Alkene Questions in English

(C) The heat of hydrogenation is inversely proportional to the stability of the alkene.
More substituted alkenes are more stable due to hyperconjugation and inductive effects.
$1$. $CH_3-CH=CH-CH_3$ ($2$-butene) is a disubstituted alkene.
$2$. $CH_3-CH=CH-C_2H_5$ ($2$-pentene) is a disubstituted alkene.
$3$. $C_2H_5-CH=CH_2$ ($1$-butene) is a monosubstituted alkene.
$4$. $C_2H_5-CH=CH-C_2H_5$ ($3$-hexene) is a disubstituted alkene.
Since monosubstituted alkenes are less stable than disubstituted alkenes,$C_2H_5-CH=CH_2$ has the lowest stability and therefore the maximum heat of hydrogenation.

(A) The reaction of an alkene with $D_2$ in the presence of a metal catalyst like $Pd$ (palladium) is a syn-addition reaction.
In syn-addition,both deuterium atoms add to the same face of the double bond.
For cyclopentene,this results in the formation of cis$-1,2-$dideuteriocyclopentane.
Therefore,the correct option is $A$.

(D) According to Markownikoff's rule,in the addition of a protic acid to an asymmetric alkene,the hydrogen atom $(H^+)$ attaches to the carbon atom with the greater number of hydrogen atoms.
In option $(d)$,the addition of $H_2O$ (catalyzed by $H^+$) to $2-methylpropene$ $(CH_3-C(CH_3)=CH_2)$ follows this rule,where $H^+$ adds to the terminal $CH_2$ group to form a stable tertiary carbocation,$(CH_3)_3C^+$,which then reacts with $H_2O$ to form $tert-butanol$.

(B) The first reaction is Hydroboration-Oxidation of propene. This reaction follows anti-Markovnikov addition of water,resulting in the formation of $A = CH_3-CH_2-CH_2OH$ (propan$-1-$ol).
The second reaction is Oxymercuration-Demercuration of propene. This reaction follows Markovnikov addition of water,resulting in the formation of $B = CH_3-CH(OH)-CH_3$ (propan$-2-$ol).
Comparing $A$ and $B$,both are alcohols with the same molecular formula $C_3H_8O$,but the position of the hydroxyl $(-OH)$ group is different. Therefore,they are positional isomers.

(C) Baeyer's reagent is defined as a cold,dilute,alkaline solution of potassium permanganate $(KMnO_4)$.
It is used as an oxidizing agent to detect the presence of unsaturation (double or triple bonds) in organic compounds.
The reaction involves the conversion of an alkene into a vicinal diol,accompanied by the disappearance of the pink color of $KMnO_4$ and the formation of a brown precipitate of manganese dioxide $(MnO_2)$.
Therefore,the correct composition is $OH^{-} + KMnO_4$.

(A) The reaction of $isobutylene$ $(2-methylpropene)$ with $O_3$ followed by reductive workup $(H_2O/Zn)$ is ozonolysis.
$Isobutylene$ is $(CH_3)_2C=CH_2$.
Ozonolysis of $isobutylene$ results in the cleavage of the $C=C$ double bond.
The products formed are $Acetone$ $(CH_3COCH_3)$ and $Formaldehyde$ $(HCHO)$.

(B) The reaction is the reductive ozonolysis of but$-2-$ene.
In this process,the carbon-carbon double bond is cleaved,and oxygen atoms are added to each carbon atom of the original double bond.
The reaction proceeds via the formation of an ozonide intermediate,which is then reduced by $Zn/H_2O$ to form carbonyl compounds.
$CH_3-CH=CH-CH_3 + O_3 \xrightarrow{Zn/H_2O} 2 CH_3-CHO$.
The product formed is acetaldehyde (ethanal).

(B) The reaction of $1,3$-butadiene with $1 \text{ eq}$ of $Br_2$ at high temperature (indicated by $\Delta$) is a thermodynamic control reaction.
In thermodynamic control,the more stable product is formed.
$1,3$-butadiene reacts with $Br_2$ to form a carbocation intermediate.
Electrophilic addition of $Br_2$ to $1,3$-butadiene can occur at the $1,2$-position or the $1,4$-position.
The $1,4$-addition product,$1,4$-dibromobut-$2$-ene,is more stable than the $1,2$-addition product,$3,4$-dibromobut-$1$-ene,because the double bond in the $1,4$-product is disubstituted,whereas the double bond in the $1,2$-product is monosubstituted.
Therefore,at higher temperatures,the $1,4$-addition product is the major product.

