Organic Chemistry – Some Basic Principles and Techniques MCQs with Answers – Part 2 (Class 11 Chemistry)

Explanation: Saturated open-chain hydrocarbons contain only single carbon-carbon bonds. In such compounds, carbon atoms carry the maximum possible number of hydrogen atoms for that open-chain skeleton. Alkanes are the common saturated hydrocarbons introduced at this level. The presence of \( \mathrm{C=C} \) or \( \mathrm{C\equiv C} \) makes a hydrocarbon unsaturated. Oxygen is not part of the definition of a hydrocarbon.

Explanation: Unsaturated hydrocarbons contain at least one carbon-carbon multiple bond. A \( \mathrm{C=C} \) double bond gives an alkene-type compound, while a \( \mathrm{C\equiv C} \) triple bond gives an alkyne-type compound. These multiple bonds reduce the number of hydrogens compared with the corresponding open-chain alkane. A compound with only \( \mathrm{C-C} \) single bonds is saturated in the hydrocarbon classification. The word unsaturated is connected with bond type, not with whether the compound is liquid or solid.

Explanation: Alkanes are saturated hydrocarbons and have only single \( \mathrm{C-C} \) bonds in the carbon skeleton. Alkenes contain at least one \( \mathrm{C=C} \) double bond. Alkynes contain at least one \( \mathrm{C\equiv C} \) triple bond. These three families are first separated by the type of carbon-carbon bonding. The suffixes \( \mathrm{-ane} \), \( \mathrm{-ene} \), and \( \mathrm{-yne} \) later become useful in naming these hydrocarbons.

Explanation: \( \textbf{Given formula:} \) The hydrocarbon is \( \mathrm{C_5H_{12}} \). \( \textbf{Open-chain alkane relation:} \) Alkanes follow \( \mathrm{C_nH_{2n+2}} \). \( \textbf{Substitute \(n=5\):} \) \[ \mathrm{H}=2(5)+2=12 \] \( \textbf{Formula match:} \) This gives \( \mathrm{C_5H_{12}} \). \( \textbf{Comparison with alkene:} \) An open-chain alkene with \(5\) carbons would be \( \mathrm{C_5H_{10}} \). \( \textbf{Comparison with alkyne:} \) An open-chain alkyne with \(5\) carbons would be \( \mathrm{C_5H_8} \). \( \textbf{Final answer:} \) \( \mathrm{C_5H_{12}} \) fits an alkane. The formula relation applies here because the question states an open-chain hydrocarbon.

Explanation: An open-chain alkane has only single carbon-carbon bonds and follows \( \mathrm{C_nH_{2n+2}} \). An open-chain alkene with one double bond follows \( \mathrm{C_nH_{2n}} \). An open-chain alkyne with one triple bond follows \( \mathrm{C_nH_{2n-2}} \). These formulas are useful for simple acyclic hydrocarbons in the introductory classification. They should not be applied blindly to rings or aromatic compounds without checking the structural condition.

Explanation: \( \textbf{Given formula:} \) The hydrocarbon is \( \mathrm{C_4H_8} \). \( \textbf{Open-chain alkene relation:} \) An open-chain alkene follows \( \mathrm{C_nH_{2n}} \). \( \textbf{Substitute \(n=4\):} \) \[ \mathrm{H}=2(4)=8 \] \( \textbf{Formula match:} \) \( \mathrm{C_4H_8} \) matches the alkene formula. \( \textbf{Alkane comparison:} \) An open-chain alkane with \(4\) carbons would be \( \mathrm{C_4H_{10}} \). \( \textbf{Alkyne comparison:} \) An open-chain alkyne with \(4\) carbons would be \( \mathrm{C_4H_6} \). \( \textbf{Final answer:} \) The compound is best placed in the alkene family. The phrase open-chain is needed because rings can also give formulas that resemble alkene formulas.

Explanation: \( \mathrm{C_3H_8} \) fits the open-chain alkane formula \( \mathrm{C_nH_{2n+2}} \). \( \mathrm{C_3H_6} \), when specified as open-chain with one \( \mathrm{C=C} \), is an alkene. \( \mathrm{C_3H_4} \), when specified as open-chain with one \( \mathrm{C\equiv C} \), is an alkyne. \( \mathrm{C_6H_6} \) as a benzene ring is an arene, which is an aromatic hydrocarbon. The structural information beside the formula prevents confusing an alkene-type formula with a cyclic possibility.

