Alkene Questions in English

(B) $1$. Calculate the molar mass of the hydrocarbon:
At $STP$,$22400 \ mL$ corresponds to $1 \ mole$ of gas.
Given $488 \ mL$ weighs $1.68 \ g$,so $22400 \ mL$ weighs: $\frac{1.68 \times 22400}{488} \approx 77.14 \ g/mol$.
$2$. Determine the empirical formula:
Carbon = $86\%$,Hydrogen = $100 - 86 = 14\%$.
Moles of $C = \frac{86}{12} \approx 7.17$.
Moles of $H = \frac{14}{1} = 14$.
Ratio $C:H = 7.17:14 \approx 1:2$.
Empirical formula is $CH_2$.
$3$. Determine molecular formula:
Empirical formula mass = $12 + 2 = 14 \ g/mol$.
$n = \frac{\text{Molar mass}}{\text{Empirical mass}} = \frac{77.14}{14} \approx 5.5$.
Since the hydrocarbon is a gas at $STP$ and follows $C_nH_{2n}$,it belongs to the alkene series.

(B) In ethylene $(CH_2=CH_2)$,both carbon atoms are $sp^2$-hybridized.
The geometry around each carbon atom is trigonal planar,which corresponds to a bond angle of $120^o$.

(C) The chemical formula of ethylene is $C_2H_4$ or $CH_2=CH_2$.
In the ethylene molecule,there is a double bond between the two carbon atoms.
$A$ double bond consists of $2$ shared electron pairs.
Since each shared pair contains $2$ electrons,the total number of shared electrons between the two carbon atoms is $2 \times 2 = 4$.

(C) $H_2O_2$ reacts with ethene $(CH_2=CH_2)$ to form ethylene glycol $(CH_2(OH)-CH_2(OH))$.
This is a hydroxylation reaction.

(B) The test for unsaturation in alkenes is known as the Baeyer's test.
In this test,alkenes react with cold dilute alkaline $KMnO_4$ solution (Baeyer's reagent).
The purple/pink colour of the $KMnO_4$ solution disappears,and a brown precipitate of $MnO_2$ is formed,indicating the presence of a double bond (unsaturation).
Therefore,cold dilute alkaline $KMnO_4$ is the correct reagent.

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

(C) The general formula for an alkene is $C_nH_{2n}$.
For $n = 10$,the molecular formula is $C_{10}H_{20}$.
The prefix for $10$ carbon atoms is 'dec-'.
Since it is an alkene,the suffix is '-ene'.
Therefore,the name of the compound is $Decene$.

(B) The structure of ethene is $CH_2 = CH_2$.
In this molecule,each carbon atom is bonded to two hydrogen atoms and one other carbon atom via a double bond.
Since each carbon atom is attached to three sigma bonds and zero lone pairs,the hybridization of each carbon atom is $sp^2$.
Therefore,there are $2$ $sp^2$ hybridized carbon atoms.

(C) Cyclohexene has the molecular formula $C_6H_{10}$.
It contains one double bond between two carbon atoms.
The two carbon atoms involved in the double bond are $sp^2$-hybridized.
The remaining $4$ carbon atoms are bonded only by single bonds and are therefore $sp^3$-hybridized.
Thus,there are $4$ $sp^3$-hybridized carbon atoms in cyclohexene.

(C) The structure of ethylene is $CH_2=CH_2$.
In this molecule,there are $4$ $C-H$ sigma bonds and $1$ $C-C$ sigma bond,totaling $5$ sigma bonds.
There is also $1$ $C-C$ pi bond.
Therefore,ethylene contains $5$ sigma and $1$ pi bond.

