Mix Examples-Hydrocarbon Questions in English

(B) The structure of $but-1-ene-3-yne$ is $CH_2=CH-C \equiv CH$.
Counting the bonds:
Sigma $( \sigma )$ bonds: There are $3$ $C-H$ bonds,$1$ $C-C$ single bond,$1$ $C=C$ double bond (contains $1$ $ \sigma $),and $1$ $C \equiv C$ triple bond (contains $1$ $ \sigma $). Total $ \sigma $ bonds = $3 + 1 + 1 + 1 + 1 = 7$.
Pi $( \pi )$ bonds: The $C=C$ bond contains $1$ $ \pi $ bond,and the $C \equiv C$ bond contains $2$ $ \pi $ bonds. Total $ \pi $ bonds = $1 + 2 = 3$.
Thus,the molecule has $7$ $ \sigma $ bonds and $3$ $ \pi $ bonds.

(A) The enolic form of acetone is propen$-2-$ol,which has the structure $CH_2=C(OH)-CH_3$.
Counting the bonds: there are $3$ $C-H$ bonds in the methyl group,$2$ $C-H$ bonds in the methylene group,$1$ $O-H$ bond,$1$ $C-O$ bond,$1$ $C-C$ bond,and $1$ $C=C$ bond (which consists of $1$ $\sigma$ and $1$ $\pi$ bond).
Total $\sigma$ bonds = $3 + 2 + 1 + 1 + 1 + 1 = 9$.
Total $\pi$ bonds = $1$.
Oxygen atom in the hydroxyl group has $2$ lone pairs.
Thus,the structure contains $9$ $\sigma$ bonds,$1$ $\pi$ bond,and $2$ lone pairs.

(A) The combustion of hydrocarbons is an exothermic process.
In an exothermic reaction,heat is released into the surroundings.
Therefore,the change in enthalpy,$\Delta H$,is always negative.

(B) The formation of a vast number of carbon compounds is primarily due to its unique property of $catenation$,which is the ability of atoms of the same element to link together to form long chains or rings.
Carbon exhibits the highest $catenation$ tendency among all elements in group $IV$ due to its small size and strong $C-C$ bond energy.

(D) The reaction between $(NH_4)_2SO_4$ and $KCNO$ proceeds as follows:
$(NH_4)_2SO_4 + 2KCNO \to 2NH_4CNO + K_2SO_4$
Upon heating,the ammonium cyanate $(NH_4CNO)$ undergoes rearrangement to form urea:
$NH_4CNO \xrightarrow{\Delta} NH_2CONH_2$
Therefore,the final product is urea.

(B) In laboratory burners,oil gas is generally used. Oil gas is a mixture of $CH_4$,$CO$,$CO_2$,and $H_2$.

(D) The given hybridisation sequence is $sp^3, sp^2, sp^2, sp^3, sp^2, sp^2, sp, sp$.
Counting the carbons:
$1(sp^3): CH_3-$
$2(sp^2): -CH=$
$3(sp^2): =CH-$
$4(sp^3): -CH_2-$
$5(sp^2): -CH=$
$6(sp^2): =CH-$
$7(sp): -C\equiv$
$8(sp): \equiv CH$
Combining these,we get $CH_3-CH=CH-CH_2-CH=CH-C\equiv CH$.
This matches option $D$.

(C)

$\text{Nature of bond}$	$\text{Number of sigma and } \pi \text{ bonds}$
$\text{Single bond}$	$1 \sigma, 0 \pi$
$\text{Double bond}$	$1 \sigma, 1 \pi$
$\text{Triple bond}$	$1 \sigma, 2 \pi$

In the given compound $CH_2=CH-CH=CH-C\equiv CH$:
There are two double bonds,each containing $1 \pi$ bond ($2 \times 1 = 2 \pi$ bonds).
There is one triple bond,which contains $2 \pi$ bonds ($1 \times 2 = 2 \pi$ bonds).
Total number of $\pi$ bonds = $2 + 2 = 4$.

(A) The octane number of a fuel is defined as the percentage of iso-octane in a mixture of iso-octane and $n$-heptane that has the same knocking characteristics as the fuel being tested.
Therefore,a petrol with an octane number of $80$ contains $80\%$ iso-octane and $20\% \, n$-heptane.

(A) Petroleum refining is the process of separating crude oil into various useful products such as gasoline,diesel,kerosene,and lubricating oils. This is primarily achieved through fractional distillation,where the components are separated based on their different boiling points.

