Alkene Questions in English

(D) The conversion of $Propene$ $(CH_3-CH=CH_2)$ to $Propane$ $(CH_3-CH_2-CH_3)$ involves the addition of hydrogen across the double bond.
This process is known as catalytic hydrogenation.
In this reaction,$Propene$ reacts with hydrogen gas in the presence of a metal catalyst like $Ni$,$Pd$,or $Pt$ to form $Propane$.

(A) The addition of $HI$ to propene follows Markovnikov's rule.
In the first step,the electrophile $H^+$ attacks the double bond to form a carbocation intermediate.
Two possible carbocations can be formed:
$1$. $CH_3-CH^+-CH_3$ (secondary carbocation,more stable)
$2$. $CH_3-CH_2-CH_2^+$ (primary carbocation,less stable)
Since the secondary carbocation is more stable due to the inductive effect and hyperconjugation,the reaction proceeds through this intermediate.
Finally,the nucleophile $I^-$ attacks the secondary carbocation to form $CH_3-CH(I)-CH_3$ (isopropyl iodide).

(C) Under thermodynamic control conditions (high temperature),the $1, 4$-addition product is the major product in the reaction of conjugated dienes with electrophiles.
For $1, 3$-butadiene,the $1, 4$-addition of $HCl$ results in the formation of $1$-chloro-$2$-butene.

(B) The reaction of an alkene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule (Kharasch effect or peroxide effect).
In this reaction,the $Br^-$ radical attacks the less substituted carbon atom of the double bond.
The reaction proceeds as follows:
${H_2}C = CH-(CH_2)_6-CH_3 + HBr \xrightarrow{peroxide} Br-CH_2-CH_2-(CH_2)_6-CH_3$.
Therefore,the product is $1-bromodecane$.

(D) The reaction of $2$-methylpropene $(CH_3)_2C=CH_2$ with $HBr$ follows Markovnikov's rule.
According to Markovnikov's rule,the electrophile $H^+$ adds to the carbon atom with more hydrogen atoms,and the nucleophile $Br^-$ adds to the carbon atom with fewer hydrogen atoms.
This leads to the formation of a stable tertiary carbocation intermediate $(CH_3)_3C^+$.
Finally,the bromide ion attacks the tertiary carbocation to form $2$-bromo-$2$-methylpropane $(CH_3)_3CBr$ as the major product.

(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
Given products are $CH_3CHO$ (acetaldehyde) and $CH_3CH_2CHO$ (propanal).
By joining the oxygen atoms of these two carbonyl groups,we can reconstruct the original alkene:
$CH_3CH=O + O=CHCH_2CH_3 \rightarrow CH_3CH=CHCH_2CH_3 + H_2O$.
The resulting alkene is $CH_3CH=CHCH_2CH_3$,which is $2$-pentene.
Therefore,the correct option is $A$.

(D) The general reaction for the ozonolysis of a symmetrical alkene is: $R-CH=CH-R + O_3 \rightarrow 2R-CHO$.
Given that the product is an aldehyde with a molar mass of $44 \ u$.
The general formula for an aldehyde is $R-CHO$.
The molar mass of $CHO$ is $12 + 1 + 16 = 29 \ u$.
Therefore,the mass of the alkyl group $R$ is $44 - 29 = 15 \ u$.
An alkyl group with a mass of $15 \ u$ is a methyl group $(-CH_3)$.
Thus,the aldehyde is acetaldehyde $(CH_3CHO)$.
Substituting this back into the reaction: $CH_3-CH=CH-CH_3 + O_3 \rightarrow 2CH_3CHO$.
The reactant is $2$-butene.

(D) The reaction is an ozonolysis of but$-2-$ene.
Step $1$: $CH_3CH = CHCH_3$ reacts with $O_3$ to form an ozonide intermediate $(A)$.
Step $2$: The ozonide intermediate $(A)$ undergoes reductive cleavage with $H_2O/Zn$ to produce carbonyl compounds.
Reaction: $CH_3CH = CHCH_3 + O_3$ $\rightarrow \text{ozonide}$ $\xrightarrow{H_2O/Zn} 2CH_3CHO$.
Thus,the product $B$ is acetaldehyde $(CH_3CHO)$.

