Alkene Questions in English

(C) The reaction of propene $(CH_3-CH=CH_2)$ with $HI$ in the presence of peroxide is an electrophilic addition reaction.
However,the peroxide effect (Kharasch effect or anti-Markovnikov addition) is only applicable to $HBr$.
$HI$ does not show the peroxide effect because the step involving the addition of the iodine radical is endothermic and reversible.
Therefore,the reaction follows Markovnikov's rule,where the electrophile $H^+$ adds to the carbon with more hydrogen atoms,and the nucleophile $I^-$ adds to the carbon with fewer hydrogen atoms.
Thus,the product is $2$-iodopropane $(CH_3-CH(I)-CH_3)$.

(D) The reaction of $1$-methylcyclohexene with $B_2H_6$ followed by oxidation with $H_2O_2/OH^-$ is a hydroboration-oxidation reaction.
This reaction follows anti-Markovnikov addition of water across the double bond.
Hydroboration is a syn-addition,meaning the $H$ and $OH$ groups add to the same side of the double bond.
In $1$-methylcyclohexene,the $OH$ group adds to the less substituted carbon $(C_2)$ and the $H$ atom adds to the more substituted carbon $(C_1)$.
Since the addition is syn,the $H$ and $OH$ groups will be on the same side of the ring,resulting in cis-$2$-methylcyclohexanol.

(B) $NBS$ ($N$-bromosuccinimide) is a reagent used for allylic bromination.
In the reaction of propene $(CH_3-CH=CH_2)$ with $NBS$ in $CCl_4$,the hydrogen atom at the allylic position is replaced by a bromine atom.
The reaction is: $CH_3-CH=CH_2 + NBS \xrightarrow{CCl_4} Br-CH_2-CH=CH_2 + \text{succinimide}$.
The product formed is $3$-bromopropene (also known as allyl bromide).

(A) The reaction of propene $(CH_3-CH=CH_2)$ with bromine $(Br_2)$ in the presence of carbon tetrachloride $(CCl_4)$ is an electrophilic addition reaction.
The bromine molecule adds across the double bond to form a vicinal dihalide.
The reaction proceeds as follows:
$CH_3-CH=CH_2 + Br_2 \xrightarrow{CCl_4} CH_3-CH(Br)-CH_2Br$
The product formed is $1,2$-dibromopropane.

(B) The reaction of ethylene with $Br_2$ proceeds via the formation of a cyclic bromonium ion intermediate.
When $Br_2$ approaches the double bond of ethylene,the electron-rich double bond attacks one of the bromine atoms,displacing the other as a bromide ion $(Br^-)$.
This results in the formation of a three-membered cyclic bromonium ion.
The subsequent nucleophilic attack by the bromide ion $(Br^-)$ occurs from the side opposite to the cyclic bromonium ion,which leads to anti-addition.

(C) According to $Markownikoff's$ rule,when an unsymmetrical reagent (like $HX$,$H_2O$,etc.) adds to an unsymmetrical alkene,the negative part of the reagent attaches to the carbon atom of the double bond that carries the lesser number of hydrogen atoms.

(B) The reaction of $3$-phenylpropene $(C_6H_5CH_2CH=CH_2)$ with $HBr$ follows Markovnikov's rule.
In the first step,the electrophile $H^+$ attacks the terminal carbon $(C_3)$ to form a more stable carbocation.
The intermediate formed is a secondary benzylic carbocation $(C_6H_5CH_2CH^+CH_3)$,which is stabilized by resonance with the phenyl ring.
Subsequently,the nucleophile $Br^-$ attacks the carbocation to form the major product,$1$-phenyl-$2$-bromopropane $(C_6H_5CH_2CH(Br)CH_3)$.

(B) The reaction of propene $(CH_3-CH=CH_2)$ with $HOCl$ (hypochlorous acid) is an electrophilic addition reaction.
$HOCl$ dissociates to provide the electrophile $Cl^{+}$ and the nucleophile $OH^{-}$.
In the first step,the electrophile $Cl^{+}$ attacks the double bond of propene to form a cyclic chloronium ion intermediate.
Therefore,the reaction proceeds via the addition of $Cl^{+}$ in the first step.

