CBSE Class 10 Chemistry Carbon and Its Compounds MCQs Set 07

Answer: (d) 3-Methyl-1-butene

Explanation:
1. The longest chain containing the double bond has 4 carbons: CH2=CH–CH–CH3, so the parent is butene.
2. Number from the end nearer the double bond: double bond at C1, methyl at C3.
3. Name: 3-methylbut-1-ene.

Teacher's Note:
1. Choose the longest chain that includes the double bond and give the double bond the lowest number.
2. Names such as 'vinyl' or 'isopropyl' are common (trivial) names, not IUPAC.

Answer: (b) C₅H₈

Explanation:
1. Alkynes have the general formula \( C_nH_{2n-2} \).
2. For n = 5 this is C₅H₈. C₄H₈ fits CnH2n (alkene), and C₇H₁₉ is not a possible hydrocarbon.

Teacher's Note:
1. Test each formula with CnH2n+2 (alkane), CnH2n (alkene), CnH2n−2 (alkyne).
2. A hydrocarbon can never have more than 2n + 2 hydrogens.

Answer: (b) 2-Ethyl-3-methylbut-1-ene

Explanation:
1. The main chain is four carbons with the double bond at the end: CH2=C–CH–CH3, a butene.
2. Numbering from the double-bond end gives the double bond at C1, ethyl at C2 and methyl at C3.
3. Substituents are listed alphabetically (ethyl before methyl): 2-ethyl-3-methylbut-1-ene.

Teacher's Note:
1. The double bond takes the lowest locant, so it is never numbered 3 when 1 is possible.
2. List substituents alphabetically.

Answer: (a) urea

Explanation:
1. Wohler prepared urea from ammonium cyanate in 1828, which disproved the vital force theory.

Teacher's Note:
1. Ammonium cyanate rearranges on heating to urea.
2. This was the first synthesis of an organic compound from an inorganic one.

Answer: (d) all of these

Explanation:
1. Propane, CH3–CH2–CH3, has an open chain of carbon atoms, so it is acyclic.
2. 'Open chain', 'acyclic' and 'aliphatic' all describe it.

Teacher's Note:
1. Aliphatic means open-chain (or non-aromatic cyclic) hydrocarbons.
2. Acyclic simply means without a ring.

Answer: (b) Berzelius

Explanation:
1. Berzelius proposed that organic compounds are formed only in living organisms under the influence of a vital force.
2. Wohler's synthesis of urea later disproved it.

Teacher's Note:
1. Do not confuse the proposer (Berzelius) with the person who disproved the theory (Wohler).
2. The theory is now only of historical interest.

Answer: (a) ammonia

Explanation:
1. Hexane, urea and ethyl alcohol all contain carbon and hydrogen with covalent bonds, so they are organic.
2. Ammonia, NH₃, has no carbon, so it is inorganic.

Teacher's Note:
1. Organic compounds are hydrocarbons and their derivatives.
2. Urea is organic even though it was once made from inorganic substances.

Answer: (b) the property of catenation

Explanation:
1. Carbon atoms link with each other in long chains, branched chains and rings; this is catenation.
2. Together with its tetravalency this gives millions of compounds.
3. Carbon's valency is fixed at four, so 'variable valency' is wrong.

Teacher's Note:
1. Catenation is the unique ability of carbon (strong C–C bonds) to form chains and rings.
2. It is more pronounced in carbon than in silicon.

Answer: (a) ethanal

Explanation:
1. The compound has two carbons, so the parent is ethane; the aldehyde group changes the ending to 'al'.
2. IUPAC name: ethanal. Acetaldehyde is its common name.

Teacher's Note:
1. Aldehydes end in –al, ketones in –one, alcohols in –ol and acids in –oic acid.
2. Count carbons including the CHO carbon.

Answer: (b) 2-methylbutanoic acid

Explanation:
1. The chain containing the COOH carbon is four carbons long, so the parent is butanoic acid.
2. The COOH carbon is C1, so the methyl group on the next carbon is at position 2.
3. Name: 2-methylbutanoic acid.

Teacher's Note:
1. The carboxyl carbon is always C1, so it never gets a locant in the name.
2. Find the longest chain through the functional group, then locate the branch.