(B) Petroleum ether is the correct answer because it has the lowest boiling point among the given fractions of petroleum.
During the fractional distillation of petroleum,components with lower boiling points vaporize first.

(D) Cyclohexane $(C_6H_{12})$ is a non-polar hydrocarbon.
It is immiscible with water and has a lower density than water $(d_{cyclohexane} \approx 0.779 \ g/cm^3 < d_{water} = 1.0 \ g/cm^3)$.
Therefore,it floats on the surface of water.

(D) Natural petroleum (crude oil) is a complex mixture of various hydrocarbons and other organic compounds.
It primarily consists of saturated hydrocarbons (alkanes),cyclic saturated hydrocarbons (cycloalkanes or naphthenes),and aromatic hydrocarbons.
Additionally,it contains non-hydrocarbon organic compounds that include elements like nitrogen,oxygen,and sulfur,along with trace amounts of metals such as iron,nickel,copper,and vanadium.
Therefore,all the given options are present in natural petroleum.

(B) The addition of $HCl$ to isobutylene $(CH_3-C(CH_3)=CH_2)$ follows Markovnikov's rule,where the electrophile $H^+$ adds to the terminal carbon to form a stable tertiary carbocation,which then reacts with $Cl^-$ to form $2$-chloro-$2$-methylpropane $(CH_3-CCl(CH_3)-CH_3)$.
For acetylene $(CH \equiv CH)$,the addition of $HCl$ occurs in two steps. The first step yields vinyl chloride $(CH_2=CHCl)$. The second step follows Markovnikov's rule,where the $H^+$ adds to the carbon already bearing the hydrogen,resulting in the formation of $1,1$-dichloroethane $(CH_3-CHCl_2)$.

(C) Gasoline,also known as petrol,is a mixture of volatile,flammable liquid hydrocarbons derived from petroleum. It is primarily used as a fuel for internal-combustion engines and also serves as a solvent for oils and fats.

(C) Petroleum refining involves the fractional distillation of crude oil based on boiling point ranges.
Kerosene is a middle distillate fraction obtained from petroleum.
The boiling range for the kerosene fraction is typically between $200 - 300 \ ^circ C$.
Therefore,the correct option is $C$.

(A) During the fractional distillation of crude oil,components are separated based on their boiling points.
As the temperature increases,the fractions with lower boiling points are collected first,followed by those with higher boiling points.
The boiling point ranges are approximately:
$1$. Gasoline: $40-175 \ ^\circ C$
$2$. Kerosene oil: $175-250 \ ^\circ C$
$3$. Diesel: $250-350 \ ^\circ C$
Therefore,the order of appearance with rising temperature is Gasoline,kerosene oil,and diesel.

(C) The complete oxidation (combustion) of any hydrocarbon,whether saturated or unsaturated,results in the formation of carbon dioxide and water as the final products.
For example,for methane $(CH_4)$:
$CH_4 + 2O_2 \to CO_2 + 2H_2O$
For ethene $(C_2H_4)$:
$C_2H_4 + 3O_2 \to 2CO_2 + 2H_2O$
Therefore,the correct option is $C$.

(B) The inorganic origin of petroleum,also known as the abiogenic theory,suggests that hydrocarbons were formed deep within the Earth's mantle long before the existence of life. One of the arguments for this theory is that $C$ and $H$ can combine by the absorption of solar energy to produce hydrocarbons.

(A) . Fractional distillation is the process used to separate hydrocarbons from petroleum based on the difference in their boiling points.

(B) The octane number of a fuel is defined as the percentage of iso-octane in a mixture of iso-octane and $n$-heptane that has the same knocking characteristics as the fuel sample.
Since the given sample contains $70\%$ iso-octane and $30\%$ $n$-heptane,the octane number of the sample is $70$.

(C) Petroleum fractions are separated based on their boiling points during fractional distillation.
Gasoline (also known as petrol) typically consists of hydrocarbons with $6-10$ carbon atoms and has a boiling range of approximately $70-200 \ ^\circ C$.
Natural gas consists of lower hydrocarbons $(C_1-C_4)$.
Gas oil and Kerosene have higher boiling ranges and higher carbon counts compared to gasoline.

