
Concept explainers
a) -CH3, -Br, -H, -I.
Interpretation:
To rank -CH3, -Br, -H, -I according to Cahn-Ingold-Prelog sequence rules.
Concept introduction:
According to Cahn-Ingold-Prelog sequence rules the member that ranks higher can be determined by considering the
To rank:
The substituents, -CH3, -Br, -H, -I, according to Cahn-Ingold-Prelog sequence rules.

Answer to Problem 45AP
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -I, -Br, -CH3, -H.
Explanation of Solution
The first atoms in the substituents given are I, Br, C and CH. Their atomic numbers are 53, 35, 12, 1 respectively. Based on these atomic numbers, the substituents can be ranked in the order -I, -Br, -CH3, -H.
According to the sequence rules the substituents can be ranked in the order -I, -Br, -CH3, -H.
b) -OH, -OCH3, -H, -CO2H.
Interpretation:
To rank -OH, -OCH3, -H, -CO2H according to Cahn-Ingold-Prelog sequence rules.
Concept introduction:
According to Cahn-Ingold-Prelog sequence rules the member that ranks higher can be determined by considering the atomic number of the first atom in each substituent. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms can be considered until the first difference is found. Multiple bonded atoms are equivalent to the same number of single bonded atoms.
To rank:
The substituents, -OH, -OCH3, -H, -CO2H according to Cahn-Ingold-Prelog sequence rules.

Answer to Problem 45AP
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -OCH3, -OH, -CO2H, -H.
Explanation of Solution
Two groups have oxygen as first atom, third one has carbon as the first atom and the fourth one is H. Their atomic numbers are 8, 1, 12 respectively. Both -OH, -OCH3 have O as the first atom. Hence the second, third and fourth atoms to which they are bonded are to be considered. Oxygen in -OH is bonded to H, while the oxygen in -OCH3 is bonded to carbon. Hence -OCH3 ranks higher than -OH. The substituent -CO2H ranks third and -H the fourth.
According to the sequence rules the substituents can be ranked in the order -OCH3, -OH, -CO2H,-H.
c) -CO2H, -CO2CH3, -CH2OH, -CH3.
Interpretation:
To rank -CO2H, -CO2CH3, -CH2OH, -CH3 according to Cahn-Ingold-Prelog sequence rules.
Concept introduction:
The member that ranks higher can be determined by considering the atomic number of the first atom in each substituent. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms can be considered until the first difference is found. Multiple bonded atoms are equivalent to the same number of single bonded atoms.
To rank:
The substituents -CO2H, -CO2CH3, -CH2OH, -CH3 according to Cahn-Ingold-Prelog sequence rules.

Answer to Problem 45AP
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -CO2CH3, -CO2H, -CH2OH, -CH3.
Explanation of Solution
The first atom in all the substituents given is C. Hence the second, third and fourth atoms to which they are bonded are to be considered. Carbon in -CO2CH3 is bonded to three oxygens(double bond) out of which one is attached to a carbon. Carbon in COOH is also bonded to three oxygen atom (double bond) out of which one is attached to a hydrogen. Carbon in –CH2OH is bonded to H, H, O. Carbon in methyl is bonded to H, H, H. Based on the atomic numbers of the differentiating atoms, the substituents can be ranked in the order -CO2CH3, -CO2H, -CH2OH,-CH3.
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -CO2CH3, -CO2H, -CH2OH,-CH3.
d) -CH3, -CH2CH3, -CH2CH2OH, -COCH3.
Interpretation:
To rank -CH3, -CH2CH3, -CH2CH2OH, -COCH3 according to Cahn-Ingold-Prelog sequence rules.
Concept introduction:
The member that ranks higher can be determined by considering the atomic number of the first atom in each substituent. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms can be considered until the first difference is found. Multiple bonded atoms are equivalent to the same number of single bonded atoms.
To rank:
The substituents -CH3, -CH2CH3, -CH2CH2OH, -COCH3 according to Cahn-Ingold-Prelog sequence rules.

Answer to Problem 45AP
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -COCH3, -CH2CH2OH, -CH2CH3, -CH3.
Explanation of Solution
The first atom in all the substituents given is C. Hence the second, third and fourth atoms to which they are bonded are to be considered.
Carbon in -COCH3 is bonded to O, O(double bond), C.
Carbon in -CH2CH2OH is bonded to H, H & C attached to H, H, O.
Carbon in –CH2CH3 is bonded to H, H & C attached to three hydrogens.
Carbon in methyl is bonded to H, H, H.
Based on the atomic numbers of the differentiating atoms, the substituents can be ranked in the order -COCH3, -CH2CH2OH, -CH2CH3, -CH3.
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -COCH3, -CH2CH2OH, -CH2CH3, -CH3.
e) -CH=CH2, -CN, -CH2NH2, -CH2Br.
Interpretation:
To rank -CH=CH2, -CN, -CH2NH2, -CH2Br according to Cahn-Ingold-Prelog sequence rules.
Concept introduction:
The member that ranks higher can be determined by considering the atomic number of the first atom in each substituent. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms can be considered until the first difference is found. Multiple bonded atoms are equivalent to the same number of single bonded atoms.
To rank:
The substituents -CH=CH2, -CN, -CH2NH2, -CH2Br according to Cahn-Ingold-Prelog sequence rules.

