JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 08

Answer: (a) sp², sp

Explanation:
1. In CH₂=CH–C≡CH the carbons are C1 (sp²), C2 (sp²), C3 (sp) and C4 (sp).
2. The only C–C single bond is between C2 and C3.
3. So the hybrid states are sp² and sp.

Teacher's Note:
1. A carbon with a double bond is sp², with a triple bond is sp, with only single bonds is sp³.
2. Count the sigma bonds plus pi bonds to find the type.

Answer: (a) CH₃CH₂CHO

Explanation:
1. The aliphatic aldehydes in order are HCHO, CH₃CHO and CH₃CH₂CHO.
2. The third member is CH₃CH₂CHO (propanal).

Teacher's Note:
1. Successive members differ by one CH₂ group.
2. Do not confuse propanal with the ketone CH₃COCH₃.

Answer: (d) an acid and an ester also

Explanation:
1. The formula fits the saturated carboxylic acids and esters, C_nH_2nO₂, with n = 4.
2. Examples: butanoic acid (CH₃CH₂CH₂COOH) and ethyl ethanoate (CH₃COOC₂H₅).
3. So it can be an acid as well as an ester.

Teacher's Note:
1. Acids and esters of the same carbon number are functional isomers.
2. Both have the general formula C_nH_2nO₂.

Answer: (c) CH₃–NH–CH₂–CH₃

Explanation:
1. A higher homologue has one more CH₂ group and the same functional group, here a secondary amine (–NH–).
2. CH₃–NH–CH₂–CH₃ is a secondary amine with one more CH₂.
3. The other options are primary or tertiary amines.

Teacher's Note:
1. Homologues share the functional group exactly.
2. The type of amine (primary, secondary, tertiary) must stay the same.

Answer: (c) C₆H₈

Explanation:
1. Alkenynes contain one double and one triple bond; the first member is CH₂=CH–C≡CH, C₄H₄.
2. The second is C₅H₆ and the third is C₆H₈.

Teacher's Note:
1. Each next member adds CH₂.
2. The general formula of an alkenyne is C_nH_2n−4.

Answer: (a) 2

Explanation:
1. Olefinic bonds are carbon-carbon double bonds.
2. There are two: CH₂=CH and CH=CH.
3. The C=O and C≡N bonds are not olefinic.

Teacher's Note:
1. Olefinic means C=C only.
2. Do not count the carbonyl C=O.

Answer: (c) 1

Explanation:
1. Acetylenic bonds are carbon-carbon triple bonds.
2. There is one, HC≡C–.
3. The C≡N is a nitrile bond, not an acetylenic one.

Teacher's Note:
1. Acetylenic means C≡C.
2. A C≡N bond is called a cyano or nitrile bond.

Answer: (b) 3

Explanation:
1. The first ester is methyl methanoate, HCOOCH₃ (2 carbons).
2. The second member has one more CH₂, for example CH₃COOCH₃, with 3 carbons.

Teacher's Note:
1. The smallest ester has 2 carbon atoms.
2. Count carbons of both the acid and the alcohol parts.

Answer: (d) (CH₃)₂CH–O–CH(CH₃)₂

Explanation:
1. A simple ether has the same alkyl group on both sides of the oxygen.
2. (CH₃)₂CH–O–CH(CH₃)₂ is diisopropyl ether.
3. CH₃COCH₃ is a ketone and the other two ethers are mixed ethers.

Teacher's Note:
1. Look for identical groups on both sides of the C–O–C unit.
2. Isopropyl on both sides gives a symmetrical ether.

Answer: (c) 5

Explanation:
1. A primary carbon is joined to only one other carbon.
2. Neopentane, C(CH₃)₄, has four CH₃ groups on one central carbon, so four primary carbons.
3. It has 5 carbon atoms and no smaller alkane has four primary carbons.

Teacher's Note:
1. Four methyl groups are needed.
2. They are attached to a quaternary carbon, so 4 + 1 = 5.

Answer: (c) CH₃CH₂CN

Explanation:
1. The carbon of a nitrile group, –C≡N, forms one triple bond and one single bond, so it is sp hybridised.
2. CH₃COOH and CH₃COCH₃ have sp² and sp³ carbons only, and butadiene has sp² carbons only.

Teacher's Note:
1. sp hybridisation appears in triple bonds and in cumulated double bonds.
2. C≡N counts as a triple bond.

