Alkene Questions in English

(B) The reaction of $trans-2-butene$ with $BH_3$ followed by $H_2O_2/OH^-$ is a hydroboration-oxidation process,which proceeds via $syn-addition$.
When $syn-addition$ of $OH$ and $H$ occurs across the double bond of $trans-2-butene$,it results in the formation of $butan-2-ol$.
Since the product $butan-2-ol$ has a chiral center,the reaction produces a pair of enantiomers in equal amounts,resulting in a $racemic$ mixture.

(C) The reaction shows the hydration of an alkene to form a secondary alcohol.
$1$. Acid-catalyzed hydration $(H_2O/H^{+})$ would involve a carbocation intermediate,which would rearrange to form a more stable benzylic carbocation,leading to a different product.
$2$. Hydroboration-oxidation $(BH_3 \cdot THF/H_2O_2-OH^{-})$ follows anti-Markovnikov addition,which would yield a primary alcohol.
$3$. Oxymercuration-demercuration $(Hg(OCOCH_3)_2 \cdot H_2O/NaBH_4 \cdot NaOH)$ follows Markovnikov's rule without carbocation rearrangement,yielding the observed secondary alcohol.
Therefore,the correct reagent is $(C)$.

(A) The reaction of propene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ proceeds via a free radical mechanism.
This is known as the Kharasch effect or Peroxide effect.
According to this effect,the addition of $HBr$ to an unsymmetrical alkene follows Anti-Markovnikov's rule.
Therefore,the bromine atom attaches to the carbon atom with the greater number of hydrogen atoms.
The reaction is: $CH_3 - CH = CH_2 + HBr \xrightarrow{Peroxide} CH_3 - CH_2 - CH_2 - Br$ ($1$-bromopropane).
Thus,the correct option is $A$.

(B) The reaction of bromine with ethene proceeds through the formation of a cyclic bromonium ion intermediate.
In a solvent like methyl alcohol $(CH_3OH)$,the cyclic bromonium ion can be attacked by either the bromide ion $(Br^-)$ or the solvent molecule $(CH_3OH)$ acting as a nucleophile.
Attack by $Br^-$ leads to the formation of $1, 2-dibromoethane$,while attack by $CH_3OH$ followed by deprotonation leads to the formation of $BrCH_2CH_2OCH_3$.

(B) The reaction proceeds via the electrophilic addition of $HI$ to the alkene.
$1$. Protonation of the double bond occurs to form the most stable carbocation. The initial carbocation formed is a secondary carbocation adjacent to a cyclopentyl ring.
$2$. To relieve ring strain and form a more stable tertiary carbocation,a ring expansion occurs,converting the five-membered ring into a six-membered ring.
$3$. The resulting tertiary carbocation is then attacked by the iodide ion $(I^-)$ to form the final product,which is $1$-iodo-$1,2$-dimethylcyclohexane.

(A) Hydroboration-oxidation of alkenes with $BH_3/THF$ follows anti-Markovnikov addition,where the boron atom attaches to the less substituted carbon atom of the double bond.
The product shown is tris($2$-methylbutyl)borane,which contains three $2$-methylbutyl groups attached to the boron atom.
This indicates that the starting alkene must be $2$-methylbut-$1$-ene $(CH_2=C(CH_3)CH_2CH_3)$.
Upon reaction with $BH_3$,the boron atom adds to the terminal $CH_2$ group,resulting in the formation of the trialkylborane derivative shown.

(D) The reaction proceeds via the following steps:
$1$. Protonation of the alkene ($3,3$-Dimethyl-$1$-butene) by $HBr$ occurs to form a secondary carbocation $(CH_3-C(CH_3)_2-CH^+-CH_3)$.
$2$. This secondary carbocation undergoes a $1,2$-methyl shift to form a more stable tertiary carbocation $(CH_3-C^+(CH_3)-CH(CH_3)_2)$.
$3$. Finally,the bromide ion $(Br^-)$ attacks the tertiary carbocation to yield the final product,$2$-bromo-$2,3$-dimethylbutane.

(C) The conversion of propene $(CH_3CH=CH_2)$ to propan$-1-$ol $(CH_3CH_2CH_2OH)$ is an anti-Markovnikov hydration reaction.
This transformation is achieved via the hydroboration-oxidation process.
In the first step,diborane $(B_2H_6)$ adds to the double bond,and in the second step,oxidation with alkaline hydrogen peroxide $(H_2O_2/OH^-)$ yields the primary alcohol.
Therefore,the correct set of reagents is $B_2H_6$ and alkaline $H_2O_2$.

