Versatile Nature of Carbon (Hydrocarbons, Functional Groups, Homologous Series, Nomenclature) Questions in English

(B) Three structural isomers are possible for pentane $(C_{5}H_{12})$:
$(i)$ $n$-pentane: $CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{3}$
$(ii)$ Isopentane ($2$-methylbutane): $CH_{3}-CH(CH_{3})-CH_{2}-CH_{3}$
$(iii)$ Neopentane ($2$,$2$-dimethylpropane): $C(CH_{3})_{4}$

(A) The two unique properties of carbon that lead to the formation of a vast number of carbon compounds are:
$(i)$ Catenation: It is the unique ability of carbon atoms to form strong covalent bonds with other carbon atoms,resulting in the formation of long chains,branched chains,or ring structures.
$(ii)$ Tetravalency: Carbon has a valency of $4$,which allows it to form covalent bonds with four other atoms of carbon or atoms of other monovalent elements like hydrogen,oxygen,nitrogen,or halogens.

(N/A) $(i)$ Butanone: $CH_{3}CH_{2}COCH_{3}$
Structure: $CH_{3}-CH_{2}-C(=O)-CH_{3}$
$(ii)$ Hexanal: $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CHO$
Structure: $CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-CHO$

(C) Butanone is a $4$-carbon ketone with the chemical formula $CH_3COCH_2CH_3$.
The suffix '-one' in the name 'butanone' indicates the presence of a ketone functional group $(>C=O)$.

(N/A) homologous series is a group of organic compounds that have the same functional group and similar chemical properties,and in which successive members differ by a $-CH_2$ unit.
For example,the alkane series is a homologous series with the general formula $C_nH_{2n+2}$.
$1$. Methane: $CH_4$
$2$. Ethane: $C_2H_6$ (or $CH_3CH_3$)
$3$. Propane: $C_3H_8$ (or $CH_3CH_2CH_3$)
$4$. Butane: $C_4H_{10}$ (or $CH_3CH_2CH_2CH_3$)
In this series,each successive member differs by a $-CH_2$ group,which corresponds to a molecular mass difference of $14 \ u$.

(A) The $IUPAC$ naming rules are applied as follows:
$(i)$ $CH_3-CH_2-Br$: This is a two-carbon chain with a bromine substituent. The prefix for bromine is 'bromo' and the alkane is 'ethane'. Thus,it is named Bromoethane.
$(ii)$ $H-CHO$: This is a one-carbon aldehyde. The prefix for one carbon is 'meth-' and the suffix for aldehyde is '-al'. Thus,it is named Methanal.
$(iii)$ $CH_3-CH_2-CH_2-CH_2-C \equiv CH$: This is a six-carbon chain with a triple bond at the first position. The prefix for six carbons is 'hex-' and the suffix for an alkyne is '-yne'. Since the triple bond starts at the first carbon,it is named Hex$-1-$yne.

(A) Butane has the molecular formula $C_4H_{10}$.
Structural isomers are compounds with the same molecular formula but different structural arrangements.
$(i)$ $n$-butane is a straight-chain isomer of butane $(C_4H_{10})$.
(ii) Isobutane ($2$-methylpropane) is a branched-chain isomer of butane $(C_4H_{10})$.
(iii) Cyclobutane has the formula $C_4H_8$,so it is not an isomer of butane.
(iv) The structure shown is not a valid isomer of butane.
Therefore,$(i)$ and $(ii)$ are the correct structural isomers of butane.

(D) The functional group $-OH$ represents an alcohol.
In the given options:
$1$. Butanone is a ketone with the functional group $C=O$.
$2$. Butanal is an aldehyde with the functional group $-CHO$.
$3$. Butanoic acid is a carboxylic acid with the functional group $-COOH$.
$4$. Butanol is an alcohol with the functional group $-OH$.
Therefore,the correct option is $D$.

(C) Ethyne,also known as acetylene,is the simplest alkyne with the chemical formula $C_2H_2$.
It consists of two carbon atoms connected by a triple bond,with each carbon atom bonded to one hydrogen atom.
The structural formula is represented as $H-C\equiv C-H$.

(D) Unsaturated compounds are those hydrocarbons that contain at least one carbon-carbon double bond $(C=C)$ or triple bond $(C≡C)$.
$(i)$ Propane $(C_3H_8)$ is an alkane,which is a saturated hydrocarbon.
$(ii)$ Propene $(C_3H_6)$ contains a double bond,making it an unsaturated hydrocarbon.
$(iii)$ Propyne $(C_3H_4)$ contains a triple bond,making it an unsaturated hydrocarbon.
$(iv)$ Chloropropane $(C_3H_7Cl)$ is a substituted alkane,which is saturated.
Therefore,$(ii)$ and $(iii)$ are unsaturated compounds.

