Alkene Questions in English

(C) When but$-2-$ene $(CH_3-CH=CH-CH_3)$ reacts with alkaline $KMnO_4$ at an elevated temperature,the double bond undergoes oxidative cleavage.
Each $=CH-CH_3$ fragment is oxidized to $CH_3COOH$ (acetic acid).
Therefore,two molecules of acetic acid are formed.
$CH_3-CH=CH-CH_3$ $\xrightarrow[\Delta]{alk. KMnO_4} 2CH_3COOK$ $\xrightarrow{H^+} 2CH_3COOH$

(A) Reductive ozonolysis of a cyclic alkene involves breaking the double bond and forming two carbonyl groups at the cleavage sites.
To obtain $3, 5-$dimethyl$-6-$oxooctanal,we need to identify the cyclic precursor.
The product $3, 5-$dimethyl$-6-$oxooctanal is an open-chain dialdehyde/keto-aldehyde.
Specifically,the structure is $CH_3-CH_2-CH(CH_3)-C(=O)-CH_2-CH(CH_3)-CH_2-CHO$.
This indicates that the original cyclic compound must have been a cyclohexene derivative where the double bond cleavage results in this specific chain.
Analyzing the options,$1-$ethyl$-2, 4-$dimethylcyclohex$-1-$ene (Option $A$) upon ozonolysis yields the desired product by breaking the $C1=C2$ bond.
Thus,the correct compound is $1-$ethyl$-2, 4-$dimethylcyclohex$-1-$ene.

(C) The reaction of $CH_3-CH_2-CH=CH_2$ with $HBr$ in the presence of $H_2O_2$ (peroxide) follows the Kharasch effect or peroxide effect,which is an anti-Markovnikov addition.
$1$. The major product formed is $CH_3-CH_2-CH_2-CH_2Br$ ($1$-bromobutane).
$2$. $1$-bromobutane does not have a chiral carbon,so it does not show optical isomerism. Thus,option $A$ is correct.
$3$. The mechanism involves free radical intermediates,so option $B$ is correct.
$4$. The peroxide effect is specific to $HBr$ because the bond dissociation energy of $HCl$ is too high for the radical chain reaction to proceed,and $HI$ is too unstable. Therefore,$HCl$ does not undergo anti-Markovnikov addition in the presence of peroxide. Thus,option $C$ is incorrect.

(C) The given reactant is an erythro-isomer. In an $E_2$ elimination reaction,the base abstracts a proton that is anti-periplanar to the leaving group $(-Br)$.
For the erythro-isomer,the anti-elimination leads to the formation of the $trans$-alkene as the major product.
In the given structure,the $Ph$ groups are on opposite sides in the $trans$ configuration,which corresponds to the product shown in option $C$.

(B) Step $1$: Dehydration of cyclohexanol. When cyclohexanol is heated with concentrated $H_2SO_4$,it undergoes acid-catalyzed dehydration to form cyclohexene $(A)$.
Step $2$: Allylic bromination. Cyclohexene $(A)$ reacts with $N$-bromosuccinimide $(NBS)$ in the presence of light or heat. $NBS$ is a specific reagent for allylic bromination,which replaces an allylic hydrogen atom with a bromine atom. Thus,the product $(B)$ is $3$-bromocyclohexene.

(C) $1$. Cyclohexanol reacting with $SOCl_2$ is a nucleophilic substitution reaction where $-OH$ is replaced by $-Cl$.
$2$. $1-$Nitro$-4-$ethylcyclohexane reacting with $Br_2$ in the presence of $h\nu$ is a free-radical substitution reaction.
$3$. $1-$Methylcyclohexene reacting with $HI$ is an electrophilic addition reaction,not a substitution reaction.
$4$. The reaction $CH_2Br-CH_2Br \xrightarrow{NaI} CH_2=CH_2 + NaBr + HI$ is an elimination reaction.
Therefore,both $(C)$ and $(D)$ are not substitution reactions. However,based on the provided solution image,the question specifically highlights the electrophilic addition reaction of $1-$methylcyclohexene with $HI$ as the intended answer.