(C) The reaction involves the addition of $HBr$ to an alkene in the presence of a peroxide,which follows the Anti-Markovnikov addition rule (Kharasch effect).
In this mechanism,the $Br^{\bullet}$ radical attacks the terminal carbon atom of the double bond to form a more stable secondary radical.
Consequently,the hydrogen atom attaches to the internal carbon atom.
The final product is $BrCH_2-CH_2(CH_2)_8COOH$.

(D) Catalytic hydrogenation $(H_2/Ni)$ of alkenes involves the syn-addition of hydrogen across the double bond.
$(a)$ $4-$methylmethylenecyclohexane on hydrogenation yields both $cis-1,4-$dimethylcyclohexane and $trans-1,4-$dimethylcyclohexane due to the approach of $H_2$ from either face of the ring.
$(b)$ $1,4-$dimethylcyclohex$-1-$ene similarly yields both $cis$ and $trans$ isomers of $1,4-$dimethylcyclohexane.
$(c)$ $3,3,6-$trimethylcyclohex$-1-$ene also produces stereoisomeric products upon hydrogenation.
Since all the given alkenes can produce $cis$ and $trans$ isomers (or diastereomers) upon catalytic hydrogenation, the correct answer is 'All of these'.

(B) The reaction of $HBr$ with an alkene follows electrophilic addition mechanism.
The first step involves the attack of the electrophile $(H^+)$ on the double bond of the alkene to form a carbocation.
This step is the rate-determining step because it involves the breaking of the $\pi$-bond and is therefore a slow step.
Thus,the correct answer is the slow addition of an electrophile.

(D) In the electrophilic addition of $Cl_2$ to an alkene,the first step involves the formation of a cyclic halonium ion intermediate. For cyclohexene,the reaction with $Cl_2$ leads to the formation of a cyclic chloronium ion. This species is more stable than a simple open-chain carbocation because the lone pair on the chlorine atom can be donated into the empty $p$-orbital of the carbocation,effectively delocalizing the positive charge. Therefore,the cyclic chloronium ion is the most stable intermediate.

(C) The given reaction involves the attack of a free radical $(Br^{\bullet})$ on an alkene to form a new carbon-centered radical.
In a radical chain reaction,a step that consumes one radical and produces another radical is known as a propagation step.
Therefore,the correct answer is $(C)$.

(B) The catalytic hydrogenation of an alkene using $H_2/Pt$ is a syn-addition reaction.
For the product to be a meso compound,the starting alkene must be capable of forming a molecule with a plane of symmetry upon the addition of two hydrogen atoms to the same side of the double bond.
Among the given options,cis$-3,5-$dimethylcyclohex$-1-$ene is optically active due to the absence of a plane of symmetry in the alkene.
Upon catalytic hydrogenation,the two hydrogen atoms add to the same face of the double bond,resulting in cis$-1,3-$dimethylcyclohexane,which possesses a plane of symmetry $(POS)$ and is therefore a meso compound.
Thus,the correct structure is cis$-3,5-$dimethylcyclohex$-1-$ene.

(B) The reaction is the anti-Markovnikov addition of $HBr$ to an alkene in the presence of peroxide $(R_2O_2)$.
The reactant is $3,4$-dimethylhex-$3$-ene.
Upon addition of $HBr$ via a free radical mechanism,the bromine radical adds to one of the carbons of the double bond,and the hydrogen adds to the other.
This creates two chiral centers in the product,$3$-bromo-$3,4$-dimethylhexane.
Since there are two chiral centers,there are $2^n = 2^2 = 4$ possible stereoisomers (optical isomers) formed as products.
These are the $(R,R)$,$(S,S)$,$(R,S)$,and $(S,R)$ configurations.

