Alkene Questions in English

(D) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
$1$. $CH_3-CH=CH_2$ (Propene) on ozonolysis gives $CH_3CHO$ (acetaldehyde) and $HCHO$ (formaldehyde).
$2$. $CH_3-CH=CH-CH_3$ ($2-$butene) on ozonolysis gives two molecules of $CH_3CHO$ (acetaldehyde).
Since the question asks for an alkene that gives acetaldehyde,both $2-$butene (which gives two molecules of acetaldehyde) and propene (which gives one molecule of acetaldehyde along with formaldehyde) are technically sources of acetaldehyde. However,in standard textbook contexts,$2-$butene is the primary answer for producing acetaldehyde as the sole product. Given the options,$(d)$ is the most comprehensive choice.

(A) The reaction is an electrophilic addition of acetyl chloride to the alkene catalyzed by a Lewis acid $(ZnCl_2)$.
The acetyl cation $(CH_3CO^+)$ attacks the terminal carbon of $2-$methylpropene to form a stable tertiary carbocation intermediate,$(CH_3)_2C^+-CH_2-COCH_3$.
This intermediate then reacts with the chloride ion $(Cl^-)$ to give $4-$chloro$-4-$methylpentan$-2-$one,$CH_3-C(Cl)(CH_3)-CH_2-COCH_3$.

(C) The hydrogenation of olefins (alkenes) is a catalytic process.
$(C)$: $CH_2=CH_2 + H_2 \xrightarrow{Raney \ Ni, \Delta} CH_3-CH_3$.
Raney $Ni$ acts as a catalyst in the presence of hydrogen gas to facilitate the addition of hydrogen across the double bond.

(C) The electrolysis of potassium succinate $(KOOC-CH_2-CH_2-COOK)$ via Kolbe's electrolytic method results in the formation of ethene $(CH_2=CH_2)$ at the anode.
The chemical reaction is as follows:
$KOOC-CH_2-CH_2-COOK + 2H_2O \xrightarrow{\text{Electrolysis}} CH_2=CH_2 + 2CO_2 + H_2 + 2KOH$.

(C) The hydrocarbon is Ethylene $(CH_2=CH_2)$.
$1$. Sabatier and Senderen's reaction involves the catalytic hydrogenation of unsaturated hydrocarbons using $Ni$ catalyst at $300\ ^oC$. Ethylene undergoes this reaction: $CH_2=CH_2 + H_2 \xrightarrow[300\ ^oC]{Ni} CH_3-CH_3$.
$2$. Ammoniacal silver nitrate (Tollens' reagent) reacts only with terminal alkynes (having acidic hydrogen) to form a white precipitate. Ethylene does not contain acidic hydrogen,so it does not form a precipitate.

(B) In this reaction,$HBr$ undergoes heterolytic fission to produce an electrophile $H^{+}$:
$HBr \rightarrow H^{+} + Br^{-}$
The electrophile $H^{+}$ attacks the double bond of propene to form a stable carbocation intermediate $(CH_3-\stackrel{+}{C}H - CH_3)$.
Subsequently,the nucleophile $Br^{-}$ attacks the carbocation to form $2$-bromopropane.
Since the initial step involves the attack of an electrophile on the alkene,this is an electrophilic addition reaction.

(A) The reaction of $1,3-$butadiene with $Br_2$ is an electrophilic addition reaction.
At low temperatures,$1,2-$addition is favored,but at higher temperatures or under thermodynamic control,$1,4-$addition is the major product.
The reaction is: $CH_2=CH-CH=CH_2 + Br_2 \rightarrow Br-CH_2-CH=CH-CH_2-Br$.
Thus,the product is $1,4-$dibromobut$-2-$ene.

