JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 19

Answer: (b) 3, 5, 0

Explanation:
1. A hydrogen is primary, secondary or tertiary according to whether the carbon holding it is attached to one, two or three other carbons.
2. The CH3 group carbon is attached to one carbon (the ring), so its 3 hydrogens are primary.
3. The five ring CH carbons are each attached to two ring carbons, so their 5 hydrogens are secondary. The ring carbon bearing CH3 has no hydrogen, so there are no tertiary hydrogens.

Teacher's Note:
1. Classify hydrogens by the type of carbon they are bonded to.
2. Count carefully, since the ring carbon with the substituent carries no hydrogen.

Answer: (b) sp, sp, sp², sp², sp³

Explanation:
1. Carbon with a triple bond (C1 and C2) has two sigma bonds and is sp.
2. Carbon with a double bond (C3 and C4) has three sigma bonds and is sp².
3. The CH3 carbon has four single bonds and is sp³.

Teacher's Note:
1. Count the sigma bonds and lone pairs to find hybridisation.
2. A triple bond gives sp, a double bond gives sp², single bonds only give sp³.

Answer: (c) Benzene

Explanation:
1. A heterocyclic compound has at least one atom other than carbon (a hetero atom such as N, O or S) in the ring.
2. Thiophene has S, furan has O and pyridine has N in the ring.
3. Benzene has only carbon atoms in its ring, so it is not heterocyclic.

Teacher's Note:
1. Look for a non-carbon atom inside the ring.
2. Benzene is the standard carbocyclic aromatic compound.

Answer: (a) 1

Explanation:
1. The ring consists of CH=CH, two C=O carbons and one ring oxygen.
2. The only atom other than carbon in the ring is the oxygen.
3. So there is 1 hetero atom in the ring (the carbonyl oxygens are outside the ring).

Teacher's Note:
1. Hetero atoms are counted inside the ring only.
2. Atoms in substituents on the ring do not count.

Answer: (c) \( \ce{C6H5-} \)

Explanation:
1. A radical is primary, secondary or tertiary according to the number of carbons attached to the carbon carrying the free valency.
2. The vinyl carbon (a) and the n-butyl carbon (d) are bonded to one carbon (primary).
3. The tert-butyl carbon (b) is bonded to three carbons (tertiary).
4. The phenyl carbon (c) is bonded to two ring carbons, so it is secondary.

Teacher's Note:
1. Count the carbons attached to the radical carbon.
2. A ring carbon in benzene has two carbon neighbours.

Answer: (d) \( \ce{CH3-CH=CH-COOH} \)

Explanation:
1. Crotonic acid is the four-carbon unsaturated acid, but-2-enoic acid.
2. Its structure is \( \ce{CH3-CH=CH-COOH} \).
3. Option (a) is acrylic acid and (c) is butyric acid.

Teacher's Note:
1. Crotonic acid is the trans but-2-enoic acid.
2. Remember acrylic acid has three carbons.

Answer: (c) Tertiary butyl carbinol

Explanation:
1. Neopentyl alcohol is \( \ce{(CH3)3C-CH2-OH} \).
2. In the carbinol system the CH2OH part is carbinol, and it is attached to a tertiary butyl group.
3. So the name is tertiary butyl carbinol.

Teacher's Note:
1. Carbinol means the CH2OH group.
2. Name the group attached to it as an alkyl prefix.

Answer: (c) Ethyl-2-bromopropanoate

Explanation:
1. The compound is an ester. The alkyl group on oxygen (ethyl) is named first, then the acid part.
2. The acid part is a three-carbon chain with Br on C2 (next to the carbonyl carbon C1), giving 2-bromopropanoate.
3. The name is ethyl 2-bromopropanoate.

Teacher's Note:
1. In an ester name the alkyl group on oxygen comes first, as a separate word.
2. Number the acid chain from the carbonyl carbon.

Answer: (d) Ethanoic anhydride

Explanation:
1. The compound is a symmetrical acid anhydride made from two ethanoic acid units.
2. The IUPAC name replaces acid with anhydride, so it is ethanoic anhydride.
3. (Acetic anhydride is the common name.)

Teacher's Note:
1. IUPAC names avoid the common word acetic.
2. Symmetrical anhydrides take the acid name plus anhydride.

Answer: (a) \( \ce{CH3CH2CHO} \)

Explanation:
1. The aliphatic aldehydes in order are methanal \( \ce{HCHO} \), ethanal \( \ce{CH3CHO} \) and propanal \( \ce{CH3CH2CHO} \).
2. So the third member is propanal.
3. Options (c) and (d) are ketones.

Teacher's Note:
1. Count HCHO as the first member.
2. Each member adds one CH2.