(B) conjugated system is defined as a system of connected $p-$orbitals with delocalized electrons in a molecule,which in general lowers the overall energy of the molecule and increases stability.
In conjugated dienes,two double bonds are separated by exactly one single bond,represented as $C=C-C=C$.
$1, 3-$butadiene $(CH_2=CH-CH=CH_2)$ and $1, 3-$pentadiene $(CH_2=CH-CH=CH-CH_3)$ both contain conjugated double bonds.
However,in standard multiple-choice questions of this type,$1, 3-$butadiene is the fundamental example of a conjugated diene.
Note: Both $B$ and $C$ are technically correct,but $1, 3-$butadiene is the most common textbook example.

(C) The reaction of $CH_3-CH=CH_2$ with $HBr$ in the presence of peroxide (Kharasch effect) proceeds via a free radical mechanism.
In this reaction,the peroxide undergoes homolytic cleavage to form radicals,which then initiate the chain reaction,leading to anti-Markovnikov addition.

(D) The addition of $HCl$ to vinyl chloride $(CH_2=CHCl)$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(Cl^-)$ attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.
In $CH_2=CHCl$,the $CH$ group has one hydrogen atom,while the $CH_2$ group has two.
Therefore,the $Cl^-$ ion attacks the $CH$ carbon,leading to the formation of $1, 1-$dichloroethane $(CH_3-CHCl_2)$.

(D) The reaction of ethene $(CH_2=CH_2)$ with bromine $(Br_2)$ is a classic example of electrophilic addition.
In this reaction,the $\pi$-bond of the alkene acts as a nucleophile and attacks the electrophilic bromine molecule,leading to the formation of a cyclic bromonium ion intermediate,followed by the attack of the bromide ion to form $CH_2Br-CH_2Br$.

(A) The catalytic reduction of unsaturated hydrocarbons (alkenes or alkynes) to alkanes is typically carried out using finely divided metals like $Pt$,$Pd$,or $Ni$ as catalysts. Among the given options,$Pt/Ni$ represents the commonly used catalysts for this process.

(C) The reaction of ethylene $(CH_2=CH_2)$ with bromine $(Br_2)$ is an electrophilic addition reaction.
In this reaction,the $\pi$-bond of the alkene breaks,and two bromine atoms add across the double bond.
The reaction proceeds as follows:
$CH_2=CH_2 + Br_2 \rightarrow Br-CH_2-CH_2-Br$
The product formed is $1,2-dibromoethane$.

(A) The conversion of propene to propane involves the addition of hydrogen across the double bond,which is known as catalytic hydrogenation.
$CH_3-CH=CH_2 + H_2 \xrightarrow{Ni, 300 \ ^oC} CH_3-CH_2-CH_3$
Thus,the correct option is $(A)$.

(C) The preparation of propane-$1$-ol from propene $(CH_3-CH=CH_2)$ is an example of hydroboration-oxidation reaction.
In this reaction,propene reacts with diborane $(B_2H_6)$ in the presence of an ether solvent to form a trialkylborane intermediate.
This intermediate is then oxidized using hydrogen peroxide $(H_2O_2)$ in the presence of an aqueous base $(NaOH)$ to yield the anti-Markovnikov product,propane-$1$-ol $(CH_3-CH_2-CH_2OH)$.

(C) The electrophilic addition of bromine to $1,3-$butadiene proceeds through the formation of a resonance-stabilized allylic carbocation intermediate.
This intermediate can be attacked by the bromide ion at the $C-2$ position to form the $1,2-$addition product ($3,4-$dibromobut-$1-$ene) or at the $C-4$ position to form the $1,4-$addition product ($1,4-$dibromobut-$2-$ene).
Therefore,both $1,2-$ and $1,4-$addition products are formed.

(B) When ethylene bromide $(Br-CH_2-CH_2-Br)$ is treated with $Zn$ dust,it undergoes dehalogenation to form ethylene $(CH_2=CH_2)$,which is an alkene.
Reaction: $Br-CH_2-CH_2-Br + Zn \rightarrow CH_2=CH_2 + ZnBr_2$

(A) The reaction of ethene with bromine in the presence of carbon tetrachloride $(CCl_4)$ is an electrophilic addition reaction.
$CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2(Br)-CH_2(Br)$
This reaction results in the formation of $1, 2-$dibromoethane.