(A) The reaction of cis-$2$-butene with $HBr$ in the presence of peroxide follows a free radical addition mechanism.
First,the peroxide generates a bromine radical $(Br^{\bullet})$.
This radical adds to the double bond of cis-$2$-butene to form a secondary alkyl radical intermediate $(CH_3-CH^{\bullet}-CH(Br)-CH_3)$.
Since the radical carbon is $sp^2$ hybridized and planar,the hydrogen atom can abstract from either side of the radical center with equal probability.
This leads to the formation of both $(R)$-$2$-bromobutane and $(S)$-$2$-bromobutane in equal amounts.
$A$ $50:50$ mixture of enantiomers is known as a racemic mixture.

(C) The reductive ozonolysis of an alkene with the structure $R_2C=CHR'$ involves the cleavage of the carbon-carbon double bond.
The reaction proceeds via the formation of an ozonide intermediate,which is then reduced by $Zn/H_2O$ or dimethyl sulfide to yield carbonyl compounds.
The cleavage of $R_2C=CHR'$ results in the formation of a ketone $(R_2C=O)$ and an aldehyde $(R'-CHO)$.
Therefore,the product $A$ is a mixture of $R_2C=O$ and $R'-CHO$.

(B) The reaction of an alkene with $MCPBA$ (meta-chloroperbenzoic acid) in $CH_2Cl_2$ is an epoxidation reaction.
This reaction is stereospecific,meaning the stereochemistry of the starting alkene is preserved in the product.
Since the starting material is $cis-but-2-ene$,the product will be $cis-2,3-dimethyloxirane$ (cis-epoxide).
Therefore,the correct product is the one where the two methyl groups are on the same side of the epoxide ring.

(B) The reaction is a hydroboration-oxidation of $1$-methylcyclopentene.
Hydroboration-oxidation is a syn-addition reaction that follows anti-Markovnikov regioselectivity.
In $1$-methylcyclopentene,the double bond is between $C_1$ and $C_2$.
The $OH$ group adds to the less substituted carbon $(C_2)$ and the $H$ atom adds to the more substituted carbon $(C_1)$.
Since it is a syn-addition,both the $H$ and $OH$ add from the same face of the ring.
Therefore,the product is $2$-methylcyclopentanol where the $H$ at $C_1$ and the $OH$ at $C_2$ are cis to each other.
This corresponds to the cis-isomer of $2$-methylcyclopentanol.

(C) The reaction is a hydroboration-oxidation (or reduction) sequence using $BD_3$ and $CH_3CO_2T$.
Step $1$: Hydroboration of propene with $BD_3$ in $THF$ follows anti-Markovnikov addition,where $D$ adds to the more substituted carbon and $B$ adds to the less substituted carbon,forming $(CH_3-CH(D)-CH_2)_3B$.
Step $2$: Treatment with $CH_3CO_2T$ (deuterated acetic acid) replaces the boron atom with a tritium $(T)$ atom via protodeboronation.
Therefore,the final product $(A)$ is $CH_3-CHD-CH_2T$.

(D) The reaction is a Prins reaction,which involves the electrophilic addition of an aldehyde (in this case,formaldehyde,$HCHO$) to an alkene in the presence of an acid catalyst $(H^+)$.
$1$. The protonated formaldehyde acts as an electrophile and attacks the double bond of $1$-methylcyclohexene to form a carbocation intermediate.
$2$. The carbocation is then attacked by water $(H_2O)$ to form the final product,which is $1$-methyl-$2$-(hydroxymethyl)cyclohexan-$1$-ol.

(B) The reaction of an alkene with $Br_2$ in the presence of $H_2O$ is a halohydrin formation reaction.
This reaction proceeds through the formation of a cyclic bromonium ion intermediate.
Water acts as a nucleophile and attacks the more substituted carbon atom of the bromonium ion due to its higher partial positive charge.
This results in an anti-addition of the $-Br$ and $-OH$ groups across the double bond.
For the given substrate,$1$-ethylcyclohexene,the $-OH$ group will attach to the carbon bearing the ethyl group,and the $-Br$ atom will attach to the adjacent carbon,resulting in the formation of $2-$bromo$-1-$ethylcyclohexan$-1-$ol.

(A) The reaction sequence $1. \text{Hg(OAc)}_2, \text{H}_2\text{O}; 2. \text{NaBH}_4$ is the Oxymercuration-Demercuration process.
This reaction involves the anti-addition of $\text{H}$ and $\text{OH}$ across the double bond,following Markovnikov's rule.
For $1$-methylcyclohexene,the $\text{OH}$ group adds to the more substituted carbon (the one with the methyl group),resulting in $1$-methylcyclohexanol.
Since the addition is anti,the $\text{OH}$ and $\text{H}$ add from opposite faces,resulting in the formation of $1$-methylcyclohexanol where the $\text{OH}$ and $\text{CH}_3$ groups are in a specific stereochemical relationship.
Based on the provided options,the correct product is $1$-methylcyclohexanol.

