IB Topic 10 Organic Chemistry

The correct answer is 3 because C5H12 represents a pentane molecule, which can exist in three different isomeric forms. These isomers include n-pentane, isopentane, and neopentane. Each isomer has a different arrangement of carbon atoms, resulting in distinct chemical and physical properties.

HCOOCH3 is an ester because it contains the ester functional group, which is formed by the condensation reaction between an alcohol and a carboxylic acid. In HCOOCH3, the methoxy group (CH3O) acts as the alcohol component, and the formic acid group (HCOO) acts as the carboxylic acid component.

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

When CH2=CH2 (ethylene) reacts with Br2 (bromine), it undergoes a halogenation reaction. In this reaction, one bromine atom replaces one hydrogen atom in the ethylene molecule, resulting in the formation of CH2BrCH2Br (1,2-dibromoethane). This product is formed by the addition of one bromine atom to each carbon atom in the ethylene molecule.

The molecular formula C6H14 represents a saturated hydrocarbon with the general formula CnH2n+2. In this case, n is equal to 6, so the general formula becomes C6H14. Structural isomers are compounds with the same molecular formula but different structural arrangements. For C6H14, there are five possible structural isomers: hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane. Therefore, the correct answer is 5.

During the incomplete combustion of a hydrocarbon, carbon and carbon monoxide can be formed. Incomplete combustion occurs when there is a limited supply of oxygen, resulting in the formation of carbon instead of carbon dioxide. Carbon monoxide is also produced as a byproduct of incomplete combustion. Hydrogen is not formed during incomplete combustion, as it is already present in the hydrocarbon and does not undergo any chemical changes. Therefore, the correct answer is I and III only.

Neighbouring members of all homologous series 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 units in its molecular structure. This results in an increase in molecular size and mass as you move from one member to the next in the series.

The given formulas represent butane or its isomer because all three formulas have the same molecular formula, C4H10, which is the molecular formula for butane. Formula I represents butane, formula II represents 2-methylpropane (an isomer of butane), and formula III represents 2,2-dimethylpropane (another isomer of butane). Therefore, all three formulas represent butane or its isomer.

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.

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. This is because ethanal is an aldehyde that is formed when one hydrogen atom from the alcohol group of ethanol is removed and replaced with a double bond to an oxygen atom. Ethanal has a lower boiling point compared to other products listed, such as ethanoic acid, ethene, and ethane 1,2 diol, allowing it to be easily separated through distillation.

The given correct answer, 2,2-dimethylbutane, is the correct name for the compound. This is because it accurately describes the structure of the molecule, which consists of a butane backbone with two methyl groups attached to the second carbon atom. The prefix "2,2-dimethyl" indicates the positions of the methyl groups on the butane backbone.

1-chlorobutane is a member of the same homologous series as 1-chloropropane because both compounds belong to the alkyl halide family, which is characterized by a halogen atom (in this case, chlorine) bonded to a carbon chain. The only difference between the two compounds is the length of the carbon chain, with 1-chloropropane having three carbon atoms and 1-chlorobutane having four carbon atoms.

The correct answer is "it involves homolytic fission and Cl. radicals." In the reaction between methane and chlorine, homolytic fission occurs, which means that the bond between the carbon and hydrogen atoms in methane is broken equally, resulting in the formation of methyl radicals (CH3•). Similarly, the bond between the chlorine atoms in chlorine (Cl2) is also broken equally, resulting in the formation of chlorine radicals (Cl•).

A ketone is a type of compound that must contain a minimum of three carbon atoms. Ketones have a carbonyl group (C=O) in the middle of the carbon chain, with two additional carbon atoms bonded to it. This structure requires at least three carbon atoms in order to form a ketone. Aldehydes, carboxylic acids, and esters can have fewer than three carbon atoms and are not necessarily required to have a minimum of three carbon atoms.

CH3OH and C2H5OH both belong to the same homologous series because they have the same functional group, which is the hydroxyl group (-OH). In both compounds, the hydroxyl group is attached to a carbon atom. The only difference between the two compounds is the number of carbon atoms in the chain. Both compounds are alcohols, with CH3OH being methanol and C2H5OH being ethanol.

Butan-1-ol is converted to butanal by acidified potassium dichromate (VI) solution. This is because acidified potassium dichromate (VI) is a strong oxidizing agent that can oxidize alcohols to aldehydes. In the case of butan-1-ol, the alcohol functional group (-OH) is oxidized to an aldehyde functional group (-CHO), resulting in the formation of butanal.

The reaction (CH3)3Br + OH- involves the formation of a positive ion. In this reaction, the bromine atom in (CH3)3Br is replaced by the hydroxide ion (OH-), resulting in the formation of (CH3)3OH, which has a positive charge due to the transfer of a lone pair of electrons from the hydroxide ion. On the other hand, the reaction CH3CH2CH2Br + OH- does not involve the formation of a positive ion since the bromine atom is replaced by the hydroxide ion, resulting in the formation of CH3CH2CH2OH, which does not have a positive charge. Therefore, the correct answer is II only.

The compound CH3CH2CHO is a member of the aldehyde homologous series. Aldehydes are organic compounds that contain a carbonyl group (C=O) bonded to at least one hydrogen atom. In CH3CH2CHO, the carbonyl group is bonded to a hydrogen atom and an ethyl group (CH3CH2-), making it an aldehyde. CH3COCH3 is a ketone, as it has two alkyl groups bonded to the carbonyl group. CH3CH2CH2OH is an alcohol, as it has a hydroxyl group (-OH) bonded to the carbon atom. CH3CH2COOH is a carboxylic acid, as it has a carboxyl group (-COOH) bonded to the carbon atom.

The formula (CH3)2CHCH2Br represents the secondary halogenoalkane. This is because it contains a secondary carbon atom (attached to two other carbon atoms) bonded to a halogen (Br). In contrast, the other options either have primary carbon atoms (attached to only one other carbon atom) or tertiary carbon atoms (attached to three other carbon atoms). Therefore, the correct answer is (CH3)2CHCH2Br.

Alkanes are a group of hydrocarbons that have single bonds between carbon atoms. They form an homologous series with the general formula CnH2n+2, which means that the number of carbon atoms in the molecule determines the number of hydrogen atoms. However, alkane molecules with different carbon chain lengths will have different physical properties such as boiling points and melting points. Therefore, statement I is true but statement II is false. Alkanes do have similar chemical properties, as they undergo similar reactions such as combustion and substitution reactions. Therefore, statement III is true.