Answer: (d) an acid and an ester also

Explanation:
1. \( \ce{C4H8O2} \) fits the general formula \( C_nH_{2n}O_2 \) of saturated monocarboxylic acids and esters.
2. For example it can be butanoic acid \( \ce{CH3CH2CH2COOH} \) or ethyl ethanoate \( \ce{CH3COOC2H5} \).
3. So it represents both an acid and an ester.

Teacher's Note:
1. Carboxylic acids and esters are functional isomers.
2. Both have the general formula CnH2nO2.

Answer: (c) \( \ce{C6H8} \)

Explanation:
1. Alkenynes contain both a double and a triple bond. The first member is \( \ce{CH2=CH-C#CH} \) (\( \ce{C4H4} \)).
2. Each next member adds CH2: second member \( \ce{C5H6} \), third member \( \ce{C6H8} \).

Teacher's Note:
1. Find the first member of the family first.
2. Then add CH2 for each next member.

Answer: (a) 2

Explanation:
1. The compound has the double bonds CH2=CH and CH=CH, so two C=C bonds.
2. The C=O is a carbonyl and the C#N is a nitrile, neither of which is olefinic.

Teacher's Note:
1. Olefinic means a carbon-carbon double bond.
2. Do not count C=O or C#N.

Answer: (c) 1

Explanation:
1. The compound has only one carbon-carbon triple bond, \( \ce{HC#C-} \).
2. The C#N bond is a carbon-nitrogen triple bond, which is a nitrile bond and not acetylenic.

Teacher's Note:
1. Acetylenic means a carbon-carbon triple bond only.
2. The nitrile triple bond does not count.

Answer: (d) \( \ce{(CH3)2CH-O-CH(CH3)2} \)

Explanation:
1. A symmetrical ether has the same alkyl group on both sides of the oxygen.
2. Option (d) is di-isopropyl ether.
3. (a) is a ketone, and (b) and (c) are unsymmetrical (mixed) ethers.

Teacher's Note:
1. Symmetrical ethers are R-O-R.
2. Check that both groups are identical.

Answer: (c) 5

Explanation:
1. A primary carbon is attached to only one other carbon.
2. The smallest alkane with four terminal CH3 groups is neopentane, \( \ce{C(CH3)4} \), with 5 carbons.
3. Isobutane has only three primary carbons.

Teacher's Note:
1. A branch carbon bearing four CH3 groups gives the most primary carbons per carbon.
2. Neopentane is the answer.

Answer: (c) \( \ce{CH3CH2CN} \)

Explanation:
1. A carbon with a triple bond or two double bonds is sp-hybridised.
2. In \( \ce{CH3CH2C#N} \) the nitrile carbon is joined by a triple bond to N, so it is sp.
3. The other compounds have only sp³ and sp² carbons.

Teacher's Note:
1. Triple bond means sp.
2. Nitrile carbons are sp hybridised.

Answer: (c) sp - sp³

Explanation:
1. C2 is part of the triple bond, so it is sp hybridised.
2. C3 is the CH2 carbon with four single bonds, so it is sp³.
3. The C2-C3 bond is therefore sp - sp³.

Teacher's Note:
1. Number the chain from the triple bond end.
2. Assign hybridisation to the two carbons separately.

Answer: (a) 120° and 109.5°

Explanation:
1. In tetrachloroethene \( \ce{Cl2C=CCl2} \) each carbon is sp² hybridised and planar, so the Cl-C-Cl angle is about 120°.
2. In tetrachloromethane the carbon is sp³ and tetrahedral, so the angle is 109.5°.

Teacher's Note:
1. sp² gives about 120 degrees and sp³ gives 109.5 degrees.
2. The double-bonded carbon is trigonal planar.

Answer: (a) 2

Explanation:
1. The simplest ester is methyl methanoate \( \ce{HCOOCH3} \).
2. It has one carbon in the acid part and one in the alkyl part, so a minimum of 2 carbons.

Teacher's Note:
1. An ester needs at least one carbon in each of the two parts.
2. The first member is HCOOCH3.

Answer: (c) a and c

Explanation:
1. A tertiary radical has its free valency on a carbon attached to three other carbons.
2. In (a) and (c) the radical carbon carries three alkyl groups.
3. (b) is secondary, and (d) is primary because the CH2 carbon is attached to only one carbon.

Teacher's Note:
1. Count the carbon atoms attached to the carbon with the free valency.
2. Neopentyl is a primary radical.

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

Explanation:
1. Amines are classified by the number of alkyl groups on nitrogen, not by the type of carbon.
2. A primary amine has one alkyl group on NH2, and that alkyl carbon can be primary, secondary or tertiary.
3. Examples are ethylamine, isopropylamine and tert-butylamine, all primary amines.

Teacher's Note:
1. Alcohols are classified by the carbon type, but amines by the nitrogen.
2. tert-Butylamine is still a primary amine.