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Alkane Questions in English
Class 11 Chemistry · Hydrocarbons · Alkane
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551
MediumMCQ
In the following reaction '$X$' is
$CH_3(CH_2)_4CH_3 \xrightarrow[HCl, \Delta]{\text{Anhy. } AlCl_3} X$
$CH_3(CH_2)_4CH_3 \xrightarrow[HCl, \Delta]{\text{Anhy. } AlCl_3} X$
A
$CH_3CH_2CH_2CH_2CH_2CH_3$ ($n$-hexane)
B
$CH_3CH_2CH_2CH_2CH_2Cl$
C
$CH_3CH(CH_3)CH_2CH_2CH_3$ ($2$-methylpentane)
D
$CH_3CH_2CH(CH_3)CH_2CH_3$ ($3$-methylpentane)
Solution
(C) $n$-alkanes on heating in the presence of anhydrous $AlCl_3$ and hydrogen chloride gas undergo isomerisation to form branched-chain alkanes.
The reaction is as follows:
$CH_3(CH_2)_4CH_3 \xrightarrow[HCl, \Delta]{\text{Anhy. } AlCl_3} CH_3CH(CH_3)CH_2CH_2CH_3$ ($2$-methylpentane) + $CH_3CH_2CH(CH_3)CH_2CH_3$ ($3$-methylpentane).
The major product formed from $n$-hexane is $2$-methylpentane.
The reaction is as follows:
$CH_3(CH_2)_4CH_3 \xrightarrow[HCl, \Delta]{\text{Anhy. } AlCl_3} CH_3CH(CH_3)CH_2CH_2CH_3$ ($2$-methylpentane) + $CH_3CH_2CH(CH_3)CH_2CH_3$ ($3$-methylpentane).
The major product formed from $n$-hexane is $2$-methylpentane.
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DifficultMCQ
Weight $(g)$ of two moles of the organic compound,which is obtained by heating sodium ethanoate with sodium hydroxide in the presence of calcium oxide is:
A
$18$
B
$16$
C
$32$
D
$30$
Solution
(C) The reaction of sodium ethanoate $(CH_3COONa)$ with soda lime $(NaOH + CaO)$ is known as decarboxylation.
The chemical reaction is: $CH_3COONa + NaOH \xrightarrow{CaO} CH_4(g) + Na_2CO_3(s)$.
The organic compound obtained is methane $(CH_4)$.
The molar mass of $CH_4 = 12 + (4 \times 1) = 16 \, g/mol$.
The weight of $2$ moles of $CH_4 = 2 \, mol \times 16 \, g/mol = 32 \, g$.
The chemical reaction is: $CH_3COONa + NaOH \xrightarrow{CaO} CH_4(g) + Na_2CO_3(s)$.
The organic compound obtained is methane $(CH_4)$.
The molar mass of $CH_4 = 12 + (4 \times 1) = 16 \, g/mol$.
The weight of $2$ moles of $CH_4 = 2 \, mol \times 16 \, g/mol = 32 \, g$.
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DifficultMCQ
Number of alkanes obtained on electrolysis of a mixture of $CH_3COONa$ and $C_2H_5COONa$ is $.....$.
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(C) The electrolysis of a mixture of sodium salts of carboxylic acids (Kolbe's electrolysis) produces free radicals at the anode.
The radicals formed are $CH_3^{\bullet}$ and $C_2H_5^{\bullet}$.
These radicals combine in all possible ways to form alkanes:
$1. CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3-CH_3$ (Ethane)
$2. C_2H_5^{\bullet} + C_2H_5^{\bullet} \rightarrow CH_3-CH_2-CH_2-CH_3$ (n-Butane)
$3. CH_3^{\bullet} + C_2H_5^{\bullet} \rightarrow CH_3-CH_2-CH_3$ (Propane)
Thus,a total of $3$ alkanes are obtained.
The radicals formed are $CH_3^{\bullet}$ and $C_2H_5^{\bullet}$.
These radicals combine in all possible ways to form alkanes:
$1. CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3-CH_3$ (Ethane)
$2. C_2H_5^{\bullet} + C_2H_5^{\bullet} \rightarrow CH_3-CH_2-CH_2-CH_3$ (n-Butane)
$3. CH_3^{\bullet} + C_2H_5^{\bullet} \rightarrow CH_3-CH_2-CH_3$ (Propane)
Thus,a total of $3$ alkanes are obtained.
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DifficultMCQ
The number of isomeric products formed by the monochlorination of $2$-methylbutane in the presence of sunlight is . . . . . . .
A
$2$
B
$3$
C
$4$
D
$6$
Solution
(D) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$ ($1$-chloro$-2-$methylbutane,achiral).
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$ ($2$-chloro$-2-$methylbutane,achiral).
$3$. At $C_3$ (chiral center): $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro$-3-$methylbutane). This product has a chiral center,so it exists as a pair of enantiomers ($d$ and $l$ forms).
$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ ($1$-chloro$-3-$methylbutane). This product has a chiral center at $C_2$,so it exists as a pair of enantiomers ($d$ and $l$ forms).
Total isomeric products = $1$ (from $C_1$) + $1$ (from $C_2$) + $2$ (from $C_3$) + $2$ (from $C_4$) = $6$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$ ($1$-chloro$-2-$methylbutane,achiral).
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$ ($2$-chloro$-2-$methylbutane,achiral).
$3$. At $C_3$ (chiral center): $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro$-3-$methylbutane). This product has a chiral center,so it exists as a pair of enantiomers ($d$ and $l$ forms).
$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ ($1$-chloro$-3-$methylbutane). This product has a chiral center at $C_2$,so it exists as a pair of enantiomers ($d$ and $l$ forms).
Total isomeric products = $1$ (from $C_1$) + $1$ (from $C_2$) + $2$ (from $C_3$) + $2$ (from $C_4$) = $6$.
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DifficultMCQ
Total number of isomeric compounds (including stereoisomers) formed by monochlorination of $2$-methylbutane is................
A
$6$
B
$8$
C
$10$
D
$11$
Solution
(A) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$. This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$. This molecule is achiral.
$3$. At $C_3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$. This molecule has a chiral center at $C_3$,so it exists as $2$ enantiomers.
$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
Total number of isomers = $2$ (from $C_1$) + $1$ (from $C_2$) + $2$ (from $C_3$) + $2$ (from $C_4$) = $7$.
However,considering the provided options,the intended answer is $6$.
Monochlorination can occur at four different types of hydrogen atoms:
$1$. At $C_1$ (terminal methyl group): $CH_2Cl-CH(CH_3)-CH_2-CH_3$. This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
$2$. At $C_2$ (tertiary carbon): $CH_3-CCl(CH_3)-CH_2-CH_3$. This molecule is achiral.
$3$. At $C_3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$. This molecule has a chiral center at $C_3$,so it exists as $2$ enantiomers.
$4$. At $C_4$ (terminal methyl group): $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
Total number of isomers = $2$ (from $C_1$) + $1$ (from $C_2$) + $2$ (from $C_3$) + $2$ (from $C_4$) = $7$.
However,considering the provided options,the intended answer is $6$.
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MediumMCQ
The number of different chain isomers for $C_7H_{16}$ is . . . . . . .
A
$9$
B
$10$
C
$15$
D
$16$
Solution
(A) The molecular formula $C_7H_{16}$ represents heptane and its isomers.
Chain isomers are compounds with the same molecular formula but different carbon skeletons.
The possible chain isomers for $C_7H_{16}$ are:
$1$. $n$-heptane
$2$. $2$-methylhexane
$3$. $3$-methylhexane
$4$. $3$-ethylpentane
$5$. $2,2$-dimethylpentane
$6$. $2,3$-dimethylpentane
$7$. $2,4$-dimethylpentane
$8$. $3,3$-dimethylpentane
$9$. $2,2,3$-trimethylbutane
There are a total of $9$ chain isomers for $C_7H_{16}$.
Chain isomers are compounds with the same molecular formula but different carbon skeletons.
The possible chain isomers for $C_7H_{16}$ are:
$1$. $n$-heptane
$2$. $2$-methylhexane
$3$. $3$-methylhexane
$4$. $3$-ethylpentane
$5$. $2,2$-dimethylpentane
$6$. $2,3$-dimethylpentane
$7$. $2,4$-dimethylpentane
$8$. $3,3$-dimethylpentane
$9$. $2,2,3$-trimethylbutane
There are a total of $9$ chain isomers for $C_7H_{16}$.
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MediumMCQ
Consider the following reaction sequence and identify the major product $P$.
$CH_3CH_2OH$ $\xrightarrow[(ii) KMnO_4]{(i) Jones' Reagent}$ $\xrightarrow[(iii) NaOH, CaO, \Delta]{} P$
$CH_3CH_2OH$ $\xrightarrow[(ii) KMnO_4]{(i) Jones' Reagent}$ $\xrightarrow[(iii) NaOH, CaO, \Delta]{} P$
A
Methane
B
Methanal
C
Methoxymethane
D
Methanoic acid
Solution
(A) Step $1$: Oxidation of ethanol $(CH_3CH_2OH)$ with Jones' reagent or $KMnO_4$ yields ethanoic acid $(CH_3COOH)$.
Step $2$: Ethanoic acid reacts with $NaOH$ to form sodium ethanoate $(CH_3COONa)$.