(B) The given reactant is $2,3$-dichlorobutane. When treated with zinc $(Zn)$ dust in the presence of heat $(\Delta)$,a dehalogenation reaction occurs. This is a classic method for preparing alkenes from vicinal dihalides.
The reaction is:
$CH_3-CH(Cl)-CH(Cl)-CH_3 + Zn \xrightarrow{\Delta} CH_3-CH=CH-CH_3 + ZnCl_2$
In the given Fischer projection,the two chlorine atoms are on the same side (erythro form). Elimination of $Cl_2$ from this specific stereoisomer leads to the formation of the cis-alkene.
Therefore,the product $(P)$ is cis-but$-2-$ene.

(C) The reaction of $3,3-\text{dimethyl-1-butene}$ with $HCl$ proceeds via a carbocation intermediate.
First,the protonation of the double bond occurs to form a secondary carbocation: $CH_3-C(CH_3)_2-CH^+-CH_3$.
This secondary carbocation is less stable and undergoes a $1,2-\text{methyl shift}$ to form a more stable tertiary carbocation: $CH_3-C^+(CH_3)-CH(CH_3)-CH_3$.
Finally,the chloride ion $(Cl^-)$ attacks this tertiary carbocation to form $2-\text{chloro-2,3-dimethylbutane}$ as the major product.

(B) The reaction is the electrophilic addition of $HBr$ to an alkene.
According to Markovnikov's rule,the hydrogen atom of $HBr$ adds to the carbon atom of the double bond that already has more hydrogen atoms,while the bromine atom adds to the more substituted carbon atom.
The starting material is $3-methylhept-1-ene$.
Upon protonation of the double bond,a carbocation is formed at the $C2$ position,which is a secondary carbocation.
However,the $C3$ position is a tertiary carbon. $A$ hydride shift from $C2$ to $C3$ is not possible here as the double bond is at the terminal position.
Wait,let's re-examine the structure: The reactant is $3-methylhept-1-ene$.
Addition of $H^+$ to $C1$ gives a carbocation at $C2$.
This is a secondary carbocation.
Attack of $Br^-$ at $C2$ gives $2-bromo-3-methylheptane$ as the major product.

(D) The reaction of an alkene with $Br_2$ in aqueous $NaCl$ proceeds via the formation of a cyclic bromonium ion intermediate.
This intermediate is subsequently attacked by nucleophiles present in the reaction medium,such as $Br^-$,$Cl^-$,or $H_2O$.
Since $Br_2$ acts as the source of the electrophile $(Br^+)$,the first atom to bond with the alkene is bromine.
Consequently,the final products must contain at least one bromine atom.
Option $D$ represents a product containing only chlorine atoms,which cannot be formed in this reaction.

(A) The reaction of an alkene with $Br_2$ typically leads to the formation of a vicinal dibromide. However,in this molecule,there is an internal nucleophile (the hydroxyl group,$-OH$) present at the end of the chain.
When $Br_2$ reacts with the double bond,a cyclic bromonium ion intermediate is formed.
The oxygen atom of the $-OH$ group then performs an intramolecular nucleophilic attack on the more substituted carbon of the bromonium ion (or the carbon that best stabilizes the developing positive charge),leading to the formation of a cyclic ether (a ring closure reaction).
Specifically,for $HO-(CH_2)_4-CH=CH_2$,the oxygen attacks the carbon at the $2$-position of the original double bond,resulting in a $6$-membered ring containing an oxygen atom (a tetrahydropyran derivative) with a $-CH_2Br$ group attached.
This is an example of halocyclization,where the internal nucleophile traps the bromonium ion to form a cyclic ether.

(A) The reactivity of alkenes towards electrophilic addition reactions is directly proportional to the electron density of the double bond. Higher electron density makes the alkene more nucleophilic,facilitating the attack of an electrophile.
$1$. In $CH_3-C(CH_3)=CH_2$ $(III)$,there are two electron-donating methyl groups ($+I$ effect and hyperconjugation),which significantly increase the electron density on the double bond.
$2$. In $CH_3-CH=CH_2$ $(II)$,there is one electron-donating methyl group,which increases the electron density compared to ethene.
$3$. In $CH_2=CH_2$ $(I)$,there are no electron-donating groups.
$4$. In $CH_2=CH-Cl$ $(IV)$,the chlorine atom exerts a strong $-I$ (inductive) effect,which decreases the electron density on the double bond,making it the least reactive.
Therefore,the order of reactivity is $III > II > I > IV$.