(C) straight chain hydrocarbon is one where carbon atoms are linked in a single continuous line without any side chains or branches.
Option $B$ represents $n$-hexane,which is a straight chain.
Option $A$ is $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$,which is $n$-hexane (straight chain).
Option $C$ is $2$-methylpentane,which has a methyl group attached to the second carbon,making it a branched chain hydrocarbon.
Option $D$ is also a representation of a straight chain hydrocarbon.
Therefore,option $C$ is the correct answer as it is a branched chain hydrocarbon.

(D) Unsaturated hydrocarbons are those that contain at least one carbon-carbon double bond $(C=C)$ or triple bond $(C \equiv C)$.
$(i) H_3C - CH_2 - CH_2 - CH_3$ is Butane,a saturated hydrocarbon (alkane).
$(ii) H_3C - C \equiv C - CH_3$ is But$-2-$yne,which contains a carbon-carbon triple bond,making it unsaturated.
$(iii) H_3C - CH(CH_3) - CH_3$ is Isobutane ($2$-methylpropane),a saturated hydrocarbon (alkane).
$(iv) H_3C - C(CH_3) = CH_2$ is $2-$methylpropene,which contains a carbon-carbon double bond,making it unsaturated.
Therefore,$(ii)$ and $(iv)$ are unsaturated hydrocarbons.

(A) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $CH_2$ unit.
Compounds with the general formula $C_nH_{2n+2}$ belong to the alkane series.
Checking the given options:
$B) C_2H_6$ follows $C_nH_{2n+2}$ ($n=2$,$2(2)+2=6$).
$C) C_3H_8$ follows $C_nH_{2n+2}$ ($n=3$,$2(3)+2=8$).
$D) CH_4$ follows $C_nH_{2n+2}$ ($n=1$,$2(1)+2=4$).
$A) C_4H_8$ follows the general formula $C_nH_{2n}$,which represents the alkene series.
Therefore,$C_4H_8$ does not belong to the same homologous series as the others.

(B) The given compound is $CH_{3}-CH_{2}-CHO$.
This molecule contains a chain of $3$ carbon atoms.
The functional group present is the aldehyde group $(-CHO)$.
For an aldehyde with $3$ carbon atoms,the parent alkane is propane.
Replacing the terminal '$e$' of propane with '$al$' gives the name propanal.
Therefore,the correct name is propanal.

(C) In organic chemistry,atoms other than carbon and hydrogen present in a molecule are known as heteroatoms.
In the given compound $CH_{3}-CH_{2}-O-CH_{2}-CH_{2}Cl$:
$1$. Carbon $(C)$ and Hydrogen $(H)$ are the basic elements forming the hydrocarbon chain.
$2$. Oxygen $(O)$ and Chlorine $(Cl)$ are atoms other than carbon and hydrogen.
Therefore,Oxygen and Chlorine are the heteroatoms present in the molecule.
Thus,the correct option is $(i)$ and $(iv)$.

(A) The general formula for the alkyne homologous series is $C_nH_{2n-2}$,where $n$ is the number of carbon atoms.
For the first member,we set $n = 2$ because a triple bond requires at least two carbon atoms.
Substituting $n = 2$ into the formula: $C_2H_{2(2)-2} = C_2H_2$.
$C_2H_2$ is known as ethyne (also called acetylene).
Therefore,the first member of the alkyne series is ethyne.

(C) The compound $CH_3-CH_2-CH_2-CH_2-COOH$ contains $5$ carbon atoms in the main chain and a carboxylic acid functional group $(-COOH)$. Therefore,its $IUPAC$ name is Pentanoic acid.
$(b)$ The compound $CH_3-CH_2-C \equiv CH$ contains $4$ carbon atoms in the main chain and a triple bond between the $3^{rd}$ and $4^{th}$ carbon atoms. Therefore,its $IUPAC$ name is But$-1-$yne or simply Butyne.

(D) The given structure is $CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CHO$. It contains $7$ carbon atoms and an aldehyde functional group $(-CHO)$. Therefore,the $IUPAC$ name is Heptanal.
$(b)$ The given structure is $CH_3-CH_2-CH_2-CH_2-CH_2-OH$. It contains $5$ carbon atoms and an alcohol functional group $(-OH)$. Therefore,the $IUPAC$ name is Pentanol.

(A) The structure $CH_3-CH_2-CH_2-OH$ contains the $-OH$ group,which is the functional group for alcohols. Therefore,it is an alcohol (propanol).
$(b)$ The structure $CH_3-CH_2-COOH$ contains the $-COOH$ group,which is the functional group for carboxylic acids. Therefore,it is a carboxylic acid (propanoic acid).

