Alkene Questions in English

(A) When ethylene dibromide $(Br-CH_2-CH_2-Br)$ is heated with granulated zinc in the presence of alcohol,it undergoes a dehalogenation reaction to form ethylene $(CH_2=CH_2)$.
The reaction is as follows:
$Br-CH_2-CH_2-Br + Zn \xrightarrow{\Delta, \text{alcohol}} CH_2=CH_2 + ZnBr_2$

(C) When ethyl iodide $(CH_3-CH_2-I)$ reacts with alcoholic $KOH$,it undergoes dehydrohalogenation to form ethene $(C_2H_4)$.
Ethene is an unsaturated hydrocarbon that decolorizes alkaline $KMnO_4$ solution (Baeyer's reagent).
The reaction is: $CH_3-CH_2-I + KOH (\text{alc.}) \rightarrow CH_2=CH_2 + KI + H_2O$.

(D) Markownikoff's rule is applicable to the addition of unsymmetrical reagents to unsymmetrical alkenes.
In the case of $CH_2 = CHBr$ (vinyl bromide),the alkene is unsymmetrical.
According to Markownikoff's rule,the positive part of the reagent $(H^+)$ attaches to the carbon atom of the double bond that has a greater number of hydrogen atoms,while the negative part $(Br^-)$ attaches to the carbon atom with fewer hydrogen atoms.
Thus,the reaction proceeds as: $CH_2 = CHBr + HBr \rightarrow CH_3 - CH(Br)_2$.

(A) The dehydration of ethyl alcohol $(CH_3-CH_2-OH)$ in the presence of concentrated $H_2SO_4$ at $443 \ K$ yields ethylene $(CH_2=CH_2)$.
$CH_3-CH_2-OH \xrightarrow[Conc. \ H_2SO_4, \ 443 \ K]{Dehydration} CH_2=CH_2 + H_2O$

(C) Alkenes contain a carbon-carbon double bond $(C=C)$.
The $\pi$ electrons of the double bond are loosely held and act as a source of electrons,making alkenes susceptible to attack by electrophiles.
Therefore,electrophilic addition reactions are the most common reactions for alkenes.

(D) The reaction of ethylene $(CH_2=CH_2)$ with cold,dilute alkaline $KMnO_4$ (Baeyer's reagent) is an oxidation reaction.
This process results in the hydroxylation of the double bond to form a vicinal diol.
The reaction is: $CH_2=CH_2 + H_2O + [O] \xrightarrow{Cold, \ alk. \ KMnO_4} CH_2(OH)-CH_2(OH)$ (Ethylene glycol).

(C) . Ethane $(C_2H_6)$ and Methane $(CH_4)$ are saturated hydrocarbons and do not decolourise $KMnO_4$ solution.
$B$. Acetylene $(C_2H_2)$ is a terminal alkyne; it decolourises $KMnO_4$ solution and forms a red precipitate with ammoniacal cuprous chloride $(Cu_2Cl_2)$.
$C$. Ethene $(C_2H_4)$ is an alkene; it decolourises $KMnO_4$ solution (Baeyer's reagent) due to the presence of a double bond,but it does not react with ammoniacal cuprous chloride because it lacks acidic hydrogen atoms.
$D$. Therefore,the correct answer is Ethene.

(A) The reaction of an alkene with hypochlorous acid $(HOCl)$ follows electrophilic addition.
$CH_2=CH_2 + HOCl \to CH_2(OH)-CH_2Cl$
This product is $1-$chloro$-2-$hydroxyethane.
Therefore,the hydrocarbon is ethylene $(CH_2=CH_2)$.

(D) The debromination of $meso-2,3-dibromobutane$ proceeds via an $E2$ mechanism,which is stereospecific.
In the $meso$ form,the two bromine atoms are anti to each other in the staggered conformation.
Elimination of $Br_2$ from this conformation leads to the formation of $trans-2-butene$.
$trans-2-butene$ is more stable than $cis-2-butene$ due to reduced steric hindrance between the methyl groups.