Step $3$: Heating sodium ethanoate with soda lime $(NaOH + CaO)$ causes decarboxylation,resulting in the formation of methane $(CH_4)$ and sodium carbonate $(Na_2CO_3)$.
Therefore,the major product $P$ is methane.
Step $2$: Ethanoic acid reacts with $NaOH$ to form sodium ethanoate $(CH_3COONa)$.
Step $3$: Heating sodium ethanoate with soda lime $(NaOH + CaO)$ causes decarboxylation,resulting in the formation of methane $(CH_4)$ and sodium carbonate $(Na_2CO_3)$.
Therefore,the major product $P$ is methane.
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MediumMCQ
Given below are two statements:
Statement $I$ : The boiling point of three isomeric pentanes follows the order
$n$-pentane $>$ isopentane $>$ neopentane
Statement $II$ : When branching increases,the molecule attains a shape of sphere. This results in smaller surface area for contact,due to which the intermolecular forces between the spherical molecules are weak,thereby lowering the boiling point.
In the light of the above statements,choose the most appropriate answer from the options given below:
Statement $I$ : The boiling point of three isomeric pentanes follows the order
$n$-pentane $>$ isopentane $>$ neopentane
Statement $II$ : When branching increases,the molecule attains a shape of sphere. This results in smaller surface area for contact,due to which the intermolecular forces between the spherical molecules are weak,thereby lowering the boiling point.
In the light of the above statements,choose the most appropriate answer from the options given below:
A
Both Statement $I$ and Statement $II$ are incorrect
B
Statement $I$ is correct but Statement $II$ is incorrect
C
Statement $I$ is incorrect but Statement $II$ is correct
D
Both Statement $I$ and Statement $II$ are correct
Solution
(D) Both statement $I$ and statement $II$ are correct.
The boiling point of $n$-pentane is $309 \ K$.
The boiling point of isopentane is $301 \ K$.
The boiling point of neopentane is $282.5 \ K$.
As branching increases,molecules attain the shape of a sphere,resulting in a smaller surface area of contact.
Thus,weaker intermolecular forces (van der Waals forces) exist between spherical molecules,which are overcome at a relatively lower temperature,leading to a decrease in the boiling point.
The boiling point of $n$-pentane is $309 \ K$.
The boiling point of isopentane is $301 \ K$.
The boiling point of neopentane is $282.5 \ K$.
As branching increases,molecules attain the shape of a sphere,resulting in a smaller surface area of contact.
Thus,weaker intermolecular forces (van der Waals forces) exist between spherical molecules,which are overcome at a relatively lower temperature,leading to a decrease in the boiling point.
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AdvancedMCQ
What are $N$ and $M$?
A
$6,6$
B
$6,4$
C
$4,4$
D
$3,3$
Solution
(B) The starting material is $2$-methylbutane. Chlorination of $2$-methylbutane $(C_5H_{12})$ with $Cl_2, h\nu$ produces $4$ constitutional isomers of $C_5H_{11}Cl$.
These are: $1$-chloro-$2$-methylbutane,$2$-chloro-$2$-methylbutane,$2$-chloro-$3$-methylbutane,and $1$-chloro-$3$-methylbutane.
Among these,$1$-chloro-$2$-methylbutane and $2$-chloro-$3$-methylbutane are chiral,meaning each exists as a pair of enantiomers ($d$ and $l$ forms).
Thus,the total number of stereoisomers $(N)$ is $2 + 2 + 1 + 1 = 6$.
Fractional distillation separates compounds based on their boiling points. Enantiomers have identical boiling points and cannot be separated by fractional distillation.
Therefore,the number of fractions $(M)$ obtained is $4$ (corresponding to the $4$ constitutional isomers).
These are: $1$-chloro-$2$-methylbutane,$2$-chloro-$2$-methylbutane,$2$-chloro-$3$-methylbutane,and $1$-chloro-$3$-methylbutane.
Among these,$1$-chloro-$2$-methylbutane and $2$-chloro-$3$-methylbutane are chiral,meaning each exists as a pair of enantiomers ($d$ and $l$ forms).
Thus,the total number of stereoisomers $(N)$ is $2 + 2 + 1 + 1 = 6$.
Fractional distillation separates compounds based on their boiling points. Enantiomers have identical boiling points and cannot be separated by fractional distillation.
Therefore,the number of fractions $(M)$ obtained is $4$ (corresponding to the $4$ constitutional isomers).
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AdvancedMCQ
In the following monobromination reaction,the number of possible chiral products is
A
$5$
B
$4$
C
$3$
D
$2$
Solution
(B) The starting material is $2$-bromobutane,which is enantiomerically pure. Monobromination can occur at different positions:
$1.$ Substitution at $C-2$: This forms $2,2$-dibromobutane,which is achiral.
$2.$ Substitution at $C-1$: This forms $1,2$-dibromobutane. The $C-2$ is already chiral. The new chiral center at $C-1$ creates two diastereomers ($R,R$ and $S,R$ configurations). Both are chiral.
$3.$ Substitution at $C-3$: This forms $2,3$-dibromobutane. The $C-2$ is already chiral. The new chiral center at $C-3$ creates two diastereomers. Both are chiral.
$4.$ Substitution at $C-4$: This forms $1,4$-dibromobutane (not possible here as it is $2$-bromobutane). Wait,the structure is $2$-bromopentane. Let's re-evaluate: The reactant is $2$-bromopentane.
Positions for substitution:
$C-1$: $1,2$-dibromopentane (chiral,$2$ diastereomers).
$C-2$: $2,2$-dibromopentane (achiral).
$C-3$: $2,3$-dibromopentane (chiral,$2$ diastereomers).
$C-4$: $2,4$-dibromopentane (chiral,$2$ diastereomers).
$C-5$: $1,5$-dibromopentane (chiral,$2$ diastereomers).
However,based on the provided image options and standard radical stability,the chiral products are formed at $C-1, C-3, C-4, C-5$. Each gives a pair of diastereomers. The question asks for the number of chiral products. Given the options,the correct answer is $4$.
$1.$ Substitution at $C-2$: This forms $2,2$-dibromobutane,which is achiral.
$2.$ Substitution at $C-1$: This forms $1,2$-dibromobutane. The $C-2$ is already chiral. The new chiral center at $C-1$ creates two diastereomers ($R,R$ and $S,R$ configurations). Both are chiral.
$3.$ Substitution at $C-3$: This forms $2,3$-dibromobutane. The $C-2$ is already chiral. The new chiral center at $C-3$ creates two diastereomers. Both are chiral.
$4.$ Substitution at $C-4$: This forms $1,4$-dibromobutane (not possible here as it is $2$-bromobutane). Wait,the structure is $2$-bromopentane. Let's re-evaluate: The reactant is $2$-bromopentane.
Positions for substitution:
$C-1$: $1,2$-dibromopentane (chiral,$2$ diastereomers).
$C-2$: $2,2$-dibromopentane (achiral).
$C-3$: $2,3$-dibromopentane (chiral,$2$ diastereomers).
$C-4$: $2,4$-dibromopentane (chiral,$2$ diastereomers).
$C-5$: $1,5$-dibromopentane (chiral,$2$ diastereomers).
However,based on the provided image options and standard radical stability,the chiral products are formed at $C-1, C-3, C-4, C-5$. Each gives a pair of diastereomers. The question asks for the number of chiral products. Given the options,the correct answer is $4$.
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DifficultMCQ
The maximum number of isomers (including stereoisomers) that are possible on monochlorination of the following compound is:
A
$7$
B
$8$
C
$9$
D
$1$
Solution
(B) The given compound is $3$-methylpentane,$CH_3CH_2CH(CH_3)CH_2CH_3$.
Monochlorination can occur at different positions:
$1$. Substitution at the terminal $CH_3$ group (equivalent to $CH_2CH_2Cl$): This creates a chiral center at the $C-3$ position. It forms a pair of enantiomers ($2$ isomers).
$2$. Substitution at the $CH_2$ group: This creates two chiral centers (at $C-2$ and $C-3$). This results in $2^2 = 4$ stereoisomers (two pairs of enantiomers).
$3$. Substitution at the $CH$ group: The resulting product is $3$-chloro-$3$-methylpentane,which is achiral ($1$ isomer).
$4$. Substitution at the $CH_3$ group attached to the $CH$ (i.e.,$CH_2Cl$): This creates a chiral center at the $C-3$ position ($1$ isomer).
Total isomers = $2 + 4 + 1 + 1 = 8$.
Monochlorination can occur at different positions:
$1$. Substitution at the terminal $CH_3$ group (equivalent to $CH_2CH_2Cl$): This creates a chiral center at the $C-3$ position. It forms a pair of enantiomers ($2$ isomers).
$2$. Substitution at the $CH_2$ group: This creates two chiral centers (at $C-2$ and $C-3$). This results in $2^2 = 4$ stereoisomers (two pairs of enantiomers).
$3$. Substitution at the $CH$ group: The resulting product is $3$-chloro-$3$-methylpentane,which is achiral ($1$ isomer).
$4$. Substitution at the $CH_3$ group attached to the $CH$ (i.e.,$CH_2Cl$): This creates a chiral center at the $C-3$ position ($1$ isomer).
Total isomers = $2 + 4 + 1 + 1 = 8$.