(C) Ozonolysis of alkenes involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
Acetone $(CH_3COCH_3)$ is a ketone with three carbon atoms.
For acetone to be formed,the alkene must have a $(CH_3)_2C=$ group.
Isobutylene $(CH_2=C(CH_3)_2)$ undergoes ozonolysis to produce acetone $(CH_3COCH_3)$ and formaldehyde $(HCHO)$.
Therefore,the correct option is $C$.

(A) Ozonolysis of an alkene followed by reduction with $Zn/H_2O$ breaks the $C=C$ double bond and adds an oxygen atom to each carbon atom.
If the product is two moles of the same aldehyde,the alkene must be symmetric.
For $2$-Butene $(CH_3-CH=CH-CH_3)$,ozonolysis yields:
$CH_3-CH=CH-CH_3 + O_3 \rightarrow 2CH_3CHO$ (Acetaldehyde).
Thus,$X$ is $2$-Butene.

(A) The reaction of $cis-2-butene$ with cold dilute $KMnO_4$ (Baeyer's reagent) involves a $syn$-hydroxylation of the double bond.
Since the addition is $syn$ and the starting material is $cis-2-butene$,the product formed is $meso-2,3-butanediol$.

(A) The reaction of $CH_2 = CH(CH_2)_2COOH$ with $HBr$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom of the double bond that has fewer hydrogen atoms.
In the given molecule $CH_2 = CH(CH_2)_2COOH$,the double bond is between $C_1$ and $C_2$.
$C_1$ has $2$ hydrogens and $C_2$ has $1$ hydrogen.
Therefore,$Br^-$ will attach to $C_2$ and $H^+$ will attach to $C_1$.
The resulting product is $CH_3-CH(Br)-(CH_2)_2COOH$.

(A) The reaction of $1,3$-butadiene with $HBr$ at $40^{\circ}C$ (higher temperature) leads to the thermodynamic product.
Under thermodynamic control,the more stable $1,4$-addition product,$1$-bromo-$2$-butene,is the major product.
At lower temperatures (e.g.,$-80^{\circ}C$),the kinetic product,$3$-bromo-$1$-butene,is favored.

(B) The conversion of propene $(CH_3-CH=CH_2)$ to $1$-propanol $(CH_3-CH_2-CH_2-OH)$ requires anti-Markovnikov addition of water.
This is achieved via hydroboration-oxidation using $B_2H_6$ followed by $H_2O_2$ in the presence of $OH^-$.

(A) The molecular formula of $(Y)$ is $C_6H_{14}$,which corresponds to an alkane. Since $(X)$ is a hydrocarbon that undergoes hydrogenation to form $C_6H_{14}$,$(X)$ must be an alkene with the formula $C_6H_{12}$.
$(X)$ is optically active,meaning it must contain a chiral center. In $3$-methyl-$1$-pentene $(CH_2=CH-CH(CH_3)-CH_2-CH_3)$,the carbon at position $3$ is chiral because it is bonded to a hydrogen atom,a methyl group,an ethyl group,and a vinyl group.
Upon catalytic hydrogenation,$3$-methyl-$1$-pentene forms $3$-methylpentane. In $3$-methylpentane,the carbon at position $3$ is bonded to two identical ethyl groups $(-CH_2CH_3)$,making it achiral. Thus,the product is optically inactive.

(A) The dehydration of $1$-butanol $(CH_3CH_2CH_2CH_2OH)$ with concentrated $H_2SO_4$ proceeds via an $E1$ mechanism.
First,the hydroxyl group is protonated to form a good leaving group,which leaves to form a primary carbocation $(CH_3CH_2CH_2CH_2^+)$.
This primary carbocation undergoes a $1,2$-hydride shift to form a more stable secondary carbocation $(CH_3CH_2CH^+CH_3)$.
Elimination of a proton from the secondary carbocation yields $2$-butene as the major product,which exists primarily as the more stable trans-$2$-butene isomer.

(A) The reaction of ethene $(CH_2 = CH_2)$ with cold,dilute alkaline potassium permanganate $(KMnO_4)$ solution is known as Baeyer's reagent test.
This reaction results in the hydroxylation of the double bond to form a vicinal diol.
$CH_2 = CH_2 + H_2O + [O] \xrightarrow{KMnO_4/OH^-} HO-CH_2-CH_2-OH$ (Ethylene glycol).
Thus,the product $X$ is ethylene glycol.