Explanation: \( \mathrm{CH_2=CH_2} \) contains a carbon-carbon double bond. Hydrocarbons with \( \mathrm{C=C} \) or \( \mathrm{C\equiv C} \) bonds are classified as unsaturated. The presence of hydrogen does not by itself make a compound saturated. Saturation depends on whether the carbon skeleton has only single carbon-carbon bonds. Ethene is organic and unsaturated, so the correction must focus on the double bond.

Explanation: The formula \( \mathrm{C_nH_{2n+2}} \) is the general formula for open-chain alkanes. It assumes a saturated acyclic hydrocarbon framework. Rings reduce the number of hydrogens compared with open-chain alkanes, so cyclic compounds require separate attention. Compounds containing oxygen are not hydrocarbons and should not be classified by alkane formulas alone. The condition “open-chain saturated hydrocarbon” is part of the meaning of the formula, not an optional detail.

Explanation: \( \textbf{Given carbon count:} \) The hydrocarbon has \(6\) carbon atoms, so \(n=6\). \( \textbf{Corresponding open-chain alkane:} \) An alkane follows \( \mathrm{C_nH_{2n+2}} \). \[ \mathrm{H}_{\text{alkane}}=2(6)+2=14 \] \( \textbf{Meaning of hydrogenation data:} \) Adding one molecule of \( \mathrm{H_2} \) increases the hydrogen count by \(2\). \( \textbf{Hydrogen added:} \) The compound adds \(2\) molecules of \( \mathrm{H_2} \), so total hydrogen increase is \(2\times2=4\). \( \textbf{Original hydrogen count:} \) \[ 14-4=10 \] \( \textbf{Original formula:} \) The hydrocarbon was \( \mathrm{C_6H_{10}} \). \( \textbf{Unsaturation interpretation:} \) A deficiency of \(4\) hydrogens compared with the alkane corresponds to two units of unsaturation in an open-chain hydrocarbon. \( \textbf{Final answer:} \) The original molecular formula is \( \mathrm{C_6H_{10}} \). This reasoning uses both the alkane formula and the hydrogenation change, not only direct substitution into one family formula.

Explanation: \( \textbf{Family condition:} \) An open-chain alkyne with one triple bond follows \( \mathrm{C_nH_{2n-2}} \). \( \textbf{Molecular formula expression:} \) \[ \mathrm{C_nH_{2n-2}} \] \( \textbf{Molar mass expression:} \) \[ 12n+(2n-2)(1)=68 \] \( \textbf{Simplify:} \) \[ 12n+2n-2=68 \] \[ 14n-2=68 \] \[ 14n=70 \] \[ n=5 \] \( \textbf{Hydrogen count:} \) \[ 2n-2=2(5)-2=8 \] \( \textbf{Formula obtained:} \) The molecular formula is \( \mathrm{C_5H_8} \). \( \textbf{Final answer:} \) The hydrocarbon is \( \mathrm{C_5H_8} \). The same-carbon alkene alternative fits an open-chain alkene formula, not a one-triple-bond alkyne.

Explanation: \( \textbf{Given:} \) The compound is an open-chain hydrocarbon with \(5\) carbon atoms. \( \textbf{Test alkene formula:} \) For an open-chain alkene, \( \mathrm{C_nH_{2n}} \). \[ \mathrm{C_5H_{10}} \] \( \textbf{Molar mass of \( \mathrm{C_5H_{10}} \):} \) \[ 5(12)+10(1)=60+10=70\,\text{g mol}^{-1} \] \( \textbf{Hydrogen percentage:} \) \[ \frac{10}{70}\times100=14.29\% \] \( \textbf{Compare with given data:} \) \(14.29\%\) is closest to \(14.3\%\). \( \textbf{Check alkyne alternative:} \) \( \mathrm{C_5H_8} \) would have hydrogen percentage \( \frac{8}{68}\times100=11.76\% \). \( \textbf{Check alkane alternative:} \) \( \mathrm{C_5H_{12}} \) would not have a carbon-carbon multiple bond. \( \textbf{Final answer:} \) The compound is best classified as an alkene. The percentage data supports the same conclusion as the open-chain \( \mathrm{C_nH_{2n}} \) relation.