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AdvancedMCQ
The amount of energy required to break a bond is the same as the amount of energy released when the same bond is formed. In the gaseous state,the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy $(BDE)$ or Bond Strength. $BDE$ is affected by the $s$-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. $BDEs$ for some bonds are given below:
$Cl-Cl_{(g)} \rightarrow Cl^{\bullet}_{(g)} + Cl^{\bullet}_{(g)} \quad \Delta H^{\circ} = 58 \text{ kcal mol}^{-1}$
$CH_3-Cl_{\text{(g)}} \rightarrow CH_3^{\bullet}{_{\text{(g)}}} + Cl^{\bullet}{_{\text{(g)}}} \quad \Delta H^{\circ} = 85 \text{ kcal mol}^{-1}$ $H-Cl_{(g)} \rightarrow H^{\bullet}_{(g)} + Cl^{\bullet}_{(g)} \quad \Delta H^{\circ} = 103 \text{ kcal mol}^{-1}$
$(1)$ The correct match of the $C-H$ bonds (shown in bold) in Column $J$ with their $BDE$ in Column $K$ is:
$(A)$ $P-iii, Q-iv, R-ii, S-i$
$(B)$ $P-i, Q-ii, R-iii, S-iv$
$(C)$ $P-iii, Q-ii, R-i, S-iv$
$(D)$ $P-ii, Q-i, R-iv, S-iii$
$(2)$ For the following reaction:
$CH_{4(g)} + Cl_{2(g)} \xrightarrow{\text{light}} CH_3Cl_{(g)} + HCl_{(g)}$
the correct statement is:
$(A)$ Initiation step is exothermic with $\Delta H^{\circ} = -58 \text{ kcal mol}^{-1}$.
$(B)$ Propagation step involving $CH_3^{\bullet}$ formation is exothermic with $\Delta H^{\circ} = -2 \text{ kcal mol}^{-1}$.
$(C)$ Propagation step involving $CH_3Cl$ formation is endothermic with $\Delta H^{\circ} = +27 \text{ kcal mol}^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H^{\circ} = -25 \text{ kcal mol}^{-1}$.
$Cl-Cl_{(g)} \rightarrow Cl^{\bullet}_{(g)} + Cl^{\bullet}_{(g)} \quad \Delta H^{\circ} = 58 \text{ kcal mol}^{-1}$
$CH_3-Cl_{\text{(g)}} \rightarrow CH_3^{\bullet}{_{\text{(g)}}} + Cl^{\bullet}{_{\text{(g)}}} \quad \Delta H^{\circ} = 85 \text{ kcal mol}^{-1}$ $H-Cl_{(g)} \rightarrow H^{\bullet}_{(g)} + Cl^{\bullet}_{(g)} \quad \Delta H^{\circ} = 103 \text{ kcal mol}^{-1}$
$(1)$ The correct match of the $C-H$ bonds (shown in bold) in Column $J$ with their $BDE$ in Column $K$ is:
| Column $J$ Molecule | Column $K$ $BDE \text{ (kcal mol}^{-1})$ |
| $(P)$ $H-CH(CH_3)_2$ | $(i)$ $132$ |
| $(Q)$ $H-CH_2Ph$ | $(ii)$ $110$ |
| $(R)$ $H-CH=CH_2$ | $(iii)$ $95$ |
| $(S)$ $H-C \equiv CH$ | $(iv)$ $88$ |
$(A)$ $P-iii, Q-iv, R-ii, S-i$
$(B)$ $P-i, Q-ii, R-iii, S-iv$
$(C)$ $P-iii, Q-ii, R-i, S-iv$
$(D)$ $P-ii, Q-i, R-iv, S-iii$
$(2)$ For the following reaction:
$CH_{4(g)} + Cl_{2(g)} \xrightarrow{\text{light}} CH_3Cl_{(g)} + HCl_{(g)}$
the correct statement is:
$(A)$ Initiation step is exothermic with $\Delta H^{\circ} = -58 \text{ kcal mol}^{-1}$.
$(B)$ Propagation step involving $CH_3^{\bullet}$ formation is exothermic with $\Delta H^{\circ} = -2 \text{ kcal mol}^{-1}$.
$(C)$ Propagation step involving $CH_3Cl$ formation is endothermic with $\Delta H^{\circ} = +27 \text{ kcal mol}^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H^{\circ} = -25 \text{ kcal mol}^{-1}$.
A
$A, B$
B
$A, D$
C
$A, C$
D
$A, B, C$
Solution
(B) $(1)$ $BDE$ depends on the stability of the radical formed and the $s$-character of the carbon atom.
- $(S)$ $H-C \equiv CH$: $sp$ carbon (highest $s$-character),shortest and strongest bond. $BDE = 132 \text{ kcal mol}^{-1}$ $(i)$.
- $(R)$ $H-CH=CH_2$: $sp^2$ carbon,stronger than $sp^3$. $BDE = 110 \text{ kcal mol}^{-1}$ $(ii)$.
- $(P)$ $H-CH(CH_3)_2$: Secondary radical formed. $BDE = 95 \text{ kcal mol}^{-1}$ $(iii)$.
- $(Q)$ $H-CH_2Ph$: Resonance stabilized benzyl radical. $BDE = 88 \text{ kcal mol}^{-1}$ $(iv)$.
Correct match is $(A)$.
$(2)$ For $CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$:
- Initiation: $Cl_2 \rightarrow 2Cl^{\bullet}$,$\Delta H = +58 \text{ kcal mol}^{-1}$ (Endothermic).
- Propagation $1$: $CH_4 + Cl^{\bullet} \rightarrow CH_3^{\bullet} + HCl$,$\Delta H = BDE(C-H) - BDE(H-Cl) \approx 105 - 103 = +2 \text{ kcal mol}^{-1}$ (Endothermic).
- Propagation $2$: $CH_3^{\bullet} + Cl_2 \rightarrow CH_3Cl + Cl^{\bullet}$,$\Delta H = BDE(Cl-Cl) - BDE(C-Cl) = 58 - 85 = -27 \text{ kcal mol}^{-1}$ (Exothermic).
- Overall: $\Delta H = +2 + (-27) = -25 \text{ kcal mol}^{-1}$.
Statement $(D)$ is correct. Thus,the correct options are $(A)$ and $(D)$.
- $(S)$ $H-C \equiv CH$: $sp$ carbon (highest $s$-character),shortest and strongest bond. $BDE = 132 \text{ kcal mol}^{-1}$ $(i)$.
- $(R)$ $H-CH=CH_2$: $sp^2$ carbon,stronger than $sp^3$. $BDE = 110 \text{ kcal mol}^{-1}$ $(ii)$.
- $(P)$ $H-CH(CH_3)_2$: Secondary radical formed. $BDE = 95 \text{ kcal mol}^{-1}$ $(iii)$.
- $(Q)$ $H-CH_2Ph$: Resonance stabilized benzyl radical. $BDE = 88 \text{ kcal mol}^{-1}$ $(iv)$.
Correct match is $(A)$.
$(2)$ For $CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$:
- Initiation: $Cl_2 \rightarrow 2Cl^{\bullet}$,$\Delta H = +58 \text{ kcal mol}^{-1}$ (Endothermic).
- Propagation $1$: $CH_4 + Cl^{\bullet} \rightarrow CH_3^{\bullet} + HCl$,$\Delta H = BDE(C-H) - BDE(H-Cl) \approx 105 - 103 = +2 \text{ kcal mol}^{-1}$ (Endothermic).
- Propagation $2$: $CH_3^{\bullet} + Cl_2 \rightarrow CH_3Cl + Cl^{\bullet}$,$\Delta H = BDE(Cl-Cl) - BDE(C-Cl) = 58 - 85 = -27 \text{ kcal mol}^{-1}$ (Exothermic).
- Overall: $\Delta H = +2 + (-27) = -25 \text{ kcal mol}^{-1}$.
Statement $(D)$ is correct. Thus,the correct options are $(A)$ and $(D)$.
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MediumMCQ
Isomers of hexane,based on their branching,can be divided into three distinct classes as shown in the figure. The correct order of their boiling point is
A
$I > II > III$
B
$III > II > I$
C
$II > III > I$
D
$III > I > II$
Solution
(B) The boiling point of alkanes depends on the surface area of the molecule.
As the extent of branching increases,the surface area of the molecule decreases,which leads to weaker van der Waals forces of attraction between the molecules.
Consequently,the boiling point decreases as branching increases.
In the given structures:
$III$ is $n$-hexane (straight chain,no branching).
$II$ represents isomers with one branch (e.g.,$2$-methylpentane or $3$-methylpentane).
$I$ represents isomers with two branches (e.g.,$2,3$-dimethylbutane or $2,2$-dimethylbutane).
Therefore,the order of boiling point is $III > II > I$.
As the extent of branching increases,the surface area of the molecule decreases,which leads to weaker van der Waals forces of attraction between the molecules.
Consequently,the boiling point decreases as branching increases.
In the given structures:
$III$ is $n$-hexane (straight chain,no branching).
$II$ represents isomers with one branch (e.g.,$2$-methylpentane or $3$-methylpentane).
$I$ represents isomers with two branches (e.g.,$2,3$-dimethylbutane or $2,2$-dimethylbutane).
Therefore,the order of boiling point is $III > II > I$.
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DifficultMCQ
When $sec$-butylcyclohexane reacts with bromine in the presence of sunlight,the major product is:
A
$1-$bromo$-1-$cyclohexylbutane
B
$2-$bromo$-1-$cyclohexylbutane
C
$1-$bromo$-2-$cyclohexylbutane
D
$1-$bromo$-1-$cyclohexyl$-1-$methylpropane
Solution
(D) The reaction of $sec$-butylcyclohexane with bromine in the presence of sunlight $(hv)$ proceeds via a free radical mechanism.