(C) The chemical formula for but-$1$-ene is $CH_2=CH-CH_2-CH_3$.
To find the number of sigma $(\sigma)$ bonds,we count the bonds in the structure:
$1$. $C=C$ bond contains $1$ $\sigma$ bond and $1$ $\pi$ bond.
$2$. $C-C$ single bonds: $2$ bonds.
$3$. $C-H$ bonds: $2$ (on $C_1$) + $1$ (on $C_2$) + $2$ (on $C_3$) + $3$ (on $C_4$) = $8$ $\sigma$ bonds.
Total $\sigma$ bonds = $1$ (from $C=C$) + $2$ (from $C-C$) + $8$ (from $C-H$) = $11$ $\sigma$ bonds.

(A) In the absence of peroxide,the addition of $HBr$ to propene follows the electrophilic addition mechanism (Markovnikov's rule).
The first step is the attack of the electrophile,which is the proton $(H^+)$,on the double bond of the alkene to form a carbocation intermediate.
Therefore,the species involved in the first step is $H^+$.

(B) The addition of $HI$ to propene $(CH_3-CH=CH_2)$ follows Markovnikov's rule.
In the first step,the electrophile $H^+$ attacks the double bond to form a carbocation intermediate.
Two possible carbocations can be formed: $CH_3-CH^+-CH_3$ (secondary carbocation) and $CH_3-CH_2-CH_2^+$ (primary carbocation).
The secondary carbocation is more stable due to the inductive effect and hyperconjugation of the two methyl groups.
Therefore,the nucleophile $I^-$ attacks the secondary carbocation to form isopropyl iodide $(CH_3-CH(I)-CH_3)$.

(D) The conversion of an alkene (butene) to an alkane (butane) is a hydrogenation reaction.
This reaction involves the addition of hydrogen across the double bond in the presence of a metal catalyst such as $Pd$,$Pt$,or $Ni$.
The reaction is: $CH_3-CH_2-CH=CH_2 + H_2 \xrightarrow{Pd} CH_3-CH_2-CH_2-CH_3$.
Therefore,$Pd/H_2$ is the correct reagent for this reduction.

(B) The reaction of propene $(CH_3 - CH = CH_2)$ with $HBr$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom with the fewer number of hydrogen atoms.
Therefore,the major product is $2$-bromopropane: $CH_3 - CH(Br) - CH_3$.

(C) The reaction of $1,3$-butadiene with bromine $(Br_2)$ is an electrophilic addition reaction.
Due to the resonance stabilization of the allylic carbocation intermediate,the addition can occur at the $1,2$-position (kinetic product) or the $1,4$-position (thermodynamic product).
Therefore,a mixture of $3,4$-dibromobut-$1$-ene ($1,2$-addition product) and $1,4$-dibromobut-$2$-ene ($1,4$-addition product) is formed.

(A) The reaction of propene $(CH_3-CH=CH_2)$ with $HBr$ in the presence of an organic peroxide (like benzoyl peroxide) follows the anti-Markovnikov addition rule (also known as the Kharasch effect or peroxide effect).
In this reaction,the bromine atom attaches to the carbon atom of the double bond that has more hydrogen atoms.
Thus,the reaction proceeds as: $CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2Br$.
The final product is $1$-bromopropane.

(C) The structural formula of $1$-butene $(CH_2=CH-CH_2-CH_3)$ is:
$H-C(H)=C(H)-C(H)_2-C(H)_3$
Counting the bonds:
There are $11$ $\sigma$ bonds (single bonds) and $1$ $\pi$ bond (the second bond in the double bond).
Total $\sigma$ bonds = $11$.

(C) The stability of alkenes is determined by the number of alkyl groups attached to the double-bonded carbons (hyperconjugation) and steric hindrance.
$1-\text{butene}$ is a monosubstituted alkene,which is less stable than disubstituted alkenes.
Between $\text{cis-}2-\text{butene}$ and $\text{trans-}2-\text{butene}$,the $\text{trans}$ isomer is more stable due to lower steric repulsion between the methyl groups.
Therefore,the order of stability is: $\text{trans-}2-\text{butene} > \text{cis-}2-\text{butene} > 1-\text{butene}$.