(A) The structure $CH_3-CO-CH_2-CH_3$ contains a carbonyl group bonded to two carbon atoms,which is characteristic of a $\text{Ketone}$ functional group.
$(b)$ The structure $CH_3-CH_2-CH=CH_2$ contains a carbon-carbon double bond,which is characteristic of an $\text{Alkene}$ functional group.

(A) $(a) CH_3COCH_2CH_2CH_2CH_3$ contains a ketone group $(-CO-)$.
$(b) CH_3CH_2CH_2COOH$ contains a carboxylic acid group $(-COOH)$.
$(c) CH_3CH_2CH_2CH_2CHO$ contains an aldehyde group $(-CHO)$.
$(d) CH_3CH_2OH$ contains an alcohol group $(-OH)$.

(N/A) Hexane $(C_6H_{14})$ has five structural isomers:
$(a)$ $n$-Hexane: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$
$(b)$ $2$-Methylpentane: $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$
$(c)$ $3$-Methylpentane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$
$(d)$ $2,2$-Dimethylbutane: $CH_3-C(CH_3)_2-CH_2-CH_3$
$(e)$ $2,3$-Dimethylbutane: $CH_3-CH(CH_3)-CH(CH_3)-CH_3$

(N/A) Hydrocarbons are organic compounds consisting entirely of carbon and hydrogen atoms. They serve as the fundamental building blocks of organic chemistry.
Examples include:
$1$. Methane $(CH_4)$
$2$. Ethane $(C_2H_6)$
$3$. Propane $(C_3H_8)$

(N/A) Saturated hydrocarbons contain only carbon-carbon single bonds $(C-C)$.
Unsaturated hydrocarbons contain at least one carbon-carbon double $(C=C)$ or triple $(C\equiv C)$ bond.
Examples of Saturated hydrocarbons:
$1$. Methane $(CH_4)$: $H-C(H)(H)-H$
$2$. Ethane $(C_2H_6)$: $H_3C-CH_3$
Examples of Unsaturated hydrocarbons:
$1$. Ethene $(C_2H_4)$: $H_2C=CH_2$
$2$. Ethyne $(C_2H_2)$: $HC\equiv CH$

(N/A) functional group is an atom or a group of atoms joined in a specific manner that is responsible for the characteristic chemical properties of organic compounds.
Examples of four different functional groups are:
$1$. Hydroxyl group: $-OH$
$2$. Aldehyde group: $-CHO$
$3$. Carboxylic acid group: $-COOH$
$4$. Ketone group: $>C=O$

(N/A) The molecular formula $C_3H_6O$ corresponds to two functional isomers:
$1$. Propanone $(CH_3COCH_3)$: $A$ ketone.
$2$. Propanal $(CH_3CH_2CHO)$: An aldehyde.
The electron dot structures are as follows:
- For Propanone: The central carbon atom is double-bonded to an oxygen atom and single-bonded to two methyl groups.
- For Propanal: The terminal carbon atom is double-bonded to an oxygen atom and single-bonded to a hydrogen atom and an ethyl group.

(B) The chemical formula $C_{3}H_{6}$ corresponds to the general formula $C_{n}H_{2n}$,which is characteristic of alkenes.
Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond.
Bromine water is a standard test for unsaturation; it turns from reddish-brown to colourless in the presence of alkenes or alkynes.
Since the compound is an alkene with $3$ carbon atoms,its name is propene $\left(H_{3}C-CH=CH_{2}\right)$.

(N/A) The aldehyde with four carbon atoms is known as Butanal.
Its structural formula is $CH_{3}-CH_{2}-CH_{2}-CHO$.

(N/A) Methanoic acid is the simplest carboxylic acid with the chemical formula $HCOOH$.
Its structural formula consists of a carbon atom double-bonded to an oxygen atom,single-bonded to a hydroxyl group $(-OH)$,and single-bonded to a hydrogen atom.
The structure is represented as:
$H - C(=O) - OH$

(N/A) The functional group $-COOH$ present in citric acid is known as the Carboxylic acid group. It is a characteristic functional group found in organic acids,where a carbon atom is double-bonded to an oxygen atom and single-bonded to a hydroxyl group $(-OH)$.

(N/A) The general formula for an alkane is $C_{n}H_{2n+2}$,where $n$ is the number of carbon atoms.
Given that the molecule contains $n = 4$ carbon atoms.
Substituting $n = 4$ into the formula: $C_{4}H_{2(4)+2} = C_{4}H_{8+2} = C_{4}H_{10}$.
The alkane with four carbon atoms is named Butane.