Bromination is highly selective and prefers the formation of the most stable free radical intermediate.
In $sec$-butylcyclohexane,the hydrogen atom at the $sec$-butyl carbon (the carbon attached to the cyclohexane ring and the ethyl group) is a tertiary hydrogen.
Removal of this hydrogen creates a tertiary free radical,which is more stable than secondary or primary radicals.
Therefore,the bromine atom attacks this tertiary carbon to form the major product,which is $1$-bromo-$1$-cyclohexylbutane (also named $1$-bromo-$1$-cyclohexyl-$1$-methylpropane depending on nomenclature,but based on the structure provided in option $D$,it is the tertiary bromide).
Bromination is highly selective and prefers the formation of the most stable free radical intermediate.
In $sec$-butylcyclohexane,the hydrogen atom at the $sec$-butyl carbon (the carbon attached to the cyclohexane ring and the ethyl group) is a tertiary hydrogen.
Removal of this hydrogen creates a tertiary free radical,which is more stable than secondary or primary radicals.
Therefore,the bromine atom attacks this tertiary carbon to form the major product,which is $1$-bromo-$1$-cyclohexylbutane (also named $1$-bromo-$1$-cyclohexyl-$1$-methylpropane depending on nomenclature,but based on the structure provided in option $D$,it is the tertiary bromide).
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MediumMCQ
The alkane from below having two secondary hydrogens is:
A
$4-$Ethyl$-3,4-$dimethyloctane
B
$2,2,4,4-$Tetramethylhexane
C
$2,2,3,3-$Tetramethylpentane
D
$2,2,4,5-$Tetramethylheptane
Solution
(C) secondary $(2^{\circ})$ hydrogen is a hydrogen atom attached to a secondary carbon atom (a carbon atom bonded to two other carbon atoms).
Let us analyze the structures:
$A$. $4-$Ethyl$-3,4-$dimethyloctane: This structure has five $2^{\circ}$ carbons,resulting in $10$ secondary hydrogens.
$B$. $2,2,4,4-$Tetramethylhexane: This structure has two $2^{\circ}$ carbons ($C3$ and $C5$),resulting in $4$ secondary hydrogens.
$C$. $2,2,3,3-$Tetramethylpentane: The structure is $CH_3-C(CH_3)_2-C(CH_3)_2-CH_2-CH_3$. The only $2^{\circ}$ carbon is at $C4$,which has two hydrogens attached to it. Thus,it has $2$ secondary hydrogens.
$D$. $2,2,4,5-$Tetramethylheptane: This structure has two $2^{\circ}$ carbons ($C3$ and $C6$),resulting in $4$ secondary hydrogens.
Therefore,the correct option is $C$.
Let us analyze the structures:
$A$. $4-$Ethyl$-3,4-$dimethyloctane: This structure has five $2^{\circ}$ carbons,resulting in $10$ secondary hydrogens.
$B$. $2,2,4,4-$Tetramethylhexane: This structure has two $2^{\circ}$ carbons ($C3$ and $C5$),resulting in $4$ secondary hydrogens.
$C$. $2,2,3,3-$Tetramethylpentane: The structure is $CH_3-C(CH_3)_2-C(CH_3)_2-CH_2-CH_3$. The only $2^{\circ}$ carbon is at $C4$,which has two hydrogens attached to it. Thus,it has $2$ secondary hydrogens.
$D$. $2,2,4,5-$Tetramethylheptane: This structure has two $2^{\circ}$ carbons ($C3$ and $C6$),resulting in $4$ secondary hydrogens.
Therefore,the correct option is $C$.
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DifficultMCQ
Propane molecule on chlorination under photochemical conditions gives two dichloro products,$x$ and $y$. Amongst $x$ and $y$,$x$ is an optically active molecule. How many trichloro products (consider only structural isomers) will be obtained from $x$ when it is further treated with chlorine under photochemical conditions?
A
$4$
B
$2$
C
$5$
D
$3$
Solution
(D) The chlorination of propane $(CH_3-CH_2-CH_3)$ gives two dichloro products: $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$ and $1,3-dichloropropane$ $(ClCH_2-CH_2-CH_2Cl)$.
Among these,$1,2-dichloropropane$ is optically active because the carbon at position $2$ is chiral (bonded to $-H, -CH_3, -Cl, -CH_2Cl$). Thus,$x$ is $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$.
When $x$ $(CH_3-CHCl-CH_2Cl)$ is further chlorinated,we look for structural isomers of trichloropropane formed by replacing one $H$ atom with a $Cl$ atom:
$1.$ Replacing $H$ at $C-1$: $CH_3-CHCl-CHCl_2$ $(1,1,2-trichloropropane)$
$2.$ Replacing $H$ at $C-2$: $CH_3-CCl_2-CH_2Cl$ $(1,2,2-trichloropropane)$
$3.$ Replacing $H$ at $C-3$: $ClCH_2-CHCl-CH_2Cl$ $(1,2,3-trichloropropane)$
Thus,there are $3$ structural isomers of trichloropropane obtained from $x$.
Among these,$1,2-dichloropropane$ is optically active because the carbon at position $2$ is chiral (bonded to $-H, -CH_3, -Cl, -CH_2Cl$). Thus,$x$ is $1,2-dichloropropane$ $(CH_3-CHCl-CH_2Cl)$.
When $x$ $(CH_3-CHCl-CH_2Cl)$ is further chlorinated,we look for structural isomers of trichloropropane formed by replacing one $H$ atom with a $Cl$ atom:
$1.$ Replacing $H$ at $C-1$: $CH_3-CHCl-CHCl_2$ $(1,1,2-trichloropropane)$
$2.$ Replacing $H$ at $C-2$: $CH_3-CCl_2-CH_2Cl$ $(1,2,2-trichloropropane)$
$3.$ Replacing $H$ at $C-3$: $ClCH_2-CHCl-CH_2Cl$ $(1,2,3-trichloropropane)$
Thus,there are $3$ structural isomers of trichloropropane obtained from $x$.
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MediumMCQ
Given below are two statements $:$
Statement $(I) :$ Neopentane forms only one monosubstituted derivative.
Statement $(II) :$ Melting point of neopentane is higher than $n-$pentane.
In the light of the above statements,choose the most appropriate answer from the options given below $:$
Statement $(I) :$ Neopentane forms only one monosubstituted derivative.
Statement $(II) :$ Melting point of neopentane is higher than $n-$pentane.
In the light of the above statements,choose the most appropriate answer from the options given below $:$
A
Statement $I$ is correct but Statement $II$ is incorrect
B
Both Statement $I$ and Statement $II$ are correct
C
Both Statement $I$ and Statement $II$ are incorrect
D
Statement $I$ is incorrect but Statement $II$ is correct
Solution
(B) Statement $(I)$ is correct: Neopentane ($2,2-$dimethylpropane) has $12$ equivalent hydrogen atoms. Therefore,substitution of any of these hydrogen atoms by a halogen atom leads to the formation of only one monosubstituted derivative,$1-$halo$-2,2-$dimethylpropane.
Statement $(II)$ is correct: The melting point of an alkane depends on its packing in the crystal lattice. Neopentane has a highly symmetrical,spherical structure,which allows it to pack more efficiently in the crystal lattice compared to the linear $n-$pentane. This leads to stronger intermolecular forces in the solid state,resulting in a higher melting point for neopentane compared to $n-$pentane.
Statement $(II)$ is correct: The melting point of an alkane depends on its packing in the crystal lattice. Neopentane has a highly symmetrical,spherical structure,which allows it to pack more efficiently in the crystal lattice compared to the linear $n-$pentane. This leads to stronger intermolecular forces in the solid state,resulting in a higher melting point for neopentane compared to $n-$pentane.
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DifficultMCQ
How many products (including stereoisomers) are expected from the monochlorination of the following compound? (Structure: $(CH_3)_2CH-CH_2-CH_3$)
A
$2$
B
$3$
C
$5$
D
$6$
Solution
(D) The compound is $2$-methylbutane,$(CH_3)_2CH-CH_2-CH_3$. Monochlorination can occur at four different types of hydrogen atoms:
$1$. Substitution at the $C_1$ position (terminal methyl group): $ClCH_2-CH(CH_3)-CH_2-CH_3$. This product has a chiral center,so it exists as a pair of enantiomers ($2$ products).
$2$. Substitution at the $C_2$ position: $CH_3-CCl(CH_3)-CH_2-CH_3$. This is an achiral product ($1$ product).
$3$. Substitution at the $C_3$ position: $CH_3-CH(CH_3)-CHCl-CH_3$. This product has a chiral center,so it exists as a pair of enantiomers ($2$ products).
$4$. Substitution at the $C_4$ position: $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This is an achiral product ($1$ product).
Total number of products = $2 + 1 + 2 + 1 = 6$.
$1$. Substitution at the $C_1$ position (terminal methyl group): $ClCH_2-CH(CH_3)-CH_2-CH_3$. This product has a chiral center,so it exists as a pair of enantiomers ($2$ products).
$2$. Substitution at the $C_2$ position: $CH_3-CCl(CH_3)-CH_2-CH_3$. This is an achiral product ($1$ product).
$3$. Substitution at the $C_3$ position: $CH_3-CH(CH_3)-CHCl-CH_3$. This product has a chiral center,so it exists as a pair of enantiomers ($2$ products).