Answer to Problem 45AP
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -CH2Br, -CN, -CH2NH2, -CH=CH2.
Explanation of Solution
The first atom in all the substituents given is C. Hence the second, third and fourth atoms to which they are bonded are to be considered.
Carbon in -CH=CH2, is bonded to C, C(double bond), H.
Carbon in -CN, is bonded to three nitrogens.
Carbon in -CH2NH2 is bonded to N, H, H.
Carbon in -CH2Br is bonded to Br, H, H.
Based on the atomic numbers of the differentiating atoms, the substituents can be ranked in the order -CH2Br, -CN, -CH2NH2, -CH=CH2.
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -CH2Br, -CN, -CH2NH2, -CH=CH2.
f) -CH=CH2, -CH2CH3, -CH2OCH3, -CH2OH.
Interpretation:
To rank -CH=CH2, -CH2CH3, -CH2OCH3, -CH2OH according to Cahn-Ingold-Prelog sequence rules.
Concept introduction:
The member that ranks higher can be determined by considering the atomic number of the first atom in each substituent. The atom with highest atomic number gets a higher rank. If a decision cannot be made by considering the atomic number of the first atom in each substituent then the second, third, fourth atoms can be considered until the first difference is found. Multiple bonded atoms are equivalent to the same number of single bonded atoms.
To rank:
The substituents -CH=CH2, -CH2CH3, -CH2OCH3, -CH2OH according to Cahn-Ingold-Prelog sequence rules.

Answer to Problem 45AP
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -CH2OCH3, -CH2OH, -CH=CH2, CH2CH3.
Explanation of Solution
The first atom in all the substituents given is C. Hence the second, third and fourth atoms to which they are bonded are to be considered.
Carbon in -CH=CH2, is bonded to C, C(Double bond), H.
Carbon in -CH2CH3 is bonded to C, H, H.
Carbon in -CH2OCH3 is bonded to H, H, O attached to a carbon.
Carbon in -CH2OH is bonded to H, H, O attached to a hydrogen.
Based on the atomic numbers of the differentiating atoms, the substituents can be ranked in the order -CH2OCH3, -CH2OH, -CH=CH2, CH2CH3.
According to Cahn-Ingold-Prelog sequence rules the substituents can be ranked in the order -CH2OCH3, -CH2OH, -CH=CH2, CH2CH3.
Want to see more full solutions like this?
Chapter 7 Solutions
OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card for McMurry's Organic Chemistry, 9th
- PROBLEMS Q1) Label the following salts as either acidic, basic, or neutral a) Fe(NOx) c) AlBr b) NH.CH COO d) HCOON (1/2 mark each) e) Fes f) NaBr Q2) What is the pH of a 0.0750 M solution of sulphuric acid?arrow_forward8. Draw all the resonance forms for each of the fling molecules or ions, and indicate the major contributor in each case, or if they are equivalent (45) (2) -PH2 سمة مدarrow_forwardA J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3arrow_forward
- 1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.arrow_forward3. For the reactions below, draw the arrows corresponding to the transformations and draw in the boxes the reactants or products as indicated. Note: Part A should have arrows drawn going from the reactants to the middle structure and the arrows on the middle structure that would yield the final structure. For part B, you will need to draw in the reactant before being able to draw the arrows corresponding to product formation. A. B. Rearrangement ΘΗarrow_forward2. Draw the arrows required to make the following reactions occur. Please ensure your arrows point from exactly where you want to exactly where you want. If it is unclear from where arrows start or where they end, only partial credit will be given. Note: You may need to draw in lone pairs before drawing the arrows. A. B. H-Br 人 C Θ CI H Cl Θ + Br Oarrow_forward
- 4. For the reactions below, draw the expected product. Be sure to indicate relevant stereochemistry or formal charges in the product structure. a) CI, H e b) H lux ligh Br 'Harrow_forwardArrange the solutions in order of increasing acidity. (Note that K (HF) = 6.8 x 10 and K (NH3) = 1.8 × 10-5) Rank solutions from least acidity to greatest acidity. To rank items as equivalent, overlap them. ▸ View Available Hint(s) Least acidity NH&F NaBr NaOH NH,Br NaCIO Reset Greatest acidityarrow_forward1. Consider the following molecular-level diagrams of a titration. O-HA molecule -Aion °° о ° (a) о (b) (c) (d) a. Which diagram best illustrates the microscopic representation for the EQUIVALENCE POINT in a titration of a weak acid (HA) with sodium. hydroxide? (e)arrow_forward
- Answers to the remaining 6 questions will be hand-drawn on paper and submitted as a single file upload below: Review of this week's reaction: H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O ---> H₂NC(=NH)N(CH3)CH2COOH (creatine) Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts) Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts) Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3 pts) NH2(C=NH)-N(CH)CH2COOH This bond Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem Q9 is valid). (4 pts) Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should be ready to understand the first half of one of the Grignard reactions presented in a past…arrow_forwardPropose a synthesis pathway for the following transformations. b) c) d)arrow_forwardThe rate coefficient of the gas-phase reaction 2 NO2 + O3 → N2O5 + O2 is 2.0x104 mol–1 dm3 s–1 at 300 K. Indicate whether the order of the reaction is 0, 1, or 2.arrow_forward