(N/A) $(i)$ The compound $CH_{3}-CH_{2}-CH_{2}-COOH$ contains the $-COOH$ group,which is a Carboxylic acid functional group.
$(ii)$ The compound $CH_{3}-CO-CH_{2}-CH_{3}$ contains the $C=O$ group bonded to two carbon atoms,which is a Ketone functional group.

(N/A) $(i)$ The molecular formula of benzene is $C_{6}H_{6}$.
$(ii)$ It contains three double bonds in its structure.

(A) For $A$ $(C_{10}H_{18})$: The general formula for an alkyne is $C_nH_{2n-2}$. Here,$n = 10$,so $2n - 2 = 2(10) - 2 = 18$. Thus,$A$ belongs to the alkyne homologous series.
For $B$ $(C_{18}H_{36})$: The general formula for an alkene is $C_nH_{2n}$. Here,$n = 18$,so $2n = 2(18) = 36$. Thus,$B$ belongs to the alkene homologous series.

(N/A) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $CH_{2}$ unit.
$1$. For $CH_{3}CH_{2}OH$ (Ethanol),the next homologue is obtained by adding a $CH_{2}$ group,which results in $CH_{3}CH_{2}CH_{2}OH$ (Propan$-1-$ol).
$2$. For $HCOOH$ (Methanoic acid),the next homologue is obtained by adding a $CH_{2}$ group,which results in $CH_{3}COOH$ (Ethanoic acid).
Therefore,the next homologues are Propanol and Ethanoic acid respectively.

(A-D) $(i)$ The compound $CH_{3}-CH_{2}-CH=CH_{2}$ has four carbon atoms and a double bond,so it is named But-$1$-ene.
$(ii)$ The compound $CH_{3}-CH_{2}-CH_{2}-C(CH_{3})_{2}-CH_{3}$ has a five-carbon chain with two methyl groups at the second position from the end,so it is named $2,2$-Dimethylpentane.
$(iii)$ The compound $CH_{3}-CH_{2}-CHO$ has three carbon atoms and an aldehyde functional group,so it is named Propanal.
$(iv)$ The compound $CH_{3}-CH(OH)-CH_{3}$ has three carbon atoms and an alcohol functional group at the second carbon,so it is named Propan-$2$-ol.

(N/A) The simplest ketone is acetone,also known as propanone. Its chemical formula is $CH_3COCH_3$. The structure consists of a central carbonyl group $(C=O)$ bonded to two methyl groups $(CH_3)$.
Structure:
$CH_3 - C(=O) - CH_3$

(B) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $CH_{2}$ unit.
$1$. $C_{2}H_{6}O$ can represent ethanol $(C_{2}H_{5}OH)$.
$2$. $CH_{4}O$ represents methanol $(CH_{3}OH)$.
Both compounds contain the same functional group,which is the hydroxyl group $(-OH)$.
Since they belong to the alcohol homologous series,$C_{2}H_{6}O$ and $CH_{4}O$ are the correct pair.

(N/A) In a homologous series of alkanes,each successive member differs from the previous one by a $-CH_2-$ group.
$(i)$ The difference in molecular mass is calculated as: $(1 \times 12) + (2 \times 1) = 14 \ u$.
$(ii)$ The difference in the number of atoms is one carbon atom and two hydrogen atoms,represented by the group $-CH_2-$.

(N/A) $(i)$ $A$ homologous series is a group of organic compounds having the same functional group and similar chemical properties,in which all successive members differ by a $-CH_{2}$ unit.
$(ii)$ The melting and boiling points of $C_{5}H_{10}$ are higher than those of $C_{4}H_{8}$ because the molecular mass of $C_{5}H_{10}$ is greater than that of $C_{4}H_{8}$. As molecular mass increases,the intermolecular forces of attraction increase,requiring more energy to overcome them.
$(iii)$ The order of increasing boiling points is: $C_{3}H_{6} < C_{4}H_{8} < C_{5}H_{10}$.

(C) $(i)$ Compounds $A, B$ and $C$ are unsaturated hydrocarbons containing a carbon-carbon double bond.
$(ii)$ The boiling point of members of a homologous series increases with an increase in molecular mass (i.e.,an increase in the number of carbon atoms). Since compound $C$ has the lowest boiling point,it must have the minimum number of carbon atoms.
$(iii)$ The general formula $C_{n}H_{2n}$ represents the homologous series of alkenes.