(A) The reaction of $trans-but-2-ene$ with $IBr$ involves the electrophilic addition of $Br^+$ to the alkene,forming a cyclic bromonium ion intermediate.
Since $IBr$ is a polar interhalogen compound,$Br$ is more electronegative than $I$,making $Br$ the electrophile $(Br^{\delta+})$ and $I$ the nucleophile $(I^{\delta-})$.
The attack of the nucleophile $(I^-)$ on the cyclic bromonium ion occurs from the side opposite to the bromonium bridge (anti-addition).
For $trans-but-2-ene$,the anti-addition of $IBr$ results in the formation of a pair of enantiomers.
Statement $A$ is incorrect because the reaction produces a racemic mixture of enantiomers,which are two distinct stereoisomers,but the question asks for the wrong statement regarding the mechanism or product nature. Actually,looking at the options,$A$ is correct (two products are formed),$B$ is correct (it is anti-addition),$C$ is correct (cyclic bromonium ion is formed),and $D$ is correct (the products are enantiomers). However,in the context of $IBr$ addition to an unsymmetrical alkene or specific stereochemistry,if the reaction were to produce meso compounds or diastereomers,that would be the error. Given the standard interpretation,all statements are technically true for this specific reaction. If forced to choose,$A$ might be considered 'wrong' if one considers the racemic mixture as a single product,but usually,it is two enantiomers. Let's re-evaluate: $trans-but-2-ene + IBr \rightarrow$ racemic mixture of $2-bromo-3-iodobutane$. This consists of two enantiomers. Thus,all statements are correct. Assuming a typo in the question,we identify the most likely intended answer.

(B) The reactant is $1,4$-dimethylcyclohex-$1$-ene.
The addition of $HBr$ to the double bond follows Markovnikov's rule.
The double bond is between $C_1$ and $C_2$.
Upon protonation of the double bond,a carbocation is formed at $C_1$ (tertiary) or $C_2$ (secondary).
The tertiary carbocation at $C_1$ is more stable.
Since the molecule has a chiral center at $C_4$,the addition of $Br^-$ to the carbocation at $C_1$ creates a new chiral center at $C_1$.
This results in the formation of two diastereomers (cis and trans isomers) because the methyl group at $C_4$ is already present.
Therefore,$2$ products are formed.

(C) According to Markovnikov's rule,the electrophile $(H^+)$ adds to the carbon with more hydrogen atoms,and the nucleophile $(Cl^-)$ adds to the more substituted carbon.
$1-$Methylcyclohexene reacts with $HCl$ to form a tertiary carbocation at the $C1$ position,which then reacts with $Cl^-$ to give $1-$chloro$-1-$methylcyclohexane.
$3-$Methylcyclohexene reacts with $HCl$ to form a secondary carbocation at the $C3$ position,which can undergo a $1,2-$hydride shift to form a more stable tertiary carbocation at the $C1$ position,eventually yielding $1-$chloro$-1-$methylcyclohexane.
Methylenecyclohexane reacts with $HCl$ to form a tertiary carbocation at the $C1$ position,which then reacts with $Cl^-$ to give $1-$chloro$-1-$methylcyclohexane.
Since all three alkenes can lead to the formation of $1-$chloro$-1-$methylcyclohexane,the question implies identifying which of the given options are valid. Given the options,$1-$methylcyclohexene,$3-$methylcyclohexene,and methylenecyclohexane all produce the same product. However,if the question asks for the most direct route,$1-$methylcyclohexene and methylenecyclohexane are primary candidates. Based on standard chemistry problems of this type,all listed alkenes can yield the product.

