JEE Chemistry Organic Chemistry Some Basic Principles and Techniques MCQs Set 48

Answer: (a) Cyclohexylbenzene

Explanation:
1. The compound has a phenyl ring bonded to a saturated six-membered ring.
2. The aromatic ring is the parent, and the saturated ring is the cyclohexyl substituent.
3. Name: cyclohexylbenzene.

Teacher's Note:
1. Biphenyl would need two benzene rings.
2. Hexylbenzene has an open chain of six carbons.

Answer: (b) (CH₃)₃CCH₂CH₃

Explanation:
1. A tertiary hydrogen is on a carbon attached to three other carbons.
2. Option (a) has CH(CH₃)₂ attached to the quaternary carbon, so that CH carbon carries three carbon neighbours.
3. Option (b), 2,2-dimethylbutane, has only CH₃, CH₂ and quaternary carbons, so it has no tertiary hydrogen.

Teacher's Note:
1. Count carbon neighbours of each hydrogen-bearing carbon.
2. A quaternary carbon has no hydrogen at all.

Answer: (c) 3

Explanation:
1. Propene is CH₂=CH–CH₃.
2. Chlorine can replace a hydrogen on C1 (1-chloropropene), on C2 (2-chloropropene) or on the methyl group (3-chloropropene).
3. So there are 3 structural isomers (geometrical isomers of 1-chloropropene are not structural).

Teacher's Note:
1. Count the sets of equivalent hydrogens: =CH₂, =CH– and –CH₃.
2. Stereo-isomers are not counted among structural isomers.

Answer: (b) Planar

Explanation:
1. The two double-bonded carbons are sp² hybridised with bond angles of about 120 degrees.
2. The two carbons and the four atoms directly attached to them lie in one plane.
3. So the shape around the double bond is planar.

Teacher's Note:
1. sp² gives trigonal planar geometry.
2. Rotation about a double bond is restricted, which causes cis-trans isomerism.

Answer: (c) 2-Propenenitrile

Explanation:
1. The nitrile carbon counts in the main chain, so the chain has three carbons: C1 (the CN carbon), C2 and C3.
2. The double bond is between C2 and C3.
3. Name: prop-2-enenitrile (written 2-propenenitrile).

Teacher's Note:
1. Count the nitrile carbon in the chain length.
2. The suffix -nitrile takes locant 1.

Answer: (b) 5

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

Teacher's Note:
1. Start with the longest chain and shorten it step by step.
2. Check for repeated structures by symmetry.

Answer: (b) CH₃–CH=CH₂

Explanation:
1. An unsaturated hydrocarbon contains only carbon and hydrogen and has a multiple bond.
2. Propene contains only C and H and has a double bond.
3. The nitrile and the aldehyde contain N and O respectively, so they are not hydrocarbons.

Teacher's Note:
1. Hydrocarbon means only C and H.
2. Unsaturated means at least one multiple bond.

Answer: (c) 2

Explanation:
1. Primary alcohols have the form R–CH₂OH with R = C₃H₇.
2. R can be n-propyl or isopropyl, giving butan-1-ol and 2-methylpropan-1-ol.
3. So there are 2.

Teacher's Note:
1. Primary means the carbon bearing OH is attached to only one other carbon.
2. The other butanols are secondary (butan-2-ol) and tertiary (2-methylpropan-2-ol).

Answer: (b) 3

Explanation:
1. The molecule has a two-fold axis and a mirror plane, so positions 1 and 4 are equivalent, positions 5 and 8 are equivalent, and positions 6 and 7 are equivalent.
2. The three distinct positions are 1(4), 5(8) and 6(7).
3. So 3 mononitro isomers are possible.

Teacher's Note:
1. Use symmetry to group equivalent positions.
2. The two rings are different because only one carries chlorine.

Answer: (b) sp to sp²

Explanation:
1. The nitrile nitrogen is bonded through a triple bond, so it is sp hybridised.
2. In the amide N–CO–CH₃, the nitrogen lone pair is delocalised into the carbonyl group, which makes the nitrogen sp².
3. So the change is sp to sp².

Teacher's Note:
1. Amide nitrogen is planar because of delocalisation of its lone pair.
2. Nitrile nitrogen is linear.

Answer: (b) 3-Methyl-2-buten-1-ol

Explanation:
1. The OH group is the principal group, so the carbon bearing OH is C1.
2. C2=C3 is the double bond, and C3 carries a methyl group.
3. Name: 3-methylbut-2-en-1-ol.

Teacher's Note:
1. Give OH the lowest locant, then the double bond.
2. The methyl group is a substituent on C3.

Answer: (d) 10

Explanation:
1. Naphthalene has two types of position: alpha (1,4,5,8) and beta (2,3,6,7).
2. Counting distinct pairs of positions by symmetry gives 10 dinitronaphthalenes.

Teacher's Note:
1. Use the D_2h symmetry of naphthalene to remove duplicates.
2. The result 10 is the well-known count.

Answer: (d) CH₃CH(CH₃)CHClCH₂OH

Explanation:
1. Butan-1-ol is HOCH₂CH₂CH₂CH₃, with C1 carrying OH.
2. Chlorine goes on C2 and a methyl on C3, giving HOCH₂–CHCl–CH(CH₃)–CH₃.
3. This matches option (a).

Teacher's Note:
1. Build the chain from the parent name and add substituents at the stated locants.
2. Check carbon count: four in the chain plus one methyl.