(B) The reaction of an alkene with a peroxyacid (like $mCPBA$) followed by acid-catalyzed hydrolysis $(H_3O^+)$ is a two-step process that results in anti-dihydroxylation.
$1$. The alkene reacts with $mCPBA$ to form an epoxide.
$2$. The epoxide undergoes acid-catalyzed ring opening by water,which proceeds via an $S_N2$-like mechanism,leading to the anti-addition of two hydroxyl groups.
For $1,2-dimethylcyclopent-1-ene$,this results in the formation of the trans$-1,2-$dimethylcyclopentane$-1,2-$diol.

(B) The reaction sequence is as follows:
$1$. The starting material is but$-1-$ene $(CH_3-CH_2-CH=CH_2)$.
$2$. The addition of $HCl$ in the presence of peroxide is a radical addition. However,$HCl$ does not undergo anti-Markovnikov addition with peroxide because the $H-Cl$ bond is too strong to be cleaved by radicals. Thus,it follows Markovnikov's rule,yielding $2-$chlorobutane $(CH_3-CH_2-CHCl-CH_3)$ as $(A)$.
$3$. The reaction of $(A)$ $(CH_3-CH_2-CHCl-CH_3)$ with $EtONa/\Delta$ (a strong base) undergoes an $E2$ elimination reaction.
$4$. According to Zaitsev's rule,the more substituted alkene is the major product. Thus,but$-2-$ene $(CH_3-CH=CH-CH_3)$ is formed as the major product $(Z)$.

(C) In Kolbe's electrolysis,the electrolysis of potassium salts of vicinal dicarboxylic acids results in the formation of an alkene.
The reaction proceeds via a free radical mechanism:
$CH_3-CH(CO_2K)-CH(CO_2K)-CH_3 \xrightarrow{\text{electrolysis}} CH_3-CH=CH-CH_3 + 2CO_2 + H_2 + 2KOH$
Between the possible geometric isomers,the trans-isomer is more stable due to reduced steric hindrance between the methyl groups.
Thus,the major product is trans-But$-2-$ene.

(B) $1$. The reduction of $Me-C \equiv C-Et$ with $Na/liq. NH_3$ (Birch reduction/dissolving metal reduction) yields the trans-alkene,$trans-pent-2-ene$ $(P)$.
$2$. The subsequent reaction of $trans-pent-2-ene$ with $Br_2$ in $CCl_4$ proceeds via anti-addition.
$3$. Anti-addition of $Br_2$ to a trans-alkene results in the formation of a pair of enantiomers,which constitutes a racemic mixture.
$4$. $A$ racemic mixture is optically inactive due to external compensation,as the optical rotation of one enantiomer is cancelled by the equal and opposite rotation of the other.

(D) The rate of catalytic hydrogenation of alkenes is inversely proportional to their stability.
More substituted alkenes are more stable and thus react more slowly.
Comparing the substitution of the double bonds:
- Bond $a$: Trisubstituted alkene.
- Bond $b$: Trisubstituted alkene.
- Bond $c$: Trisubstituted alkene.
- Bond $d$: Monosubstituted (terminal) alkene.
Since the monosubstituted alkene $(d)$ is the least substituted and therefore the least stable,it will undergo catalytic hydrogenation most readily (first).

(D) The reaction in option $D$ is the acid-catalyzed hydration of $1-$methylcyclohexene.
In this reaction,the proton $(H^+)$ adds to the less substituted carbon of the double bond to form a more stable tertiary carbocation at the carbon bearing the methyl group.
Water then attacks this tertiary carbocation,followed by deprotonation to yield $1-$methylcyclohexanol as the major product.
This follows Markovnikov's rule.

