Q43.The edge length of a face centered cubic cell of an ionic substance is 508pm. If the radius of the cation is 110pm, the radius of the anion is (1) 288pm (2) 398pm (3) 618pm (4) 144pm
What This Question Tests
This question tests the understanding of the E1 dehydration mechanism of alcohols, including carbocation formation, rearrangement via hydride shift to form a more stable carbocation, and subsequent proton elimination following Saytzeff's rule.
Concepts Tested
📚 NCERT Sections This Tests
4.2 — Ionic Or Electrovalent Bond Other Factors. The Crystal Structure Of Sodium
Chemistry Class 11 · Chapter 4
4.2 Ionic or Electrovalent Bond other factors. The crystal structure of sodium chloride, NaCl (rock salt), for example isFrom the Kössel and Lewis treatment of the shown below.formation of an ionic bond, it follows that the formation of ionic compounds would primarily depend upon: • The ease of formation of the positive and negative ions from the respective neutral atoms; • The arrangement of the positive and negative ions in the solid, that is, the lattice of the crystalline compound. The formation of a positive ion involves ionization, i.e., removal of electron(s) from the neutral atom and that of the negative ion involves the addition of electron(s) to the Rock salt structure neutral atom. In ionic solids, the sum of the electron gain M(g) → M+(g) + e– ; enthalpy and the ionization enthalpy may be Ionization enthalpy positive but still the crystal structure gets X(g) + e– → X – (g) ; stabilized due to the energy released in the Electron gain enthalpy formation of the crystal lattice. For example: the ionization enthalpy for Na+(g) formation M+(g) + X –(g) → MX(s) from Na(g) is 495.8 kJ mol–1 ; while the electron The electron gain enthalpy, ∆egH, is the gain enthalpy for the change Cl(g) + e–→ enthalpy change (Unit 3), when a gas phase Cl– (g) is, – 348.7 kJ mol–1 only. The sum of the atom in its ground state gains an electron. two, 147.1 kJ mol-1 is more than compensated The electron gain process may be exothermic for by the enthalpy of lattice formation of or endothermic. The ionization, on the other NaCl(s) (–788 kJ mol–1). Therefore, the energy hand, is always endothermic. Electron released in the processes is more than the Reprint 2025-26 Chemical Bonding And Molecular Structure 107 energy absorbed. Thus a qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state. Since lattice enthalpy plays a key role in the formation of ionic compounds, it is important that we learn more about it. 4.2.1 Lattice Enthalpy The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions. For example, the lattice enthalpy of NaCl is 788 kJ mol–1. This means that 788 Fig. 4.1 The bond length in a covalent kJ of energy is required to separate one mole molecule AB. of solid NaCl into one mole of Na+ (g) and one R = rA + rB (R is the bond length and rA and rB are mole of Cl– (g) to an infinite distance. the covalent radii of atoms A and B respectively) This process involves both the attractive forces between ions of opposite charges in the same molecule. The van der Waals and the repulsive forces between ions of radius represents the overall size of the like charge. The solid crystal being three- atom which includes its valence shell in a dimensional; it is not possible to calculate nonbonded situation. Further, the van der lattice enthalpy directly from the interaction Waals radius is half of the distance between of forces of attraction and repulsion only. two similar atoms in separate molecules in Factors associated with the crystal geometry a solid. Covalent and van der Waals radii of have to be included. chlorine are depicted in Fig. 4.2.
7.23 — Give Iupac Names Of The Following Ethers:
Chemistry Class 12 · Chapter 7
7.23 Give IUPAC names of the following ethers: 7.24 Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (i) 1-Propoxypropane (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane 7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers. 7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction. 7.27 Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason. 7.28 Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether. 7.29 Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring. 7.30 Write the mechanism of the reaction of HI with methoxymethane. 7.31 Write equations of the following reactions: (i) Friedel-Crafts reaction – alkylation of anisole. (ii) Nitration of anisole. (iii) Bromination of anisole in ethanoic acid medium. (iv) Friedel-Craft’s acetylation of anisole. 7.32 Show how would you synthesise the following alcohols from appropriate alkenes? CH3 OH (i) OH (ii) OH (iii) (iv) OH 7.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom. Chemistry 224 Reprint 2025-26 Answers to Some Intext Questions 7.1 Primary alcohols (i), (ii), (iii) Secondary alcohols (iv) and (v) Tertiary alcohols (vi) 7.2 Allylic alcohols (ii) and (vi) 7.3 (i) 4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol (ii) 2, 5-Dimethylhexane-1,3-diol (iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol 7.4 OH CH2 C OCH3 7.5 (i) CH3 CH CH3 (ii) O OH (iii) CH3 CH2 CH CH2 OH CH3 7.7 (i) 1-Methylcyclohexene (ii) A Mixture of but-1-ene and but-2-ene. But-2-ene is the major product formed due to rearrangement to give secondary carbocation.
7.7 — Predict The Major Product Of Acid Catalysed Dehydration Of
Chemistry Class 12 · Chapter 7
7.7 Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol
📋 Question Details
- Chapter
- Alcohols Phenols Ethers
- Topic
- Dehydration of alcohols, Carbocation rearrangement, Saytzeff's rule
- Year
- 2010
- Shift
- Unknown
- Q Number
- Q43
- Type
- MCQ
- NCERT Ref
- Class 12 Chemistry Ch 11: Alcohols Phenols Ethers
More from this Chapter
Q68.Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives (1) 2,4,6-trinitrobenzene (2) o-nitrophenol (3) p-nitrophenol (4) nitrobenzene
Q41.The number of stereoisomers possible for a compound of the molecular formula CH3 −CH = CH −CH(OH) −Me is : (1) 3 (2) 2 (3) 4 (4) 6
Q56.The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is : (1) benzoic acid (2) salicylaldehyde (3) salicylic acid (4) phthalic acid
Q56.From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2 , is (1) 2-Butanol (2) 2-Methylpropan-2-ol (3) 2-Methylpropanol (4) 1-Butanol