Class 11 Chemistry Chapter 8: Organic Chemistry – Some Basic Principles and Techniques worksheet with answer key
A 20-mark worksheet on Organic Chemistry – Some Basic Principles and Techniques for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
Show answer
Answer: (c) ammonium cyanate
Wöhler heated ammonium cyanate: NH₄CNO → NH₂CONH₂. The other three are inorganic salts, but not the one he used.
Show answer
Answer: (c) space-filling model
The space-filling model shows the relative size of each atom from its van der Waals radius and no bonds. The framework model shows only bonds; ball-and-stick and ball-and-spring show both atoms and bonds.
Show answer
Answer: (a) HC≡CH
Electronegativity of carbon rises with the s character of its hybrid orbitals. The sp carbon of ethyne has 50% s character, more than sp² (33%) or sp³ (25%).
Show answer
Answer: (d) a heterocyclic alicyclic (non-aromatic) compound
Its ring holds an atom other than carbon, so it is heterocyclic, and it is saturated, so it is alicyclic, not aromatic. Furan, its unsaturated relative, is the heterocyclic aromatic one.
Show answer
Answer: (d) 53.3%
Each mole of O₂ in the compound gives 2 mol CO₂, so 88 g CO₂ comes from 32 g O₂. % O = (32 × 0.44 × 100)/(88 × 0.30) = 53.3%. Using 44 g gives an impossible 106.7%; 40.0% is the carbon fraction 12/44; 26.7% halves the right answer.
Show answer
Ethyne < ethene < ethane. The carbons are sp, sp² and sp³ hybridised. The more s character the hybrid orbital has, the closer it is to the nucleus, and the shorter and stronger the bond it forms; sp has 50%, sp² 33% and sp³ 25% s character.
- order ethyne < ethene < ethane (1)
- reason from s character of sp, sp², sp³ (1)
Show answer
Only acetamide (its nitrogen is in an amide group). In nitrobenzene the nitrogen is in a nitro group and in pyridine it is in the ring; under Kjeldahl conditions such nitrogen does not change to ammonium sulphate, so the ammonia found is too low. Dumas' method is used for them.
- acetamide only (1)
- nitro and ring nitrogen are not converted to ammonium sulphate (1)
Show answer
(a) The mixture boils when p(oil) + p(water) = p(atmosphere): p(oil) = 760 – 733 = 27 mm. (b) The oil is steam volatile and is immiscible with water.
- p(oil) = 760 – 733 = 27 mm (1)
- steam volatile and immiscible with water (1)
Show answer
(i) The longest chain has six carbons; numbering from the right gives the methyls 2, 4, 4 (from the left 3, 3, 5): 2,4,4-trimethylhexane. (ii) –COOH is principal, its carbon is C-1, Br on C-3: 3-bromobutanoic acid. (iii) –CN outranks the keto group; the nitrile carbon is C-1 and the C=O is on C-4 of a five-carbon chain: 4-oxopentanenitrile.
- 2,4,4-trimethylhexane (1)
- 3-bromobutanoic acid (1)
- 4-oxopentanenitrile (1)
Show answer
(a) C⁺H₃: carbon has a sextet; sp² hybridised; trigonal planar, with an empty p orbital perpendicular to the plane. H₃C:⁻: carbon has an octet (three bond pairs and a lone pair); generally sp³; a distorted tetrahedron. (b) ĊH₃ < ĊH₂CH₃ < ĊH(CH₃)₂ < Ċ(CH₃)₃: stability rises from primary to tertiary.
- carbocation: sextet, sp², trigonal planar (1)
- carbanion: octet, sp³, distorted tetrahedron (1)
- radical order methyl < ethyl < isopropyl < tert-butyl (1)
Show answer
(a) I: CH₂=CH–Ö–CH₃. An arrow from a lone pair on O to the O–C bond and an arrow from the C=C π bond onto the terminal CH₂ give II: ⁻:CH₂–CH=O⁺–CH₃. (b) I contributes more: it has no separation of charge, while II puts the negative charge on carbon and the positive charge on the more electronegative oxygen. (c) +R: the lone pair moves away from O into the π system, so the terminal CH₂ carbon becomes electron rich.
- both structures with correct curved arrows (1)
- I is major: no charge separation; II has + on O (1)
- +R, terminal CH₂ carbon electron rich (1)