(D) The boiling point of hydrocarbons depends on molecular weight,molecular shape,and branching.
For isomers,the boiling point decreases as the degree of branching increases because the surface area decreases,leading to weaker van der Waals forces.
Comparing the given compounds: $1-$Butene $(C_4H_8)$,$1-$Butyne $(C_4H_6)$,$n-$Butane $(C_4H_{10})$,and Isobutane $(C_4H_{10})$.
Isobutane ($2-$methylpropane) is a branched isomer of $n-$butane.
Among the alkanes,branched isomers have lower boiling points than straight-chain isomers.
Comparing the boiling points: $n-$Butane $(\approx 272.5 \ K) > \text{Isobutane} (\approx 261.4 \ K)$.
$1-$Butyne and $1-$Butene have higher boiling points due to the presence of $\pi$ bonds and higher molecular polarity compared to alkanes.
Therefore,Isobutane has the minimum boiling point among the given options.

(D) The octane number of fuel can be improved by increasing the percentage of branched-chain alkanes,alkenes,and aromatic hydrocarbons.
Thus,the octane number can be changed by processes such as isomerisation (reforming),alkylation,and aromatisation (cyclisation).

(C) The approximate composition of gasoline is $C_6-C_{11}$.
It has a boiling point range of $70-200\, ^\circ C$.
It is primarily used as motor fuel,in dry cleaning,and for the production of petrol gas.

(D) Acrylonitrile is an organic compound with the formula $CH_2=CHCN$.
It is also known as vinyl cyanide,cyanoethene,or prop-$2$-ene nitrile.
All these names refer to the same chemical structure where a vinyl group $(CH_2=CH-)$ is attached to a nitrile group $(-CN)$.
Therefore,all the given options are correct names for acrylonitrile.

(B) Both ethene $(CH_2=CH_2)$ and ethyne $(CH \equiv CH)$ are unsaturated hydrocarbons.
They both react with Baeyer's reagent (cold alkaline $KMnO_4$ solution),which results in the decolourisation of the purple solution and the formation of a brown precipitate of manganese dioxide $(MnO_2)$.

(A) Ethene is an alkene $(C_2H_4)$ and ethane is an alkane $(C_2H_6)$.
$A.$ Iodine in $CCl_4$ is generally not used for unsaturation tests because the reaction is reversible and slow.
$B.$ Bromine in $CCl_4$ (Bromine water test) is used to distinguish alkenes from alkanes. Alkenes decolorize the reddish-brown color of bromine,while alkanes do not.
$C.$ Alkaline $KMnO_4$ (Baeyer's reagent) is used to distinguish alkenes from alkanes. Alkenes decolorize the purple color of $KMnO_4$ to form a brown precipitate of $MnO_2$,while alkanes do not react.
$D.$ Ammoniacal $Cu_2Cl_2$ (Cuprous chloride) is used specifically to identify terminal alkynes (forming red precipitates of copper acetylides). It does not react with ethene or ethane.
Since both $A$ and $D$ are not used to distinguish ethene from ethane,but $D$ is a specific reagent for terminal alkynes,$A$ is the most appropriate choice in the context of standard laboratory tests for unsaturation where iodine is ineffective.

(D) Unsaturated hydrocarbons,such as alkenes ($CH_2=CH_2$,$CH_3-CH=CH_2$) and alkynes $(CH\equiv CH)$,undergo electrophilic addition reactions with halogens like chlorine $(Cl_2)$.
In these reactions,the $\pi$-bond breaks,and chlorine atoms are added across the double or triple bond.

(D) The general formula for the alkene homologous series is $C_nH_{2n}$.
$Ethene$ $(C_2H_4)$,$1-$butene $(C_4H_8)$,and $2-$butene $(C_4H_8)$ all belong to the alkene series.
$2-$butyne $(C_4H_6)$ belongs to the alkyne series with the general formula $C_nH_{2n-2}$.
Therefore,$2-$butyne is not a member of the alkene homologous series.

(C) The reactivity of hydrocarbons towards electrophilic addition reactions depends on the degree of unsaturation.
$C_2H_6$ (ethane) is a saturated hydrocarbon and is the least reactive.
$C_2H_4$ (ethene) and $C_2H_2$ (ethyne) are unsaturated.
Generally,the reactivity order for electrophilic addition is $C_2H_4 > C_2H_2$ because the $\pi$-electron density in the double bond of ethene is more accessible than in the triple bond of ethyne.
However,in the context of general chemical reactivity (acidity and addition),$C_2H_2$ is often considered more reactive than $C_2H_4$ due to the presence of acidic hydrogens and the high energy of the $\pi$-bonds.
Thus,the correct order is $C_2H_2 > C_2H_4 > C_2H_6$.