(D) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
Given products are $CH_3CH_2CHO$ (propanal) and $CH_3CHO$ (ethanal).
By joining these two fragments by removing the oxygen atoms and forming a double bond between the carbonyl carbons,we get the structure:
$CH_3CH_2CH=O + O=CHCH_3 \rightarrow CH_3CH_2CH=CHCH_3$.
The resulting alkene is $pent-2-ene$ $(CH_3CH_2CH=CHCH_3)$.

(B) In an ethylene $(C_2H_4)$ molecule, the carbon atoms are connected by a double bond.
The bond length of a carbon-carbon single bond is $1.54 \ \mathring{A}$.
The bond length of a carbon-carbon double bond is $1.34 \ \mathring{A}$ (or $134 \ pm$).
The bond length of a carbon-carbon triple bond is $1.20 \ \mathring{A}$ (or $120 \ pm$).
Therefore, the correct bond length for ethylene is $1.34 \ \mathring{A}$.

(B) The reaction of ethene $(CH_2=CH_2)$ with bromine $(Br_2)$ is an electrophilic addition reaction.
The bromine water loses its characteristic red-brown colour to form a colourless liquid.
The reaction is: $CH_2=CH_2 + Br_2 \rightarrow CH_2Br-CH_2Br$.
The product formed is $1,2$-dibromoethane,which is commonly known as ethylene dibromide.
This reaction is used as a standard test for the presence of a carbon-carbon double bond.

(B) The peroxide effect (anti-Markovnikov addition) is only applicable to $HBr$.
$HCl$ and $HI$ do not show the peroxide effect because the radical chain steps are not both exothermic.
Therefore,$HCl$ adds to propene $(CH_3-CH=CH_2)$ according to Markovnikov's rule to give $2-$chloropropane $(CH_3-CHCl-CH_3)$.

(B) The addition of $HI$ to alkenes in the presence of peroxide does not follow the anti-Markovnikov rule because the free radical mechanism is not feasible for $HI$ due to the endothermic nature of the iodine radical formation step.
Therefore,the reaction follows the Markovnikov addition rule,where the electrophile $H^+$ adds to the carbon with more hydrogen atoms.
$CH_3-CH=CH_2 + HI \rightarrow CH_3-CH(I)-CH_3$
The product formed is $2-$iodopropane (also known as isopropyl iodide).

(C) The reaction of $1-$butene $(C_2H_5CH = CH_2)$ with hydrogen halide $(HX)$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(X^-)$ attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.
In $C_2H_5 - CH = CH_2$,the $CH$ group has one hydrogen and the $CH_2$ group has two hydrogens.
Therefore,the $X$ atom attaches to the $CH$ group,forming $C_2H_5 - CHX - CH_3$ as the major product.
This is the more stable product because it involves the formation of a more stable secondary carbocation intermediate.

(A) When alkyl halides are heated with strong bases in an alcoholic solvent,they undergo a $1,2$-elimination reaction (specifically a dehydrohalogenation reaction) to form alkenes.
Alcoholic $KOH$ (Alc. $KOH$) is a standard reagent used for this purpose because the ethoxide or hydroxide ion acts as a strong base to abstract a proton from the $\beta$-carbon,leading to the elimination of $HX$.

(C) The reaction of $2-$chlorobutane with alcoholic $NaOH$ is a dehydrohalogenation reaction,which follows $Saytzeff$ rule.
According to $Saytzeff$ rule,the more substituted alkene is the major product.
$CH_3-CH_2-CH(Cl)-CH_3 \xrightarrow{alc. NaOH} CH_3-CH=CH-CH_3$ ($2-$Butene,major product) + $CH_3-CH_2-CH=CH_2$ ($1-$Butene,minor product).
Thus,$2-$Butene is formed in a larger amount.