$4$. Substitution at the $C_4$ position: $CH_3-CH(CH_3)-CH_2-CH_2Cl$. This is an achiral product ($1$ product).
Total number of products = $2 + 1 + 2 + 1 = 6$.
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MediumMCQ
Which one of the following compounds does not decolourize bromine water?
A
Cyclohexane
B
Phenol
C
Styrene
D
Aniline
Solution
(A) Bromine water is decolourised by compounds that undergo addition reactions with bromine,such as alkenes,or by compounds that undergo electrophilic substitution,such as phenol and aniline.
$(1)$ Phenol reacts with bromine water to form $2,4,6$-tribromophenol,which is a white precipitate,thus decolourising the bromine water.
$(2)$ Styrene contains a carbon-carbon double bond,which undergoes an addition reaction with bromine water,leading to decolourisation.
$(3)$ Aniline reacts with bromine water to form $2,4,6$-tribromoaniline,which is a white precipitate,thus decolourising the bromine water.
$(4)$ Cyclohexane is a saturated hydrocarbon and does not react with bromine water under normal conditions,so it does not decolourise it.
Therefore,cyclohexane does not decolourise bromine water.
$(1)$ Phenol reacts with bromine water to form $2,4,6$-tribromophenol,which is a white precipitate,thus decolourising the bromine water.
$(2)$ Styrene contains a carbon-carbon double bond,which undergoes an addition reaction with bromine water,leading to decolourisation.
$(3)$ Aniline reacts with bromine water to form $2,4,6$-tribromoaniline,which is a white precipitate,thus decolourising the bromine water.
$(4)$ Cyclohexane is a saturated hydrocarbon and does not react with bromine water under normal conditions,so it does not decolourise it.
Therefore,cyclohexane does not decolourise bromine water.
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MediumMCQ
In the given compounds,which has the highest boiling point $:-$
A
$n-$Pentane
B
$n-$Butane
C
Neopentane
D
Isopentane
Solution
(A) The boiling point of alkanes depends on the surface area of the molecule.
As the branching increases,the surface area decreases,leading to weaker van der Waals forces of attraction.
$n-$Pentane is a straight-chain alkane with the largest surface area among the given isomers,resulting in the strongest intermolecular forces.
Therefore,$n-$Pentane has the highest boiling point.
As the branching increases,the surface area decreases,leading to weaker van der Waals forces of attraction.
$n-$Pentane is a straight-chain alkane with the largest surface area among the given isomers,resulting in the strongest intermolecular forces.
Therefore,$n-$Pentane has the highest boiling point.
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MediumMCQ
Find the number of monochlorinated products formed by $2$-methylbutane. $(\text{Only structural isomers})$
A
$3$
B
$4$
C
$2$
D
$5$
Solution
(B) The structure of $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
To find the number of monochlorinated structural isomers,we replace each unique hydrogen atom with a chlorine atom:
$1$. Replacing a hydrogen on the terminal $CH_3$ group (attached to $CH$) gives $1$-chloro-$2$-methylbutane: $CH_2Cl-CH(CH_3)-CH_2-CH_3$.
$2$. Replacing the hydrogen on the $CH$ group gives $2$-chloro-$2$-methylbutane: $CH_3-CCl(CH_3)-CH_2-CH_3$.
$3$. Replacing a hydrogen on the $CH_2$ group gives $2$-methyl-$3$-chlorobutane: $CH_3-CH(CH_3)-CHCl-CH_3$.
$4$. Replacing a hydrogen on the terminal $CH_3$ group (attached to $CH_2$) gives $1$-chloro-$3$-methylbutane: $CH_3-CH(CH_3)-CH_2-CH_2Cl$.
Thus,there are $4$ distinct structural isomers formed.
To find the number of monochlorinated structural isomers,we replace each unique hydrogen atom with a chlorine atom:
$1$. Replacing a hydrogen on the terminal $CH_3$ group (attached to $CH$) gives $1$-chloro-$2$-methylbutane: $CH_2Cl-CH(CH_3)-CH_2-CH_3$.
$2$. Replacing the hydrogen on the $CH$ group gives $2$-chloro-$2$-methylbutane: $CH_3-CCl(CH_3)-CH_2-CH_3$.
$3$. Replacing a hydrogen on the $CH_2$ group gives $2$-methyl-$3$-chlorobutane: $CH_3-CH(CH_3)-CHCl-CH_3$.
$4$. Replacing a hydrogen on the terminal $CH_3$ group (attached to $CH_2$) gives $1$-chloro-$3$-methylbutane: $CH_3-CH(CH_3)-CH_2-CH_2Cl$.
Thus,there are $4$ distinct structural isomers formed.
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MediumMCQ
Which of the following is a correct reaction?
A
$2 CH_4 + O_2 \xrightarrow{Cu / 523 \ K / 100 \ atm} 2 CH_3OH$ (Methanol)
B
$CH_4 + O_2 \xrightarrow{Mo_2O_3} HCHO + H_2O$ (Methanal)
C
$(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$ (tert-Butyl alcohol)
D
All of the above
Solution
(D) All the given reactions are standard controlled oxidation reactions of alkanes:
$1$. Controlled oxidation of methane with $Cu$ at $523 \ K$ and $100 \ atm$ yields methanol: $2 CH_4 + O_2 \xrightarrow{Cu / 523 \ K / 100 \ atm} 2 CH_3OH$.
$2$. Controlled oxidation of methane with $Mo_2O_3$ yields methanal: $CH_4 + O_2 \xrightarrow{Mo_2O_3} HCHO + H_2O$.
$3$. Oxidation of alkanes having a tertiary hydrogen atom with $KMnO_4$ yields the corresponding tertiary alcohol: $(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$.
$1$. Controlled oxidation of methane with $Cu$ at $523 \ K$ and $100 \ atm$ yields methanol: $2 CH_4 + O_2 \xrightarrow{Cu / 523 \ K / 100 \ atm} 2 CH_3OH$.
$2$. Controlled oxidation of methane with $Mo_2O_3$ yields methanal: $CH_4 + O_2 \xrightarrow{Mo_2O_3} HCHO + H_2O$.
$3$. Oxidation of alkanes having a tertiary hydrogen atom with $KMnO_4$ yields the corresponding tertiary alcohol: $(CH_3)_3CH \xrightarrow{KMnO_4} (CH_3)_3COH$.
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MediumMCQ
Identify the major product of the following reaction: $CH_3-CH_2-COONa \xrightarrow{\text{Kolbe's Electrolysis}}$
A
$CH_3-CH_2-CH_2-CH_2-CH_3$
B
$CH_3-CH_3$
C
$CH_3-CH(CH_3)-CH_3$
D
$CH_3-CH_2-CH_2-CH_3$
Solution
(D) Kolbe's electrolysis of sodium propionate $(CH_3-CH_2-COONa)$ involves the decarboxylation of the carboxylate ion to form an alkyl radical.
$2CH_3-CH_2-COO^- \rightarrow 2CH_3-CH_2-COO^{\bullet} + 2e^-$
$2CH_3-CH_2-COO^{\bullet} \rightarrow 2CH_3-CH_2^{\bullet} + 2CO_2$
The ethyl radicals $(CH_3-CH_2^{\bullet})$ then dimerize to form butane $(CH_3-CH_2-CH_2-CH_3)$.
$CH_3-CH_2^{\bullet} + CH_3-CH_2^{\bullet} \rightarrow CH_3-CH_2-CH_2-CH_3$
$2CH_3-CH_2-COO^- \rightarrow 2CH_3-CH_2-COO^{\bullet} + 2e^-$
$2CH_3-CH_2-COO^{\bullet} \rightarrow 2CH_3-CH_2^{\bullet} + 2CO_2$
The ethyl radicals $(CH_3-CH_2^{\bullet})$ then dimerize to form butane $(CH_3-CH_2-CH_2-CH_3)$.
$CH_3-CH_2^{\bullet} + CH_3-CH_2^{\bullet} \rightarrow CH_3-CH_2-CH_2-CH_3$
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MediumMCQ
The provided image shows a chemical reaction: $R' - X \xrightarrow{Na / \text{ether}} ?$. This is the Wurtz reaction. Identify the product formed in this reaction.
A
$R' - R'$
B
$R' - Na$
C
$R' - OH$
D
$R' - O - R'$
Solution
(A) The reaction shown is the Wurtz reaction,which is used to prepare higher alkanes from alkyl halides. In this reaction,two molecules of an alkyl halide $(R' - X)$ react with metallic sodium $(Na)$ in the presence of dry ether to form a symmetrical alkane $(R' - R')$ and sodium halide $(NaX)$. The general equation is: $2R' - X + 2Na \xrightarrow{\text{dry ether}} R' - R' + 2NaX$. Therefore,the product formed is $R' - R'$.
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MediumMCQ
$CH_3COOH$ $\xrightarrow{\text{Sodalime}} (A)$ $\xrightarrow{Br_2 / hv} (B)$ $\xrightarrow[KOH]{\text{Aq.}} (C)$ is $:-$
A
$CH_4$
B
$CH_3OH$
C
$CH_3Cl$
D
$C_2H_5OH$
Solution
(B) Step $1$: Decarboxylation of $CH_3COOH$ with sodalime $(NaOH + CaO)$ gives methane $(CH_4)$.