(C) The octane number is defined as the percentage by volume of iso-octane in a mixture of iso-octane and $n$-heptane that has the same knocking characteristics as the fuel being tested.
An octane number of $40$ means the mixture contains $40\%$ iso-octane and $(100 - 40) = 60\%$ $n$-heptane.
Therefore,the correct option is $C$.

(D) $CO$ (carbon monoxide) is a highly poisonous gas that binds to hemoglobin in the blood,preventing oxygen transport.

(A) The destructive distillation of wood involves heating wood in the absence of air.
This process yields wood gas,tar,charcoal,$CH_3OH$ (commonly known as wood alcohol),and $CH_3COOH$ (acetic acid or vinegar).
Therefore,the correct answer is $Wood$.

(C) Glycerol $(CH_2(OH)CH(OH)CH_2OH)$ undergoes dehydration when heated with $KHSO_4$ (a dehydrating agent).
This reaction involves the loss of two molecules of water to form $CH_2 = CH - CHO$,which is known as $Acrolein$.

(B) Power alcohol is a mixture of $80\%$ petrol and $20\%$ ethyl alcohol (ethanol). It is used as a fuel in internal combustion engines.

(C) Ketene $(CH_2=C=O)$ is produced by the thermal decomposition (pyrolysis) of acetone $(CH_3-CO-CH_3)$ at high temperatures $(700-800^\circ C)$.
The reaction is:
$CH_3-CO-CH_3 \xrightarrow{800^\circ C} CH_2=C=O + CH_4$.

(A) The chemical structure of $Pent-1-en-4-yne$ is $CH_2=CH-CH_2-C\equiv CH$.
Counting the bonds:
- $C-H$ bonds: $2+1+2+1 = 6$ $\sigma$ bonds.
- $C-C$ bonds: $4$ $\sigma$ bonds.
- Total $\sigma$ bonds = $6 + 4 = 10$.
- $\pi$ bonds: One in the double bond and two in the triple bond = $1 + 2 = 3$ $\pi$ bonds.
Therefore,there are $10$ $\sigma$ bonds and $3$ $\pi$ bonds.

(B) The dipole moment of a molecule depends on its symmetry and the polarity of its bonds.
$CH_3-C \equiv C-CH_3$ (but$-2-$yne) is a symmetric,linear molecule.
Due to its high degree of symmetry,the bond dipoles cancel each other out,resulting in a net dipole moment of zero.
Therefore,it has the lowest dipole moment among the given options.

(B) The chemical formula for isocyanic acid is $HNCO$.
When the hydrogen atom is replaced by a methyl group $(-CH_3)$,we get methyl isocyanate $(CH_3-N=C=O)$.
This compound is stable and is a well-known chemical intermediate.
Therefore,the product formed is methyl isocyanate.

(D) The structure of allyl isocyanide is $CH_2=CH-CH_2-N\equiv C:$.
Counting the bonds:
$1$. $C-H$ bonds: $5$
$2$. $C-C$ bonds: $2$
$3$. $C-N$ bond: $1$
$4$. $N\equiv C$ bond: $1$
Total $\sigma$ bonds = $5 + 2 + 1 + 1 = 9$.
Total $\pi$ bonds = $1$ (in $C=C$) + $2$ (in $N\equiv C$) = $3$.
There is $1$ lone pair on the terminal carbon atom of the isocyanide group.

(C) The reaction is an electrophilic aromatic substitution (bromination) of a bicyclic system containing an amide group. The $-NH-CO-$ group is an ortho/para-directing group. In the given structure,the positions ortho to the $-NH-$ group are blocked by the $-Me$ group or the ring fusion. The para position relative to the $-NH-$ group is available for substitution. Therefore,the bromine atom will be substituted at the position para to the $-NH-$ group,which corresponds to the position shown in option $C$.

(A) Baeyer's test (alkaline $KMnO_4$) is given by both alkenes and alkynes,resulting in the decolorization of the purple solution.
Ammoniacal silver nitrate (Tollens' reagent) gives a white precipitate only with terminal alkynes (alkynes with an acidic hydrogen).
$C_2H_4$ (ethene) is an alkene,so it gives a positive Baeyer's test but does not react with Tollens' reagent.
$C_3H_4$ (propyne) and $C_2H_2$ (ethyne) are terminal alkynes,so they give positive results for both tests.
$C_2H_6$ (ethane) is an alkane and does not give either test.