(A) Alkenes follow the general formula $C_{n}H_{2n}$.
$C_{2}H_{4}$ $(n=2)$ and $C_{4}H_{8}$ $(n=4)$ fit this formula,so they are alkenes.
Alkynes follow the general formula $C_{n}H_{2n-2}$.
$C_{3}H_{4}$ $(n=3)$ and $C_{2}H_{2}$ $(n=2)$ fit this formula,so they are alkynes.

(N/A) $(i)$ The functional group $-COOH$ is known as the Carboxylic acid group.
$(ii)$ The functional group $-COOR'$ is known as the Ester group.
$(iii)$ The functional group $-OH$ is known as the Alcohol group.
$(iv)$ The functional group $-X$ (where $X = F, Cl, Br, I$) is known as the Halide group,and the compound is called an Alkyl halide or Haloalkane.

(N/A) $(i)$ All members of a homologous series contain the same functional group.
$(ii)$ They exhibit similar chemical properties due to the presence of the same functional group.
$(iii)$ There is a regular gradation in physical properties such as melting point,boiling point,and density with an increase in molecular mass.
$(iv)$ Any two successive members of the series differ by a $-CH_2$ unit or a molecular mass of $14u$.

(C) homologous series is a group of organic compounds having the same functional group and similar chemical properties,in which the successive members differ by a $-CH_{2}$ group.
$1$. Alkynes: These follow the general formula $C_{n}H_{2n-2}$. The compounds $C_{3}H_{4}, C_{4}H_{6}, C_{5}H_{8}$ belong to this series.
$2$. Alkenes: These follow the general formula $C_{n}H_{2n}$. The compounds $C_{3}H_{6}, C_{4}H_{8}, C_{5}H_{10}$ belong to this series.

(N/A)

Saturated hydrocarbons	Unsaturated hydrocarbons
$1.$ Saturated hydrocarbons burn with a clean blue flame because they have a higher hydrogen-to-carbon ratio,allowing for complete combustion.	$1.$ Unsaturated hydrocarbons burn with a yellow,sooty flame due to a higher carbon-to-hydrogen ratio,which leads to incomplete combustion.
$2.$ They undergo complete combustion,producing only $CO_{2}$ and $H_{2}O$ as products.	$2.$ They undergo incomplete combustion,producing unburnt carbon particles (soot) in addition to $CO_{2}$ and $H_{2}O$.

$(ii)$ $(a)$ The chemical name of $C_{4}H_{10}$ is Butane. Its structure is: $CH_{3}-CH_{2}-CH_{2}-CH_{3}$.
$(b)$ The chemical name of $C_{2}H_{2}$ is Ethyne. Its structure is: $H-C\equiv C-H$.

(N/A) $(i)$ Structural isomerism is the phenomenon in which compounds have the same molecular formula but different structural arrangements of atoms.
$(ii)$ Propane $(C_{3}H_{8})$ cannot exhibit structural isomerism because it contains only three carbon atoms. With only three carbon atoms,there is only one possible way to arrange them in a chain,meaning no alternative structural arrangement can be formed.
$(iii)$ Butane $(C_{4}H_{10})$ has two structural isomers:
$(a)$ $n$-butane: $A$ straight-chain structure where four carbon atoms are linked in a single row.
$(b)$ Isobutane ($2$-methylpropane): $A$ branched-chain structure where three carbon atoms form the main chain and one carbon atom is attached as a branch to the central carbon atom.

(N/A) homologous series is a group of organic compounds having the same functional group and similar chemical properties,in which all successive members differ by a $-CH_2$ unit.
Two consecutive members of the homologous series of alcohols are:
$(i)$ Methanol: $CH_3OH$
(ii) Ethanol: $C_2H_5OH$
$(i)$ The physical properties (such as melting point,boiling point,and solubility) of the compounds in a homologous series are determined by the length of the carbon chain (molecular mass).
(ii) The chemical properties of the compounds in a homologous series are determined by the functional group present in the molecule.

(N/A) Hydrocarbons are organic compounds consisting entirely of carbon and hydrogen atoms linked by covalent bonds.
$(b)$ The structural difference is that saturated hydrocarbons contain only single covalent bonds between carbon atoms (e.g.,alkanes),whereas unsaturated hydrocarbons contain at least one double or triple covalent bond between carbon atoms (e.g.,alkenes or alkynes).

(N/A) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $CH_{2}$ unit.
$1$. For $CH_{3}OH$ (Methanol),the next homologue is obtained by adding a $CH_{2}$ group,which results in $C_{2}H_{5}OH$ (Ethanol).
$2$. For $HCOOH$ (Methanoic acid),the next homologue is obtained by adding a $CH_{2}$ group,which results in $CH_{3}COOH$ (Ethanoic acid).