(B) The reaction of ethene $(CH_2=CH_2)$ with hypochlorous acid $(HOCl)$ follows an electrophilic addition mechanism to form ethylene chlorohydrin $(Cl-CH_2-CH_2-OH)$,which is molecule $M$.
Ethylene chlorohydrin then undergoes hydrolysis in the presence of a mild base like aqueous sodium bicarbonate $(aq. NaHCO_3)$,which acts as reagent $R$,to produce ethylene glycol $(HO-CH_2-CH_2-OH)$.

(A) Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond $(C=C)$.
Due to the presence of the $\pi$-bond,they are electron-rich and generally undergo electrophilic addition reactions.

(D) The reaction of propene $(CH_3-CH=CH_2)$ with $HBr$ follows Markovnikov's rule.
According to this rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom having a lesser number of hydrogen atoms.
Thus,the major product formed is $2-$bromopropane $(CH_3-CH(Br)-CH_3)$.

(D) The reaction shows the oxidative cleavage of an alkene,$2,3-dimethylbut-2-ene$,into two molecules of acetone $(propanone)$.
This is a strong oxidation reaction typically carried out using an oxidizing agent like alkaline $KMnO_4$ under heating conditions.
Therefore,$X$ is $KMnO_4$.

(B) The reaction of an alkene with cold,dilute,alkaline $KMnO_4$ solution (Baeyer's reagent) results in the formation of a vicinal diol.
During this reaction,the purple colour of $KMnO_4$ disappears,which serves as a qualitative test for the presence of a double or triple bond (unsaturation).
This specific test is known as $Baeyer's$ test.

(B) The addition of $HCl$ to propene follows Markovnikov's rule,even in the presence of peroxide.
Peroxide effect (Kharasch effect) is only applicable to $HBr$ and not to $HCl$ or $HI$.
In the reaction,the electrophile $H^+$ attacks the double bond to form the more stable secondary carbocation,$CH_3-CH^{+}-CH_3$,as the reaction intermediate.
Therefore,the correct intermediate is $CH_3-CH^{+}-CH_3$.

(A) The reaction of ethene $(CH_2=CH_2)$ with cold,dilute alkaline $KMnO_4$ (Baeyer's reagent) is an oxidation reaction.
$CH_2 = CH_2 + [O] + H_2O \xrightarrow{KMnO_4 / OH^-} HO-CH_2-CH_2-OH$
The product formed is $Ethane-1,2-diol$,commonly known as $Ethylene \ glycol$.
This reaction is used as a test for unsaturation,where the purple color of $KMnO_4$ disappears.

(B) Ozonolysis of alkenes involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For $but-2-ene$ $(CH_3-CH=CH-CH_3)$,the reaction with $O_3$ followed by $Zn/H_2O$ proceeds as follows:
$CH_3-CH=CH-CH_3 + O_3 \rightarrow CH_3-CH(O_2)CH-CH_3$ (ozonide intermediate)
$CH_3-CH(O_2)CH-CH_3 + Zn/H_2O \rightarrow 2CH_3CHO$ (acetaldehyde or ethanal).
Thus,$but-2-ene$ yields two molecules of acetaldehyde.

(C) The Anti-Markownikoff addition of $HBr$ (also known as the peroxide effect or Kharasch effect) is only observed in unsymmetrical alkenes.
$But-2-ene$ $(CH_3-CH=CH-CH_3)$ is a symmetrical alkene.
In a symmetrical alkene,the addition of $HBr$ results in the same product regardless of whether it follows Markownikoff's rule or the Anti-Markownikoff mechanism.
Therefore,the Anti-Markownikoff addition is not observed for $But-2-ene$.