(A) The acidity of hydrocarbons depends on the percentage of $s$-character in the hybrid orbitals of the carbon atom attached to the hydrogen.
$s$-character $\propto$ Electronegativity of carbon $\propto$ Acidity.
For $1$-Alkyne ($sp$ hybridization): $50\% \ s$-character.
For Alkene ($sp^2$ hybridization): $33.3\% \ s$-character.
For Alkane ($sp^3$ hybridization): $25\% \ s$-character.
Therefore,the correct order of acidity is: $1$-Alkyne > Alkene > Alkane.

(A) $1$. Hydroxylation of ethyne $(C_2H_2)$ typically yields glyoxal or oxalic acid,not an alkene.
$2$. $C_2H_5OH \xrightarrow{HI/red, P, 150 ^\circ C}$ produces ethane $(C_2H_6)$,an alkane.
$3$. Butanone $\xrightarrow{Zn/Hg, HCl, \text{reflux}}$ is a Clemmensen reduction,producing butane $(C_4H_{10})$,an alkane.
$4$. Electrolysis of sodium propionate $(CH_3CH_2COONa)$ via Kolbe's electrolysis produces butane $(C_4H_{10})$,an alkane.
Since the question asks for a reaction that does not produce an alkene,and all options provided result in products other than alkenes,the question implies identifying the reaction that is chemically distinct or incorrectly framed. However,based on standard chemistry,none of these produce an alkene. Given the context of such multiple-choice questions,option $A$ is the most distinct as it involves oxidation,whereas others are reduction or coupling reactions.

(A) Reaction $(a)$ is an electrophilic addition of $HBr$ to an alkene,which follows the ionic mechanism (Markovnikov addition).
Reaction $(b)$ is the addition of $HBr$ to an alkene in the presence of peroxide,known as the Kharasch effect or peroxide effect,which proceeds via a free radical mechanism.
Reaction $(c)$ is the free radical substitution of methane with chlorine in the presence of ultraviolet light $(hv)$,which proceeds via a free radical mechanism.
Therefore,reactions $(b)$ and $(c)$ proceed via a free radical mechanism.

(C) $1$. An optically active alkane must contain at least one chiral carbon atom (a carbon bonded to four different groups).
$2$. Let's analyze the options:
- Option $A$: $CH_3-CH_2-C\equiv CH$ is an alkyne,not an alkane.
- Option $B$: $CH_3-CH_2-CH(CH_3)-CH_3$ is $2-methylbutane$. It is achiral because it has two identical methyl groups attached to the chiral center.
- Option $C$: The structure is $1-cyclopropyl-1-methylethane$. The central carbon is bonded to $H$,$CH_3$,$C_2H_5$,and a cyclopropyl group. Since all four groups are different,it is chiral and optically active. Its formula is $C_6H_{12}$,molecular weight = $72+12 = 84$.
- Option $D$: $CH_3-CH_2-CH_2-CH_3$ is $n-butane$,which is achiral.
$3$. Thus,the optically active alkane with the lowest molecular weight among the choices is the one in option $C$.

(B) Propene is an alkene $(CH_3-CH=CH_2)$ and contains a carbon-carbon double bond,which makes it unsaturated. Propane is an alkane $(CH_3-CH_2-CH_3)$ and is saturated.
Bromine water is a reddish-brown solution. When added to an alkene,the bromine adds across the double bond,causing the solution to decolorize.
Since propane is saturated,it does not react with bromine water,and the color remains unchanged.
Therefore,bromine water can be used to distinguish and separate (by reaction) propene from propane.

(B) hydrocarbon is an organic compound consisting entirely of hydrogen and carbon atoms.
$1$. Benzene $(C_6H_6)$ is a hydrocarbon.
$2$. Urea $(NH_2CONH_2)$ contains nitrogen and oxygen in addition to carbon and hydrogen,so it is not a hydrocarbon.
$3$. Ammonium cyanate $(NH_4OCN)$ is an inorganic compound.
$4$. Phenol $(C_6H_5OH)$ contains an oxygen atom,so it is not a hydrocarbon.
Since the question asks for a substance that is not a hydrocarbon,both $B$,$C$,and $D$ fit the criteria. However,in the context of introductory chemistry,$Urea$ $(NH_2CONH_2)$ is the most common example cited as a non-hydrocarbon organic compound. Given the options,$B$ is the most appropriate answer.

(A) In the catalytic hydrogenation (reduction) of unsaturated hydrocarbons like alkenes and alkynes,finely divided metals such as $Pt$,$Pd$,or $Ni$ are commonly used as catalysts. $Ni$ is often used in the form of Raney nickel. Therefore,$Pt/Ni$ is the most appropriate choice among the given options for general catalytic reduction.