(B) The reaction of $1$-butene $(CH_3-CH_2-CH=CH_2)$ with water in the presence of an acid catalyst $(H^+)$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(OH^-)$ attaches to the carbon atom with fewer hydrogen atoms.
In $1$-butene,the double bond is between $C_1$ and $C_2$. $C_2$ has one hydrogen atom,while $C_1$ has two.
Therefore,the $OH$ group attaches to $C_2$,resulting in the formation of $2$-butanol $(CH_3-CH_2-CH(OH)-CH_3)$.
Since the $OH$ group is attached to a secondary carbon atom,the product is a secondary alcohol.

(C) The reaction of $Br_2$ with an alkene involves the formation of a cyclic bromonium ion intermediate.
In the case of $cis-2-butene$,the $anti$-addition of $Br_2$ leads to the formation of the $meso-2,3-dibromobutane$ isomer because the molecule has a plane of symmetry.

(C) The Markovnikov rule states that when an unsymmetrical reagent adds to an unsymmetrical alkene,the negative part of the reagent attaches to the carbon atom with fewer hydrogen atoms.
But-$2$-ene $(CH_3-CH=CH-CH_3)$ is a symmetrical alkene.
Because it is symmetrical,the addition of $HBr$ results in the same product regardless of which carbon the bromine atom attaches to.
Therefore,the concept of the Markovnikov rule is not applicable to symmetrical alkenes like but-$2$-ene.

(A) $1$. The molecular formula $C_6H_{10}$ corresponds to the general formula $C_nH_{2n-2}$,indicating the presence of either one triple bond or two double bonds or a ring with one double bond.
$2$. Since it absorbs only one molecule of $H_2$,it must contain exactly one double bond (or one triple bond that is reduced to a double bond,but the ozonolysis product suggests a cyclic structure).
$3$. Ozonolysis of the hydrocarbon yields $CHO(CH_2)_4CHO$,which is hexanedial. This indicates that the original molecule was a cyclic compound with a double bond in the ring.
$4$. $A$ cyclic $C_6$ hydrocarbon with one double bond is cyclohexene. Ozonolysis of cyclohexene breaks the double bond to form the open-chain dialdehyde $CHO(CH_2)_4CHO$.

(C) The anti-Markovnikov addition (also known as the peroxide effect or Kharasch effect) is specifically observed in the addition of $HBr$ to unsymmetrical alkenes like propene $(C_3H_6)$ in the presence of organic peroxides.
In this reaction,the bromine atom attaches to the carbon atom with more hydrogen atoms.
$C_3H_6 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2Br$ ($1$-bromopropane).
Other hydrogen halides like $HCl$ and $HI$ do not show this effect due to the energetics of the reaction steps.

(C) The reaction of ethyl iodide $(C_2H_5I)$ with alcoholic $KOH$ undergoes a dehydrohalogenation reaction (elimination reaction) to form ethene $(C_2H_4)$.
$C_2H_5I + KOH (\text{alc.}) \rightarrow C_2H_4 + KI + H_2O$
Ethene $(C_2H_4)$ is an unsaturated hydrocarbon containing a double bond.
Unsaturated hydrocarbons react with Baeyer's reagent (alkaline $KMnO_4$ solution) to decolorize the purple color of $KMnO_4$ due to the formation of ethylene glycol.

(A) The reaction is an oxymercuration-demercuration process.
$1$. Oxymercuration with $Hg(OAc)_2$ and $H_2O$ adds $OH$ to the more substituted carbon (Markovnikov addition) and $HgOAc$ to the less substituted carbon.
$2$. Demercuration with $NaBD_4$ replaces the $HgOAc$ group with a deuterium atom $(D)$.
$3$. The reaction follows Markovnikov's rule,placing the $OH$ group on the $C-2$ position and the $D$ atom on the $C-1$ position.
$4$. The final product is $CH_3CH_2CH(OH)CH_2D$.

(C) Markovnikov's rule states that in the addition of a polar reagent (like $HX$) to an unsymmetrical alkene,the negative part of the reagent attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.
$C_2H_4$ is a symmetrical alkene,so it does not follow Markovnikov's rule.
$C_3H_6$ (propene) is an unsymmetrical alkene.
$Br_2$ and $I_2$ are symmetrical reagents,so they do not follow Markovnikov's rule.
$HBr$ is an unsymmetrical reagent.
Therefore,the reaction between $C_3H_6$ and $HBr$ is the correct choice where Markovnikov's rule is applicable.