(B) The reaction between propene $(CH_3-CH=CH_2)$ and $HCl$ is an example of an electrophilic addition reaction.
In the first step,the electrophile $H^+$ attacks the double bond to form a more stable carbocation (isopropyl carbocation).
In the second step,the nucleophile $Cl^-$ attacks the carbocation to form isopropyl chloride $(CH_3-CHCl-CH_3)$.
This follows Markovnikov's rule.

(A) The stability of alkenes increases with the number of alkyl groups attached to the double-bonded carbon atoms due to the hyperconjugation and inductive effect.
$R_2C = CR_2$ is a tetra-substituted alkene,which is the most stable among the given options.

(D) The $IUPAC$ name $2$-methyl-$2$-butene indicates a parent chain of $4$ carbon atoms (butene) with a double bond at the $2^{nd}$ position and a methyl group at the $2^{nd}$ position.
The structure is $CH_3-C(CH_3)=CH-CH_3$.

(A) The peroxide effect (Kharasch effect) is only applicable to $HBr$ and not to $HCl$ or $HI$. $HCl$ does not show the anti-Markovnikov addition in the presence of peroxide because the $H-Cl$ bond is too strong to be broken by free radicals. Therefore,the reaction follows the standard Markovnikov addition mechanism,forming the more stable secondary carbocation intermediate: $CH_3-CH^+-CH_3$.

(A) Markownikoff's rule is applicable to unsymmetrical alkenes where the negative part of the addendum adds to the carbon atom with fewer hydrogen atoms.
$CH_2 = CH_2$ (Ethene) is a symmetrical alkene,so it does not follow Markownikoff's rule.
$CH_3 - CH = CH - CH_3$ (But$-2-$ene) is also a symmetrical alkene.
However,in the context of standard multiple-choice questions,$CH_2 = CH_2$ is the simplest symmetrical alkene that does not show regioselectivity,making it the primary answer.

(B) The reaction of $CH_3CH=CH-C_6H_4-OH$ with $HBr$ is an electrophilic addition reaction across the double bond.
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.
The carbocation formed at the benzylic position (adjacent to the phenyl ring) is stabilized by resonance with the benzene ring.
Therefore,the $Br^-$ attacks the carbon adjacent to the benzene ring,resulting in the product $CH_3CH_2CHBr-C_6H_4-OH$.

(A) Conjugated alkadienes are the most stable due to resonance stabilization and delocalization of $\pi$-electrons.
Isolated alkadienes are less stable than conjugated ones but more stable than cumulated ones.
Cumulated alkadienes are the least stable due to the presence of $sp$-hybridized carbon atoms and steric strain.
Therefore,the order of stability is: $\text{Conjugated} > \text{Isolated} > \text{Cumulated}$.

(D) The heat of hydrogenation is inversely proportional to the stability of the alkene.
More substituted alkenes are more stable.
Among the given options,trans-$2$-Butene is the most stable alkene due to minimum steric hindrance and maximum hyperconjugation.
Therefore,it releases the minimum amount of energy upon hydrogenation.

(B) The heat of hydrogenation is inversely proportional to the stability of the alkene.
More substituted alkenes are more stable.
Among the given options,trans-$2$-Butene is the most stable isomer due to minimal steric hindrance.
Therefore,trans-$2$-Butene has the lowest heat of hydrogenation per mole.

(B) The reaction of bromine $(Br_2)$ with an alkene $(R-CH=CH-R)$ is a classic example of an electrophilic addition reaction.
In this process,the $\pi$-bond of the alkene acts as a nucleophile and attacks the bromine molecule,leading to the formation of a cyclic bromonium ion intermediate.
Subsequently,the bromide ion $(Br^-)$ attacks the cyclic intermediate to form a vicinal dibromide $(R-CH(Br)-CH(Br)-R)$.