Explanation: \( \textbf{Reference alkane:} \) The open-chain saturated hydrocarbon with \(4\) carbons is \( \mathrm{C_4H_{10}} \). \( \textbf{Given deficiency:} \) The compound has \(4\) fewer hydrogen atoms. \[ 10-4=6 \] \( \textbf{Formula of compound:} \) The compound is \( \mathrm{C_4H_6} \). \( \textbf{Alkene comparison:} \) A one-double-bond open-chain hydrocarbon with \(4\) carbons would be \( \mathrm{C_4H_8} \), only \(2\) hydrogens fewer than the alkane. \( \textbf{Alkyne comparison:} \) A one-triple-bond open-chain hydrocarbon with \(4\) carbons follows \( \mathrm{C_nH_{2n-2}} \). \[ \mathrm{C_4H_{2(4)-2}}=\mathrm{C_4H_6} \] \( \textbf{Bonding interpretation:} \) A triple bond accounts for a four-hydrogen deficiency compared with the open-chain alkane. \( \textbf{Final answer:} \) The most reasonable feature is one \( \mathrm{C\equiv C} \) bond. This comparison should be made against the saturated open-chain formula, not by counting only the number of carbon atoms.

Explanation: \( \mathrm{C_4H_{10}} \) fits \( \mathrm{C_nH_{2n+2}} \), so P is an alkane. \( \mathrm{C_4H_8} \) has two fewer hydrogens than the alkane and can add one molecule of \( \mathrm{H_2} \), so Q fits an alkene. \( \mathrm{C_4H_6} \) has four fewer hydrogens than the alkane and can add two molecules of \( \mathrm{H_2} \), so R fits an alkyne with one triple bond. The hydrogenation observation confirms the formula-based classification. The comparison works because all three are specified as open-chain hydrocarbons.

Explanation: Benzene is an aromatic hydrocarbon, and aromatic hydrocarbons are called arenes. Its formula \( \mathrm{C_6H_6} \) alone should not be used to classify it as an ordinary open-chain alkyne. The structure contains a benzene ring, which gives it a special aromatic classification. Alkanes, alkenes, and alkynes are first introduced through open-chain bonding patterns. Structural information is essential here because formula resemblance can be misleading.

Explanation: A functional group is an atom or group of atoms that gives an organic compound its characteristic chemical properties. For example, \( \mathrm{-OH} \), \( \mathrm{-CHO} \), and \( \mathrm{-COOH} \) are functional groups in different families of organic compounds. The carbon chain provides the skeleton, but the functional group often decides the main type of reaction. Molecular mass and empirical formula give composition-related information but do not directly name the reactive group. Recognising the functional group is a major step in classifying and naming organic compounds.

Explanation: \( \mathrm{CH_3CH_2OH} \) contains a two-carbon hydrocarbon part, \( \mathrm{CH_3CH_2-} \), and a hydroxy group, \( \mathrm{-OH} \). The \( \mathrm{-OH} \) group is the functional group because it gives the compound alcohol character. The carbon-hydrogen part is the skeleton to which the functional group is attached. It is not correct to treat oxygen alone as the full functional group in this representation; the group is read as \( \mathrm{-OH} \). Separating skeleton and functional group helps in both classification and naming.

Explanation: Both \( \mathrm{CH_3OH} \) and \( \mathrm{CH_3CH_2OH} \) contain the \( \mathrm{-OH} \) group. This group places them in the alcohol family and gives them similar chemical behaviour. Their carbon chains are not identical because one has one carbon and the other has two carbons. They also do not have the same molecular formula. The similar reactions come mainly from the common functional group, while physical properties may change gradually with chain length.

Explanation: Row P is suitable because both compounds contain the \( \mathrm{-OH} \) group and belong to the alcohol family. Row Q is suitable because both compounds contain the \( \mathrm{-COOH} \) group and belong to the carboxylic acid family. Row R is unsuitable because \( \mathrm{CH_3CH_2OH} \) contains an \( \mathrm{-OH} \) group, while \( \mathrm{CH_3OCH_3} \) contains a \( \mathrm{C-O-C} \) ether linkage. These two formulas have the same molecular formula \( \mathrm{C_2H_6O} \), but their connectivity and functional groups are different. Functional-group recognition must be based on structure, not just on element count.

Explanation: The group \( \mathrm{-COOH} \) is called the carboxyl group. Organic compounds containing this group are classified as carboxylic acids. An alcohol contains \( \mathrm{-OH} \) attached to carbon, but it does not contain the full \( \mathrm{-COOH} \) group. An aldehyde contains \( \mathrm{-CHO} \), and an ether contains a \( \mathrm{C-O-C} \) linkage. The extra \( \mathrm{C=O} \) part in \( \mathrm{-COOH} \) makes it different from a simple hydroxy group.