$CH_3COOH + NaOH \xrightarrow{\Delta} CH_4 + Na_2CO_3$ $(A = CH_4)$
Step $2$: Free radical bromination of methane gives methyl bromide $(CH_3Br)$.
$CH_4 + Br_2 \xrightarrow{hv} CH_3Br + HBr$ $(B = CH_3Br)$
Step $3$: Nucleophilic substitution of methyl bromide with aqueous $KOH$ gives methanol $(CH_3OH)$.
$CH_3Br + KOH_{(aq)} \rightarrow CH_3OH + KBr$ $(C = CH_3OH)$
Thus,the final product $(C)$ is $CH_3OH$.
$CH_3COOH + NaOH \xrightarrow{\Delta} CH_4 + Na_2CO_3$ $(A = CH_4)$
Step $2$: Free radical bromination of methane gives methyl bromide $(CH_3Br)$.
$CH_4 + Br_2 \xrightarrow{hv} CH_3Br + HBr$ $(B = CH_3Br)$
Step $3$: Nucleophilic substitution of methyl bromide with aqueous $KOH$ gives methanol $(CH_3OH)$.
$CH_3Br + KOH_{(aq)} \rightarrow CH_3OH + KBr$ $(C = CH_3OH)$
Thus,the final product $(C)$ is $CH_3OH$.
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EasyMCQ
Which of the following compounds is least soluble in water at $STP$?
A
$C_2H_5OH$
B
$CH_3OH$
C
$CH_3NH_2$
D
$CH_4$
Solution
(D) Solubility in water depends on the ability of a molecule to form hydrogen bonds with water molecules.
$C_2H_5OH$,$CH_3OH$,and $CH_3NH_2$ are polar molecules capable of forming hydrogen bonds with water,making them soluble.
$CH_4$ (methane) is a non-polar hydrocarbon.
Non-polar molecules do not form hydrogen bonds with water and have very low solubility in water.
Therefore,$CH_4$ is the least soluble compound among the given options.
$C_2H_5OH$,$CH_3OH$,and $CH_3NH_2$ are polar molecules capable of forming hydrogen bonds with water,making them soluble.
$CH_4$ (methane) is a non-polar hydrocarbon.
Non-polar molecules do not form hydrogen bonds with water and have very low solubility in water.
Therefore,$CH_4$ is the least soluble compound among the given options.
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EasyMCQ
Which among the following has the lowest boiling point?
A
$CH_3-O-CH_2-CH_3$
B
$CH_3-COOH$
C
$CH_3-CH_2-CH_2-CH_3$
D
$CH_3-CH_2-CH_2-OH$
Solution
(C) To determine the lowest boiling point,we analyze the intermolecular forces present in each molecule:
$1$. $CH_3-COOH$ (Ethanoic acid) exhibits strong hydrogen bonding and forms dimers,leading to a very high boiling point.
$2$. $CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol) exhibits hydrogen bonding,resulting in a relatively high boiling point.
$3$. $CH_3-O-CH_2-CH_3$ (Methoxyethane) is an ether,which exhibits dipole-dipole interactions but no hydrogen bonding.
$4$. $CH_3-CH_2-CH_2-CH_3$ (n-Butane) is an alkane,which only exhibits weak London dispersion forces.
Comparing these,$n$-butane has the weakest intermolecular forces,thus it has the lowest boiling point.
$1$. $CH_3-COOH$ (Ethanoic acid) exhibits strong hydrogen bonding and forms dimers,leading to a very high boiling point.
$2$. $CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol) exhibits hydrogen bonding,resulting in a relatively high boiling point.
$3$. $CH_3-O-CH_2-CH_3$ (Methoxyethane) is an ether,which exhibits dipole-dipole interactions but no hydrogen bonding.
$4$. $CH_3-CH_2-CH_2-CH_3$ (n-Butane) is an alkane,which only exhibits weak London dispersion forces.
Comparing these,$n$-butane has the weakest intermolecular forces,thus it has the lowest boiling point.
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MediumMCQ
Identify the product obtained in the following reaction:
$n CH_3 MgI + n H_2 O \xrightarrow{\text{dry ether}} \text{product}$
$n CH_3 MgI + n H_2 O \xrightarrow{\text{dry ether}} \text{product}$
A
$n MgI$ and $n CH_4$
B
$\frac{n}{2} C_2 H_6$
C
$n CH_3 OH$ and $n MgI$
D
$n CH_4$ and $n MgI(OH)$
Solution
(D) Grignard reagents $(RMgX)$ react with compounds containing active hydrogen atoms (like water,alcohols,or amines) to form the corresponding hydrocarbon.
The reaction is as follows:
$n CH_3 MgI + n H_2 O \rightarrow n CH_4 + n Mg(OH)I$
Here,$CH_3 MgI$ (Methyl magnesium iodide) reacts with $H_2 O$ to produce $CH_4$ (Methane) and $Mg(OH)I$ (Hydroxy magnesium iodide).
The reaction is as follows:
$n CH_3 MgI + n H_2 O \rightarrow n CH_4 + n Mg(OH)I$
Here,$CH_3 MgI$ (Methyl magnesium iodide) reacts with $H_2 O$ to produce $CH_4$ (Methane) and $Mg(OH)I$ (Hydroxy magnesium iodide).
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MediumMCQ
Identify the product $Z$ in the following series of reactions: $CH_3CH_2OH$ $\xrightarrow{SOCl_2, \Delta} X$ $\xrightarrow{Mg, \text{Dry ether}} Y$ $\xrightarrow{NH_3} Z + Mg(Cl)(NH_2)$
A
Ethyl chloride
B
Ethyl magnesium chloride
C
Ethyl amine
D
Ethane
Solution
(D) Step $1$: Reaction of ethanol with $SOCl_2$ (thionyl chloride) gives ethyl chloride $(X)$.
$CH_3CH_2OH + SOCl_2 \rightarrow CH_3CH_2Cl + SO_2 + HCl$
So,$X = CH_3CH_2Cl$ (Ethyl chloride).
Step $2$: Reaction of ethyl chloride with $Mg$ in dry ether gives ethyl magnesium chloride $(Y)$,which is a Grignard reagent.
$CH_3CH_2Cl + Mg \xrightarrow{\text{Dry ether}} CH_3CH_2MgCl$
So,$Y = CH_3CH_2MgCl$ (Ethyl magnesium chloride).
Step $3$: Reaction of Grignard reagent $(Y)$ with ammonia $(NH_3)$ yields an alkane $(Z)$ and a magnesium salt.
$CH_3CH_2MgCl + NH_3 \rightarrow CH_3CH_3 + Mg(Cl)(NH_2)$
So,$Z = CH_3CH_3$ (Ethane).
Therefore,the correct option is $D$.
$CH_3CH_2OH + SOCl_2 \rightarrow CH_3CH_2Cl + SO_2 + HCl$
So,$X = CH_3CH_2Cl$ (Ethyl chloride).
Step $2$: Reaction of ethyl chloride with $Mg$ in dry ether gives ethyl magnesium chloride $(Y)$,which is a Grignard reagent.
$CH_3CH_2Cl + Mg \xrightarrow{\text{Dry ether}} CH_3CH_2MgCl$
So,$Y = CH_3CH_2MgCl$ (Ethyl magnesium chloride).
Step $3$: Reaction of Grignard reagent $(Y)$ with ammonia $(NH_3)$ yields an alkane $(Z)$ and a magnesium salt.
$CH_3CH_2MgCl + NH_3 \rightarrow CH_3CH_3 + Mg(Cl)(NH_2)$
So,$Z = CH_3CH_3$ (Ethane).
Therefore,the correct option is $D$.
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View Solution580
MediumMCQ
Identify the product '$Z$' in the following series of reactions.
$C_2H_5OH$ $\xrightarrow[\Delta]{SOCl_2} X$ $\xrightarrow[\text{Dry ether}]{Mg} Y$ $\xrightarrow{NH_3} Z$
$C_2H_5OH$ $\xrightarrow[\Delta]{SOCl_2} X$ $\xrightarrow[\text{Dry ether}]{Mg} Y$ $\xrightarrow{NH_3} Z$
A
Ethyl chloride
B
Ethyl magnesium chloride
C
Ethyl amine
D
Ethane
Solution
(D) The reaction sequence is as follows:
$1$. $C_2H_5OH + SOCl_2 \xrightarrow{\Delta} C_2H_5Cl (X) + SO_2 + HCl$
$2$. $C_2H_5Cl + Mg \xrightarrow{\text{Dry ether}} C_2H_5MgCl (Y)$ (Grignard reagent)
$3$. $C_2H_5MgCl + NH_3 \rightarrow C_2H_6 (Z) + Mg(NH_2)Cl$
Grignard reagents react with compounds containing active hydrogen atoms (like $NH_3$) to form the corresponding alkane. Therefore,the product $Z$ is ethane.
$1$. $C_2H_5OH + SOCl_2 \xrightarrow{\Delta} C_2H_5Cl (X) + SO_2 + HCl$
$2$. $C_2H_5Cl + Mg \xrightarrow{\text{Dry ether}} C_2H_5MgCl (Y)$ (Grignard reagent)
$3$. $C_2H_5MgCl + NH_3 \rightarrow C_2H_6 (Z) + Mg(NH_2)Cl$
Grignard reagents react with compounds containing active hydrogen atoms (like $NH_3$) to form the corresponding alkane. Therefore,the product $Z$ is ethane.