(B) The reaction involves the oxidative cleavage of propene$-1,2-$diol using $KMnO_4$ in an acidic medium.
The initial step is the hydroxylation of propene to form propane$-1,2-$diol $(CH_3-CH(OH)-CH_2OH)$.
Further oxidation of propane$-1,2-$diol with $KMnO_4$ leads to the cleavage of the $C-C$ bond between the two hydroxyl-bearing carbons.
The cleavage of $CH_3-CH(OH)-CH_2OH$ results in the formation of acetic acid $(CH_3COOH)$ and formic acid $(HCOOH)$.
Therefore,$X$ is $CH_3COOH$.

(C) The addition of hydrogen halides to unsymmetrical alkenes in the presence of organic peroxides follows the anti-Markovnikov rule,also known as the peroxide effect or Kharasch effect.
This effect is only observed with $HBr$.
For $HF$ and $HCl$,the bond dissociation energy is too high for the free radical mechanism to proceed.
For $HI$,the iodine free radical $(I^{\bullet})$ is formed,but it tends to combine with another iodine radical to form $I_2$ rather than adding to the double bond.
Therefore,only $HBr$ undergoes this reaction.

(D) $1$. In reaction $A$,$CH_3CH=CH_2 + HBr \rightarrow CH_3CH(Br)CH_3$ ($2$-bromopropane),which is achiral.
$2$. In reaction $B$,$CH_2=CH_2 + HOBr \rightarrow HO-CH_2-CH_2-Br$ ($2$-bromoethanol),which is achiral.
$3$. In reaction $C$,$CH_3CH_2CH=CH_2 + HBr \xrightarrow{H_2O_2} CH_3CH_2CH_2CH_2Br$ ($1$-bromobutane),which is achiral.
$4$. In reaction $D$,$CH_3CH_2CH=CH_2 + HBr \rightarrow CH_3CH_2CH(Br)CH_3$ ($2$-bromobutane). The carbon atom attached to the bromine atom is bonded to four different groups $(-H, -Br, -CH_3, -CH_2CH_3)$,making it a chiral center.

(C) The reaction of an alkene with $B_2H_6$ followed by oxidation with alkaline $H_2O_2$ is known as Hydroboration-Oxidation.
This reaction follows the Anti-Markovnikov rule,where the $OH$ group is added to the less substituted carbon atom.
The reaction proceeds as follows: $RCH = CH_2 + B_2H_6$ $\rightarrow (RCH_2CH_2)_3B$ $\xrightarrow{H_2O_2/OH^-} RCH_2CH_2OH$.
Thus,the product is a primary alcohol,$RCH_2CH_2OH$.

(D) The reaction of an alkene with $NaIO_4$ in the presence of a catalytic amount of $KMnO_4$ is a method of oxidative cleavage of the double bond.
For the given compound $CH_3-C(CH_3)=CH-CH_3$ ($2$-methylbut$-2-$ene):
$1$. The double bond breaks between the $C_2$ and $C_3$ carbons.
$2$. The $C_2$ carbon is bonded to two methyl groups,forming acetone $(CH_3COCH_3)$.
$3$. The $C_3$ carbon is bonded to one hydrogen and one methyl group,forming acetaldehyde $(CH_3CHO)$.
Thus,the reaction is: $CH_3-C(CH_3)=CH-CH_3 \xrightarrow{NaIO_4/KMnO_4} CH_3COCH_3 + CH_3CHO$.

(A) $Baeyer's$ reagent is an alkaline solution of potassium permanganate $(KMnO_4)$.
It is used in the laboratory to test for the presence of unsaturation (double or triple bonds) in organic compounds.
When an alkene reacts with $Baeyer's$ reagent,the purple color of $KMnO_4$ disappears,and a brown precipitate of manganese dioxide $(MnO_2)$ is formed,indicating the presence of a double bond.

(A) The reaction of isobutylene $(CH_3)_2C=CH_2$ with $HOCl$ (hypochlorous acid) is an electrophilic addition reaction.
$HOCl$ dissociates into $OH^-$ and $Cl^+$.
The electrophile $Cl^+$ attacks the double bond to form a more stable carbocation.
In isobutylene,the tertiary carbocation $(CH_3)_2C^+-CH_2Cl$ is more stable than the primary one.
Then,the nucleophile $OH^-$ attacks the tertiary carbocation to form the final product,$1$-chloro-$2$-methylpropan-$2$-ol,which is $(CH_3)_2C(OH)CH_2Cl$.