(A) The reaction of $1,3$-cyclohexadiene with $1 \text{ mole}$ of $HBr$ is an electrophilic addition reaction.
$H^+$ adds to one of the double bonds to form the most stable carbocation.
In $1,3$-cyclohexadiene,the protonation of one double bond leads to an allylic carbocation,which is resonance-stabilized.
$Br^-$ then attacks the carbocation.
The major product formed is $3$-bromocyclohexene,as it is the kinetic product of the electrophilic addition to the conjugated diene system.

(A) The reaction of methylenecyclohexane with $BH_3$ followed by oxidation with $H_2O_2/OH^-$ is a Hydroboration-Oxidation $(HBO)$ reaction.
This reaction proceeds via the anti-Markovnikov addition of water $(H-OH)$ across the double bond.
In methylenecyclohexane,the exocyclic double bond is between the ring carbon and the terminal $CH_2$ group.
The $OH$ group attaches to the less substituted carbon (the terminal $CH_2$ group),resulting in the formation of cyclohexylmethanol.

(B) Hydroboration-oxidation is a reaction where $BH_3$ adds across a double bond.
$1$. It involves $Syn$ addition of $H$ and $BH_2$ across the double bond.
$2$. The reaction proceeds through a concerted mechanism involving a four-membered cyclic transition state.
$3$. It follows Anti-Markovnikov regioselectivity,where the boron atom attaches to the less substituted carbon atom.
Therefore,statements $I$ and $II$ are correct. Statement $III$ is incorrect because it follows Anti-Markovnikov rule,and statement $IV$ is incorrect because it is a $Syn$ addition.

(A) The reaction proceeds via a carbocation intermediate.
First,the addition of $H^{+}$ to $CH_3-C(CH_3)_2-CH=CH_2$ follows Markovnikov's rule to form a secondary carbocation: $CH_3-C(CH_3)_2-C^{+}H-CH_3$.
Next,a $1,2$-methyl shift occurs to convert the secondary carbocation into a more stable tertiary carbocation: $CH_3-C^{+}(CH_3)-CH(CH_3)-CH_3$.
Finally,the attack of $H_2O$ on this tertiary carbocation followed by deprotonation yields $2,3$-dimethyl-$2$-butanol,which is $CH_3-C(OH)(CH_3)-CH(CH_3)-CH_3$,as the major product $A$.

(A) The given reactant is $2,3-dibromobutane$. The reaction is a debromination reaction,which typically proceeds via an $E2$ mechanism using zinc dust $(Zn)$ in an alcoholic solvent.
In the given Fischer projection,the two bromine atoms are on the same side (erythro form). When this specific stereoisomer undergoes $E2$ elimination,the hydrogen and bromine atoms must be in an anti-periplanar conformation.
For the erythro$-2,3-$dibromobutane,the anti-elimination of $Br_2$ leads to the formation of $cis-but-2-ene$ as the major product.
Therefore,the correct option is $A$.

(A) The reaction proceeds in two steps:
$(i)$ Dehydration of cyclopentanol with $H_3PO_4$ at $150^{\circ}C$ leads to the formation of cyclopentene.
(ii) Reaction of cyclopentene with perbenzoic acid $(C_6H_5COOOH)$ is an epoxidation reaction,which converts the alkene into an epoxide (cyclopentene oxide).

(A) The reaction of cyclopentene with $N$-bromosuccinimide $(NBS)$ in the presence of heat is an example of allylic bromination.
$NBS$ provides a low concentration of $Br_2$,which undergoes free radical substitution at the allylic position.
The allylic position in cyclopentene is the carbon atom adjacent to the double bond.
Therefore,the major product formed is $3$-bromocyclopent-$1$-ene.

(B) The reaction of cyclohexene with $Br_2$ in $CCl_4$ is an electrophilic addition reaction.
This reaction proceeds via the formation of a cyclic bromonium ion intermediate.
The bromide ion $(Br^-)$ attacks the cyclic bromonium ion from the side opposite to the bromonium bridge,resulting in anti-addition.
For cyclohexene,this anti-addition of bromine atoms results in the formation of trans-$1,2$-dibromocyclohexane.
Since the product is a pair of enantiomers (trans-$1,2$-dibromocyclohexane exists as $(1R, 2R)$ and $(1S, 2S)$ forms),the resulting mixture is a racemic mixture.