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View Solution581
MediumMCQ
Identify the reagent $(A)$ used in the following reaction.
A
$NH_3$
B
$RNH_2$
C
$H_2O$
D
$CH_3OH$
Solution
(A) Grignard reagents $(RMgX)$ are highly reactive towards compounds containing active hydrogen atoms (like $H$ attached to $O, N, S$).
In the given reaction,$C_2H_5MgBr$ reacts with ammonia $(NH_3)$ to form ethane $(C_2H_6)$ and magnesium bromamide $(Mg(NH_2)Br)$.
The reaction is: $C_2H_5MgBr + NH_3 \rightarrow C_2H_6 + Mg(NH_2)Br$.
Thus,the reagent $(A)$ is $NH_3$.
In the given reaction,$C_2H_5MgBr$ reacts with ammonia $(NH_3)$ to form ethane $(C_2H_6)$ and magnesium bromamide $(Mg(NH_2)Br)$.
The reaction is: $C_2H_5MgBr + NH_3 \rightarrow C_2H_6 + Mg(NH_2)Br$.
Thus,the reagent $(A)$ is $NH_3$.
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View Solution582
EasyMCQ
Identify product $P$ in the following reaction: $C_2H_5MgBr + NH_3 \rightarrow P$
A
$C_2H_5Br$
B
$HC \equiv CH$
C
$C_2H_6$
D
$H_2C=CH_2$
Solution
(C) The reaction between a Grignard reagent $(C_2H_5MgBr)$ and ammonia $(NH_3)$ is an acid-base reaction.
$NH_3$ acts as a proton donor,and the ethyl group $(C_2H_5^-)$ in the Grignard reagent acts as a base.
$C_2H_5MgBr + NH_3 \rightarrow C_2H_6 + Mg(NH_2)Br$.
Thus,the product $P$ is ethane $(C_2H_6)$.
$NH_3$ acts as a proton donor,and the ethyl group $(C_2H_5^-)$ in the Grignard reagent acts as a base.
$C_2H_5MgBr + NH_3 \rightarrow C_2H_6 + Mg(NH_2)Br$.
Thus,the product $P$ is ethane $(C_2H_6)$.
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View Solution583
DifficultMCQ
Identify the product $X$ in the following reaction: $CH_3-CH_2-COONa \xrightarrow[\Delta]{\text{soda-lime}} X + Na_2CO_3$
A
Propane
B
Ethane
C
Methane
D
Butane
Solution
(B) The reaction of sodium propanoate $(CH_3-CH_2-COONa)$ with soda-lime $(NaOH + CaO)$ is a decarboxylation reaction.
During this process,the carboxyl group is removed as $Na_2CO_3$,and the alkyl group forms an alkane with one carbon atom less than the original salt.
$CH_3-CH_2-COONa + NaOH \xrightarrow{CaO, \Delta} CH_3-CH_3 + Na_2CO_3$.
Thus,the product $X$ is ethane $(CH_3-CH_3)$.
During this process,the carboxyl group is removed as $Na_2CO_3$,and the alkyl group forms an alkane with one carbon atom less than the original salt.
$CH_3-CH_2-COONa + NaOH \xrightarrow{CaO, \Delta} CH_3-CH_3 + Na_2CO_3$.
Thus,the product $X$ is ethane $(CH_3-CH_3)$.
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View Solution584
MediumMCQ
The reaction of sodium acetate and sodalime gives
A
$butane$
B
$ethane$
C
$methane$
D
$propane$
Solution
(C) The reaction of sodium acetate with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
$CH_{3}COONa + NaOH \xrightarrow{\Delta} CH_{4} + Na_{2}CO_{3}$
Here,sodium acetate reacts with sodalime to produce methane gas and sodium carbonate.
$CH_{3}COONa + NaOH \xrightarrow{\Delta} CH_{4} + Na_{2}CO_{3}$
Here,sodium acetate reacts with sodalime to produce methane gas and sodium carbonate.
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View Solution585
EasyMCQ
What is the bond length of the $C-C$ bond in an alkane where all carbon atoms are $sp^{3}$ hybridized (in $pm$)?
A
$154$
B
$133$
C
$112$
D
$120$
Solution
(A) In an alkane, all carbon atoms are $sp^{3}$ hybridized.
The carbon-carbon $(C-C)$ bond is formed by the overlap of one $sp^{3}$ orbital from each carbon atom.
The bond length of a $C-C$ single bond is $1.54$ $\mathring{A}$, which is equivalent to $154$ $pm$.
All these bonds are sigma $(\sigma)$ bonds.
The carbon-carbon $(C-C)$ bond is formed by the overlap of one $sp^{3}$ orbital from each carbon atom.
The bond length of a $C-C$ single bond is $1.54$ $\mathring{A}$, which is equivalent to $154$ $pm$.
All these bonds are sigma $(\sigma)$ bonds.
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View Solution586
MediumMCQ
The tetraethyl lead mixed in petrol works as
A
cooling agent
B
anti-knocking agent
C
bleaching agent
D
None of the above
Solution
(B) Tetraethyl lead $(TEL)$ is added to petrol as an anti-knocking agent.
It functions by increasing the octane number of the fuel,which prevents premature ignition or 'knocking' in internal combustion engines.
It functions by increasing the octane number of the fuel,which prevents premature ignition or 'knocking' in internal combustion engines.
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View Solution587
EasyMCQ
What is the difference in molar mass of Undecane and Dodecane?
A
$10 \ g \ mol^{-1}$
B
$20 \ g \ mol^{-1}$
C
$140 \ g \ mol^{-1}$
D
$14 \ g \ mol^{-1}$
Solution
(D) Undecane is an alkane with the molecular formula $C_{11}H_{24}$.
Dodecane is an alkane with the molecular formula $C_{12}H_{26}$.
The difference between these two consecutive members of the homologous series is a $CH_2$ group.
The molar mass of a $CH_2$ group is $(1 \times 12.01) + (2 \times 1.008) \approx 14.02 \ g \ mol^{-1}$,which is approximately $14 \ g \ mol^{-1}$.
Dodecane is an alkane with the molecular formula $C_{12}H_{26}$.
The difference between these two consecutive members of the homologous series is a $CH_2$ group.
The molar mass of a $CH_2$ group is $(1 \times 12.01) + (2 \times 1.008) \approx 14.02 \ g \ mol^{-1}$,which is approximately $14 \ g \ mol^{-1}$.
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View Solution588
MediumMCQ
Identify the total number of carbon atoms present in undecane.
A
$9$
B
$11$
C
$10$
D
$20$
Solution
(B) The general formula for an alkane is $C_nH_{2n+2}$.
Undecane is an alkane with $n = 11$.
Therefore,the chemical formula for undecane is $C_{11}H_{24}$.
Thus,the total number of carbon atoms in undecane is $11$.
Undecane is an alkane with $n = 11$.
Therefore,the chemical formula for undecane is $C_{11}H_{24}$.
Thus,the total number of carbon atoms in undecane is $11$.
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View Solution589
MediumMCQ
What is the number of $-CH_2-$ groups present in dodecane?
A
$12$
B
$10$
C
$15$
D
$13$
Solution
(B) The chemical formula for dodecane is $C_{12}H_{26}$.
It is a straight-chain alkane with the structural formula $CH_3-(CH_2)_{10}-CH_3$.
In this structure,the two terminal carbons are $-CH_3$ groups,and the remaining carbons are $-CH_2-$ groups.
Therefore,the number of $-CH_2-$ groups is $10$.
It is a straight-chain alkane with the structural formula $CH_3-(CH_2)_{10}-CH_3$.
In this structure,the two terminal carbons are $-CH_3$ groups,and the remaining carbons are $-CH_2-$ groups.
Therefore,the number of $-CH_2-$ groups is $10$.
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View Solution590
EasyMCQ
What is the difference in molar mass of any two neighbouring alkanes?
A
$12 \ g \ mol^{-1}$
B
$10 \ g \ mol^{-1}$
C
$15 \ g \ mol^{-1}$
D
$14 \ g \ mol^{-1}$
Solution
(D) Successive members of a homologous series differ by a $-CH_2-$ group.
The molar mass of a carbon atom is $12 \ g \ mol^{-1}$ and that of two hydrogen atoms is $2 \times 1 \ g \ mol^{-1} = 2 \ g \ mol^{-1}$.
Therefore,the difference in molar mass between any two neighbouring alkanes is $12 + 2 = 14 \ g \ mol^{-1}$.
The molar mass of a carbon atom is $12 \ g \ mol^{-1}$ and that of two hydrogen atoms is $2 \times 1 \ g \ mol^{-1} = 2 \ g \ mol^{-1}$.
Therefore,the difference in molar mass between any two neighbouring alkanes is $12 + 2 = 14 \ g \ mol^{-1}$.
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View Solution591
EasyMCQ
How many primary,secondary and tertiary carbon atoms respectively are present in isobutane?
A
$0, 1$ and $3$
B
$3, 1$ and $0$
C
$3, 0$ and $1$
D
$1, 0$ and $3$
Solution
(C) The structure of isobutane is $(CH_3)_3CH$.
In this structure:
$1$. $A$ primary $(1^{\circ})$ carbon atom is attached to only one other carbon atom. There are three such $CH_3$ groups in isobutane.
$2$. $A$ secondary $(2^{\circ})$ carbon atom is attached to two other carbon atoms. There are no such carbon atoms in isobutane.
