
Interpretation:
Products formed for the reaction have to be predicted. The given reaction is shown as,
Concept Introduction:
Deactivators are electron withdrawing groups attached to the benzenes that have either positive charge or an atom with high electronegativity. They are meta directors.
- Strong deactivators: They are very strong electron withdrawing groups.
- Moderate deactivators: It contains a pi bond that is attached to a strong electronegative group.
- Weak deactivators: It consists of halogens.
Activators are electron donating groups attached to the benzenes that have either electron density that is able to push into benzene ring or a lone pair of electrons. They are ortho-para directing.
- Strong activators: It contains a lone pair next to the
aromatic ring. - Moderate activators: It has a lone pair next to the aromatic ring that can take part in the resonance outside the ring as well.
- Weak activators: It consists of alkyl groups.
Halogens are deactivators that are ortho-para directing.
Rules to identify the group that dominates the directing effects:
- Ortho-para directors will always dominate meta directors.
- Strong activators will always dominate weak activators
Nitration: In nitration reaction, one nitro group

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Chapter 4 Solutions
ORGANIC CHEMISTRY-NEXTGEN+BOX (2 SEM.)
- Decide whether these proposed Lewis structures are reasonable. proposed Lewis structure Yes. Is the proposed Lewis structure reasonable? Cl- : 2: :Z: :Z: N—N : 0: C C1: O CO No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* ☐ Yes. No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | Yes. No, it has the wrong number of valence electrons. The correct number is: No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | If two or more atoms of the same element don't satisfy the octet rule, just enter the chemical symbol as many times as necessary. For example, if two oxygen atoms don't satisfy the octet rule, enter "0,0". ☑arrow_forwardUse the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions ΔΗ is (pick one) A This reaction is faster above 103. °C than below. AS is (pick one) ΔΗ is (pick one) B This reaction is spontaneous only above -9. °C. AS is (pick one) ΔΗ is (pick one) C The reverse of this reaction is always spontaneous. AS is (pick one) 18 Ararrow_forwardUse the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions A The reverse of this reaction is always spontaneous but proceeds slower at temperatures below 41. °C. ΔΗ is (pick one) AS is (pick one) ΔΗ is (pick one) B This reaction is spontaneous except above 94. °C. AS is (pick one) This reaction is always spontaneous, but ΔΗ is (pick one) C proceeds slower at temperatures below −14. °C. AS is (pick one) Х 00. 18 Ar 무ㅎ B 1 1arrow_forward
- Draw the product of the reaction shown below. Ignore inorganic byproducts. + H CH3CH2OH HCI Drawingarrow_forwardplease explain this in simple termsarrow_forwardK Most Reactive Na (3 pts) Can the metal activity series (shown on the right) or a standard reduction potential table explain why potassium metal can be prepared from the reaction of molten KCI and Na metal but sodium metal is not prepared from the reaction of molten NaCl and K metal? Show how (not). Ca Mg Al с Zn Fe Sn Pb H Cu Ag Au Least Reactivearrow_forward
- (2 pts) Why is O2 more stable as a diatomic molecule than S2?arrow_forwardDraw the Lewis structure for the polyatomic phosphite (PO¾³¯) a anion. Be sure to include all resonance structures that satisfy the octet rule. C I A [ ]¯arrow_forwardDecide whether these proposed Lewis structures are reasonable. proposed Lewis structure Is the proposed Lewis structure reasonable? Yes. :0: Cl C C1: 0=0: : 0 : : 0 : H C N No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* ☐ Yes. No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* Yes. ☐ No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | * If two or more atoms of the same element don't satisfy the octet rule, just enter the chemical symbol as many times as necessary. For example, if two oxygen atoms don't satisfy the octet rule, enter "0,0".arrow_forward
- Draw the Lewis structure for the polyatomic trisulfide anion. Be sure to include all resonance structures that satisfy the octet rule. с [ ] - Garrow_forward1. Calculate the accurate monoisotopic mass (using all 1H, 12C, 14N, 160 and 35CI) for your product using the table in your lab manual. Don't include the Cl, since you should only have [M+H]*. Compare this to the value you see on the LC-MS printout. How much different are they? 2. There are four isotopic peaks for the [M+H]* ion at m/z 240, 241, 242 and 243. For one point of extra credit, explain what each of these is and why they are present. 3. There is a fragment ion at m/z 184. For one point of extra credit, identify this fragment and confirm by calculating the accurate monoisotopic mass. 4. The UV spectrum is also at the bottom of your printout. For one point of extra credit, look up the UV spectrum of bupropion on Google Images and compare to your spectrum. Do they match? Cite your source. 5. For most of you, there will be a second chromatographic peak whose m/z is 74 (to a round number). For one point of extra credit, see if you can identify this molecule as well and confirm by…arrow_forwardPlease draw, not just describe!arrow_forward
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