(C) The reaction is an ozonolysis of an alkene,$Me_2C=CHCH_3$.
In the presence of $O_3$ followed by $H_2O$ (oxidative workup,as $Zn$ is absent),the double bond is cleaved.
The fragments formed are $Me_2C=O$ (acetone) and $CH_3CH=O$ (acetaldehyde).
Since $Zn$ is absent,the aldehyde $(CH_3CHO)$ is further oxidized to the corresponding carboxylic acid,$CH_3COOH$.
Therefore,the final products are $Me_2CO$ and $CH_3COOH$.

(C) Ethylene reacts with sulphur monochloride $(S_2Cl_2)$ to form mustard gas $(Cl-CH_2-CH_2-S-CH_2-CH_2-Cl)$.
The chemical reaction is:
$2CH_2=CH_2 + S_2Cl_2 \rightarrow (Cl-CH_2-CH_2)_2S + S$.

(A) The stability of alkenes is directly proportional to the number of $\alpha$-hydrogen atoms (hyperconjugation) and steric factors.
$V$ (Ethene): $CH_2=CH_2$ $(0 \ \alpha-H)$
$I$ (Propene): $CH_3-CH=CH_2$ $(3 \ \alpha-H)$
$II$ (cis-but-$2$-ene): $CH_3-CH=CH-CH_3$ ($6 \ \alpha-H$,less stable due to steric hindrance)
$III$ (trans-but-$2$-ene): $CH_3-CH=CH-CH_3$ ($6 \ \alpha-H$,more stable due to less steric hindrance)
$IV$ ($2,3$-dimethylbut-$2$-ene): $(CH_3)_2C=C(CH_3)_2$ $(12 \ \alpha-H)$
Increasing order of stability: $V < I < II < III < IV$.

(B) The reaction of propene $(CH_3-CH=CH_2)$ with $HOCl$ involves the formation of a chloronium ion intermediate.
$HOCl$ dissociates to provide an electrophilic $Cl^{+}$ species.
In the first step,the $\pi$-bond of the alkene attacks the electrophilic $Cl^{+}$ to form a cyclic chloronium ion intermediate.
This is followed by the nucleophilic attack of $\overset{\Theta}{O}H$ on the more substituted carbon atom of the chloronium ion to yield the final product,$1-chloro-2-propanol$ (or $2-chloro-1-propanol$ depending on regioselectivity,but the electrophile is $Cl^{+}$).
Therefore,the reaction proceeds via the addition of $Cl^{+}$ in the first step.

(B) Ozonolysis involves the cleavage of double bonds to form carbonyl compounds.
For $cyclohex-1,3-diene$,the two double bonds at positions $1$ and $3$ are cleaved.
The cleavage of the double bond between $C_1-C_2$ and $C_3-C_4$ in the six-membered ring results in the formation of $CHO-CHO$ (glyoxal) and $CHO-CH_2-CH_2-CHO$ (butanedial).
Therefore,the correct olefin is $cyclohex-1,3-diene$.

(A) The hydration of isobutene ($2-$methylprop$-1-$ene) in the presence of dilute $H_2SO_4$ follows Markovnikov's addition.
In the first step,the proton $(H^+)$ from the acid attacks the terminal $CH_2$ group of the alkene to form a stable $tert-$butyl carbocation $((CH_3)_3C^+)$.
In the second step,water acts as a nucleophile and attacks the $tert-$butyl carbocation.
Finally,deprotonation yields $tert-$butyl alcohol ($2-$methylpropan$-2-$ol) as the major product.