(C) The reaction of an alkene with dilute aqueous $KMnO_4$ (Baeyer's reagent) at room temperature is a syn-hydroxylation reaction.
$R-CH=CH-R \xrightarrow[\text{room temp.}]{\text{dil. aqueous } KMnO_4} R-CH(OH)-CH(OH)-R$ (vicinal diol).
In contrast,concentrated $KMnO_4$ with heat causes oxidative cleavage of the double bond to form carboxylic acids.

(B) The reaction of $2-$methyl$-1-$butene $(CH_2=C(CH_3)CH_2CH_3)$ with aqueous $H_2SO_4$ is an acid-catalyzed hydration.
It follows Markovnikov's rule,where the $H^{+}$ adds to the carbon with more hydrogens to form a stable tertiary carbocation $(CH_3-C^{+}(CH_3)CH_2CH_3)$.
Subsequent attack by $H_2O$ and loss of a proton yields $2-$methyl$-2-$butanol $(CH_3-C(OH)(CH_3)CH_2CH_3)$ as the major product.

(C) When ethanol is heated in the presence of conc. $H_2SO_4$,dehydration takes place to give ethene.
This process involves the removal of a water molecule from the ethanol molecule,which is why it is called a dehydration reaction.
$\underset{\text{Ethanol}}{C_2H_5OH}$ $\xrightarrow[160^{\circ}C - 170^{\circ}C]{\text{Conc. } H_2SO_4} \underset{\text{Ethene}}{CH_2=CH_2} + \underset{\text{Water}}{H_2O}$

(A) Baeyer's reagent is an alkaline solution of cold $KMnO_4$.
It is primarily used for the detection of unsaturation (double or triple bonds) in a molecule,as it causes the purple color of $KMnO_4$ to disappear.

(B) The hydration of alkenes in the presence of concentrated $H_2SO_4$ follows $Markownikoff's$ rule.
Propylene $(CH_3-CH=CH_2)$ reacts with water in the presence of concentrated $H_2SO_4$ to form isopropyl alcohol $(CH_3-CH(OH)-CH_3)$.
Reaction: $CH_3-CH=CH_2 + H_2O \xrightarrow[Conc. H_2SO_4]{Markownikoff's \, rule} CH_3-CH(OH)-CH_3$.

(D) $(D)$ In the presence of organic peroxides, the addition of $HBr$ to an alkene follows the Anti-Markovnikov rule (Kharasch effect).
The bromine atom adds to the carbon atom of the double bond that has more hydrogen atoms.
Therefore, $Br$ adds to the terminal $CH_2$ group of $CH_2 = CH - (CH_2)_2COOH$, resulting in $BrCH_2CH_2(CH_2)_2COOH$.

(B) The heat of hydrogenation is inversely proportional to the stability of the alkene.
More substituted or more stable alkenes release less heat upon hydrogenation.
Among the given options,$Trans-2-$butene is the most stable isomer due to minimal steric hindrance and greater symmetry compared to $Cis-2-$butene and $1-$butene.
Therefore,$Trans-2-$butene releases the lowest quantity of heat upon catalytic hydrogenation.

(C) $1$. The hydrocarbon $X$ decolourises bromine water,indicating it is an unsaturated hydrocarbon (alkene).
$2$. $X$ reacts with acidic $KMnO_4$ (a strong oxidizing agent) to produce two moles of the same carboxylic acid. This indicates that the double bond is located in the middle of a symmetrical alkene.
$3$. The reaction is: $CH_3CH_2CH=CHCH_2CH_3 + [O] \xrightarrow{H^+} 2 CH_3CH_2COOH$ (Propanoic acid).
$4$. Among the given options,$CH_3CH_2CH=CHCH_2CH_3$ (Hex$-3-$ene) is the only symmetrical alkene that cleaves to form two identical carboxylic acid molecules.

(C) The reaction between $1, 3-$butadiene and ethylene (ethene) is a Diels-Alder reaction.
This is a $[4+2]$ cycloaddition reaction.
$CH_2=CH-CH=CH_2 + CH_2=CH_2 \xrightarrow{\Delta} \text{Cyclohexene}$
The product formed is cyclohexene.