(C) The electrophilic addition of bromine $(Br_2)$ to $1, 3-$butadiene $(CH_2=CH-CH=CH_2)$ is a classic example of conjugated diene reactivity.
The reaction proceeds through a common allylic carbocation intermediate.
Depending on the reaction conditions (temperature),two products are formed:
$1$. $1, 2-$addition product: $3, 4-$dibromobut-$1-$ene.
$2$. $1, 4-$addition product: $1, 4-$dibromobut-$2-$ene.
At low temperatures,the $1, 2-$addition product is the kinetic product (formed faster),while at higher temperatures,the $1, 4-$addition product is the thermodynamic product (more stable due to the more substituted double bond).

(C) The anti-Markovnikov addition of $HX$ to alkenes proceeds via a free radical mechanism.
For $HCl$,the bond dissociation energy is very high,making the homolytic cleavage of the $H-Cl$ bond endothermic.
For $HI$,the addition of the iodine radical to the alkene is an endothermic step.
Since one of the propagation steps is endothermic for both $HCl$ and $HI$,the overall reaction does not proceed via the anti-Markovnikov pathway in the presence of peroxide.

(C) The correct answer is $C$.
$CH_3-CH=CH-CH_2-CH_3$ is an alkene (pent$-2-$ene).
Alkenes contain a carbon-carbon double bond,which undergoes oxidation with cold dilute alkaline $KMnO_4$ (Baeyer's reagent) to form vicinal diols,resulting in the decolourization of the purple $KMnO_4$ solution.
Option $A$ is an alkane,option $B$ is naphthalene (an aromatic hydrocarbon),and option $D$ is an alkane,none of which react with cold dilute $KMnO_4$ under standard conditions.

(C) Anti-Markovnikov's addition (peroxide effect) involves the formation of free radicals.
For the reaction to proceed via a free radical mechanism,the $H-Cl$ bond must undergo homolytic cleavage.
However,the $H-Cl$ bond has a very high bond dissociation energy.
Therefore,the energy required to form $H^{\bullet}$ and $Cl^{\bullet}$ free radicals is not available,making the initiation step endothermic and unfavorable.
Consequently,$HCl$ does not show the peroxide effect.

(D) $1, 3-$butadiene $(CH_2=CH-CH=CH_2)$ contains two double bonds separated by a single bond,which is the definition of conjugated double bonds.
It undergoes electrophilic addition reactions with $HBr$ to form products like $3-$bromo-$1-$butene and $1-$bromo-$2-$butene.
It is a key monomer used in the production of synthetic rubber (polybutadiene),thus it forms polymers.
Since all three statements are correct,the correct option is $D$.

(C) $2,2$-dimethylpropane (neopentane) has the molecular formula $C_5H_{12}$. Isomers of $C_5H_{12}$ are $n$-pentane,isopentane ($2$-methylbutane),and neopentane ($2,2$-dimethylpropane).
$(1)$ $CH_3-CH=C(CH_3)_2$ ($2$-methylbut-$2$-ene) has the formula $C_5H_{10}$. On catalytic hydrogenation,it gives $2$-methylbutane $(C_5H_{12})$,which is an isomer of $2,2$-dimethylpropane.
$(2)$ $CH_3-CH=CH-CH_3$ (but-$2$-ene) has the formula $C_4H_8$. On hydrogenation,it gives butane $(C_4H_{10})$,which is not an isomer of $C_5H_{12}$.
$(3)$ $CH_3-CH=CH-CH_2-CH_3$ (pent-$2$-ene) has the formula $C_5H_{10}$. On catalytic hydrogenation,it gives $n$-pentane $(C_5H_{12})$,which is an isomer of $2,2$-dimethylpropane.
$(4)$ $CH_3-C(CH_3)=C(CH_3)-CH_3$ ($2,3$-dimethylbut-$2$-ene) has the formula $C_6H_{12}$. On hydrogenation,it gives $2,3$-dimethylbutane $(C_6H_{14})$,which is not an isomer of $C_5H_{12}$.
Thus,compounds $(1)$ and $(3)$ yield isomers of $2,2$-dimethylpropane.