(C) $1$. The starting material is $1,3$-butadiene $(CH_2=CH-CH=CH_2)$.
$2$. Addition of $HCl$ at $40^{\circ}C$ (thermodynamic control) to $1,3$-butadiene primarily yields the $1,4$-addition product,which is $1$-chloro$-2-$butene $(CH_3-CH=CH-CH_2Cl)$.
$3$. Subsequent addition of $HI$ to $1$-chloro$-2-$butene follows Markovnikov's rule. The proton $(H^+)$ adds to the carbon with more hydrogens,and the iodide $(I^-)$ adds to the more substituted carbon.
$4$. The reaction is: $CH_3-CH=CH-CH_2Cl + HI \rightarrow CH_3-CH(I)-CH_2-CH_2Cl$.
$5$. The final product $C$ is $1$-chloro$-3-$iodobutane.

(C) The reactivity of an alkene towards $EAR$ depends on the stability of the carbocation intermediate formed after the attack of the electrophile.
$EDG$ (Electron Donating Groups) increase the electron density on the double bond and stabilize the carbocation intermediate,thereby increasing the reactivity.
In option $C$,the $-OCH_3$ group is a strong electron-donating group by resonance ($+M$ effect),which significantly stabilizes the carbocation formed compared to the inductive $(+I)$ effects of alkyl groups in other options.
Therefore,$CH_3-CH=CH-OCH_3$ is the most reactive towards $EAR$.

(D) The stability of alkenes is determined by the number of hyperconjugative structures (alpha-hydrogens) and steric hindrance.
$(X)$ is $2$-ethyl$-1$-butene, which has $5$ alpha-hydrogens.
$(Y)$ is $trans$-$3$-hexene, which is a disubstituted alkene with $4$ alpha-hydrogens and less steric hindrance.
$(Z)$ is $cis$-$3$-hexene, which is a disubstituted alkene with $4$ alpha-hydrogens but more steric hindrance than the trans isomer.
According to Saytzeff's rule and stability trends, highly substituted alkenes are more stable.
$(X)$ has $5$ alpha-hydrogens, making it more stable than the disubstituted isomers $(Y)$ and $(Z)$.
Between $(Y)$ and $(Z)$, the $trans$ isomer $(Y)$ is more stable than the $cis$ isomer $(Z)$ due to lower steric repulsion.
Therefore, the stability order is $X > Y > Z$.

(C) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the formation of two carbonyl groups at the cleavage sites.
For cyclic alkenes,ozonolysis results in the opening of the ring to form a dicarbonyl compound.
$2,6-$heptanedione is a seven-carbon chain with ketone groups at positions $2$ and $6$.
This indicates that the original alkene was a six-membered ring with two methyl groups attached to the double-bonded carbons,specifically $1,2-$dimethylcyclopentene.
When $1,2-$dimethylcyclopentene undergoes ozonolysis,the double bond breaks,resulting in a seven-carbon chain with ketones at the ends of the original double bond,which corresponds to $2,6-$heptanedione.

(D) The reaction shown is the ozonolysis of an alkene.
$1$. The ozonide intermediate formed is $CH_3-CH-O-CH_2$ (with an $O-O$ bridge).
$2$. Reductive ozonolysis with $Zn + H_2O$ cleaves the $C=C$ bond.
$3$. The products are $B$ and $HCHO$ (methanal).
$4$. The structure of the ozonide indicates the original alkene was propene $(CH_3-CH=CH_2)$.
$5$. Cleavage of propene $(CH_3-CH=CH_2)$ yields ethanal $(CH_3CHO)$ and methanal $(HCHO)$.
$6$. Therefore,$A$ is propene and $B$ is ethanal.

(C) The acid-catalyzed hydration of an alkene follows Markownikoff's rule.
According to this rule,the negative part of the addendum (the $OH^{-}$ group from water) attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.

(B) The reaction of $3$-methyl-$1$-butene with $HBr$ follows the electrophilic addition mechanism.
$1.$ Protonation of the double bond forms a secondary carbocation: $CH_3-CH(CH_3)-CH=CH_2 + H^{+} \to CH_3-CH(CH_3)-C^{+}H-CH_3$ ($2^{\circ}$ carbocation).
$2.$ $A$ $1,2$-hydride shift occurs to form a more stable tertiary carbocation: $CH_3-CH(CH_3)-C^{+}H-CH_3 \xrightarrow{1,2-H \text{ shift}} CH_3-C^{+}(CH_3)-CH_2-CH_3$ ($3^{\circ}$ carbocation).
$3.$ Nucleophilic attack by $Br^{-}$ gives the major product: $CH_3-C^{+}(CH_3)-CH_2-CH_3 + Br^{-} \to CH_3-C(Br)(CH_3)-CH_2-CH_3$ ($2$-bromo-$2$-methylbutane).