(B) The rate of catalytic hydrogenation of alkenes is inversely proportional to the steric hindrance around the double bond.
More substituted alkenes have greater steric hindrance,which makes it difficult for the alkene to approach the metal catalyst surface.
Therefore,the order of reactivity is: monosubstituted > disubstituted > trisubstituted > tetrasubstituted.
Among the given options,the cis-disubstituted alkene $(R-CH=CH-R)$ is generally more reactive than the trans-disubstituted alkene due to less steric crowding on one side,but in terms of general substitution,the least substituted alkene is the fastest. Comparing the options provided,the disubstituted alkenes are more reactive than tri- and tetra-substituted ones. Between cis and trans,the cis-isomer is often more reactive in hydrogenation due to the accessibility of the double bond.

(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.

(B) The $2$-butenyl radical is derived from $2$-butene $(CH_3-CH=CH-CH_3)$ by removing a hydrogen atom from the carbon adjacent to the double bond.
Specifically,the $2$-butenyl group is represented by the structure $CH_3-CH=CH-CH_2-$.
This corresponds to option $B$.

(C) The conversion of propene to propan-$1$-ol is an anti-Markovnikov addition of water across the double bond.
This is achieved via the hydroboration-oxidation reaction.
$CH_3-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^-]{(i) B_2H_6} CH_3-CH_2-CH_2OH$
Propene $\rightarrow$ Propan-$1$-ol.

(B) The reaction sequence is as follows:
$CH_3CH_2CH_2OH \xrightarrow{PCl_5} CH_3CH_2CH_2Cl (A)$
$CH_3CH_2CH_2Cl \xrightarrow{alc. KOH} CH_3CH=CH_2 (B)$
Here,$A$ is $1$-chloropropane and $B$ is propene. The alcoholic $KOH$ causes dehydrohalogenation (elimination reaction) to form an alkene.

(D) The reaction of an alkene with $B_2H_6$ followed by oxidation with $H_2O_2/OH^-$ is known as hydroboration-oxidation.
This reaction follows anti-Markovnikov addition of water across the double bond.
$R-CH=CH_2 + (i) B_2H_6, (ii) H_2O_2/OH^- \rightarrow R-CH_2-CH_2OH$.

(C) When glycerol is heated with concentrated $H_2SO_4$,it undergoes dehydration to form $CH_2=CH-CHO$,which is known as $Prop-2-enal$ (acrolein). This compound has a characteristic pungent,irritating smell.
$CH_2OH-CHOH-CH_2OH \xrightarrow{H_2SO_4} CH_2=CH-CHO + 2H_2O$

(D) The reaction of ethene with $HOCl$ (hypochlorous acid) follows electrophilic addition to form ethylene chlorohydrin $(CH_2Cl-CH_2OH)$,which is $A$.
$CH_2=CH_2 + HOCl \rightarrow CH_2Cl-CH_2OH$ $(A)$
Subsequently,the treatment of $CH_2Cl-CH_2OH$ with an aqueous base like $NaHCO_3$ $(R)$ leads to the hydrolysis of the alkyl chloride to form ethylene glycol $(CH_2OH-CH_2OH)$.

(C) The synthesis of glycerol from propene involves the following steps:
$1$. Propene is chlorinated at $500 \ ^\circ C$ to form allyl chloride: $CH_3-CH=CH_2 + Cl_2 \xrightarrow{500 \ ^\circ C} ClCH_2-CH=CH_2 + HCl$.
$2$. Allyl chloride is hydrolyzed with $Na_2CO_3$ at $150 \ ^\circ C$ to form allyl alcohol: $ClCH_2-CH=CH_2 + H_2O \xrightarrow{Na_2CO_3} HOCH_2-CH=CH_2$.
$3$. Allyl alcohol reacts with $HOCl$ to form glycerol-$\beta$-chlorohydrin: $HOCH_2-CH=CH_2 + HOCl \rightarrow HOCH_2-CHCl-CH_2OH$.
$4$. Finally,glycerol-$\beta$-chlorohydrin is treated with $NaOH$ to yield glycerol: $HOCH_2-CHCl-CH_2OH + NaOH \rightarrow HOCH_2-CH(OH)-CH_2OH + NaCl$.