(C) The reaction of ethylene $(CH_2=CH_2)$ with ozone $(O_3)$ leads to the formation of an unstable intermediate known as ethylene ozonide.
The reaction is:
$CH_2=CH_2 + O_3 \rightarrow \text{Ethylene ozonide}$
This ozonide further undergoes reductive cleavage to form formaldehyde $(HCHO)$.

(C) $(C)$ Oils are unsaturated esters containing double bonds in their fatty acid chains. They are converted into saturated fats by the process of catalytic hydrogenation, where hydrogen is added across the double bonds in the presence of a catalyst like $Ni$ or $Pd$.

(C) The process of adding hydrogen to an unsaturated hydrocarbon (olefin) to convert it into a saturated hydrocarbon (paraffin) is known as hydrogenation.
The reaction is represented as:
$R-CH=CH-R + H_2 \xrightarrow{Ni/Pd/Pt} R-CH_2-CH_2-R$
Thus,the correct option is $(C)$.

(D) The reaction of $3-methyl-1-butene$ $(CH_2=CH-CH(CH_3)_2)$ with $DCl$ proceeds via the formation of a carbocation intermediate.
$1$. Protonation (or deuteration) of the double bond occurs to form the most stable carbocation.
$2$. Addition of $D^+$ to the terminal carbon $(C_1)$ leads to a secondary carbocation: $CH_2D-CH^+-CH(CH_3)_2$.
$3$. Subsequent attack by $Cl^-$ yields the product $CH_2D-CHCl-CH(CH_3)_2$ (Option $a$).
$4$. Alternatively,a $1,2-$hydride shift can occur in the secondary carbocation to form a more stable tertiary carbocation: $CH_3-CH^+-C(CH_3)_2$.
$5$. Attack by $Cl^-$ on this tertiary carbocation yields $CH_3-CHCl-C(CH_3)_2$ (not listed).
$6$. If $D^+$ adds to $C_2$,it forms a primary carbocation,which is unstable. However,if the mechanism involves rearrangement or specific pathways,$CH_3-CDCl-CH(CH_3)_2$ (Option $c$) can be formed via different intermediates or pathways involving hydride shifts.
Given the standard electrophilic addition mechanism,both $(a)$ and $(c)$ are considered possible products due to carbocation rearrangements.

(B) The reaction of $2\text{-bromobutane}$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).
According to $Saytzeff's$ rule,the more substituted alkene is the major product.
$CH_3-CH(Br)-CH_2-CH_3 + KOH (\text{alc.}) \to CH_3-CH=CH-CH_3 (\text{But-2-ene}) + KBr + H_2O$.
Since $But-2-ene$ is more stable than $But-1-ene$,it is the major product.

(C) The reaction of alkenes with cold,dilute alkaline $KMnO_4$ (Baeyer's reagent) is a syn-hydroxylation reaction.
In this reaction,two hydroxyl $(-OH)$ groups are added to the same side of the double bond.
Therefore,cyclopentene reacts with alkaline $KMnO_4$ to form cis-$1, 2-$cyclopentanediol.

(A) The stability of an alkene is directly proportional to the number of alkyl groups attached to the double-bonded carbon atoms due to the hyperconjugation effect and inductive effect.
$A$ represents a tetra-substituted alkene $(R_2C=CR_2)$,which has the maximum number of alkyl groups.
$B$ represents a di-substituted alkene.
$C$ is chemically incorrect as written.
$D$ represents ethene,which is unsubstituted.
Therefore,$R_2C=CR_2$ is the most stable alkene.

(C) The reaction of ethene with acidic $KMnO_4$ is an oxidative cleavage reaction.
$CH_2=CH_2 + 2[O] \xrightarrow{KMnO_4/H^+} HCHO + HCHO$
Since ethene is a terminal alkene,the oxidative cleavage of the double bond results in the formation of formaldehyde $(HCHO)$.