(C) In the first step,propyne reacts with excess $HCl$ to form $2,2$-dichloropropane $(A)$ via Markovnikov addition.
In the second step,alkaline hydrolysis of $2,2$-dichloropropane replaces the two chlorine atoms with hydroxyl groups,forming an unstable gem-diol which loses water to give propanone $(x)$.
$CH_3-C \equiv CH + 2HCl \rightarrow CH_3-CCl_2-CH_3 (A)$
$CH_3-CCl_2-CH_3$ $\xrightarrow{H_2O/OH^{-}} [CH_3-C(OH)_2-CH_3]$ $\rightarrow CH_3-CO-CH_3 (x) + H_2O$

(A) The heat of hydrogenation $(HOH)$ is inversely proportional to the stability of the alkene. More substituted alkenes are generally more stable,thus having a lower $HOH$.
In option $(A)$,the structures are: $1,2-$dimethylcyclohexene (trisubstituted),$2$,$3$-dimethylcyclohexene (tetrasubstituted),and $1-$methyl$-2-$methylenecyclohexane (disubstituted). The stability order is tetrasubstituted > trisubstituted > disubstituted. Thus,the $HOH$ order should be disubstituted > trisubstituted > tetrasubstituted. The given order in $(A)$ is incorrect.
In option $(B)$,the order is $1-$methylcyclohexene (trisubstituted) < methylenecyclohexane (disubstituted) < $1-$methylenecyclohexene (disubstituted,but with different substitution patterns). This is generally consistent with stability trends.
In option $(C)$,the order is trans$-2-$butene < cis$-2-$butene < $1-$butene. This is the correct order of $HOH$ because stability is trans > cis > terminal alkene.
In option $(D)$,the order is bicyclo[$4.2$.$0$]oct$-1-$ene < bicyclo[$4.2$.$0$]oct$-7-$ene. This is correct due to Bredt's rule and ring strain considerations.
Therefore,the incorrect order is $(A)$.

(D) The acidity order of the conjugate acids is: $CH_3-OH > CH_3-C \equiv C-H > CH_3-NH_2 > CH_4 > CH_2=CH_2$.
An acid-base reaction is possible if a stronger acid reacts with a stronger base to form a weaker acid and a weaker base.
In option $D$,$CH_2=CH_2$ is a much weaker acid $(pK_a \approx 44)$ than $CH_4$ ($pK_a \approx 50$ is incorrect,actually $CH_4$ is $pK_a \approx 50$ and $CH_2=CH_2$ is $pK_a \approx 44$,but the reaction requires the conjugate acid of the base to be weaker than the reactant acid).
Specifically,$CH_3^-$ is a much stronger base than $CH_2=CH^-$. Therefore,the reaction $CH_2=CH_2 + CH_3^- \rightarrow CH_2=CH^- + CH_4$ is not feasible as it would produce a stronger base from a weaker one.

(A) The rate of catalytic hydrogenation of alkenes depends on the stability of the alkene.
More substituted alkenes are more stable and have lower heat of hydrogenation,thus they react slower.
Less substituted alkenes are less stable and have higher heat of hydrogenation,thus they react faster.
Comparing the given structures:
$A$: $CH_3-CH_2-CH=CH_2$ (monosubstituted)
$B$: $CH_3-CH=CH-CH_3$ (disubstituted,cis)
$C$: $CH_3-CH=CH-CH_3$ (disubstituted,trans)
$D$: $(CH_3)_2C=CH_2$ (disubstituted)
Among these,the monosubstituted alkene $(A)$ is the least stable and therefore undergoes hydrogenation at the fastest rate.

(C) The reaction is Hydroboration-Oxidation.
It involves the anti-Markovnikov addition of $H_2O$ across the double bond without rearrangement.
$CH_3-C(CH_3)_2-CH=CH_2 \xrightarrow[H_2O_2/OH^{-}]{B_2H_6/THF} CH_3-C(CH_3)_2-CH_2-CH_2OH$
The product is $3,3-\text{dimethylbutan-1-ol}$,which is a primary $(1^o)$ alcohol and lacks a chiral center,making it optically inactive.

