
(a)
Interpretation:
The names of
Concept introduction:
Functional groups are atom or groups of atoms which are responsible for special chemical or physical properties of the molecule. IUPAC purposed a complete list of different functional groups with their prefix and suffix which can be used in the naming of the molecule. Some of the atoms can act as functional group such as −Cl, -F, -Br etc whereas some of them are groups of atoms like −NH2, -NO2, -OH etc.
(b)
Interpretation:
Identify the hybridization of
Concept introduction:
Hybridization is the process of mixing of atomic orbitals to form same energy and same shape hybrid orbitals which form covalent bond with overlapping of atomic orbitals of other atoms. The number of
(c)
Interpretation:
Identify the chiral C atoms in ephedrine.
Concept introduction:
A chiral carbon atom is the carbon atom which is bonded with four different atoms or group of atoms. The presence of chiral carbon atom makes that molecule optically active so it can rotate the plane polarized light either clockwise or anti-clockwise. Such molecules are called as optically active molecule and two different configuration of same molecule are called as enantiomers. They are non-superimposable mirror images of each other.
(d)
Interpretation:
Calculate the pKb value for ephedrine with the help of given information.
Concept introduction:
Ephedrine is a weak base which is mainly used as a center nervous system stimulant. Since it acts as a weak base therefore it gives OH- ions in its aqueous solution.
The base dissociation constant
The relation between

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Chapter 26 Solutions
GENERAL CHEMISTRY(LL)-W/MASTERINGCHEM.
- Heating an alcohol in the presence of sulfuric or phosphoric acid will cause a dehydration to occur: the removal of the elements of water from a molecule, forming an alkene. The reaction usually follows an E1 mechanism. The SN1 pathway is suppressed by using a strong acid whose conjugate base is a poor nucleophile. Further, heating the reaction mixture causes a greater increase in the rate of E1 compared to the rate of Sy1. 3rd attempt h Draw curved arrow(s) to show how the alcohol reacts with phosphoric acid. TH © 1 0 0 +1% # 2nd attempt Feedback H Ju See Periodic Table See Hint H Jud See Periodic Table See Hintarrow_forwardPart 2 (0.5 point) 0- Draw the major organic product with the correct geometry. 10 1: 70000 х く 1st attempt Part 1 (0.5 point) Feedback Please draw all four bonds at chiral centers. P See Periodic Table See Hintarrow_forwardHeating an alcohol in the presence of sulfuric or phosphoric acid will cause a dehydration to occur: the removal of the elements of water from a molecule, forming an alkene. The reaction usually follows an E1 mechanism. The SN1 pathway is suppressed by using a strong acid whose conjugate base is a poor nucleophile. Further, heating the reaction mixture causes a greater increase in the rate of E1 compared to the rate of S№1. 2nd attempt 0 See Periodic Table See Hint Draw the organic intermediate from the first step (no byproducts) and draw curved arrow(s) to show how it reacts. TH +11: 1st attempt Feedback H H H C F F See Periodic Table See Hintarrow_forward
- This molecule undergoes an E1 mechanism when stirred in methanol. 3rd attempt CH₂OH CH₂OH 6148 O See Periodic Table. See Hint Draw 3 chemical species including formal charges and lone pairs of electrons. Add the missing curved arrow notation. H N O O SA 3 Br Iarrow_forwardComplete the mechanism for the E1 reaction below by following the directions written above each of the five boxes. Be sure to include lone pair electrons and nonzero formal charges. 1st attempt Y 0 + Provide the missing curved arrow notation. 01: See Periodic Table See Hint H C Br Iarrow_forwardPlease help answer number 2. Thanks in advance.arrow_forward
- How do I explain this? Thank you!arrow_forwardWhen an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M* = 163 m/z) is formed. In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm. The 13C NMR and DEPT spectra are provided. Draw the structure of the product as the resonance contributor lacking any formal charges. 13C NMR DEPT 90 200 160 120 80 40 0 200 160 120 80 40 0 DEPT 135 T 200 160 120 80 40 0 Draw the unknown amide. Select Dow Templates More Fragearrow_forwardIdentify the unknown compound from its IR and proton NMR spectra. C4H6O: 'H NMR: 82.43 (1H, t, J = 2 Hz); 8 3.41 (3H, s); 8 4.10 (2H, d, J = 2 Hz) IR: 2125, 3300 cm¹ The C4H6O compound liberates a gas when treated with C2H5 MgBr. Draw the unknown compound. Select Draw с H Templates Morearrow_forward
- Please help with number 6 I got a negative number could that be right?arrow_forward1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.arrow_forwardGive the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.arrow_forward
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