(A) Alkaline $KMnO_4$ is known as Baeyer's reagent. It is used for the detection of unsaturation in organic compounds.

(C) $1\%$ cold alkaline $KMnO_4$ is known as Baeyer's reagent.
It reacts with ethylene $(CH_2=CH_2)$ to undergo syn-hydroxylation,resulting in the formation of ethylene glycol $(CH_2OH-CH_2OH)$.
The reaction is: $CH_2=CH_2 + H_2O + [O] \xrightarrow{KMnO_4} CH_2OH-CH_2OH$.

(C) The acid-catalyzed hydration of alkenes follows $Markownikoff's$ rule,where the hydroxyl group $(-OH)$ attaches to the more substituted carbon atom of the double bond.
In $2-$phenylpropene $(C_6H_5-C(CH_3)=CH_2)$,the carbon atom at position $2$ is more substituted (bonded to a phenyl group and a methyl group) than the terminal carbon atom.
Therefore,the electrophilic addition of $H^+$ followed by the nucleophilic attack of $H_2O$ results in the formation of $2-$phenylpropan$-2-$ol as the major product.

(C) When glycerol $(CH_2OH-CHOH-CH_2OH)$ reacts with excess $HI$,it first forms $1,2,3-triiodopropane$ $(CH_2I-CHI-CH_2I)$.
This compound is unstable and loses a molecule of $I_2$ to form allyl iodide $(CH_2=CH-CH_2I)$.

(B) The reaction is a Wittig reaction between a carbonyl compound and an alkylidene phosphorane (phosphorus ylide).
The reactants are $CH_3CH_2C(=O)CH_3$ (butanone) and $Ph_3P=CHCH_2CH_2CH_3$ (propylidene triphenylphosphorane).
The mechanism involves the formation of an oxaphosphetane intermediate,which decomposes to form the alkene and triphenylphosphine oxide $(Ph_3P=O)$.
The reaction is: $CH_3CH_2C(=O)CH_3 + Ph_3P=CHCH_2CH_2CH_3 \rightarrow CH_3CH_2C(CH_3)=CHCH_2CH_2CH_3 + Ph_3P=O$.
The product is $3-\text{methylhept-3-ene}$.

(A) Ozonolysis involves the cleavage of the $C=C$ double bond to form carbonyl compounds. The reaction is represented as: $R_1R_2C=CR_3R_4 \xrightarrow{(i)O_3, (ii)Zn, H_2O} R_1R_2C=O + O=CR_3R_4$.
For the given products $CH_3CH_2CHO$ (propanal) and $CH_3COCH_3$ (acetone),the alkene must be $CH_3CH_2CH=C(CH_3)_2$.
Upon ozonolysis: $CH_3CH_2CH=C(CH_3)_2 \xrightarrow{(i)O_3, (ii)Zn, H_2O} CH_3CH_2CHO + CH_3COCH_3$.

(D) Vigorous oxidation of alkenes with hot alkaline $KMnO_4$ (permanganate solution) leads to the cleavage of the double bond.
For the alkene $(CH_3)_2C=CHCH_2CH_3$:
$1$. The double bond breaks between the $C_2$ and $C_3$ carbons.
$2$. The $(CH_3)_2C=$ part is oxidized to acetone,$(CH_3)_2C=O$.
$3$. The $=CHCH_2CH_3$ part is oxidized to propanoic acid,$CH_3CH_2COOH$.
Therefore,the products are acetone and propanoic acid.

(A) . Alkenes react with water in the presence of an acid catalyst to form alcohols. This process is known as hydration.
$H_2C=CH_2 + H_2O \xrightarrow{H_3PO_4, 300 \ ^\circ C/60 \ atm} C_2H_5OH$

(A) The reaction of ethene $(CH_2=CH_2)$ with bromine water $(Br_2/H_2O)$ is an electrophilic addition reaction.
In the presence of water,the bromonium ion intermediate is attacked by the water molecule (nucleophile) instead of the bromide ion.
The reaction is: $CH_2=CH_2 + Br_2 + H_2O \rightarrow CH_2(Br)-CH_2(OH) + HBr$.
The product $CH_2(Br)-CH_2(OH)$ is known as ethylene bromohydrin (or $2-$bromoethanol).
Therefore,the correct option is $A$.