(A) Step $1$: Reaction of propene with $Cl_2$ at $700 \ K$ is an allylic substitution reaction. $CH_3-CH=CH_2 + Cl_2 \xrightarrow{700 \ K} ClCH_2-CH=CH_2 (A) + HCl$.
Step $2$: Reaction of allyl chloride $(A)$ with $Na_2CO_3$ at $420 \ K$ and $12 \ atm$ is a hydrolysis reaction. $ClCH_2-CH=CH_2 + Na_2CO_3 + H_2O \rightarrow HOCH_2-CH=CH_2 (B) + NaCl + NaHCO_3$.
Step $3$: Reaction of allyl alcohol $(B)$ with $HOCl$ followed by $NaOH$ is a chlorohydrin formation followed by epoxide formation or direct addition. However,in this specific sequence,$HOCl$ adds across the double bond to form $HOCH_2-CH(OH)-CH_2Cl$,and subsequent treatment with $NaOH$ leads to the formation of glycerol or related diols. Given the options,the product $C$ is $CH_2(OH)-CH(OH)-CH_2(OH)$ (glycerol) or a related derivative. Re-evaluating the standard industrial process for glycerol synthesis from propene,the correct product $C$ is $CH_2(OH)-CH(OH)-CH_2(OH)$.

(D) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
If one of the products is formaldehyde $(HCHO)$,it implies that the original compound must have a terminal double bond,specifically a $CH_2=C < $ group.
This structural unit is known as a vinyl group $(CH_2=CH-)$ or a terminal methylene group.
Therefore,the formation of formaldehyde during ozonolysis is a characteristic test for the presence of a terminal vinyl group.

(C) When $1$-butene $(CH_3CH_2CH=CH_2)$ is heated with $Al_2(SO_4)_3$ at $500 \ K$,it undergoes isomerization to form a more stable alkene.
The terminal alkene $1$-butene isomerizes to the more stable branched alkene,$2$-methylpropene (isobutylene),which is $(CH_3)_2C=CH_2$.

(A) The acid-catalyzed hydration of alkenes follows $Markovnikov's$ rule.
For alkenes other than ethene (e.g.,propene),the addition of water occurs such that the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom.
This leads to the formation of secondary or tertiary alcohols depending on the structure of the alkene.
For example,propene $(CH_3-CH=CH_2)$ reacts with water in the presence of an acid to form propan$-2-$ol,which is a secondary alcohol.

(A) The reaction follows $Markovnikov's$ rule,where the electrophile $H^+$ adds to the carbon atom with more hydrogen atoms,and the nucleophile $Cl^-$ adds to the more substituted carbon atom (forming a more stable carbocation).
$1$. The reactant is $2-methylbut-2-ene$.
$2$. The addition of $HCl$ leads to the formation of $2-chloro-2-methylbutane$ as the major product.
$3$. The structure is $CH_3-CH_2-C(Cl)(CH_3)_2$.

(B) The reaction of ethene with dilute alkaline $KMnO_4$ (Baeyer's reagent) is a syn-hydroxylation reaction.
In this mechanism,the two oxygen atoms added to the double bond are derived from the permanganate ion $(MnO_4^-)$.
Therefore,the statement that both oxygen atoms are obtained from $KMnO_4$ is correct.
Statements $B$,$C$,and $D$ are incorrect as they suggest the oxygen atoms come from $H_2O$ or $NaOH$.

(B) The reaction of $1,2$-dibromoethane $(BrCH_2-CH_2Br)$ with zinc dust in an alcoholic solution leads to a dehalogenation reaction.
In this reaction,zinc removes two bromine atoms from adjacent carbon atoms,resulting in the formation of a double bond between them.
The chemical equation is: $BrCH_2-CH_2Br + Zn \rightarrow CH_2=CH_2 + ZnBr_2$.
Thus,the product obtained is ethene $(CH_2=CH_2)$.