(B) The reaction sequence is as follows:
$1.$ $CH_3-CH_2-CH_2-Br + NaOH(aq.) \rightarrow CH_3-CH_2-CH_2-OH (X)$ (Nucleophilic substitution)
$2.$ $CH_3-CH_2-CH_2-OH \xrightarrow[Heat]{Al_2O_3} CH_3-CH=CH_2 (Y)$ (Dehydration)
$3.$ $CH_3-CH=CH_2 + HOCl \rightarrow CH_3-CH(OH)-CH_2-Cl (Z)$
In the addition of $HOCl$ to propene,the electrophilic $Cl^{+}$ attacks the terminal carbon to form a more stable secondary carbocation intermediate,which is then attacked by the nucleophilic $OH^{-}$ at the second carbon. This follows Markovnikov's rule for the addition of $HOCl$,where the $OH$ group attaches to the more substituted carbon atom.

(A) The reaction is an alkoxymercuration-demercuration of $1$-methylcyclohexene using mercuric acetate $Hg(OAc)_2$ and methanol $CH_3OH$,followed by reduction with $NaBH_4$.
This reaction follows Markovnikov's rule,where the nucleophile (methoxy group,$-OCH_3$) attacks the more substituted carbon atom of the double bond.
In $1$-methylcyclohexene,the carbon at position $1$ is more substituted than the carbon at position $2$.
Therefore,the methoxy group attaches to the $C-1$ position,resulting in $1$-methoxy-$1$-methylcyclohexane as the major product.

(B) The reaction shows the conversion of cyclohexene to trans-cyclohexane$-1,2-$diol.
This is an anti-dihydroxylation reaction.
Reagent $(i) CF_3CO_3H$ followed by $(ii) H_3O^{\oplus}$ (acid-catalyzed hydrolysis of an epoxide) results in anti-addition of two hydroxyl groups to the double bond.
Cold $KMnO_4$ and $OsO_4$ are reagents for syn-dihydroxylation.
Ozonolysis $(O_3)$ followed by reductive workup leads to cleavage of the double bond to form dicarbonyl compounds.

(A) The reaction $O_3/Zn$ is reductive ozonolysis.
Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds (aldehydes or ketones).
In the given product $OHC-CH_2-C(=O)-C(=O)-CH_2-CHO$,we observe a structure derived from the cleavage of a double bond.
Specifically,the reactant $X$ is bicyclopropylidene.
When bicyclopropylidene undergoes ozonolysis,the double bond between the two cyclopropyl rings is cleaved,resulting in the formation of the dicarbonyl product shown.

(A) The reaction $CH_3-CH=CH_2 + NOCl$ is an Electrophilic Addition Reaction $(EAR)$.
$NOCl$ dissociates to form $NO^+$ (electrophile) and $Cl^-$ (nucleophile).
The electrophile $NO^+$ attacks the double bond to form the more stable carbocation,which is the secondary carbocation $(CH_3-CH^+-CH_2-NO)$.
Subsequently,the nucleophile $Cl^-$ attacks the carbocation to form the major product: $CH_3-CHCl-CH_2-NO$ ($1$-nitroso$-2-$chloropropane).

(C) The reaction proceeds via a carbocation intermediate.
$1.$ Protonation of $CH_3-C(CH_3)_2-CH=CH_2$ gives a secondary carbocation $CH_3-C(CH_3)_2-C^{+}H-CH_3$.
$2.$ $A$ $1,2$-methyl shift occurs to form a more stable tertiary carbocation $CH_3-C^{+}(CH_3)-CH(CH_3)-CH_3$.
$3.$ Nucleophilic attack by $H_2O$ on the tertiary carbocation followed by deprotonation yields the major product $CH_3-C(OH)(CH_3)-CH(CH_3)-CH_3$ ($2,3$-dimethyl-$2$-butanol).

(B) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For a compound to yield acetone $(CH_3COCH_3)$ upon ozonolysis,it must contain the structural unit $(CH_3)_2C=$.
$2-$Methyl$-2-$butene has the structure $(CH_3)_2C=CH-CH_3$.
Upon ozonolysis,the double bond breaks to form $(CH_3)_2C=O$ (acetone) and $CH_3CHO$ (acetaldehyde).
Therefore,$2-$Methyl$-2-$butene is the correct compound.