(A) Hydroboration-oxidation is a syn-addition reaction.
In the hydroboration of $1-$methylcyclopentene with $B_2D_6,$ the boron atom and the deuterium atom add across the double bond in a syn-fashion.
Since the methyl group is at the $C_1$ position,the boron atom adds to the less hindered $C_2$ position,and the deuterium atom adds to the $C_1$ position.
Because it is a syn-addition,both the deuterium and the boron atom add from the same face of the cyclopentene ring.
Subsequent oxidation with alkaline $H_2O_2$ replaces the boron group with a hydroxyl $(-OH)$ group with retention of configuration.
Therefore,the $-OH$ group and the deuterium atom will be on the same side (cis) of the ring,and the $-OH$ group will be at the $C_2$ position while the deuterium is at the $C_1$ position.
Looking at the options,the structure where $-OH$ and $D$ are cis to each other is represented in option $A$.

(D) An unsaturated compound contains at least one carbon-carbon double bond $(C=C)$ or triple bond $(C\equiv C)$.
$A$. Acetonitrile $(CH_3CN)$ contains a carbon-nitrogen triple bond,but the carbon-carbon bond is single.
$B$. Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ contains only single carbon-carbon bonds.
$C$. Malic acid $(HOOC-CH_2-CH(OH)-COOH)$ contains only single carbon-carbon bonds.
$D$. Acrylic acid $(CH_2=CH-COOH)$ contains a carbon-carbon double bond $(C=C)$,making it an unsaturated compound.

(C) The reaction is an oxymercuration-demercuration process.
Step $1$: Oxymercuration involves the electrophilic attack of $Hg(OAc)^+$ on the double bond of propene $(CH_3-CH=CH_2)$,followed by the attack of water $(H_2O)$ to form a mercurial alcohol. The $OH$ group attaches to the more substituted carbon (Markovnikov addition),resulting in $CH_3-CH(OH)-CH_2-HgOAc$.
Step $2$: Demercuration with $NaBD_4$ replaces the $-HgOAc$ group with a deuterium atom $(D)$.
Therefore,the final product is $CH_3-CH(OH)-CH_2D$.

(A) Diastereomers are stereoisomers that are not mirror images of each other.
In the reaction of $3$-methylpent-$1$-ene with $HBr$ (electrophilic addition),the $H^+$ adds to the terminal carbon to form a carbocation at the $C_2$ position. This carbocation is planar. The $Br^-$ ion can attack from either side of the planar carbocation. Since the molecule already contains a chiral center at the $C_3$ position,the formation of a new chiral center at the $C_2$ position results in two products that have different configurations at the new chiral center but the same configuration at the original chiral center. These are diastereomers.

(B) The reductive ozonolysis of $1-$methylcyclohexene involves the oxidative cleavage of the $C=C$ double bond.
In $1-$methylcyclohexene,the double bond is between $C1$ (which has a methyl group) and $C2$.
Upon ozonolysis,the $C1$ atom is converted into a ketone group $(CH_3-CO-)$ and the $C2$ atom is converted into an aldehyde group $(-CHO)$.
The ring opens to form a linear chain with $6$ carbon atoms in total.
The structure is $CH_3-CO-CH_2-CH_2-CH_2-CH_2-CHO$,which can be written as $CH_3CO(CH_2)_4CHO$.

(B) The reactivity of alkenes towards electrophilic addition of $HBr$ depends on the stability of the intermediate carbocation formed after the protonation of the double bond.
Greater the electron-donating ability of the substituent attached to the double bond,the more stable the resulting carbocation will be.
In compound $(b)$,the $-OCH_3$ group is directly attached to the double bond and exerts a strong $+M$ (mesomeric) effect,which significantly stabilizes the carbocation.
In compound $(a)$,the $-CH_3$ group exerts a $+I$ (inductive) effect and hyperconjugation,which provides moderate stability.
In compound $(c)$,the $-CH_3OCH_2-$ group exerts a weak $+I$ effect,but it is further away from the double bond compared to the groups in $(a)$ and $(b)$,making it the least effective at stabilizing the carbocation.
Therefore,the order of stability of the carbocations,and thus the reactivity of the alkenes,is $b > a > c$.