(D) The products of ozonolysis are $2$ moles of $HCHO$,$1$ mole of $CO_2$,and $1$ mole of $CH_3COCHO$.
By joining the carbonyl groups of these products with double bonds,we can reconstruct the original alkene structure.
The fragments are: $H_2C=O$,$O=C=O$,$O=C(CH_3)-CH=O$,and $O=CH_2$.
Removing the oxygen atoms and connecting the carbon atoms at the carbonyl sites with double bonds results in the structure: $CH_2=C=C(CH_3)-CH=CH_2$.
Thus,the correct structure is $CH_2=C=C(CH_3)-CH=CH_2$.

(B) The reaction involves the oxidation of propene to propane-$1,2$-diol,followed by oxidative cleavage.
The cleavage of propane-$1,2$-diol $(CH_3-CH(OH)-CH_2(OH))$ using an oxidizing agent leads to the breaking of the $C-C$ bond between the two hydroxyl-bearing carbons.
This process yields ethanoic acid $(CH_3COOH)$ and methanoic acid $(HCOOH)$.
Therefore,$X$ is $CH_3COOH$.

(D) Oxidation of alkenes with acidic potassium permanganate $(KMnO_4)$ leads to the cleavage of the double bond.
For $2-$butene $(CH_3-CH=CH-CH_3)$,the double bond breaks at the center,resulting in two molecules of acetic acid $(CH_3COOH)$.
The reaction is: $CH_3-CH=CH-CH_3 + 4[O] \xrightarrow{KMnO_4} 2CH_3COOH$.

(D) The conversion of an alkene to an alkane is a hydrogenation reaction.
$But-1-ene$ $(CH_3CH_2CH=CH_2)$ reacts with hydrogen gas in the presence of a metal catalyst like $Pd$,$Pt$,or $Ni$ to form $butane$ $(CH_3CH_2CH_2CH_3)$.
This process is known as catalytic hydrogenation.
Therefore,the correct reagent is $Pd / H_2$.

(A) In the presence of peroxide,the addition of $HBr$ to an unsymmetrical alkene like propene follows the Anti-Markownikoff's rule.
According to this rule,the negative part of the reagent $(Br^-)$ attaches to the carbon atom with the greater number of hydrogen atoms.
Reaction: $CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2-Br$ ($n-$propyl bromide).
Therefore,the major product is $n-$propyl bromide.

(A) The reaction of ethylene $(CH_2=CH_2)$ with ozone $(O_3)$ followed by reductive cleavage with zinc and water $(Zn/H_2O)$ is known as ozonolysis.
$CH_2=CH_2 + O_3 \rightarrow CH_2CH_2O_3 \text{ (Ethylene ozonide)}$
$CH_2CH_2O_3 \xrightarrow{Zn/H_2O} 2HCHO \text{ (Formaldehyde)}$
Thus,the final product obtained is formaldehyde.

(D) . $C_2H_4$ (ethene),$C_3H_6$ (propene),and $C_4H_8$ (butene) are all alkenes.
Alkenes contain a carbon-carbon double bond $(C=C)$,which undergoes an electrophilic addition reaction with bromine $(Br_2)$.
This reaction results in the decolourisation of the reddish-brown bromine solution,forming a vicinal dibromide.
Since all the given compounds are alkenes,they will all discharge the red colour of bromine.

(A) $HI \to H^{+} + I^{-}$
$CH_3-CH=CH_2 + H^{+} \to CH_3-CH_2-CH_2^{+}$ ($1^o$ carbonium ion,less stable) and $CH_3-CH^{+}-CH_3$ ($2^o$ carbonium ion,more stable).
The reaction proceeds through the formation of a more stable $2^o$ carbonium ion intermediate.
$CH_3-CH^{+}-CH_3 + I^{-} \to CH_3-CH(I)-CH_3$ (isopropyl iodide,major product).

(A) The reaction of an alkene with bromine $(Br_2)$ is an electrophilic addition reaction.
When $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$ reacts with $Br_2$,the bromine atoms add across the double bond.
$CH_3-CH_2-CH=CH_2 + Br_2 \to CH_3-CH_2-CH(Br)-CH_2Br$.
The product formed is $1, 2-dibromobutane$.