Answer: (c) ester

Explanation:
1. CH3COOC2H5 is ethyl ethanoate, with the group –COO– joining two alkyl groups.
2. That is the ester functional group.

Teacher's Note:
1. An ester is formed from a carboxylic acid and an alcohol.
2. Ketones have C=O between two carbons; esters have C(=O)–O–C.

Answer: (c) 5

Explanation:
1. The isomers are: hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane.
2. So there are 5.

Teacher's Note:
1. Systematically shorten the main chain: 6, 5, 4 carbons.
2. Remember that C₄H₁₀ has 2, C₅H₁₂ has 3 and C₆H₁₄ has 5 isomers.

Answer: (b) position isomerism

Explanation:
1. Both are C₃H₈O alcohols with the same carbon chain, but the –OH group is on C1 in one and on C2 in the other.
2. That is position isomerism.

Teacher's Note:
1. Same functional group, different position: position isomerism.
2. Different functional groups: functional isomerism.

Answer: (c) 14

Explanation:
1. Consecutive members differ by one –CH₂– group.
2. Mass of CH₂ = 12 + 2 = 14.

Teacher's Note:
1. A homologous series is a family with the same functional group and a CH₂ difference.
2. Their physical properties vary gradually.

Answer: (b) 2,4-Dimethylpentan-2-ol

Explanation:
1. The longest chain containing the carbon bearing OH has five carbons: C1 (CH3), C2 (C–OH, with a methyl), C3 (CH2), C4 (CH with a methyl) and C5 (CH3).
2. Number so that –OH gets the lowest number: it is at C2.
3. Methyl groups are at C2 and C4: 2,4-dimethylpentan-2-ol.

Teacher's Note:
1. The principal functional group (–OH) gets the lowest locant.
2. Then number the substituents.

Answer: (d) 1-bromo-3-methylbutane

Explanation:
1. The main chain is four carbons: BrCH2–CH2–CH(CH3)–CH3, a butane.
2. Number from the end that gives the lowest locants: Br at C1, methyl at C3.
3. Name: 1-bromo-3-methylbutane.

Teacher's Note:
1. Halogens are treated as prefixes and are alphabetised with the other substituents.
2. Pick the longest continuous chain, not the one drawn horizontally.

Answer: (c) C₃H₄

Explanation:
1. Alkanes obey \( C_nH_{2n+2} \).
2. C₂H₆, C₄H₁₀ and C₅H₁₂ fit; C₃H₄ has the formula of an alkyne (CnH2n−2).

Teacher's Note:
1. Check the hydrogen count against 2n + 2.
2. C₃H₄ is propyne.

Answer: (c) the same molecular formula and different structures

Explanation:
1. Isomers are compounds that have the same molecular formula but different structural arrangements of atoms, and so different properties.

Teacher's Note:
1. Same formula, different structure is the definition of isomerism.
2. Different formulas mean different compounds altogether.

Answer: (a) They have high melting and boiling points.

Explanation:
1. Organic compounds are covalent and are held by weak intermolecular forces, so they generally have low melting and boiling points.
2. They are mostly insoluble in water but soluble in organic solvents, and isomerism is common.

Teacher's Note:
1. High melting and boiling points are typical of ionic compounds.
2. Exceptions exist, such as sugar and urea.

Answer: (c) 3,3-Dimethyl-1-butene

Explanation:
1. The longest chain containing the double bond has four carbons: CH2=CH–C–CH3, so the parent is butene.
2. Number from the end nearer the double bond: the double bond is at C1 and both methyl groups are on C3.
3. The name is 3,3-dimethylbut-1-ene (written in the older style as 3,3-dimethyl-1-butene).

Teacher's Note:
1. Choose the longest chain that includes the double bond and give the double bond the lowest number.
2. Locants for identical substituents are repeated: 3,3-dimethyl.

Answer: (c) Cyclobutane

Explanation:
1. Aromatic compounds contain a benzene ring.
2. Benzene, naphthalene and phenol all contain a benzene ring.
3. Cyclobutane is a saturated four-membered ring with no benzene ring, so it is not aromatic.

Teacher's Note:
1. Aromatic compounds are based on the benzene ring with its delocalised electrons.
2. Cyclobutane is an alicyclic (non-aromatic) compound.