(C) Compound $(A)$ is $(CH_3)_3C-CH=CH-CH_3$ ($4,4$-dimethylpent-$2$-ene).
Upon ozonolysis,it yields $(CH_3)_3C-CHO$ (pivalaldehyde) and $CH_3CHO$ (acetaldehyde).
$(CH_3)_3C-CH=CH-CH_3 \xrightarrow[Zn]{O_3} (CH_3)_3C-CHO + CH_3CHO$
Pivalaldehyde $(B)$ lacks $\alpha$-hydrogens and thus undergoes the Cannizzaro reaction.
Additionally,$(A)$ exhibits geometrical isomerism because the groups attached to each carbon of the double bond are different ($H$ and $t$-butyl on one,$H$ and methyl on the other).

(D) The reaction is a hydroboration-oxidation of $1$-Methylcyclopentene using deuterated reagents.
$1$. The first step involves the addition of $B-D$ across the double bond via syn-addition. The boron atom attaches to the less substituted carbon,and the deuterium atom attaches to the more substituted carbon.
$2$. The second step is the oxidation with $D_2O_2$ in the presence of $NaOD$,which replaces the boron group with an $-OD$ group with retention of configuration.
This results in the anti-Markovnikov addition of $D$ and $OD$ across the double bond with syn-stereochemistry.
Therefore,the correct reagent is $1. B_2D_6, 2. D_2O_2, NaOD$.

(C) $HCl$ does not show the peroxide effect (Kharasch effect). The reaction proceeds via Electrophilic Addition Reaction $(EAR)$.
$1$. Protonation of the alkene leads to the formation of a more stable carbocation.
$2$. The $1$-methylcyclopentyl carbocation undergoes ring expansion to form a more stable $6$-membered ring (cyclohexyl carbocation).
$3$. The chloride ion $(Cl^-)$ then attacks the carbocation to form $1$-chloro-$1$-methylcyclohexane as the major product.

(A) The reaction of propene with nitrosyl chloride $(NOCl)$ is an Electrophilic Addition Reaction $(EAR)$.
In $NOCl$,the bond breaks heterolytically to form $NO^+$ (nitrosonium ion) as the electrophile and $Cl^-$ as the nucleophile.
$NO^+$ attacks the double bond of propene to form a more stable carbocation intermediate (secondary carbocation).
$CH_3-CH=CH_2 + NO^+ \to CH_3-CH^+-CH_2-NO$
Then,the nucleophile $Cl^-$ attacks the carbocation to form the final product:
$CH_3-CH(Cl)-CH_2-NO$
Thus,the product is $1-nitro-2-chloropropane$ (or $2-chloro-1-nitropropane$ depending on nomenclature,but based on the structure,it is $CH_3-CH(Cl)-CH_2(NO)$).

(D) The reaction proceeds as follows:
$1$. Cyclohexene reacts with $N$-bromosuccinimide $(NBS)$ to undergo allylic bromination,forming $3-$bromocyclohexene.
$2$. $3-$bromocyclohexene reacts with magnesium $(Mg)$ in the presence of dry ether to form the Grignard reagent,cyclohex$-2-$en$-1-$ylmagnesium bromide.
$3$. The Grignard reagent then reacts with carbon dioxide $(CO_2)$ followed by acidic workup $(H^+)$ to form the corresponding carboxylic acid,which is cyclohex$-2-$ene$-1-$carboxylic acid.
Therefore,the final product $(x)$ is cyclohex$-2-$ene$-1-$carboxylic acid.

(B) The reaction involves the electrophilic addition of $HCl$ to $2$-methylbut-$1,3$-diene (isoprene).
$1$. Protonation of the conjugated diene occurs at the terminal carbon to form the most stable carbocation,which is a resonance-stabilized allylic carbocation.
$2$. The resonance hybrid allows for the attack of the chloride ion $(Cl^-)$ at different positions.
$3$. The thermodynamic product $(TCP)$ is formed by the addition of $Cl^-$ to the more substituted end of the allylic system,resulting in $1$-chloro-$3$-methylbut-$2$-ene.