(C) Wilkinson's catalyst $(W.C.)$,$[RhCl(PPh_3)_3]$,is a homogeneous catalyst used for the selective hydrogenation of alkenes.
It is highly sensitive to steric hindrance.
It preferentially reduces less sterically hindered alkenes (e.g.,terminal alkenes) over more substituted or sterically hindered ones.
In option $C$,the endocyclic double bond in $1$-methylcyclohex-$1$-ene is more substituted and sterically hindered than a terminal alkene,so it is not the preferred site for reduction by $W.C.$ compared to terminal alkenes. Thus,the reaction shown in option $C$ is incorrect.

(B) The reaction of an alkene with $HBr$ in the presence of a peroxide $(R_2O_2)$ and light $(hv)$ proceeds via a free radical mechanism,which follows the Anti-Markovnikov rule.
In this mechanism,the bromine radical $(Br^{\bullet})$ attacks the double bond to form the more stable radical intermediate.
For the given substrate,the $Br^{\bullet}$ radical attacks the less substituted carbon of the double bond,leading to a more stable tertiary $(3^{\circ})$ radical intermediate.
Subsequently,this radical abstracts a hydrogen atom from $HBr$ to form the final product,which is $1$-methyl-$2$-bromobicyclo$[4.2.0]$octane.

(D) The reaction of but$-2-$ene with $H_3O^{+}$ involves the formation of a planar carbocation intermediate,$CH_3-CH_2-C^{+}H-CH_3$.
Since the carbocation is planar,the nucleophile $(H_2O)$ can attack from either side of the plane with equal probability.
This results in the formation of a $50:50$ mixture of $(R)$ and $(S)$ enantiomers of butan$-2-$ol,which is a racemic mixture.

(A) The reaction involves the electrophilic addition of $HBr$ to $3-bromocyclohexene$.
First,the protonation of the double bond occurs to form a stable carbocation at the $C-3$ position relative to the existing $Br$ atom.
Since the reaction proceeds through a classical carbocation,the $Br^-$ ion can attack the carbocation from either face.
However,due to steric hindrance and the geometry of the cyclohexane ring,the attack of the bromide ion results in the formation of $trans-1,3-dibromocyclohexane$ as the major product,as shown in the reaction mechanism.

(C) The reaction of pent$-4-$en$-1-$ol with $Br_2$ in $H_2O$ proceeds via the formation of a cyclic bromonium ion intermediate.
The internal hydroxyl group $(-OH)$ then attacks the more substituted carbon of the bromonium ion in an intramolecular $S_N2$ fashion to form a five-membered tetrahydrofuran ring.
Since the bromonium ion can form on either face of the double bond,the resulting product is a racemic mixture of the two enantiomers shown in $(v)$ and $(vi)$.

(D) The reaction of alkenes with $HBr$ proceeds via the formation of a carbocation intermediate. The rate of the reaction is directly proportional to the stability of the carbocation formed in the rate-determining step.
$(A)$ Forms a secondary $(2^{\circ})$ carbocation.
$(B)$ Forms a primary $(1^{\circ})$ carbocation.
$(C)$ Forms an allylic carbocation (stabilized by resonance).
$(D)$ Forms a tertiary $(3^{\circ})$ carbocation.
Since a tertiary $(3^{\circ})$ carbocation is more stable than a secondary $(2^{\circ})$,primary $(1^{\circ})$,or even an allylic carbocation in this specific context of alkyl substitution,the reaction in option $(D)$ proceeds at the fastest rate.

(A) The addition of $Br_2$ to an alkene is an anti-addition reaction. When an alkene already contains a chiral center,the addition of $Br_2$ creates new chiral centers. The resulting products will be diastereomers because the configuration at the existing chiral center remains unchanged,while the new chiral centers are formed in different relative configurations. $3-Methylcyclopentene$ contains a chiral center at the $C-3$ position. Upon reaction with $Br_2$ in $CCl_4$,the anti-addition of bromine across the double bond results in the formation of two diastereomeric products,as the existing chiral center at $C-3$ is not involved in the reaction.

