Concept explainers
Interpretation:
The nucleophilic substitution reaction of
Concept introduction:
Electrophiles are electron-deficient species, which has positive or partially positive charge. Lewis acids are electrophiles, which accept electron pair.
Nucleophiles are electron-rich species, which has negative or partially negative charge. Lewis bases are nucleophiles, which donate electron pair.
Substitution reaction: A reaction in which one of the hydrogen atoms of a hydrocarbon or a functional group is substituted by any other functional group is called substitution reaction.
Nucleophilic substitution reaction is a reaction in which an electron rich nucleophile attacks the positive or partial positive charge of an atom or a group of atoms to replace a leaving group.
The
The
In double bond or cyclic compounds, if two same
If the two functional groups are present on the different sides of the double bond or cyclic compound, the given compound can be labeled as Trans.
Cis-trans isomerism exists in the compounds in which similar groups are present on the adjacent carbon atoms.
Chair conformations: It is the most stable conformation, which accurately shows the spatial arrangement of atoms.
Equatorial bonds are parallel to the average plane of the ring, while axial bonds are perpendicular to the average plane of the ring.
The conformation having bonds at the equatorial positions are more stable than those with bonds at the axial position.
On flipping the cyclohexane ring, axial bonds become equatorial bonds and equatorial bonds becomes axial bond.
Bulkier group acquires equatorial positions to form stable conformer due to steric factors.
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Organic Chemistry
- Review Problem 5.2 Construct handheld models of the 2-butanols represented in Fig. 5.3 and demonstrate for yourself that they are not mutually superposable. (a) Make similar models of 2-bromo- propane. Are they superposable? (b) Is a molecule of 2-bromopropane chiral? (c) Would you expect to find enantiomeric forms of 2-bromopropane? H но OH H HO HH OH OH HO H CH CH CH C CH CH II II (a) (b) (c) Figure 5.3 (a) Three-dimensional drawings of the 2-butanol enantiomers I and II. (b) Models of the 2-butanol enantiomers. (c) An unsuccessful attempt to superpose models of I and II.arrow_forward• PRACTICE PROBLEM 8.13 Specify the appropriate alkene and reagents for synthesis of each of the following alcohols by hydroboration–oxidation. (a) (c) OH (e) CH3 OH AH OH no mobe OH ( (b) (d) (f) OH HT H3 D OH OH HO OHarrow_forwardE In the equilibrium reaction of the conformations of 1,3-butadiene, what adequately explains the 15 kJ/mol energy barrier between the s-cis and s-trans conformation? 15 kJ/mol Sterics cause the s-cis conformation to be more stable and, therefore, higher in energy. (B) The л-bonds are not conjugated during the conversion of one conformation to the other. The s-trans is more stable because it is aromatic. The s-cis is less stable because it is non-aromatic. (E) Both the s-cis and s-trans are conjugated because all carbons are sp³-hybridized. 0° 90° Dihedral angle 180°arrow_forward
- 6.51b Explain the observed stereochemistry. When the second intermediate is redrawn from the following perspective, it becomes clear that O the electrons in the carbon-carbon-bond come from above the carbon-axygen x bond, allowing for the chirality center to be generated with the stereochemistry shown. the electrons in the carbon-carbon-bond come from below the carbon-oxygen x bond, allowing for the chirality center to be generated with the stereochemistry shown. the electrons in the carbon-exygen-bond come from below the carbon-cartion a bond, allowing for the chirality center to be generated with the stereochemistry shown. the electrons in the carbon-exygen x-band come from above the carbon-carbon bond, allowing for the chirality center to be generated with the stereochemistry shown. OOarrow_forward• PRACTICE PROBLEM 8.24 A, B, and C are alkynes. Elucidate their structures and that of D using the following reaction roadmap. H2, Pt H,, Pt A (C3H14) (C3H14) IR: 3300 cm (1) O3 (2) HOẠC HO, H2, Pt (C3H12) (C3H16) (1) O3 (2) HOAC hol bian vd beeollot Но. ОН AOHarrow_forwardThe compounds drawn should each contain a cyclohexane ring. For all three compounds draw a wedge and dash structure, Chair I, and Chair II conformations. Formula: C9H18 with substitution 1,1- disubstituted with stereochemistry of (R,S) Formula: C7H13Cl with substitution 1,3- disubstituted with stereochemistry of (R,R) Formula: C7H14O with substitution 1,4- disubstituted with stereochemistry (S,S)arrow_forward
- The energy difference between cis- and trans-but-2-ene is about 4 kJ/mol; however, the trans isomer of 4,4-dimethylpent2-ene is nearly 16 kJ/mol more stable than the cis isomer. Explain this large difference.7arrow_forwardjust problem 5.23 pleasearrow_forwardRank the following alkenes in order of increasing stabilityarrow_forward
- Can you help me answer and explain 3.55arrow_forward· Using chair conformational structure, show the 5-) nucleophilic substitution reaction that would take place when trans-1-bromo-4-tert-butylcyclohexane react with iodide ion. (Show the most stable conformation of the reactant and product)arrow_forward• PRACTICE PROBLEM 5.22 Write three-dimensional formulas for all of the stereoisomers of each of the following compounds. Label pairs of enantiomers and label meso compounds. enslg a nsudomotaib esvierd ofni oluelom oreovb dase to ope.n 1oim s ier 1erto (a) ibaong (b) A boemo OH (c) CI ОН CI F F OH win vle (f) HO,C. (d) (e) OH CI CO,H Br im slm (b) OH Tartaric acidarrow_forward