Answer: (c) sp – sp³

Explanation:
1. C2 is part of the triple bond, so it is sp.
2. C3 is the CH₂ carbon, so it is sp³.
3. The bond is sp – sp³.

Teacher's Note:
1. Number the chain from the end with the triple bond.
2. Name the bond by the hybrid state of the two carbons.

Answer: (a) H₂C=CH–C≡CH

Explanation:
1. In H₂C=CH–C≡CH the first two carbons carry a double bond (sp², sp²) and the last two carry a triple bond (sp, sp).
2. The sequence is sp²–sp²–sp–sp.

Teacher's Note:
1. Write the hybrid state under every carbon, left to right.
2. A cumulated diene has sp centre carbons.

Answer: (d) Acetic acid and butyric acid

Explanation:
1. Homologues have the same functional group and differ by one or more CH₂ units.
2. Acetic acid (C₂) and butyric acid (C₄) are both carboxylic acids differing by two CH₂ units.
3. The other pairs are isomers or have different functional groups.

Teacher's Note:
1. Isomers have the same formula, homologues do not.
2. Check that the functional group is the same.

Answer: (c) Alkanal: C_nH_2n+1O

Explanation:
1. Alkanals have the general formula C_nH_2nO, for example CH₃CHO is C₂H₄O.
2. The other three formulae are correct.

Teacher's Note:
1. Check with the first member: HCHO is CH₂O, which fits C_nH_2nO.
2. Saturated acids and esters share C_nH_2nO₂.

Answer: (a) 120° and 109.5°

Explanation:
1. In tetrachloroethene each carbon is sp² hybridised and trigonal planar, so Cl–C–Cl is about 120°.
2. In CCl₄ the carbon is sp³, giving the tetrahedral angle of 109.5°.

Teacher's Note:
1. sp² gives about 120° and sp³ gives 109.5°.
2. The angle depends on the hybrid state of the central carbon.

Answer: (a) 2

Explanation:
1. The simplest ester is methyl methanoate, HCOOCH₃.
2. It contains 2 carbon atoms, one from the acid part and one from the alcohol part.

Teacher's Note:
1. An ester needs a carbon in the acyl group and at least one in the alkoxy group.
2. Methyl methanoate is the smallest ester.

Answer: (d) Alkenes and cyclic alkanes

Explanation:
1. Alkenes (one C=C) and cycloalkanes (one ring) both have two hydrogens fewer than an alkane, C_nH_2n+2.
2. So both fit C_nH_2n.

Teacher's Note:
1. One double bond or one ring removes two hydrogens from the alkane formula.
2. Alkenes and cycloalkanes are therefore isomers.

Answer: (c) C₉H₁₆

Explanation:
1. Alkynes follow C_nH_2n−2.
2. For C₉H₁₆: 2 \(\times\) 9 − 2 = 16, which fits.
3. C₇H₁₄ and C₁₆H₃₂ fit alkenes, C₁₀H₂₂ fits an alkane.

Teacher's Note:
1. A formula with two hydrogens fewer than C_nH_2n may contain an alkyne, a diene or a ring with a double bond.
2. Only C₉H₁₆ is of that form here.

Answer: (c) a and c

Explanation:
1. A tertiary radical has its free valence on a carbon that is joined to three other carbon atoms.
2. (a) tert-butyl and (c) tert-pentyl satisfy this.
3. (b) isopropyl is secondary and (d) neopentyl is primary.

Teacher's Note:
1. Count the carbon atoms attached to the carbon carrying the free valence.
2. One carbon neighbour gives primary, two secondary, three tertiary.

Answer: (a) vinyl acetate

Explanation:
1. The compound is the ester of acetic acid (CH₃COO–) with the vinyl group (CH₂=CH–).
2. Its common name is vinyl acetate.
3. (Vinyl ethanoate is the IUPAC-style name, not the common name.)

Teacher's Note:
1. In ester names the alkyl or alkenyl group is named first, then the acid part.
2. Vinyl acetate is the monomer of polyvinyl acetate.

Answer: (d) a primary, secondary or tertiary carbon atom

Explanation:
1. The class of an amine depends on the number of alkyl groups on the nitrogen, not on the type of carbon.
2. In a primary amine the nitrogen carries one alkyl group, and that carbon may be primary, secondary or tertiary.
3. Examples: CH₃CH₂NH₂, (CH₃)₂CHNH₂ and (CH₃)₃CNH₂.

Teacher's Note:
1. Amines are classified by substitution on nitrogen.
2. Alcohols and alkyl halides are classified by the carbon.