(A) The enthalpy of hydrogenation is inversely proportional to the stability of the alkene.
Stability of alkenes is determined by the number of $\alpha$-hydrogens (hyperconjugation).
Compound $I$ is a tetrasubstituted alkene with $8$ $\alpha$-hydrogens.
Compound $III$ is a trisubstituted alkene with $5$ $\alpha$-hydrogens.
Compound $II$ is a disubstituted alkene with $4$ $\alpha$-hydrogens.
Since stability order is $I > III > II$,the enthalpy of hydrogenation order is $II > III > I$.

(A) The structure of $1-$phenylbut$-1-$ene is $CH_3-CH_2-CH=CH-Ph$.
Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form two carbonyl compounds.
$CH_3-CH_2-CH=CH-Ph \xrightarrow{O_3, Zn/H_2O} CH_3-CH_2-CHO + Ph-CHO$.
The products formed are propanal $(CH_3-CH_2-CHO)$ and benzaldehyde $(Ph-CHO)$.

(A) The reaction of $1$-butene with $Br_2$ in carbon tetrachloride $(CCl_4)$ is an electrophilic addition reaction.
In this reaction,the bromine molecule adds across the carbon-carbon double bond.
This results in the formation of a vicinal dihalide,which is $1,2$-dibromobutane.
The chemical equation is:
$CH_3-CH_2-CH=CH_2 + Br_2 \xrightarrow{CCl_4} CH_3-CH_2-CH(Br)-CH_2Br$

(D) $P$: Acid-catalyzed hydration of $Ph-CH_2-CH=CH_2$ involves the formation of a more stable carbocation. The initial carbocation $Ph-CH_2-CH^+-CH_3$ rearranges to the more stable benzylic carbocation $Ph-CH^+-CH_2-CH_3$,which then reacts with $H_2O$ to give $Ph-CH(OH)-CH_2-CH_3$.
$Q$: Oxymercuration-demercuration using $Hg(OAc)_2/H_2O$ followed by $NaBD_4$ results in Markovnikov addition of $H$ and $OD$ (from $D$ in $NaBD_4$) across the double bond without rearrangement. The product is $Ph-CH_2-CH(OH)-CH_2D$.
$R$: Hydroboration-oxidation using $BD_3/THF$ followed by $H_2O_2/OH^-$ results in anti-Markovnikov addition of $D$ and $OH$ across the double bond. The boron adds to the less substituted carbon,and subsequent oxidation replaces boron with $OH$. The product is $Ph-CH_2-CH(D)-CH_2OH$.

(A) The reaction involves an elimination process where the leaving group is removed to form a double bond.
In the given substrate,the iodine atom is a better leaving group than the chlorine atom.
Therefore,the elimination reaction will preferentially occur at the carbon atom bonded to the iodine,leading to the formation of the corresponding alkene product $Z$.

(B) Step $1$: Reaction of $1,2$-dimethylcyclohexane with $Br_2/h\nu$ (free radical bromination) leads to the formation of $1$-bromo-$1,2$-dimethylcyclohexane (compound $A$).
Step $2$: Treatment of $A$ with $CH_3ONa$ (a strong base) results in dehydrobromination via an $E2$ mechanism to form $1,2$-dimethylcyclohex$-1-$ene (compound $B$).
Step $3$: Ozonolysis of $1,2$-dimethylcyclohex$-1-$ene $(B)$ followed by reductive workup with $Zn/H_2O$ cleaves the double bond to yield $2,7$-octanedione (compound $C$).

(B) The reaction of $1,3$-butadiene with $1$ equivalent of $HBr$ is an electrophilic addition reaction.
At low temperature $(-80^{\circ}C)$,the reaction is under kinetic control,favoring the formation of the $1,2$-addition product,which is $3$-bromo-but-$1$-ene $(B)$.
At higher temperature $(25^{\circ}C)$,the reaction is under thermodynamic control,favoring the formation of the more stable $1,4$-addition product,which is $1$-bromo-but-$2$-ene $(A)$.
In $3$-bromo-but-$1$-ene,the bromine atom is at the $3$-position,while in $1$-bromo-but-$2$-ene,the bromine atom is at the $1$-position.
Since the position of the functional group (bromine) changes relative to the carbon chain,$A$ and $B$ are position isomers.