(B) The hydroboration-oxidation of alkenes follows anti-Markovnikov regioselectivity to yield primary alcohols.
The reaction proceeds as follows:
$3R-CH=CH_2 + 1/2 B_2H_6 \rightarrow (R-CH_2-CH_2)_3B$
$(R-CH_2-CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3R-CH_2-CH_2-OH + H_3BO_3$
Thus,the final product is a primary alcohol,$R-CH_2-CH_2-OH$.

(C) Baeyer's reagent is an alkaline solution of cold $KMnO_4$.
It is used for the detection of unsaturation (double or triple bonds) in a molecule,as it undergoes a color change from purple to brown $MnO_2$ precipitate.

(A) At high temperatures $(500^\circ C)$,chlorine reacts with propene via free radical allylic substitution to produce allyl chloride $(CH_2Cl-CH=CH_2)$.
At such high temperatures,the electrophilic addition reaction across the double bond is suppressed in favor of the free radical substitution at the allylic carbon.

(C) The reaction of ethylene $(CH_2=CH_2)$ with $Br_2$ in the presence of $NaCl$ involves the formation of a cyclic bromonium ion intermediate.
This intermediate can be attacked by either the bromide ion $(Br^-)$ or the chloride ion $(Cl^-)$ present in the solution.
Therefore,the products formed are $1,2$-dibromoethane $(BrCH_2-CH_2Br)$ and $1$-bromo-$2$-chloroethane $(ClCH_2-CH_2Br)$.

(B) The reaction of propene $(CH_3-CH=CH_2)$ with hydrogen bromide $(HBr)$ 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 the lesser number of hydrogen atoms.
Therefore,the reaction is: $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$.
The product formed is $2-$bromopropane.

(A) The reaction of propene with $HBr$ in the presence of a peroxide follows the anti-Markovnikov addition rule (also known as the Kharasch effect or peroxide effect).
In this reaction,the $Br$ atom attaches to the carbon atom with more hydrogen atoms.
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2Br$ ($1$-bromopropane).

(C) The ozonolysis of $2$-methylbut-$2$-ene $(CH_3-C(CH_3)=CH-CH_3)$ proceeds as follows:
$(1) \ O_3$
$(2) \ Zn/H_2O$
The reaction breaks the double bond to form $CH_3COCH_3$ (acetone,a ketone) and $CH_3CHO$ (acetaldehyde,an aldehyde).
Therefore,the products are both an aldehyde and a ketone.

(C) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For the compound $RCH=CR_2$,the cleavage results in $RCHO$ and $R_2CO$.
$RCH=CR_2 \xrightarrow[2. Zn/H_2O]{1. O_3} RCHO + R_2CO$

(B) An unsaturated compound contains at least one carbon-carbon double or triple bond.
$C_6H_{14}$ is an alkane $(C_nH_{2n+2})$,which is saturated.
$C_4H_8$ follows the general formula $C_nH_{2n}$,indicating it is an alkene,which is an unsaturated compound.
$C_3H_7OH$ and $CH_3OH$ are alcohols,which are saturated compounds.

(B) Olefin,because a carbon-carbon double bond is present in ethylene $(CH_2=CH_2)$.

(B) Step $1$: $CH_3-CH_2-CH_2-OH$ reacts with $PCl_3$ to form $1$-chloropropane $(A)$ via nucleophilic substitution.
$CH_3-CH_2-CH_2-OH + PCl_3 \rightarrow CH_3-CH_2-CH_2-Cl (A) + H_3PO_3$
Step $2$: $1$-chloropropane $(A)$ reacts with alcoholic $KOH$ to undergo dehydrohalogenation (elimination reaction) to form propene $(B)$.
$CH_3-CH_2-CH_2-Cl + Alco. KOH \rightarrow CH_3-CH=CH_2 (B) + KCl + H_2O$
Therefore,the compound $B$ is propene.

(C) The statement describes the mechanism of electrophilic addition to an asymmetric alkene,which is governed by Markownikoff's rule.
According to Markownikoff's rule,when a protic acid $HX$ adds to an asymmetric alkene,the hydrogen atom $(H)$ attaches to the carbon atom of the double bond that already has a greater number of hydrogen atoms,while the halide $(X)$ attaches to the more substituted carbon atom.
This occurs because the reaction proceeds through the formation of the most stable carbocation intermediate.

