B$.

(D) The reaction of an alkene with cold,dilute,aqueous $KMnO_4$ (Bayer's reagent) results in syn-dihydroxylation,where a double bond is converted into a vicinal diol (two $-OH$ groups are added across the double bond).
In the given compound,$5-$allylcyclohex$-3-$ene$-1-$ol,there are two double bonds: one in the cyclohexene ring and one in the allyl side chain.
$1$. The double bond in the cyclohexene ring undergoes dihydroxylation,adding two $-OH$ groups.
$2$. The double bond in the allyl side chain also undergoes dihydroxylation,adding two more $-OH$ groups.
$3$. The original molecule already contains one $-OH$ group at the $1-$position of the ring.
Total $-OH$ groups = $1$ (original) $+ 2$ (from ring double bond) $+ 2$ (from side chain double bond) = $5$ hydroxyl groups.

(C) The reaction $CH_2=CH_2 + H_2 \xrightarrow{Ni, 300^{\circ}C} CH_3-CH_3$ is the catalytic hydrogenation of ethene to ethane.
This specific process,where alkenes are reduced to alkanes using a metal catalyst like $Ni$,$Pt$,or $Pd$,is known as the Sabatier-Senderens reaction.

(B) Ozonolysis of $2-$methylbuta$-1,3-$diene $(CH_2=C(CH_3)-CH=CH_2)$ involves the cleavage of both double bonds.
Reductive ozonolysis using $O_3$ followed by $Zn-H_2O$ converts the alkene carbons into carbonyl groups (aldehydes or ketones).
The reaction proceeds as follows:
$CH_2=C(CH_3)-CH=CH_2 \xrightarrow[Zn-H_2O]{O_3} HCHO + CH_3COCHO + HCHO$
Thus,the products are $HCHO$ (formaldehyde) and $CH_3COCHO$ (methylglyoxal).
Therefore,option $(b)$ is correct.

(D) Addition of bromine can be used as the test for unsaturation with regard to colour change of reaction.
For example,when an unsaturated compound reacts with $Br_2$,the reddish-brown colour of $Br_2$ rapidly disappears.
This process is known as the decolourisation of bromine water.

(D) Step $1$: Debromination of $2,3$-dibromo-$5$-methylhexane with $Zn$ dust leads to the formation of $5$-methylhex-$2$-ene.
Step $2$: Ozonolysis $(O_3)$ of $5$-methylhex-$2$-ene forms an ozonide intermediate.
Step $3$: Reductive hydrolysis of the ozonide with $Zn/H_2O$ cleaves the double bond to yield two carbonyl compounds.
The reaction is: $(CH_3)_2CH-CH_2-CH=CH-CH_3 + O_3 \rightarrow (CH_3)_2CH-CH_2-CHO + CH_3-CHO$.
Thus,the products are $3$-methylbutanal $(P)$ and ethanal $(Q)$.

(C) The reaction of propene with hydrogen halides $(HX)$ is an electrophilic addition reaction.
In this reaction,the rate-determining step involves the protonation of the alkene by the hydrogen halide.
The reactivity of hydrogen halides depends on the strength of the $H-X$ bond.
The bond dissociation energy decreases in the order $HCl > HBr > HI$.
Therefore,$HI$ is the most reactive because it has the weakest $H-X$ bond,making it the easiest to break to provide a proton $(H^+)$.
The correct order of reactivity is $HI > HBr > HCl$.

(B) In the given reactions:
$1$. $CH_3CH=CH_2 \xrightarrow{HBr} CH_3CH(Br)CH_3$ $(P)$
This is an electrophilic addition reaction following Markovnikov's rule,where the nucleophile $(Br^-)$ attaches to the more substituted carbon.
$2$. $CH_3CH=CH_2 \xrightarrow[(BH_3)_2]{H_2O, H_2O_2 / OH^-} CH_3CH_2CH_2OH$ $(Q)$
This is a hydroboration-oxidation reaction,which follows anti-Markovnikov's rule,resulting in the anti-Markovnikov addition of water ($H$ and $OH$) across the double bond.
Therefore,the correct products are $P = CH_3CH(Br)CH_3$ and $Q = CH_3CH_2CH_2OH$.