Answer: (a) 4

Explanation:
1. The alcohols are butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol.
2. That gives 4.

Teacher's Note:
1. Replace one H of each butane skeleton by OH.
2. n-Butane gives 2 and isobutane gives 2.

Answer: (c) 120°

Explanation:
1. Each carbon in ethylene is sp² hybridised and trigonal planar.
2. The ideal angle in such an arrangement is 120°.

Teacher's Note:
1. sp gives 180°, sp² gives 120°, sp³ gives 109°28'.
2. The real H–C–H angle is slightly less than 120°.

Answer: (b) Ethyl 2-(chlorocarbonyl)benzoate

Explanation:
1. In the priority order carboxylic acid, anhydride, ester, acid halide, the ester ranks above the acid chloride, so the ester is the principal group.
2. The parent is ethyl benzoate, with the ester carbon at C1.
3. The COCl group at C2 is named chlorocarbonyl: ethyl 2-(chlorocarbonyl)benzoate.

Teacher's Note:
1. Priority order: carboxylic acid, anhydride, ester, acid halide, amide.
2. Name the ester as alkyl benzoate with the other group as a prefix.

Answer: (d) 6 sigma and 2 pi

Explanation:
1. There are two C–C double bonds, each with one sigma and one pi bond.
2. Four C–H bonds add four sigma bonds.
3. Total sigma = 2 + 4 = 6 and pi = 2.

Teacher's Note:
1. Every single bond is one sigma; each double bond is one sigma plus one pi.
2. Propadiene has two cumulated double bonds.

Answer: (c) 3

Explanation:
1. The ethers of C₄H₁₀O are methoxypropane, 2-methoxypropane (methyl isopropyl ether) and ethoxyethane.
2. That gives 3.

Teacher's Note:
1. Split the four carbons as 1+3 and 2+2.
2. Count isomeric propyl groups for the 1+3 split.

Answer: (b) 4-Bromo-3-methylpent-2-ene

Explanation:
1. The longest chain containing the double bond has five carbons.
2. Numbering from the end that gives the double bond the lower locant puts the double bond at C2=C3.
3. The methyl is on C3 and bromine on C4: 4-bromo-3-methylpent-2-ene.

Teacher's Note:
1. The double bond gets lowest locant before the halogen.
2. Pentene chain: C1 to C5.

Answer: (b) n-Propyl propanoate

Explanation:
1. The alkyl group on oxygen is propyl.
2. The acyl part CH₃CH₂C(=O)– comes from propanoic acid.
3. Name: propyl propanoate.

Teacher's Note:
1. Read the group on O first, then the acid part.
2. A symmetrical-looking name can still describe an unsymmetrical ester.

Answer: (b) 12

Explanation:
1. Both rings are equivalent. Place the two bromine atoms either on the same ring or on different rings.
2. Same ring: 6 possible position pairs after symmetry (2,3; 2,4; 2,5; 2,6; 3,4; 3,5).
3. Different rings: 6 pairs (2,2'; 2,3'; 2,4'; 3,3'; 3,4'; 4,4'). Total = 12.

Teacher's Note:
1. Treat the biphenyl linkage as carbon 1 of each ring.
2. Count same-ring and different-ring cases separately.

Answer: (c) Ethyl 4-methoxybutanoate

Explanation:
1. The ester is the principal group; the alkyl group on oxygen is ethyl.
2. The acid chain has four carbons counting the carbonyl carbon as C1, and the methoxy group is on C4.
3. Name: ethyl 4-methoxybutanoate.

Teacher's Note:
1. Count the carbonyl carbon as C1.
2. Ether groups become alkoxy prefixes when a higher priority group is present.

Answer: (b) 5-Bromo-3-(bromomethyl)pent-1-ene

Explanation:
1. The longest chain containing the double bond runs C1=C2–C3–C4–C5Br, five carbons.
2. A bromomethyl group is attached at C3 and a bromine at C5.
3. Name: 5-bromo-3-(bromomethyl)pent-1-ene.

Teacher's Note:
1. The chain must contain the double bond.
2. Complex substituents such as bromomethyl are placed in brackets.

Answer: (b) 2-Ethynylbut-3-en-1-amide

Explanation:
1. The amide is the principal group, so the chain must contain the carbonyl carbon as C1.
2. Through the vinyl group the chain is C1(=O)NH₂, C2, C3=C4; through the ethynyl group it is C1, C2, C3≡C4. Both have four carbons and one multiple bond.
3. When lengths tie, the chain with the greater number of double bonds is chosen, so the vinyl chain is the parent and ethynyl is a prefix.
4. Name: 2-ethynylbut-3-en-1-amide (but-3-enamide).

Teacher's Note:
1. Amides take the suffix -amide and C1 is the carbonyl carbon.
2. Chain selection order: length, then number of multiple bonds, then number of double bonds.

Answer: (a) 3-(Bromomethyl)-5-chloropent-1-ene

Explanation:
1. The longest chain containing the double bond is C1=C2–C3–C4–C5Cl, which has five carbons.
2. The bromomethyl group is on C3 and chlorine is on C5.
3. Name: 3-(bromomethyl)-5-chloropent-1-ene (substituents in alphabetical order, bromomethyl before chloro).

Teacher's Note:
1. Number from the end nearer the double bond.
2. Alkoxy and halo groups are only prefixes.