(B) The reaction is an electrophilic addition of $HCl$ to vinyl chloride $(CH_2=CHCl)$.
According to Markownikoff's rule,the electrophile $(H^+)$ attaches to the carbon atom with more hydrogen atoms,and the nucleophile $(Cl^-)$ attaches to the carbon atom with fewer hydrogen atoms.
The reaction proceeds as follows:
$CH_2=CHCl + HCl \rightarrow CH_3-CHCl_2$ ($1$,$1$-dichloroethane).
Therefore,the correct option is $(B)$.

(B) The addition of $HBr$ to unsymmetrical alkenes follows Markownikoff's rule in the absence of peroxides,leading to the formation of the more stable carbocation intermediate.
In the presence of peroxides $(R_2O_2)$,the reaction proceeds via a free radical mechanism,which results in anti-Markownikoff addition.
For example,the reaction of propene with $HBr$ in the presence of peroxide yields $1$-bromopropane instead of $2$-bromopropane.
Therefore,the correct option is $(B)$.

(A) The peroxide effect,also known as the $Kharasch$ effect or anti-$Markownikoff$ addition,is specifically applicable only to $HBr$ in the presence of organic peroxides.
For $HCl$ and $HI$,the peroxide effect is not observed because the bond dissociation energy of $HCl$ is too high to be cleaved by free radicals,and the iodine radical formed in the case of $HI$ tends to combine to form $I_2$ rather than adding to the double bond.
Therefore,the correct option is $A$.

(C) The reaction of $3$-methyl-$2$-pentene $(CH_3-CH=C(CH_3)-CH_2-CH_3)$ with $HOCl$ proceeds via the formation of a cyclic chloronium ion intermediate.
$HOCl$ dissociates to provide $Cl^+$ (electrophile) and $OH^-$ (nucleophile).
The $Cl^+$ ion attacks the double bond to form a cyclic chloronium ion.
The nucleophile $(OH^-)$ then attacks the more substituted carbon atom $(C3)$ due to its higher carbocation-like character in the transition state,following Markovnikov's rule.
Therefore,the major product is $3$-chloro-$3$-methylpentan-$2$-ol,which corresponds to the structure $CH_3-CH(Cl)-C(OH)(CH_3)-CH_2-CH_3$.

(A) . Ethylene $(CH_2=CH_2)$ contains a carbon-carbon double bond,which consists of one strong $\sigma$-bond and one weak $\pi$-bond.
Due to the presence of the electron-rich $\pi$-bond,ethylene is highly susceptible to electrophilic attack,making addition reactions the characteristic and most easily occurring reactions for alkenes.

(B) Butadiene $CH_2=CH-CH=CH_2$
$A$ system where two double bonds are separated by a single bond is known as a conjugated double bond system.

(A) Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond.
The general formula for acyclic alkenes is $C_nH_{2n}$,where $n$ is the number of carbon atoms $(n \geq 2)$.

(A) The given reaction is Oxymercuration-Demercuration $(OMDM)$.
$1$. This reaction involves the Markovnikov addition of water ($H$ and $OH$) across the double bond.
$2$. Crucially,no carbocation rearrangement occurs in this process because a bridged mercurinium ion intermediate is formed.
$3$. The $-OH$ group adds to the more substituted carbon of the double bond,and the $-H$ atom adds to the less substituted carbon.
$4$. Therefore,$3,3$-dimethyl-$1$-butene $(CH_3-C(CH_3)_2-CH=CH_2)$ reacts to form $3,3$-dimethyl-$2$-butanol $(CH_3-C(CH_3)_2-CH(OH)-CH_3)$.

(B) The alkaline $KMnO_4$ solution is known as $Baeyer's$ reagent.
It is used to test for the presence of unsaturation (double or triple bonds) in organic compounds.
$C_3H_6$ (propene) is an alkene,which contains a carbon-carbon double bond.
Therefore,it reacts with alkaline $KMnO_4$ to undergo oxidation,resulting in the decolourization of the purple $KMnO_4$ solution.