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For a compound '$X$' to yield two molecules of formaldehyde $(HCHO)$ upon ozonolysis,the structure must be ethene $(CH_2=CH_2)$.
The reaction is: $CH_2=CH_2 \xrightarrow[(2) Zn/H_2O]{(1) O_3} 2HCHO$.
Therefore,the compound '$X$' is $C_2H_4$.

(B) The addition of $HOCl$ to propene follows Markovnikov's rule.
$HOCl$ dissociates as $HO^{\delta-} - Cl^{\delta+}$.
The electrophile $Cl^{+}$ attacks the double bond to form a cyclic chloronium ion intermediate.
The nucleophile $OH^{-}$ then attacks the more substituted carbon atom $(C-2)$ to open the ring.
Thus,the reaction is: $CH_3-CH=CH_2 + HOCl \to CH_3-CH(OH)-CH_2Cl$.
Therefore,Option $B$ is the correct answer.

(C) The reaction is the catalytic hydrogenation of $2,3-\text{dimethylbut-2-ene}$.
$(CH_3)_2C=C(CH_3)_2 + H_2 \xrightarrow{\text{Catalyst}} (CH_3)_2CH-CH(CH_3)_2$.
The product formed is $2,3-\text{dimethylbutane}$.
In $2,3-\text{dimethylbutane}$,there are no chiral carbon atoms present because the molecule is symmetric and each carbon atom is bonded to at least two identical groups.
Since there are no chiral centers,the molecule is achiral and exhibits $0$ optical isomers.

(B) In the presence of peroxide,the addition of $HBr$ to an unsymmetrical alkene like isobutene $(CH_2=C(CH_3)_2)$ follows the Anti-Markovnikov rule (Kharasch effect).
The bromine atom attaches to the less substituted (primary) carbon atom.
$CH_2=C(CH_3)_2 + HBr \xrightarrow{\text{Peroxide}} Br-CH_2-CH(CH_3)_2$ (Isobutyl bromide).

(C) The given statement is the definition of $Markownikoff's \ rule$.
According to this rule,during the electrophilic addition of an unsymmetrical reagent to an unsymmetrical alkene,the negative part of the addendum attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.
Example: $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$ $(2-Bromopropane)$.

(D) The stability of alkenes is determined by the number of alkyl groups attached to the double-bonded carbon atoms (hyperconjugation effect).
$1-$butene $(CH_3-CH_2-CH=CH_2)$ has $1$ alkyl group.
$2-$butene $(CH_3-CH=CH-CH_3)$ has $2$ alkyl groups.
$1-$pentene $(CH_3-CH_2-CH_2-CH=CH_2)$ has $1$ alkyl group.
$2-$pentene $(CH_3-CH_2-CH=CH-CH_3)$ has $2$ alkyl groups.
Among the options,$2-$pentene is the most stable because it has a higher degree of substitution compared to $1-$butene and $1-$pentene,and the alkyl groups provide greater stabilization through hyperconjugation compared to $2-$butene due to the longer chain length.

(B) Markownikoff’s rule is applicable to unsymmetrical alkenes where the addition of a polar reagent occurs across the double bond.
$B$ $(CH_3 - CH = CH - CH_3)$ is a symmetrical alkene (but$-2-$ene),therefore,the addition of a polar reagent to it does not require Markownikoff’s rule as the product formed is the same regardless of the orientation of the addition.

(C) Acid catalyzed hydration of alkenes follows Markovnikov's rule.
$2-$phenylpropene $(CH_3-C(Ph)=CH_2)$ reacts with $H_3O^{+}$ to form a stable carbocation intermediate.
The proton $(H^{+})$ adds to the terminal carbon $(CH_2)$ to form a tertiary benzylic carbocation $(CH_3-C^{+}(Ph)-CH_3)$.
Water then attacks this carbocation to form $2-$phenylpropan$-2-$ol $(CH_3-C(OH)(Ph)-CH_3)$.