(D) The given compound is $3-$ethyl$-3-$methylpenta$-1, 4-$diene. Its structure is $CH_2=CH-C(CH_3)(C_2H_5)-CH=CH_2$.
Ozonolysis followed by reductive workup $(Zn/H_2O)$ cleaves the double bonds and replaces them with carbonyl groups.
Cleaving the two terminal double bonds $(CH_2=)$ results in the formation of $2$ moles of formaldehyde $(HCHO)$.
The remaining central part of the molecule becomes a dicarbonyl compound: $OHC-C(CH_3)(C_2H_5)-CHO$.
Thus,the products are $2$ moles of $HCHO$ and $1$ mole of $2-$ethyl$-2-$methylpropanedial.

(A) The given reaction is the oxidative cleavage of cyclopentene to form glutaric acid $(HOOC-(CH_2)_3-COOH)$.
Strong oxidizing agents like hot acidic potassium permanganate $(KMnO_4-H_2SO_4 / \Delta)$ cause the oxidative cleavage of the carbon-carbon double bond in alkenes to produce carboxylic acids.
Thus,option $(A)$ is correct.

(C) The reaction is an oxymercuration-demercuration of an alkene.
$1$. In the first step,$Hg(OAc)_2$ in aqueous $THF$ reacts with the alkene to form an organomercurial intermediate via an electrophilic addition of $Hg(OAc)^+$ followed by the attack of water.
$2$. In the second step,$NaBH_4$ reduces the $C-Hg$ bond to a $C-H$ bond,resulting in the Markovnikov addition of water across the double bond.
$3$. For methylenecyclohexane,the addition of water follows Markovnikov's rule,where the $OH$ group attaches to the more substituted carbon (the tertiary carbon of the ring).
$4$. Thus,the product formed is $1$-methylcyclohexanol.
Hence,option $(C)$ is the correct answer.

(A) The reaction involves the electrophilic addition of $Br_2$ to an alkene in the presence of $CCl_4$ at low temperature $(253 \ K)$.
Alkenes are more reactive towards electrophilic addition than alkynes.
Therefore,$Br_2$ selectively adds across the double bond to form a vicinal dibromide.
The reaction is: $HC \equiv C-CH_2-CH=CH_2 + Br_2 \xrightarrow{CCl_4} HC \equiv C-CH_2-CH(Br)-CH_2Br$.
Thus,the major product is $HC \equiv C-CH_2-CH(Br)-CH_2Br$.

(D) According to Markownikoff's rule,the addition of reagents such as $HX$ to unsymmetrical alkenes occurs in such a way that the negative part of the adding molecule attaches to the carbon atom of the double bond which carries the lesser number of hydrogen atoms.
For the reaction of but$-1-$ene $(CH_3CH_2CH=CH_2)$ with $HBr$,the major product is $2-$bromobutane $(CH_3CH_2CH(Br)CH_3)$,not $1-$bromobutane $(CH_3CH_2CH_2CH_2Br)$.
Therefore,the assertion $(A)$ is incorrect because it states $1-$bromobutane is the major product.
The reason $(R)$ is correct as it correctly states the rule governing the addition of hydrogen halides to alkenes.
Thus,$A$ is incorrect and $R$ is correct.

(A) To determine the structure of the alkene that undergoes ozonolysis,place the carbonyl oxygen atoms of the products face to face and replace the $O$ atoms with a double bond $(C=C)$.
Acetaldehyde is $CH_3CHO$ and acetone is $(CH_3)_2CO$.
Aligning them: $CH_3(H)C=O + O=C(CH_3)_2$.
Removing the oxygen atoms and joining the carbons with a double bond gives $CH_3(H)C=C(CH_3)_2$.
The structure is $CH_3-CH=C(CH_3)_2$.
The longest chain contains $4$ carbon atoms with a methyl group at the $2$-position,so the $IUPAC$ name is $2$-methyl-$2$-butene.