(C) The anti-Markovnikov addition (peroxide effect) is only observed with $HBr$ in the presence of organic peroxides.
This is because the $H-Cl$ bond is very strong $(431 \ kJ \ mol^{-1})$,and the enthalpy change for the homolytic cleavage of $HCl$ by free radicals is endothermic,making the reaction energetically unfavorable.
In contrast,the $H-Br$ bond is weaker,allowing for the formation of bromine free radicals which facilitate the anti-Markovnikov addition.

(B) The reaction of an alkene with $KMnO_4$ in the presence of $NaIO_4$ is a form of oxidative cleavage,similar to ozonolysis.
In this reaction,the double bond is broken,and each carbon atom of the double bond is oxidized to a carbonyl group.
For the alkene $CH_3-C(CH_3)=CH-CH_3$:
$1$. The bond between $C_2$ and $C_3$ breaks.
$2$. The $C_2$ atom (which is bonded to two methyl groups) is oxidized to acetone $(CH_3COCH_3)$.
$3$. The $C_3$ atom (which is bonded to one hydrogen and one methyl group) is oxidized to acetaldehyde $(CH_3CHO)$,which is further oxidized to acetic acid $(CH_3COOH)$ in the presence of $KMnO_4$.
Therefore,the products are $CH_3COCH_3$ and $CH_3COOH$.

(A) The reaction of ethene $(CH_2=CH_2)$ with bromine $(Br_2)$ in an inert solvent like $CCl_4$ is an electrophilic addition reaction.
The double bond breaks,and one bromine atom attaches to each carbon atom.
The reaction is: $CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2Br-CH_2Br$.
The product formed is $1,2$-dibromoethane.

(A) The reaction of $NOCl$ (Tilden's reagent) with an alkene follows Markovnikov's rule.
$NOCl$ dissociates into $NO^+$ (electrophile) and $Cl^-$ (nucleophile).
In propene $(CH_3-CH=CH_2)$,the electrophilic addition of $NO^+$ occurs at the terminal carbon $(CH_2)$ to form a more stable secondary carbocation $(CH_3-CH^+-CH_2-NO)$.
Subsequently,the nucleophile $Cl^-$ attacks the carbocation to form $CH_3-CH(Cl)-CH_2(NO)$.

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
Given products are propanal $(CH_3CH_2CHO)$ and propanone $(CH_3COCH_3)$.
To find the parent alkene,remove the oxygen atoms from the carbonyl groups and join the carbon atoms with a double bond:
$CH_3CH_2CH = O + O = C(CH_3)_2 \rightarrow CH_3CH_2CH = C(CH_3)_2$.
The parent alkene is $2\text{-methylpent-2-ene}$ $(CH_3CH_2CH = C(CH_3)_2)$.

(C) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
Acetone is $CH_3-CO-CH_3$ and formaldehyde is $H-CHO$.
By removing the oxygen atoms from these products and joining the carbonyl carbons with a double bond,we get the structure of the alkene:
$(CH_3)_2C=O + O=CH_2 \rightarrow (CH_3)_2C=CH_2$.
The starting substance is $2-\text{methylpropene}$,which is $CH_3-C(CH_3)=CH_2$.

(A) The reaction of isobutene $(CH_3)_2C=CH_2$ with $HBr$ follows Markovnikov's rule.
In this reaction,the electrophile $H^+$ attacks the double bond to form the more stable $3^{\circ}$ carbocation,$(CH_3)_3C^+$.
Subsequently,the nucleophile $Br^-$ attacks the carbocation to form $2$-bromo-$2$-methylpropane,which is commonly known as $tert$-butyl bromide.

(B) The catalytic hydrogenation of alkenes using $Ni$,$Pd$,or $Pt$ catalysts involves syn-addition of hydrogen (or deuterium) to the double bond.
For $trans-2$-butene,the addition of $D_2$ occurs from the same side (syn-addition).
Since the starting material is $trans-2$-butene,the syn-addition of $D_2$ results in the formation of $meso-2,3$-dideuteriobutane.
This is because the resulting molecule has a plane of symmetry,making it a $meso$ compound.

(B) The reaction of ethylene with carbon monoxide and water in the presence of a catalyst at high temperature and pressure is known as the hydrocarboxylation reaction.
The chemical equation is: $C_2H_4 + CO + H_2O \xrightarrow{\text{High temp.}} C_2H_5COOH$.
The product formed is propionic acid $(C_2H_5COOH)$.