IBSL Chemistry Topic 10 Organic Chemistry Multiple Choice Questions

The compound CH3CH2CHO is a member of the aldehyde homologous series because it contains a carbonyl group (C=O) attached to a carbon atom, which is characteristic of aldehydes. Aldehydes are organic compounds that have a carbonyl group at the end of the carbon chain, and in this case, the CHO group indicates the presence of an aldehyde functional group. The other compounds listed do not have a carbonyl group attached to a carbon atom, so they do not belong to the aldehyde homologous series.

In a homologous series, the members differ by a CH2 group. This means that each member of the series has the same functional group but differs in the number of CH2 groups in their structure. This results in a gradual increase in molecular size and weight as you move along the series.

Neon (Ne) is a noble gas, belonging to Group 18 of the periodic table. Noble gases have a full outer shell of valence electrons, making them very stable and unreactive. This stability is why they are called "noble" gases, as they don't readily interact with other elements.

CH3OH and C2H5OH both belong to the same homologous series because they have the same functional group, which is the hydroxyl group (-OH). In this case, both compounds are alcohols, with the difference being the number of carbon atoms attached to the hydroxyl group. CH3OH is methanol, while C2H5OH is ethanol.

During incomplete combustion, a hydrocarbon does not burn completely, resulting in the formation of carbon and carbon monoxide. Incomplete combustion occurs when there is a limited supply of oxygen, causing the hydrocarbon to break down incompletely. Carbon is formed due to the insufficient oxygen supply, while carbon monoxide is produced as a result of the incomplete oxidation of carbon. Hydrogen, on the other hand, does not play a role in the incomplete combustion process and is not formed. Therefore, the correct answer is I and III only.

When CH2=CH2 reacts with Br2, the double bond between the carbon atoms is broken and each carbon atom forms a single bond with a bromine atom. This results in the formation of CH2BrCH2Br, which is the correct answer.

HCOOCH3 is the correct answer because it is the only compound that fits the definition of an ester. Esters are organic compounds formed by the reaction between an alcohol and an acid, resulting in the formation of an ester bond. In HCOOCH3, the -COO- group represents the ester bond, with the methoxy group (CH3O-) acting as the alcohol component and the formate group (HCOO-) acting as the acid component. The other compounds listed are not esters; CH3COOH is acetic acid, C2H5CHO is ethanal, and CH3OC2H5 is ethyl methyl ether.

When ethanol is partially oxidized by an acidified solution of potassium dichromate (VI), the product that can be obtained by distillation as soon as it is formed is ethanal. Ethanol can be oxidized to ethanal by the action of an oxidizing agent such as potassium dichromate (VI) in an acidic solution. Ethanal is a volatile compound that can be easily separated from the reaction mixture by distillation. Ethanoic acid, ethene, and ethane 1,2 diol are not formed as immediate products of this oxidation reaction and would require further reactions or conditions for their formation.

The reaction between methane and chlorine involves homolytic fission, which means that the bond between the carbon and hydrogen in methane is broken equally, resulting in the formation of methyl radicals (CH3·) and chlorine radicals (Cl·). This process occurs because the bond between carbon and hydrogen is relatively weak, and the reaction is initiated by the absorption of energy, such as heat or light. The formation of Cl· radicals allows for further reactions to occur, leading to the overall reaction between methane and chlorine.

Butan-1-ol is converted to butanal by acidified potassium dichromate (VI) solution.

1-chlorobutane is a member of the same homologous series as 1-chloropropane because they both have the same functional group (chlorine atom) attached to a carbon chain. In both compounds, the carbon chain contains three carbon atoms, but in 1-chlorobutane, there is an additional carbon atom, making it a butane. Therefore, both compounds belong to the same series of compounds with similar chemical properties and increasing carbon chain length.

When CH3CH2OH is refluxed with acidified potassium dichromate(VI), the final product formed is CH3COOH, which is acetic acid. This reaction is known as the oxidation of ethanol. The acidified potassium dichromate(VI) acts as an oxidizing agent, converting the ethanol into acetic acid.

The correct answer is 3 because C5H12 is the molecular formula for pentane, which is an alkane. Alkanes have a general formula of CnH2n+2, where n represents the number of carbon atoms. In the case of pentane, there are 5 carbon atoms, so the formula is C5H12. Isomers are different compounds that have the same molecular formula but different structural arrangements. For C5H12, there are three different structural isomers: n-pentane, isopentane, and neopentane. Therefore, the number of different isomers of C5H12 is 3.

Ketones are compounds that contain a carbonyl group (C=O) bonded to two carbon atoms. In order to have a carbonyl group, there must be at least three carbon atoms present in the compound. Therefore, a ketone must contain a minimum of three carbon atoms.

The IUPAC name for CH3CH2CH(CH3)2 is 2-methylbutane. This is because the longest carbon chain in the molecule contains four carbons, so it is a butane. The methyl group attached to the second carbon atom is indicated by the prefix "2-methyl."

The formulas I, II, and III all represent different forms of butane or its isomer. Formula I, CH3(CH2)2CH3, is the straight-chain form of butane. Formula II, CH3CH(CH3)CH3, is the branched isomer of butane called 2-methylpropane or isobutane. Formula III, (CH3)3CH, is also a branched isomer of butane called 2,2-dimethylpropane or neopentane. Therefore, all three formulas represent either butane or its isomer.

The molecular formula C6H14 represents a saturated hydrocarbon with 6 carbon atoms and 14 hydrogen atoms. To determine the number of structural isomers, we need to consider the different ways in which these atoms can be arranged. The five possible structural isomers for C6H14 are: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane. These isomers have different structural arrangements of carbon and hydrogen atoms, resulting in distinct chemical properties.

The formation of a positive ion involves the loss of electrons, resulting in a species with a net positive charge. In reaction II, (CH3)3Br + OH-, the bromine atom loses an electron to form a bromine cation (positive ion). In reaction I, CH3CH2CH2Br + OH-, there is no formation of a positive ion as the bromine atom does not lose an electron. Therefore, only reaction II involves the formation of a positive ion.

Alkanes form a homologous series with the general formula CnH2n+2 (I is correct). While they do not all have identical physical properties, as these properties such as boiling point and melting point vary with molecular size (II is incorrect), they do all have similar chemical properties, such as being relatively unreactive compared to other hydrocarbon families, with similar types of reactions like combustion (III is correct).

The formula (CH3)2CHCH2Br represents the secondary halogenoalkane. This is because the carbon atom attached to the bromine atom is bonded to two other carbon atoms, making it a secondary carbon. The presence of the two methyl groups ((CH3)2) on one of the carbon atoms further confirms that it is a secondary halogenoalkane.