(A) From the given reactions:
$1$. $X + HCl \rightarrow C_2H_5Cl$ (Addition reaction indicates $X$ is an alkene,$C_2H_4$).
$2$. $Y + HCl \rightarrow C_2H_5Cl$ (Substitution reaction indicates $Y$ is an alcohol,$C_2H_5OH$).
$3$. The conversion of $Y$ $(C_2H_5OH)$ to $X$ $(C_2H_4)$ is a dehydration reaction.
$4$. Dehydration of ethanol to ethene occurs by heating with $Al_2O_3$ at $350^{\circ}C$.
$\underset{(Y)}{C_2H_5OH} \xrightarrow[350^{\circ}C]{Al_2O_3} \underset{(X)}{C_2H_4} + H_2O$

(A) The given reaction sequence is $A$ $\xrightarrow{HBr} C_2H_5Br$ $\xrightarrow{B} A$.
Step $1$: $A$ reacts with $HBr$ to form $C_2H_5Br$ (ethyl bromide). This indicates that $A$ is ethene $(C_2H_4)$.
Reaction: $CH_2=CH_2 HBr \rightarrow CH_3-CH_2Br$.
Step $2$: $C_2H_5Br$ reacts with reagent $B$ to regenerate $A$ $(C_2H_4)$.
This is a dehydrohalogenation reaction,which requires an alcoholic solution of $KOH$ and heat $(\Delta)$.
Reaction: $CH_3-CH_2Br \text{alc. } KOH \xrightarrow{\Delta} CH_2=CH_2 KBr H_2O$.
Therefore,$A$ is $C_2H_4$ and $B$ is alcoholic $KOH / \Delta$.

(A) The reaction sequence is as follows:
$1$. The reaction of a vicinal dibromide with $Zn$ dust (dehalogenation) leads to the formation of an alkene.
$2$. $X = (CH_3)_2CBrCH_2Br$ reacts with $Zn$ to form $Y = (CH_3)_2C=CH_2$ (isobutylene).
$3$. Ozonolysis of isobutylene $(CH_3)_2C=CH_2$ in the presence of $Zn/H_2O$ yields acetone $(CH_3)_2CO$ and formaldehyde $(CH_2O)$.

(A) The conversion of $1, 2$-dibromoethane $(BrCH_2-CH_2Br)$ to ethylene $(H_2C=CH_2)$ is a dehalogenation reaction.
This reaction involves the removal of two bromine atoms from adjacent carbon atoms using zinc dust in the presence of alcohol under heating $(\Delta)$.
The chemical equation is: $BrCH_2-CH_2Br + Zn \xrightarrow{\text{alcohol}, \Delta} H_2C=CH_2 + ZnBr_2$.
Thus,the correct reaction conditions are $Zn$,alcohol,$\Delta$.

(D) The stability of alkenes in fused ring systems is governed by Bredt's rule and the degree of substitution at the double bond.
Structure $III$ is a tetrasubstituted alkene (specifically,a tetrasubstituted double bond at the bridgehead position),which is highly stable due to hyperconjugation and inductive effects.
Structure $I$ is a trisubstituted alkene.
Structure $II$ is a disubstituted alkene.
Greater substitution leads to higher stability.
Therefore,the correct order of stability is $III > I > II$.

(A) The Diels-Alder reaction requires a conjugated diene and a dienophile (an alkene or alkyne).
In option $A$,the reactant $CH_2=CH-CH_2-CH=CH_2$ is $penta-1,4-diene$.
This is an isolated diene,not a conjugated diene.
Therefore,it cannot participate in the Diels-Alder reaction.

(C) The compound is $C_8H_8$ (Styrene).
$1$. Ammoniacal silver nitrate (Tollens' reagent) reacts with terminal alkynes to form a white precipitate. Since the compound does not produce this precipitate,it does not contain a terminal $-C \equiv CH$ group.
$2$. Cold dilute alkaline $KMnO_4$ (Baeyer's reagent) is a test for unsaturation (alkenes or alkynes). Decolorization indicates the presence of a double or triple bond.
$3$. Styrene $(C_6H_5-CH=CH_2)$ has the molecular formula $C_8H_8$. It contains a double bond,which reacts with $KMnO_4$ to decolorize it,but it lacks a terminal alkyne group,so it does not react with Tollens' reagent.