(D) . Solution of bromine in carbon tetrachloride is used to test for unsaturation of alkene.
The red colour of bromine disappears due to the formation of a colourless dibromoalkane (e.g.,for ethene,$C_2H_4Br_2$ is formed).

(B) The reaction of propylene with sulphuric acid follows Markovnikov's rule.
$CH_3-CH=CH_2 + H_2O \xrightarrow{H_2SO_4} CH_3-CH(OH)-CH_3$
Thus,in this reaction,isopropyl alcohol is formed.

(B) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For propene $(CH_2=CH-CH_3)$:
$CH_2=CH-CH_3 + O_3 \rightarrow \text{ozonide intermediate}$
$\text{Ozonide} + Zn/H_2O \rightarrow HCHO + CH_3CHO$
Formaldehyde $(HCHO)$ and acetaldehyde $(CH_3CHO)$ are produced.
Therefore,the correct option is $(B)$.

(A) The reaction of ethene $(H_2C=CH_2)$ with cold alkaline $KMnO_4$ (Baeyer's reagent) is an oxidation reaction.
This process results in the syn-hydroxylation of the double bond to form a vicinal diol.
The product formed is ethylene glycol $(HOCH_2-CH_2OH)$.

(A) The Anti-Markownikoff's rule (also known as the peroxide effect or Kharasch effect) is applicable only to unsymmetrical alkenes.
$2-$butene is a symmetrical alkene because the double bond is located between the $2^{nd}$ and $3^{rd}$ carbon atoms,resulting in identical groups on both sides of the double bond $(CH_3-CH=CH-CH_3)$.
Therefore,$2-$butene does not show Anti-Markownikoff's addition.
$1-$butene,$2-$pentene,and $2-$hexene are unsymmetrical alkenes and can follow the rule.

(B) The reaction of $HBr$ with propene in the presence of peroxide follows the Anti-Markovnikov rule (Kharasch effect).
The bromine atom attaches to the terminal carbon atom.
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2Br$ ($n-propyl$ bromide).

(C) $C_2H_4$ (ethene) is an alkene.
Alkenes react with $KMnO_4$ (Baeyer's reagent) to undergo oxidation,resulting in the decolorization of the purple solution.
However,they do not react with ammoniacal $AgNO_3$ (Tollens' reagent) because they lack acidic terminal hydrogen atoms.
Terminal alkynes like $C_2H_2$ react with both $KMnO_4$ and ammoniacal $AgNO_3$.
Alkanes like $CH_4$ and $C_2H_6$ are saturated hydrocarbons and do not react with either reagent.

(B) The reaction of $1,3-$butadiene with $HBr$ produces two isomers: $3-$bromobutene ($1,2-$addition product) and $1-$bromo$-2-$butene ($1,4-$addition product).
At low temperatures (e.g.,$-80\,^{\circ}C$),the reaction is kinetically controlled,favoring the $1,2-$addition product ($3-$bromobutene) due to the proximity of the intermediate carbocation.
At higher temperatures (e.g.,$40\,^{\circ}C$),the reaction is thermodynamically controlled,favoring the more stable $1,4-$addition product ($1-$bromo$-2-$butene) because the double bond in the $1,4-$product is more substituted and thus more stable.
Therefore,at $40\,^{\circ}C$,$1-$bromo$-2-$butene is the major product.

(C) In the presence of peroxide,the addition of $HBr$ to an unsymmetrical alkene follows the Anti-Markovnikov rule (also known as the Kharasch effect).
In this reaction,the bromine atom attaches to the carbon atom of the double bond that has more hydrogen atoms.
Reaction: $CH_3-C(CH_3)=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH(CH_3)-CH_2Br$.