(B) The reaction sequence is as follows:
$1.$ $C_2H_5I \xrightarrow{Alco. KOH} CH_2=CH_2$ $(X)$
$2.$ $CH_2=CH_2 \xrightarrow{Br_2} Br-CH_2-CH_2-Br$ $(Y)$
$3.$ $Br-CH_2-CH_2-Br \xrightarrow{KCN} NC-CH_2-CH_2-CN$ $(Z)$
Therefore,$Z$ is $NC-CH_2-CH_2-CN$.

(A) The reaction of alkenes with bromine is an electrophilic addition reaction.
Among the given options,$C_3H_6$ (propene) is an alkene with an electron-donating methyl group $(-CH_3)$ attached to the double bond.
This methyl group increases the electron density of the double bond via the inductive effect,making it more nucleophilic and thus more reactive towards the electrophilic bromine compared to $C_2H_4$ (ethene) or $C_2H_2$ (ethyne).
$C_4H_{10}$ (butane) is an alkane and does not undergo addition reactions with bromine under normal conditions.

(C) Markovnikov's rule states that in the addition of a protic acid to an asymmetric alkene,the acid hydrogen attaches to the carbon with the smaller number of hydrogen atoms and the halide group attaches to the carbon with the larger number of hydrogen atoms. However,the product $1-$chloro$-1-$methylcyclohexane can be formed from both methylenecyclohexane and $1-$methylcyclohexene.
For methylenecyclohexane: The protonation of the double bond leads to a $3^{\circ}$ carbocation at the $1-$position,which then reacts with $Cl^-$ to give $1-$chloro$-1-$methylcyclohexane.
For $1-$methylcyclohexene: The protonation leads to a $2^{\circ}$ carbocation,which undergoes rearrangement to form a more stable $3^{\circ}$ carbocation,which then reacts with $Cl^-$ to give $1-$chloro$-1-$methylcyclohexane.
Since both alkenes yield the same product,the correct answer is $(C)$.

(A) $3,3-$Dimethyl$-1-$butene $(CH_3-C(CH_3)_2-CH=CH_2)$ on treatment with a strong acid undergoes protonation to form a secondary carbocation: $CH_3-C(CH_3)_2-C^+H-CH_3$.
This carbocation undergoes a $1,2-$methyl shift to form a more stable tertiary carbocation: $CH_3-C^+(CH_3)-CH(CH_3)-CH_3$.
Finally,it loses a proton (deprotonation) to form the more substituted alkene,$2,3-$dimethyl$-2-$butene $(CH_3-C(CH_3)=C(CH_3)-CH_3)$,as the major product.

(A) Ozonolysis of a cyclic alkene involves the cleavage of the double bond to form a dicarbonyl compound. The given product is $CH_3-CO-CH_2-CH_2-CH_2-CHO$. This is a $6$-carbon chain with a ketone group at position $2$ and an aldehyde group at position $6$. This implies the original cyclic compound must have been a $6$-membered ring,specifically $1-methylcyclohex-1-ene$. However,looking at the options provided,we must evaluate which cyclic compound yields this specific structure. The structure $CH_3-CO-CH_2-CH_2-CH_2-CHO$ is $6-oxohexanal$. This is formed by the oxidative cleavage of $1-methylcyclohex-1-ene$. Given the options are all $5$-membered rings,there might be a discrepancy in the question's structure. Re-evaluating the product $CH_3-CO-CH_2-CH_2-CH_2-CHO$,it corresponds to the ozonolysis of $1-methylcyclohex-1-ene$. If the question implies a $5$-membered ring,the product would be different. Based on standard chemistry problems of this type,$1-methylcyclohex-1-ene$ is the correct precursor for this specific linear dicarbonyl.