(A) The reaction of $cis-but-2-ene$ with Baeyer's reagent (cold,dilute alkaline $KMnO_4$) involves syn-hydroxylation of the double bond.
This process adds two hydroxyl $(-OH)$ groups to the same side of the alkene.
For $cis-but-2-ene$,syn-addition results in the formation of $meso-butane-2,3-diol$.
The structure of $meso-butane-2,3-diol$ is represented by option $(A)$.

(A) The addition of $HBr$ to isobutylene $(2-\text{methylpropene})$ in the presence of peroxide $(R_2O_2)$ follows the Anti-Markovnikov rule,resulting in the formation of isobutyl bromide $(X = CH_3-CH(CH_3)-CH_2-Br)$.
Subsequent reaction of $X$ with sodium methoxide $(CH_3ONa)$,which is a strong nucleophile,leads to an $S_N2$ reaction to form isobutyl methyl ether $(Y = CH_3-CH(CH_3)-CH_2-OCH_3)$.

(C) The reaction of $2,3$-dichlorobutane with $Zn$ dust is a dehalogenation reaction.
In the given stereoisomer (meso-form),the two $Cl$ atoms are anti to each other,leading to the formation of $trans-but-2-ene$.
$trans-but-2-ene$ has a center of symmetry,which makes its dipole moment equal to zero.
$trans-isomers$ are more symmetrical than $cis-isomers$,allowing them to pack better in the crystal lattice,resulting in a higher melting point.
Therefore,the statement that the melting point is greater than its geometrical isomer $(cis-but-2-ene)$ is correct.

(A) $N$-Bromosuccinimide $(N.B.S.)$ is a reagent used for allylic bromination in the presence of light $(h\nu)$.
In the given substrate,$1,5,5-\text{trimethylcyclohex-1-ene}$,there are two types of allylic positions:
$1$. The allylic position at $C-3$ (adjacent to the double bond,containing two hydrogen atoms).
$2$. The allylic position at the methyl group attached to $C-1$.
Allylic bromination occurs via a free radical mechanism,forming the most stable allylic radical intermediate.
The radical formed at the $C-3$ position is more stable due to the hyperconjugation and inductive effects of the alkyl groups compared to the radical formed on the methyl group.
Therefore,the bromine atom replaces one of the hydrogen atoms at the $C-3$ position.
The major product is $3-\text{bromo}-1,5,5-\text{trimethylcyclohex-1-ene}$.

(C) The pyrolysis of quaternary ammonium hydroxides follows Hofmann's rule,where the less substituted alkene is the major product. This is because the bulky leaving group $(-N(CH_3)_3)$ makes the transition state leading to the more substituted alkene sterically hindered.
Reaction:
$[CH_3-CH_2-CH(CH_3)-N^{+}(CH_3)_3]OH^{-} \xrightarrow{\Delta} CH_3-CH_2-CH=CH_2 + N(CH_3)_3 + H_2O$
Thus,$1-Butene$ is the major product.

(C) The given reaction is the ozonolysis of $2\text{-methylprop-1-ene}$ (isobutylene) using $(i) O_3$ followed by $(ii) H_2O/Zn$.
Ozonolysis involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
The reaction is:
$(CH_3)_2C=CH_2 + O_3$ $\rightarrow \text{ozonide intermediate}$ $\xrightarrow{H_2O/Zn} (CH_3)_2C=O + HCHO$
Here,$(CH_3)_2C=O$ is acetone and $HCHO$ is formaldehyde.
Therefore,the products are acetone and formaldehyde.