(B) The decolourization of the pink colour of alkaline $KMnO_4$ (Baeyer's reagent) is used to detect the presence of unsaturation (double or triple bonds) in organic compounds.
Both alkenes (olefinic) and alkynes (acetylenic) undergo this reaction,but the term 'olefinic' is the most common classification associated with this test in general contexts.
Since both $B$ and $C$ are unsaturated,but the question typically refers to the Baeyer's test for alkenes,$B$ is the primary answer.

(C) Ozonolysis of $2,3$-dimethyl-$2$-butene $(CH_3-C(CH_3)=C(CH_3)-CH_3)$ involves the cleavage of the $C=C$ double bond and the addition of oxygen atoms to each carbon atom,resulting in the formation of two molecules of acetone $(CH_3-CO-CH_3)$.
$CH_3-C(CH_3)=C(CH_3)-CH_3 + O_3 \xrightarrow{Zn/H_2O} 2 CH_3-CO-CH_3$.

(A) The reactivity of alkenes towards electrophilic addition of $HCl$ depends on the stability of the intermediate carbocation formed.
$(1)$ $CH_2 = CH_2$ forms a primary carbocation $(CH_3-CH_2^+)$,which is the least stable.
$(3)$ $CH_3CH = CHCH_3$ forms a secondary carbocation $(CH_3-CH_2-CH^+-CH_3)$,which is more stable than the primary one.
$(2)$ $(CH_3)_2C = CH_2$ forms a tertiary carbocation $((CH_3)_2C^+-CH_3)$,which is the most stable.
Since the stability of the carbocation intermediate follows the order: $1^\circ < 2^\circ < 3^\circ$,the reactivity towards $HCl$ follows the same order: $(1) < (3) < (2)$.

(C) The given substance is $2-$methylpropene,which has the structure $CH_2=C(CH_3)_2$.
When $HCl$ is added to $2-$methylpropene,the reaction follows Markovnikov's rule.
According to Markovnikov's rule,the electrophile $(H^+)$ adds to the carbon atom with more hydrogen atoms,and the nucleophile $(Cl^-)$ adds to the carbon atom with fewer hydrogen atoms.
$CH_2=C(CH_3)_2 + HCl \rightarrow CH_3-C(Cl)(CH_3)_2$.
The product formed is $2-$chloro$-2-$methylpropane.

(A) . Ozonolysis is a chemical process used to locate the position of a double bond in an alkene.
In this reaction,the alkene reacts with ozone $(O_3)$ followed by reductive cleavage (using $Zn/H_2O$) to form carbonyl compounds.
The position of the original double bond can be determined by joining the carbonyl carbon atoms of the resulting products.
For example,if the products of ozonolysis are two molecules of ethanal $(CH_3CHO)$,the original alkene was $CH_3-CH=CH-CH_3$ ($2$-Butene).

(A) Bayer's reagent $(1\% \text{ alkaline } KMnO_4 \text{ solution})$ is used to detect the presence of unsaturation (double or triple bonds) in organic compounds. Both alkenes and alkynes decolourise it.
Tollen's reagent (ammoniacal silver nitrate) reacts only with terminal alkynes (alkynes having a hydrogen atom attached to a triply bonded carbon) to form a white precipitate of silver acetylide.
$1.$ Ethene $(CH_2=CH_2)$ is an alkene. It decolourises Bayer's reagent due to the presence of a double bond but does not react with Tollen's reagent because it is not a terminal alkyne.
$2.$ Ethyne $(CH \equiv CH)$ is a terminal alkyne. It decolourises Bayer's reagent and also reacts with Tollen's reagent.
$3.$ Ethane and Methane are saturated hydrocarbons (alkanes) and do not react with either reagent.
Therefore,the gas that decolourises Bayer's reagent but does not react with Tollen's reagent is Ethene.