(D) Oxidative ozonolysis of alkenes using $O_3 / H_2O_2$ leads to the cleavage of the double bond.
If the alkene has a terminal $=CH_2$ group,it is oxidized to $CO_2$ and $H_2O$.
If it has a $=CHR$ group,it is oxidized to a carboxylic acid $(RCOOH)$.
If it has a $=CR_2$ group,it is oxidized to a ketone $(R_2C=O)$.
$(A)$ Methylenecyclohexane has a terminal $=CH_2$ group,so it evolves $CO_2$.
$(B)$ $2-$Cyclohexylprop$-1-$ene has a terminal $=CH_2$ group,so it evolves $CO_2$.
$(C)$ Buta$-1,3-$diene $(H_2C=CH-CH=CH_2)$ has terminal $=CH_2$ groups,so it evolves $CO_2$.
$(D)$ Cyclohexene has an internal double bond $(=CH-)$,which upon oxidative ozonolysis yields adipic acid $(HOOC-(CH_2)_4-COOH)$,not $CO_2$.
Therefore,cyclohexene does not evolve $CO_2$ gas.

(B) $1$. $cis-3-hexene$ undergoes $syn-addition$ of $OH$ groups using cold alkaline $KMnO_4$ or $OsO_4$ to form $meso-3,4-hexanediol$.
$2$. $trans-3-hexene$ undergoes $anti-addition$ of $OH$ groups using $RCO_3H$ followed by acid-catalyzed hydrolysis $(H_3O^{\oplus})$ to form $meso-3,4-hexanediol$.
$3$. Therefore,$(a)$ is cold $KMnO_4$ (or $OsO_4$) and $(b)$ is $RCO_3H / H_3O^{\oplus}$.

(C) The reaction is an intramolecular oxymercuration-demercuration $(OMDM)$ reaction.
$1$. The alkene reacts with $Hg(OAc)_2$ to form a cyclic mercurinium ion intermediate.
$2$. The internal hydroxyl group $(-OH)$ acts as a nucleophile and attacks the more substituted carbon of the mercurinium ion,leading to ring closure.
$3$. The subsequent treatment with $NaBH_4$ replaces the $-HgOAc$ group with a hydrogen atom.
$4$. The starting material is $6$-methylhept-$5$-en-$1$-ol. The cyclization forms a six-membered ring containing an oxygen atom (a tetrahydropyran derivative) with two methyl groups at the $2$-position.
$5$. The final product is $2,2$-dimethyltetrahydro-$2H$-pyran.

(A) The reaction proceeds via electrophilic addition of $HBr$ to the alkene.
$1$. Protonation of the double bond occurs to form the most stable carbocation. The initial carbocation formed is a tertiary carbocation on the cyclobutane ring.
$2$. The four-membered ring undergoes ring expansion to form a more stable five-membered ring carbocation.
$3$. $A$ $1,2$-methyl shift occurs to generate an even more stable tertiary carbocation.
$4$. Finally,the bromide ion $(Br^-)$ attacks this tertiary carbocation to form the major product,$1$-bromo-$1$-methylcyclopentane.

(B) The reaction is an alkoxymercuration reaction.
First,a cyclic mercurinium ion intermediate is formed across the double bond.
The nucleophile $(EtOH)$ attacks the more substituted carbon atom (Markovnikov's rule) because it can better stabilize the partial positive charge in the transition state.
Therefore,the $-OEt$ group attaches to the tertiary carbon,and the $-HgOAc$ group attaches to the secondary carbon.
The final product is $CH_3-C(CH_3)(OEt)-CH(HgOAc)-CH_3$.

(D) Markownikoff's rule states that in the electrophilic addition of $HX$ to an unsymmetrical alkene,the negative part of the addendum $(X^-)$ attaches to the carbon atom with the fewer number of hydrogen atoms.
This rule is based on the formation of the most stable carbocation intermediate.
In the reaction of $O_2N - CH = CH_2$ with $HBr$,the $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect).
When $H^+$ attacks the terminal $CH_2$ group,it forms a carbocation at the carbon attached to the $-NO_2$ group $(CH_3 - CH^+ - NO_2)$.
This carbocation is highly unstable due to the electron-withdrawing nature of the $-NO_2$ group.
Consequently,the addition occurs in a way that avoids this unstable intermediate,effectively violating the standard Markownikoff's rule as the $Br^-$ attaches to the carbon with more hydrogen atoms to avoid the highly destabilized carbocation.

(A) At low temperature $(0^\circ C)$,the $1,2$-addition product $(CH_2=CH-CH(Br)-CH_3)$ is the major product because it forms faster due to a lower activation energy; this is known as kinetic control.
At higher temperatures $(25^\circ C)$,the reaction is reversible and reaches equilibrium,favoring the more stable $1,4$-addition product $(Br-CH_2-CH=CH-CH_3)$ which has a more substituted double bond; this is known as thermodynamic control.