(A) $1$. The starting material is butan-$2$-one $(CH_3CH_2COCH_3)$.
$2$. Reduction with $LiAlH_4$ converts the ketone to a secondary alcohol,$X$,which is butan-$2$-ol $(CH_3CH_2CH(OH)CH_3)$.
$3$. Dehydration of butan-$2$-ol with $conc. H_2SO_4$ at high temperature follows Saytzeff's rule to form the more substituted alkene,but-$2$-ene $(CH_3CH=CHCH_3)$,as the major product $Y$.
$4$. In but-$2$-ene $(CH_3CH=CHCH_3)$,there are two $\alpha$-carbons adjacent to the double bond,each having $3$ hydrogen atoms.
$5$. Therefore,the total number of $\alpha$-hydrogen atoms available for hyperconjugation is $3 + 3 = 6$. However,checking the options provided,let's re-evaluate the structure. If $Y$ is $CH_3CH=CHCH_3$,it has $6$ $\alpha$-hydrogens. Given the options,if the major product is considered to be pent$-2-$ene or if there is a typo in the question's structure,we look for the closest match. Re-examining the starting material $CH_3CH_2COCH_3$ (butan$-2-$one),the product is indeed but$-2-$ene. With $6$ hydrogens,and $5$ being the closest option,there might be a slight discrepancy in the question's provided options. Assuming the intended structure leads to $5$ hydrogens (e.g.,if the product was a different isomer),but based on standard chemistry,$6$ is the correct count. Given the options,$A$ $(5)$ is the most likely intended answer if the structure was slightly different.

(A) The reaction of $Methylenecyclohexane$ with $Br_2$ in the presence of $H_2O$ is a halohydrin formation reaction.
$1$. The double bond attacks the $Br_2$ molecule to form a cyclic bromonium ion intermediate.
$2$. Water $(H_2O)$,acting as a nucleophile,attacks the more substituted carbon of the bromonium ion because it can better stabilize the partial positive charge in the transition state.
$3$. This leads to the formation of $1-(bromomethyl)cyclohexan-1-ol$ as the major product.
The reaction is:
$C_6H_{10}=CH_2 + Br_2 + H_2O \rightarrow C_6H_{10}(OH)(CH_2Br) + HBr$

(A) According to Markovnikov's rule,the negative part of the addendum $(Br^-)$ adds to the carbon atom of the double bond which contains the least number of hydrogen atoms.
$CH_2=CH-CH_3 + HBr \to CH_3-CH(Br)-CH_3$
The product formed is $2$-bromopropane.

(A) The reaction is the electrophilic addition of $HBr$ to an alkene $(allylbenzene)$.
According to Markovnikov's rule,the electrophile $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms,forming the more stable carbocation.
In $C_6H_5-CH_2-CH=CH_2$,the carbocation formed after the addition of $H^+$ to the terminal $CH_2$ group is $C_6H_5-CH_2-CH^+-CH_3$.
This is a secondary $(2^{\circ})$ carbocation,which is relatively stable.
Then,the nucleophile $(Br^-)$ attacks this carbocation to form the product $C_6H_5-CH_2-CH(Br)-CH_3$ ($1$-phenyl$-2-$bromopropane).
Therefore,the correct option is $A$.

(A) The reaction of cyclohexene with $Br_2$ in the presence of water $(H_2O)$ is a halohydrin formation reaction.
In this reaction,the electrophilic bromine atom adds to the double bond to form a cyclic bromonium ion intermediate.
Water,acting as a nucleophile,then attacks the more substituted carbon atom of the bromonium ion from the side opposite to the bromine atom (anti-addition).
This results in the formation of $2$-bromocyclohexanol,where the $-Br$ and $-OH$ groups are in a trans configuration.
The product is a racemic mixture of the two enantiomers of $trans-2$-bromocyclohexanol.

(B) The heat of hydrogenation $(\Delta H_{H_2})$ is inversely proportional to the stability of the alkene.
More substituted alkenes are more stable due to hyperconjugation and inductive effects,thus they have lower heat of hydrogenation.
Comparing the structures:
$A$ is a terminal alkene.
$B$ is a tetrasubstituted alkene (most substituted).
$C$ is a trisubstituted alkene.
$D$ is a monosubstituted terminal alkene.
Among the given options,the compound in option $B$ is the most substituted alkene,which makes it the most stable and therefore it has the least heat of hydrogenation.

(C) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond and the addition of an oxygen atom to each carbon atom. To obtain only ketones as products,both carbon atoms of the double bond must be disubstituted (i.e.,they should not have any hydrogen atoms attached).
For the alkene in option $(c)$ ($1$-isopropylidenecyclopentane),the ozonolysis reaction is:
$C_5H_8=C(CH_3)_2 \xrightarrow{O_3, Zn/H_2O} C_5H_8=O + O=C(CH_3)_2$
Here,the products are cyclopentanone and acetone,both of which are ketones. Therefore,option $(c)$ is the correct answer.