$3$. $A$ tertiary $(3^{\circ})$ carbon atom is attached to three other carbon atoms. The central carbon atom is attached to three $CH_3$ groups,so it is a tertiary carbon.
Thus,there are $3$ primary,$0$ secondary,and $1$ tertiary carbon atom in isobutane.
In this structure:
$1$. $A$ primary $(1^{\circ})$ carbon atom is attached to only one other carbon atom. There are three such $CH_3$ groups in isobutane.
$2$. $A$ secondary $(2^{\circ})$ carbon atom is attached to two other carbon atoms. There are no such carbon atoms in isobutane.
$3$. $A$ tertiary $(3^{\circ})$ carbon atom is attached to three other carbon atoms. The central carbon atom is attached to three $CH_3$ groups,so it is a tertiary carbon.
Thus,there are $3$ primary,$0$ secondary,and $1$ tertiary carbon atom in isobutane.
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View Solution592
MediumMCQ
What is the molecular formula of undecane?
A
$C_{11}H_{24}$
B
$C_9H_{18}$
C
$C_{12}H_{26}$
D
$C_{20}H_{42}$
Solution
(A) Undecane is an alkane with $11$ carbon atoms. The general formula for alkanes is $C_nH_{2n+2}$.
For $n = 11$,the molecular formula is $C_{11}H_{2(11)+2} = C_{11}H_{24}$.
The structural formula is $CH_3-(CH_2)_9-CH_3$.
For $n = 11$,the molecular formula is $C_{11}H_{2(11)+2} = C_{11}H_{24}$.
The structural formula is $CH_3-(CH_2)_9-CH_3$.
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View Solution593
MediumMCQ
What is the number of tertiary carbon atoms in a molecule of isobutane?
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(A) The structure of isobutane is $CH_3-CH(CH_3)-CH_3$.
$A$ tertiary carbon atom is a carbon atom bonded to three other carbon atoms.
In isobutane,the central carbon atom is bonded to three methyl groups $(-CH_3)$.
Therefore,there is $1$ tertiary carbon atom in a molecule of isobutane.
$A$ tertiary carbon atom is a carbon atom bonded to three other carbon atoms.
In isobutane,the central carbon atom is bonded to three methyl groups $(-CH_3)$.
Therefore,there is $1$ tertiary carbon atom in a molecule of isobutane.
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View Solution594
MediumMCQ
Identify the decreasing order of boiling point of the following alkanes:
$(i)$ $n$-pentane
$(ii)$ Isopentane
$(iii)$ Neopentane
$(i)$ $n$-pentane
$(ii)$ Isopentane
$(iii)$ Neopentane
A
Isopentane $>$ $n$-pentane $>$ Neopentane
B
Neopentane $>$ Isopentane $>$ $n$-pentane
C
$n$-pentane $>$ Isopentane $>$ Neopentane
D
Isopentane $>$ Neopentane $>$ $n$-pentane
Solution
(C) The boiling point of alkanes depends on the surface area of the molecule.
As the branching in the alkane chain increases,the surface area decreases,leading to weaker van der Waals forces of attraction.
Therefore,the boiling point decreases as branching increases.
$n$-pentane is a straight-chain alkane with the largest surface area,followed by Isopentane (one branch),and Neopentane (two branches) has the smallest surface area.
Thus,the decreasing order of boiling point is: $n$-pentane $>$ Isopentane $>$ Neopentane.
As the branching in the alkane chain increases,the surface area decreases,leading to weaker van der Waals forces of attraction.
Therefore,the boiling point decreases as branching increases.
$n$-pentane is a straight-chain alkane with the largest surface area,followed by Isopentane (one branch),and Neopentane (two branches) has the smallest surface area.
Thus,the decreasing order of boiling point is: $n$-pentane $>$ Isopentane $>$ Neopentane.
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View Solution595
EasyMCQ
$LPG$ mainly consists of the following:
A
$methane$
B
$hydrogen$
C
$acetylene$
D
$butane$
Solution
(D) $LPG$ (Liquefied Petroleum Gas) is a mixture of hydrocarbon gases used as a fuel.
It mainly consists of $butane$ $(C_4H_{10})$ and $propane$ $(C_3H_8)$.
Among the given options,$butane$ is the primary component.
It mainly consists of $butane$ $(C_4H_{10})$ and $propane$ $(C_3H_8)$.
Among the given options,$butane$ is the primary component.
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View Solution596
DifficultMCQ
What is the number of moles of carbon atoms present in $n$ mole molecules of an alkane if it exhibits five structural isomers?
A
$4$
B
$3$
C
$5$
D
$6$
Solution
(D) An alkane exhibits five structural isomers when it contains $6$ carbon atoms (hexane,$C_6H_{14}$).
Therefore,in $n$ moles of such alkane molecules,the number of moles of carbon atoms is $6 \times n = 6n$.
The structural isomers of $C_6H_{14}$ are:
$1$. $CH_3-(CH_2)_4-CH_3$ (Hexane)
$2$. $CH_3-CH(CH_3)-(CH_2)_2-CH_3$ ($2$-Methylpentane)
$3$. $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ ($3$-Methylpentane)
$4$. $CH_3-C(CH_3)_2-CH_2-CH_3$ ($2$,$2$-Dimethylbutane)
$5$. $CH_3-CH(CH_3)-CH(CH_3)-CH_3$ ($2$,$3$-Dimethylbutane)
Therefore,in $n$ moles of such alkane molecules,the number of moles of carbon atoms is $6 \times n = 6n$.
The structural isomers of $C_6H_{14}$ are:
$1$. $CH_3-(CH_2)_4-CH_3$ (Hexane)
$2$. $CH_3-CH(CH_3)-(CH_2)_2-CH_3$ ($2$-Methylpentane)
$3$. $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ ($3$-Methylpentane)
$4$. $CH_3-C(CH_3)_2-CH_2-CH_3$ ($2$,$2$-Dimethylbutane)
$5$. $CH_3-CH(CH_3)-CH(CH_3)-CH_3$ ($2$,$3$-Dimethylbutane)
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View Solution597
EasyMCQ
Identify the molecular formula of an alkane that exhibits only two different structural isomers.
A
$C_2H_6$
B
$C_3H_8$
C
$C_4H_{10}$
D
$C_6H_{14}$
Solution
(C) An alkane with the molecular formula $C_4H_{10}$ (butane) exhibits exactly two structural isomers: $n$-butane (butane) and $2$-methylpropane (isobutane).
| $2$ Carbon atoms | Ethane: $0$ structural isomers |
| $3$ Carbon atoms | Propane: $0$ structural isomers |
| $4$ Carbon atoms | Butane: $2$ structural isomers ($n$-butane and $2$-methylpropane) |
| $6$ Carbon atoms | Hexane: $5$ structural isomers |
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View Solution598
MediumMCQ
What is the number of moles of '$C$' atoms present in $n$ mole of an alkane molecule if it exhibits three structural isomers (in $n$)?
A
$3$
B
$4$
C
$5$
D
$6$
Solution
(C) An alkane with three structural isomers is pentane $(C_5H_{12})$.
The three isomers are:
$1$. $n$-Pentane: $CH_3-CH_2-CH_2-CH_2-CH_3$
$2$. Isopentane ($2$-methylbutane): $CH_3-CH(CH_3)-CH_2-CH_3$
$3$. Neopentane ($2$,$2$-dimethylpropane): $CH_3-C(CH_3)_2-CH_3$
In one molecule of pentane,there are $5$ carbon atoms.
Therefore,in $n$ moles of pentane,the number of moles of '$C$' atoms is $5 \times n = 5n$.
The three isomers are:
$1$. $n$-Pentane: $CH_3-CH_2-CH_2-CH_2-CH_3$
$2$. Isopentane ($2$-methylbutane): $CH_3-CH(CH_3)-CH_2-CH_3$
$3$. Neopentane ($2$,$2$-dimethylpropane): $CH_3-C(CH_3)_2-CH_3$
In one molecule of pentane,there are $5$ carbon atoms.
Therefore,in $n$ moles of pentane,the number of moles of '$C$' atoms is $5 \times n = 5n$.
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View Solution599
MediumMCQ
What is the molecular formula of an alkane if it exhibits three structural isomers?
A
$C_3H_8$
B
$C_4H_{10}$
C
$C_5H_{12}$
D
$C_6H_{14}$
Solution
(C) The number of structural isomers for alkanes increases with the number of carbon atoms. Based on the data:
$1$. $C_3H_8$ (Propane): $0$ structural isomers.
$2$. $C_4H_{10}$ (Butane): $2$ structural isomers ($n$-butane and isobutane).
$3$. $C_5H_{12}$ (Pentane): $3$ structural isomers ($n$-pentane,isopentane,and neopentane).
$4$. $C_6H_{14}$ (Hexane): $5$ structural isomers.
Therefore,the alkane with three structural isomers is $C_5H_{12}$.
$1$. $C_3H_8$ (Propane): $0$ structural isomers.
$2$. $C_4H_{10}$ (Butane): $2$ structural isomers ($n$-butane and isobutane).
$3$. $C_5H_{12}$ (Pentane): $3$ structural isomers ($n$-pentane,isopentane,and neopentane).
$4$. $C_6H_{14}$ (Hexane): $5$ structural isomers.
Therefore,the alkane with three structural isomers is $C_5H_{12}$.
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View SolutionHydrocarbons — Alkane · Frequently Asked Questions
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