(C) In the hydroboration-oxidation reaction,$3 \ mol$ of alkene reacts with $1 \ mol$ of $BH_3$ to form a trialkylborane intermediate.
Therefore,the stoichiometry is $3 \text{ moles of alkene} : 1 \text{ mole of } BH_3$.
For $2 \ mol$ of $1$-pentene,the moles of $BH_3$ required = $\frac{1}{3} \times 2 = 0.666... \approx 0.67 \ mol$.

(D) Under conditions of low concentration of $Br_2$ and high temperature,the reaction proceeds via a free radical mechanism.
This mechanism favors allylic substitution over electrophilic addition.
The allylic radical formed is $CH_2=CH-CH_2^{\bullet}$ (where the label $^{14}C$ is at the terminal position).
Due to resonance,the radical can be represented as $CH_2=CH-CH_2^{\bullet} \leftrightarrow ^{\bullet}CH_2-CH=CH_2$.
Since the radical is delocalized,the bromine atom can attach to either terminal carbon,resulting in a mixture of $^{14}CH_2=CH-CH_2-Br$ and $CH_2=CH-^{14}CH_2-Br$.
Therefore,both $(A)$ and $(B)$ are formed.

(A) The reaction involves allylic bromination using $NBS$ ($N$-Bromosuccinimide),which proceeds via a free radical mechanism.
The reaction proceeds through the formation of the most stable allylic free radical.
Removing a hydrogen atom from position $a$ generates a $3^o$ allylic free radical,which is stabilized by resonance with the adjacent double bond.
Since the $3^o$ allylic free radical is more stable than other possible radicals,bromination occurs preferentially at position $a$.

(B) The reagent $m-CPBA$ (meta-chloroperbenzoic acid) is an electrophilic reagent used for the epoxidation of alkenes.
Epoxidation occurs more readily at the alkene with higher electron density,which is the more substituted alkene.
In the given molecule,the internal double bond $(CH_3-C(CH_3)=C(CH_3)-)$ is more substituted (tetrasubstituted) than the terminal double bond ($-CH=CH_2$,monosubstituted).
Therefore,the $m-CPBA$ will selectively epoxidize the more substituted internal double bond.

(D) The given reactant is $CH_3O-CH(CO_2H)-CH_2-CH_2-CH=CH-CH_2-CH_3$.
Ozonolysis ($O_3$ followed by $(CH_3)_2S$) cleaves the $C=C$ double bond to form two carbonyl compounds.
The cleavage results in:
$CH_3O-CH(CO_2H)-CH_2-CH_2-CHO \ (A)$
and
$CH_3-CH_2-CHO \ (B)$.
Compound $(A)$ is optically active due to the chiral center at the carbon atom attached to the $-OCH_3$ and $-CO_2H$ groups.
Therefore,the correct option is $(D)$.

(B) The molecule is limonene,which has two $(C=C)$ double bonds: one is a trisubstituted double bond inside the ring,and the other is a disubstituted terminal double bond (exocyclic).
Catalytic hydrogenation using $PtO_2$ is sensitive to the steric hindrance and the stability of the alkene.
The less substituted (and thus less stable) double bond is more reactive towards hydrogenation.
The terminal exocyclic double bond is less substituted and more accessible than the endocyclic trisubstituted double bond.
Therefore,the terminal double bond is reduced first to form $1$-methyl-$4$-isopropylcyclohex-$1$-ene.

(C) $1$. The reaction of cyclohexene with $HOCl$ (hypochlorous acid) proceeds via the formation of a chloronium ion intermediate,followed by the attack of water (nucleophile) from the side opposite to the chlorine atom. This results in the formation of $trans-2-chlorocyclohexanol$ as product $(A)$.
$2$. In the presence of a base like $NaOH$,the $trans-2-chlorocyclohexanol$ undergoes an intramolecular $S_N2$ reaction. The hydroxyl group is deprotonated to form an alkoxide ion,which then attacks the carbon atom bearing the chlorine atom from the backside,displacing the chloride ion. This process is facilitated by Neighbouring Group Participation ($N$.$G$.$P$.).
$3$. This intramolecular cyclization leads to the formation of $cyclohexene oxide$ (an epoxide) as product $(B)$.