Concept explainers
a.
Interpretation:The given compound is to be named.
Concept Introduction: The naming of carbon compounds is done according to IUPAC principles.
a.
Answer to Problem 6RQ
The name of the given compound is 1,2-dimethylbenzene.
Explanation of Solution
The main carbon group is benzene. The two methyl groups are attached on the first and second carbon which adds the 1,2-dimethyl- prefix.
b.
Interpretation:The given compound is to be named.
Concept Introduction: The naming of carbon compounds is done according to IUPAC principles.
b.
Answer to Problem 6RQ
The name of the given compound is 1,3,5-tribromobenzene.
Explanation of Solution
The main carbon group is benzene. The three bromo groups are attached to the first, third and fifth carbon which gives the 1,3,5-tribromo- prefix.
d.
Interpretation:The given compound is to be named.
Concept Introduction: The naming of carbon compounds is done according to IUPAC principles.
d.
Answer to Problem 6RQ
The name of the given compound is naphthalene.
Explanation of Solution
The given compound’s actual IUPAC convention name is quite long.So to avoid problems it has been given the standard name of naphthalene.
d.
Interpretation:The given compound is to be named.
Concept Introduction: The naming of carbon compounds is done according to IUPAC principles.
d.
Answer to Problem 6RQ
The name of the given compound is naphthalene.
Explanation of Solution
The given compound’s actual IUPAC convention name is quite long.So to avoid problems it has been given the standard name of naphthalene.
Chapter 20 Solutions
World of Chemistry, 3rd edition
- 6. Consider the following exothermic reaction below. 2Cu2+(aq) +41 (aq)2Cul(s) + 12(aq) a. If Cul is added, there will be a shift left/shift right/no shift (circle one). b. If Cu2+ is added, there will be a shift left/shift right/no shift (circle one). c. If a solution of AgNO3 is added, there will be a shift left/shift right/no shift (circle one). d. If the solvent hexane (C6H14) is added, there will be a shift left/shift right/no shift (circle one). Hint: one of the reaction species is more soluble in hexane than in water. e. If the reaction is cooled, there will be a shift left/shift right/no shift (circle one). f. Which of the changes above will change the equilibrium constant, K?arrow_forwardShow work. don't give Aiarrow_forwardShow work with explanation needed. don't give Ai generated solutionarrow_forward
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- 3. Provide the missing compounds or reagents. 1. H,NNH КОН 4 EN MN. 1. HBUCK = 8 хно Panely prowseful kanti-chuprccant fad, winddively, can lead to the crading of deduc din-willed, tica, The that chemooices in redimi Грин. " like (for alongan Ridovi MN نيا . 2. Cl -BuO 1. NUH 2.A A -BuOK THE CF,00,H Ex 5)arrow_forward2. Write a complete mechanism for the reaction shown below. NaOCH LOCH₁ O₂N NO2 CH₂OH, 20 °C O₂N NO2arrow_forward4. Propose a synthesis of the target molecules from the respective starting materials. a) b) LUCH C Br OHarrow_forward
- The following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forwardPlease correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forward
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