(C) $1$. $KMnO_4$ solution (Baeyer's reagent) is decolourised by unsaturated hydrocarbons like alkenes and alkynes due to the presence of multiple bonds.
$2$. Ammoniacal cuprous chloride $(Cu_2Cl_2 + NH_4OH)$ reacts specifically with terminal alkynes to form a red precipitate of copper acetylide.
$3$. Ethane and methane are saturated hydrocarbons (alkanes) and do not decolourise $KMnO_4$.
$4$. Acetylene (ethyne) is a terminal alkyne and gives a red precipitate with ammoniacal cuprous chloride.
$5$. Ethene $(CH_2=CH_2)$ is an alkene; it decolourises $KMnO_4$ but does not react with ammoniacal cuprous chloride to form a precipitate.
$6$. Therefore,the correct answer is Ethene.

(A) The reaction of cyclohexene with $OsO_4$ (osmium tetroxide) is a syn-hydroxylation reaction.
$OsO_4$ adds to the double bond of the alkene to form a cyclic osmate ester intermediate.
Subsequent treatment with $NaHSO_3$ (sodium bisulfite) or $H_2O$ hydrolyzes the osmate ester to yield a $cis$-diol.
Therefore,the correct product is a $cis$-diol.

(C) The hydrocarbon $C_6H_{10}$ has a degree of unsaturation of $2$.
Since it absorbs only one molecule of $H_2$,it contains one double bond and one ring.
Ozonolysis of cyclohexene breaks the $C=C$ bond to form hexane$-1,6-$dial $(OHC-CH_2-CH_2-CH_2-CH_2-CHO)$.
Therefore,the hydrocarbon is cyclohexene.

(B) Alkenes undergo electrophilic addition reactions.
$HOCl$ acts as a source of the electrophile $Cl^{+}$.
In the first step,the electrophile $Cl^{+}$ attacks the double bond of propene to form a cyclic chloronium ion intermediate.
Subsequently,the nucleophile $OH^{-}$ attacks the more substituted carbon atom to form the final product,chlorohydrin.

(C) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For a cyclic alkene like $C_6H_{10}$ (cyclohexene),the ring opens during ozonolysis.
The reaction is: $\text{Cyclohexene} + O_3$ $\rightarrow \text{Ozonide}$ $\xrightarrow{Zn/H_2O} OHC-(CH_2)_4-CHO$ (hexane-$1,6-$dial).
Thus,the starting alkene is cyclohexene.

(D) The reaction of an alkene with $NaIO_4$ in the presence of $KMnO_4$ (Lemieux-von Rudloff reagent) results in the oxidative cleavage of the $C=C$ double bond.
In the compound $CH_3-C(CH_3)=CH-CH_3$,the double bond is cleaved.
The substituted carbon $CH_3-C(CH_3)=$ is oxidized to a ketone,$CH_3COCH_3$ (acetone).
The $=CH-CH_3$ part is oxidized to an aldehyde,which is further oxidized to a carboxylic acid,$CH_3COOH$ (acetic acid),under these conditions.
Therefore,the final products are $CH_3COCH_3 + CH_3COOH$.

(A) The addition reaction across a double bond involves the breaking of the $\pi$-bond and the formation of two new $\sigma$-bonds.
In the reaction $ >C = C < + XY \to >C(X)-C(Y) < $,the reagent $XY$ adds across the double bond,resulting in the saturated product where $X$ and $Y$ are attached to the adjacent carbon atoms.

(D) The addition of $HBr$ to alkenes is an electrophilic addition reaction,which proceeds via the formation of a carbocation intermediate.
Greater stability of the carbocation leads to a faster reaction rate.
In $(CH_3)_2C = CH_2$,the addition of $H^+$ leads to the formation of a tertiary carbocation,$(CH_3)_3C^+$,which is the most stable among the given options due to the inductive effect and hyperconjugation of six $\alpha$-hydrogens.
Therefore,$(CH_3)_2C = CH_2$ reacts most easily with $HBr$.