(A) The reaction of propene $(CH_3-CH=CH_2)$ with $HBr$ is an electrophilic addition reaction.
According to Markovnikov's rule,the electrophile $H^+$ adds to the carbon atom having more hydrogen atoms to form the more stable carbocation.
Here,the secondary carbocation $(CH_3-CH^+-CH_3)$ is more stable than the primary carbocation $(CH_3-CH_2-CH_2^+)$.
Therefore,the bromide ion $(Br^-)$ attacks the secondary carbocation to form $2$-bromopropane $(CH_3-CH(Br)-CH_3)$ as the major product.

(A) The reaction of an alkene with hot acidic $KMnO_4$ causes oxidative cleavage of the double bond.
An alkene of the form $R_2C=CH-R'$ will be cleaved to form a ketone $(R_2C=O)$ and a carboxylic acid $(R'-COOH)$.
The products given are butan$-2-$one $(CH_3-CH_2-CO-CH_3)$ and pentanoic acid $(CH_3-CH_2-CH_2-CH_2-COOH)$.
Combining these fragments at the carbonyl carbons,the original alkene must be $3-$methylhept$-2-$ene $(CH_3-CH_2-C(CH_3)=CH-CH_2-CH_2-CH_3)$.

(D) The reaction proceeds as follows:
$1$. Cyclohexene reacts with $NBS$ ($N$-Bromosuccinimide) to undergo allylic bromination,forming $3$-bromocyclohexene.
$2$. $3$-bromocyclohexene reacts with $Mg$ in the presence of ether to form the Grignard reagent,$3$-cyclohexenylmagnesium bromide.
$3$. The Grignard reagent then reacts with $CO_2$ followed by acidic workup $(H^+)$ to form the carboxylic acid,cyclohex$-2-$ene$-1-$carboxylic acid.
Thus,the final product $X$ is cyclohex$-2-$ene$-1-$carboxylic acid.

(C) Kolbe electrolysis of potassium succinate involves the decarboxylation of the salt of a dicarboxylic acid.
Potassium succinate is $K_2C_4H_4O_4$ (or $KOOC-CH_2-CH_2-COOK$).
The reaction is: $K_2C_4H_4O_4 + 2H_2O \xrightarrow{\text{electrolysis}} C_2H_4 + 2CO_2 + 2KOH + H_2$.
Thus,the products are ethene $(C_2H_4)$,carbon dioxide $(CO_2)$,potassium hydroxide $(KOH)$,and hydrogen gas $(H_2)$.

(C) The reaction involves the electrophilic addition of $HCl$ to vinylcyclobutane.
When $HCl$ adds to the double bond,the proton $(H^+)$ attacks the terminal carbon of the vinyl group to form a secondary carbocation at the carbon adjacent to the cyclobutane ring.
This secondary carbocation is unstable due to ring strain and undergoes a ring expansion rearrangement to form a more stable tertiary carbocation (a cyclohexyl cation).
Finally,the chloride ion $(Cl^-)$ attacks this tertiary carbocation to form $1$-chlorocyclohexane as the major product.

(D) $1$. Bayer's reagent $(1\% \text{ alkaline } KMnO_4)$ reacts with propene $(CH_3-CH=CH_2)$ to form propane$-1,2-$diol $(CH_3-CH(OH)-CH_2OH)$ as product $X$.
$2$. Periodic acid $(HIO_4)$ is a glycol-cleaving agent that breaks vicinal diols into carbonyl compounds.
$3$. The reaction is: $CH_3-CH(OH)-CH_2OH + HIO_4 \rightarrow CH_3-CHO + HCHO + H_2O$.
$4$. Thus,the products are acetaldehyde $(CH_3-CHO)$ and formaldehyde $(HCHO)$.

(B) Step $1$: Dehydrohalogenation of $2$-chloro-$3$-methylbutane with $C_2H_5ONa / C_2H_5OH$ follows the $E2$ mechanism. According to Saytzeff's rule,the more substituted alkene is the major product $(B)$.
$(CH_3)_2CH-CHCl-CH_3 \xrightarrow{C_2H_5ONa} (CH_3)_2C=CH-CH_3$ ($2$-methylbut-$2$-ene).
Step $2$: Ozonolysis of $2$-methylbut-$2$-ene with oxidative workup $(O_3 / H_2O_2)$ cleaves the double bond to form a ketone and a carboxylic acid.
$(CH_3)_2C=CH-CH_3 \xrightarrow{O_3 / H_2O_2} (CH_3)_2C=O + CH_3COOH$.
Thus,the products are Acetone and Ethanoic acid.