(C) Let us analyze each reaction:
$(A)$ Ozonolysis of propene $(CH_3-CH=CH_2)$ with oxidative workup $(O_3/H_2O)$ yields acetic acid and formic acid. Formic acid $(HCOOH)$ further oxidizes to $CO_2$ and $H_2O$. Thus,$CO_2$ is released.
$(B)$ Methylenecyclohexane on oxidation with warm $KMnO_4$ undergoes cleavage of the double bond. The terminal $=CH_2$ group is oxidized to $CO_2$ and $H_2O$. Thus,$CO_2$ is released.
$(C)$ Ozonolysis of allylcyclohexane with reductive workup $(O_3/Zn)$ yields cyclohexylacetaldehyde and formaldehyde $(HCHO)$. No $CO_2$ is produced in this reaction.
$(D)$ Alkaline $KMnO_4$ oxidation of propyne $(CH_3-C \equiv CH)$ leads to oxidative cleavage,producing acetic acid and $CO_2$. Thus,$CO_2$ is released.
Therefore,option $(C)$ does not release $CO_2$.

(A) The given reaction is Hydroboration-Oxidation of an alkene.
$1$. The reagent $(i) B_2H_6(THF)$ followed by $(ii) H_2O_2, NaOH$ performs anti-Markovnikov addition of water across the double bond.
$2$. The hydroxyl group $(-OH)$ attaches to the less substituted carbon atom of the double bond.
$3$. In the reactant,vinylcyclopentane,the terminal carbon of the vinyl group is less substituted than the carbon attached to the cyclopentyl ring.
$4$. Therefore,the $-OH$ group adds to the terminal carbon,resulting in the formation of $2-cyclopentylethanol$.

(C) The stability of an alkene is directly proportional to the number of $\alpha$-hydrogen atoms attached to the double-bonded carbon atoms (hyperconjugation effect).
$1$. $(CH_3)_3C-CH=CH_2$: The double-bonded carbons have $0$ $\alpha$-hydrogen atoms.
$2$. $CH_2=CH_2$: This has $0$ $\alpha$-hydrogen atoms.
$3$. $(CH_3)_2C=CH_2$: The double-bonded carbon has two methyl groups attached,providing $6$ $\alpha$-hydrogen atoms.
$4$. $CH_3-CH=CH_2$: This has $3$ $\alpha$-hydrogen atoms.
Since $(CH_3)_2C=CH_2$ has the highest number of $\alpha$-hydrogen atoms $(6)$,it is the most stable alkene among the given options.

(C) The reaction involves the acid-catalyzed dehydration of a secondary alcohol $(1-cyclohexylethanol)$.
$1$. Protonation of the hydroxyl group: The $-OH$ group is protonated by $H^+$ from $Conc. H_2SO_4$ to form a good leaving group $(-OH_2^+)$.
$2$. Formation of carbocation: Loss of water molecule generates a secondary carbocation at the carbon atom attached to the cyclohexane ring.
$3$. Rearrangement: The secondary carbocation undergoes a $1,2-hydride$ shift to form a more stable tertiary carbocation (the positive charge moves to the carbon on the ring).
$4$. Elimination: Loss of a proton from the adjacent carbon leads to the formation of the most stable alkene,which is the $Saytzeff$ product,$1-ethylcyclohex-1-ene$.

(B) Markovnikov's rule states that the electrophilic part of the reagent (e.g.,$H^+$) adds to the carbon atom of the double bond that already has a greater number of hydrogen atoms.
In the case of $CCl_3 - CH = CH_2$,the $CCl_3$ group is a strong electron-withdrawing group due to the inductive effect ($-I$ effect).
This effect destabilizes the formation of a carbocation at the $CH$ position,and the polarization of the double bond is reversed compared to simple alkenes.
As a result,the electrophile $(H^+)$ adds to the $CH_2$ group instead of the $CH$ group,which is an anti-Markovnikov addition.