(D) The reaction of propene with concentrated acidic $KMnO_4$ is an oxidative cleavage reaction.
$CH_3-CH=CH_2$ undergoes cleavage at the double bond.
The terminal $CH_2$ group is oxidized to $CO_2$ and $H_2O$,but under specific conditions or depending on the representation of the cleavage of the terminal carbon,it yields $HCOOH$ (formic acid).
The $CH_3-CH=$ part is oxidized to acetic acid $(CH_3COOH)$.
Therefore,the reaction is: $CH_3-CH=CH_2 + 4[O] \xrightarrow{KMnO_4/H^+} CH_3COOH + HCOOH$.

(D) The reaction of $2-methylpropene$ $(CH_2=C(CH_3)-CH_3)$ with $HBr$ follows Markovnikov's rule.
$H^{+}$ adds to the carbon with more hydrogen atoms,forming a stable tertiary carbocation $(CH_3-C^{+}(CH_3)-CH_3)$.
Subsequently,$Br^{-}$ attacks this carbocation to yield $2-bromo-2-methylpropane$ $(CH_3-C(Br)(CH_3)-CH_3)$.

(B) The reaction involves the electrophilic addition of $HBr$ to an alkene. According to Markovnikov's rule,the proton $(H^+)$ adds to the carbon atom with more hydrogen atoms,and the bromide ion $(Br^-)$ adds to the more substituted carbon atom to form a more stable carbocation intermediate.
In $CH_2=CH-CCl_3$,the $CCl_3$ group is strongly electron-withdrawing due to the inductive effect. However,the addition follows the standard Markovnikov rule where $H^+$ attaches to the terminal $CH_2$ group to form the stable carbocation $CH_3-CH^+-CCl_3$.
Then,the $Br^-$ ion attacks this carbocation to yield the final product: $CH_3-CH(Br)-CCl_3$.

(A) The reaction of ethylene with bromine in the presence of a nucleophilic solvent like methanol $(CH_3OH)$ proceeds through the formation of a cyclic bromonium ion intermediate.
This cyclic bromonium ion is electrophilic in nature.
It can be attacked by the bromide ion $(Br^-)$ to form $1, 2-dibromoethane$ or by the solvent molecule $(CH_3OH)$ to form $BrCH_2CH_2OCH_3$ after the loss of a proton.
Thus,the product depends on which nucleophile attacks the intermediate.

(C) The addition of hydrogen halides $(HX)$ to alkenes in the presence of peroxide follows the anti-Markownikoff rule only for $HBr$.
For $HCl$,the $H-Cl$ bond is too strong to be broken by free radicals.
For $HI$,the iodine free radical $(I^{\bullet})$ formed is stable,but the step involving the formation of the $C-I$ bond is endothermic,and $I^{\bullet}$ radicals tend to combine to form $I_2$ instead of adding to the alkene.
Thus,only $HBr$ undergoes free radical addition to give the anti-Markownikoff product.

(A) The reaction of propene $(CH_3-CH=CH_2)$ with $HBr$ follows Markownikoff's rule.
According to this rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom with the lesser number of hydrogen atoms.
$CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$ (Isopropyl bromide).

(B) Ethylene $(CH_2=CH_2)$ reacts with bromine $(Br_2)$ in an addition reaction to form ethylene dibromide $(CH_2Br-CH_2Br)$.

(A) $NBS$ ($N$-bromosuccinimide) is a selective brominating reagent that performs allylic bromination in the presence of light or a radical initiator.
In methylenecyclohexane,the carbon atom adjacent to the double bond (the allylic position) is the carbon of the exocyclic methylene group.
Therefore,the bromine atom replaces one of the hydrogen atoms on the exocyclic carbon,resulting in the formation of $1-$(bromomethyl)cyclohex$-1-$ene.

(A) Dehydrochlorination of allyl chloride $(CH_2=CHCH_2Cl)$ involves the removal of a hydrogen atom from the $C_2$ position and the chlorine atom from the $C_3$ position,leading to the formation of propadiene $(CH_2=C=CH_2)$.
$CH_2=CHCH_2Cl \xrightarrow{\text{dehydrochlorination}} CH_2=C=CH_2$